# When does a finite ring become a finite field?

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 :)

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.

• 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. 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

• 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. Dec 23, 2017 at 15:37
• It is not true, the link contains the complete article. Specifically, it is Theorem 3, whose proof is on pages 36 and 37. Dec 23, 2017 at 16:44
• This theorem is more general in that it does not assume the ring is commutative like Number does in his question. Dec 23, 2017 at 17:41