Let $k$ be a commutative ring, and let $M$ and $N$ be $k$-modules. Let $\mathrm{End}(M) = \mathrm{Hom}_k (M,M)$ be the endomorphism algebra.
Is it true that $\mathrm{End}(M) \otimes \mathrm{End}(N) \cong \mathrm{End}(M \otimes N)$?
I saw this referenced on a blog but I can't find a proof of it. I tried to show that $\mathrm{End}(M \otimes N)$ satisfies the universal property of the tensor product of $\mathrm{End}(M)$ and $\mathrm{End}(N)$ but to no avail.
Thanks for any help.
Best Answer
Seems you need for this at least one of the modules $M,N$ to be a finitely projective $k$-module. See Bourbaki, Algebra, chapter II, $\S$ 4.4, proposition 4.
EDIT. But if you just want a map
$$ \mathrm{End} (M) \otimes \mathrm{End}(N) \longrightarrow \mathrm{End}(M\otimes N) $$
you've got it! :-)
Just send $\phi \otimes \psi \in \mathrm{End} (M) \otimes \mathrm{End}(N)$ to $\phi \otimes \psi \in \mathrm{End}(M\otimes N)$. :-D
I mean: the "second" $\phi\otimes \psi$ is
$$ (\phi\otimes\psi)(m\otimes n) = \phi(m)\otimes \psi(n) \ . $$