All rings are commutative (additively)

02 Jul 2017

Normally a ring is defined to be an abelian group (written additively) with an extra multiplication operation under which everything is a semigroup. Why do we restrict to an abelian group additively though? What happens if we try and make the same definition with an arbitrary group?

Well in fact we would get exactly the same objects! Indeed the rest of the ring axioms are enough to force $a + b = b+a$ always, hence removing that axiom is harmless except that it requires you to prove this to get the theory off the ground.

We can prove this very explicitly by observing that

$$a + a + b + b = (1 +1 )(a+b) = (a+b) + (a+b) = a+b+a+b,\,\forall a,b\in R$$

hence

$$a+b = b+a \in R.$$

While this calculation shows that the base group must be abelian in order to have a ring structure on top, I think a more conceptual explanation for why this is the case would be nice also. The best I’ve got so far in that vein is the following:

The ring structure on a group $G$ defines an injective function $L_{\bullet}\colon G \to \operatorname{End}(G)$ via the left multiplication map $g\mapsto (h\mapsto g\times h)$, the ring axioms guarantee that each such map is a group endomorphism. Additionally distribution implies that $L_{g+h} = L_g + L_h$. The existence of a unit for multiplication implies that the identity map $\operatorname{id}_G\colon g\mapsto g$ is in the image of $G$ inside $\operatorname{End}(G)$, hence the doubling/squaring map is in the image $G$, and so is an endomorphism of $G$. This fact that squaring is a group morphism then guarantees that the group is abelian. (For a different example of this this statement consider a group where squaring sends everything to the identity, i.e. in which every non-identity element has order 2, this group must then also be abelian).

This still feels a little convoluted to me and the calculation is of course a lot more revealing, but trying to unpack what’s going on in the calculation and why it turns out that way is quite interesting, and who knows maybe it will lend itself to some kind of generalisation?

This little fact is literally the first exercise in Lam’s A first course in noncommutative rings, but it had certainly escaped my attention until now.