$L\otimes_{\Delta}\text{Hom}_{\Delta}(M,\Delta)\cong \text{Hom}_{\Delta}(M,L)$

Question

This is exercise 5 in maximal orders by I.Reiner. This is not homework though.

Let $\Delta$ be a ring $L_{\Delta}$ be any module, and let $M_{\Delta}$ be a finitely generated and projective. Prove that
$$L\otimes_{\Delta}\text{Hom}_{\Delta}(M,\Delta)\cong \text{Hom}_{\Delta}(M,L)$$

So far I have proved this to be true for finitely generated free modules. I am having a hard time extending to arbitrary free modules because Hom doesn't commute with arbitrary direct sum in the first slot. Then given any projective module $M$ there exists a $F$ free and $M'$ such that $F=M\oplus M'$. Then we have
$$L\otimes_{\Delta}\text{Hom}_{\Delta}(M,\Delta)\oplus L\otimes_{\Delta}\text{Hom}_{\Delta}(M,\Delta)\cong \text{Hom}_{\Delta}(M,L)\oplus \text{Hom}_{\Delta}(M',L).$$ From here I am stuck. Any ideas or hints would be much appreciated. Thanks.

Answer 1

It (usually) doesn't hold when $M$ is free of infinite rank. But when $M$ is f.g. projective, then $M$ is a direct summand of a f.g. free module, and you are done:

Show that the class of $M$s which satisfy the claim is closed under finite direct sums, direct summands, and contains $\Delta$. It follows that the class contains all f.g. projective modules. You may replace the map by any natural transformation between additive functors between abelian categories.

Edit: See the comments for more details.