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I have a question regarding $\mathrm{Hom}(L, M \otimes N)$, where $L,M,N$ are $A$-modules over a commutative ring $A$. I know, that $$\mathrm{Hom}_A(M \otimes_A N, L) \cong \mathrm{Hom}_A(M, \mathrm{Hom}_A(N,L))$$

Is there an analogue for $\mathrm{Hom}(L, M \otimes N)$? Something like $$\mathrm{Hom}(L, M \otimes N) \cong \mathrm{Hom}(\mathrm{Hom}(L,M), N)$$ I know the proof cannot be analogue, as the first isomorphism is given by the isomorphism between $\mathrm{Bil}(M \times N,L) \cong \mathrm{Hom}(M, \mathrm{Hom}(N,L))$ and you can obviously not speak of $\mathrm{Bil}(L, M \times N)$.

Thanks in advance

Watson
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Jonas W.
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1 Answers1

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The precise question could be: does the functor $F : M \mapsto M \otimes_A N$ have a left adjoint (i.e. is it a right adjoint of some other functor)? It doesn't, because a right adjoint is left exact, and $F$ is not left exact (unless $N$ is a flat $A$-module).

Further details: a functor $F : \mathsf{Mod}_A \to \mathsf{Mod}_A$ is a right adjoint if there is a functor $G : \mathsf{Mod}_A \to \mathsf{Mod}_A$ such that there are natural bijections $$\mathrm{Hom}_A(G(L), M) = \mathrm{Hom}_A(L, F(M)).$$

However, you might be interested by the following related question: (1).

Watson
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    nice and elegant answer, thanks! – Jonas W. Jun 25 '18 at 15:46
  • The functor $F$ might have a left adjoint in some cases. We saw that flatness of $N$ is necessary. If $N$ is a free module of finite rank, then $F$ has a left adjoint, namely $F$ itself (see here). – Watson Jun 26 '18 at 12:25
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    This question tells you when $F$ has a left adjoint: exactly when $N$ is finitely generated and projective (if $A$ is a domain, this is the same as "finitely generated and flat").

    In that case, the left adjoint of $F$ is $N^* \otimes_A -$, i.e. we have natural bijections $$ \mathrm{Hom}_A(N^* \otimes_A X, Y) \cong \mathrm{Hom}_A(X, Y \otimes_A N).$$

     – Watson Jun 26 '18 at 12:38
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    In the above bijections are deduced from $$\mathrm{Hom}_A(N^* \otimes_A X, Y) \cong \mathrm{Hom}_A(X, \mathrm{Hom}_A(N^*, Y)) \cong \mathrm{Hom}_A(X, N \otimes_A Y),$$ using this question and the fact that $N$ is finitely generated projective. – Watson Jun 26 '18 at 12:58