How can one see that $\operatorname{tr}(f\otimes g)=\operatorname{tr}f\operatorname{ tr }g$?

Question

Suppose you have two free modules $M$ and $N$ of finite rank over a commutative ring $R$. Let's also take some $f\in\operatorname{End}_R(M)$ and $g\in\operatorname{End}_R(N)$, which gives a corresponding $f\otimes g\in\operatorname{End}_R(M\otimes N)$.

Apparently, the trace of these endomorphisms "distributes" over the tensor product, in that $$ \operatorname{tr}(f\otimes g)=\operatorname{tr}f\operatorname{tr}g. $$

This is not clear to me. Is there an explanation why this is true?

Answer 1

What is your definition of trace? Best one is the evaluation map: $$ \phi:M\rightarrow M \text{ is an element of } M\otimes M^\vee $$ because for $M$ free $$ M\otimes M^\vee \rightarrow \operatorname{End(M)} \quad m\otimes \phi \mapsto \phi(-)m $$ is an isomorphism. Trace is the evaluation $$ \operatorname{tr}: M\otimes M^\vee \rightarrow R $$

Answer 2

Hint:

Let $\{a_i\}$ be a basis for $M$ and $\{b_j\}$ a basis for $N$. Prove then that $[f\otimes g]_{\{a_i\otimes b_j\}}=[f]_{\{a_i\}}\otimes [g]_{\{b_j\}}$ where by $[\cdot]_\mathscr{B}$ I mean matrix representation with respect to $\mathscr{B}$ and the tensor product on the right hand side is the Kronecker product.