R-linear categories

Question

Wikipedia "Preadditive category" and nLab "algebroid" seem to define $R$-linear category as one whose hom-sets are $R$-modules, where those modules are over a COMMUTATIVE ring. Does anybody knows why they require the ring to be commutative? When defining preadditive categories, categories whose hom-sets are abelian groups, no such requirement is made. What is the difference? I anticipate the objection that abelian groups are modules over $\mathbb Z$ which is commutative too. But Wikipedia in the same article, includes the category of (left) modules over a ring $R$ as an example of a preadditive category, and that ring is not required to be commutative.

Answer 1

The source of the reason is in module theory. For $R$-modules $M$ and $N$, the set $\mathrm{Hom}_R(M,N)$ is naturally an abelian group. You'd prefer it to be an $R$-module since that's the category from which $M$ and $N$ came. This is true if $R$ is commutative. In particular, if $M$ is an $(R,S)$-bimodule then $\mathrm{Hom}_R(M,N)$ has a natural $S$-module structure, and there is a similar story in the $N$-variable. But, if $R$ is commutative, everything works out to be the same in both variables and (a) $\mathrm{Hom}_R(M,N)$ is an $R$-module and (b) this structure is compatible in both slots.

Answer 2

We require composition of morphisms to be R-bilinear, i.e. passing through $\hom(A,B)\,\otimes_R\,\hom(B,C)$.

If $R$ were not commutative, we need $\hom(A,B)$ be a right $R$-module and $\hom(B,C)$ a left $R$-module to interpret $R$-bilinearity (or tensor product).
So that, all homsets need be an $R$-$R$-bimodule, which indeed seem to make sense, and is satisfied automatically if $R$ is commutative.
But then, one feels that it still should generalize further, allowing $R$-$S$-bimodules in some way..

Now, to keep things in a certain level of generality, we often assume $R$ is commutative.

Answer 3

There is a general notion of what it means for a category $C$ to be enriched over another category $M$, meaning at a minimum that all of the homs in $C$ are objects in $M$. But you need more than this: to define composition you need to explain what it means to write down a morphism called "composition," which means you need to explain what its source and target are, as objects in $M$. This means $M$ needs to be a monoidal category, equipped with some notion of "tensor product" $\otimes$ of its objects, so that you can define composition as a morphism

$$\text{Hom}(a, b) \otimes \text{Hom}(b, c) \to \text{Hom}(a, c)$$

in $M$.

If $R$ is a commutative ring, then the category $\text{Mod}(R)$ of $R$-modules naturally acquires a (symmetric) monoidal structure given by the tensor product $\otimes_R$ over $R$. However, if $R$ is noncommutative, and we consider, say, left modules, there is not even a monoidal structure, since one can't take the tensor product of a left module and a left module. So it doesn't make sense to consider enrichments over $\text{Mod}(R)$ when $R$ is noncommutative, at least without specifying some other monoidal structure.

But Wikipedia in the same article, includes the category of (left) modules over a ring $R$ as an example of a preadditive category, and that ring is not required to be commutative.

That's irrelevant; preadditive categories are only required to be enriched over abelian groups, or $\mathbb{Z}$-modules, and $\mathbb{Z}$ is always commutative.