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This question is a follow up to Sasha's comment in Duals and Tensor products.

In the comment there, it is claimed that the given a ring $A$ and modules $M$, $N$, there is an isomorphism $$ RHom(M,A)\otimes^L RHom(N,A) = RHom(M\otimes^L N, A). $$

Question 1. Shouldn't we make assumption on $A$ to make sure the RHom's are bounded above in order for the (derived) tensor product to exist?

Then it is claimed that if $M$ is finite projective, then $RHom(M,A) = Hom(M,A)$ (obvious) and that the zero-th homology group of $Hom(M,A)\otimes^L RHom(N,A)$ is $Hom(M,A)\otimes Hom(N,A)$...

Question 2: Why is this last statement true? It would be true if $Hom(M,A)$ is flat, but why is it true in general?


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up vote 4 down vote accepted

In order to get the first isomorphism you probably need $M$ and $N$ to be perfect, i.e. they must have a projective resolution of finite length by projective modules of finite type. In that case $RHom(-,A)$ behaves as the usual duality functor on finite-dimensional vector spaces.

As for your second question, $Hom(M,A)$ is projective (you can readily see that it is free if $M$ is free of finite type and then argue with retractions), therefore the cohomology of $Hom(M,A)\otimes^L RHom(N,A)$ is $Hom(M,A)\otimes Ext^n(N,A)$.

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Thanks for your answer, Fernando, that does it. – unknown Jun 2 '11 at 17:10

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