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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.

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Well, it's true when $k$ is a field and $M$ and $N$ are finite-dimensional: this is basically the Kronecker product construction. I don't immediately see how this generalises. –  Zhen Lin Sep 17 '12 at 13:06

1 Answer 1

up vote 2 down vote accepted

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

$$ (\phi\otimes\psi)(m\otimes n) = \phi(m)\otimes \psi(n) \ . $$

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Thank you. The source I was reading must have been incorrect then. Luckily I only needed a map $\mathrm{End}(M) \otimes \mathrm{End}(N) \to \mathrm{End}(M \otimes N)$. –  Paul Slevin Sep 17 '12 at 13:15

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