add a useful identity

Author: darijgr

notes.tex

@@ -32671,6 +32671,43 @@ \subsection{$\mathbb{K}$-algebras}
 $b$ of this $\mathbb{K}$-algebra).
 \end{definition}
 
+The following property of $\mathbb{K}$-algebras is easy to check but quite useful:
+
+\begin{proposition}
+\label{prop.K-alg.prods-scal}Let $A$ be a $\mathbb{K}$-algebra. Let
+$k\in\mathbb{N}$.
+
+\textbf{(a)} Any $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in\mathbb{K}$ and
+$a_{1},a_{2},\ldots,a_{k}\in A$ satisfy%
+\[
+\left(  \lambda_{1}a_{1}\right)  \left(  \lambda_{2}a_{2}\right)
+\cdots\left(  \lambda_{k}a_{k}\right)  =\left(  \lambda_{1}\lambda_{2}%
+\cdots\lambda_{k}\right)  \left(  a_{1}a_{2}\cdots a_{k}\right)  .
+\]
+
+
+\textbf{(b)} Any $\lambda\in\mathbb{K}$ and $a\in A$ satisfy $\left(  \lambda
+a\right)  ^{k}=\lambda^{k}a^{k}$.
+\end{proposition}
+
+\begin{proof}
+[Proof of Proposition \ref{prop.K-alg.prods-scal}.] \textbf{(a)} This can
+easily be proven by induction on $k$. (The induction base boils down to the
+obvious fact that $1_{A}=1_{\mathbb{K}}\cdot1_{A}$. The induction step uses
+the identity
+\[
+\left(  \lambda a\right)  \left(  \mu b\right)  =\left(  \lambda\mu\right)
+\left(  ab\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\lambda,\mu\in
+\mathbb{K}\text{ and }a,b\in A;
+\]
+this identity can be easily proven using \textquotedblleft Scale-invariance of
+multiplication\textquotedblright\ and axiom \textbf{(h)} from Definition
+\ref{def.module.module}.)
+
+\textbf{(b)} This is a particular case of Proposition
+\ref{prop.K-alg.prods-scal} \textbf{(a)}.
+\end{proof}
+
 \subsection{\label{sect.la.modiso}Module isomorphisms}
 
 In analogy to Definition \ref{def.riiso.riiso}, we define: