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Suppose S a commutative, projective, $R$-algebra of finite type (with $R$ a ring). We want to show that, if $M$ is any $S$-module, then there exists a canonical isomorphism: $$ Hom_{S\otimes_R S}(S \otimes_R M,M\otimes_R S)\to Hom_{S\otimes_R S}(End_R(S), End_R(M))$$ given by $f\mapsto\tilde{f}$ where, if $f(1\otimes m)=\sum_i t_i\otimes m_i$ then $\tilde{f}(s\otimes\varphi)(m)=\sum_i sm_i\varphi(t_i)$. Note that we are using the well known fact that $End_R(S)\cong S\otimes_R S^\ast$.

I'm still stuck trying to prove the injectivity, but I think that both the injectivity and the surjectivity will follow from the fact that, being $S$ projective and of finite type, every element $s\in S$ can be written as $s=\sum_{i=1}^n \psi_i(s)s_i $ (note that the sum is finite!) where the $s_i$'s and the $\varphi_i$'s do not depend on $s$. Could you help me with that? Thank you.

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There is nothing homological in the question :) –  Mariano Suárez-Alvarez Nov 20 '11 at 18:21
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First, you can simply a bit the sets you are working with : if $A$ is a commutative ring, $B$ a ring, $M$ is a $B$-module and $P$ is an $A\otimes B$-module, there is an isomorphism of $A\otimes B$-modules between $Hom_{A\otimes B}(A \otimes M, P)$ and $Hom_B(M,P)$.

So in your case you are looking at the canonical isomorphism $Hom_S(M,M\otimes_R S)\to Hom_S(S^*, End_R(M))$

given by $f \mapsto \tilde{f}$ where if $f(m) = \sum_i m_i \otimes t_i$ then $\tilde{f}(\varphi)(m) = \sum_i \varphi(t_i) m_i$.

If $S$ is a projective $R$-module, there is an injective map $\pi : S \to R^n$ Denote by $\pi_k : S \to R$ the $k$-th coordinate. For every $m$, we have $(id\otimes \pi) \circ f (m) = \sum \tilde{f}(\pi_k)(m) \otimes e_k$

Suppose $\tilde{f} = 0$. Then $(id\otimes \pi) \circ f (m) = 0 \in M \otimes R^n$, so $f(m) = 0 \in M \otimes S$. This is true for every $m \in M$, so $f = 0$.

Now I don't know how to prove that this canonical morphism is surjective in general, though you can find an inverse when $S$ is a free $R$-module : If $(s_1 \ldots s_n)$ is a basis for $S$, and $(s_i^* \ldots s_n^*)$ is its dual basis, define $\hat{f}(m) = \sum f(s_i^*)(m) \otimes s_i$.

Then it is straightforward to check that $\bar{\hat{f}} = f$ and $\hat{\bar{f}} = f$

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