# Every nonzero element in a finite ring is either a unit or a zero divisor

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

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.

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

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.

Hint $$\,\ \overbrace{|R|<\infty\ \Rightarrow\ r^j=r^k}^{\rm\large pigeonhole},\: j>k\$$ $$\Rightarrow\ (r^{j-k}-1)\,\color{#0a0}{r^k}=0\$$ $$\overset{\!\large \color{#0a0}{r\ \nmid\ 0}}\Longrightarrow\ \overbrace{r^{j-k}=1}^{\!\!\!\!\textstyle\color{#c00}r\,(r^i)\!=\!1^{\phantom{|^|}}\!\!\!\!\!\!}\,$$ $$\,\Rightarrow\, \color{#c00}r\,$$ is a unit

Remark $$\$$ The idea generalizes: if a non-zero-divisor $$\,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 zero-divisor or a unit of $$R\,$$ (he deems this a "rough zero-divisor dual" of fraction / localization extensions)

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 left-hand 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 $$x(x^{n-1}-x^nr)=0=(x^{n-1}-x^nr)x$$ since $$r$$ commutes with $$x$$. By minimality of $$n$$, $$x^{n-1}-x^nr\neq 0$$. Thus, $$x$$ has been demonstrated to be a two-sided zero divisor.

I. Kaplansky, Topological representations of algebras II, Trans. Amer. Math. Soc. 68 (1950), 62-75. MR 11:317

Let $a$ in $R$ be non-zero and suppose that $a$ is not a zero-divisor.

First I will prove the cancellation property just for $a$. If $ab = ac$, then $ab-ac = 0$ and $a(b-c) = 0$. Since $a$ is not a zero-divisor, then $b-c = 0$ 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 $a^{i-j}a^j = 1a^j$ (remember we have unity), then cancellation gives us $a^{i-j} = 1$. If $a = 1$ then $a$ is clearly a unit.

If $a\ne 1$, then $i-j \gt 1$ so we can factor out one copy of $a$ to get $a^{i-j-1}a^1 = 1$.

Thus the element $a^{i-j-1}$ is the multiplicative inverse of $a$, so $a$ is a unit.

Thus every nonzero element of this ring that is not a zero-divisor is a unit. In other words, every nonzero element is either a zero-divisor 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 zero-divisor nor a unit.

• Hello new user! This might help you in the future. – user228113 Apr 20 '15 at 6:50

Let $a\not=0$

Because $R$is finite then

$a^j=a$ , then $(a^j -a )=0$

$a (a^{j-1}-1) =0$

If $a\not=0$ then $a$ is zero divisor and $a^j a^{-1} = a^{j-2}a=1$ so $a$ is unit

• This is not correct. It is not necessarily true that $a^j = a$ for some $j > 1$. e.g. In the ring $\mathbb{Z}/4\mathbb{Z}$, we do not have that $2^j = 2$ for any $j > 1$, since $2^j = 0$ for every $j > 1$. Your proof also claims to show that $a$ is always both a zero-divisor, and a unit, but this is not true. Finally, $a(a^{j - 1} - 1)$ doesn't imply that $a$ is a zero-divisor. It only implies that $a$ is a zero-divisor if we know that $a$ and $a^{j-1} - 1$ are both not equal to $0$. – Dylan Mar 27 '18 at 13:04
• A good way to start here is “suppose $a$ isn’t a zero divisor.” Then you can go at least two ways to prove $a$ is a unit, as outlined above, using the ideal the multiplication by $a$ is an injective map. There doesn’t seem to be any way to conclude that $a=a^j$ with $j\geq 2$ like you want, but certainly there are two distinct powers which have to be equal in the ring, which is like what you’re doing. That’s one of the solutions above. – rschwieb Mar 27 '18 at 16:09