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

Beweis

Es ist

${}{\begin{aligned}\det A\det B&={\left(\sum _{\sigma \in S_{n}}\operatorname {sgn} (\sigma )\prod _{i=1}^{n}a_{i,\sigma (i)}\right)}\cdot {\left(\sum _{\tau \in S_{n}}\operatorname {sgn} (\tau )\prod _{j=1}^{n}b_{j,\tau (j)}\right)}\\&=\sum _{\sigma \in S_{n}}\sum _{\tau \in S_{n}}\operatorname {sgn} (\sigma )\operatorname {sgn} (\tau )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{i,\tau (i)},\end{aligned}}$

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

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

Andererseits ist

${}{\begin{aligned}\det \left(A\cdot B\right)&=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{i=1}^{n}{\left(\sum _{j=1}^{n}a_{ij}b_{j\pi (i)}\right)}\\&=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\sum _{j_{1},\ldots ,j_{n}}\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\pi (i)}\\&=\sum _{j_{1},\ldots ,j_{n}}\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\pi (i)},\end{aligned}}$

wobei ${}j,i\in {\{1,\ldots ,n\}}$ und ${}{\left(j_{1},\ldots ,j_{n}\right)}\in {\{1,\ldots ,n\}}^{n}$ alle möglichen Kombinationen durchläuft. Sind im Tupel ${}{\left(j_{1},\ldots ,j_{n}\right)}$ zwei Komponenten gleich, so ist:

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

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

Und somit letztendlich:

${}\det \left(A\circ B\right)=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{i=1}^{n}a_{i,j_{i}}b_{j_{i},\pi (i)}=\sum _{\sigma \in S_{n}}\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{i=1}^{n}a_{i,\sigma (i)}b_{\sigma (i),\pi (i)}=\det A\cdot \det B\,.$

Zur bewiesenen Aussage