Let $R$ be a finite ring with unity. Prove that every nonzero element of $R$ is either a unit or a zerodivisor.

To fix lhf's answer for the noncommutative case see e.g. [here](https://math.stackexchange.com/a/3889508/242) and [here](https://math.stackexchange.com/q/3656572/242) and [here](https://web.archive.org/web/20150512111012/https://crazyproject.wordpress.com/2010/08/14/basicpropertiesofleftandrightunitsandleftandrightzerodivisors/) – Bill Dubuque Nov 01 '20 at 10:05
5 Answers
In a finite commutative ring with unity, every element is either a unit or a zerodivisor. 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 $a(uv)=0$ and $uv\ne0$ and so $a$ is a zero divisor.
For the noncommutative case, see this answer.
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2If you want to know when the converse holds, see http://mathoverflow.net/questions/42647/ringsinwhicheverynonunitisazerodivisor, which is related to Pete's answer. – lhf Aug 31 '11 at 16:29

1@Pete, you answer [here](http://mathoverflow.net/questions/42647/ringsinwhicheverynonunitisazerodivisor/42650#42650) is nice as well. – lhf Aug 31 '11 at 16:31

12@Pete: Most of your answers are overkill, which is why I enjoy them so much! :) – Asaf Karagila Aug 31 '11 at 16:47

5Perhaps 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

1@lhf: To be honest I had by now completely forgotten about that MO answer. $@$Asaf: um, okay, glad you're enjoying them. – Pete L. Clark Sep 01 '11 at 14:23

4I think for a complete solution, you still have to state that no element can be both a unit and a zero divisor. That's not a big deal, of course. – azimut Sep 13 '13 at 10:50

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@VHP, the map used in my answer is very useful and used in several arguments in both group and ring theory. – lhf Jun 30 '14 at 20:38


5The answer of lhf have the assumption ``$R$ is commutative'', but the condition is not necessarily. See http://homepages.math.uic.edu/~radford/math516f06/WH4Sol.pdf – bfhaha Jun 15 '14 at 17:39

Where was the commutativity used? Surely not to show $ax=xa$ since it will follow anyway? – Hrit Roy Dec 13 '19 at 13:36

@bfhaha See [here](https://math.stackexchange.com/a/3889508/242) for how to extend this answer to the noncommutative case. At lhf: please add this link to the answer since this question is being used as a dupe target in this case and it is already causing confusion (this remark is unlikely to be seen being buried deep down in comments). – Bill Dubuque Nov 01 '20 at 09:43



1Essentially the same proof also shows that this is true if $R$ is a finitedimensional vector over a field. – Aryaman Maithani Jul 04 '21 at 10:35
Your question is incomplete: you say you want to prove that every nonzero element of $R$ is "either a zerodivisor?" If one inserts a unit or before zerodivisor 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 zerodivisor. 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 zerodivisor, 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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10+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

2@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 $\ne 2$.](http://math.stackexchange.com/questions/30556/afiniteringisafieldifitsunitscup0compriseafieldofcharacte/30566#30566) – Bill Dubuque Sep 03 '11 at 23:00
Hint $\,\ \overbrace{R<\infty\ \Rightarrow\ r^j=r^k}^{\rm\large pigeonhole},\: j>k\ $ $\Rightarrow\ (r^{jk}1)\,\color{#0a0}{r^k}=0\ $ $\overset{\!\large \color{#0a0}{r\ \nmid\ 0}}\Longrightarrow\ \overbrace{r^{jk}=1}^{\!\!\!\!\textstyle\color{#c00}r\,(r^i)\!=\!1^{\phantom{^}}\!\!\!\!\!\!}\, $ $\,\Rightarrow\, \color{#c00}r\, $ is a unit
Remark $\ $ The idea generalizes: if a nonzerodivisor $\,r\,$ 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 $\:r.\:$ 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 $\rm\:X^n =\: X\:.$
P. M. Cohn has shown that every commutative ring $R$ can be embedded in a ring $S$ where every element of $S$ is either a zerodivisor or a unit of $R\,$ (he deems this a "rough zerodivisor dual" of fraction / localization extensions)
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Since one good cannonball deserves another, I'd like to provide a solution using right Artinian rings that aren't necessarily commutative.
Definitions:
A ring $R$ is called strongly $\pi$regular if for all $x\in R$, chains of the form $xR\supseteq x^2R\supseteq x^3R\supseteq\dots \supseteq x^iR\supseteq\dots$ become stationary.
A ring is called Dedekind finite if $xy=1$ implies $yx=1$ for all $x,y\in R$.
Strongly $\pi$regular rings were introduced by Kaplansky in the citation at the bottom. The definition is usually given in terms of "$\forall x\exists r(x^n=x^{n+1}r)$", but this is equivalent.
Moreover, it's been shown that $r$ can be chosen to commute with $x$, and so the lefthand version of this definition is equivalent to this one.
It's obvious right Artinian rings are strongly $\pi$regular, and it turns out they are Dedekind finite too.
Proposition: In a strongly $\pi$regular Dedekind finite ring (in particular, right or left Artinian rings), each element is a unit or a zero divisor. (Zero being counted as a zero divisor.)
Proof: Let $x\in R$ be a nonunit, and let $n$ be minimal such that there exists $r$ that commutes with $x$ and $x^n=x^{n+1}r$. Since $x$ isn't a unit, $n\geq 1$. (Because if $1=xr$, $x$ would be a unit by Dedekind finiteness.)
Rearranging, we get $x(x^{n1}x^nr)=0=(x^{n1}x^nr)x$ since $r$ commutes with $x$. By minimality of $n$, $x^{n1}x^nr\neq 0$. Thus, $x$ has been demonstrated to be a twosided zero divisor.
I. Kaplansky, Topological representations of algebras II, Trans. Amer. Math. Soc. 68 (1950), 6275. MR 11:317
Let $a$ in $R$ be nonzero and suppose that $a$ is not a zerodivisor.
First I will prove the cancellation property just for $a$. If $ab = ac$, then $abac = 0$ and $a(bc) = 0$. Since $a$ is not a zerodivisor, then $bc = 0$ so $b = c$.
Consider the set $\{a^n\mid n \in\mathbb N\}=\{1,a^1,a^2,...\}$
Since $R$ is finite, we must have $a^i = a^j$ for some $i$, $j$ with $i \gt j$. Then since we have the cancellation property for $a$ and we have $a^{ij}a^j = 1a^j$ (remember we have unity), then cancellation gives us $a^{ij} = 1$. If $a = 1$ then $a$ is clearly a unit.
If $a\ne 1$, then $ij \gt 1$ so we can factor out one copy of $a$ to get $a^{ij1}a^1 = 1$.
Thus the element $a^{ij1}$ is the multiplicative inverse of $a$, so $a$ is a unit.
Thus every nonzero element of this ring that is not a zerodivisor is a unit. In other words, every nonzero element is either a zerodivisor or a unit.
If we drop the finite condition then the result does not hold true. For example, $\mathbb Z$ is a commutative ring with unity, but $2$ is neither a zerodivisor nor a unit.

2Hello new user! [This](http://meta.math.stackexchange.com/questions/5020/mathjaxbasictutorialandquickreference) might help you in the future. – Apr 20 '15 at 06:50

1This is the same as the proof I posted $4$ years prior here. Why repeat it? – Bill Dubuque Nov 01 '20 at 09:22