I would be very grateful if you give me a hint on it:

Suppose $R$ is a finite commutative ring with identity such that $ x^3 = x $ for all elements $x$ of $R$. Then $R$ is a finite direct product of fields of order $2$ or $3$.

Should I try to incorporate Artin-Wedderburn?


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    $\begingroup$ Is that $R$ an Artinian ring ? $\endgroup$ – Hamou Jul 27 '14 at 23:31
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    $\begingroup$ Some hypothesis is missing. An infinite direct product of fields of order two is not a finite direct product of fields of order two and of order three. $\endgroup$ – Mariano Suárez-Álvarez Jul 27 '14 at 23:31
  • $\begingroup$ You're deff right; R is finite, which I had forgotten to include $\endgroup$ – Sean Jul 27 '14 at 23:48

I'd like to point out that the commutativity assumption can be removed. A theorem of Jacobson says that any ring with unit satisfying $a^{n(a)}=a$, with $n(a)>1$ an integer depending on $a$, is commutative. This is part of a large collection of ring theory results assuring commutativity, and can be found in Herstein's book Noncommutative Rings.

Indeed, the first step of the proof shows that $R$ is semisimple by considering $a(1-a^{a(n)-1})=0$. Thus we can apply Artin-Wedderburn to write $R$ as a sum of matrices over division rings. Commutativity implies the matrices are all $1\times 1$, so $R$ is a sum of division rings. Commutativity, or finiteness and an overpowered use of Wedderburn's theorem, implies the division rings are fields. The orders of the fields are then restricted by the specific relation $x^3=x$, as desired.

Indeed, we don't have to invoke Jacobson. We have semisimple by the same argument, Artin-Wedderburn applies, and $a^{n(a)}=a$ means we have no nilpotent elements. Any $n\times n$ matrix ring has nilpotents if $n>1$, so $R$ is a sum of division rings. By finiteness, Wedderburn's theorem applies to conclude the division rings are fields (whence $R$ is commutative).

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    $\begingroup$ Very interesting (although it might have been better as a comment to the question, not an answer). $\endgroup$ – Jeppe Stig Nielsen Jul 28 '14 at 9:38
  • $\begingroup$ @Jeppe Comments are erasable, and I believe it is contrary to the site intent to put useful information in unsearchable comments. As it didn't seem a germane addition to the other answers, here we are. $\endgroup$ – zibadawa timmy Jul 28 '14 at 9:43
  • $\begingroup$ @Jeppe After checking Herstein to be sure, the observation itself naturally yields a complete answer. I have edited in the details. $\endgroup$ – zibadawa timmy Jul 28 '14 at 10:13
  • $\begingroup$ @user The answer already notes you can get away without it, with essentially the same proof (except wedderburn is now useful). Though Ehsan's answer on your link, combined with an answer linked in another answer here, shows that in this case we can drop both the unit and commutativity assumptions, which is very interesting. Didn't think you could drop both. $\endgroup$ – zibadawa timmy Jul 29 '14 at 5:43

Since $R$ is a finite ring, it is Artinian, hence is a finite product of Artinian local rings. Moreover $x^3 = x$ for all $x$ implies $R$ is reduced ($0 = x^m \implies 0 = x^{3^k} = x$ for some $k$). Thus $R \cong \prod_{i=1}^n R_i$, where each $R_i$ is finite, local, and reduced, hence is a finite field. Then each $R_i$ also satisfies $x^3 = x$ for all $x \in R_i$, but the only finite fields satisfying this are $\mathbb{F}_2$ and $\mathbb{F}_3$ (recall that a finite field of order $p^n$ is the splitting field of $x^{p^n} - x$).

Edit: Since one of the anwers mentioned that commutativity can be dropped, let me just add that the hypothesis of a unity can also be dropped: see e.g. this question.

  • $\begingroup$ Why must finite local reduced rings be fields? $\endgroup$ – Nishant Jul 28 '14 at 1:27
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    $\begingroup$ @Nishant: More generally, $0$-dimensional local reduced rings are the same as fields, since $0$-dimensional + local $\iff$ unique prime ideal, and reduced $\iff$ intersection of all primes is $0$ $\endgroup$ – zcn Jul 28 '14 at 1:28
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    $\begingroup$ +1 since this method also works for other equations of the type $x^n=x$. I've worked on this a couple of years ago and have found some cute equivalences of categories after dropping the finiteness condition (generalizing the well-known case $n=2$). $\endgroup$ – Martin Brandenburg Jul 28 '14 at 10:31

For any $x$, we have that $x^2$ is idempotent. Pick an $x \neq \pm 1, 0$ (if none exists we're done) and decompose $R$ into the direct sum of $x^2R$ and $(1-x^2)R$. These are smaller finite rings (caution: the unit element of these rings is not the $1$ of the big ring) so this process cannot continue forever and so you win.


One remark not in any of the other answers: although you don't need to observe it directly (it comes out later in the other arguments presents), you can notice that $2^3 = 2$, and hence that $6 = 0$, in $R$, so that by the CRT, the ring $R$ is a product of an $\mathbb F_2$-algebra and an $\mathbb F_3$-algebra.

This doesn't particularly simplify the rest of the arguments, but it might simplify your conceptualization of the problem and its solution, since it makes it clear from that the beginning that $R$ is the product of something of char. $2$ and something of char. $3$.


Here is a collection of facts which may or may not serve as hints:

  1. Every prime ideal of $R$ is maximal
  2. The zero ideal is the intersection of all the prime ideals in $R$
  3. $R \simeq R / (0)$


As requested, I'll expand this into a more complete answer.

Since $R$ is finite, for any prime ideal $\mathfrak{p}$, $R / \mathfrak{p}$ is a finite integral domain, and thus is a field, and so $\mathfrak{p}$ is maximal. In fact, since every element of $R/\mathfrak{p}$ satisfies $x^3 - x = 0$, $R/\mathfrak{p}$ is a field with at most 3 elements, and so is either $\Bbb{F}_2$ or $\Bbb{F}_3$.

The condition that $x^3 = x$ prevents $R$ from having a nilradical that isn't the zero ideal. Since $R$ is finite, there are only a finite number of prime ideals, say $\mathfrak{p}_1, \dots, \mathfrak{p_n}$. It follows that $$ \prod_{i=1}^n \mathfrak{p}_i = \bigcap_{i=1}^n \mathfrak{p}_i= (0) .$$

Since each $\mathfrak{p}_i$ is maximal, the prime ideals are pairwise comaximal, so the Chinese Remainder Theorem gives $$R \simeq R/(0) \simeq \prod_{i=1}^n R/\mathfrak{p}_i.$$

The comment before shows that the terms in the product are fields of order 2 or 3.

  • $\begingroup$ i am trying to solve a similar problem, with the same hints that you mentioned here, could you explain some more about $\endgroup$ – Dimitri Aug 13 '14 at 7:49
  • $\begingroup$ I should add that this argument was adapted from a similar one in Paul Pollack's Paper "On Polynomial Rings With a Goldbach Property" which is attributed to Pierre Samuel. $\endgroup$ – Dan Z Aug 13 '14 at 21:26

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