# Finite Rings: Units or Zero divisors

Let $R$ be a finite commutative ring with unity. I have to prove that every nonzero element of $R$ is either a unit or a zero-divisor.

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In a finite commutative ring with unity, every element is either a unit or a zero-divisor. Indeed, let $a\in R$ and consider the map on $R$ given by $x \mapsto ax$. If this map is injective then it has to be surjective, because $R$ is finite. Hence, $1=ax$ for some $x\in R$ and $a$ is a unit. If the map is not injective then there are $u,v\in R$, with $u\ne v$, such that $au=av$. But then $a(u-v)=0$ and $u-v\ne0$ and so $a$ is a zero divisor.

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+1: My answer looks like overkill compared to this. – Pete L. Clark Aug 31 '11 at 16:25
If you want to know when the converse holds, see mathoverflow.net/questions/42647/…, which is related to Pete's answer. – lhf Aug 31 '11 at 16:29
@Pete, you answer here is nice as well. – lhf Aug 31 '11 at 16:31
@Pete: Most of your answers are overkill, which is why I enjoy them so much! :-) – Asaf Karagila Aug 31 '11 at 16:47
Perhaps my answer can be better phrased as: If the map is surjective then $a$ is a unit. Otherwise, the map cannot be injective, because $R$ is finite, and so $a$ is a zero divisor. – lhf Aug 31 '11 at 18:32

Your question is incomplete: you say you want to prove that every nonzero element of $R$ is "either a zero-divisor?" If one inserts a unit or before zero-divisor then you get a true statement, so I'll assume for now that's what you meant.

First, following a comment by Gerry Myerson on a recent related answer, let me divulge that for me zero is a zero-divisor. I claim that this is just a convention that you should be able to translate back from if you see fit.

Next, note that if you have a family $\{R_i\}_{i \in I}$ of rings in which every element is either a unit or a zero-divisor, the same holds in the Cartesian product $R = \prod_{i \in I} R_i$.

In your case you can use the structure theorem for Artinian rings: $R$ is a finite product of local Artinian rings -- to reduce to the case in which $R$ is local Artinian. Then the maximal ideal is nilpotent, so every nonunit is nilpotent and in particular a zero divisor.

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+1: Funny! It's like killing birds with cannon balls. :D I had never thought that a finite ring is Artinian. – Andrea Aug 31 '11 at 18:36
@Andrea Another application of $\rm\:R\:$ finite $\rm\:\Rightarrow\: R\:$ Artinian is in this prior answer: A finite ring is a field if its units $\cup\ \{0\}$ comprise a field of characteristic $\ne 2$. – Gone Sep 3 '11 at 23:00

HINT $\rm\ |R|<\infty\ \Rightarrow\ r^j=r^k,\: j<k\ \Rightarrow\ r^j\:(1-r^{k-j})=0\ \Rightarrow\ 1 = r^{k-j}\:$ if $\rm\:r\:$ not a zero-divisor.

NOTE $\$ The idea generalizes: if a non-zero-divisor $\rm\:r\:$ is algebraic then it divides the least degree coefficient of any polynomial of which it is a root. When said coefficient is a unit then so too is $\rm\:r\:.\:$ Hence the result holds more generally for any ring satisfying a polynomial identity whose least degree coefficient is unit, e.g. for Jacobson's famous rings satisfying the identity $\rm\:X^n =\: X\:.$

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 I've removed my comments then. Sorry for the noise. – lhf Sep 1 '11 at 16:08