# Matrices over noncommutative rings?

In chapter 1, section 2 of Categories for the Working Mathematician, Mac Lane says:

For each commutative ring $K$, the set $\mathbf{Matr_K}$ of all rectangular matrices with entries in $K$ is a category; the objects are all positive integers $m,n,...$, and each $m\times n$ matrix $A$ is regarded as an arrow $A:n\rightarrow m$, with composition the usual matrix product.

This is undoubtedly true. But why restrict the statement to commutative rings? Surely, matrices over noncommutative rings also form a category. Or am I missing something?

The answer to this question suggests the noncommutative rings are just not that well studied, so perhaps Perhaps Mac Lane was going for familiarly over generality. Another question addresses the same passage in the book but does not raise the question of commutativity.

To be specific about my questions

1. Is my claim "For each noncommutative ring $K$, the set of all rectangular matrices with entries in $K$ is a category" true?
2. Is the category of rectangular matrices over a commutative ring somehow nicer than that over all (commutative or otherwise) rings?

EDIT: Strikeout the claim that noncommutative rings are not well studied, given the consensus in the comments that they are. In that case, why stipulate commutative rings? Were they less well studied back in the 1970s?

• 1. Yes. I think "familiarity over generality" is exactly what happened here. – darij grinberg Sep 18 '18 at 16:04
• 2. The former category has a few properties that the latter doesn't have (e.g., a left-invertible endomorphism is always right-invertible, and vice versa), but I wouldn't really say it is "nicer" in any fundamental way. – darij grinberg Sep 18 '18 at 16:05
• I like this question and hope someone better-versed in category theory will answer it. I do know an answer to your question 2, but I am sure that McLane would strongly disagree with me. Namely: in the commutative case (and especially in the field case, which the suggestive notation $K$ forces us to think about even if the word says "ring") we can imagine the objects of the categories as actual concrete flesh-and-blood mathematical objects (vectorspaces/free modules) which is trickier in the non-commutative case (even if it can be done). (Ctd in next comment) – Vincent Sep 18 '18 at 16:06
• But MacLane tries to steer us away from thinking about the objects as actual living things with properties and a character etc and only do that for the arrows. (Hence the arrows being matrices and the objects 'just' numbers rather than vectorspaces.) So, as I said, he would probably do not share my notion of nice here. – Vincent Sep 18 '18 at 16:08
• You got the wrong impression in the first linked question. I tried to explain why I thought so. . I agree with darij's comment. – rschwieb Sep 18 '18 at 18:49

## 1 Answer

The construction goes through just fine for noncommutative rings, and in fact slightly more generally: you don't need additive inverses, so you can do this construction for semirings or rigs, and there are a few interesting and important examples of this.

It is very slightly trickier to say how to make this construction "concrete" in the noncommutative case, but not by much: this category is equivalent to the category of finitely generated free right $K$-modules and $K$-module homomorphisms. When $K$ is a field this reduces to the familiar case of finite-dimensional $K$-vector spaces.

You get some bonus structure in the commutative case: namely, tensor product defines a symmetric monoidal structure on this category.