For commutative rings $A$ with identity $1\ne0$, nilpotent elements are zero-divisors. The converse is false, i.e. there is a commutative ring $A$ with identity $1\ne0$ and a zero-divisor $x$ in $A$ which is not nilpotent. Where is such an example?

Is it meaningful to ask the "percentage" of commutative rings $A$ with identity $1\ne0$ for which the converse holds, i.e. zero-divisors are nilpotent?

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    $\begingroup$ Have you thought about the integers modulo $n$? $\endgroup$ Dec 22, 2018 at 12:45
  • $\begingroup$ You can also search yourself at this site, e.g. here, for useful links. $\endgroup$ Dec 22, 2018 at 12:51
  • $\begingroup$ @Lord Shark the Unknown: Thank you. Right, in $\mathbb{Z}/10\mathbb{Z}$, $2$ is a zero-divisor but not nilpotent. How did you came up with this example? How about the next question? $\endgroup$
    – user584333
    Dec 22, 2018 at 12:52
  • $\begingroup$ Sai, have a look at this post, which gives an answer which rings have this property. $\endgroup$ Dec 22, 2018 at 12:56
  • $\begingroup$ @Dietrich Burde: Thank you for the reference. $\endgroup$
    – user584333
    Dec 22, 2018 at 13:01

3 Answers 3


An example is in the integers mod $6$, where $2\cdot 3=0$, but no power of either of these individually is zero.

If we had some sort of measure on a set of rings (which could not possibly be all rings, because there are too many) with finite total measure, we could ask about proportion. I know of no such commonly used measure, but it's possible such a thing has been considered.

In my intuition, the proportion would be small, but I can't think of a quick reason to see why, other than that being nilpotent seems like a special property while being a zero divisor seems very common. For example, the only time this is true for rings of integers modulo $n$ is when $n$ is a prime power, and there are vanishingly few of those compared to all integers.

If you look at the answer by rschwieb in the question Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? linked above in a comment by Dietrich Burde, which is not the accepted answer, it characterizes these rings. A ring has this property if and only if it is the quotient of an arbitrary commutative ring by a primary ideal. An ideal $I$ is said to be primary if whenever we have that $a,b\in R$ satisfy $ab\in I$, then we have either $a\in I$ or $b^n\in I$ for some $n>0$. This answers part of your question. It is nicely illustrated in the $\Bbb Z_n$ case: an ideal $n\Bbb Z$ of $\Bbb Z$ is primary if and only if $n=p^k$ for some prime $p$ and integer $k>0$.

So rings like these are actually "close" to integral domains, which are quotients of arbitrary commutative rings by prime ideals. Again, no quantitative argument for why they should be rare, only an intuitive one.

  • $\begingroup$ Couldn't one equip a ring $R$ with the Zariski topology, then produce a Borel sigma algebra, and equip it with a pre-measure, which can then be extended to a measure by Caratheodory's extension theorem, to at least begin discussing "percentages" when the ring itself is of finite measure? $\endgroup$ Dec 22, 2018 at 21:44
  • $\begingroup$ @Chickenmancer that wouldn't give a percentage of rings, I suspect you mean to give a percentage of zero divisors that are nilpotent? $\endgroup$
    – jgon
    Dec 22, 2018 at 21:50
  • $\begingroup$ Ah, I misinterpreted the question as "what percentage of a ring $A$." Thanks for clarifying, @jgon. $\endgroup$ Dec 22, 2018 at 22:15
  • $\begingroup$ Curious about the downvote. An explanation would be helpful. $\endgroup$ Dec 23, 2018 at 17:43

Matt Samuel already gave a good answer with both an example and some thoughts on why this should be rare. This answer intends to add a different perspective on the proportion of the rings with all zero divisors nilpotent. $\newcommand\AA{\Bbb{A}}\newcommand\PP{\mathbb{P}}\newcommand\Disc{\operatorname{Disc}}$

One perspective on what proportion of rings have no nonnilpotent zero divisors is the following. Note that this is really rough first thought.

First let's be careful though, rings with no zero-divisors trivially have this property, so I'll try to consider what rings with zero-divisors have the property that all zero-divisors are nilpotent.

Step 1: Set up a space to parametrize a nice class of rings

Let $k$ be an algebraically closed field. Let $n,d_1,d_2 > 0$. Let $x=(x_1,x_2,\ldots,x_n)$ be coordinates on $\AA^n_k$. Let $r_1=\binom{n+d_1}{d_2}$, which is the number of monomials in $k[x_1,\ldots,x_n]$ of degree at most $d_1$. Thus there are $r_1$ multiindices $I$ of degree at most $d_1$. Let $r_2=\binom{n+d_2}{d_2}$ as well. Let $c=(c_I)$ be coordinates on $\PP^{r_1-1}_k$, and let $d=(d_J)$ be coordinates on $\PP^{r_2-1}_k$ as $I$ ranges over multiindices of degree at most $d_1$ and $J$ ranges over multiindices of degree at most $d_2$.

Then consider the subvariety, $V$, of $\AA^n_k\times_k\PP^{r_1-1}_k\times_k \PP^{r_2-1}_k$ cut out by the polynomial $$f_{c,d}(x)=g_c(x)g_d(x):=\left(\sum_I c_Ix^I\right)\left(\sum_J d_Jx^J\right).$$

$V$ is equipped with a natural map $V\to \PP^{r_1-1}_k\times_k\PP^{r_2-1}$, and the fiber over a point $(a,b)$ in $\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$ is the subvariety of $\AA^n_k$ cut out by the degree at most $d_1+d_2$ reducible polynomial $$f_{a,b}(x)=g_a(x)g_b(x).$$

This subvariety of $\AA^n_k$ can be thought of as corresponding to the ring $k[x_1,\ldots,x_n]/(g_ag_b)$, so we can think of our variety $V$ as parametrizing a certain nice class of rings all (except for a Zariski closed subset where $g_a=0$ or $g_b=0$) of which are guaranteed to have zero-divisors, since $g_a$ and $g_b$ are zero divisors.

Step 2: Investigate the points corresponding to rings with the desired property and related properties

We can then ask if we can characterize the subset of $V$ corresponding to rings in which all zero-divisors are nilpotent.

Well, what are the zero-divisors in $k[x_1,\ldots,x_n]/(f_{a,b})$? Since $k[x_1,\ldots,x_n]$ is a UFD, handily enough, the zero-divisors are precisely the factors of $f_{a,b}$. Moreover $k[x_1,\ldots,x_n]/(f_{a,b})$ has nilpotents if and only if $f_{a,b}$ is not square free (take the radical of $f_{a,b}$, it will be nilpotent if and only if it is nonzero, if and only if $f_{a,b}$ is not square free). However the ring $k[x_1,\ldots,x_n]/(f_{a,b})$ satisfies our property that all zero-divisors are nilpotent if and only if $f_{a,b}$ is a power of an irreducible polynomial.

Step 3: Conclude

Note that if $f_{a,b}$ is square-free if and only if it is relatively prime to $f_{a,b}'$, which is true if and only if $\Disc(f_{a,b})\ne 0$. Thus every point $(a,b)\in\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$ corresponding to a ring with any nilpotents at all, let alone one in which every zero-divisor is nilpotent, satisfies a polynomial equation, $\Disc(f_{a,b})=0$. Thus rings in our parametrized class with any nilpotents at all correspond to a (proper) Zariski closed subset of $\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$, which is irreducible (see here).

Thus rings in our parametrized class satisfying your property (all zero-divisors are nilpotent) are contained in a codimension 1 class of rings (any nilpotents). Hence almost all rings in our parametrized class have zero-divisors that are not nilpotent.


This is really rough. The parametrized class of rings is a very specific, very nice subset of $k$-algebras. Nonetheless, hopefully it will give you (or other readers) intuition on why very few rings should have the property you want (as long as they have some zero divisors of course).

I needed to eliminate integral domains because I chose to parametrize hypersurfaces, and most hypersurfaces are irreducible. (Check out the link, Qiaochu Yuan gives a really nice quick proof of this fact). There's a reasonable chance that I wouldn't have needed to eliminate integral domains if I'd chosen e.g. codimension 2 subvarieties, but those are much harder to characterize.

  • $\begingroup$ I was thinking of excluding integral domains as well, but notice for integers that isn't necessary. A ring of integers modulo $n$ is an integral domain only if $n$ is prime. $\endgroup$ Dec 23, 2018 at 0:41
  • $\begingroup$ @MattSamuel, I originally didn't think to do so, but unfortunately, the particular class of rings I chose to parametrize requires me to do so, since most hypersurfaces in dimensions $\ge 2$ are irreducible. I suspect I wouldn't have to eliminate integral domains if I'd chosen e.g. codimension 2 subspaces, but those are harder to work with. $\endgroup$
    – jgon
    Dec 23, 2018 at 1:07
  • $\begingroup$ I'll add a link to a great answer by Qiaochu Yuan justifying the claim that most hypersurfaces are irreducible when I eventually get back on a computer and I remember. $\endgroup$
    – jgon
    Dec 23, 2018 at 1:08

Another class of examples are matrix rings over a field. A $n\times n$ matrix $A$ is a zero-divisor if and only if $A$ is not invertible.

Certainly if $A$ is a zero divisor then $A$ cannot be invertible. If $A$ is not invertible then $\ker A \ne 0$ and taking any non-zero matrix $B$ with $\operatorname{col}B \subseteq \ker A$ you have $AB = 0$. For example, let $v$ be a non-zero vector with $Av = 0$ and let $B$ be the matrix whose columns are all $v$.

The nilpotent matrices are rare among zero-divisors. Specifically, nilpotent matrices are solutions to the equation $A^n = 0$. As a rule, there are always more non-solutions to a polynomial equation than there are solutions.

In precise terms, we look at the zero set $\{A : \det A = 0\}$ (the set of zero-divisors). This zero-set is a manifold if the underlying field is $\mathbf{C}$ and is a variety in the general case. The non-nilpotents are dense in the Zariski topology looking over a general ring, and dense in the classical topology looking over $\mathbf C$.


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