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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.)2012-03-25

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