I recently rememebered the following theorem by Ganesan:

Let $R$ be a commutative ring with $0<n<\infty$ non-zero zero divisors. Then $\operatorname{card}(R)\leq(n+1)^2.$

The proof proceeds by considering the annihilator of one of the zero divisors. Let $z_1,\ldots,z_n$ be the non-zero zero divisors of $R.$ Let $A_1=\{x\in R\,|\,xz_1=0\}.$ It is a well-known fact that this is an ideal in $R.$ We also have $\operatorname{card}(A_1)<n+1$ because $z_1\neq 0.$

Now let's pick one representative $r_x$ from each coset $x\in R/A_1.$ Consider the map $$x\mapsto r_xz_1.$$ Suppose $r_xz_1=r_yz_1.$ Then $(r_x-r_y)z_1=0,$ so $(r_x-r_y)\in A_1,$ which means $x=y.$ Therefore the map is injective. But for any $r\in R,$ the element $rz_i$ is a (possibly zero) zero-divisor. It follows that there are at most $n+1$ cosets in $R/A_1$ and since each coset has cardinality equal to $\operatorname{card}(A_1)\leq n+1,$ we have $\operatorname{card}(R) \leq (n+1)^2.\square$

So we have that if $R$ has exactly one non-zero zero divisor $x,$ then $R$ has at most $4$ elements. I thought it would be a cool thing to determine which rings (with or without unity, since the theorem doesn't require its existence) have this property. It is possible to do this by checking all rngs with at most $4$ elements. But I found that I can't determine all non-isomorphic commutative rngs with at most $4$ elements. Wikipedia says there are $11$ rngs with four elements counting non-commutative ones. There can't be many commutative ones then, but checking all possible operation tables to find them is tedious since I don't know what the $11$ rings are.

Is there a nice way to do this? The condition that there is exactly one-nonzero zero divisor seems pretty strong and I suspect the only such rng is $\mathbb Z/2\mathbb Z$ with zero multiplication, but this is only a suspicion. I've checked that there are no such rngs with $3$ elements, but what about the four-element ones?

  • $\begingroup$ Do you restrict to unitary rings or is the existence of a multiplicative identity not required? $\endgroup$ Mar 28, 2012 at 14:58
  • $\begingroup$ It's not. But if you have an argument about unitary rings, I would be glad to see it too. $\endgroup$
    – user23211
    Mar 28, 2012 at 15:01

2 Answers 2


Note: The above comments were posted while I was writing this answer - I've assumed $R$ is unitary. In particular, my comment about $\mathbb{Z}/2\mathbb{Z}$ with the zero multiplication not being a ring no longer stands ("ring" to me includes unitary).

If $|R|=4$, then as the characteristic of $R$ divides $4$, it must be either $1$, $2$ or $4$. If the characteristic is $1$, you have the zero ring, so that's not an example. If the characteristic is $4$, then you have $\mathbb{Z}/4\mathbb{Z}$, which has exactly one non-zero zero divisor.

In the case that the characteristic is $2$, we can write the elements as $0$, $1$, $a$ and $1+a$; if $1+a=0$ then $a=-1=1$, if $1+a=1$ then $a=0$ and if $1+a=a$ then $0=1$, so $1+a$ is different from the first three. Then (as we're assuming $R$ is commutative), the multiplication table is completely determined by the choice of $a^2$.

There are four possible choices, although $a^2=0$ and $a^2=1$ give isomorphic rings by swapping $a$ and $a+1$, so there are three different possibilities. Only the $a^2=0$ choice gives a ring with exactly one non-zero zero divisor (namely $a$). This ring is isomorphic to $\mathbb{F}_2(x)/(x^2)$.

Note that you can't have the zero multiplication on $\mathbb{Z}/2\mathbb{Z}$ and still get a ring, as then you don't have an identity element, so that isn't in fact an example. As the zero ring isn't an example, $\mathbb{Z}/2\mathbb{Z}$ (the only ring of two elements) isn't an example, and as you've checked the ring of $3$ elements isn't an example either, the two order four examples are the only ones.

  • 4
    $\begingroup$ $\mathbb{Z}/4\mathbb{Z}$ has only 1 nonzero zero divisor. $\endgroup$ Mar 28, 2012 at 15:19
  • $\begingroup$ Yes, you're right. Edited accordingly. $\endgroup$
    – mdp
    Mar 28, 2012 at 16:08
  • $\begingroup$ Thank you for the answer. I still have some hope that perhaps someone might come up with the solution for non-unitary rings so will wait a bit with accepting. $\endgroup$
    – user23211
    Mar 29, 2012 at 13:24
  • $\begingroup$ Sure, that makes sense. I feel somehow that the answer for non-unitary rings could be arrived at along the same lines, although the case analysis will be more unpleasant, and I don't really have the time or the inclination to work through it. If somebody has a really slick answer I'd be interested to see it though. $\endgroup$
    – mdp
    Mar 29, 2012 at 14:25
  • $\begingroup$ Actually it turns out that there are no other examples among non-unitary rings--please see my answer. $\endgroup$
    – mathmandan
    Oct 2, 2015 at 1:28

It turns out that all such rings have a multiplicative unit, so there are no further examples besides the ones already listed in Matthew Pressland's answer.

Proof: Suppose $R$ is a commutative ring (with or without multiplicative identity) of four elements, with a unique non-zero zero divisor $x$. Then there exists an element $y\in R$ such that $xy = 0$ and $y\neq 0$. By uniqueness of the zero divisor, we conclude $x=y$. So $x^2 = 0$.

Suppose $t\in R$ and $t\neq x$ and $t\neq 0$. Then by commutativity $(tx)^2 = txtx = ttxx = t^2\cdot 0 = 0$. Therefore either $tx = 0$ or $tx = x$ (since otherwise $tx$ would be another zero divisor), and further $tx \neq 0$ since otherwise $t$ would be another zero divisor. So $tx=x$ for all $t\in R$ such that $t\neq x$ and $t\neq 0$.

The underlying group $(R, +, 0)$ is isomorphic (as groups) either to $\mathbb{Z}/4 \mathbb{Z}$ or to $\mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$.

Case I: $(R,+,0)$ is cyclic. We claim $x$ cannot be the group generator. Proof of Claim: Suppose $x$ generates $R$ as a group, so $R = \{ 0, x, x+x, x+x+x\}$. Then $(x+x)\cdot x = x^2 + x^2 = 0+0 = 0$, so $(x+x)$ is a zero divisor, contradicting uniqueness.

So let $R = \{ 0, x, a, b\}$, where $a$ and $b$ are generators of the group $R\cong \mathbb{Z}/4\mathbb{Z}$. Then $a+a = x = b+b$. We also have $b+b+b=a$ and $a+a+a = b$.

From above we know that $ax=x=bx$. Hence $(ab)x = a(bx) = ax = x$, so either $ab = a$ or $ab = b$ (the other possibilities $ab=x$ and $ab=0$ are excluded since they would imply $abx=0$). Assume without loss of generality that $ab=b$. Now $ab=b$, and $ax = x$, and $a\cdot 0 = 0$. Finally $a^2 = a(b+b+b) = ab+ab+ab = b+b+b = a$, so $a$ is a multiplicative identity for $R$.

The value of $b^2$ remains to be determined, but $b = (a+a+a)$ so $b^2 = (a+a+a)\cdot(a+a+a) = 9a^2 = 9a = a$. So in Case I, $R \cong \mathbb{Z}/4 \mathbb{Z}$ as rings, by the correspondence $0 \mapsto 0$, $a \mapsto 1$, $x \mapsto 2$ and $b \mapsto 3$.

Case II: $R \cong \mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$ as groups, so letting $R = \{0, x, a, b\}$ as before, we have $x+x = a+a = b+b = 0$, and the sum of any two nonzero elements is the remaining nonzero element.

Now $b^2 = (a+x)^2 = a^2 + 2ax + x^2 = a^2+0+0$, so $a^2=b^2$. Further, we can't have $a^2 = x$ since that would imply $a^4 = x^2 = 0$ and $a$ would be a zero divisor. Thus, either $a^2 = b^2 = a$, or $a^2 = b^2 = b$. Without loss of generality, let us assume that $a^2 = b^2 = a$.

From above we also already know $ax=x=bx$, and finally $ab = a(a+x) = a^2 + ax = a + x = b$. Thus $a$ is again a multiplicative identity on $R$, and the multiplication table is determined.

Indeed, we now see that in Case II, $R \cong \mathbb{F}_2[X] / (X^2)$, by the correspondence $0 \mapsto 0, a\mapsto 1, x\mapsto X$, and $b\mapsto (X+1)$.

Thus Cases I and II are the two examples listed in Matthew Pressland's answer, and there are no others. $\Box$


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