Abelian categories that are not monoidal

Question

Forgive me if this turns out to be a naive question. I'm quite convinced that not all abelian categories admit (symmetric?) monoidal structure (of course, with the tensor product being additive, meaning that it is an additive bifunctor), but I can't think of an abelian category that doesn't! Maybe I just know of too few abelian categories. Can someone give me an example?

Answer 1

In the paper

Hovey, Mark, Additive closed symmetric monoidal structures on R-modules, J. Pure Appl. Algebra 215, No. 5, 789-805 (2011). ZBL1223.18005.

Hovey shows the following theorem (Theorem 3.3)

Suppose k is a division ring. If k is not a field, then there is no additive closed symmetric monoidal structure on the category of k-modules. If k is a field, there is a unique additive closed symmetric monoidal category structure on the category of k-modules, up to symmetric monoidal equivalence.

Hovey gives a characterization of such monoidal structures via bimodules by exploting the Eilenberg-Watts theorem. He shows that if R is the unit of an additive closed monoidal category structure on R-Mod then R has to be commutative and is the usual such structure ( Proposition 3.1 ).

Combined with this, a simple dimension argument gives you the proof for the theorem.

Answer 2

Let $\mathcal{A}$ be an additive category with a monoidal structure such that the maps $$ \otimes \colon \mathcal{A}(A,B)\times\mathcal{A}(C,D) \to \mathcal{A}(A\otimes C,B\otimes D) $$ are biadditive. Let $U$ be the unit object. If $n$ is an integer with $n.1_U=0$, it follows (by identifying $A$ with $U\otimes A$) that $n.1_A=0$ for all objects $A$. From this we see that the abelian category of finite abelian groups admits no monoidal structure. (Of course the tensor product is commutative and associative, but there is no unit.)

I can't immediately think of an example where the category admits infinite coproducts. Note that the category of torsion abelian groups is symmetric monoidal under $\text{Tor}(-,-)$, with unit $\mathbb{Q}/\mathbb{Z}$. (Indeed, for finite abelian $A,B$, a pair of elements $\alpha\in A^*=\text{Hom}(A,\mathbb{Q}/\mathbb{Z})$ and $\beta\in B^*$ give a map $\text{Tor}(A,B)\xrightarrow{\text{Tor}(\alpha,\beta)}\text{Tor}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z})=\mathbb{Q}/\mathbb{Z}$. This construction gives a map $A^*\otimes B^*\to \text{Tor}(A,B)^*$, and by adjunction we get a map $\text{Tor}(A,B)\to(A^*\otimes B^*)^*$, which is an isomorphism. From this point of view it is easy to see that $\text{Tor}(-,-)$ is commutative associative on finite abelian groups, and thus also on the $\text{Ind}$-completion of that category, which is the category of torsion abelian groups.)