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GEO proofread 2.4 fertig

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TODO: chapter 2.5
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Henry Haustein 2018-12-25 13:27:50 +01:00
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commit 446b7737ee
6 changed files with 128 additions and 31 deletions
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10
3. Semester/GEO/TeX_files/Kommutative_Ringe/Ringe_von_Bruechen.tex
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@ -23,7 +23,7 @@ Sei $R$ ein Ring.
\begin{definition}
Sei $S \subseteq R\setminus \{0\}$ multiplikativ und ohne Nullteiler. Definiere Äquivalenzrelation $\sim$ auf $R \times S$:
\begin{align}
(r,s) \sim (r^{'}, s^{'}) \Leftrightarrow r s^{'} = r^{'}s\notag
(r,s) \sim (r', s') \Leftrightarrow r s' = r's\notag
\end{align}
Schreibe $\sfrac{r}{s}$ für die $\sim$-Äquivalenzklasse von $(r,s)$ und
\begin{align}
@ -45,10 +45,10 @@ Sei $R$ ein Ring.
\item $\sim$ ist Äquivalenzrelation: reflexiv, transitiv, symmetrisch (siehe Analysis Kapitel ? konstruktion rationaler Zahlen)
\item Multiplikation ist wohldefiniert:
\begin{align}
& &\frac{r_1}{s_1} \cdot \frac{r_2}{s_2} &= \frac{r_1 r_2}{s_1 s_2}, \frac{r_1}{s_1} = \frac{r_1^{'}}{s_1^{'}}\notag \\
&\Rightarrow &\frac{r_1^{'}}{s_1^{'}} \cdot \frac{r_2}{s_2} &= \frac{r_1^{'} r_2}{s_1^{'} s_2}\notag \\
&\Rightarrow &r_1 r_2 s_1^{'} s_2 = r_1^{'} r_2 s_1 s_2\notag \\
&\Rightarrow &\frac{r_1 r_2}{s_1 s_2} = \frac{r_1^{'} r_2}{s_^{'} s_2}\notag
& &\frac{r_1}{s_1} \cdot \frac{r_2}{s_2} &= \frac{r_1 r_2}{s_1 s_2}, \frac{r_1}{s_1} = \frac{r_1'}{s_1'}\notag \\
&\Rightarrow &\frac{r_1'}{s_1'} \cdot \frac{r_2}{s_2} &= \frac{r_1' r_2}{s_1' s_2}\notag \\
&\Rightarrow &r_1 r_2 s_1' s_2 = r_1' r_2 s_1 s_2\notag \\
&\Rightarrow &\frac{r_1 r_2}{s_1 s_2} = \frac{r_1' r_2}{s_1' s_2}\notag
\end{align}
\item Addition ist wohldefiniert: analog
\item $(S^{-1}R, + ,\cdot)$ ist ein Ring: Übung

93
3. Semester/GEO/TeX_files/Kommutative_Ringe/Teilbarkeit.tex
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@ -85,7 +85,7 @@ Sei $R$ ein \textbf{nullteilfreier} Ring.
\begin{remark}
In jedem Hauptidealring gilt:
\begin{align}
(x) + (y) = (\ggT(x,y,)) \notag
(x) + (y) = (\ggT(x,y)) \notag
\end{align}
anders gesagt
\begin{align}
@ -95,11 +95,11 @@ Sei $R$ ein \textbf{nullteilfreier} Ring.
\end{remark}
\begin{proposition}[Erweiterter euklidischer Algorithmus]
Sei $R$ euklidisch mit euklidischer Gradfunktion $\delta$, und seien $x,y \in R$. Man setze $x_0 = x, x_1 = y, b_0 = 0, a_1 = 0, b_1 = 1$ und berechne iterativ $x_{i+1}, q_{i+1}, a_{i+1}, b_{i+1}$ für $i \ge 1$ als
Sei $R$ euklidisch mit euklidischer Gradfunktion $\delta$, und seien $x,y \in R$. Man setze $x_0 = x$, $x_1 = y$, $a_0=1$ $b_0 = 0$, $a_1 = 0$, $b_1 = 1$ und berechne iterativ $x_{i+1}$, $q_{i+1}$, $a_{i+1}$, $b_{i+1}$ für $i \ge 1$ als
\begin{align}
x_{i-1} &= q_{i+1}x_i + x_{i+1} &\qquad &x_{i+1} = 0 \text{ oder } \delta(x_{i+1}) < \delta(x_i) \notag \\
a_{i+1} &= a_{i-1} + q_{i+1}a_i & & \notag \\
b_{i+1} = b_{i-1} - q_{i+1}b_i & & \notag
b_{i+1} &= b_{i-1} - q_{i+1}b_i & & \notag
\end{align}
solange bis $x_{k+1} = 0$. Dann ist
\begin{align}
@ -108,7 +108,7 @@ Sei $R$ ein \textbf{nullteilfreier} Ring.
\end{proposition}
\begin{proof}
Da $\delta(x_1) > \delta(x_2) > \dots >$ wird $x_{k+1} = 0$ irgendwann erreicht. Für jedes $i \le k$ ist
Da $\delta(x_1) > \delta(x_2) > \dots$ wird $x_{k+1} = 0$ irgendwann erreicht. Für jedes $i \le k$ ist
\begin{align}
\ggT(x_{i-1}, x_i) &= \ggT(q_{i+1}x_i + x_{i+1}, x_i)\notag \\
&=\ggT(x_{i+1}, x_i) \notag \\
@ -116,41 +116,84 @@ Sei $R$ ein \textbf{nullteilfreier} Ring.
\end{align}
somit im Allgemeinen:
\begin{align}
\ggT(x,y) &= \ggT(x_0,x_1) = \ggT(x_0,x_1) = \dots = \ggT(x_k, \underbrace{x_{k+1}}_{=0})\notag \\
&= x_k \notag
\ggT(x,y) &= \ggT(x_0,x_1) = \dots = \ggT(x_k, \underbrace{x_{k+1}}_{=0}) = x_k \notag
\end{align}
Per Induktion sieht man, dass $x_i = a_i x + b_i y$ für alle \\
$i \le k: x_{i-1} = a_{i-1} x + b_{i-1}y, x_i = a_i x + b_i y$ sowie\\
$x_{i-1} = q_{i+1 x_i + x_{i+1}}$\\
Per Induktion sieht man, dass $x_i = a_i x + b_i y$ für alle $i \le k: x_{i-1} = a_{i-1} x + b_{i-1}y$, $x_i = a_i x + b_i y$ sowie $x_{i-1} = q_{i+1} x_i + x_{i+1}$\\
\begin{align}
\Rightarrow x_{i-1} &= x_{i-1}q_{i+1}x_i = (a_{i+1} - q_{i+1} q_i)x + (b_{i-1}-q_{i+1}b_i)y \notag\\
\Rightarrow x_{i+1} &= x_{i-1}q_{i+1}x_i = (a_{i+1} - q_{i+1} a_i)x + (b_{i-1}-q_{i+1}b_i)y \notag\\
&= a_{i+1}x + b_{i+1}y\notag
\end{align}
\end{proof}
\begin{example}
$R = \whole, x = 5, y= 13$
\begin{align}
5 &= 0\cdot 13 + 5 \notag \\
13 &= 2\cdot 5 + 3 \notag \\
5 &= 1\cdot 3 + 2 \notag \\
3 &= 1\cdot 2 + 1 \notag \\
5 &= 2\cdot \underline{1} + 0 \notag \\
\end{align}
$\Rightarrow \ggT(5,13) = 1 = 2\cdot 13 - 5 \cdot 5$ %TODO somewhere here is an error!
$R = \whole$, $x = \textcolor{red}{5}$, $y= \textcolor{blue}{13}$
\begin{center}\begin{tikzpicture}
\node at (0,0) (gleich) {$=$};
\node at (0,-1) (gleich) {$=$};
\node at (0,-2) (gleich) {$=$};
\node at (0,-3) (gleich) {$=$};
\node at (0,-4) (gleich) {$=$};
\node at (1,0) (mal) {$\cdot$};
\node at (1,-1) (mal) {$\cdot$};
\node at (1,-2) (mal) {$\cdot$};
\node at (1,-3) (mal) {$\cdot$};
\node at (1,-4) (mal) {$\cdot$};
\node at (2,0) (plus) {$+$};
\node at (2,-1) (plus) {$+$};
\node at (2,-2) (plus) {$+$};
\node at (2,-3) (plus) {$+$};
\node at (2,-4) (plus) {$+$};
\node[red] at (-0.5,0) (erg1) {5};
\node at (-0.5,-1) (erg2) {13};
\node at (-0.5,-2) (erg3) {5};
\node at (-0.5,-3) (erg4) {3};
\node at (-0.5,-4) (erg5) {2};
\node at (0.5,0) (fak1) {0};
\node at (0.5,-1) (fak2) {2};
\node at (0.5,-2) (fak3) {1};
\node at (0.5,-3) (fak4) {1};
\node at (0.5,-4) (fak5) {2};
\node[blue] at (1.5,0) (quot1) {13};
\node at (1.5,-1) (quot2) {5};
\node at (1.5,-2) (quot3) {3};
\node at (1.5,-3) (quot4) {2};
\node at (1.5,-4) (quot5) {\textbf{1}};
\node at (2.5,0) (plus1) {5};
\node at (2.5,-1) (plus2) {3};
\node at (2.5,-2) (plus3) {2};
\node at (2.5,-3) (plus4) {1};
\node at (2.5,-4) (plus5) {0};
\draw[->] (quot1) -- (erg2);
\draw[->] (quot2) -- (erg3);
\draw[->] (quot3) -- (erg4);
\draw[->] (quot4) -- (erg5);
\draw[->] (plus1) -- (quot2);
\draw[->] (plus2) -- (quot3);
\draw[->] (plus3) -- (quot4);
\draw[->] (plus4) -- (quot5);
\end{tikzpicture}\end{center}
$\Rightarrow \ggT(5,13) = 1 = 2\cdot 13 - 5 \cdot 5$
\end{example}
\begin{example}
Bestimme $x \in \whole$ mit $x \in 1 \mod 5$ und $x \equiv 2 \mod 13$.\\
Bestimme $x \in \whole$ mit $x \equiv 1 \mod 5$ und $x \equiv 2 \mod 13$.\\
$b_1 = 2 \cdot 13 = 26$
\begin{align}
b_1 &= 1 \mod 5 \notag \\
b_1 &= 0 \mod 13\notag
b_1 &\equiv 1 \mod 5 \notag \\
b_1 &\equiv 0 \mod 13\notag
\end{align}
$b_2 = -5\cdot 5 = -25$
\begin{align}
b_2 &= 0 \mod 5\notag \\
b_2 &= 1 \mod 13\notag
b_2 &\equiv 0 \mod 5\notag \\
b_2 &\equiv 1 \mod 13\notag
\end{align}
Setze $x = 1\cdot b_1 + 2 \cdot b_2 = -24$. Die anderen sind $x = 25\cdot 13 = 41 + 65\whole$. %TODO somewhere here is an error!
Setze $x = 1\cdot b_1 + 2 \cdot b_2 = -24$. Die anderen sind $x+5\cdot 13\whole=41+65\whole$.
\end{example}

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3. Semester/GEO/Vorlesung GEO.pdf
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Material/orthogonaler_UVR.pdf
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54
Material/orthogonaler_UVR.tex
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@ -38,4 +38,58 @@
\node at (2.2,3) (annulator) {$L^0$};
\node at (5.7,-1) (punkt2) {$a=(a_1,a_2)$};
\end{tikzpicture}\end{center}
\begin{center}\begin{tikzpicture}
\node at (0,0) (gleich) {$=$};
\node at (0,-1) (gleich) {$=$};
\node at (0,-2) (gleich) {$=$};
\node at (0,-3) (gleich) {$=$};
\node at (0,-4) (gleich) {$=$};
\node at (1,0) (mal) {$\cdot$};
\node at (1,-1) (mal) {$\cdot$};
\node at (1,-2) (mal) {$\cdot$};
\node at (1,-3) (mal) {$\cdot$};
\node at (1,-4) (mal) {$\cdot$};
\node at (2,0) (plus) {$+$};
\node at (2,-1) (plus) {$+$};
\node at (2,-2) (plus) {$+$};
\node at (2,-3) (plus) {$+$};
\node at (2,-4) (plus) {$+$};
\node[red] at (-0.5,0) (erg1) {5};
\node at (-0.5,-1) (erg2) {13};
\node at (-0.5,-2) (erg3) {5};
\node at (-0.5,-3) (erg4) {3};
\node at (-0.5,-4) (erg5) {2};
\node at (0.5,0) (fak1) {0};
\node at (0.5,-1) (fak2) {2};
\node at (0.5,-2) (fak3) {1};
\node at (0.5,-3) (fak4) {1};
\node at (0.5,-4) (fak5) {2};
\node[blue] at (1.5,0) (quot1) {13};
\node at (1.5,-1) (quot2) {5};
\node at (1.5,-2) (quot3) {3};
\node at (1.5,-3) (quot4) {2};
\node at (1.5,-4) (quot5) {\textbf{1}};
\node at (2.5,0) (plus1) {5};
\node at (2.5,-1) (plus2) {3};
\node at (2.5,-2) (plus3) {2};
\node at (2.5,-3) (plus4) {1};
\node at (2.5,-4) (plus5) {0};
\draw[->] (quot1) -- (erg2);
\draw[->] (quot2) -- (erg3);
\draw[->] (quot3) -- (erg4);
\draw[->] (quot4) -- (erg5);
\draw[->] (plus1) -- (quot2);
\draw[->] (plus2) -- (quot3);
\draw[->] (plus3) -- (quot4);
\draw[->] (plus4) -- (quot5);
\end{tikzpicture}\end{center}
\end{document}

2
texmf/tex/latex/mathoperators/mathoperators.sty
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@ -141,6 +141,6 @@
\DeclareMathOperator{\Syl}{Syl}
\DeclareMathOperator{\Typ}{Typ}
\DeclareMathOperator{\LC}{LC}
\DeclareMathOperator{\Quot}{\Quot}
\DeclareMathOperator{\Quot}{Quot}
\endinput