It is a well-known result that if a ring $R$ satisfies $a^2=a$ for each $a\in R$, then $R$ must be commutative. See here for proof.

I am wondering whether the same result holds for finite rings if we only assume sufficiently many (but not necessarily all) elements of the ring are idempotents. Recall that an element $a\in R$ is called idempotent if $a^2=a$. For example, suppose $R$ is a finite ring in which at least $80$% elements are idempotents. Can we conclude that $R$ is commutative? More generally,

Does there exist an absolute constant $0<k<1$, such that whenever a finite ring $R$ satisfies $$\frac{\textrm{Number of idempotents}}{|R|}\ge k$$ then $R$ is commutative.

Motivation: We know that if every element of a group $G$ satisfies $a^2=1$, then $G$ must be abelian. This is a relatively easy exercise. However, it turns out that we only need 75% percent of elements to satisfy $a^2=1$ in order force $G$ to be abelian. See here. For reference, $a\in G$ is called involution if $a^2=1$.

  • $\begingroup$ Presumably $R$ is finite here. $\endgroup$ Jul 16, 2013 at 22:38
  • $\begingroup$ @QiaochuYuan: Yup, corrected. $\endgroup$
    – Prism
    Jul 16, 2013 at 22:40

2 Answers 2


Here is an example showing that $k > \frac{3}{4}$. Recall that given a ring $R$, a ring $S$, and an $(R, S)$-bimodule $M$, there is a triangular ring whose elements can be written formally as

$$\left[ \begin{array}{cc} r & m \\\ 0 & s \end{array} \right]$$

where $r \in R, m \in M, s \in S$ and multiplication is matrix multiplication. An element written as above is idempotent iff $r, s$ are idempotent and $m = rm + ms$.

Let $R = S = \mathbb{F}_2^n$ and let $M = \mathbb{F}_2$ with $R, S$ both acting on $M$ by multiplication by the first coordinate. The resulting triangular ring has $2^{2n+1}$ elements and is not commutative. $2^{2n}$ elements are diagonal, and all of these are idempotent. The non-diagonal matrices are idempotent iff their diagonal entries $r, s$ have different first coordinates. There are $2^{2n-1}$ such matrices, so in total

$$\frac{2^{2n} + 2^{2n-1}}{2^{2n+1}} = \frac{3}{4}$$

of the elements of this ring are idempotent.

Edit: Here are some quick observations. All rings are unital. (I wouldn't need to point this out except that the Wikipedia article doesn't make this assumption, so a cursory examination of that article will appear to contradict what I write below.)

Lemma: Every finite ring is a direct product of finite rings of prime power order.

Proof. Let $R$ be a finite ring. The underlying abelian group of $R$ is the direct product of its Sylow $p$-subgroups as a group. Let $a$ be an element of $p$-power additive order and let $b$ be an element of $q$-power additive order, where $p \neq q$ are distinct primes. Then $ab$ has both $p$-power and $q$-power additive order and hence $ab = 0$. On the other hand, if $p = q$ then $ab$ has $p$-power additive order. So $R$ is also the direct product of its Sylow $p$-subgroups as a ring. $\Box$

Since the density of idempotents in a finite ring is multiplicative under direct product, it follows that to look for rings where the density of idempotents is as large as possible it suffices to restrict our attention to rings of prime power order.

Recall that in any ring the smallest subring containing the unit (the prime subring) is $\mathbb{Z}/n\mathbb{Z}$ for some uniquely determined natural number $n$, the characteristic of the ring.

Lemma: $\mathbb{F}_p$ is the only ring of order $p$ for $p$ prime.

Proof. Let $R$ be a ring of order $p$. Then $R$ must have characteristic $p$. $\Box$

Lemma: $\mathbb{Z}/p^2\mathbb{Z}, \mathbb{F}_{p^2}, \mathbb{F}_p[x]/x^2, \mathbb{F}_p \times \mathbb{F}_p$ are the only rings of order $p^2$ for $p$ prime. In particular, every ring of order $p^2$ is commutative.

Proof. Let $R$ be a ring of order $p^2$. Then $R$ has characteristic either $p$ or $p^2$. In the latter case it can only be $\mathbb{Z}/p^2\mathbb{Z}$. In the former case it is an algebra over $\mathbb{F}_p$. Let $x \in R$ be an element not contained in the prime subring. Then the elements $ax + b, a \in \mathbb{F}_p^{\times}, b \in \mathbb{F}_p$ are also not contained in the prime subring (since if they were then so would $x$), from which it follows that every element of $R$ can be written uniquely in the form $ax + b$. In particular, $x^2 = ax + b$ for some $a, b$. This polynomial is either irreducible, reducible with two distinct roots, or reducible with a repeated root, which gives $R \cong \mathbb{F}_{p^2}, R \cong \mathbb{F}_p \times \mathbb{F}_p$, and $R \cong \mathbb{F}_p[x]/x^2$ respectively. $\Box$

Corollary: Let $n$ be a cubefree positive integer (a positive integer not divisible by the cube of any positive integer greater than $1$; equivalently, a positive integer whose prime factorization has only exponents of $1$ or $2$). Then every ring of order $n$ is commutative.

It follows that the smallest possible order of a noncommutative ring is $8$; above and in the comments the example $\left[ \begin{array}{cc} \mathbb{F}_2 & \mathbb{F}_2 \\\ 0 & \mathbb{F}_2 \end{array} \right]$ has this order. More generally, we should restrict our attention to rings of order $p^k$ for $p$ prime and $k \ge 3$.

Edit #2: More observations.

Lemma: Let $R$ be a ring of odd order. Then at most half of the nonzero elements of $R$ are idempotent.

Proof. Let $a$ be a nonzero idempotent element of $R$. Then $a^2 = a$, hence $(-a)^2 = a^2 = -a \neq a$ (since $a$ cannot have additive order $2$), hence $-a$ is not idempotent. $\Box$

Corollary: Let $R$ be a noncommutative ring of odd order. Then the proportion of idempotent elements of $R$ is at most $\frac{14}{27}$.

Proof. By the lemma, the proportion of idempotent elements of $R$ is at most

$$\frac{ \frac{|R| - 1}{2} + 1}{|R|} = \frac{1}{2} + \frac{1}{2|R|}$$

and by previous results, a noncommutative ring of odd order has order at least $3^3 = 27$. $\Box$

Since we already have an example where the proportion of idempotent elements is $\frac{3}{4}$ it follows that we should restrict our attention to rings of order $2^k$ for $k \ge 3$.

  • $\begingroup$ Thank you! That was ridiculously fast. That said, I shall need some time in understanding this answer. :) $\endgroup$
    – Prism
    Jul 16, 2013 at 22:55
  • $\begingroup$ @Prism: it might help to know that if a finite ring consists only of idempotents then it is isomorphic to $\mathbb{F}_2^n$ for some $n$. This is a simple case of Stone's representation theorem (en.wikipedia.org/wiki/…). $\endgroup$ Jul 16, 2013 at 22:57
  • $\begingroup$ @Prism: it might also help to specialize this example to $n = 1$, where it just gives the ring of upper triangular $2 \times 2$ matrices over $\mathbb{F}_2$. This ring has $8$ elements and $6$ of them are idempotent. $\endgroup$ Jul 16, 2013 at 22:59
  • $\begingroup$ Ah yes, that is an exercise I blindly did in Dummit & Foote a while ago. And, I really like the specialization; it gives a quick way to literally check this answer (in special case). Thanks again! (+1) $\endgroup$
    – Prism
    Jul 16, 2013 at 23:01

The other answer is getting large and difficult to edit; furthermore it turns out that the extra observations in the edits aren't necessary.

First let's introduce some useful terminology. The idempotent density of a finite ring $R$ is the number of idempotents in $R$ divided by $|R|$.

Theorem: Let $R$ be a finite ring with idempotent density strictly greater than $\frac{3}{4}$. Then $R$ is commutative, and in fact is isomorphic to $\mathbb{F}_2^n$ for some $n$.

Proof. Since more than $\frac{3}{4}$ of the elements of $R$ are idempotent, more than $\frac{1}{2}$ of the elements of $R$ have the property that $a$ and $a + 1$ are simultaneously idempotent. Hence $(a + 1)^2 = a^2 + 2a + 1 = a + 2a + 1 = a + 1$, so $2a = 0$. The set of elements satisfying $2a = 0$ is an additive subgroup of $R$ and hence has order dividing $|R|$, but a divisor larger than $\frac{|R|}{2}$ can only be equal to $|R|$. It follows that $R$ has characteristic $2$.

Suppose that $R$ has a nontrivial idempotent $a$ (not equal to $0$ or $1$). Consider its centralizer or commutant $\{ b \in R : ab = ba \}$. Since greater than $\frac{3}{4}$ of the elements of $R$ are idempotent, greater than $\frac{1}{2}$ of the elements of $R$ have the property that $b$ and $a + b$ are simultaneously idempotent. But then

$$(a + b)^2 = a^2 + ab + ba + b^2 = a + ab + ba + b = a + b$$

hence $ab + ba = 0$, hence $ab = ba$. It follows that the centralizer of $R$ has order greater than $\frac{|R|}{2}$. As above, since the centralizer is an additive subgroup, it also has order dividing $|R|$, so must be all of $R$. We conclude that $a$ is central.

But since $a$ is arbitrary, it follows that every idempotent in $R$ is central. Moreover, any central idempotent $a$ of a ring $R$ gives rise to a direct product decomposition

$$R \cong aR \times (1-a)R$$

and $aR$ and $(1-a)R$ inherit the property that every idempotent is central. Keep taking direct product decompositions until we have written $R$ as the direct product of indecomposable rings $R_i$ (rings not admitting a nontrivial direct product decomposition; equivalently, rings with no central idempotents). Since all of these rings also have the property that every idempotent is central, they have no nontrivial idempotents, hence have exactly $2$ idempotents. But since idempotent density is multiplicative under direct products and $R$ has idempotent density greater than $\frac{2}{3}$, each $R_i$ must have only $2$ elements, and the conclusion follows. $\Box$

(There is an arguably cleaner way to write this argument: assume $R$ is the smallest counterexample and then derive a contradiction, since either $aR$ or $(1-a)R$ is a smaller counterexample.)

  • $\begingroup$ The "thanks" is "in advance for pointing out what I'm overlooking." I just didnt' have enough space... $\endgroup$
    – rschwieb
    Jul 17, 2013 at 13:05
  • $\begingroup$ @rschwieb: it's elements for which $a$ and $a + 1$ are simultaneously idempotent. $\endgroup$ Jul 17, 2013 at 19:07
  • $\begingroup$ Urgh! Blind in the stupidest ways... thanks. $\endgroup$
    – rschwieb
    Jul 17, 2013 at 19:40
  • $\begingroup$ Dear Qiaochu, I think you have completely solved the problem! $\endgroup$
    – Prism
    Jul 18, 2013 at 15:50

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