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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?

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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 $R$.

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  • $\begingroup$ Thanks. The non-commutative counterexample based logic is really cool. $\endgroup$
    – Keeley Hoek
    Dec 6, 2020 at 16:56
  • $\begingroup$ Do you agree though that every such unit preserving $F$ (to which Eilenberg-Watts applies) has a unique strong monoidal structure which extends the data of a fixed choice of isomorphism $F(A) \to R$? $\endgroup$
    – Keeley Hoek
    Dec 6, 2020 at 16:56
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    $\begingroup$ That's correct. This is basically because the monoidal structure can be defined in terms of the unit: to compute $X\otimes Y$, you write $X$ and $Y$ using colimits of copies of $1$ (i.e., take presentations of them as modules), and then combine them using the fact that $1\otimes 1=1$ and $\otimes$ preserves colimits in both variables. $\endgroup$
    – Eric Wofsey
    Dec 6, 2020 at 16:59
  • $\begingroup$ Sorry, one more thing. Is it clear that functors F and G of the above kind represent the same morphism $A \to R$ exactly if they are monoidally naturally isomorphic? $\endgroup$
    – Keeley Hoek
    Dec 6, 2020 at 20:10
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    $\begingroup$ Yes. A natural isomorphism $F\to G$ is the same thing as an isomorphism of bimodules $F(A)\to G(A)$. If you require the natural isomorphism to be monoidal, that means it commutes with the identifications $F(A)\cong R$ and $G(A)\cong R$, so it becomes the identity map $R\to R$ so the $A$-module structures on both $R$s are the same since it is a bimodule map. $\endgroup$
    – Eric Wofsey
    Dec 6, 2020 at 20:23

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