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?


2 Answers 2


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.

  • 1
    $\begingroup$ This is funny, I just added this coment to my answer. Telepathy! Anyway: +1 is in order. $\endgroup$ Aug 14, 2018 at 13:39

Yes, and this is the usual way to state it (answer to your last question).

As for your first question: let $0\neq a\in R$ be any element. Multiply $a$ by all elements of $R$. When multiplying with two different elements, the two products are different, as there are no zero divisors. So we obtain every element as a product, in particular, $a=ax$ for some $x\in R$.

Let $b\in R$ be arbitrary. Then $bxa=bax=ba$, so again, as there are no zero divisors, we have $bx=b$. Thus $x$ is a unit element.

In fact, with a little more care, it is also possible to get rid of the commutativity condition. Check my calculation, locate the place where I used it, and then you can fix it so that it works for arbitrary finite rings.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.