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Could someone provide me a link to the proof of the adjointness of Hom and Tensor. I did an extensive google search but could not find anything self contained that presented the proof in full generality (or at least the generality I know). Let $R\to S$ be a ring homomorphism, let $M,N$ be $S$-modules and $Q$ an $R$-module. Then, we have $$\textrm{Hom}_R(M\otimes_S N,Q) \cong \textrm{Hom}_S(M,\textrm{Hom}_R(N,Q)$$

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What definition of the tensor product are you working with? –  Qiaochu Yuan Apr 9 '11 at 17:35
    
Rotman, An introduction to Homological Algebra, 2nd edition, proves it, and calls it the "adjoint isomorphism theorem". –  lentic catachresis Apr 9 '11 at 17:38

1 Answer 1

Let $f \in Hom_R(M\otimes_S N,Q)$. We define $g \in Hom_S(M,Hom_R(N,Q))$ by:

$$g(m)(n)=f(m \otimes n)$$

Similarly, if $g$ is defined, we can easily define $f$.

I'll leave it to you to prove that this map between $f$ and $g$ actually goes to the appropriate sets, but this is the basic argument. As mentioned by one of the comments, it does depend on how you define the tensor product.

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