Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A ring $R$ is a Boolean ring if $x^2=x$ for all $x\in R$. By Stone representation theorem a Boolean ring is isomorphic to a subring of the ring of power set of some set.

My question is what is an example of a ring $R$ with $x^3=x$ for all $x\in R$ that is not a Boolean ring? (Obviously every Boolean ring satisfies this condition.)

share|cite|improve this question
Would $\mathbb{Z}_3$ work? – Isaac Jan 6 '11 at 18:04
You are right. I should have thought about that before I ask. Note that the same question can be asked for $x^n=x, n\ge 4$. – TCL Jan 6 '11 at 18:13

It is a theorem of Jacobson that any ring whose elements satisfy $x^n = x$ for a particular $n$ is commutative. For $n = 3$, if we restrict ourselves to examples of characteristic $3$ then I suspect all examples are given by the ring of continuous functions $T \to \mathbb{F}_3$ for a Stone space $T$ (I think you can more or less go through the proof of Stone's representation theorem). I do not know whether Mariano's example has characteristic $3$ or not.

Edit: some details. If $P$ is a prime ideal of a ring $R$ all of whose elements satisfy $x^n = x$, then $R/P$ is an integral domain with the same property. This implies that $R/P$ is a finite integral domain, hence a finite field $\mathbb{F}_q$. Hence all prime ideals of $R$ are maximal (so $R$ is zero-dimensional). Moreover, since $R$ has no nilpotents by assumption, its nilradical is zero, and since its nilradical coincides with its Jacobson radical, its Jacobson radical is zero. (So elements of $R$ are faithfully represented as functions on $\text{Spec } R$.)

The elements of any residue field $\mathbb{F}_q$ are all the roots of $x^q = x$, so we have $q - 1 | n - 1$. If we restrict to $n = p$ prime and require that $R$ has characteristic $p$, then $\mathbb{F}_p$ is the only possible residue field. It follows that $R$ embeds into the ring of continuous functions $\text{Spec } R \to \mathbb{F}_p$, where $\text{Spec } R$ is a Stone space, and from here (following, for example, the approach used in this blog post) I think we can show that $R$ is isomorphic to this ring.

More generally, $\text{Spec } R$ can be divided into subspaces according to residue field, and then $R$ embeds into a product of rings of the above form for the various possible $p$ (prime power instead of prime). I don't know if every example is of this form.

share|cite|improve this answer
If my examples are all of that form, then all finite examples are of that form. – Mariano Suárez-Alvarez Jan 6 '11 at 18:33
@Mariano: retracted. I don't know whether your examples have characteristic 3. – Qiaochu Yuan Jan 6 '11 at 18:34
Since $\mathbb Z_6$ also satisfies the identity, the characteristic is not $3$. Notice that $\gcd\{k^3-k:k\in\mathbb N \}=6$, so I guess the characteristic is $6$. – Mariano Suárez-Alvarez Jan 6 '11 at 18:37
Amazingly, brute force computation and the OEIS suggest that the sequence given by $\gcd\{k^n-k:k\in\mathbb N\}$ is, and for prime exponents, the list of denominators of Bernoulli numbers. Cute. – Mariano Suárez-Alvarez Jan 6 '11 at 18:43
@Mariano: ah, right. Well, Z_6 is the product of an example of characteristic 2 and an example of characteristic 3. It's not clear to me whether your example has this property. – Qiaochu Yuan Jan 6 '11 at 18:46

Actually, not only would $\mathbb{Z}_3$ work, but it's the only solution that's an integral domain not of characteristic 2 (since, in such a case, $x^3-x=0\,\Rightarrow\,x\in {0,1,-1}$).

Another solution would be $R:=\mathbb{Z}_3[\mathbb{Z}_2]$, the group ring of $\mathbb{Z}_3$ over $\mathbb{Z}_2$ (ie the ring of "polynomials" over $\mathbb{Z}_3$, except that exponents are in $\mathbb{Z}_2$). To see why, given an element $f(x)\in R$ with

$f(x)=a_0+a_1 x^{b_1} + \cdots + a_n x^{b_n}$

with $a_i\in \mathbb{Z}_3$ and $b_j \in \mathbb{Z}_2$. Since $R$ is a ring of characteristic 3, the Freshman's Dream implies that

$(f(x))^3=a_0^3+a_1^3 x^{3b_1} + \cdots + a_n^3 x^{3b_n} = a_0 + a_1 x^{b_1} + \cdots + a_n x^{b_n}=f(x)$.

In fact, by the same argument, if you're given any ring $T$ of characteristic 3 with $x^3=x$ for all $x\in T$, then $T[\mathbb{Z}_2]$ satisfies this property as well.

I can't think of another class of examples off the top of my head, but I'd be surprised if boolean rings and the class of examples above were the only examples of rings of this type.

EDIT: Yes, the solution of modding out a free algebra by an appropriate ideal would also work nicely.

share|cite|improve this answer
T[Z_2] is isomorphic to T x T. More generally the direct product of any collection of examples is an example. – Qiaochu Yuan Jan 6 '11 at 18:29
Right you are. Good catch! – user5137 Jan 6 '11 at 18:32

You can always pick a set $X$, consider the free $\mathbb Z$-algebra $A=\mathbb Z\langle X\rangle$, and divide by the ideal generated by all the elements $x^3-x$ for $x\in A$ to que a ring $B_X$, no?

(Since the quotient is going to be commutative, you can also start from the polynomial ring...)

To show that the resulting quotient is non-trivial, consider the many maps $B_X\to\mathbb Z_3$, which you get from functions $X\to\mathbb Z_3$. Indeed, for every example of a ring $R$ satisfying the condition, you can pick an $X$ and a surjective map $B_X\to R$. So these examples are universally complicated :)

Later: Notice that $\mathbb Z_6$ is an example of a ring satisfying the identity, so $B_X$ is not a product of copies of $\mathbb Z_3$.

share|cite|improve this answer
Yes, but you need to show that the result isn't the trivial ring (which is Boolean!). – Qiaochu Yuan Jan 6 '11 at 18:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.