diff --git a/3. Semester/GEO/TeX_files/Kommutative_Ringe/Ringe_von_Bruechen.tex b/3. Semester/GEO/TeX_files/Kommutative_Ringe/Ringe_von_Bruechen.tex index b6e17e8..b3e95bc 100644 --- a/3. Semester/GEO/TeX_files/Kommutative_Ringe/Ringe_von_Bruechen.tex +++ b/3. Semester/GEO/TeX_files/Kommutative_Ringe/Ringe_von_Bruechen.tex @@ -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 diff --git a/3. Semester/GEO/TeX_files/Kommutative_Ringe/Teilbarkeit.tex b/3. Semester/GEO/TeX_files/Kommutative_Ringe/Teilbarkeit.tex index be4958c..0b91f3b 100644 --- a/3. Semester/GEO/TeX_files/Kommutative_Ringe/Teilbarkeit.tex +++ b/3. Semester/GEO/TeX_files/Kommutative_Ringe/Teilbarkeit.tex @@ -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} \ No newline at end of file diff --git a/3. Semester/GEO/Vorlesung GEO.pdf b/3. Semester/GEO/Vorlesung GEO.pdf index fb121b3..1cffe98 100644 Binary files a/3. Semester/GEO/Vorlesung GEO.pdf and b/3. Semester/GEO/Vorlesung GEO.pdf differ diff --git a/Material/orthogonaler_UVR.pdf b/Material/orthogonaler_UVR.pdf index b56408d..818587f 100644 Binary files a/Material/orthogonaler_UVR.pdf and b/Material/orthogonaler_UVR.pdf differ diff --git a/Material/orthogonaler_UVR.tex b/Material/orthogonaler_UVR.tex index 19e4baf..fb4a45c 100644 --- a/Material/orthogonaler_UVR.tex +++ b/Material/orthogonaler_UVR.tex @@ -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} \ No newline at end of file diff --git a/texmf/tex/latex/mathoperators/mathoperators.sty b/texmf/tex/latex/mathoperators/mathoperators.sty index 2d29d1f..be8294d 100644 --- a/texmf/tex/latex/mathoperators/mathoperators.sty +++ b/texmf/tex/latex/mathoperators/mathoperators.sty @@ -141,6 +141,6 @@ \DeclareMathOperator{\Syl}{Syl} \DeclareMathOperator{\Typ}{Typ} \DeclareMathOperator{\LC}{LC} -\DeclareMathOperator{\Quot}{\Quot} +\DeclareMathOperator{\Quot}{Quot} \endinput