When is $M \otimes_A -$ representable?

Let $A$ be a commutative ring, $M$ be a $A$-module. When is $M \otimes_A - : A\text{-mod} \rightarrow A\text{-mod}$ representable? In other words, when will there exist a $A$-module $P$ s.t $M \otimes_A -=Hom(P,-)$ ?

A neccessary condition is that $M$ is flat, a sufficient condition is that $M$ is free of finite rank. I wonder whether flatness or projectivity is sufficient.

Claim: Let $M$ be a right $A$-module. The functor $M \otimes_A (-)$ from left $A$-modules to abelian groups is representable if and only if $M$ is a finitely generated projective $A$-module, in which case it can be written $\text{Hom}(M^{\ast}, -)$ where $M^{\ast} = \text{Hom}_A(M, A)$ is the dual (a left $A$-module).

Proof. $\Leftarrow$: the conclusion clearly holds if $M$ is finite free. If $M$ is finitely generated projective, then writing it as a retract of a finite free module, the conclusion again holds, because retracts are absolute and commute with every functor (see this blog post, search for "the facts of life").

$\Rightarrow$: if $M \otimes_A (-) \cong \text{Hom}_A(N, -)$ then substituting $(-) = A$ gives $M \cong \text{Hom}_A(N, A)$, so $M$ is the $A$-linear dual of a module $N$ such that $\text{Hom}_A(N, -)$ commutes with colimits. This is true if and only if $N$ is finitely generated projective (see this blog post), hence a retract of a finite free module, and again using that retracts are absolute we find that $M$ is also finitely generated projective with dual $N$ (because this fact holds for finite free modules and is preserved by retracts). $\Box$

Here is some discussion of the relationship between the hypothesis that $M \otimes_A (-)$ is representable and the hypothesis that it preserves limits.

Lemma: $M \otimes_A (-)$ is representable iff it preserves limits.

Proof. One direction is clear. In the other direction, since $M \otimes_A (-)$ preserves colimits it is accessible, so by the presentable adjoint functor theorem, if $M \otimes_A (-)$ commutes with limits then it has a left adjoint

$$L : \text{Ab} \to \text{Mod}(A).$$

This functor commutes with colimits, so by the Eilenberg-Watts theorem it must be given by $N \otimes (-)$ for some left $A$-module $N$. Taking right adjoints again, using the tensor-hom adjunction, we conclude that we have a natural isomorphism

$$M \otimes_A (-) \cong \text{Hom}_A(N, -)$$

as desired. $\Box$

Next, here is a proof of $\Leftarrow$ using only the hypothesis that $M \otimes_A (-)$ commutes with limits.

Claim: $M \otimes_A (-)$ commutes with limits if and only if $M$ is finitely generated projective.

Proof. One direction we've already shown above. In the other direction, $M$ must in particular be flat. Assuming that $M$ is flat, $M \otimes_A (-)$ commutes with limits iff it commutes with infinite products. So consider the natural map

$$M \otimes_A \prod_{i \in I} A \to \prod M \otimes_A A \cong \prod_{i \in I} M$$

for any index set $I$. Setting $I = M$, the RHS has a natural element $\prod_{m \in M} m$ which lists every element of $M$, and by hypothesis the map above is an isomorphism, so there must be an element

$$\sum_{j=1}^n m_j \otimes (\prod_i a_{ij})$$

mapping to it. This element expresses every element $m \in M$ as a linear combination of a finite collection of elements $m_j$, from which it follows that $M$ is finitely generated.

So far we've used the fact that this map is surjective; now let's also use the fact that it's injective. An element in its kernel is an element of the form $\sum_{j=1}^n m_j \otimes (\prod_i b_{ij})$ (the $m_j$ are the generators we just identified) such that

$$\forall i \in I, \sum_{j=1}^n m_j b_{ij} = 0$$

so, in other words, it is a collection of relations the generators satisfy, and injectivity means any such element must in fact be zero in the tensor product $M \otimes_A \prod A$.

Now set $I$ to index every relation that the generators satisfy, and $\sum m_j \otimes (\prod b_{ij})$ to be a list of all such relations. The condition that this element is zero in the tensor product means that it must be possible to transform it into zero using a finite sequence of applications of bilinearity, together with relations in $M$ and $\prod A$. In the process of doing this a finite number of relations in $M$ get used to transform everything to zero; these relations must then generate every relation satisfied by the generators, from which it follows that $M$ is finitely presented. Now it's enough to prove the following.

Lemma: A finitely presented flat module is projective.

Corollary: A module is finitely presented flat iff it's finitely generated projective.

Proof. There is a proof of this in the Stacks project which I haven't looked at in detail.

A more categorical proof is as follows: by Lazard's theorem, a flat module $M$ is a filtered colimit of free modules. If $M$ is finitely presented as a module then it is a compact object, meaning that $\text{Hom}(M, -)$ commutes with filtered colimits, so writing $M \cong \text{colim}_i F_i$ as a filtered colimit of free modules, we conclude that the identity $M \to M$ factors through one of the $F_i$, hence that $M$ is a retract of a free module, hence projective. $\Box$