If $A$ is a finite-dimensional algebra over a field $k$, the usual norm $N_{A/k}: A \to k$ maps $a$ to the determinant of the $k$-linear endomorphism of $A$ given by $x \mapsto ax$. For $A \in \textbf{CS}_k$, what is the relationship between $N_{A/k}$ and the reduced norm $\text{nr}_{A/k}$?
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$\begingroup$ $\textbf{CS}_k$ standing for the finite-dimensional central simple algebras over $k$, or...? $\endgroup$– GNiklaschCommented Aug 18, 2015 at 7:31
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We have $A \otimes k^s \cong \text{M}_n(k^s)$ for some $n \ge 1$, namely $n = \sqrt{[A:k]}.$ We claim that $N_{A/k}(a) = \text{nr}_{A/k}(a)^n$ for all $a \in A$. It suffices to prove this after tensoring with $k^s$, thus we may reduce to the case $A = \text{M}_n(k)$ (and $k$ separably closed). Then the formula follows from the fact that $\text{M}_n(k)$ as a left $\text{M}_n(k)$-module is a direct sum of $n$ spaces (the columns), each isomorphic to $k^n$ with its usual left $\text{M}_n(k)$-module structure.
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$\begingroup$ Well, not every finite-dimensional algebra is central simple. $\endgroup$ Commented Aug 18, 2015 at 7:14