Does every finitary monad with this propery arise as a free module monad? Let $T:\mathbf{Set} \rightarrow \mathbf{Set}$ denote a finitary monad such that $T(\emptyset) \cong 1$ and $T(A \sqcup B) \cong T(A) \times T(B)$, naturally in $A$ and $B$.

Question. Does $T$ necessarily arise as the free module monad for some unital semiring?


I think the answer is Yes.
Since $T$ is finitary, we may restrict to finite sets in the domain. We have natural isomorphisms $T(A) \cong T(\{1\})^A$. The set $T(\{1\})$ has a semiring structure defined as follows:


*

*The unique map $\emptyset \to \{1\}$ induces a map $\{1\} \cong T(\emptyset) \to T(\{1\})$, i.e. an element $0 \in T(\{1\})$.

*The monad unit $\{1\} \to T(\{1\})$ induces another element $1 \in T(\{1\})$.

*The codiagonal $\{1\} + \{1\} \to \{1\}$  induces the addition map $T(\{1\}) \times T(\{1\}) \to T(\{1\} + \{1\}) \to T(\{1\}).$

*There is a natural map  $T(A) \times B \to T(A \times B)$, a so-called strength, induced by the natural map $B \to \hom(A,A \times B) \to \hom(T(A),T(A \times B))$. This induces a multiplication map
$$T(A) \times T(A) \to T(A \times T(A)) \cong T(T(A) \times A) \to T(T(A \times A)) \to T(A).$$
In the last step, we have used the monad multiplication.


Of course one has to check that the semiring axioms are satisfied (this requires some work), and that the isomorphism $T(A) \cong T(\{1\})^A$ respects the monad structure (this should be easy). I think the proof may appear somewhere in Durov's thesis.
Edit: It is 4.8.6 in Durov's thesis. But we have to require that the natural maps $T(A + B) \cong T(A) \times T(B)$ are not arbtirary, but rather induced by the monad structure in a certain way, see 4.8.1. Specifically, the projection $T(A + B) \to T(A)$ (and likewise $T(A + B) \to T(B)$) should be the $T$-module map which is induced by the map $A + B \to T(A)$ which is the monad unit on $A$ and the "zero map" $B \to \{1\} \to T(\emptyset) \to T(A)$ on $B$. 
