Special case of the Eilenberg-Watts theorem for the base ring Let $A$ and $R$ be commutative rings and let $F$ be a functor $\text{Mod}(A) \to \text{Mod}(R)$ from $A$-modules to $R$-modules. The Eilenberg-Watts theorem says that if $F$ is right exact and preserves small coproducts then $F$ is isomorphic to the functor $- \mapsto M \otimes_A - $ given by tensoring with an $(R, A)$-bimodule $M$.
My question is: Is anyone aware of a (hopefully abelian category flavored) condition on such $F$ which characterizes when the underlying $(R, A)$-bimodule is just $R$ equipped with a ring morphism $f : A \to R$?
I think one answer is when $F$ is a strong monoidal functor, but this is a little bit clumsy because a monoidal structure on $F$ is extra data. Can anyone see a better answer?
 A: First, let me point out that you shouldn't really expect there to be such a characterization that does not involve some extra data, because the $R$-module isomorphism from $M$ to $R$ is not unique.  So, you are going to need some sort of extra data to encode the choice of that isomorphism.  A related issue is that if you allow noncommutative rings, the property you are asking about is not Morita-invariant: if you identify $\text{Mod}(R)$ with $\text{Mod}(M_2(R))$ in the canonical way, then a functor $F:\text{Mod}(A)\to\text{Mod}(R)$ with your property would not have that property as a functor $\text{Mod}(A)\to\text{Mod}(M_2(R))$.  So you are going to need to use something about the commutativity of your rings, which basically means using the monoidal structure.
You don't need to go all the way to giving $F$ a strong monoidal structure, though: it suffices for it just to preserve the unit.  That is, the data of an isomorphism from $F$ to a functor $R\otimes_A -$ for some $A$-algebra structure on $R$ is equivalent to the data of an isomorphism $F(1_{\text{Mod}(A)})\to 1_{\text{Mod}(R)}$.  Indeed, identifying $F$ with $M\otimes_A-$ for some bimodule $M$, an isomorphism $F(1_{\text{Mod}(A)})\to 1_{\text{Mod}(R)}$ is just an isomorphism $M\to R$ as an $R$-module.  It is then automatic that the $A$-module structure on $M$ comes from a homomorphism $A\to R$, since the $A$-module structure must commute with the $R$-module structure and every $R$-module endomorphism of $M$ is multiplication by an element of $R$.
