I'm relatively new to ring theory so this is probably a simple question.

Its easy to see that a finite ring that is commutative and has no-zero divisors (i.e. an integral domain) must have multiplicative inverses.

I am wondering if we can rearrange these properties and always get implication or produce finite rings with only 2 of the above properties.

Explicitly my questions are:

1) Does a finite ring that is commutative with multiplicative inverses always have no-zero divisors? (EDIT: this one is pretty easy too, let $xy=0$ and assume $x\neq 0$. then $x^{-1}xy=x^{-1}0$. Hence, $y=0$ as desired.)

2) Is a finite ring that has multiplicative inverses and no zero divisors always commutative?

If these are actually simple exercises, I'd just like a hint to get started with the proof or any counterexamples.

Thanks :)


2 Answers 2


HINT 1) If $\;\rm R\;$ is finite then $\;\rm x\to r\:x\;$ is onto iff 1-1, so $\;\rm R\;$ is a field iff $\;\rm R\;$ is a domain.

2) is Wedderburn's little theorem.

  • $\begingroup$ Thanks! I see (1) now. Just say x,y are in R and then do xy=0. WLOG assume x not 0. then x has an inverse so (x^-1)xy=(x^-1)0, hence y=0. I think I need some more time with Weddernburn's little theorem, since I haven't really heard of a lot of the terms used in the link. $\endgroup$
    – WWright
    Aug 31, 2010 at 2:39

Des MacHale has a nice and simple answer to the question When is a finite ring a field?

Theorem: Let $(R,+,\cdot)$ be a finite non-zero ring with the property that if $a$ and $b$ in $R$ satisfy $ab=0$, then either $a=0$ or $b=0$. Then $(R,+,\cdot)$ is a field.

MacHale demonstrates that 1) and 2) are particular cases of this theorem. The article is published online by the Irish Mathematical Society here

  • $\begingroup$ It would be better to post the proof directly here, or at least to outline it. The link is only to the first page, which only states the theorem. $\endgroup$
    – rogerl
    Dec 23, 2017 at 15:37
  • $\begingroup$ It is not true, the link contains the complete article. Specifically, it is Theorem 3, whose proof is on pages 36 and 37. $\endgroup$
    – G al Cubo
    Dec 23, 2017 at 16:44
  • $\begingroup$ This theorem is more general in that it does not assume the ring is commutative like Number does in his question. $\endgroup$
    – nombre
    Dec 23, 2017 at 17:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .