Homomorphisms and tensor product concerning $\mathrm{Hom}(L, M \otimes N)$

Question

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

Answer 1

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).

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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 $$\mathrm{Hom}_A(G(L), M) = \mathrm{Hom}_A(L, F(M)).$$

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