Finite integral domains are commutative?

Question

Here, integral domain is a non-zero ring $R$ (not necessarily commutative, and not necessarily contains unity), in which $ab=0$ implies $a=0$ or $b=0$.

Question If $R$ is a finite integral domain, is it necessarily commutative?

Answer 1

The pieces of this proof are already strewn about the site. I think I'll tidy it into a single argument though.

A nonzero finite ring with no nonzero zero divisors has identity.

First of all, left multiplication by a nonzero element makes an injective map. (This is shown multiple times throughout the site, and follows easily from the hypothesis.) Since an injective map on a finite set is surjective, we can solve $ax=b$ whenever $a$ is nonzero.

Given nonzero $a$, there us a $b$ such that $ab=a$. Then $a(b-b^2)=0$ implies $b=b^2$.

For an arbitrary $c$, $(cb-c)b=0$ implies $cb=c$ for all $c$. Thus the ring has a right identity. Symmetrically, it has a left identity, and these are necessarily equal and are the identity if the ring.

The surjective try argument given above now allows you to show the ring is a division ring.

Finally, the hard part is showing that it is commutative, a result known as Wedderburn's Little Theorem. You can find proofs online and in many books. It takes a bit of work that is not worth reproducing here.