There is an easy argument which shows that a finite integral domain (commutative unital ring with no zero divisors) is a field. Here I wonder whether this result still stands if the term "unital" is dropped.
In other words, can a finite commutative ring with no zero divisors always contain a multiplicative identity? More generally, if this is true, can we even generalize Wedderburn's little theorem: every finite ring with no zero divisors is a field?
Best Answer
With care, you can do it for finite, nonzero, noncommutative rings with no nonzero zero divisors, the last part meaning that it is left and right cancellative.
Let $a\in R$ be nonzero. Then left multiplication by $a$ on elements of $R$ is injective, and since $R$ is finite $a=ax$ for some $x\in R$. Then it also follows that $aa=axa$ and $a=xa$ by multiplication and cancellation (cancellation being possible in a ring without nonzero zero divisors.)
Then for any other $b\in R$, $bxa=ba$ implies $bx=b$ and $axb=ab$ implies $xb=b$ after cancellations.
At this point we're looking at a finite ring with nonzero identity with no nonzero zero divisors, and Wedderburn's little theorem would carry through to show us it is commutative and a field.