Is the existence of finite biproducts a strengthening of commutativity?

Question

Let $T$ denote a monosorted Lawvere theory (call its distinguished object $G$) equipped with a distinguished constant $0 : G \leftarrow 1$ that is "idempotent" in the following sense: for all arrows $f : G \leftarrow G^n$ of $T$, we have $0 = f \circ \mathrm{diag}_{n,1} \circ 0,$ where $\mathrm{diag}_{n,1} : G^n \leftarrow G$ is the diagonal inclusion. In more classical notation, this is just saying that

$$f(0,\ldots,0) = 0$$

for all arrows $f : G \leftarrow G^n$ of $T$. Then the category of models of $T$ in $\mathbf{Set}$ necessarily has a $0$ object, and thus we can make sense of biproducts.

Write $\mathbf{C}$ for the aforementioned category of models in $\mathbf{Set}$. The only cases I know of where $\mathbf{C}$ has all finite biproducts are the case where $T$ is the Lawvere theory of $S$-modules for some commutative semiring $S$. (This covers abelian groups, commutative monoids, real vector spaces, etc.) In these cases, it also holds that $T$ is commutative.

Now unfortunately, $T$ being commutative doesn't imply the existence of biproducts; e.g. take $T$ equal to the theory of pointed medial magmas satisfying $0+0=0$. It seems natural to ask whether the converse holds:

Question. Does the existence of all finite biproducts in $\mathbf{C}$ imply that $T$ is commutative? That is, does the existence of biproducts represent a strengthening of commutativity? And if so, is the existence of all finite biproducts in $\mathbf{C}$ equivalent to a straightforwardly expressed condition on the structure of $T$?

Answer 1

No, not every algebraic theory with finite biproducts is commutative. For instance, the category of $R$-modules for any ring $R$ has finite biproducts, but the theory of $R$-modules is commutative if and only if $R$ is commutative.

But that's basically the only exception. Indeed:

Theorem. Let $\mathbb{T}$ be a Lawvere theory. The following are equivalent:

• $\mathbf{Mod} (\mathbb{T})$ is a semiadditive category (i.e. has finite biproducts).
• $\mathbb{T}$ is a semiadditive category.
• $\mathbb{T}$ is the theory of $S$-modules for some semiring $S$.

Indeed, since $\mathbb{T}^\mathrm{op}$ is a full subcategory of $\mathbf{Mod} (\mathbb{T})$ that is closed under finite products, if $\mathbf{Mod} (\mathbb{T})$ is semiadditive, then so too must $\mathbb{T}$. Moreover, when $\mathbb{T}$ is semiadditive, then a functor $\mathbb{T} \to \mathbf{Set}$ preserves finite products if and only if it factors as a semiadditive (i.e. preserves finite biproducts) functor $\mathbb{T} \to \mathbf{CMon}$ followed by the forgetful functor $\mathbf{CMon} \to \mathbf{Set}$. Thus, if $\mathbb{T}$ is semiadditive, then $\mathbf{Mod} (\mathbb{T})$ is (isomorphic to) the category of semiadditive functors $\mathbb{T} \to \mathbf{CMon}$.

Now, suppose $\mathbb{T}$ is semiadditive. It is a fact that semiadditive categories are $\mathbf{CMon}$-enriched categories (but not all of them), and semiadditive functors are the same as $\mathbf{CMon}$-enriched functors. Let $\mathbb{S}$ be the full $\mathbf{CMon}$-enriched subcategory of $\mathbb{T}$ spanned by the generating object $G$. Of course, $S = \mathbb{S} (G, G)$ is a semiring, and an $S$-module is the same thing as a $\mathbf{CMon}$-enriched functor $\mathbb{S} \to \mathbf{CMon}$. It is not hard to see that every $\mathbf{CMon}$-enriched functor $\mathbb{S} \to \mathbf{CMon}$ extends to a semiadditive functor $\mathbb{T} \to \mathbf{CMon}$ in an essentially unique way, so it follows that $\mathbf{Mod} (\mathbb{T})$ is equivalent to $\mathbf{Mod} (S)$.

Of course, it is clear that the category of $S$-modules for any semiring $S$ is semiadditive. This completes the proof of the theorem.