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 $\text{nr}\_{A/k}$?



Relationship between $N_{A/K}$ and the reduced norm $\text{nr}_{A/k}$?
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.


