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NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 371823, 14736]*) (*NotebookOutlinePosition[ 372641, 14764]*) (* CellTagsIndexPosition[ 372597, 14760]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[StyleBox["II.4. Moduln", FontColor->GrayLevel[0]]], "Subtitle"], Cell["\<\ Als Vorlage f\[UDoubleDot]r den gr\[ODoubleDot]\[SZ]ten Teil dieses Textes \ diente ein Skript von Professor R\[UDoubleDot]ck zur Linearen Algebra II im \ Sommersemester 2002. Vielen Dank. \ \>", "Text"], Cell[CellGroupData[{ Cell[TextData[StyleBox[" Grundlegende Begriffe ", FontWeight->"Plain", FontVariations->{"CompatibilityType"->0}]], "Section"], Cell[TextData[{ "Am Anfang zur Erinnerung die Definitionen einiger grundlegender \ algebraischer Strukturen und zweier Schreibweisen:\nF\[UDoubleDot]r ", StyleBox["n", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalN] sei \[DoubleStruckN] := {1,\ \[Ellipsis], ", StyleBox["n", FontSlant->"Italic"], "} . Entsprechend werden die Bezeichnungen \[DoubleStruckK], \ \[DoubleStruckL], \[DoubleStruckM], \[Ellipsis] benutzt.\nStatt \ \[DoubleStruckCapitalZ] / ", StyleBox["n", FontSlant->"Italic"], "\[DoubleStruckCapitalZ] wird auch ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]], " geschrieben." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.1:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nGRUPPE (", FontFamily->"Arial"], StyleBox["group", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nABELSCHE GRUPPE (", FontFamily->"Arial"], StyleBox["abelian group", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], "\nEine nichtleere Menge ", StyleBox["G", FontSlant->"Italic"], " mit einer Verkn\[UDoubleDot]pfung \[CirclePlus] hei\[SZ]t ", StyleBox["GRUPPE", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind:\n(G ", StyleBox["i", FontSlant->"Italic"], ")\tAssoziativgesetz: \n \tF\[UDoubleDot]r alle ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(g\_1\), "TraditionalForm"], ",", \(g\_2\)}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`g\_3\)]], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " gilt: ", Cell[BoxData[ \(TraditionalForm\`g\_1\)]], "\[CirclePlus] (", Cell[BoxData[ \(TraditionalForm\`g\_2\)]], "\[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`g\_3\)]], ") = (", Cell[BoxData[ \(TraditionalForm\`g\_1\)]], "\[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`g\_2\)]], ") \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`g\_\(\(3\)\(\ \)\)\)]], " .\n(G ", StyleBox["ii", FontSlant->"Italic"], ") \tExistenz eines neutralen Elementes:\n\tEs gibt ein ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " mit ", StyleBox["n", FontSlant->"Italic"], " \[CirclePlus] ", StyleBox["g", FontSlant->"Italic"], " = ", StyleBox["g", FontSlant->"Italic"], " \[CirclePlus] ", StyleBox["n", FontSlant->"Italic"], " = ", StyleBox["g", FontSlant->"Italic"], " f\[UDoubleDot]r alle ", StyleBox["g", FontSlant->"Italic"], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " .\n(G ", StyleBox["iii", FontSlant->"Italic"], ") \tExistenz von Inversen:\n\tZu jedem ", StyleBox["g", FontSlant->"Italic"], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " gibt es ein ", StyleBox["g'", FontSlant->"Italic"], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " mit ", StyleBox["g", FontSlant->"Italic"], " \[CirclePlus] ", StyleBox["g'", FontSlant->"Italic"], " = ", StyleBox["g'", FontSlant->"Italic"], " \[CirclePlus] ", StyleBox["g", FontSlant->"Italic"], " = ", StyleBox["n .\n\n", FontSlant->"Italic"], "Eine Gruppe hei\[SZ]t ", StyleBox["ABELSCH", FontFamily->"Arial"], ", wenn die Verkn\[UDoubleDot]pfung \[CirclePlus] kommutativ ist, d.h. f\ \[UDoubleDot]r alle ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(g\_1\), "TraditionalForm"], ",", \(g\_2\)}], TraditionalForm]]], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " gilt: \n", Cell[BoxData[ \(TraditionalForm\`g\_1\)]], "\[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`g\_2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`g\_2\)]], "\[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`g\_1\)]], " ." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.2:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nRING (", FontFamily->"Arial"], StyleBox["ring", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Eine nichtleere Menge ", StyleBox["R", FontSlant->"Italic"], " mit zwei Verkn\[UDoubleDot]pfungen + und \[CenterDot] hei\[SZ]t ", StyleBox["RING", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind:\n(R ", StyleBox["i", FontSlant->"Italic"], ")\t(", StyleBox["R", FontSlant->"Italic"], ", +) ist abelsche Gruppe; das neutrale Element bzgl. + wird Null \ genannt.\n(R ", StyleBox["ii", FontSlant->"Italic"], ")\tAssoziativgesetz f\[UDoubleDot]r die Multiplikation: \n \tF\ \[UDoubleDot]r alle ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], ",", \(r\_2\)}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gilt: ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] (", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], ") = (", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_\(\(3\)\(\ \)\)\)]], ".\n(R ", StyleBox["iii", FontSlant->"Italic"], ") \tDistributivgesetze:\n\t F\[UDoubleDot]r alle ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], ",", \(r\_2\)}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gilt: \n\t ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] (", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], ") = (", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") + (", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], ") , (", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], " = (", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], ") + (", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_3\)]], ") .\n\nEin Ring ", StyleBox["R", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["KOMMUTATIV", FontFamily->"Arial"], ", wenn die Verkn\[UDoubleDot]pfung \[CenterDot] kommutativ ist, d.h. f\ \[UDoubleDot]r alle ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ FormBox["r", "TraditionalForm"], "1"], " ", ",", \(r\_2\)}], TraditionalForm]]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gilt: \n", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " . \nEin Element ", StyleBox["e", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["EINSELEMENT (", FontFamily->"Arial"], StyleBox["multiplcative identity", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], ", wenn ", StyleBox["e", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["r", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["e", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " f\[UDoubleDot]r alle ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], "gilt; oft wird ", StyleBox["e", FontSlant->"Italic"], " als 1 geschrieben." }], "Text"], Cell[TextData[{ StyleBox["Beispiele:", FontFamily->"Arial"], "\nWichtige Ringe sind :\n\t\[DoubleStruckCapitalZ] , der Ring der ganzen \ Zahlen mit Addition und Multiplikation,\n\t", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]], ", der Restklassenring modulo ", StyleBox["n", FontSlant->"Italic"], " .\n\t2", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\)]], " , die Menge der geraden ganzen Zahlen; sie bildet einen kommutativen Ring \ ohne Einselement.\n\t\[DoubleStruckCapitalZ][", StyleBox["X", FontSlant->"Italic"], "] , der Ring der Polynome mit Koeffizienten aus \[DoubleStruckCapitalZ] .\n\ \t", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] , der Ring der Polynome mit Koeffizienten aus einem K\[ODoubleDot]rper \ ", StyleBox["K", FontSlant->"Italic"], " .\n\t", StyleBox["K", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ", ", StyleBox["n", FontSlant->"Italic"], ") , der Ring der ", StyleBox["n", FontSlant->"Italic"], "\[Times]", StyleBox["n", FontSlant->"Italic"], "-Matrizen mit Komponenten aus einem K\[ODoubleDot]rper ", StyleBox["K .", FontSlant->"Italic"] }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.3:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nIDEAL (", FontFamily->"Arial"], StyleBox["ideal", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Eine nichtleere Teilmenge ", StyleBox["I", FontSlant->"Italic"], " eines Ringes ", StyleBox["R", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["IDEAL", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind:\n(I ", StyleBox["i", FontSlant->"Italic"], ")\t(", StyleBox["I", FontSlant->"Italic"], ", +) ist Unterguppe der Gruppe (", StyleBox["R", FontSlant->"Italic"], ", +) .\n(I ", StyleBox["ii", FontSlant->"Italic"], ")\tF\[UDoubleDot]r alle ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " und alle ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " gilt ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Beispiel:", FontFamily->"Arial"], "\nEs sei ", StyleBox["R", FontSlant->"Italic"], " = 2", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\)]], " . 4", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\)]], " ist Ideal von 2", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\)]], " ." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.4:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nHAUPTIDEAL (", FontFamily->"Arial"], StyleBox["principal ideal", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Ein Ideal ", StyleBox["I", FontSlant->"Italic"], " eines Ringes ", StyleBox["R", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["HAUPTIDEAL", FontFamily->"Arial"], ", wenn es ein ", StyleBox["y", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " gibt, so dass zu jedem ", StyleBox["z", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " ein ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " existiert mit ", StyleBox["z", FontSlant->"Italic"], " = ", StyleBox["y", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["r", FontSlant->"Italic"], " . Das Element ", StyleBox["y", FontSlant->"Italic"], " erzeugt das Ideal ", StyleBox["I", FontSlant->"Italic"], " und wird auch Erzeuger ", StyleBox["(", FontFamily->"Arial"], StyleBox["generator", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[") ", FontFamily->"Arial"], "genannt.\nEs werden dann folgende Schreibweisen benutzt: ", StyleBox["I", FontSlant->"Italic"], " = \[LeftAngleBracket]", StyleBox["y", FontSlant->"Italic"], "\[RightAngleBracket] oder ", StyleBox["I", FontSlant->"Italic"], " = [", StyleBox["y", FontSlant->"Italic"], "] ." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.5:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nINTEGRIT\[CapitalADoubleDot]TSBEREICH (", FontFamily->"Arial"], StyleBox["integral domain", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], StyleBox["Ein kommutativer Ring ", FontFamily->"Times New Roman"], StyleBox["R", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" mit Einselement hei\[SZ]t ", FontFamily->"Times New Roman"], StyleBox["INTEGRIT\[CapitalADoubleDot]TSBEREICH", FontFamily->"Arial"], StyleBox[", wenn f", FontFamily->"Times New Roman"], "\[UDoubleDot]r alle ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[NotEqual] 0 , \n", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[NotEqual] 0 folgt ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[NotEqual] 0 .\n\nManchmal werden Integrit\[ADoubleDot]tsbereiche auch \ Integrit\[ADoubleDot]tsringe genannt." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.6:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nNULLTEILER (", FontFamily->"Arial"], StyleBox["zero divisor", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es sei ", StyleBox["R", FontSlant->"Italic"], " ein Ring. Ein ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], "hei\[SZ]t", StyleBox[" ", FontSlant->"Italic"], StyleBox["NULLTEILER", FontFamily->"Arial"], ", wenn ", StyleBox["a", FontSlant->"Italic"], " \[NotEqual] 0 und es ein ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gibt mit ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 und ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " = 0 . " }], "Text"], Cell[TextData[{ StyleBox["Beispiel:", FontFamily->"Arial"], "\nIst ", StyleBox["n", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalN] keine Primzahl, etwa ", StyleBox["n", FontSlant->"Italic"], " = ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalN] \\ {1} , so gibt es in ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_n\)]], " Nullteiler, n\[ADoubleDot]mlich z.B. ", Cell[BoxData[ \(TraditionalForm\`a\&_\)]], " und ", Cell[BoxData[ \(TraditionalForm\`b\&_\)]], " , denn ", Cell[BoxData[ \(TraditionalForm\`\(a\[CenterDot]b\)\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`n\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " ." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.7:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nK\[CapitalODoubleDot]RPER (", FontFamily->"Arial"], StyleBox["field", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], StyleBox["Ein kommutativer Ring mit Einselement (", FontFamily->"Times New Roman"], StyleBox["K", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[",", FontFamily->"Times New Roman"], StyleBox[" +", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[", \[CenterDot]) hei\[SZ]t ", FontFamily->"Times New Roman"], StyleBox["K\[CapitalODoubleDot]RPER", FontFamily->"Arial"], StyleBox[", wenn (", FontFamily->"Times New Roman"], StyleBox["K", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" \\ {0}, \[CenterDot]) abelsche Gruppe ist.\n\nAlso besitzt \ jedes Element ", FontFamily->"Times New Roman"], StyleBox["k", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" \[Element] ", FontFamily->"Times New Roman"], StyleBox["K", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" \\ {0} ein multiplikatives Inverses. Es gibt in ", FontFamily->"Times New Roman"], StyleBox["K", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" keine Nullteiler.", FontFamily->"Times New Roman"] }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.8:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nVEKTORRAUM (", FontFamily->"Arial"], StyleBox["vector space", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es sei ", StyleBox["K", FontSlant->"Italic"], " ein K\[ODoubleDot]rper. Eine nichtleere Menge ", StyleBox["V", FontSlant->"Italic"], " mit einer Verkn\[UDoubleDot]pfung (Addition) + : ", StyleBox["V", FontSlant->"Italic"], " \[Times] ", StyleBox["V", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["V", FontSlant->"Italic"], " und einer \[ADoubleDot]u\[SZ]eren Verkn\[UDoubleDot]pfung (skalare \ Multiplikation) \[CenterDot] : ", StyleBox["K", FontSlant->"Italic"], " \[Times] ", StyleBox["V", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["V", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["VEKTORRAUM", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind:\n(V ", StyleBox["i", FontSlant->"Italic"], ")\t(", StyleBox["V", FontSlant->"Italic"], ", +) ist abelsche Gruppe.\nF\[UDoubleDot]r alle \[Alpha], \[Beta] \ \[Element] ", StyleBox["K", FontSlant->"Italic"], " und alle ", StyleBox["v", FontSlant->"Italic"], ", ", StyleBox["w", FontSlant->"Italic"], " \[Element] ", StyleBox["V", FontSlant->"Italic"], " gelten:\n(V ", StyleBox["ii", FontSlant->"Italic"], ")\tDistributivgesetze:\n\t(\[Alpha] + \[Beta]) \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " = (\[Alpha]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") + (\[Beta]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") , \[Alpha] \[CenterDot] (", StyleBox["v", FontSlant->"Italic"], " + ", StyleBox["w", FontSlant->"Italic"], ") = (\[Alpha]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") + (\[Alpha]\[CenterDot]", StyleBox["w", FontSlant->"Italic"], ") ,\n(V ", StyleBox["iii", FontSlant->"Italic"], ")\tAssoziativgesetz:\n\t\[Alpha] \[CenterDot] (\[Beta]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") = (\[Alpha]\[CenterDot]\[Beta]) \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " ,\n(V ", StyleBox["iv", FontSlant->"Italic"], ")\t1 \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " = ", StyleBox["v", FontSlant->"Italic"], " .\t" }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.9:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nR-MODUL (", FontFamily->"Arial"], StyleBox["module", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es sei ", StyleBox["R", FontSlant->"Italic"], " ein Ring mit Eins. Eine nichtleere Menge ", StyleBox["M", FontSlant->"Italic"], " mit einer Verkn\[UDoubleDot]pfung (Addition) + : ", StyleBox["M", FontSlant->"Italic"], " \[Times] ", StyleBox["M ", FontSlant->"Italic"], "\[RightArrow] ", StyleBox["M", FontSlant->"Italic"], " und einer \[ADoubleDot]u\[SZ]eren Verkn\[UDoubleDot]pfung (skalare \ Multiplikation) \[CenterDot] : ", StyleBox["R", FontSlant->"Italic"], " \[Times] ", StyleBox["M", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["R-MODUL", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind:\n(M ", StyleBox["i", FontSlant->"Italic"], ")\t(", StyleBox["M", FontSlant->"Italic"], ", +) ist abelsche Gruppe.\nF\[UDoubleDot]r alle \[Alpha], \[Beta] \ \[Element] ", StyleBox["R", FontSlant->"Italic"], " und alle ", StyleBox["v", FontSlant->"Italic"], ", ", StyleBox["w", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " gelten:\n(M ", StyleBox["ii", FontSlant->"Italic"], ")\t Distributivgesetze:\n \t (\[Alpha] + \[Beta] ) \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " = (\[Alpha]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") + (\[Beta]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") , \[Alpha] \[CenterDot] (", StyleBox["v", FontSlant->"Italic"], " + ", StyleBox["w", FontSlant->"Italic"], ") = (\[Alpha]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") + (\[Alpha]\[CenterDot]", StyleBox["w", FontSlant->"Italic"], ") ,\n(M ", StyleBox["iii", FontSlant->"Italic"], ")\tAssoziativgesetz:\n\t \[Alpha] \[CenterDot] (\[Beta]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ") = (\[Alpha]\[CenterDot]\[Beta]) \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " ,\n(M ", StyleBox["iv", FontSlant->"Italic"], ")\t 1 \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " = ", StyleBox["v", FontSlant->"Italic"], " .\t" }], "Text"], Cell[TextData[{ StyleBox["Beispiele:", FontFamily->"Arial"], " \n1)\nVektorr\[ADoubleDot]ume sind Module.\n\n2)\nEs sei ", StyleBox["R", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ] mit einem ", StyleBox["m", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalZ] sei ", StyleBox["M", FontSlant->"Italic"], " := ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_m\)]], " der Restklassenring modulo ", StyleBox["m", FontSlant->"Italic"], " mit der dort \[UDoubleDot]blichen Addition + . Die skalare \ Multiplikation \[Mu] sei (in naheliegender Weise) definiert durch: \[Mu] : \ \[DoubleStruckCapitalZ] \[Times] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_m\)]], " \[RightArrow] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_m\)]], " durch \[Mu](", StyleBox["r", FontSlant->"Italic"], ", ", Cell[BoxData[ \(TraditionalForm\`a\&_\)]], ") := ", Cell[BoxData[ \(TraditionalForm\`\(r\[CenterDot]a\)\&_\)]], " . Es ist zu \[UDoubleDot]berpr\[UDoubleDot]fen, ob \[Mu] wohldefiniert \ ist: Es sei also mit ", StyleBox["a", FontSlant->"Italic"], ",", StyleBox[" b", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalZ] ", Cell[BoxData[ \(TraditionalForm\`a\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`b\&_\)]], " , d.h. ", StyleBox["m", FontSlant->"Italic"], " | (", StyleBox["a", FontSlant->"Italic"], " \[Dash] ", StyleBox["b", FontSlant->"Italic"], ") . Es folgt ", StyleBox["m", FontSlant->"Italic"], " | ", StyleBox["r", FontSlant->"Italic"], "\[CenterDot](", StyleBox["a", FontSlant->"Italic"], " \[Dash] ", StyleBox["b", FontSlant->"Italic"], ") , also auch ", Cell[BoxData[ \(TraditionalForm\`\(r\[CenterDot]a\)\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(r\[CenterDot]b\)\&_\)]], " , w.z.b.w. . \[CapitalUDoubleDot]blicherweise wird statt \"\[Mu]\" nur \ \"\[CenterDot]\" geschrieben.\nAls ", "\[CapitalUDoubleDot]bungsaufgabe ", "ist zu beweisen, dass ", StyleBox["M", FontSlant->"Italic"], " ein ", StyleBox["R", FontSlant->"Italic"], "-Modul ist. \n\n", StyleBox["R", FontSlant->"Italic"], " ist eine unendliche Menge, aber ", StyleBox["M", FontSlant->"Italic"], " ist endlich. Es gilt f\[UDoubleDot]r ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["m", FontSlant->"Italic"], " \[NotEqual] 0 und ", Cell[BoxData[ \(TraditionalForm\`1\&_\)]], " \[NotEqual] ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " aber ", StyleBox["m", FontSlant->"Italic"], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`1\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " . Also gibt es hier nicht \"so etwas\", wie lineare \ Unabh\[ADoubleDot]gigkeit." }], "Text"], Cell[TextData[{ "3)\nEs seien ", StyleBox["R", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ] und ", StyleBox["M", FontSlant->"Italic"], " := \[DoubleStruckCapitalQ] mit der \[UDoubleDot]blichen Addition und \ Multiplikation. Wieder l\[ADoubleDot]\[SZ]t sich leicht zeigen, dass ", StyleBox["M", FontSlant->"Italic"], " ein ", StyleBox["R", FontSlant->"Italic"], "-Modul ist. \nIn diesem Modul l\[ADoubleDot]\[SZ]t sich die Null als \ nichttriviale Linearkombination zweier beliebiger Elemente ungleich Null \ darstellen:\nEs seien ", Cell[BoxData[ \(TraditionalForm\`m\_v\)]], " = ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\/b\_\[Nu]\)]], " f\[UDoubleDot]r \[Nu] = 1, 2 und ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b\_\[Nu]\)]], " \[Element] \[DoubleStruckCapitalZ] \\ {0} . Dann folgt ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\(a\_2\), "TraditionalForm"], "\[CenterDot]", \(b\_1\)}], "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], " + (", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SubscriptBox[ StyleBox[ RowBox[{"\[Dash]", StyleBox["a", FontSlant->"Italic"]}]], "1"], "TraditionalForm"], "\[CenterDot]", \(b\_2\)}], TraditionalForm]]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " = 0 .\n", StyleBox["M", FontSlant->"Italic"], " l\[ADoubleDot]\[SZ]t sich nicht \"endlich erzeugen\", denn angenommen es \ g\[ADoubleDot]be nun f\[UDoubleDot]r \[Nu] \[Element] \[DoubleStruckN] ", Cell[BoxData[ \(TraditionalForm\`m\_v\)]], " = ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\/b\_\[Nu]\)]], ", die das leisten, dann hat ein ", StyleBox["m", FontSlant->"Italic"], " mit ", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[Lambda]\_1\), "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], " + \[CenterEllipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[Lambda]\_n\), "TraditionalForm"], "\[CenterDot]", \(m\_n\)}], TraditionalForm]]], " h\[ODoubleDot]chstens das Produkt ", Cell[BoxData[ \(TraditionalForm\`b\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`b\_n\)]], " im Nenner. " }], "Text"], Cell[TextData[{ "4)\nEs sei ", StyleBox["V", FontSlant->"Italic"], " ein Vektorraum \[UDoubleDot]ber einem K\[ODoubleDot]rper ", StyleBox["K", FontSlant->"Italic"], " . Es seien \[CurlyPhi] \[Element] hom(", StyleBox["V", FontSlant->"Italic"], ", ", StyleBox["V", FontSlant->"Italic"], ") und ", StyleBox["p", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") \[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] . Dann sei die skalare Multiplikation \[CenterDot] : ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] \[Times] ", StyleBox["V", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["V", FontSlant->"Italic"], " definiert durch ", StyleBox["p", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " := (", StyleBox["p", FontSlant->"Italic"], "(\[CurlyPhi]))(", StyleBox["v", FontSlant->"Italic"], ") . Dann ist ", StyleBox["V", FontSlant->"Italic"], " ein ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Modul.\nBeweis als \[CapitalUDoubleDot]bung.\nDieses Beispiel ist sehr \ grundlegend f\[UDoubleDot]r die Anwendung eines zentralen Satzes der \ Modultheorie auf die Darstellung von linearen Abbildungen. " }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.10:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nUNTERMODUL (", FontFamily->"Arial"], StyleBox["submodule", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Eine nichtleere Teilmenge ", StyleBox["N ", FontSlant->"Italic"], " eines ", StyleBox["R", FontSlant->"Italic"], "-Moduls ", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["UNTERMODUL", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind: \n(UM ", StyleBox["i", FontSlant->"Italic"], ") ", StyleBox["N", FontSlant->"Italic"], " ist Untergruppe der additiven Gruppe von ", StyleBox["M", FontSlant->"Italic"], " .\n(UM ", StyleBox["ii", FontSlant->"Italic"], ") F\[UDoubleDot]r alle ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " und alle ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["N ", FontSlant->"Italic"], "ist auch ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.1", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEine nichtleere Menge ", StyleBox["N ", FontSlant->"Italic"], " eines ", StyleBox["R", FontSlant->"Italic"], "-Moduls ", StyleBox["M", FontSlant->"Italic"], " ist Untermodul, wenn gilt:\nF\[UDoubleDot]r alle ", Cell[BoxData[ \(TraditionalForm\`n\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`n\_2\)]], " \[Element] ", StyleBox["N ", FontSlant->"Italic"], "ist auch ", Cell[BoxData[ \(TraditionalForm\`n\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`n\_2\)]], " \[Element] ", StyleBox["N ", FontSlant->"Italic"], ".\nF\[UDoubleDot]r alle ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " und alle ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["N ", FontSlant->"Italic"], "ist auch ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " . \n", "Beweis als \[CapitalUDoubleDot]bung.", StyleBox["\n\n", FontColor->RGBColor[1, 0, 0]], "Dieses Kriterium wird meist zum Nachweis der Untermoduleigenschaft \ benutzt." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.11:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nMODULHOMOMORPHISMUS (", FontFamily->"Arial"], StyleBox["homomorphism of modules", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es seien ", StyleBox[" M", FontSlant->"Italic"], ", ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["R", FontSlant->"Italic"], "-Moduln. Eine Abbildung ", StyleBox["f", FontSlant->"Italic"], " : ", StyleBox["M", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["N", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["MODULHOMOMORPHISMUS", FontFamily->"Arial"], ", oder auch ", StyleBox["R-LINEARE ABBILDUNG", FontFamily->"Arial"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind: \n(H ", StyleBox["i", FontSlant->"Italic"], ")\tF\[UDoubleDot]r alle ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], "\[Element] ", StyleBox["M", FontSlant->"Italic"], " gilt: ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") + ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") .\n(H ", StyleBox["ii", FontSlant->"Italic"], ")\tF\[UDoubleDot]r alle ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " und alle ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gilt ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], ") = ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["m", FontSlant->"Italic"], ") .\n\nDie Vorsilben Auto-, Endo-, Iso- werden sinngem\[ADoubleDot]\[SZ] \ auch bei diesen -homomorphismen gebraucht." }], "Text"], Cell[TextData[{ StyleBox["Bemerkung:", FontFamily->"Arial"], " Das Zeichen \[TildeEqual] soll im Folgenden f\[UDoubleDot]r die \ Isomorphie von Moduln benutzt werden." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.12:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nKERN, BILD (", FontFamily->"Arial"], StyleBox["kernel, image", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es seien ", StyleBox[" M", FontSlant->"Italic"], ", ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["R", FontSlant->"Italic"], "-Moduln und ", StyleBox["f", FontSlant->"Italic"], " : ", StyleBox["M", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["N", FontSlant->"Italic"], " ein Homomorphismus. \nkern ", StyleBox["f", FontSlant->"Italic"], " := { ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M ", FontSlant->"Italic"], " | ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["m", FontSlant->"Italic"], ") = 0 } hei\[SZ]t der ", StyleBox["KERN", FontFamily->"Arial"], " von ", StyleBox["f ", FontSlant->"Italic"], "und ist ein ", StyleBox["R", FontSlant->"Italic"], "-Modul.\nbild ", StyleBox["f", FontSlant->"Italic"], " := { ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["m", FontSlant->"Italic"], ") \[Element] ", StyleBox["N", FontSlant->"Italic"], " | ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " } hei\[SZ]t das ", StyleBox["BILD", FontFamily->"Arial"], " von ", StyleBox["f", FontSlant->"Italic"], " und ist ein ", StyleBox["R", FontSlant->"Italic"], "-Modul.\n\n", StyleBox["Beweis", FontFamily->"Arial"], " f\[UDoubleDot]r die ", StyleBox["R", FontSlant->"Italic"], "-Modul-Eigenschaft:\nEs seien f\[UDoubleDot]r \[Nu] = 1, 2 ", Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " und ", StyleBox["r ", FontSlant->"Italic"], "\[Element] ", StyleBox["R", FontSlant->"Italic"], " .\nZum Bild: ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") + ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") \[Element] bild ", StyleBox["f", FontSlant->"Italic"], " , ", StyleBox["r ", FontSlant->"Italic"], "\[CenterDot] ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["r", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") \[Element] bild ", StyleBox["f", FontSlant->"Italic"], " .\nZum Kern: Es seien ", Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], " \[Element] kern ", StyleBox["f ", FontSlant->"Italic"], ". Mit ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") + ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = 0 + 0 = 0 folgt auch ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " \[Element] kern ", StyleBox["f ", FontSlant->"Italic"], ".\n ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["r", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") = ", StyleBox["r ", FontSlant->"Italic"], "\[CenterDot] ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ") = ", StyleBox["r", FontSlant->"Italic"], "\[CenterDot]0 = 0 ." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.13:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nSUMME, DIREKTE SUMME, DIREKTES PRODUKT (", FontFamily->"Arial"], StyleBox["sum, direct sum, direct product", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es seien ", StyleBox[" M", FontSlant->"Italic"], ", ", StyleBox["N", FontSlant->"Italic"], " Untermodulen eines ", StyleBox["R-", FontSlant->"Italic"], "Moduls. \nDie Menge { ", StyleBox["n", FontSlant->"Italic"], " + ", StyleBox["m", FontSlant->"Italic"], " | ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " , ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " } =: ", StyleBox["M", FontSlant->"Italic"], " + ", StyleBox["N ", FontSlant->"Italic"], "hei\[SZ]t die ", StyleBox["SUMME", FontFamily->"Arial"], " von ", StyleBox["N", FontSlant->"Italic"], " und ", StyleBox["M ", FontSlant->"Italic"], "und ist ein ", StyleBox["R", FontSlant->"Italic"], "-Untermodul.\nDie Summe ", StyleBox["M", FontSlant->"Italic"], " + ", StyleBox["N", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["DIREKT", FontFamily->"Arial"], ", wenn ", StyleBox["M", FontSlant->"Italic"], " ", StyleBox["\[Intersection]", FontSize->9], " ", StyleBox["N", FontSlant->"Italic"], " = {0} ; Schreibweise: ", StyleBox["M ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\[CirclePlus]\)]], " ", StyleBox["N ", FontSlant->"Italic"], ".\nDie Menge {(", StyleBox["n,", FontSlant->"Italic"], " ", StyleBox["m", FontSlant->"Italic"], ") | ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " , ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " } =: ", StyleBox["M", FontSlant->"Italic"], " \[Times] ", StyleBox["N ", FontSlant->"Italic"], "hei\[SZ]t ", StyleBox["DIREKTES PRODUKT", FontFamily->"Arial"], " von ", StyleBox["N", FontSlant->"Italic"], " und ", StyleBox["M ", FontSlant->"Italic"], "und ist ein ", StyleBox["R", FontSlant->"Italic"], "-Modul." }], "Text"], Cell[TextData[{ StyleBox["Folgerung:", FontFamily->"Arial"], " \nSind ", Cell[BoxData[ \(TraditionalForm\`M\_\[Nu]\)]], " , \[Nu] \[Element] \[DoubleStruckN] , Untermoduln eines ", StyleBox["R-", FontSlant->"Italic"], "Moduls ", StyleBox["M", FontSlant->"Italic"], " , so ist ", StyleBox["M", FontSlant->"Italic"], " direkte Summe der ", Cell[BoxData[ \(TraditionalForm\`M\_\[Nu]\)]], " , geschrieben \n", StyleBox["M", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "\[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`M\_n\)]], " , wenn ", StyleBox["M", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "+ \[CenterEllipsis] + ", Cell[BoxData[ \(TraditionalForm\`M\_n\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Nu] = 1\)\%n\)]], " ", Cell[BoxData[ \(TraditionalForm\`M\_\[Nu]\)]], " und ", Cell[BoxData[ \(TraditionalForm\`M\_\[Mu]\)]], StyleBox["\[Intersection] ", FontSize->9], Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Nu] \[NotEqual] \[Mu]\)\)]], Cell[BoxData[ \(TraditionalForm\`M\_\[Nu]\)]], " = {0} f\[UDoubleDot]r \[Mu] \[Element] \[DoubleStruckN] ." }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.2", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", Cell[BoxData[ \(TraditionalForm\`M\_\[Nu]\)]], " , \[Nu] \[Element] \[DoubleStruckN] , Untermoduln eines ", StyleBox["R-", FontSlant->"Italic"], "Moduls ", StyleBox["M", FontSlant->"Italic"], " . 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Es bezeichne ", StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["N", FontSlant->"Italic"], " die Menge der zugeh\[ODoubleDot]rigen \ \[CapitalADoubleDot]quivalenzklassen ", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " = ", StyleBox["m", FontSlant->"Italic"], " + ", StyleBox["N", FontSlant->"Italic"], " . \n", StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["N", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["QUOTIENTENMODUL", FontFamily->"Arial"], " und ist ein ", StyleBox["R", FontSlant->"Italic"], "-Modul.", StyleBox["\n", FontFamily->"Arial"], StyleBox["Statt des Namens Quotientenmodul wird auch der Name Faktormodul \ benutzt.\n\nDie Addition von Restklassen ", FontFamily->"Times New Roman"], "ist wohldefiniert", StyleBox[" durch ", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`\(m\_1\)\&_\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\(m\_2\)\&_\)]], " := ", Cell[BoxData[ \(TraditionalForm\`\(m\_1 + \ m\_2\)\&_\)]], " , denn f\[UDoubleDot]r \[Nu] = 1, 2 , ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\[Nu]\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["N", FontSlant->"Italic"], " und \n", Cell[BoxData[ \(TraditionalForm\`\(m\_\[Nu]\)\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(\[Mu]\_\[Nu]\)\&_\)]], " folgt ", Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], " \[Dash]", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\[Nu]\)]], " \[Element]", StyleBox[" ", FontFamily->"Times New Roman"], StyleBox["N", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" , also auch (", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") \[Dash]", StyleBox[" (", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`\[Mu]\_1\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_2\)]], ") \[Element]", StyleBox[" ", FontFamily->"Times New Roman"], StyleBox["N", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" .\nDie skalare Multiplikation wird durch ", FontFamily->"Times New Roman"], StyleBox["r", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["\[CenterDot]", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`m\&_\)]], StyleBox[" := ", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`\(r\[CenterDot]m\)\&_\)]], " definiert." }], "Text"], Cell[TextData[{ StyleBox["Beispiele:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "1) \nEs seien ", StyleBox["M", FontSlant->"Italic"], " der \[DoubleStruckCapitalZ]-Modul \[DoubleStruckCapitalQ] und ", StyleBox["N", FontSlant->"Italic"], " der \[DoubleStruckCapitalZ]-Untermodul mit ", StyleBox["N", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ] .\n\[DoubleStruckCapitalQ] / \ \[DoubleStruckCapitalZ] = { ", StyleBox["q", FontSlant->"Italic"], " + \[DoubleStruckCapitalZ] | ", StyleBox["q", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalQ] } . Zwei Elemente ", Cell[BoxData[ \(TraditionalForm\`p\&_\)]], " und ", Cell[BoxData[ \(TraditionalForm\`q\&_\)]], " aus \[DoubleStruckCapitalQ] / \[DoubleStruckCapitalZ] sind genau dann \ gleich, wenn ", StyleBox["p", FontSlant->"Italic"], " \[Dash] ", StyleBox["q", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalZ] .\n\n2)\nEs seien ", StyleBox["M", FontSlant->"Italic"], " der \[DoubleStruckCapitalQ]-Modul mit ", StyleBox["M", FontSlant->"Italic"], " := \[DoubleStruckCapitalQ] \[Times] \[DoubleStruckCapitalQ] und ", StyleBox["N", FontSlant->"Italic"], " der Untermodul mit ", StyleBox["N", FontSlant->"Italic"], " := { (", StyleBox["r", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], ") | ", StyleBox["r", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalQ] } . \nDann gilt ", StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["N", FontSlant->"Italic"], " = \[DoubleStruckCapitalQ] \[Times] \[DoubleStruckCapitalQ] / ", StyleBox["N", FontSlant->"Italic"], " = { (", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") + ", StyleBox["N ", FontSlant->"Italic"], " | (", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") \[Element] ", StyleBox["M", FontSlant->"Italic"], " } . 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Damit sind (4, 3) und (2, 1) Elemente der gleichen Restklasse { \ (2, 1) + (", StyleBox["r", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], ") | ", StyleBox["r", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalQ] } , also einer Geraden in \ \[DoubleStruckCapitalQ] \[Times] \[DoubleStruckCapitalQ] ." }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.3", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n\[Pi]", StyleBox[" : ", FontFamily->"Times New Roman"], StyleBox["M", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" \[RightArrow] ", FontFamily->"Times New Roman"], StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["mit \[Pi](", FontFamily->"Times New Roman"], StyleBox["m", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[") := ", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " ist ein ", StyleBox["R", FontSlant->"Italic"], "-Modul-Homomorphismus.\n\n", StyleBox["Beweis", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], " trivial!" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.1", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" ", FontFamily->"Arial"], StyleBox["1. 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Es seien ", StyleBox["f", FontSlant->"Italic"], " : ", StyleBox["M", FontSlant->"Italic"], " \[RightArrow]", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["N", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic"], "-Modul-Homomorphismus und \[Pi]", StyleBox[" : ", FontFamily->"Times New Roman"], StyleBox["M", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" \[RightArrow] ", FontFamily->"Times New Roman"], StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["mit \[Pi](", FontFamily->"Times New Roman"], StyleBox["m", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[") := ", FontFamily->"Times New Roman"], Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " .", StyleBox["\n", FontVariations->{"CompatibilityType"->0}], StyleBox["i", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") Dann ist kern ", FontVariations->{"CompatibilityType"->0}], StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["-Untermodul von ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" und bild ", FontVariations->{"CompatibilityType"->0}], StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["-Untermodul von ", FontVariations->{"CompatibilityType"->0}], StyleBox["N", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" .\n", FontVariations->{"CompatibilityType"->0}], StyleBox["ii", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") Es gibt einen eindeutig bestimmten ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic"], "-Modul-Isomorphismus ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], " : ", StyleBox["M", FontSlant->"Italic"], " / kern ", StyleBox["f", FontSlant->"Italic"], " \[RightArrow] bild ", StyleBox["f", FontSlant->"Italic"], " mit ", StyleBox["f", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], " \[SmallCircle] \[Pi] ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nDer Beweis zu ", StyleBox["i", FontSlant->"Italic"], ") folgt mit den Definitionen von Kern und Bild .\nZu ", StyleBox["ii", FontSlant->"Italic"], ") : Es seien ", StyleBox["m", FontSlant->"Italic"], ", \[Mu] \[Element] ", StyleBox["M", FontSlant->"Italic"], " . ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], " : ", StyleBox["M", FontSlant->"Italic"], " / kern ", StyleBox["f", FontSlant->"Italic"], " \[RightArrow] bild ", StyleBox["f", FontSlant->"Italic"], " mit ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], "(", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], ") := ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\)]], ") ist wohldefiniert, denn mit ", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\&_\)]], " , was ja \n", StyleBox["m", FontSlant->"Italic"], " \[Dash] \[Mu] \[Element] kern ", StyleBox["f", FontSlant->"Italic"], " bedeutet, folgt ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\)]], ") \[Dash] 0 = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\)]], ") + ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(\[Mu] \[Dash] ", Cell[BoxData[ \(TraditionalForm\`m\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\)]], " + \[Mu] \[Dash] ", StyleBox["m", FontSlant->"Italic"], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(\[Mu]) . \nDie Surjektivit\[ADoubleDot]t von ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], " ist trivial. \nZur Injektivit\[ADoubleDot]t: Es seien ", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\&_\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " / kern ", StyleBox["f ", FontSlant->"Italic"], "mit ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], "(", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], ") = ", Cell[BoxData[ \(TraditionalForm\`f\&_\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Mu]\&_\)]], ") , also ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", Cell[BoxData[ \(TraditionalForm\`m\)]], ") = ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(\[Mu]) . Dann folgt ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->9], "(", StyleBox["m", FontSlant->"Italic"], " \[Dash] \[Mu]) = 0 , also \n", StyleBox["m", FontSlant->"Italic"], " \[Dash] \[Mu] \[Element] kern ", StyleBox["f ", FontSlant->"Italic"], ", d.h. ", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\&_\)]], " . " }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.15:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nLINEARKOMBINATION (", FontFamily->"Arial"], StyleBox["linear combination", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es seien ", StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul, ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " und ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . \n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%n\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(m\_\[Nu]\)}]}], TraditionalForm]]], " hei\[SZ]t ", StyleBox["LINEARKOMBINATION", FontFamily->"Arial"], " der ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " . Die Menge der Linearkombinationen von ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " wird mit Lin (", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], ") oder auch mit [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], "] bezeichnet." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.16:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nENDLICH ERZEUGTER R-MODUL (", FontFamily->"Arial"], StyleBox["finitely generated R-modul", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nZYKLISCHER R-MODUL (", FontFamily->"Arial"], StyleBox["cyclic R-modul", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nFREIER R-MODUL (", FontFamily->"Arial"], StyleBox["free R-modul", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es sei ", StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul. \n", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["ENDLICH ERZEUGT", FontFamily->"Arial"], ", falls es ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " gibt mit ", StyleBox["M", FontSlant->"Italic"], " = [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], "] . \n", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["ZYKLISCHER R-MODUL", FontFamily->"Arial"], ", falls es ein ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " gibt mit ", StyleBox["M", FontSlant->"Italic"], " = [", StyleBox["m", FontSlant->"Italic"], "] .\n", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["FREIER R-MODUL", FontFamily->"Arial"], ", falls es ein ", StyleBox["n", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalN] und ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " gibt, so dass zu jedem ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M ", FontSlant->"Italic"], " eindeutig bestimmte ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], "existieren mit ", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%n\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(m\_\[Nu]\)}]}], TraditionalForm]]], " ." }], "Text"], Cell[TextData[{ "Bei zyklischen Moduln wird statt ", StyleBox["M", FontSlant->"Italic"], " = [", StyleBox["m", FontSlant->"Italic"], "] h\[ADoubleDot]ufig die Schreibweise ", StyleBox["mR", FontSlant->"Italic"], " benutzt." }], "Text"], Cell[TextData[{ StyleBox["Beispiele:", FontFamily->"Arial"], " \n1)\nEs sei ", StyleBox[" R", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul und 1 \[Element] ", StyleBox["R", FontSlant->"Italic"], " . ", StyleBox["M", FontSlant->"Italic"], " := [1] ist zyklischer ", StyleBox["R", FontSlant->"Italic"], "-Modul.\n\n2)\nEs sei ", StyleBox[" R", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul und ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . Es sei ", StyleBox["M", FontSlant->"Italic"], " := [", StyleBox["m", FontSlant->"Italic"], "] . Es sei \[CurlyPhi] : ", StyleBox["R", FontSlant->"Italic"], " \[RightArrow] ", StyleBox["M", FontSlant->"Italic"], " definiert durch \[CurlyPhi](", StyleBox["r", FontSlant->"Italic"], ") := ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " . Dann ist \[CurlyPhi] surjektiver Modul-Homomorphismus.\nkern \ \[CurlyPhi] = { ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], "|", StyleBox[" r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m ", FontSlant->"Italic"], "= 0 } =: ", StyleBox["Y", FontSlant->"Italic"], " ist Ideal in ", StyleBox["R", FontSlant->"Italic"], " . Deshalb gilt nach Satz II.4.1 ", StyleBox["R", FontSlant->"Italic"], " / ", StyleBox["Y", FontSlant->"Italic"], " \[TildeEqual] ", StyleBox["M", FontSlant->"Italic"], " , d.h. jeder zyklische Modul ist isomorph zum Quotientenmodul des Rings \ nach einem Ideal.\n\n3)\nDer Nullmodul {0} ist ein freier Modul, der durch \ die leere Menge erzeugt wird." }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.4", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es sei ", FontVariations->{"CompatibilityType"->0}], StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul. ", StyleBox["M", FontSlant->"Italic"], " ist genau dann frei, wenn es einen", StyleBox[" R", FontSlant->"Italic"], "-Modul-Isomorphismus von ", Cell[BoxData[ \(TraditionalForm\`R\^n\)]], " nach ", StyleBox["M", FontSlant->"Italic"], " gibt." }], "Text"], Cell["Beweis als \[CapitalUDoubleDot]bung.", "Text"], Cell[TextData[{ StyleBox["Definition II.4.17:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nLINEAR UN / ABH\[CapitalADoubleDot]NGIG (", FontFamily->"Arial"], StyleBox["linear in / dependent", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es sei ", StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul. \nEs seien ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " . Die Familie ", StyleBox["F", FontSlant->"Italic"], " := (", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], ") hei\[SZ]t ", StyleBox["LINEAR UNABH\[CapitalADoubleDot]NGIG,", FontFamily->"Arial"], " wenn aus ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%n\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(m\_\[Nu]\)}]}], TraditionalForm]]], " = 0 mit ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], " auch ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " = 0 f\[UDoubleDot]r alle \[Nu] \[Element] \[DoubleStruckN] folgt. \ Andernfalls hei\[SZ]t ", StyleBox["F", FontSlant->"Italic"], " ", StyleBox["LINEAR ABH\[CapitalADoubleDot]NGIG", FontFamily->"Arial"], ". " }], "Text"], Cell[TextData[{ StyleBox["Bemerkung:", FontFamily->"Arial"], " Die Definition eines freien Moduls ist \[ADoubleDot]quivalent zur \ Existenz eines linear unabh\[ADoubleDot]ngigen Erzeugendensystems. Dieses \ wird dann wieder ", StyleBox["BASIS", FontFamily->"Arial"], " genannt." }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.2", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es seien ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein Integrit\[ADoubleDot]tsbereich, ", FontVariations->{"CompatibilityType"->0}], StyleBox["M ", FontSlant->"Italic"], "ein ", StyleBox["R-", FontSlant->"Italic"], "Modul", StyleBox[" mit einer Basis ", FontVariations->{"CompatibilityType"->0}], "(", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], ")", StyleBox[" und ", FontVariations->{"CompatibilityType"->0}], "(", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], ") eine linear unabh\[ADoubleDot]ngige Familie von Elementen aus ", StyleBox["M", FontSlant->"Italic"], " . Dann ist ", StyleBox["k", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nInduktion nach ", StyleBox["n", FontSlant->"Italic"], " :\n", StyleBox["n", FontSlant->"Italic"], " = 1 : \nEs sei ", StyleBox["M", FontSlant->"Italic"], " := [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "] und damit ", StyleBox["M", FontSlant->"Italic"], " \[TildeEqual] ", StyleBox["R", FontSlant->"Italic"], " . Angenommen es sei (", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], ") linear unabh\[ADoubleDot]ngig.\nMit ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], " und ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_2\), "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[NotEqual] 0 folgt dann ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_2\), "TraditionalForm"], "\[CenterDot]", \(x\_1\)}], TraditionalForm]]], "\[Dash] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterDot]", \(x\_2\)}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\(r\_2\), "TraditionalForm"], "\[CenterDot]", \(r\_1\)}], "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], " \[Dash] ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_2\), "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], " = 0 , d.h. ein Widerspruch; also gilt ", StyleBox["k", FontSlant->"Italic"], " \[LessEqual] 1 .\n\nAllgemeiner Induktionsschritt:\nEs sei (", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], ") linear unabh\[ADoubleDot]ngig und es sei ", StyleBox["M", FontSlant->"Italic"], " := [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], "] \[TildeEqual] ", Cell[BoxData[ \(TraditionalForm\`R\^n\)]], " . Es sei ", StyleBox["M' ", FontSlant->"Italic"], ":= [", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], "] \[TildeEqual] ", Cell[BoxData[ FormBox[ SuperscriptBox["R", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], ". Falls {", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], "} \[Subset] ", StyleBox["M'", FontSlant->"Italic"], " , dann ist nach Induktionsvoraussetzung ", StyleBox["k", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " \[Dash] 1 < ", StyleBox["n", FontSlant->"Italic"], " .\nFalls {", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], "} \[NotSubset] ", StyleBox["M'", FontSlant->"Italic"], " , dann sei o.B.d.A. ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], "\[NotElement] ", StyleBox["M'", FontSlant->"Italic"], " . \nEs gilt ", StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["M'", FontSlant->"Italic"], " \[TildeEqual] ", Cell[BoxData[ \(TraditionalForm\`R\[CenterDot]m\_1\)]], ". Von daher ist (", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "1"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "\[Nu]"], TraditionalForm]]], ") f\[UDoubleDot]r jedes \[Nu] \[GreaterEqual] 2 linear \ abh\[ADoubleDot]ngig in ", StyleBox["M", FontSlant->"Italic"], " / ", StyleBox["M'", FontSlant->"Italic"], " (analog zum Fall ", StyleBox["n", FontSlant->"Italic"], " = 1), d.h. es gibt ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`s\_\[Nu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , nicht beide gleich Null mit ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], " ", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "1"], TraditionalForm]]], " + ", Cell[BoxData[ \(TraditionalForm\`s\_\[Nu]\)]], " ", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "\[Nu]"], TraditionalForm]]], " = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " . Es ist ", Cell[BoxData[ \(TraditionalForm\`s\_\[Nu]\)]], " \[NotEqual] 0 , denn sonst w\[ADoubleDot]re ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], " ", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "1"], TraditionalForm]]], " = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " , also ", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "1"], TraditionalForm]]], " = 0 + ", StyleBox["M'", FontSlant->"Italic"], " , d.h. ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], "\[Element] ", StyleBox["M'", FontSlant->"Italic"], " .\nEs sei nun ", Cell[BoxData[ \(TraditionalForm\`y\_\[Nu]\)]], " := ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_\[Nu]\), "TraditionalForm"], \(x\_1\)}], TraditionalForm]]], "+ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(s\_\[Nu]\), "TraditionalForm"], \(x\_\[Nu]\)}], TraditionalForm]]], " \[Element] ", StyleBox["M' ", FontSlant->"Italic"], "f\[UDoubleDot]r \[Nu] = 2,\[Ellipsis], ", StyleBox["k", FontSlant->"Italic"], " . (", Cell[BoxData[ \(TraditionalForm\`y\_2\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`y\_k\)]], ") ist l.u. in ", Cell[BoxData[ FormBox[ StyleBox[\(M'\), FontSlant->"Italic"], TraditionalForm]]], " , denn sei\n ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 2\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(y\_\[Nu]\)}]}], TraditionalForm]]], " = 0 , dann ist ( ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 2\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(r\_\[Nu]\)}]}], TraditionalForm]]], " ) ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], "+ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 2\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(s\_\[Nu]\)}]}], "TraditionalForm"], "\[CenterDot]", \(x\_\[Nu]\)}], TraditionalForm]]], " = 0 . Da (", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], ") linear unabh\[ADoubleDot]ngig und ", Cell[BoxData[ \(TraditionalForm\`s\_\[Nu]\)]], " \[NotEqual] 0 , folgt \n ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " = 0 f\[UDoubleDot]r \[Nu] = 2,\[Ellipsis], ", StyleBox["k", FontSlant->"Italic"], " , da ", StyleBox["R", FontSlant->"Italic"], " Integrit\[ADoubleDot]tsbereich. Also ist nach Induktionsvoraussetzung \ ", StyleBox["k", FontSlant->"Italic"], " \[Dash] 1 \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " \[Dash] 1 , d.h. ", StyleBox["k", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Bemerkung:", FontFamily->"Arial"], "\nDie Elementeanzahl einer Basis ist daher wohlbestimmt; sie hei\[SZ]t ", StyleBox["RANG", FontFamily->"Arial"], " ", StyleBox["(", FontFamily->"Arial"], StyleBox["rank", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], " von ", StyleBox["M", FontSlant->"Italic"], " . Der Ausdruck Dimension ist bei Moduln nicht gebr\[ADoubleDot]uchlich.\n\ \n", StyleBox["Beispiele:", FontFamily->"Arial"], "\n1)\n", StyleBox["M", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ], ", StyleBox["R", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ] . ", StyleBox["M", FontSlant->"Italic"], " ist frei vom Rang 1 .\nEs sei ", StyleBox["N", FontSlant->"Italic"], " := { 2\[CenterDot]", StyleBox["z", FontSlant->"Italic"], " | ", StyleBox["z", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalZ] } = 2\[DoubleStruckCapitalZ] . Dies \ ist ein ", StyleBox["R", FontSlant->"Italic"], "-Untermodul von ", StyleBox["M", FontSlant->"Italic"], " und ", StyleBox["N", FontSlant->"Italic"], " ist frei vom Rang 1 , aber ", StyleBox["N", FontSlant->"Italic"], " \[NotEqual] ", StyleBox["M", FontSlant->"Italic"], " . \n(Wie ist das bei Untervektorr\[ADoubleDot]umen ", StyleBox["U", FontSlant->"Italic"], " \[Subset] ", StyleBox["V", FontSlant->"Italic"], " mit dim ", StyleBox["U", FontSlant->"Italic"], " = dim ", StyleBox["V", FontSlant->"Italic"], " ?)\n\n2)\nBei Moduln gilt nicht, dass jedes Erzeugendensystem eine Basis \ enth\[ADoubleDot]lt.\nDazu sei \[DoubleStruckCapitalZ] als \ \[DoubleStruckCapitalZ]-Modul betrachtet. (1) ist Basis und (2, 3) ist ein \ Erzeugendensystem, denn 3 \[Dash] 2 = 1 . Aber \n", StyleBox["X", FontSlant->"Italic"], " := (2, 3) ist keine \[DoubleStruckCapitalZ]-Basis von \ \[DoubleStruckCapitalZ] , denn wegen 3\[CenterDot]2 \[Dash] 2\[CenterDot]3 = \ 0 ist ", StyleBox["X", FontSlant->"Italic"], " linear abh\[ADoubleDot]ngig.", StyleBox["\n\n", FontColor->RGBColor[1, 0, 0]], "3)\nEs sei ", StyleBox["G", FontSlant->"Italic"], " eine endliche abelsche Gruppe mit der Ordnung ", StyleBox["n", FontSlant->"Italic"], " \[NotEqual] 0. Dann ist ", StyleBox["G", FontSlant->"Italic"], " ein \[DoubleStruckCapitalZ]-Modul. Jede nichtleere Teilfamilie von ", StyleBox["G", FontSlant->"Italic"], " ist linear abh\[ADoubleDot]ngig, denn sei ", StyleBox["g", FontSlant->"Italic"], " \[Element] ", StyleBox["G", FontSlant->"Italic"], " ; dann gilt ", StyleBox["n", FontSlant->"Italic"], "\[CenterDot]", StyleBox["g", FontSlant->"Italic"], " = 0 . Also hat dieser \[DoubleStruckCapitalZ]-Modul keine Basis!" }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Moduln \[UDoubleDot]ber Euklidischen Ringen ", FontWeight->"Plain", FontVariations->{"CompatibilityType"->0}]], "Subtitle"], Cell[TextData[{ StyleBox["Definition II.4.18:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nEUKLIDISCHER RING (", FontFamily->"Arial"], StyleBox["euclidian ring", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Ein Integrit\[ADoubleDot]tsbereich ", StyleBox[" R ", FontSlant->"Italic"], "hei\[SZ]t ", StyleBox["EUKLIDISCHER RING, ", FontFamily->"Arial"], "wenn es eine Funktion ", StyleBox["n", FontSlant->"Italic"], " : ", StyleBox["R", FontSlant->"Italic"], " \\ {0} \[RightArrow] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalN]\_0\)]], " mit folgenden Eigenschaften gibt:\n(ER ", StyleBox["i", FontSlant->"Italic"], ") ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\)]], ") \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`b\)]], ") f\[UDoubleDot]r alle ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " \\ {0} .\n(ER ", StyleBox["ii", FontSlant->"Italic"], ") Zu beliebigen ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 gibt es ", StyleBox["q", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], ", so dass gilt:\n\t ", StyleBox["a", FontSlant->"Italic"], " = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " + ", StyleBox["r", FontSlant->"Italic"], " , wobei ", StyleBox["r", FontSlant->"Italic"], " = 0 oder ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") < ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["b", FontSlant->"Italic"], ") .\n\nDie Funktion ", StyleBox["n", FontSlant->"Italic"], " wird h\[ADoubleDot]ufig ", StyleBox["NORM", FontFamily->"Arial"], " oder ", StyleBox["GRAD", FontFamily->"Arial"], " genannt." }], "Text"], Cell[TextData[{ "(ER ", StyleBox["ii", FontSlant->"Italic"], ") fordert \" < \" und nicht \" \[LessEqual] \" !\n" }], "Text"], Cell[TextData[{ StyleBox["Beispiele:", FontFamily->"Arial"], " \n1)\nIm Ring der ganzen Zahlen \[DoubleStruckCapitalZ] entspricht (ER \ ", StyleBox["ii", FontSlant->"Italic"], ") gerade der \[UDoubleDot]blichen Division mit Rest, bei der ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") := | ", StyleBox["r", FontSlant->"Italic"], " | benutzt wird.\n\n2)\nEs sei ", StyleBox["K", FontSlant->"Italic"], " ein K\[ODoubleDot]rper und ", StyleBox["R", FontSlant->"Italic"], " = ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] der Polynomring \[UDoubleDot]ber ", StyleBox["K", FontSlant->"Italic"], " . F\[UDoubleDot]r die Normfunktion wird jetzt der Grad eines Polynoms \ gesetzt. Das Nullpolynom hat den \"Grad\" \[Dash]\[Infinity] . Der Grad eines \ nicht konstanten Polynoms ungleich dem Nullpolynom ist gr\[ODoubleDot]\[SZ]er \ als Null und es gilt f\[UDoubleDot]r Polynome ", StyleBox["g", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], "), ", StyleBox["h", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") ungleich Null grad( ", StyleBox["g", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") ) \[LessEqual] grad( ", StyleBox["g", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", StyleBox["h", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") ) .\nDie (Polynom-) Division mit Rest liefert, falls ", StyleBox["h", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") \[NotEqual] 0 , Polynome ", StyleBox["q", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ")", StyleBox[" ", FontSlant->"Italic"], " und ", StyleBox["r", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") mit ", StyleBox["g", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") = ", StyleBox["q", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", StyleBox["h", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") + ", StyleBox["r", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") , wobei entweder ", StyleBox["r", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") = 0 oder grad( ", StyleBox["r", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") ) < grad( ", StyleBox["h", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") ) .\n\n3) \nPolynomdivision \[UDoubleDot]ber \[DoubleStruckCapitalZ]:\n\n\ ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " + ", Cell[BoxData[ \(TraditionalForm\`2 x\^2\)]], " + 3 ", StyleBox["x", FontSlant->"Italic"], " + 2 = ", Cell[BoxData[ \(TraditionalForm\`\((x\^2\ + \ 4)\)\)]], "(", StyleBox["x", FontSlant->"Italic"], " + 2) \[Dash] (", StyleBox["x", FontSlant->"Italic"], " + 6)\n", Cell[BoxData[ \(TraditionalForm\`\[Dash]( x\^\(\(3\)\(\ \ \ \ \ \ \ \ \)\)\ \ \ \ \ \ \ \ \ + \ 4 x)\)], FontVariations->{"CompatibilityType"->0}], "\n", StyleBox["__________________________________________________________________\ _______________________\n ", FontSize->2], " ", Cell[BoxData[ \(TraditionalForm\`2 x\^2\)]], " \[Dash]", StyleBox["x\n ", FontSlant->"Italic"], "\[Dash](", Cell[BoxData[ \(TraditionalForm\`2 x\^2\)]], " + 8) \n", StyleBox[" \ ______________________________________________________________________________\ _________\n ", FontSize->2], " \[Dash]", StyleBox["x", FontSlant->"Italic"], " \[Dash] 6 ", StyleBox[" \n\n", FontSlant->"Italic"], "4)\nPolynomdivision \[UDoubleDot]ber ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_5\)]], ":\n \n ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\(2\&_\) x\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`3\&_\)]], " ", StyleBox["x", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`2\&_\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\((x\^2\ + 4\&_)\)\)]], "(", StyleBox["x", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`2\&_\)]], ") + (", Cell[BoxData[ \(TraditionalForm\`4\&_\)]], StyleBox["x", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`4\&_\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\[Dash]( x\^\(\(3\)\(\ \ \ \ \ \ \ \ \)\)\ \ \ \ \ \ \ \ \ \ + \ \(4\&_\) x)\)], FontVariations->{"CompatibilityType"->0}], "\n", StyleBox["__________________________________________________________________\ _______________________\n ", FontSize->2], " ", Cell[BoxData[ \(TraditionalForm\`\(2\&_\) x\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`4\&_\)]], " ", StyleBox["x\n ", FontSlant->"Italic"], "\[Dash](", Cell[BoxData[ \(TraditionalForm\`\(2\&_\) x\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`8\&_\)]], ") \n", StyleBox[" \ ______________________________________________________________________________\ _________\n ", FontSize->2], " ", Cell[BoxData[ \(TraditionalForm\`4\&_\)]], StyleBox["x", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`\[Dash]6\&_\)]], " ", StyleBox[" \n ", FontSlant->"Italic"] }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.3", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nIn euklidischen Ringen sind alle Ideale Hauptideale." }], "Text"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nEs seien ", StyleBox["R", FontSlant->"Italic"], " ein euklidischer Ring und ", StyleBox["I", FontSlant->"Italic"], " ein Ideal.\n1. Fall: ", StyleBox["I", FontSlant->"Italic"], " = {0} trivial!\n2. Fall: Es sei ", StyleBox["I", FontSlant->"Italic"], " \[NotEqual] {0} . Dann gibt es ein ", StyleBox["y", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " , ", StyleBox["y", FontSlant->"Italic"], " \[NotEqual] 0 , so dass ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["y", FontSlant->"Italic"], ") minimal ist. \nJetzt wird gezeigt, dass jedes ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " ein Vielfaches von ", StyleBox["y", FontSlant->"Italic"], " ist: \nZu ", StyleBox["y", FontSlant->"Italic"], " und ", StyleBox["x", FontSlant->"Italic"], " gibt ", StyleBox["q", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], " mit ", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["y", FontSlant->"Italic"], " + ", StyleBox["r", FontSlant->"Italic"], " , wobei ", StyleBox["r", FontSlant->"Italic"], " = 0 oder ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") < ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["y", FontSlant->"Italic"], ") . Mit ", StyleBox["x", FontSlant->"Italic"], " und ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["y", FontSlant->"Italic"], " ist auch ", StyleBox["r", FontSlant->"Italic"], " = ", StyleBox["x", FontSlant->"Italic"], " \[Dash] ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["y", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " . Nun ist ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["y", FontSlant->"Italic"], ") aber minimal vorausgesetzt, dh. ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") < ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["y", FontSlant->"Italic"], ") ist nicht m\[ODoubleDot]glich, also ist ", StyleBox["r", FontSlant->"Italic"], " = 0 , also ", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["y", FontSlant->"Italic"], " . Damit ist \n", StyleBox["I", FontSlant->"Italic"], " = ", StyleBox["R", FontSlant->"Italic"], "\[CenterDot]", StyleBox["y", FontSlant->"Italic"], " , d.h. Hauptideal \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Bemerkung:", FontFamily->"Arial"], "\nRinge, in denen alle Ideale Hauptideale sind, werden ", StyleBox["Hauptidealringe", FontFamily->"Arial"], " (", StyleBox["principal ideal domain", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[" : PID", FontFamily->"Arial"], ") genannt." }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.4", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs sei", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring, ", StyleBox["M", FontSlant->"Italic"], " sei ein freier ", StyleBox["R", FontSlant->"Italic"], "-Modul und ", StyleBox["N", FontSlant->"Italic"], " ein Untermodul von ", StyleBox["M", FontSlant->"Italic"], " . Dann ist auch ", StyleBox["N", FontSlant->"Italic"], " ein freier ", StyleBox["R", FontSlant->"Italic"], "-Modul. " }], "Text"], Cell[TextData[{ "Im ", StyleBox["Beweis ", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "wird eine Basis von ", StyleBox["N ", FontSlant->"Italic"], "konstruiert", StyleBox[":", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nAls freier ", StyleBox["R", FontSlant->"Italic"], "-Modul gilt nach Lemma II.4.4 ", StyleBox["M", FontSlant->"Italic"], " := [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], "] \[TildeEqual] ", Cell[BoxData[ \(TraditionalForm\`R\^n\)]], ". \nF\[UDoubleDot]r die Untermoduln ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " := ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["\[Intersection]", FontSize->9], " [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], "] gilt ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], "\[Subset] ", Cell[BoxData[ \(TraditionalForm\`N\_2\)]], "\[Subset] \[Ellipsis] \[Subset] ", Cell[BoxData[ \(TraditionalForm\`N\_n\)]], " = ", StyleBox["N", FontSlant->"Italic"], " .\n\nJetzt wird mit Induktion nach \[Nu] gezeigt, dass ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " freier ", StyleBox["R", FontSlant->"Italic"], "-Modul ist.\n\[Nu] = 1 :\nMit ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " = ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["\[Intersection]", FontSize->9], " [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "] sei ", StyleBox["I", FontSlant->"Italic"], " := {", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`r\[CenterDot]m\_1\)]], "\[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], "} . \n", StyleBox["I", FontSlant->"Italic"], " ist Ideal in ", StyleBox["R", FontSlant->"Italic"], " , denn:\nEs seien ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " , ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], ", dann folgt (", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], ", also ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " , und \n(", StyleBox["r", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\(\(\[CenterDot]\)\(r\_1\)\)\)]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot]( ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], ") \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " , also ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " . \nNach dem vorigen Satz ist ", StyleBox["I ", FontSlant->"Italic"], "Hauptideal, also ", StyleBox["I", FontSlant->"Italic"], " = ", StyleBox["R", FontSlant->"Italic"], "\[CenterDot]", StyleBox["a ", FontSlant->"Italic"], "f\[UDoubleDot]r ein ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . \nEs gilt ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " = [", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "] , denn:\nEs sei ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " . Dann gibt es ein \[Alpha] \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["x", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[Alpha]\[CenterDot]m\_1\)]], ". Also ist \[Alpha] \[Element] ", StyleBox["I", FontSlant->"Italic"], " und damit \[Alpha] = \[Beta]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " , woraus ", StyleBox["x", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[Beta]\[CenterDot]a\[CenterDot]m\_1\)]], ", also \n", StyleBox["x", FontSlant->"Italic"], " \[Element] [", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "] folgt .\nEs sei nun ", StyleBox["x", FontSlant->"Italic"], " \[Element] [", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "] . Dann ist ", StyleBox["x", FontSlant->"Italic"], " = \[Gamma] \[CenterDot] ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\(\(\[CenterDot]\)\(\ \)\(m\_1\)\)\)]], " , und da \[Gamma] \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["I", FontSlant->"Italic"], " , ist ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " , w.z.b.w. .\n\nWenn ", StyleBox["a", FontSlant->"Italic"], " = 0 , dann ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " = {0} .\nEs sei nun ", StyleBox["a", FontSlant->"Italic"], " \[NotEqual] 0 .\nJedes ", StyleBox["x", FontSlant->"Italic"], " \[Element] [", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "] wird eindeutig dargestellt: Es sei etwa ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = ", StyleBox["x", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " . Da ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " Basiselement, folgt \n", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " und da ", StyleBox["a", FontSlant->"Italic"], " \[NotEqual] 0 und ", StyleBox["R", FontSlant->"Italic"], " Integrit\[ADoubleDot]tsbreich auch ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " .\nEs gilt ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " \[TildeEqual] ", StyleBox["R", FontSlant->"Italic"], " , denn durch ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " \[RightTeeArrow] ", StyleBox["r", FontSlant->"Italic"], " wird ein Isomorphismus definiert.\nAlso ist ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], " frei.\n\nAllgemeiner Induktionsschritt: \nInduktionsvoraussetzung: F\ \[UDoubleDot]r ein \[Nu] sei ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " freier ", StyleBox["R", FontSlant->"Italic"], "-Modul.\nEs sei ", StyleBox["I", FontSlant->"Italic"], " := {", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " | Es gibt ein ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " mit ", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(b\_1\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(b\_\[Nu]\), "TraditionalForm"], \(m\_\[Nu]\)}], TraditionalForm]]], " + ", StyleBox["r", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`m\_\(\[Nu] + 1\)\)]], " . } . \nWie oben wird gezeigt, dass ", StyleBox["I", FontSlant->"Italic"], " Hauptideal ist, also ", StyleBox["I", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`R\[CenterDot]a\_\(\[Nu] + 1\)\)]], " f\[UDoubleDot]r ein ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " .\nWenn ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " = 0 , dann folgt ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " und ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " ist daher frei.\nWenn ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " \[NotEqual] 0 , dann gibt es ein ", StyleBox["w", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " mit ", StyleBox["w", FontSlant->"Italic"], " \[NotEqual] 0 und ", StyleBox["w", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(w\_1\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(w\_\[Nu]\), "TraditionalForm"], \(m\_\[Nu]\)}], TraditionalForm]]], " + ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`m\_\(\[Nu] + 1\)\)]], " , wobei ", Cell[BoxData[ \(TraditionalForm\`w\_\[Mu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " .\nEs gilt ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " \[CirclePlus] [", StyleBox["w", FontSlant->"Italic"], "] denn:\nEs sei ", StyleBox["m", FontSlant->"Italic"], " := ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(c\_1\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(c\_\[Nu]\), "TraditionalForm"], \(m\_\[Nu]\)}], TraditionalForm]]], " + ", StyleBox["r", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`m\_\(\[Nu] + 1\)\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " beliebig und ", StyleBox["r", FontSlant->"Italic"], " = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " mit ", StyleBox["q", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . Nun ist\n", StyleBox["m", FontSlant->"Italic"], " \[Dash] ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["w", FontSlant->"Italic"], " = (", Cell[BoxData[ \(TraditionalForm\`c\_1\)]], " \[Dash] ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`w\_1\)]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " + \[Ellipsis] + (", Cell[BoxData[ \(TraditionalForm\`c\_\[Nu]\)]], " \[Dash] ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`w\_\[Nu]\)]], ") \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " . Also ist ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " \[Subset] [ ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " ", StyleBox["\[Union]", FontSize->9], " {", StyleBox["w", FontSlant->"Italic"], "} ] . \nAndererseits ist [ ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " ", StyleBox["\[Union]", FontSize->9], " {", StyleBox["w", FontSlant->"Italic"], "} ] \[Subset] ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " , also ist ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " + ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["w ", FontSlant->"Italic"], ".\nBleibt zu zeigen, dass die Summe direkt ist:\nEs sei ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " ", StyleBox["\[Intersection]", FontSize->9], " ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["w ", FontSlant->"Italic"], " , also\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(c\_1\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(c\_\[Nu]\), "TraditionalForm"], \(m\_\[Nu]\)}], TraditionalForm]]], " = ", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot]", StyleBox["w", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ( ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(w\_1\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(w\_\[Nu]\), "TraditionalForm"], \(m\_\[Nu]\)}], TraditionalForm]]], " + ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`m\_\(\[Nu] + 1\)\)]], ") . Wegen der Eindeutigkeit der Darstellung folgt \n0 =", StyleBox[" r", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Nu] + 1\)\)]], " , also ", StyleBox["r", FontSlant->"Italic"], " = 0 und damit ", StyleBox["x", FontSlant->"Italic"], " = 0 , d.h. ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[CirclePlus]\)]], " ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["w ", FontSlant->"Italic"], ".\nDeshalb ist ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " freier ", StyleBox["R", FontSlant->"Italic"], "-Modul. Die Basis von ", Cell[BoxData[ \(TraditionalForm\`N\_\(\[Nu] + 1\)\)]], " setzt sich zusammen aus der Basis von ", Cell[BoxData[ \(TraditionalForm\`N\_\[Nu]\)]], " und ", StyleBox["w", FontSlant->"Italic"], " \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Beispiel:", FontFamily->"Arial"], " \nEs seien ", StyleBox["R", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ] und ", StyleBox["M", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ] \[Times] \[DoubleStruckCapitalZ] . \n", StyleBox["M", FontSlant->"Italic"], " ist frei mit der Basis ( ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = (1, 0) , ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " = (0,1) ) . \nEs sei nun ", StyleBox["N", FontSlant->"Italic"], " := { (", Cell[BoxData[ \(TraditionalForm\`z\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`z\_2\)]], ") \[Element] ", StyleBox["M", FontSlant->"Italic"], " | 6 ", Cell[BoxData[ \(TraditionalForm\`z\_1\)]], " + 2 ", Cell[BoxData[ \(TraditionalForm\`z\_2\)]], " = 0 } . ", StyleBox["N", FontSlant->"Italic"], " ist Untermodul. \nKonstruiere Basis: ", Cell[BoxData[ \(TraditionalForm\`N\_1\)]], "= ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`\([m\_1]\)\)]], " = {(0, 0)} \[Subset] ", Cell[BoxData[ \(TraditionalForm\`N\_2\)]], " = ", StyleBox["N", FontSlant->"Italic"], " . ", Cell[BoxData[ \(TraditionalForm\`I\_2\)]], " = ", StyleBox["R", FontSlant->"Italic"], "\[CenterDot]3 , da (", Cell[BoxData[ \(TraditionalForm\`z\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`z\_2\)]], ") = ", Cell[BoxData[ \(TraditionalForm\`z\_1\)]], "(1, 0) + ", Cell[BoxData[ \(TraditionalForm\`z\_2\)]], "(0, 1) und 3 | ", Cell[BoxData[ \(TraditionalForm\`z\_2\)]], " . W\[ADoubleDot]hle ", StyleBox["w", FontSlant->"Italic"], " := (\[Dash]1, 3) . Deshalb ist ", StyleBox["N", FontSlant->"Italic"], " = [ (\[Dash]1, 3) ] \[TildeEqual] ", StyleBox["R", FontSlant->"Italic"], " . " }], "Text"], Cell[TextData[{ StyleBox["Korollar II.4.4", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es seien ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontVariations->{"CompatibilityType"->0}], "euklidischer Ring", StyleBox[", ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" freier ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["-Modul mit Basis (", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], StyleBox[") und ", FontVariations->{"CompatibilityType"->0}], StyleBox["N", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein Untermodul von ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" . Dann gibt es ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\[Mu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , so dass mit\n\t", Cell[BoxData[ \(TraditionalForm\`w\_1\)]], " := ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_11\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], ",\n\t", Cell[BoxData[ \(TraditionalForm\`w\_2\)]], " := ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_12\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_22\), "TraditionalForm"], \(m\_2\)}], TraditionalForm]]], ",\n\t \[VerticalEllipsis]\n\t", Cell[BoxData[ \(TraditionalForm\`w\_n\)]], " := ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_\(1 n\)\), "TraditionalForm"], \(m\_1\)}], TraditionalForm]]], "+ \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SubscriptBox["a", StyleBox["mn", FontSlant->"Italic"]], "TraditionalForm"], \(m\_n\)}], TraditionalForm]]], "\ndie Familie ( ", Cell[BoxData[ \(TraditionalForm\`w\_\[Nu]\)]], " | \[Nu] so, dass ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\[Nu]\)]], " \[NotEqual] 0 ) eine Basis von ", StyleBox["N", FontSlant->"Italic"], " bildet.\nWeiterhin gilt: Rang ", StyleBox["N", FontSlant->"Italic"], " = ", StyleBox["n", FontSlant->"Italic"], " genau dann, wenn ", Cell[BoxData[ \(TraditionalForm\`a\_11\)]], " \[CenterDot] \[Ellipsis] \[CenterDot] ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox["nn", FontSlant->"Italic"]], TraditionalForm]]], " \[NotEqual] 0 . " }], "Text"], Cell[TextData[{ "Der ", StyleBox["Beweis ", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "folgt direkt aus dem des vorherigen Satzes \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.19:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nTORSIONSELEMENT (", FontFamily->"Arial"], StyleBox["torsion element", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es seien ", StyleBox["R", FontSlant->"Italic"], " ein Integrit\[ADoubleDot]tsbereich und ", StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul. Ein ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["TORSIONSELEMENT", FontFamily->"Arial"], ", falls es ein ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["r", FontSlant->"Italic"], " \[NotEqual] 0 gibt, so dass ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 ." }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.5", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es seien ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic"], " ein Integrit\[ADoubleDot]tsbereich", StyleBox[" und ", FontVariations->{"CompatibilityType"->0}], StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul. Dann bildet die Menge aller Torsionselemente von ", StyleBox["M", FontSlant->"Italic"], " einen Untermodul ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " von ", StyleBox["M", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nEs seien ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " , also ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterDot]", \(m\_1\)}], TraditionalForm]]], " = 0 und ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_2\), "TraditionalForm"], "\[CenterDot]", \(m\_2\)}], TraditionalForm]]], " = 0 ", StyleBox[" ", FontSlant->"Italic"], "mit ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " \[NotEqual] 0 und ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[NotEqual] 0 . Dann ist ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterDot]", \(r\_2\)}], TraditionalForm]]], " \[NotEqual] 0 und \n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterDot]", \(r\_2\)}], TraditionalForm]]], " \[CenterDot] (", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = 0 , also ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " .\nEs seien ", Cell[BoxData[ \(TraditionalForm\`m\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " und ", Cell[BoxData[ \(TraditionalForm\`r\_0\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", Cell[BoxData[ \(TraditionalForm\`r\_0\)]], " \[NotEqual] 0 , also ", Cell[BoxData[ \(TraditionalForm\`r\_0\)]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 ; dann ist auch ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`r\_0\)]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`r\_0\)]], " \[CenterDot] ", StyleBox["r ", FontSlant->"Italic"], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 f\[UDoubleDot]r alle ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , also ", StyleBox["r ", FontSlant->"Italic"], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.20:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nTORSIONSMODUL (", FontFamily->"Arial"], StyleBox["torsion module", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nTORSIONSFREI (", FontFamily->"Arial"], StyleBox["torsion-free", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], StyleBox["Es seien ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic"], " ein Integrit\[ADoubleDot]tsbereich", StyleBox[" und", FontVariations->{"CompatibilityType"->0}], " ", StyleBox[" M", FontSlant->"Italic"], " ein ", StyleBox["R-", FontSlant->"Italic"], "Modul. \n", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["TORSIONSMODUL,", FontFamily->"Arial"], " falls ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = ", StyleBox["M", FontSlant->"Italic"], " und ", StyleBox["M", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["TORSIONSFREI,", FontFamily->"Arial"], " falls ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = {0} ." }], "Text"], Cell[TextData[{ StyleBox["Beispiel:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nEs seien ", StyleBox["R", FontSlant->"Italic"], " = \[DoubleStruckCapitalZ] und ", StyleBox["M", FontSlant->"Italic"], " := \[DoubleStruckCapitalQ] / \[DoubleStruckCapitalZ] . Dann ist ", StyleBox["M ", FontSlant->"Italic"], "Torsionsmodul, denn jedes ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " l\[ADoubleDot]\[SZ]t sich darstellen als ", StyleBox["m", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " + \[DoubleStruckCapitalZ] mit ", StyleBox["r", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalQ] , also ", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`p\/q\)]], ", ", StyleBox["q", FontSlant->"Italic"], " \[NotEqual] 0 . Dann folgt ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot](", Cell[BoxData[ \(TraditionalForm\`p\/q\)]], " + \[DoubleStruckCapitalZ] ) = ", StyleBox["p", FontSlant->"Italic"], " + \[DoubleStruckCapitalZ] = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " \[Element] \[DoubleStruckCapitalQ] / \[DoubleStruckCapitalZ] , d.h. ", StyleBox["m", FontSlant->"Italic"], " ist Torsionselement und \n", StyleBox["M", FontSlant->"Italic"], " Torsionsmodul." }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.6", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["M", FontSlant->"Italic"], " ein endlich erzeugter ", StyleBox["R", FontSlant->"Italic"], "-Modul. Wenn ", StyleBox["M", FontSlant->"Italic"], " torsionsfrei ist, dann ist ", StyleBox["M", FontSlant->"Italic"], " ein freier ", StyleBox["R", FontSlant->"Italic"], "-Modul. " }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "Es sei ", StyleBox["M", FontSlant->"Italic"], " = [", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], "] . Es sei (", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_r\)]], ") eine maximal linear unabh\[ADoubleDot]ngige Teilfamilie von (", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], ") . Dann ist \n", StyleBox["N", FontSlant->"Italic"], " := [", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_r\)]], "] freier ", StyleBox["R", FontSlant->"Italic"], "-Modul. Zu jedem \[Nu] \[Element] \[DoubleStruckN] gibt es ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], ", ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(b\_\[Nu]1\), "TraditionalForm"], ",", "\[Ellipsis]", ",", SubscriptBox["b", StyleBox[ RowBox[{"\[Nu]", StyleBox["r", FontSlant->"Italic"]}]]]}], TraditionalForm]]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , die f\[UDoubleDot]r jedes \[Nu] nicht alle Null sind und ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " \[NotEqual] 0 erf\[UDoubleDot]llen, mit ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], \(m\_\[Nu]\)}], TraditionalForm]]], "+ ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(b\_\[Nu]1\), "TraditionalForm"], \(x\_1\)}], TraditionalForm]]], " + \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SubscriptBox["b", StyleBox[ RowBox[{"\[Nu]", StyleBox["r", FontSlant->"Italic"]}]]], "TraditionalForm"], \(x\_r\)}], TraditionalForm]]], " = 0 . Daher ist ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], " = \[Dash] (", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(b\_\[Nu]1\), "TraditionalForm"], \(x\_1\)}], TraditionalForm]]], " + \[Ellipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SubscriptBox["b", StyleBox[ RowBox[{"\[Nu]", StyleBox["r", FontSlant->"Italic"]}]]], "TraditionalForm"], \(x\_r\)}], TraditionalForm]]], ") \[Element] ", StyleBox["N", FontSlant->"Italic"], " . Mit ", StyleBox["a", FontSlant->"Italic"], " := ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], "\[CenterDot] \[Ellipsis] \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gilt dann ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_\[Nu]\)]], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " f\[UDoubleDot]r \[Nu] \[Element] \[DoubleStruckN] und damit ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " f\[UDoubleDot]r alle ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " . Also ist ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["M", FontSlant->"Italic"], " \[Subset] ", StyleBox["N", FontSlant->"Italic"], " .\nNach Satz II.4.4 ist daher ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["M", FontSlant->"Italic"], " als Untermodul des freien ", StyleBox["R", FontSlant->"Italic"], "-Moduls ", StyleBox["N", FontSlant->"Italic"], " auch frei.\nNun wird aber durch ", StyleBox["m", FontSlant->"Italic"], " \[RightTeeArrow] ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " ein ", StyleBox["R", FontSlant->"Italic"], "-Modul-Isomorphismus von ", StyleBox["M", FontSlant->"Italic"], " auf ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["M", FontSlant->"Italic"], " definiert. \nDie Surjektivit\[ADoubleDot]t ist trivial, die Injektivit\ \[ADoubleDot]t folgt so: ", Cell[BoxData[ \(TraditionalForm\`a\[CenterDot]m\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`a\[CenterDot]m\_2\)]], " bedingt ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] (", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], "\[Dash] ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = 0 und da ", StyleBox["M", FontSlant->"Italic"], " torsionsfrei ist, folgt ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " .\nInsgesamt ist deshalb ", StyleBox["M", FontSlant->"Italic"], " als isomorpher Modul eines freien Modul auch frei \ \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Korollar II.4.6", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es seien ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " ", "euklidischer Ring", " und", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" endlich erzeugter ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["-Modul. Dann ist ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" / ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " freier ", StyleBox["R", FontSlant->"Italic"], "-Modul." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " ist endlich erzeugt, denn ", Cell[BoxData[ \(TraditionalForm\`x\&_\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " l\[ADoubleDot]\[SZ]t sich darstellen als ", Cell[BoxData[ \(TraditionalForm\`x\&_\)]], " = ", StyleBox["x", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " .\n", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " ist torsionsfrei, denn:\nEs seien ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], " \[NotEqual] 0 und ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\)]], " derart, dass ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](", StyleBox["m", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], ") = 0 \[Element] (", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], Cell[BoxData[ \(TraditionalForm\`)\_t\)]], " . Dann ist ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " . Also gibt es ein ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 und 0 = ", StyleBox["b", FontSlant->"Italic"], "\[CenterDot](", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], ") = (", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["a", FontSlant->"Italic"], ")\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " . Da ", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["a ", FontSlant->"Italic"], "\[NotEqual] 0 , ist ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " , d.h. ", StyleBox["m", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = 0 \[Element] ", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " . Daher ist \n", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " ist torsionsfrei. Deshalb ", StyleBox["ist ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" / ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " nach dem vorigen Satz frei \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.7", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " ", "euklidischer Ring", " und ", StyleBox["M", FontSlant->"Italic"], " ein endlich erzeugter ", StyleBox["R", FontSlant->"Italic"], "-Modul. Dann gibt es einen freien Untermodul \n", StyleBox["N", FontSlant->"Italic"], " von ", StyleBox["M", FontSlant->"Italic"], " mit ", StyleBox["M", FontSlant->"Italic"], " = ", StyleBox["N", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\[CirclePlus]\)]], " ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "Nach den Korollar II.4.6 ist ", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " frei und damit nach Lemma II.4.4 isomorph zu ", Cell[BoxData[ \(TraditionalForm\`R\^k\)]], " f\[UDoubleDot]r ein ", StyleBox["k", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalN] . Es sei also ", StyleBox["M", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = [", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "1"], TraditionalForm]]], ",\[Ellipsis], ", Cell[BoxData[ FormBox[ SubscriptBox[ FormBox[\(x\&_\), "TraditionalForm"], "k"], TraditionalForm]]], "] . Die Familie (", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], ") ist linear unabh\[ADoubleDot]ngig, denn wenn ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\_\[Nu]\)}]}], TraditionalForm]]], " = 0 , dann ist auch ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(a\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\&_\_\[Nu]\)}]}], TraditionalForm]]], " = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " und damit gilt ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " = 0 f\[UDoubleDot]r alle \[Nu] . \nEs seien ", StyleBox["N", FontSlant->"Italic"], " := [", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_k\)]], "] und ", StyleBox["y", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " . Zu ", StyleBox["y", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(y\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\_\[Nu]\)}]}], TraditionalForm]]], " gibt es ein ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["r", FontSlant->"Italic"], " \[NotEqual] 0 und \n", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["y", FontSlant->"Italic"], " = 0 = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{"r", "\[CenterDot]", FormBox[\(y\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\_\[Nu]\)}]}], TraditionalForm]]], " , woraus ", Cell[BoxData[ FormBox[ RowBox[{"r", "\[CenterDot]", FormBox[\(y\_\[Nu]\), "TraditionalForm"]}], TraditionalForm]]], " = 0 und damit ", Cell[BoxData[ \(TraditionalForm\`y\_\[Nu]\)]], " = 0 f\[UDoubleDot]r alle \[Nu] folgt. Also gilt ", StyleBox["N", FontSlant->"Italic"], " ", StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = {0} .\nEs seien ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " und ", Cell[BoxData[ \(TraditionalForm\`m\&_\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(s\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\&_\_\[Nu]\)}]}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(s\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\_\[Nu]\)}]}], TraditionalForm]]], " + ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " . Mit ", StyleBox["n", FontSlant->"Italic"], " := ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%k\), "TraditionalForm"], RowBox[{ FormBox[\(s\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(x\_\[Nu]\)}]}], TraditionalForm]]], " folgt ", Cell[BoxData[ FormBox[ OverscriptBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], StyleBox[" ", FontSlant->"Italic"], "\[Dash]", " ", StyleBox["n", FontSlant->"Italic"]}], "_"], TraditionalForm]]], " = ", Cell[BoxData[ \(TraditionalForm\`0\&_\)]], " , d.h. ", StyleBox["m", FontSlant->"Italic"], " \[Dash] ", StyleBox["n", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " , \nalso ist ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["N", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " \[FilledSmallSquare]\n" }], "Text"], Cell[TextData[{ StyleBox["Korollar II.4.7", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " ", "euklidischer Ring", " und ", StyleBox["M", FontSlant->"Italic"], " ein endlich erzeugter ", StyleBox["R", FontSlant->"Italic"], "-Modul. Dann ist auch ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], StyleBox[" endlich erzeugt.", FontVariations->{"CompatibilityType"->0}] }], "Text"], Cell[TextData[{ "Der ", StyleBox["Beweis ", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "folgt aus dem Beweis des letzten Satzes." }], "Text"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Zerlegung in euklidischen Ringen", FontFamily->"Arial", FontWeight->"Plain"]], "Section"], Cell[TextData[{ StyleBox["Definition II.4.19:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nEINHEIT (", FontFamily->"Arial"], StyleBox["unity)\n", FontFamily->"Arial", FontSlant->"Italic"], StyleBox["ASSOZIIERT (", FontFamily->"Arial"], StyleBox["associated)\n", FontFamily->"Arial", FontSlant->"Italic"], StyleBox["IRREDUZIBEL (", FontFamily->"Arial"], StyleBox["irreducible", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nTEILER (", FontFamily->"Arial"], StyleBox["divisor", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nPRIMELEMENT (", FontFamily->"Arial"], StyleBox["prime", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es sei ", StyleBox["R", FontSlant->"Italic"], " ein kommutativer Ring mit Eins. \nEin Element \[CurlyEpsilon] \ \[Element] ", StyleBox["R ", FontSlant->"Italic"], "hei\[SZ]t ", StyleBox["EINHEIT", FontFamily->"Arial"], ", wenn es ein \[Eta] \[Element] ", StyleBox["R", FontSlant->"Italic"], " gibt mit \[CurlyEpsilon] \[CenterDot] \[Eta] = 1 . \nEin Element ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], "hei\[SZ]t ", StyleBox["ASSOZIIERT ", FontFamily->"Arial"], "zu einem Element ", StyleBox["s", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], ", wenn es eine Einheit \[CurlyEpsilon] \[Element] ", StyleBox["R", FontSlant->"Italic"], " gibt, mit ", StyleBox["r", FontSlant->"Italic"], " = ", StyleBox["s", FontSlant->"Italic"], " \[CenterDot] \[CurlyEpsilon] . \nEin Element ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], StyleBox["\[Dash] ", FontSlant->"Italic"], "{0} ", StyleBox[" ", FontSlant->"Italic"], "hei\[SZ]t ", StyleBox["IRREDUZIBEL", FontFamily->"Arial"], ", wenn es keine Einheit ist und wenn aus ", StyleBox["p", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , immer folgt, dass ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " oder ", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " eine Einheit ist.\nEin Element ", StyleBox["t", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["TEILER", FontFamily->"Arial"], " von ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , wenn es ein ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], "gibt mit ", StyleBox["a", FontSlant->"Italic"], " = ", StyleBox["t", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["b", FontSlant->"Italic"], " . Sprechweise: ", StyleBox["t", FontSlant->"Italic"], " ", StyleBox["TEILT", FontFamily->"Arial"], " ", StyleBox["a", FontSlant->"Italic"], " , geschrieben ", StyleBox["t", FontSlant->"Italic"], " | ", StyleBox["a", FontSlant->"Italic"], " .\nEin Element ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], StyleBox["\[Dash] ", FontSlant->"Italic"], "{0}", StyleBox[" ", FontSlant->"Italic"], "hei\[SZ]t ", StyleBox["PRIMELEMENT ", FontFamily->"Arial"], "oder kurz ", StyleBox["PRIM", FontFamily->"Arial"], ", wenn es keine Einheit ist und wenn aus ", StyleBox["p | ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , immer folgt, dass ", StyleBox["p | ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " oder ", StyleBox["p | ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " ." }], "Text"], Cell[TextData[{ StyleBox["Achtung:", FontFamily->"Arial"], "\nDer Sprachgebrauch von irreduzibel und prim ist nicht einheitlich." }], "Text"], Cell[TextData[{ StyleBox["Beispiele:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "In ", StyleBox["R", FontSlant->"Italic"], " = \[DoubleStruckCapitalZ] sind \[PlusMinus]1 die Einheiten und die \ Primzahlen multipliziert mit \[PlusMinus]1 die irreduziblen Elemente.\nIn \ ", StyleBox["R", FontSlant->"Italic"], " = \[DoubleStruckCapitalQ] sind alle Element ungleich Null Einheiten.\nIn \ ", StyleBox["R", FontSlant->"Italic"], " = ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] sind alle Element aus ", StyleBox["K", FontSlant->"Italic"], " \\ {0} die Einheiten und die irreduziblen Polynome die irreduziblen \ Elemente." }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.5", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein Integrit\[ADoubleDot]tsbereich. Die Primelemente sind \ irreduzibel.", FontVariations->{"CompatibilityType"->0}], " " }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], StyleBox["Es sei ", FontFamily->"Times New Roman", FontVariations->{"CompatibilityType"->0}], StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], " prim und es gelte ", StyleBox["p", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . O.B.d.A. gelte ", StyleBox["p | ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " und es sei ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " = ", StyleBox["c", FontSlant->"Italic"], "\[CenterDot]", StyleBox["p", FontSlant->"Italic"], " f\[UDoubleDot]r ein ", StyleBox["c", FontSlant->"Italic"], " \[Element] ", StyleBox["R ", FontSlant->"Italic"], ". Dann folgt mit ", StyleBox["p", FontSlant->"Italic"], " = ", StyleBox["c", FontSlant->"Italic"], "\[CenterDot]", StyleBox["p", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " und ", StyleBox["p", FontSlant->"Italic"], " \[NotEqual] 0 im Integrit\[ADoubleDot]tsbereich ", StyleBox["R", FontSlant->"Italic"], " , dass 1 = ", StyleBox["c", FontSlant->"Italic"], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " , also ", Cell[BoxData[ \(TraditionalForm\`p\_2\)]], " eine Einheit ist \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Beispiel:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "In ", StyleBox["R", FontSlant->"Italic"], " := \[DoubleStruckCapitalZ][", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "] = { ", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "| ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalZ] } ist das Element 2 irreduzibel, \ aber nicht prim.\nBeweis:\nAngenommen, es sei 2 = (", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") \[CenterDot] (", StyleBox["c", FontSlant->"Italic"], " + ", StyleBox["d", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") mit ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ",", StyleBox[" c", FontSlant->"Italic"], ",", StyleBox[" d", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalZ] . Nun gilt f\[UDoubleDot]r ", StyleBox["r", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalR] und ", StyleBox["w", FontSlant->"Italic"], ", ", StyleBox["z", FontSlant->"Italic"], " \[Element] \[DoubleStruckCapitalC] mit \n", StyleBox["r", FontSlant->"Italic"], " = ", StyleBox["w", FontSlant->"Italic"], "\[CenterDot]", StyleBox["z", FontSlant->"Italic"], " auch ", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`w\&_\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`z\&_\)]], " , also 2 = (", StyleBox["a", FontSlant->"Italic"], " \[Dash] ", StyleBox["b", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") \[CenterDot] (", StyleBox["c", FontSlant->"Italic"], " \[Dash] ", StyleBox["d", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") , woraus 4 = ", Cell[BoxData[ \(TraditionalForm\`\((a\^2\ + \ 3\ b\^2)\)\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`\((c\^2\ + \ 3\ d\^2)\)\)]], " folgt. Da die Gleichung ", Cell[BoxData[ \(TraditionalForm\`a\^2\ + \ 3\ b\^2\)]], " = 2 keine ganzzahlige L\[ODoubleDot]sung hat, folgt ", Cell[BoxData[ \(TraditionalForm\`a\^2\ + \ 3\ b\^2\)]], " = 1 oder ", Cell[BoxData[ \(TraditionalForm\`c\^2\ + \ 3\ d\^2\)]], " = 1 , was wiederum\n", StyleBox["a", FontSlant->"Italic"], " = \[PlusMinus]1 und ", StyleBox["b", FontSlant->"Italic"], " = 0 oder ", StyleBox["c", FontSlant->"Italic"], " = \[PlusMinus]1 und ", StyleBox["d", FontSlant->"Italic"], " = 0 bedingt. Daher ist 2 irreduzibel in \[DoubleStruckCapitalZ][", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "] .\n2 ist auch keine Einheit in \[DoubleStruckCapitalZ][", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "] , denn die Gleichung 2 \[CenterDot] (", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") = 1 hat keine ganzzahligen L\[ODoubleDot]sungen.\n2 ist aber auch \ nicht prim, denn 2 teilt 4 = (1 + ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") \[CenterDot] (1 \[Dash] ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") , jedoch teilt 2 weder (1 + ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") noch \n(1 \[Dash] ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") in \[DoubleStruckCapitalZ][", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "] .\nAus (1 + ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") = 2\[CenterDot](", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], ") folgt in \[DoubleStruckCapitalC] 2 = | 1 + ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "| = 2\[CenterDot] |", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`\@\[Dash]3\)]], "| = 2", Cell[BoxData[ \(TraditionalForm\`\@\(a\^2\ + \ 3\ b\^2\)\)]], ", und das ist nur f\[UDoubleDot]r \n", StyleBox["a", FontSlant->"Italic"], " = 1 und ", StyleBox["b", FontSlant->"Italic"], " = 0 in \[DoubleStruckCapitalZ] l\[ODoubleDot]sbar." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.22:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nGR\[CapitalODoubleDot]SZTER GEMEINSAMER TEILER : ggT (", FontFamily->"Arial"], StyleBox["greatest common divisor ", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[":", FontFamily->"Arial"], StyleBox[" gcd", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\nTEILERFREMD (", FontFamily->"Arial"], StyleBox["relatively prime", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")\n", FontFamily->"Arial"], "Es seien ", StyleBox["R", FontSlant->"Italic"], " ", StyleBox["ein Integrit\[ADoubleDot]tsbereich und", FontVariations->{"CompatibilityType"->0}], " ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . Ein ", StyleBox["d", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " hei\[SZ]t ", StyleBox["ein", FontVariations->{"Underline"->True}], " ", StyleBox["GR\[CapitalODoubleDot]SZTER GEMEINSAMER TEILER", FontFamily->"Arial"], " (ggT) von ", StyleBox["a", FontSlant->"Italic"], " und ", StyleBox["b ", FontSlant->"Italic"], ", wenn folgende Bedingungen erf\[UDoubleDot]llt sind:\n(", StyleBox["i", FontSlant->"Italic"], ")\t", StyleBox["d", FontSlant->"Italic"], " ist Teiler sowohl von ", StyleBox["a", FontSlant->"Italic"], " als auch ", StyleBox["b", FontSlant->"Italic"], " ,\n(", StyleBox["ii", FontSlant->"Italic"], ")\tist ", StyleBox["t", FontSlant->"Italic"], " irgend ein Teiler von ", StyleBox["a", FontSlant->"Italic"], " und ", StyleBox["b", FontSlant->"Italic"], " , so ist ", StyleBox["t", FontSlant->"Italic"], " auch Teiler von ", StyleBox["d", FontSlant->"Italic"], " ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["a", FontSlant->"Italic"], " und ", StyleBox["b ", FontSlant->"Italic"], " hei\[SZ]en ", StyleBox["TEILERFREMD, ", FontFamily->"Arial"], "wenn ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = 1 ." }], "Text"], Cell[TextData[{ StyleBox["Bemerkungen:", FontFamily->"Arial"], "\nEin ggT ist nur bis auf assoziierte Elemente eindeutig bestimmt:\nIst \ z.B. ", StyleBox["R", FontSlant->"Italic"], " = \[DoubleStruckCapitalZ] und ", StyleBox["a", FontSlant->"Italic"], " = 12 und ", StyleBox["b", FontSlant->"Italic"], " = 30 , so sind dann 6 und \[Dash]6 gr\[ODoubleDot]\[SZ]te gemeinsame \ Teiler !" }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Es seien ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 , wobei noch ", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], ", ", StyleBox["b", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_b\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Beta]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Beta]\_n\)\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " paarweisen, nicht nur durch Einheiten unterschiedlichen irreduziblen \ Elementen, ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], " und ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_b\)]], " Einheiten und ", Cell[BoxData[ \(TraditionalForm\`\[Alpha]\_\[Nu]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Beta]\_\[Nu]\)]], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalN]\_0\)]], " .\n\nDann gilt: ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(min(\[Alpha]\_1, \ \[Beta]\_1)\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(min(\[Alpha]\_n, \ \[Beta]\_n)\)\)]], " ist ein ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") . \t\t" }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Zu der Schreibweise ", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], " : Wenn nicht ausdr\[UDoubleDot]cklich etwas anderes bemerkt wird, soll \ dabei ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], " eine Einheit sein und die ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " paarweise verschiedene, nicht assoziierte irreduzible Elemente aus ", StyleBox["R", FontSlant->"Italic"], " . " }], "Text"], Cell[TextData[{ StyleBox["Euklidischer Algorithmus:", FontFamily->"Arial"], "\n\nEin ggT l\[ADoubleDot]\[SZ]t sich mit dem euklidischen Algorithmus \ berechnen, also einer Kette von Divisionen mit Rest:\nEs seien also ", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], "\[Element] ", StyleBox["R", FontSlant->"Italic"], " .\n\t", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " = ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], " + ", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], " mit ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], ") < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ") , \n\t", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`q\_2\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`a\_3\)]], " mit ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_3\)]], ") < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], ") , \n\t \[VerticalEllipsis]\n\t", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ SubscriptBox["q", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot] ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " + ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " mit ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], ") < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_\(n - 1\)\)]], ") , ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " \[NotEqual] 0 ,\n\t", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " = ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " + 0 . " }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Dann ist ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " ein ggT(", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ") .\n\nDer Algorithmus endet, da ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], ") < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_\(n - 1\)\)]], ") < \[Ellipsis] < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], ") < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ") und ", StyleBox["n ", FontSlant->"Italic"], ": ", StyleBox["R", FontSlant->"Italic"], " \[RightArrow] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalN]\_0\)]], " .\n", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " ist gemeinsamer Teiler, denn ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " | ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " , ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], "| ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], " , \[Ellipsis] , ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " | ", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " . \nEs sei ", StyleBox["t", FontSlant->"Italic"], " ein weiterer gemeinsamer Teiler von ", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ". Dann folgt auch ", StyleBox["t", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], " , ", StyleBox["t", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`a\_3\)]], " ,\[CenterEllipsis] , ", StyleBox["t", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " , d.h. ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " ist ein ggT(", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], ") ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Satz II.4.8", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs sei", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring. Zu ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " existiert ein ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") =: ", StyleBox["d", FontSlant->"Italic"], " mit ", StyleBox["d", FontSlant->"Italic"], " = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + \[Mu]\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " , wobei \[Lambda], \[Mu] \[Element] ", StyleBox["R", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], StyleBox["Es sei ", FontFamily->"Times New Roman", FontVariations->{"CompatibilityType"->0}], StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" := { ", FontFamily->"Times New Roman", FontVariations->{"CompatibilityType"->0}], StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " | ", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], "\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["s", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " , ", StyleBox["r", FontSlant->"Italic"], ", ", StyleBox["s", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " } . ", StyleBox["A", FontSlant->"Italic"], " ist ein Ideal, denn:\nEs seien f\[UDoubleDot]r \[Nu] = 1, 2 ", Cell[BoxData[ \(TraditionalForm\`x\_\[Nu]\)]], " \[Element] ", StyleBox["A", FontSlant->"Italic"], " , mit ", Cell[BoxData[ \(TraditionalForm\`x\_\[Nu]\)]], " = ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], "\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`s\_\[Nu]\)]], "\[CenterDot]", StyleBox["b . ", FontSlant->"Italic"], "Dann folgt auch ", Cell[BoxData[ \(TraditionalForm\`\((x\_1\ + \ x\_2)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\((r\_1\ + \ r\_2)\)\)]], "\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`\((s\_1\ + \ s\_2)\)\)]], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["A", FontSlant->"Italic"], " . \nMit ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["A", FontSlant->"Italic"], " und ", StyleBox["t", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " folgt analog ", StyleBox["t", FontSlant->"Italic"], "\[CenterDot]", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["A", FontSlant->"Italic"], " . Nach Satz II.4.3 ist ", StyleBox["A", FontSlant->"Italic"], " ein Hauptideal. Also gibt es ein ", StyleBox["d", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , so dass \n", StyleBox["A", FontSlant->"Italic"], " = [", StyleBox["d", FontSlant->"Italic"], "] . Da ", StyleBox["d", FontSlant->"Italic"], " \[Element] ", StyleBox["A ", FontSlant->"Italic"], ", gibt es \[Lambda], \[Mu] \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["d", FontSlant->"Italic"], " = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + \[Mu]\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " . Es gilt aber mit ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[Element] ", StyleBox["A", FontSlant->"Italic"], " auch, dass ", StyleBox["a", FontSlant->"Italic"], " und ", StyleBox["b", FontSlant->"Italic"], " Vielfache von ", StyleBox["d", FontSlant->"Italic"], " sind, ", StyleBox["d", FontSlant->"Italic"], " also Teiler von ", StyleBox["a", FontSlant->"Italic"], " und ", StyleBox["b", FontSlant->"Italic"], " .\nEs sei nun ", StyleBox["t", FontSlant->"Italic"], " irgend ein Teiler von ", StyleBox["a", FontSlant->"Italic"], " und ", StyleBox["b", FontSlant->"Italic"], " , etwa ", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], "\[CenterDot]", StyleBox["t", FontSlant->"Italic"], " und ", StyleBox["b", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`b\_1\)]], "\[CenterDot] ", StyleBox["t", FontSlant->"Italic"], " . Dann folgt mit \n", StyleBox["d", FontSlant->"Italic"], " = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + \[Mu]\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " = (", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\[CenterDot]a\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\[CenterDot]b\_1\)]], ")\[CenterDot]", StyleBox["t", FontSlant->"Italic"], " , dass ", StyleBox["t", FontSlant->"Italic"], " auch Teiler von ", StyleBox["d", FontSlant->"Italic"], " ist, d.h. ", StyleBox["d", FontSlant->"Italic"], " ist ein ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.6", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ", \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 . Genau dann gilt ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") , wenn ", StyleBox["b", FontSlant->"Italic"], " eine Einheit ist.\nInsbesondere ist ", StyleBox["b", FontSlant->"Italic"], " genau dann eine Einheit, wenn ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["b", FontSlant->"Italic"], ") = ", StyleBox["n", FontSlant->"Italic"], "(1) ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "In euklidischen Ringen gilt nach Definition f\[UDoubleDot]r ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 mit der zugeh\[ODoubleDot]rigen Normfunktion ", StyleBox["n", FontSlant->"Italic"], " : ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") \n( = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["a", FontSlant->"Italic"], ") ) . Ist nun ", StyleBox["b", FontSlant->"Italic"], " eine Einheit mit ", StyleBox["b", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\(\(\[CenterDot]\)\(b\^'\)\)\)]], " = 1 , so folgt ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ")\[LessEqual] ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\(\(\[CenterDot]\)\(b\^'\)\)\)]], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") , d.h. ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") .\nUmgekehrt folgt aus ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") auch, dass ", StyleBox["b", FontSlant->"Italic"], " eine Einheit ist, denn:\nEs gibt ", StyleBox["q", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], " = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot](", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") + ", StyleBox["r", FontSlant->"Italic"], " mit ", StyleBox["r", FontSlant->"Italic"], " = 0 oder ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") < ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") . Wenn ", StyleBox["r", FontSlant->"Italic"], " \[NotEqual] 0 , folgt ", StyleBox[" r", FontSlant->"Italic"], " = ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](1 \[Dash] ", StyleBox["q\[CenterDot]b", FontSlant->"Italic"], ") , also \n", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](1 \[Dash] ", StyleBox["q\[CenterDot]b", FontSlant->"Italic"], ")) \[GreaterEqual] ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], ") , d.h. ein Widerspruch. Daher ist ", StyleBox["r", FontSlant->"Italic"], " = 0 und 1 = ", StyleBox["q", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " , d.h. ", StyleBox["b", FontSlant->"Italic"], " ist eine Einheit.\nDer zweite Teil der Behauptung des Satzes folgt aus \ dem ersten, wenn einfach ", StyleBox["a", FontSlant->"Italic"], " = 1 gesetzt wird \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.9", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs sei", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring. Dann ist jedes Element aus ", StyleBox["R", FontSlant->"Italic"], " entweder eine Einheit oder es kann als endliches Produkt von \ irreduziblen Elementen aus ", StyleBox["R", FontSlant->"Italic"], " dargestellt werden." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "Es sei ", StyleBox["r ", FontSlant->"Italic"], "\[Element] ", StyleBox["R", FontSlant->"Italic"], " . Der Beweis wird durch vollst\[ADoubleDot]ndige Induktion nach der Norm \ der Elemente ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") gef\[UDoubleDot]hrt.\nDer Induktionsanfang mit ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") = 1 folgt sofort nach Lemma II.4.6. Dann ist ", StyleBox["r", FontSlant->"Italic"], " eine Einheit.\nNach Induktionsannahme gelte also die Aussage des Satzes \ f\[UDoubleDot]r alle ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") < ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") . \nIst ", StyleBox["r", FontSlant->"Italic"], " irreduzibel, so nichts mehr zu beweisen.\nAlso sei ", StyleBox["r", FontSlant->"Italic"], " keine Einheit und aus ", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " folge, dass weder ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " noch ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " eine Einheit ist. Aus (ER ", StyleBox["i", FontSlant->"Italic"], ") und Lemma II.4.6 folgt, dass ", Cell[BoxData[ \(TraditionalForm\`n(r\_1)\)]], " < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") und ", Cell[BoxData[ \(TraditionalForm\`n(r\_2)\)]], " < ", StyleBox["n", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") = ", StyleBox["n", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], ") . Nach Induktionsvoraussetzung k\[ODoubleDot]nnen ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " und ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " jeweils als endliches Produkt von irreduziblen Elementen aus ", StyleBox["R", FontSlant->"Italic"], " dargestellt werden. Damit ist auch \n", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " ein solches \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.7", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["c", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], " | ", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["c", FontSlant->"Italic"], " , aber ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = 1 . Dann gilt ", StyleBox["a", FontSlant->"Italic"], " | ", StyleBox["c", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "Nach Satz II.4.8 und ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = 1 gibt es \[Lambda], \[Mu] \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit 1 = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + \[Mu]\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " , woraus ", StyleBox["c", FontSlant->"Italic"], " = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["c", FontSlant->"Italic"], " + \[Mu]\[CenterDot]", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["c ", FontSlant->"Italic"], " folgt. Nun gilt \n", StyleBox["a", FontSlant->"Italic"], " | \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["c", FontSlant->"Italic"], " und aus ", StyleBox["a", FontSlant->"Italic"], " | ", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["c", FontSlant->"Italic"], " folgt ", StyleBox["a", FontSlant->"Italic"], " | \[Mu]\[CenterDot]", StyleBox["b", FontSlant->"Italic"], "\[CenterDot]", StyleBox["c ", FontSlant->"Italic"], ", also insgesamt ", StyleBox["a", FontSlant->"Italic"], " | ", StyleBox["c", FontSlant->"Italic"], " \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Lemma II.4.8", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["p", FontSlant->"Italic"], " irreduzibel und ", StyleBox["p", FontSlant->"Italic"], " | ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["b", FontSlant->"Italic"], " . Dann teilt ", StyleBox["p", FontSlant->"Italic"], " wenigstens einen der Faktoren ", StyleBox["a", FontSlant->"Italic"], " oder ", StyleBox["b", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], StyleBox["Wenn ", FontFamily->"Times New Roman", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox["p", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" kein Teiler von ", FontFamily->"Times New Roman"], StyleBox["a", FontSlant->"Italic"], StyleBox[" ist, dann gilt ", FontFamily->"Times New Roman"], " ggT(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["p", FontSlant->"Italic"], ") = 1", StyleBox[" . Nach dem vorigen Lemma folgt ", FontFamily->"Times New Roman"], StyleBox["p", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" | ", FontFamily->"Times New Roman"], StyleBox["b", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" \[FilledSmallSquare]", FontFamily->"Times New Roman"] }], "Text"], Cell[TextData[{ StyleBox["Korollar II.4.8", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und f\[UDoubleDot]r \[Nu] \[Element] \ \[DoubleStruckCapitalN] ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], ", ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , mit irreduziblen ", StyleBox["p", FontSlant->"Italic"], " und ", StyleBox["p", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], StyleBox["\[CenterEllipsis]", FontSlant->"Italic"], " ", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], " . \nDann teilt ", StyleBox["p", FontSlant->"Italic"], " wenigstens einen der Faktoren ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " .\n" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.10", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs sei", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring. Dann hat jedes Element aus ", StyleBox["R", FontSlant->"Italic"], " , das keine Einheit ist, eine eindeutige Zerlegung als Produkt von \ irreduziblen Elementen aus ", StyleBox["R", FontSlant->"Italic"], " .\nDas hei\[SZ]t: Ist ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " keine Einheit, so gibt es irreduzible ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , so dass ", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " . Eindeutigkeit bedeutet hier: Wenn ", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " = ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " mit irreduziblen ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " , dann gilt ", StyleBox["m", FontSlant->"Italic"], " = ", StyleBox["n ", FontSlant->"Italic"], " und nach einer eventuell notwendigen Umnummerierung ", Cell[BoxData[ \(TraditionalForm\`q\_\[Nu]\)]], " = ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_\[Nu]\)]], " , wobei ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_\[Nu]\)]], " Einheiten sind." }], "Text"], Cell[TextData[{ StyleBox["Beispiel:", FontFamily->"Arial"], " 12 = 1\[CenterDot]2\[CenterDot]2\[CenterDot]3 = (\[Dash]1\[CenterDot]2)\ \[CenterDot]3\[CenterDot](\[Dash]1\[CenterDot]2) = \ 2\[CenterDot]2\[CenterDot]3." }], "Text"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "Nach Satz II.4.9 folgt die Existenz der Zerlegung eines Elementes in ", StyleBox["R", FontSlant->"Italic"], " . Es bleibt die \"Eindeutigkeit\" der Zerlegung zu beweisen. \nEs sei \ also ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " keine Einheit und es seien irreduzible ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], ", ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " , so dass ", StyleBox["r", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " = ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " . Da ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " | ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " gilt auch ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " | ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " . Nach Korollar II.4.8 teilt ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " ein ", Cell[BoxData[ \(TraditionalForm\`q\_\[Nu]\)]], " , also ist ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " assoziert zu ", Cell[BoxData[ \(TraditionalForm\`q\_\[Nu]\)]], " , also \n", Cell[BoxData[ \(TraditionalForm\`q\_\[Nu]\)]], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_\[Nu]\)]], " , wobei ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_\[Nu]\)]], " eine Einheit ist. Damit gilt ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\[CurlyEpsilon]\_\[Nu]\), "TraditionalForm"], "\[CenterDot]", \(p\_1\)}], TraditionalForm]]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(q\_\[Nu]\[Dash]1\), "TraditionalForm"], "\[CenterDot]", \(q\_\(\[Nu] + 1\)\)}], TraditionalForm]]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " . Auf beiden Seiten wird ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " gek\[UDoubleDot]rzt und der eben beschriebene Prozess f\[UDoubleDot]r \ jeden der Faktoren ", Cell[BoxData[ \(TraditionalForm\`p\_\[Mu]\)]], " wiederholt. Dies f\[UDoubleDot]hrt f\[UDoubleDot]r ", StyleBox["m", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " auf der linken Seite der letzten Gleichung zu 1 und auf der rechten \ Seite zu einem Produkt aus ", StyleBox["m", FontSlant->"Italic"], " Einheiten und ", StyleBox["n", FontSlant->"Italic"], " \[Dash] ", StyleBox["m", FontSlant->"Italic"], " Faktoren ", Cell[BoxData[ \(TraditionalForm\`q\&~\_\[Nu]\)]], " . Die analoge Argumentation kann mit ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], " an Stelle von ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], " auf ", Cell[BoxData[ \(TraditionalForm\`q\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`q\_n\)]], " = ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_m\)]], " angewandt werden. etc., was zu ", StyleBox["n", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["m ", FontSlant->"Italic"], "f\[UDoubleDot]hrt \[FilledSmallSquare] " }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Torsionsmoduln \[UDoubleDot]ber euklidischen Ringen", FontFamily->"Arial"], StyleBox[" ", FontColor->GrayLevel[0]] }], "Section", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Definition II.4.23:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nANNULATOR (", FontFamily->"Arial"], StyleBox["annihilator)", FontFamily->"Arial", FontSlant->"Italic"], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " Ring und ", StyleBox["M", FontSlant->"Italic"], " ein ", StyleBox["R", FontSlant->"Italic"], "-Modul. \nDie Menge Ann(", StyleBox["M", FontSlant->"Italic"], ") := { ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " | ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 f\[UDoubleDot]r alle ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " } hei\[SZ]t ", StyleBox["ANNULATOR", FontFamily->"Arial"], " von ", StyleBox["M", FontSlant->"Italic"], " .\nIst ", StyleBox["x", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " , dann hei\[SZ]t die Menge Ann(", StyleBox["x", FontSlant->"Italic"], ") := { ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " | ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["x", FontSlant->"Italic"], " = 0 } der ", StyleBox["ANNULATOR", FontFamily->"Arial"], " von ", StyleBox["x .", FontSlant->"Italic"], " " }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Bemerkungen:", FontFamily->"Arial"], "\nDer Annulator Ann(", StyleBox["M", FontSlant->"Italic"], ") ist ein Ideal in ", StyleBox["R", FontSlant->"Italic"], " ; das gilt entsprechend f\[UDoubleDot]r Ann(", StyleBox["x", FontSlant->"Italic"], ") .\nDenn: Es seien ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], " \[Element] Ann(", StyleBox["M", FontSlant->"Italic"], ") und ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . Dann folgt f\[UDoubleDot]r alle ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " \n(", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], " + ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], ") \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`r\_2\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = 0 + 0 = 0 und (", Cell[BoxData[ \(TraditionalForm\`r\[CenterDot]r\_1\)]], ")\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = ", StyleBox["r ", FontSlant->"Italic"], "\[CenterDot](", Cell[BoxData[ \(TraditionalForm\`r\_1\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], ") = ", StyleBox["r", FontSlant->"Italic"], "\[CenterDot]0 = 0 .\nWeiterhin gilt:\nIst ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["M", FontSlant->"Italic"], " := [", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`x\_n\)]], "] ein endlich erzeugter Torsionsmodul \[UDoubleDot]ber ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["M", FontSlant->"Italic"], " \[NotEqual] {0}, dann ist Ann(", StyleBox["M", FontSlant->"Italic"], ") ein Ideal in ", StyleBox["R", FontSlant->"Italic"], " ungleich {0} als auch ungleich ", StyleBox["R", FontSlant->"Italic"], " . \nDeshalb gibt es ein ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], " und Ann(", StyleBox["M", FontSlant->"Italic"], ") = ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " .\nDenn: \nIn dem Torsionsmodul gibt es zu jedem ", Cell[BoxData[ \(TraditionalForm\`x\_\[Nu]\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " ein ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], " \[NotEqual] 0 , so dass ", Cell[BoxData[ \(TraditionalForm\`r\_\[Nu]\)]], "\[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`x\_\[Nu]\)]], " = 0 . \nDaraus folgt, dass 0 \[NotEqual] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterEllipsis]", " ", SubscriptBox[ StyleBox["r", FontSlant->"Italic"], \(\(n\)\(\ \)\)]}], TraditionalForm]]], "\[Element] Ann(", StyleBox["M", FontSlant->"Italic"], ") , denn:\nEs sei ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " mit ", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%n\), "TraditionalForm"], \(a\_\[Nu]\)}], "TraditionalForm"], \(x\_\[Nu]\)}], TraditionalForm]]], " . Dann ist ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(r\_1\), "TraditionalForm"], "\[CenterEllipsis]", " ", SubscriptBox[ StyleBox["r", FontSlant->"Italic"], \(\(n\)\(\ \)\)]}], TraditionalForm]]], "\[CenterDot] ", StyleBox["m = ", FontSlant->"Italic"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[\(\[CapitalSigma]\+\(\[Nu] = 1\)\%n\), "TraditionalForm"], \(a\_\[Nu]\)}], "TraditionalForm"], "\[CenterDot]", FormBox[\(r\_1\), "TraditionalForm"]}], "\[CenterEllipsis]", " ", RowBox[{ SubscriptBox[ StyleBox["r", FontSlant->"Italic"], \(\(n\)\(\ \)\)], "\[CenterDot]", \(x\_\[Nu]\)}]}], TraditionalForm]]], " = 0 .\nAlso ist Ann(", StyleBox["M", FontSlant->"Italic"], ") \[NotEqual] {0} .\nFalls ", StyleBox["M", FontSlant->"Italic"], " \[NotEqual] {0} und Ann(", StyleBox["M", FontSlant->"Italic"], ") = ", StyleBox["R", FontSlant->"Italic"], " , w\[ADoubleDot]re 1\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = 0 f\[UDoubleDot]r alle ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " , also ", StyleBox["M", FontSlant->"Italic"], " = {0} .\nEin euklidischer Ring ist Hauptidealring. Deshalb l\[ADoubleDot]\ \[SZ]t sich das Ideal Ann(", StyleBox["M", FontSlant->"Italic"], ") darstellen als Ann(", StyleBox["M", FontSlant->"Italic"], ") = ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["a", FontSlant->"Italic"], " ." }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.24:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \np-KOMPONENTE (", FontFamily->"Arial"], StyleBox["p-part", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], StyleBox[" ", FontFamily->"Arial", FontSlant->"Italic"], "\n", StyleBox["p-PRIM\[CapitalADoubleDot]RE KOMPONENTE (", FontFamily->"Arial"], StyleBox["primary part", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], StyleBox[" ", FontFamily->"Arial", FontSlant->"Italic"], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring, ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " und ", StyleBox["M", FontSlant->"Italic"], " ein ", StyleBox["R", FontSlant->"Italic"], "-Modul. \n", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") := { ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " | Es gibt ein \[Alpha] \[Element] \[DoubleStruckCapitalN] mit ", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 .} hei\[SZ]t die ", StyleBox["p-KOMPONENTE ", FontFamily->"Arial"], "von ", StyleBox["M . ", FontSlant->"Italic"], "Ist noch ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " irreduzibel, so hei\[SZ]t ", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") die ", StyleBox["p-PRIM\[CapitalADoubleDot]RE KOMPONENTE", FontFamily->"Arial"], " von ", StyleBox["M", FontSlant->"Italic"], " . " }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Bemerkungen:", FontFamily->"Arial"], "\n1) \n", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") ist Untermodul von ", StyleBox["M", FontSlant->"Italic"], " :\nEs seien ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") mit ", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = 0 = ", Cell[BoxData[ \(TraditionalForm\`p\^\[Beta]\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], " , wobei o.B.d.A. \[Alpha] \[LessEqual] \[Beta] . Damit folgt ", Cell[BoxData[ \(TraditionalForm\`p\^\[Beta]\)]], " \[CenterDot] (", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " + ", Cell[BoxData[ \(TraditionalForm\`m\_2\)]], ") = 0 und \n", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " \[CenterDot] ", StyleBox["r", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], " = 0 f\[UDoubleDot]r ", StyleBox["r", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . \n2) \nEs sei ", StyleBox["M", FontSlant->"Italic"], " wie in der vorletzten Bemerkung und Ann(", StyleBox["M", FontSlant->"Italic"], ") = ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["a ", FontSlant->"Italic"], "mit ", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], ". Es sei noch ", StyleBox["p", FontSlant->"Italic"], " \[NotEqual] ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " \nf\[UDoubleDot]r alle \[Nu] . Dann gilt ", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") = {0} , denn: \nEs sei ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") . Dann gibt es ein \[Alpha] mit ", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 . Da ", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " und ", StyleBox["a", FontSlant->"Italic"], " teilerfremd, gibt es \[Lambda], \[Mu] \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit \n1 = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\ \[CenterDot]p\^\[Alpha]\)]], " . Damit folgt ", StyleBox["m", FontSlant->"Italic"], " = \[Lambda]\[CenterDot]", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["m ", FontSlant->"Italic"], "+ ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\ \[CenterDot]p\^\[Alpha]\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = 0 + 0 .\n\n" }], "Text", FontWeight->"Plain"], Cell["\<\ Allgemeiner gilt:\ \>", "Text"], Cell[TextData[{ StyleBox["Lemma II.4.9", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["p", FontSlant->"Italic"], ", ", StyleBox["q", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " teilerfremd und ungleich Null. Dann gilt f\[UDoubleDot]r jeden ", StyleBox["R", FontSlant->"Italic"], "-Modul ", StyleBox["M", FontSlant->"Italic"], " , dass ", Cell[BoxData[ \(TraditionalForm\`M(p)\)]], " ", StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`M(q)\)]], " = {0} . " }], "Text"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nEs sei ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M(p)\)]], " ", StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`M(q)\)]], " , also ", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 = ", Cell[BoxData[ \(TraditionalForm\`q\^\[Beta]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " . F\[UDoubleDot]r alle \[Alpha], \[Beta] \[Element] \ \[DoubleStruckCapitalN] gilt ggT(", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`q\^\[Beta]\)]], ") = 1 . Daher gibt es von \n\[Alpha], \[Beta] abh\[ADoubleDot]ngige \ \[Lambda], \[Mu] \[Element] ", StyleBox["R", FontSlant->"Italic"], " mit 1 = \[Lambda]\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], " + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(\[Mu]\)\(\ \)\(\[CenterDot]\)\), "TraditionalForm"], \(q\^\[Beta]\)}], TraditionalForm]]], " . Daraus folgt ", StyleBox["m", FontSlant->"Italic"], " = \[Lambda]\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\^\[Alpha]\)]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(\[Mu]\)\(\ \)\(\[CenterDot]\)\), "TraditionalForm"], \(q\^\[Beta]\)}], TraditionalForm]]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.11", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["M", FontSlant->"Italic"], " ein endlich erzeugter Torsionsmodul \[UDoubleDot]ber ", StyleBox["R", FontSlant->"Italic"], " . Es sei Ann(", StyleBox["M", FontSlant->"Italic"], ") = ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["a ", FontSlant->"Italic"], "mit \n", StyleBox["a", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], ", wobei ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], " eine Einheit und die ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], " paarweise verschiedene, nicht assoziierte, irreduzible Elemente aus ", StyleBox["R", FontSlant->"Italic"], " seien.\nDann gilt: ", StyleBox["M", FontSlant->"Italic"], " = ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ") \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_n\)]], ") ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nInduktion nach ", StyleBox["n", FontSlant->"Italic"], " , der Anzahl der irreduziblen Faktoren:\n", StyleBox["n", FontSlant->"Italic"], " = 1 : \nNach Voraussetzung ist Ann(", StyleBox["M", FontSlant->"Italic"], ") = ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], ". F\[UDoubleDot]r alle ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " gilt also ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 . \nDaraus folgt ", StyleBox["M ", FontSlant->"Italic"], "= { ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 } = ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ") .\n\nInduktionsschritt:\nMit ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " := { ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " | ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 } ist ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "\[Subset]", StyleBox[" ", FontColor->RGBColor[1, 0, 0]], StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ") .\n", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " ist Untermodul von ", StyleBox["M", FontSlant->"Italic"], " und Ann(", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], ") = ", StyleBox["R", FontSlant->"Italic"], " \[CenterDot] ( ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ") .\nNach Induktionsvoraussetzung gilt ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ") \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], ") .\n\nNun ist ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") = ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") f\[UDoubleDot]r \[Nu] = 1,\[Ellipsis], ", StyleBox["n", FontSlant->"Italic"], " \[Dash] 1 , denn:\nMit der Definition der ", StyleBox["p", FontSlant->"Italic"], "-Komponente ist ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") = { ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " | Es gibt ein \[Alpha] \[Element] \[DoubleStruckCapitalN] mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\[Alpha]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 .} . Mit ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "\[Subset] ", StyleBox["M", FontSlant->"Italic"], " folgt ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") \[Subset] ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") . \nUmgekehrt sei ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\[Alpha]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 und o.B.d.A. \[Alpha] \[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`\[Alpha]\_\[Nu]\)]], " , also ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") . Au\[SZ]erdem gilt ", StyleBox["a", FontSlant->"Italic"], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 .\nDa f\[UDoubleDot]r \[Nu] = 1,\[Ellipsis], ", StyleBox["n", FontSlant->"Italic"], " \[Dash] 1 jeweils ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\(\[Alpha]\_\[Nu]\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_\[Nu]\)]], "\[CenterDot]", StyleBox["a", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\[Nu]\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\[Alpha]\)]], " ein ggT(", StyleBox["a", FontSlant->"Italic"], ", ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\[Alpha]\)]], ") ist, folgt ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\(\[Alpha]\_\[Nu]\)\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_\[Nu]\)]], "\[CenterDot]", StyleBox["a\[CenterDot]m", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Mu]\_\[Nu]\)\(\ \)\)\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\[Alpha]\)]], "\[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 . Daraus folgt ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 , d.h. ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\%\[Alpha]\)]], " \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = 0 , also ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "(", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], ") . \nDeshalb folgt ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ") \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], ") .\n\nEs gilt ", StyleBox["M", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "\[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], ") , denn:\n", StyleBox["i", FontSlant->"Italic"], ") ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " + ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_n\)]], ") \[Subset] ", StyleBox["M ", FontSlant->"Italic"], "gilt trivialerweise. \nEs gilt auch ", StyleBox["M", FontSlant->"Italic"], " \[Subset] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " + ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_n\)]], ") , denn:\nEs sei ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " . Aus ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " = 0 = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon]\_a\)]], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " folgt (", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ") \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], ") und ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], "\[CenterDot]", StyleBox["m", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " . \nMit 1 = ggT(", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\[Alpha]\)]], ") folgt aus Satz II.4.8 \n", StyleBox["m", FontSlant->"Italic"], " = (\[Lambda] \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], " + \[Mu] \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], ") \[CenterDot] ", StyleBox["m", FontSlant->"Italic"], " = (\[Lambda] \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ") \[CenterDot] ", StyleBox["m +", FontSlant->"Italic"], " \[Mu] \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`p\_n\%\(\[Alpha]\_n\)\)]], " \[CenterDot] ", StyleBox["m ", FontSlant->"Italic"], "\[Element] ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_n\)]], ") + ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], " .\nDamit gilt ", StyleBox["M", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], ") .\n", StyleBox["ii", FontSlant->"Italic"], ") Die Summe ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], ") ist direkt, denn:\n", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], " und ", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], " sind teilerfremd. Mit Lemma II.4.9 folgt ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ") ", StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], ") = {0} und dann mit \n{0} \[Subset] ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], "\[Subset] ", StyleBox["M", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`p\_1\%\(\[Alpha]\_1\)\)]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SubsuperscriptBox["p", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]], SubscriptBox["\[Alpha]", StyleBox[ RowBox[{ StyleBox["n", FontSlant->"Italic"], "\[Dash]1"}]]]], TraditionalForm]]], ") auch ", Cell[BoxData[ \(TraditionalForm\`M\_1\)]], StyleBox["\[Intersection]", FontSize->9], " ", Cell[BoxData[ \(TraditionalForm\`M\)]], "(", Cell[BoxData[ FormBox[ SubscriptBox["p", StyleBox["n", FontSlant->"Italic"]], TraditionalForm]]], ") = {0} \[FilledSmallSquare]" }], "Text", FontWeight->"Plain"], Cell[TextData[{ "\nNun sollen die Bausteine ", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") auch noch zerlegt werden. Dazu ohne Beweis" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.12 ", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["M", FontSlant->"Italic"], " ein endlich erzeugter Torsionsmodul \[UDoubleDot]ber ", StyleBox["R", FontSlant->"Italic"], " . Mit einem irreduziblen ", StyleBox["p", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " gelte ", StyleBox["M", FontSlant->"Italic"], " = ", StyleBox["M", FontSlant->"Italic"], "(", StyleBox["p", FontSlant->"Italic"], ") . \nDann existieren durch ", StyleBox["M", FontSlant->"Italic"], " eindeutig bestimmte nat\[UDoubleDot]rliche Zahlen ", Cell[BoxData[ \(TraditionalForm\`k\_1\)]], "\[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`k\_2\)]], "\[GreaterEqual] \[CenterEllipsis] \[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`k\_n\)]], "\[GreaterEqual] 1 und ", Cell[BoxData[ \(TraditionalForm\`m\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`m\_n\)]], " \[Element] ", StyleBox["M", FontSlant->"Italic"], " mit\n", StyleBox["M", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["Rm", FontSlant->"Italic"], "1"], TraditionalForm]]], " \[CirclePlus] ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["Rm", FontSlant->"Italic"], "2"], TraditionalForm]]], " \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["Rm", FontSlant->"Italic"], "n"], TraditionalForm]]], " und ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["Rm", FontSlant->"Italic"], "\[Nu]"], TraditionalForm]]], " \[TildeEqual] ", StyleBox["R", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`p\^k\_\[Nu]\)]], StyleBox["R", FontSlant->"Italic"], " , \[Nu] \[Element] \[DoubleStruckN] ." }], "Text", FontWeight->"Plain"], Cell["Damit folgt nun der ", "Text"], Cell[TextData[{ StyleBox["Satz II.4.13", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" Hauptsatz \[UDoubleDot]ber endlich erzeugte Moduln \ \[UDoubleDot]ber euklidischen Ringen", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nEs seien", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["R", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ein", FontVariations->{"CompatibilityType"->0}], " euklidischer Ring und ", StyleBox["M", FontSlant->"Italic"], " ein endlich erzeugter ", StyleBox["R", FontSlant->"Italic"], "-Modul.\nDann gibt es eindeutig bestimmte irreduzible, paarweise \ nichtassoziierte Elemente ", Cell[BoxData[ \(TraditionalForm\`p\_1\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`p\_r\)]], " in ", StyleBox["R", FontSlant->"Italic"], " , ein ", StyleBox["d", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalN]\_0\)]], " und nat\[UDoubleDot]rliche Zahlen ", Cell[BoxData[ \(TraditionalForm\`k\_11\)]], "\[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`k\_12\)]], "\[GreaterEqual] \[CenterEllipsis] \[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`k\_\(1 n\_1\)\)]], "\[GreaterEqual] 1 , \[CenterEllipsis] , ", Cell[BoxData[ FormBox[ SubscriptBox["k", StyleBox[ RowBox[{ StyleBox["r", FontSlant->"Italic"], "1"}]]], TraditionalForm]]], "\[GreaterEqual] ", Cell[BoxData[ 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StyleBox["\[CirclePlus]", FontSize->16], \(\[Nu] = 1\), \(n\_r\)], TraditionalForm]]], " ", StyleBox["R", FontSlant->"Italic"], " / ", Cell[BoxData[ \(TraditionalForm\`p\_r\%\(k\_\(r\ \[Nu]\)\)\)]], " ", StyleBox["R", FontSlant->"Italic"], " ", StyleBox[") ", FontSize->16], ".\n " }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Der ", StyleBox["Beweis", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], " folgt aus \nSatz II.4.7 ", StyleBox[" ", FontSize->9], ": Zerlegung in freien und Torsionsanteil: ", StyleBox["M", FontSlant->"Italic"], " \[TildeEqual] ", StyleBox["N", FontSlant->"Italic"], " \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " , ", StyleBox["N", FontSlant->"Italic"], " \[TildeEqual] ", Cell[BoxData[ \(TraditionalForm\`R\^d\)]], " ,\nSatz II.4.11 : Zerlegung in ", Cell[BoxData[ \(TraditionalForm\`p\_\[Nu]\)]], "-prim\[ADoubleDot]re Komponenten: ", Cell[BoxData[ \(TraditionalForm\`M\_t\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{ 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Anwendungen : Normalformen von Matrizen", "Subtitle", FontWeight->"Plain"], Cell[TextData[{ "Nun sollen die Hauptergebnisse des letzten Abschnittes benutzt werden zur \ Herleitung der ", StyleBox["J", FontFamily->"Arial"], StyleBox["ORDAN", FontFamily->"Arial", FontSize->9], StyleBox["schen Normalform", FontFamily->"Arial"], " eines Endomorphismus \[CurlyPhi] (", ButtonBox["Marie Ennemond Camille Jordan", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Jordan.\ html"], None}, ButtonStyle->"Hyperlink"], ", 1838 \[Dash] 1922, franz\[ODoubleDot]sischer Mathematiker). Als erstes \ muss mit \[CurlyPhi] und dem Vektorraum ", StyleBox["V", FontSlant->"Italic"], " ein Modul konstruiert werden, damit die S\[ADoubleDot]tze II.4.12, \ II.4.13 angewandt werden k\[ODoubleDot]nnen. \nDas leistet der folgende \ Satz:" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.14", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox["K", FontSlant->"Italic"], " ein K\[ODoubleDot]rper, ", StyleBox["V", FontSlant->"Italic"], " ein ", StyleBox["K", FontSlant->"Italic"], "-Vektorraum der Dimension ", StyleBox["n", FontSlant->"Italic"], " (\[GreaterEqual] 1 ) und \[CurlyPhi]", StyleBox[" ", FontSlant->"Italic"], "\[Element] hom(", StyleBox["V", FontSlant->"Italic"], ",", StyleBox[" V", FontSlant->"Italic"], ") ", StyleBox[".", FontSlant->"Italic"], "\n", StyleBox["Mit der Verkn\[UDoubleDot]pfung ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\&~\)]], " :", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["] \[Times] ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[LongRightArrow] ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" definiert durch ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\&~\)]], "(", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["), ", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") := (", FontVariations->{"CompatibilityType"->0}], StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(\[CurlyPhi]))(", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") und der Vektorraumaddition wird \n", FontVariations->{"CompatibilityType"->0}], StyleBox["V ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["zu einem ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["]-Modul ", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" . \n", FontVariations->{"CompatibilityType"->0}], StyleBox["M", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ist endlich erzeugt und ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["]-Torsionsmodul.\nEs gilt: ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ SubscriptBox["Ann", RowBox[{ StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["]", FontVariations->{"CompatibilityType"->0}]}]], TraditionalForm]]], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`K[X]\)]], " , wobei ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") das Minimalpolynom von \[CurlyPhi] ist." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Bemerkungen:", FontFamily->"Arial"], "\nStatt ", StyleBox["M", FontSlant->"Italic"], " ist in diesem Zusammenhang die Bezeichnung ", Cell[BoxData[ \(TraditionalForm\`V\_\[CurlyPhi]\)]], " gebr\[ADoubleDot]uchlich. 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" }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.15", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es sei ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`V\_1\)]], " \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`V\_k\)]], " eine Zerlegung von ", StyleBox["V", FontSlant->"Italic"], " als ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Modul. Dann ist dies auch eine Zerlegung als \n", StyleBox["K", FontSlant->"Italic"], "-Vektorraum. Au\[SZ]erdem gilt f\[UDoubleDot]r alle \[Kappa] \[Element] \ \[DoubleStruckK] : \[CurlyPhi](", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], ") \[Subset] ", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], " ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Beweis:\n", FontFamily->"Arial"], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`V\_1\)]], " \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`V\_k\)]], " bedeutet, dass sich jedes ", StyleBox["v", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[Element] ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontVariations->{"CompatibilityType"->0}], " eindeutig als Summe von ", Cell[BoxData[ \(TraditionalForm\`v\_\[Kappa]\)]], StyleBox[" \[Element] ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], StyleBox[" ", FontVariations->{"CompatibilityType"->0}], "darstellen l\[ADoubleDot]\[SZ]t; da spielt der Skalarbereich keine Rolle!\n\ Sei \[Kappa] \[Element] \[DoubleStruckK] . ", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], " ist ein ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Untermodul, also bzgl. der skalaren Multiplikation ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\&~\)]], " abgeschlossen, d.h. f\[UDoubleDot]r alle \n ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " \[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] und beliebige ", StyleBox["v", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], " gilt ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(\[CurlyPhi])\[CenterDot]", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic"], " \[Element] ", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], " . Wird nun insbesondere ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " := ", StyleBox["X", FontSlant->"Italic"], " , so folgt \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") \[Element] ", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], " \[FilledSmallSquare]" }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Bemerkung und Erinnerung: ", FontFamily->"Arial"], "\nEs seien ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`V\_1\)]], " \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`V\_k\)]], " und (", Cell[BoxData[ \(TraditionalForm\`v\_\(\[Kappa]\ 1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(\[Kappa]\ k\_\[Kappa]\)\)]], ") Basen der ", StyleBox["K", FontSlant->"Italic"], "-Vektorr\[ADoubleDot]ume ", Cell[BoxData[ \(TraditionalForm\`V\_\[Kappa]\)]], " , dann ist \n(", Cell[BoxData[ \(TraditionalForm\`v\_\(1\ 1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(1\ k\_1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(k\ 1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(k\ k\_k\)\)]], ") eine Basis von ", StyleBox["V", FontSlant->"Italic"], " . \nEs sei ", Cell[BoxData[ \(TraditionalForm\`A\_\[Kappa]\)]], " die darstellende Matrix von \[CurlyPhi]", Cell[BoxData[ FormBox[ SubscriptBox["|", SubscriptBox[ StyleBox["V", FontSize->10], StyleBox["\[Kappa]", FontSize->10]]], TraditionalForm]]], " bzgl. (", Cell[BoxData[ \(TraditionalForm\`v\_\(\[Kappa]\ 1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(\[Kappa]\ k\_\[Kappa]\)\)]], ") .\nDann hat \[CurlyPhi] bzgl. der Basis (", Cell[BoxData[ \(TraditionalForm\`v\_\(1\ 1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(1\ k\_1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(k\ 1\)\)]], ",\[Ellipsis], ", Cell[BoxData[ \(TraditionalForm\`v\_\(k\ k\_k\)\)]], ") die Matrix" }], "Text", FontWeight->"Plain"], Cell[TextData[{ " ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", GridBox[{ { FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["1", FontFamily->"Times New Roman"]], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " ", " ", RowBox[{" ", StyleBox["0", FontSize->14]}]}, {" ", FrameBox[ StyleBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], "2"], FontFamily->"Times New Roman"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " ", " "}, { StyleBox[" ", FontSize->16], " ", StyleBox["\[DescendingEllipsis]", FontSize->16], " "}, { RowBox[{ StyleBox["0", FontSize->14], StyleBox[" ", FontSize->16]}], StyleBox[" ", FontSize->16], " ", FrameBox[ StyleBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["k", FontSlant->"Italic"]], FontFamily->"Times New Roman"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}]} }], ")"}]], FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], " .\n" }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Definition II.4.25:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nBEGLEITMATRIX (", FontFamily->"Arial"], StyleBox["companian matrix", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], "\nEs sei ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") \[Element] ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], "ein normiertes Polynom mit ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") ", FontVariations->{"CompatibilityType"->0}], ":= ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%m\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " . Die Matrix\n\n ", StyleBox["C", FontSlant->"Italic"], "( ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], ") := ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ {"0", "0", "\[CenterEllipsis]", "0", "0", StyleBox[ SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], "0"], FontFamily->"Times New Roman"]}, {"1", "0", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], StyleBox["1", FontFamily->"Times New Roman"]]}, {"0", "1", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], "2"]}, {"\[VerticalEllipsis]", "\[VerticalEllipsis]", "\[DescendingEllipsis]", \(\(\ \)\(\[VerticalEllipsis]\)\), "\[VerticalEllipsis]", "\[VerticalEllipsis]"}, {"0", "0", "\[CenterEllipsis]", "1", "0", StyleBox[ SubscriptBox["\[Dash]a", RowBox[{"m", "\[VeryThinSpace]", StyleBox[ RowBox[{"\[Dash]", StyleBox["2", FontSlant->"Plain"]}]]}]], FontFamily->"Times New Roman", FontSlant->"Italic"]}, {"0", "0", "\[CenterEllipsis]", "0", "1", StyleBox[ SubscriptBox[ StyleBox["\[Dash]a", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[ RowBox[{ StyleBox["m", FontFamily->"Times New Roman"], "\[Dash]", StyleBox["1", FontFamily->"Times New Roman", FontSlant->"Plain"]}]]], FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], "\n \nhei\[SZ]t ", StyleBox["BEGLEITMATRIX", FontFamily->"Arial"], " von ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " ." }], "Text"], Cell[TextData[{ StyleBox["Beispiel (Anfang):", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox["\n", FontFamily->"Arial"], "Es sei ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") ", FontVariations->{"CompatibilityType"->0}], ":= ", Cell[BoxData[ \(\(-4\) + 16\ t\ - 25\ t\^2 + 19\ \(t\^3\) \[Dash]\ 7\ t\^4 + t\^5\)], FontFamily->"Times New Roman"], " . Die Begleitmatrix zu ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") ", FontVariations->{"CompatibilityType"->0}], " ist dann\n\n\t ", StyleBox["C", FontSlant->"Italic"], "(", StyleBox["f", FontSlant->"Italic"], " (", StyleBox["X", FontSlant->"Italic"], "))", Cell[BoxData[ StyleBox[ RowBox[{":=", RowBox[{"(", GridBox[{ {"0", "0", "0", "0", "4"}, {"1", "0", "0", "0", \(-16\)}, {"0", "1", "0", "0", "25"}, {"0", "0", "1", "0", \(-19\)}, {"0", "0", "0", "1", "7"} }], ")"}]}], FontFamily->"Times New Roman"]]], " ." }], "Text"], Cell["Mit", "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{ RowBox[{"A", " ", "=", RowBox[{"(", GridBox[{ {"0", "0", "0", "0", "4"}, {"1", "0", "0", "0", \(-16\)}, {"0", "1", "0", "0", "25"}, {"0", "0", "1", "0", \(-19\)}, {"0", "0", "0", "1", "7"} }], ")"}]}], ";"}], " ", "\[IndentingNewLine]", " ", \(e\_\(\(1\)\(\ \)\) = {1, 0, 0, 0, 0};\)}]}]], "Input"], Cell["folgt, dass ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({v\_1 = e\_1, v\_2 = A . v\_1, v\_3 = A . v\_2, \ v\_4 = A . v\_3, \ v\_5 = A . v\_4}\)], "Input"], Cell[BoxData[ \({{1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}, {0, 0, 0, 0, 1}}\)], "Output"] }, Open ]], Cell[TextData[{ "und damit ( ", Cell[BoxData[ \(TraditionalForm\`v\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`v\_2\)]], "= ", Cell[BoxData[ \(TraditionalForm\`A(v\_1)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`v\_3\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(A\^2\)(v\_1)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`v\_4\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(A\^3\)(v\_1)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`v\_5\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(A\^4\)(v\_1)\)]], " ) eine Basis des ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^5\)]], " bildet. \nWeiterhin folgt f\[UDoubleDot]r das charakteristische Polynom \ ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_A\)]], "(", StyleBox["X", FontSlant->"Italic"], ") von ", StyleBox["A", FontSlant->"Italic"], " :" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(\(\[Chi]\_A[ X_] := \(-CharacteristicPolynomial[A, \ X]\);\)\ \[IndentingNewLine] \[Chi]\_A[X]\)\(\ \)\)\)], "Input"], Cell[BoxData[ \(\(-4\) + 16\ X - 25\ X\^2 + 19\ X\^3 - 7\ X\^4 + X\^5\)], "Output"] }, Open ]], Cell[TextData[{ "Also ist ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") ", FontVariations->{"CompatibilityType"->0}], "= ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_A\)]], "(", StyleBox["X", FontSlant->"Italic"], ") . " }], "Text"], Cell[TextData[{ StyleBox["Satz II.4.16", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es seien ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[", ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[", \[CurlyPhi] wie in Satz II.4.14.\n", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ist genau dann ein zyklischer ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Modul, wenn es ein normiertes, nicht-konstantes Polynom ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " \[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] und eine Basis ", StyleBox["B", FontSlant->"Italic"], " von ", StyleBox["V", FontSlant->"Italic"], " gibt, so dass die bez\[UDoubleDot]glich ", StyleBox["B", FontSlant->"Italic"], " darstellende Matrix von \[CurlyPhi] die Begleitmatrix von ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " ist.\nUnter diesen Bedingungen gilt:\n \t", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " = ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") , ( ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") ist das Minimalpolynom ) .\n \tMit ", StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " ist ( ", StyleBox["v", FontSlant->"Italic"], ", \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], "), ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], "),\[Ellipsis], ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ) eine Basis von ", StyleBox["V", FontSlant->"Italic"], " .\n \tBez\[UDoubleDot]glich dieser Basis wird \[CurlyPhi] durch die \ Begleitmatrix ", StyleBox["C", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")) dargestellt.\n\t", Cell[BoxData[ FormBox[ SubscriptBox["Ann", RowBox[{ StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["]", FontVariations->{"CompatibilityType"->0}]}]], TraditionalForm]]], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`K[X]\)]], " .\n \tDas charakteristische Polynom ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") ist gleich dem Minimalpolynom ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") ." }], "Text"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\n1)\nEs sei ", StyleBox["V", FontSlant->"Italic"], " ", StyleBox["zyklischer ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Modul, also ", StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " f\[UDoubleDot]r ein ", StyleBox["v", FontSlant->"Italic"], " \[Element] ", StyleBox["V", FontSlant->"Italic"], " . Wie in Satz II.4.14 folgt \n", Cell[BoxData[ FormBox[ SubscriptBox["Ann", RowBox[{ StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["]", FontVariations->{"CompatibilityType"->0}]}]], TraditionalForm]]], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`K[X]\)]], " . Also ist das gesuchte Polynom das Minimalpolynom von \[CurlyPhi] : ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") := ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%m\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " \nmit ", Cell[BoxData[ \(TraditionalForm\`a\_m\)]], " = 1 . \nJetzt soll gezeigt werden, dass ", StyleBox["B", FontSlant->"Italic"], " := ( ", StyleBox["v", FontSlant->"Italic"], ", \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], "), ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], "),\[Ellipsis], ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ) eine Basis von ", StyleBox["V", FontSlant->"Italic"], " als ", StyleBox["K", FontSlant->"Italic"], "-Vektorraum ist:\nDa ", StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " , gibt es zu jedem ", StyleBox["w", FontSlant->"Italic"], " \[Element] ", StyleBox["V ", FontSlant->"Italic"], "ein ", StyleBox["g", FontSlant->"Italic"], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " \[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] mit ", StyleBox["w", FontSlant->"Italic"], " = ", StyleBox["g", FontSlant->"Italic"], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], "\[CenterDot]", StyleBox["v = g", FontSlant->"Italic"], StyleBox["(\[CurlyPhi])\[CenterDot]", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic"], " . Zu ", StyleBox["g", FontSlant->"Italic"], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " existieren \n", StyleBox["q", FontSlant->"Italic"], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["), ", FontVariations->{"CompatibilityType"->0}], StyleBox["r", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " \[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] mit ", StyleBox["g", FontSlant->"Italic"], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " = ", StyleBox["q", FontSlant->"Italic"], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], "\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") + ", StyleBox["r", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " mit grad ", StyleBox["r", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") < grad ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")", StyleBox[" oder ", FontVariations->{"CompatibilityType"->0}], StyleBox["r", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") = 0 . Nun ist ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(\[CurlyPhi])\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " = 0 und damit ", StyleBox["w", FontSlant->"Italic"], " = ", StyleBox["g", FontSlant->"Italic"], StyleBox["(\[CurlyPhi])\[CenterDot]", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(\[CurlyPhi])\[CenterDot]", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" . Es sei ", FontVariations->{"CompatibilityType"->0}], StyleBox["r", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") ", FontVariations->{"CompatibilityType"->0}], ":= ", Cell[BoxData[ FormBox[ UnderoverscriptBox["\[CapitalSigma]", \(\[Mu] = 0\), StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\`r\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " . Dann folgt ", StyleBox["w", FontSlant->"Italic"], " = ", Cell[BoxData[ FormBox[ UnderoverscriptBox["\[CapitalSigma]", \(\[Mu] = 0\), StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\`r\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^\[Mu]\)]], "(", StyleBox["v", FontSlant->"Italic"], ") . Also ist ", StyleBox["B", FontSlant->"Italic"], " ein Erzeugendensystem. \n", StyleBox["B", FontSlant->"Italic"], " ist auch linear unabh\[ADoubleDot]ngig, denn:\nEs sei \[ScriptL](", StyleBox["X", FontSlant->"Italic"], ") := ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_0\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_1\)]], " ", StyleBox["X", FontSlant->"Italic"], " + \[CenterEllipsis] + ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SubscriptBox["\[Lambda]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], "TraditionalForm"], SuperscriptBox["X", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]]}], TraditionalForm]]], " und 0 = ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_0\)]], StyleBox["v", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_1\)]], "\[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") + \[CenterEllipsis] + ", Cell[BoxData[ FormBox[ SubscriptBox["\[Lambda]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") = \[ScriptL]", Cell[BoxData[ \(TraditionalForm\`\((\[CurlyPhi])\)\)]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " . \nDann ist \[ScriptL](", StyleBox["X", FontSlant->"Italic"], ") \[Element] Ann", Cell[BoxData[ FormBox[ SubscriptBox[" ", RowBox[{ StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], StyleBox["]", FontSlant->"Italic"]}]], TraditionalForm]]], "(", StyleBox["V", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] , also ist \[ScriptL](", StyleBox["X", FontSlant->"Italic"], ") durch ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") teilbar. Da der Grad von \[ScriptL] kleiner gleich ", StyleBox["m", FontSlant->"Italic"], "\[Dash]1 ist, folgt dann \[ScriptL](", StyleBox["X", FontSlant->"Italic"], ") = 0 , d.h. ", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_\[Mu]\)]], " = 0 f\[UDoubleDot]r \[Mu] = 0,\[Ellipsis], ", StyleBox["m", FontSlant->"Italic"], "\[Dash]1 .\nAlso ist ", Cell[BoxData[ \(TraditionalForm\`\(dim\_K\) V = \(m = grad\ \(\(m\_\[CurlyPhi]\)(X)\)\)\)]], " , d.h. das Minimalpolymon ist auch das charakteristische.\n\nZur \ Matrixdarstellung:\n \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") = 0 \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " + 1 \[CenterDot] \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") ,\n ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\)]], "(\[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ")) = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], ") = 0 \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " + 0 \[CenterDot] \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") + 1 \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], ") ,\n\t\t \[VerticalEllipsis]\n \[CurlyPhi](", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ")) \t= ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") = 0 \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " + \[CenterEllipsis] + 0 \[CenterDot] ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") + 1 \[CenterDot] ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") .\nDa ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(\[CurlyPhi])(", StyleBox["v", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%m\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^\[Mu]\)]], "(", StyleBox["v", FontSlant->"Italic"], ") = 0 mit ", Cell[BoxData[ \(TraditionalForm\`a\_m\)]], " = 1 , folgt damit nun \n \[CurlyPhi](", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ")) = ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox["m", FontSlant->"Italic"]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") = \[Dash]", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " \[Dash] ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], "\[CenterDot]\[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") \[Dash] \[CenterEllipsis] \[Dash] ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot] ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ,\nd.h., dass die bez\[UDoubleDot]glich ", StyleBox["B", FontSlant->"Italic"], " darstellende Matrix von \[CurlyPhi] die Begleitmatrix von ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") ist.\n2)\nEs sei andererseits ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " := ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%m\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`a\_m\)]], " = 1 und ", StyleBox["m", FontSlant->"Italic"], " \[GreaterEqual] 1 . Weiterhin sei ", StyleBox["B", FontSlant->"Italic"], " eine Basis bzgl. der \[CurlyPhi] durch die Begleitmatrix von ", StyleBox["f ", FontSlant->"Italic"], " dargestellt wird. Damit hat ", StyleBox["B", FontSlant->"Italic"], " genau ", StyleBox["m", FontSlant->"Italic"], " Elemente.\nEs sei ", StyleBox["B", FontSlant->"Italic"], " := ", Cell[BoxData[ FormBox[ RowBox[{"(", RowBox[{\(y\_0\), ",", "\[Ellipsis]", ",", SubscriptBox["y", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]]}], ")"}], TraditionalForm]]], " . Da ", StyleBox["B", FontSlant->"Italic"], " Begleitmatrix, folgt\n", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi](y\_0)\)]], " \t = ", Cell[BoxData[ \(TraditionalForm\`y\_1\)]], " ,\n", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi](y\_1)\)]], " \t = ", Cell[BoxData[ \(TraditionalForm\`y\_2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], Cell[BoxData[ \(TraditionalForm\`\((y\_0)\)\)]], " ,\n\t \[VerticalEllipsis]\n", Cell[BoxData[ FormBox[ RowBox[{"\[CurlyPhi]", "(", SubscriptBox["y", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ SubscriptBox["y", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\((y\_0)\)\)]], " .\nDamit ist ", StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`K[X]\[CenterDot]y\_0\)]], " , also zyklisch.\nWeiterhin gilt \n", Cell[BoxData[ FormBox[ RowBox[{"\[CurlyPhi]", "(", SubscriptBox["y", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], ")"}], TraditionalForm]]], " = \[Dash]", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`y\_0\)]], " \[Dash] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_1\), "TraditionalForm"], "\[CenterDot]", \(y\_1\)}], TraditionalForm]]], " \[Dash] \[CenterEllipsis] \[Dash] ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot] ", Cell[BoxData[ FormBox[ SubscriptBox["y", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\n\t = \[Dash]", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " \[CenterDot] ", Cell[BoxData[ \(TraditionalForm\`y\_0\)]], " \[Dash] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(a\_1\), "TraditionalForm"], "\[CenterDot]", \(\[CurlyPhi](y\_0)\)}], TraditionalForm]]], " \[Dash] \[CenterEllipsis] \[Dash] ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot] ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], "TraditionalForm"], "(", \(y\_0\), ")"}], TraditionalForm]]], "\nund ", Cell[BoxData[ FormBox[ RowBox[{"\[CurlyPhi]", "(", SubscriptBox["y", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], ")"}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox["m", FontSlant->"Italic"]], "TraditionalForm"], "(", \(y\_0\), ")"}], TraditionalForm]]], " . Damit folgt ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\((\[CurlyPhi])\)\[CenterDot]y\_0\)]], " = 0 und ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\((\[CurlyPhi])\)\)]], " = 0 . Also ist ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") Teiler von ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") . \nNun gibt es aber kein nichtkonstantes Polynom ", FontVariations->{"CompatibilityType"->0}], StyleBox["g", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") ", FontVariations->{"CompatibilityType"->0}], "\[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] ", StyleBox["mit einem Grad kleiner als ", FontVariations->{"CompatibilityType"->0}], StyleBox["m", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" und ", FontVariations->{"CompatibilityType"->0}], StyleBox["g", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\((\[CurlyPhi])\)\[CenterDot]y\_0\)]], " = 0 , denn sonst w\[ADoubleDot]re ", StyleBox["B", FontSlant->"Italic"], " nicht linear unabh\[ADoubleDot]ngig. Da ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " normiert vorausgesetzt ist, folgt damit ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", StyleBox["f", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->9, FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[")", FontVariations->{"CompatibilityType"->0}], " \[FilledSmallSquare]" }], "Text"], Cell[TextData[{ StyleBox["Definition II.4.26:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nZYKLISCH (", FontFamily->"Arial"], StyleBox["cyclic", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], "\nEs seien ", StyleBox["V", FontSlant->"Italic"], " ein ", StyleBox["K", FontSlant->"Italic"], "-Vektorraum und \[CurlyPhi] \[Element] hom(", StyleBox["V", FontSlant->"Italic"], ", ", StyleBox["V", FontSlant->"Italic"], ") . \[CurlyPhi] hei\[SZ]t ", StyleBox["ZYKLISCH", FontFamily->"Arial"], ", wenn es ", StyleBox["v", FontSlant->"Italic"], " \[Element] ", StyleBox["V", FontSlant->"Italic"], " gibt, so dass \n( ", StyleBox["v", FontSlant->"Italic"], ", \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], "), ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], "),\[Ellipsis], ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ) eine Basis von ", StyleBox["V ", FontSlant->"Italic"], "ist." }], "Text"], Cell[TextData[{ StyleBox["Korollar II.4.16", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es sei \[CurlyPhi] ein zyklischer Endomorphismus des ", FontVariations->{"CompatibilityType"->0}], StyleBox["m", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["-dimensionalen ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["-Vektorraumes ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" . \nDann hat das ", FontVariations->{"CompatibilityType"->0}], " Minimalpolynom", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") von \[CurlyPhi] den Grad ", StyleBox["m , ", FontSlant->"Italic"], "also ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%m\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`a\_m\)]], " = 1 . \nBzgl. der Basis ", StyleBox["B", FontSlant->"Italic"], " = (", StyleBox["v", FontSlant->"Italic"], ", \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], "), ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], "),\[Ellipsis], ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ) ist die \[CurlyPhi] darstellende Matrix die Begleitmatrix\n\n \ \t", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ {"0", "0", "\[CenterEllipsis]", "0", "0", StyleBox[ SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], "0"], FontFamily->"Times New Roman"]}, {"1", "0", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], StyleBox["1", FontFamily->"Times New Roman"]]}, {"0", "1", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], "2"]}, {"\[VerticalEllipsis]", "\[VerticalEllipsis]", "\[DescendingEllipsis]", \(\(\ \)\(\[VerticalEllipsis]\)\), "\[VerticalEllipsis]", "\[VerticalEllipsis]"}, {"0", "0", "\[CenterEllipsis]", "1", "0", StyleBox[ SubscriptBox["\[Dash]a", RowBox[{"m", "\[VeryThinSpace]", StyleBox[ RowBox[{"\[Dash]", StyleBox["2", FontSlant->"Plain"]}]]}]], FontFamily->"Times New Roman", FontSlant->"Italic"]}, {"0", "0", "\[CenterEllipsis]", "0", "1", StyleBox[ SubscriptBox[ StyleBox["\[Dash]a", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[ RowBox[{ StyleBox["m", FontFamily->"Times New Roman"], "\[Dash]", StyleBox["1", FontFamily->"Times New Roman", FontSlant->"Plain"]}]]], FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], " .\n \n", StyleBox["Das ", FontVariations->{"CompatibilityType"->0}], " Minimalpolynom", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") von \[CurlyPhi] ist gleich dem charakteristischen Polynom ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") von \[CurlyPhi] ." }], "Text"], Cell[TextData[{ "Der ", StyleBox["Beweis ", FontFamily->"Arial"], "folgt aus Satz II.4.16." }], "Text"], Cell[TextData[{ StyleBox["Bemerkung:", FontFamily->"Arial"], "\nNach diesen Vorbereitungen wird nun folgenderma\[SZ]en vorgegangen:\nWie \ in Satz II.4.14 wird ", StyleBox["V", FontSlant->"Italic"], " als ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Modul betrachtet und es gilt ", StyleBox[" ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ SubscriptBox["Ann", RowBox[{ StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["[", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["]", FontVariations->{"CompatibilityType"->0}]}]], TraditionalForm]]], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") = ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ")\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`K[X]\)]], " . Das Minimalpolynom ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") habe eine Zerlegung mit irreduziblen Faktoren ", Cell[BoxData[ \(TraditionalForm\`\(p\_\[Kappa]\)(X)\)]], " : ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox[\(p\_1\), "TraditionalForm"], "(", "X", ")"}], \(\[Alpha]\_1\)], TraditionalForm]]], "\[CenterEllipsis] ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox[\(p\_k\), "TraditionalForm"], "(", "X", ")"}], \(\[Alpha]\_k\)], TraditionalForm]]], " . \nNach Satz II.4.11 hat ", StyleBox["V", FontSlant->"Italic"], " dann eine Zerlegung als ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]-Modul ", StyleBox["V", FontSlant->"Italic"], " =", StyleBox[" V", FontSlant->"Italic"], "(", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(p\_1\), "TraditionalForm"], "(", "X", ")"}], TraditionalForm]]], ") \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", StyleBox["V", FontSlant->"Italic"], "(", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(p\_k\), "TraditionalForm"], "(", "X", ")"}], TraditionalForm]]], ") , wobei \n", StyleBox["V", FontSlant->"Italic"], "(", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(p\_\[Kappa]\), "TraditionalForm"], "(", "X", ")"}], TraditionalForm]]], ") = { ", StyleBox["v", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[Element] ", FontVariations->{"CompatibilityType"->0}], StyleBox["V", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" | Es gibt ein \[Mu]", FontVariations->{"CompatibilityType"->0}], " mit ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{"(", FormBox[ RowBox[{ FormBox[\(p\_\[Kappa]\), "TraditionalForm"], "(", "\[CurlyPhi]", ")"}], "TraditionalForm"], ")"}], "\[Mu]"], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") = 0 .} . Nach Satz II.4.12 hat jede Komponente ", StyleBox["V", FontSlant->"Italic"], "(", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(p\_\[Kappa]\), "TraditionalForm"], "(", "X", ")"}], TraditionalForm]]], ") eine Zerlegung in zyklische Summanden: \n", StyleBox["V", FontSlant->"Italic"], "(", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(p\_\[Kappa]\), "TraditionalForm"], "(", "X", ")"}], TraditionalForm]]], ") = ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`v\_1\)]], " \[CirclePlus] \[CenterEllipsis] \[CirclePlus] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`v\_k\)]], " , wobei ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`v\_\[Kappa]\)]], " \[TildeEqual] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] / ", Cell[BoxData[ \(TraditionalForm\`\(\(p\_\[Kappa]\)(X)\)\^k\_\[Kappa]\)]], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] mit ", Cell[BoxData[ \(TraditionalForm\`k\_1\)]], "\[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`k\_2\)]], "\[GreaterEqual] \[CenterEllipsis] \[GreaterEqual] ", Cell[BoxData[ \(TraditionalForm\`k\_k\)]], "\[GreaterEqual] 1 . \nAlso gilt ", StyleBox["V ", FontSlant->"Italic"], "= ", Cell[BoxData[ \(TraditionalForm\`\[CirclePlus]\+\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " = ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", Cell[BoxData[ \(TraditionalForm\`v\_\(\[Kappa]\ \[Mu]\)\)]], " \[TildeEqual] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] / ", Cell[BoxData[ \(TraditionalForm\`\(\(p\_\[Kappa]\)(X)\)\^k\_\(\[Kappa]\ \[Mu]\)\)]], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] . Nach Satz II.4.15 ist ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " auch \n", StyleBox["K", FontSlant->"Italic"], "-Untervektorraum und es gilt \[CurlyPhi](", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], ") \[Subset] ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " . Nach Satz II.4.16 stimmen auf den Summanden ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " das charakteristische Polynom und das Minimalpolynom \[UDoubleDot]berein \ und sind ", Cell[BoxData[ \(TraditionalForm\`\(\(p\_\[Kappa]\)(X)\)\^k\_\(\[Kappa]\ \[Mu]\)\)]], " .\n" }], "Text"], Cell[TextData[{ "Nach dem Satz von Cayley-Hamilton ist ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") Teiler von ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") . Genauer gilt:" }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Satz II.4.17", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\nEs seien ", StyleBox["K", FontSlant->"Italic"], " ein K\[ODoubleDot]rper, ", StyleBox["V", FontSlant->"Italic"], " ein ", StyleBox["K", FontSlant->"Italic"], "-Vektorraum der Dimension \[GreaterEqual] 1 und \[CurlyPhi]", StyleBox[" ", FontSlant->"Italic"], "\[Element] hom(", StyleBox["V", FontSlant->"Italic"], ",", StyleBox[" V", FontSlant->"Italic"], ")", StyleBox[" ", FontSlant->"Italic"], ".\nDas Minimalpolynom und das charakteristische Polynom haben die gleichen \ irreduziblen Faktoren." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\nEs sei ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox[\(p\_1\), "TraditionalForm"], "(", "X", ")"}], \(\[Alpha]\_1\)], TraditionalForm]]], " \[CenterEllipsis] ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox[\(p\_m\), "TraditionalForm"], "(", "X", ")"}], \(\[Alpha]\_m\)], TraditionalForm]]], " . Nach der Bemerkung vor diesem Satz ist ", StyleBox["V ", FontSlant->"Italic"], "= ", Cell[BoxData[ \(TraditionalForm\`\[CirclePlus]\+\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " und auf ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " stimmen das charakteristische Polynom und das Minimalpolynom jeweils \ \[UDoubleDot]berein und sind ", Cell[BoxData[ \(TraditionalForm\`\(\(p\_\[Kappa]\)(X)\)\^k\_\(\[Kappa]\ \[Mu]\)\)]], " . Die Vereinigung der Basen ", Cell[BoxData[ \(TraditionalForm\`B\_\(\[Kappa]\ \[Mu]\)\)]], " der ", Cell[BoxData[ \(TraditionalForm\`V\_\(\[Kappa]\ \[Mu]\)\)]], " ist eine Basis von ", StyleBox["V ", FontSlant->"Italic"], " bzgl. der \[CurlyPhi] durch eine Blockmatrix dargestellt wird: ", StyleBox["B", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\[Union]\_\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`B\_\(\[Kappa]\ \[Mu]\)\)]], " . Dann hat die bzgl. ", StyleBox["B", FontSlant->"Italic"], " darstellende Matrix von \[CurlyPhi] die Gestalt\n\n", StyleBox[" \t A = ", FontSlant->"Italic"], StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", GridBox[{ { FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["11", FontFamily->"Times New Roman"]], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " ", " ", " ", StyleBox["0", FontSize->14]}, {" ", "\[DescendingEllipsis]", " ", " ", " "}, {" ", " ", FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], RowBox[{ StyleBox["1", FontFamily->"Times New Roman"], StyleBox[ SubscriptBox[ StyleBox["k", FontSlant->"Italic"], "1"], FontFamily->"Times New Roman"], " "}]], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " ", " "}, {" ", " ", " ", "\[DescendingEllipsis]", " "}, { StyleBox["0", FontSize->14], " ", " ", " ", FrameBox[ RowBox[{ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[\(k\ k\_k\), FontFamily->"Times New Roman", FontSlant->"Italic"]], " "}], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}]} }], ")"}]], FontFamily->"Times New Roman"], " .\n\nNun gilt f\[UDoubleDot]r das charakteristische Polynom von \ \[CurlyPhi] : ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_A\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`\[CapitalPi]\+\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(A\_\(\(\[Kappa]\)\(\ \)\(\[Mu]\)\(\ \)\)\)\ \)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`\[CapitalPi]\+\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`m\_\(A\_\(\(\[Kappa]\)\(\ \)\(\[Mu]\)\(\ \)\)\)\)]], "(", StyleBox["X", FontSlant->"Italic"], ") = ", Cell[BoxData[ \(TraditionalForm\`\[CapitalPi]\+\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(\(p\_\[Kappa]\)(X)\)\^k\_\(\[Kappa]\ \[Mu]\)\)]], " .\nAlso haben das Minimalpolynom und das charakteristische Polynom von \ \[CurlyPhi] die gleichen irreduziblen Faktoren \[FilledSmallSquare]" }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Lemma II.4.10", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es sei ", FontVariations->{"CompatibilityType"->0}], StyleBox["W", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" := ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " \[TildeEqual] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] / ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox["p", "TraditionalForm"], "(", "X", ")"}], "\[Alpha]"], TraditionalForm]]], "\[CenterDot] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] eine zyklische Komponente von ", StyleBox["V", FontSlant->"Italic"], " , wobei ", StyleBox["p", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], ") \[Element] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] ein normiertes, irreduzibles Polynom ist.\nEs sei ", Cell[BoxData[ \(TraditionalForm\`\(p(X)\)\^\[Alpha]\)]], " := ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%m\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`a\_m\)]], " = 1 , dann ist ", StyleBox["B", FontSlant->"Italic"], " := ( ", StyleBox["v", FontSlant->"Italic"], ", \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], "), ", Cell[BoxData[ \(TraditionalForm\`\[CurlyPhi]\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], "),\[Ellipsis], ", Cell[BoxData[ FormBox[ SuperscriptBox["\[CurlyPhi]", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ) eine Basis von ", StyleBox["W", FontSlant->"Italic"], " als \n", StyleBox["K", FontSlant->"Italic"], "-Vektorraum und bzgl. dieser hat ", Cell[BoxData[ FormBox[ RowBox[{"\[CurlyPhi]", SubscriptBox["|", StyleBox["W", FontSize->10]]}], TraditionalForm]]], " folgende Matrix:\n\n ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ {"0", "0", "\[CenterEllipsis]", "0", "0", StyleBox[ SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], "0"], FontFamily->"Times New Roman"]}, {"1", "0", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], StyleBox["1", FontFamily->"Times New Roman"]]}, {"0", "1", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], "2"]}, {"\[VerticalEllipsis]", "\[VerticalEllipsis]", "\[DescendingEllipsis]", \(\(\ \)\(\[VerticalEllipsis]\)\), "\[VerticalEllipsis]", "\[VerticalEllipsis]"}, {"0", "0", "\[CenterEllipsis]", "1", "0", StyleBox[ SubscriptBox["\[Dash]a", RowBox[{"m", "\[VeryThinSpace]", StyleBox[ RowBox[{"\[Dash]", StyleBox["2", FontSlant->"Plain"]}]]}]], FontFamily->"Times New Roman", FontSlant->"Italic"]}, {"0", "0", "\[CenterEllipsis]", "0", "1", StyleBox[ SubscriptBox[ StyleBox["\[Dash]a", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[ RowBox[{ StyleBox["m", FontFamily->"Times New Roman"], "\[Dash]", StyleBox["1", FontFamily->"Times New Roman", FontSlant->"Plain"]}]]], FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], ".\n " }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Der ", StyleBox["Beweis", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], " verl\[ADoubleDot]uft entsprechend dem von Satz II.4.16." }], "Text"], Cell["\<\ Entsprechend der Bemerkung vor Satz II.4.17 und mit Satz II.4.16 folgt\ \>", "Text"], Cell[TextData[{ StyleBox["Satz II.4.18", FontFamily->"Arial", FontVariations->{"Underline"->True}], " ", StyleBox["Rationale Normalform bzw. J", FontFamily->"Arial"], StyleBox["ORDAN", FontFamily->"Arial", FontSize->9], StyleBox["-H", FontFamily->"Arial"], StyleBox["\[CapitalODoubleDot]LDER", FontFamily->"Arial", FontSize->9], StyleBox[" Normalform", FontFamily->"Arial"], "\nEs seien ", StyleBox["K", FontSlant->"Italic"], " ein K\[ODoubleDot]rper, ", StyleBox["V", FontSlant->"Italic"], " ein ", StyleBox["K", FontSlant->"Italic"], "-Vektorraum der Dimension ", StyleBox["n", FontSlant->"Italic"], " und \[CurlyPhi] \[Element] hom(", StyleBox["V", FontSlant->"Italic"], ", ", StyleBox["V", FontSlant->"Italic"], ") . \nDann gibt es eine Basis ", StyleBox["B", FontSlant->"Italic"], " von ", StyleBox["V", FontSlant->"Italic"], " , bzgl. der \[CurlyPhi] folgende Matrix hat" }], "Text", FontWeight->"Plain"], Cell[TextData[{ " ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", GridBox[{ { FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["1", FontFamily->"Times New Roman"]], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " ", " ", RowBox[{" ", StyleBox["0", FontFamily->"Times New Roman", FontSize->14]}]}, {" ", FrameBox[ StyleBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], "2"], FontFamily->"Times New Roman"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " ", " "}, { StyleBox[" ", FontSize->16], " ", StyleBox["\[DescendingEllipsis]", FontSize->16], " "}, { StyleBox[\(\(0\)\(\ \)\), FontFamily->"Times New Roman", FontSize->14], StyleBox[" ", FontSize->16], " ", FrameBox[ StyleBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["k", FontSlant->"Italic"]], FontFamily->"Times New Roman"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}]} }], ")"}]], FontFamily->"Times New Roman"], StyleBox[" ,", FontFamily->"Times New Roman"] }], "Text", FontWeight->"Plain"], Cell[TextData[{ "wobei die K\[ADoubleDot]stchen die Matrizen \n\n ", Cell[BoxData[ \(TraditionalForm\`A\_\[Kappa]\)]], " := ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ {"0", "0", "\[CenterEllipsis]", "0", "0", StyleBox[ SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], \(\[Kappa]\ 0\)], FontFamily->"Times New Roman"]}, {"1", "0", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], \(\[Kappa]\ 1\)]}, {"0", "1", "\[CenterEllipsis]", "0", "0", SubscriptBox[ StyleBox[ RowBox[{ StyleBox["\[Dash]", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]}]], \(\[Kappa]\ 2\)]}, {"\[VerticalEllipsis]", "\[VerticalEllipsis]", "\[DescendingEllipsis]", \(\(\ \)\(\[VerticalEllipsis]\)\), "\[VerticalEllipsis]", "\[VerticalEllipsis]"}, {"0", "0", "\[CenterEllipsis]", "1", "0", StyleBox[ SubscriptBox["\[Dash]a", RowBox[{ StyleBox["\[Kappa]", FontSlant->"Plain"], " ", RowBox[{"(", RowBox[{ SubscriptBox[\(\(m\)\(\[VeryThinSpace]\)\), StyleBox["\[Kappa]", FontSlant->"Plain"]], StyleBox[ RowBox[{"\[Dash]", StyleBox["2", FontSlant->"Plain"]}]]}], ")"}]}]], FontFamily->"Times New Roman", FontSlant->"Italic"]}, {"0", "0", "\[CenterEllipsis]", "0", "1", StyleBox[ SubscriptBox[ StyleBox["\[Dash]a", FontFamily->"Times New Roman", FontSlant->"Italic"], RowBox[{ StyleBox["\[Kappa]", FontSlant->"Plain"], " ", RowBox[{"(", RowBox[{ SubscriptBox["m", StyleBox["\[Kappa]", FontSlant->"Plain"]], StyleBox[ RowBox[{"\[Dash]", StyleBox["1", FontSlant->"Plain"]}]]}], ")"}]}]], FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], " \n\nsind, und die Polynome ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ SubscriptBox[ FormBox["p", "TraditionalForm"], "\[Kappa]"], "(", "X", ")"}], \(\[Alpha]\_\[Kappa]\)], TraditionalForm]]], " := ", Cell[BoxData[ \(TraditionalForm\`\[CapitalSigma]\+\(\[Mu] = 0\)\%\(m\_\[Kappa]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Kappa]\ \[Mu]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`X\^\[Mu]\)]], " mit ", Cell[BoxData[ \(TraditionalForm\`a\_\(\[Kappa]\ m\_\[Kappa]\)\)]], " = 1 jeweils irreduzible Polynome aus ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] sind. Bis auf die Reihenfolge der K\[ADoubleDot]stchen ist die \ Darstellung eindeutig.\n\n", StyleBox[" ", FontVariations->{"Underline"->True}], StyleBox[ButtonBox["Otto Ludwig H\[ODoubleDot]lder", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Holder.\ html"], None}, ButtonStyle->"Hyperlink"], FontVariations->{"Underline"->True}], StyleBox[", 1859 \[Dash] 1937", FontVariations->{"Underline"->True}] }], "Text", FontWeight->"Plain"], Cell["\<\ Es wird nun noch eine etwas \[UDoubleDot]bersichtlichere Struktur der Matrix \ entwickelt. Dazu als Vorbereitung\ \>", "Text"], Cell[TextData[{ StyleBox["Lemma II.4.11", FontFamily->"Arial", FontVariations->{"Underline"->True}], "\n", StyleBox["Es sei speziell ", FontVariations->{"CompatibilityType"->0}], StyleBox["p", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox["(", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[") := ", FontVariations->{"CompatibilityType"->0}], StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[Dash] ", FontVariations->{"CompatibilityType"->0}], StyleBox["a", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" mit ", FontVariations->{"CompatibilityType"->0}], StyleBox["a", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[Element] ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" und mit einem ", FontVariations->{"CompatibilityType"->0}], StyleBox["v", FontSlant->"Italic"], " \[Element] ", StyleBox["V", FontSlant->"Italic"], " ", StyleBox[" sei wieder ", FontVariations->{"CompatibilityType"->0}], StyleBox["W", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" := ", FontVariations->{"CompatibilityType"->0}], StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "]\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " \[TildeEqual] ", StyleBox["K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] / (", StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[Dash] ", FontVariations->{"CompatibilityType"->0}], StyleBox["a", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], ")", Cell[BoxData[ \(TraditionalForm\`\^m\)]], "\[CenterDot]", StyleBox[" K", FontSlant->"Italic"], "[", StyleBox["X", FontSlant->"Italic"], "] .\nDann ist ( ", StyleBox["v", FontSlant->"Italic"], ", (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ", (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ \(TraditionalForm\`\^2\)]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ",\[Ellipsis], (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " ) \nbzw. ( (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ",\[Ellipsis], (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ \(TraditionalForm\`\^2\)]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ", (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ", ", StyleBox["v", FontSlant->"Italic"], " ) \njeweils eine Basis von ", StyleBox["W", FontSlant->"Italic"], " . \nDie \[CurlyPhi] darstellenden Matrizen bzgl. dieser Basen sind\n\n \ ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ { StyleBox["a", FontSlant->"Italic"], "0", " ", " ", " ", "0"}, {"1", StyleBox["a", FontSlant->"Italic"], ".", " ", " ", " "}, {\(\(0\)\(\ \)\), ".", ".", ".", " ", " "}, {" ", ".", ".", ".", ".", " "}, {" ", " ", \(\(\ \)\(.\)\), ".", StyleBox["a", FontSlant->"Italic"], "0"}, {"0", " ", " ", "0", "1", StyleBox["a", FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], StyleBox[" bzw. ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ { StyleBox["a", FontSlant->"Italic"], "1", "0", " ", " ", "0"}, {"0", StyleBox["a", FontSlant->"Italic"], ".", ".", " ", " "}, {" ", ".", ".", ".", ".", " "}, {" ", " ", ".", ".", ".", "0"}, {" ", " ", " ", ".", StyleBox["a", FontSlant->"Italic"], "1"}, {"0", " ", " ", " ", "0", StyleBox["a", FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], " ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ StyleBox["Beweis:", FontFamily->"Arial", FontVariations->{"CompatibilityType"->0}], "\n( ", StyleBox["v", FontSlant->"Italic"], ", (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ", (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ \(TraditionalForm\`\^2\)]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ",\[Ellipsis], (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " ) ist Basis von ", StyleBox["W", FontSlant->"Italic"], " , denn:\nEs sei ", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " \[CenterDot] ", StyleBox["v", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a", FontSlant->"Italic"], ")\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " + \[CenterEllipsis] + ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a", FontSlant->"Italic"], ")", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " = 0 . Dann ist das Polynom \n", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], " + ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], "\[CenterDot](", StyleBox["X", FontSlant->"Italic"], " \[Dash] ", StyleBox["a", FontSlant->"Italic"], ") + \[CenterEllipsis] + ", Cell[BoxData[ FormBox[ SubscriptBox["a", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot](", StyleBox["X", FontSlant->"Italic"], " \[Dash] ", StyleBox["a", FontSlant->"Italic"], ")", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], " durch (", StyleBox["X", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" \[Dash] ", FontVariations->{"CompatibilityType"->0}], StyleBox["a", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], ")", Cell[BoxData[ \(TraditionalForm\`\^m\)]], " teilbar, also (wie oben) ", Cell[BoxData[ \(TraditionalForm\`a\_\[Mu]\)]], " = 0 f\[UDoubleDot]r \[Mu] = 0,\[Ellipsis], ", StyleBox["m", FontSlant->"Italic"], "\[Dash]1 .\n\nEine Familie von Vektoren bleibt eine Basis, wenn in der \ Familie nur umgeordnet wird! Damit ist auch \n( (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ",\[Ellipsis], (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ \(TraditionalForm\`\^2\)]], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ", (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)\[CenterDot]", StyleBox["v", FontSlant->"Italic"], ", ", StyleBox["v", FontSlant->"Italic"], " ) auch eine Basis von ", StyleBox["W", FontSlant->"Italic"], " .\n\nWie sieht die darstellende Matrix im ersten Fall aus?\n \ \[CurlyPhi](", StyleBox["v", FontSlant->"Italic"], ") = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot]", StyleBox["v", FontSlant->"Italic"], " ,\n \[CurlyPhi]((\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(", StyleBox["v", FontSlant->"Italic"], ")) = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(", StyleBox["v", FontSlant->"Italic"], ") = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ \(TraditionalForm\`\^2\)]], "(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(", StyleBox["v", FontSlant->"Italic"], ") ,\n \[VerticalEllipsis]\n \ \[CurlyPhi]((\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ")) = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]2"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ,\n \n \[CurlyPhi]((\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ")) = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox["m", FontSlant->"Italic"]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") = 0 + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]1"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Damit hat die darstellende Matrix von \[CurlyPhi] bzgl. der ersten Basis \ die Gestalt\n\n ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", GridBox[{ { StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["0", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox["0", FontFamily->"Times New Roman"]}, { StyleBox["1", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[\(\(.\)\(\ \)\), FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"]}, { StyleBox["0", FontFamily->"Times New Roman"], StyleBox[".", FontFamily->"Times New Roman"], StyleBox[".", FontFamily->"Times New Roman"], StyleBox[\(\(.\)\(\ \)\), FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"]}, { StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[\(\(\ \)\(.\)\), FontFamily->"Times New Roman"], StyleBox[".", FontFamily->"Times New Roman"], StyleBox[".", FontFamily->"Times New Roman"], StyleBox[\(\(.\)\(\ \)\), FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"]}, { StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[\(\(\ \)\(.\)\), FontFamily->"Times New Roman"], StyleBox[".", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["0", FontFamily->"Times New Roman"]}, { StyleBox["0", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox["0", FontFamily->"Times New Roman"], StyleBox["1", FontFamily->"Times New Roman"], StyleBox["a", FontFamily->"Times New Roman", FontSlant->"Italic"]} }], ")"}]], FontFamily->"Times New Roman"], " ." }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Wie sieht die darstellende Matrix bzgl. der zweiten Basis aus? Der Beweis \ ist jetzt etwas k\[UDoubleDot]rzer aufgeschrieben.\nEs sei ", Cell[BoxData[ \(TraditionalForm\`v\_\[Kappa]\)]], " := (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]\[Kappa]"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") , \[Kappa] \[Element] \[DoubleStruckM] ; insbesondere ist dann ", Cell[BoxData[ \(TraditionalForm\`v\_m\)]], " = ", StyleBox["v", FontSlant->"Italic"], " .\n\n \[CurlyPhi](", Cell[BoxData[ \(TraditionalForm\`v\_\[Kappa]\)]], ") = \[CurlyPhi]( (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]\[Kappa]"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") ) = (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)(\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]\[Kappa]"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]\[Kappa]"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ")\n \t= (\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", RowBox[{ StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]"}]], "(", "\[Kappa]\[Dash]1", ")"}]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ") + ", StyleBox["a", FontSlant->"Italic"], "\[CenterDot](\[CurlyPhi] \[Dash] ", StyleBox["a ", FontSlant->"Italic"], "id)", Cell[BoxData[ FormBox[ SuperscriptBox["", StyleBox[ RowBox[{ StyleBox["m", FontSlant->"Italic"], "\[Dash]\[Kappa]"}]]], TraditionalForm]]], "(", StyleBox["v", FontSlant->"Italic"], ")\n \t= ", Cell[BoxData[ FormBox[ TagBox[ RowBox[{ StyleBox["{", ShowAutoStyles->False, FontWeight->"Bold"], StyleBox[" ", ShowAutoStyles->False], StyleBox[GridBox[{ {\(v\_\(\[Kappa]\ \[Dash]\ 1\) + \ a\ v\_\[Kappa]\)}, {\(a\ v\_\[Kappa]\)} }], ShowAutoStyles->True]}], (#&)], TraditionalForm]]], " ", Cell[BoxData[ FormBox[GridBox[{ {"f\[UDoubleDot]r", \(\[Kappa]\ \[GreaterEqual] \ 2\)}, {"f\[UDoubleDot]r", \(\[Kappa]\ = \ 1\)} }], TraditionalForm]]], " .\n Damit hat die \[CurlyPhi] darstellende Matrix jetzt die Gestalt: \ \n \n ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ { StyleBox["a", FontSlant->"Italic"], "1", "0", " ", " ", "0"}, {"0", StyleBox["a", FontSlant->"Italic"], ".", ".", " ", " "}, {" ", ".", ".", ".", ".", " "}, {" ", " ", ".", ".", ".", "0"}, {" ", " ", " ", ".", StyleBox["a", FontSlant->"Italic"], "1"}, {"0", " ", " ", " ", "0", StyleBox["a", FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], " \[FilledSmallSquare]\n " }], "Text"], Cell["Die Matrizen aus dem letzten Lemma haben einen Namen:", "Text"], Cell[TextData[{ StyleBox["Definition II.4.27:", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" \nJORDANK\[CapitalADoubleDot]STCHEN (", FontFamily->"Arial"], StyleBox["Jordan block", FontFamily->"Arial", FontSlant->"Italic"], StyleBox[")", FontFamily->"Arial"], "\nEs sei ", StyleBox["a", FontSlant->"Italic"], " \[Element] ", StyleBox["R", FontSlant->"Italic"], " . Matrizen der Form\n \n \t ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ { StyleBox["a", FontSlant->"Italic"], "0", " ", " ", " ", "0"}, {"1", StyleBox["a", FontSlant->"Italic"], ".", " ", " ", " "}, {\(\(0\)\(\ \)\), ".", ".", ".", " ", " "}, {" ", ".", ".", ".", ".", " "}, {" ", " ", \(\(\ \)\(.\)\), ".", StyleBox["a", FontSlant->"Italic"], "0"}, {"0", " ", " ", "0", "1", StyleBox["a", FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], " ", StyleBox[" bzw. ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", StyleBox[GridBox[{ { StyleBox["a", FontSlant->"Italic"], "1", "0", " ", " ", "0"}, {"0", StyleBox["a", FontSlant->"Italic"], ".", ".", " ", " "}, {" ", ".", ".", ".", ".", " "}, {" ", " ", ".", ".", ".", "0"}, {" ", " ", " ", ".", StyleBox["a", FontSlant->"Italic"], "1"}, {"0", " ", " ", " ", "0", StyleBox["a", FontSlant->"Italic"]} }], FontFamily->"Times New Roman"], ")"}]], FontFamily->"Times New Roman"], " \n \n hei\[SZ]en ", StyleBox["JORDANK\[CapitalADoubleDot]STCHEN", FontFamily->"Arial"], " oder auch ", StyleBox["JORDANBL\[CapitalODoubleDot]CKE", FontFamily->"Arial"], "." }], "Text"], Cell[TextData[{ StyleBox["Bemerkung:", FontFamily->"Arial"], "\nIn den meisten B\[UDoubleDot]chern wird jeweils nur eine \"Sorte\" der \ Jordank\[ADoubleDot]stchen benutzt." }], "Text"], Cell[TextData[{ StyleBox["Beispiel (Fortsetzung):", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox["\n", FontFamily->"Arial"], "Es sei nochmals das obige Beispiel mit der Matrix\n\n\t\t ", StyleBox["A", FontSlant->"Italic"], " := ", Cell[BoxData[ StyleBox[ RowBox[{"(", GridBox[{ {"0", "0", "0", "0", "4"}, {"1", "0", "0", "0", \(-16\)}, {"0", "1", "0", "0", "25"}, {"0", "0", "1", "0", \(-19\)}, {"0", "0", "0", "1", "7"} }], ")"}], FontFamily->"Times New Roman"]]], " \n\nbetrachtet. Es sei also \[CurlyPhi]: ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^5\)]], " \[RightArrow] ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^5\)]], " und \[CurlyPhi] werde bzgl. der Einheitsbasis durch die Matrix ", StyleBox["A", FontSlant->"Italic"], " dargestellt.\nEs sei " }], "Text"], Cell[BoxData[ \(v\_1 = {4, \(-12\), 13, \(-6\), 1}; v\_2 = {\(-4\), 8, \(-5\), 1, 0}; v\_3 = {4, \(-4\), 1, 0, 0};\)], "Input"], Cell["Dann folgt", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({A . v\_1\ \[Equal] \ v\_1, \ A . v\_2\ \[Equal] \ v\_1 + v\_2, \ A . v\_3\ \[Equal] \ v\_2 + v\_3}\)], "Input"], Cell[BoxData[ \({True, True, True}\)], "Output"] }, Open ]], Cell["und mit", "Text"], Cell[BoxData[ \(w\_1 = {2, \(-7\), 9, \(-5\), 1}; \ w\_2 = {\(-1\), 3, \(-3\), 1, 0};\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \({A . w\_1\ \[Equal] 2\ w\_1, \ A . w\_2\ \[Equal] \ w\_1\ + 2 w\_2}\)], "Input"], Cell[BoxData[ \({True, True}\)], "Output"] }, Open ]], Cell[TextData[{ "Also l\[ADoubleDot]\[SZ]t sich der ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^5\)]], " als direkte Summe der beiden Teilr\[ADoubleDot]ume ", Cell[BoxData[ \(TraditionalForm\`V\_1\)]], ":= ", Cell[BoxData[ \(TraditionalForm\`\([v\_1, \ v\_2, \ v\_3]\)\)]], " und ", Cell[BoxData[ \(TraditionalForm\`V\_2\)]], " := ", Cell[BoxData[ \(TraditionalForm\`\([w\_1, \ w\_2]\)\)]], " auffassen. \nBez\[UDoubleDot]glich ", Cell[BoxData[ \(TraditionalForm\`V\_1\)]], " wird ", Cell[BoxData[ \(TraditionalForm\`\(\(\[CurlyPhi]\)\( | \_\(V\_1\)\)\)\)]], " dargestellt durch die Matrix\n\n\t\t \t", Cell[BoxData[ StyleBox[ RowBox[{"(", GridBox[{ {"1", "1", "0"}, {"0", "1", "1"}, {"0", "0", "1"} }], ")"}], FontFamily->"Times New Roman"]]], " .\n\nBez\[UDoubleDot]glich ", Cell[BoxData[ \(TraditionalForm\`V\_2\)]], " wird ", Cell[BoxData[ \(TraditionalForm\`\(\(\[CurlyPhi]\)\( | \_\(V\_2\)\)\)\)]], " dargestellt durch die Matrix\n\n \ ", Cell[BoxData[ StyleBox[ RowBox[{"(", GridBox[{ {"2", "1"}, {"0", "2"} }], ")"}], FontFamily->"Times New Roman"]]], " ." }], "Text"], Cell[TextData[{ "Nun soll noch das Minimalpolynom ", Cell[BoxData[ \(TraditionalForm\`\(\(\(m\_\[CurlyPhi]\)(X)\)\(\ \)\)\)]], " von \[CurlyPhi] gefunden werden. Dazu wird zuerst das charakteristische \ Polynom in Linearfaktoren zerlegt:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(\(\[Chi]\_A[ X_] := \(-CharacteristicPolynomial[A, \ X]\);\)\ \[IndentingNewLine] \[Chi]\_A[X]\)\(\ \)\)\)], "Input"], Cell[BoxData[ \(\(-4\) + 16\ X - 25\ X\^2 + 19\ X\^3 - 7\ X\^4 + X\^5\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Factor[\ \[Chi]\_A[X]\ ]\)], "Input"], Cell[BoxData[ \(\((\(-2\) + X)\)\^2\ \((\(-1\) + X)\)\^3\)], "Output"] }, Open ]], Cell[TextData[{ "Nach dem Satz II.4.17 haben Minimalpolynom und charakteristisches Polynom \ die gleichen irreduziblen Faktoren \n(", StyleBox["X", FontSlant->"Italic"], " \[Dash] 1) und (", StyleBox["X", FontSlant->"Italic"], " \[Dash] 2) . Mit Satz II.4.16 und dem \"Anfang\" dieses Beispiels folgt, \ dass ", Cell[BoxData[ \(TraditionalForm\`\(\(\(m\_\[CurlyPhi]\)(X)\)\(\ \)\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") .\n\nSchlie\[SZ]lich wird mit ", StyleBox["Mathematica", FontSlant->"Italic"], " die ", StyleBox["J", FontFamily->"Times New Roman"], StyleBox["ORDAN", FontFamily->"Times New Roman", FontSize->9], StyleBox["sche Normalform", FontFamily->"Times New Roman"], " berechnet:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({H, JA} = JordanDecomposition[A]; MatrixForm[JA]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "1", "0", "0", "0"}, {"0", "1", "1", "0", "0"}, {"0", "0", "1", "0", "0"}, {"0", "0", "0", "2", "1"}, {"0", "0", "0", "0", "2"} }], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell["Allgemein gilt:", "Text"], Cell[TextData[{ StyleBox["Satz II.4.19", FontFamily->"Arial", FontVariations->{"Underline"->True}], StyleBox[" J", FontFamily->"Arial"], StyleBox["ORDAN", FontFamily->"Arial", FontSize->9], StyleBox["sche Normalform", FontFamily->"Arial"], " \nEs seien ", StyleBox["K", FontSlant->"Italic"], " ein K\[ODoubleDot]rper, ", StyleBox["V", FontSlant->"Italic"], " ein ", StyleBox["K", FontSlant->"Italic"], "-Vektorraum der Dimension ", StyleBox["n", FontSlant->"Italic"], " und \[CurlyPhi] \[Element] hom(", StyleBox["V", FontSlant->"Italic"], ", ", StyleBox["V", FontSlant->"Italic"], ") . Das Minimalpolynom ", Cell[BoxData[ \(TraditionalForm\`m\_\[CurlyPhi]\)]], "(", StyleBox["X", FontSlant->"Italic"], ") von \[CurlyPhi] zerfalle in Linearfaktoren. Dann gibt es eine Basis \ ", StyleBox["B", FontSlant->"Italic"], " von ", StyleBox["V", FontSlant->"Italic"], " bzgl. der \[CurlyPhi] die folgende darstellende Matrix hat:\n\n \ ", StyleBox[" ", FontFamily->"Times New Roman"], Cell[BoxData[ RowBox[{"(", GridBox[{ { StyleBox[ FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], "1"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[\(\(\ \)\(0\)\), FontFamily->"Times New Roman", FontSize->14]}, { StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[ FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], "2"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[" ", FontFamily->"Times New Roman"]}, { StyleBox[" ", FontFamily->"Times New Roman", FontSize->16], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox["\[DescendingEllipsis]", FontFamily->"Times New Roman", FontSize->16], StyleBox[" ", FontFamily->"Times New Roman"]}, { RowBox[{ StyleBox["0", FontFamily->"Times New Roman", FontSize->14], StyleBox[" ", FontFamily->"Times New Roman", FontSize->16]}], StyleBox[" ", FontFamily->"Times New Roman", FontSize->16], StyleBox[" ", FontFamily->"Times New Roman"], StyleBox[ FrameBox[ SubscriptBox[ StyleBox["A", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox["k", FontSlant->"Italic"]], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], FontFamily->"Times New Roman"]} }], ")"}]], FontFamily->"Times New Roman"], " .\n \nDie ", Cell[BoxData[ \(TraditionalForm\`A\_\[Kappa]\)]], " sind Jordank\[ADoubleDot]stchen. Bis auf Vertauschung der \ K\[ADoubleDot]stchen ist die Darstellung eindeutig." }], "Text", FontWeight->"Plain"], Cell[TextData[{ "Der ", StyleBox["Beweis ", FontFamily->"Arial"], "folgt aus Satz II.4.18 und dem letzten Lemma \[FilledSmallSquare] " }], "Text"] }, Open ]] }, FrontEndVersion->"4.1 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 695}}, WindowToolbars->"EditBar", WindowSize->{924, 638}, WindowMargins->{{3, Automatic}, {Automatic, 9}}, PrintingCopies->1, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"FirstPageHeader"->False, "FacingPages"->True}, Magnification->1.5 ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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