tensor of Hom-sets.

Let $k$ be a field. Let $A$ be a commutative $k$-algebra. Let $M,N$ be $A$-modules, and assume that $M$ is finite over $A$.

Is it true that the map $$\mathrm{Hom}_{A}(M,N) \otimes_{k} \mathrm{Hom}_A(M,N) \to \mathrm{Hom}_{A\otimes_k A}(M\otimes_k M,N\otimes_k N)$$ is an isomorphism of $A\otimes_k A$-modules?

Thanks!

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What map? There are certainly ones that are not isomorphisms, so you probably have a particular one in mind. – Tobias Kildetoft Jan 27 '12 at 21:10
Define $\psi(\phi_1\otimes \phi_2) (a\otimes b) = \phi_1(a)\otimes \phi_2(b)$. – just me Jan 27 '12 at 21:14

More generally, let us look at the natural map

$$\Phi:\mathrm{Hom}_{A}(M,N) \otimes_{k} \mathrm{Hom}_A(M',N') \to \mathrm{Hom}_{A\otimes_k A}(M\otimes_k M',N\otimes_k N')$$

Consider a free presentation $$(\star)\qquad\qquad F_1\to F_0\to M\to 0,$$ so that this is an exact sequence with the $F_i$ free. Applying $\hom_A(-,N)$ we get an exact sequence $$0\to\hom_A(M,N)\to\hom_A(F_0,N)\to\hom_A(F_1,N)$$ of vector spaces. Tensoring it now with the vector space $\hom_A(M',N')$ we get a short exact sequence $$0\to\hom_A(M,N)\otimes\hom_A(M',N')\to\hom_A(F_0,N)\otimes\hom_A(M',N')\to\hom_A(F_1,N)\otimes\hom_A(M',N')$$

Now, start over: tensor the sequence $(\star)$ over $k$ with $M'$, to get an exact sequence of $A\otimes A$-modules $$F_1\otimes M'\to F_0\otimes M'\to M\otimes M'\to 0,$$ to which we can apply the functor $\hom_{A\otimes A}(-,N\otimes N')$ to get an exact sequence of the form $$0\to\hom_{A\otimes A}(M\otimes M',N\otimes N')\to\hom_{A\otimes A}(F_0\otimes M',N\otimes N')\to\hom_{A\otimes A}(F_1\otimes M',N\otimes N')$$

Next: the two exact sequences we got fit in a commutative diagram

0 --> hom_A(M,N) (x) hom_A(M',N') --> hom_A(F_0,N) (x) hom_A(M',N') --> hom_A(F_1,N) (x) hom_A(M',N')
|                                     |                                     |
|                                     |                                     |
V                                     V                                     V
0 --> hom_A(M (x) M',N (x) N')    --> hom_A(F_0 (x) M',N (x) N')    --> hom_A(F_1 (x) M',N (x) N')


in which the vertical arrows are various instances of $\Phi$. Commutativity and exactness of the rows shows that to show that the leftmost vertical map is an iso it is enough to show that the two other vertical maps are iso.

This means that it is enough to show that the map $\Phi$ is an iso when $M$ is free.

A similiar argument, now with $M'$, shows that it is also enough to show that $\Phi$ is an iso when both $M$ and $M'$ are free.

Can you do that?

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WHen you do that, you'll find that $M$ and $M'$ better be finitely presented... (You can assume that $A$ is noetherian, for example) – Mariano Suárez-Alvarez Jan 27 '12 at 21:28
In general, the map is not surjective. You can find examples with $A=k$. – Mariano Suárez-Alvarez Jan 27 '12 at 23:15
For example, let $A=k$ and $M=N=k^{(\mathbb N)}$ be a vector space of countable dimension with $\{e_i:i\geq1\}$ a basis. I'm pretty sure that the map $f:M\otimes M\to N\otimes N$ such that $f(e_i\otimes e_j)=\delta_{i,j}e_i\otimes e_j$ is not in the image of your map. – Mariano Suárez-Alvarez Jan 27 '12 at 23:19