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

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?

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I changed \text{tr} to \operatorname{tr}. These don't always give the same results. In particular (although it doesn't work properly on Wikipedia) operatorname results in proper spacing, so you don't need to write \operatorname{tr } with a blank space after the "r". (In coming months Wikipedia will switch to mathJax, which is used on this site, and maybe it will work right then.) –  Michael Hardy Mar 25 '12 at 21:03

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 $$\operatorname{tr}: M\otimes M^\vee \rightarrow R$$
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.