# Kommutative Algebra/Modultheorie/Multiplikativität der Determinante/Fakt/Beweis

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Beweis
{\displaystyle {}{\begin{aligned}det(A)det(B)&=\sum _{\sigma \in S_{n}}sgn(\sigma )\prod _{i=1}^{n}a_{i,\sigma (i)}+\sum _{\tau \in S_{n}}sgn(\tau )\prod _{j=1}^{n}b_{j,\tau (j)}\\&=\sum _{\sigma \in S_{n}}\sum _{\tau \in S_{n}}sgn(\sigma )sgn(\tau )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{i,\tau (i)},\end{aligned}}}

da für ${\displaystyle {}\tau ,\sigma \in S_{n}\colon \tau \mapsto \tau \circ \sigma :=\lambda }$ bijektiv ist ergibt sich:

{\displaystyle {}{\begin{aligned}\sum _{\sigma \in S_{n}}\sum _{\tau \in S_{n}}sgn(\sigma )sgn(\tau )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{i,\tau (i)}&=\sum _{\sigma \in S_{n}}\sum _{\tau \in S_{n}}sgn(\sigma )sgn(\tau )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{\sigma (i),\tau \circ \sigma (i)}\\&=\sum _{\sigma \in S_{n}}\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{\sigma (i),\lambda (i)}.\end{aligned}}}

Andererseits ist

{\displaystyle {}{\begin{aligned}det(A\cdot B)&=\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}(\sum _{j=1}^{n}a_{ij}b_{j\lambda _{j}})\\&=\sum _{\lambda \in S_{n}}sgn(\lambda )\sum _{j_{i}}(\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\lambda (i)}\\&=\sum _{j_{i}}\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\lambda (i)},\end{aligned}}}

wobei ${\displaystyle {}j,i\in \{1,...,n\}}$ und ${\displaystyle {}j_{i}\in \{1,...,n\}^{n}}$ alle möglichen Kombinationen durchgeht. Sind in ${\displaystyle {}j_{i}}$ zwei Komponenten gleich, so ist:

{\displaystyle {}{\begin{aligned}\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\lambda (i)}&=\prod _{i=1}^{n}a_{i,j_{i}}\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}b_{j_{i},\lambda (i)}\\&=\prod _{i=1}^{n}a_{i,j_{i}}det(b_{j_{i},s})_{s\in \{1,...,n\}}\\&=0,\end{aligned}}}

denn in ${\displaystyle {}b_{j_{i},s}}$ sind zwei Zeilen gleich.

Und somit letztendlich:

${\displaystyle {}det(A\cdot B)=\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\lambda (i)}=\sum _{\sigma \in S_{n}}\sum _{\lambda \in S_{n}}sgn(\lambda )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{\sigma (i),\lambda (i)}=det(A)\cdot det(B)\,.}$
Zur bewiesenen Aussage