# Proving a ring in which $r^n=r$ for all $r$ is commutative.

Let $$R$$ be a ring with identity such that there is a positive integer $$n\geq 2$$ for which $$r^n=r$$ for all $$r\in R$$. Prove $$R$$ is commutative.

I had proven before that If $$n=2$$ it is commutative as follows:

$$r+s=(r+s)^2=r^2+rs+sr+r^2=r+rs+sr+s\implies 0=rs+sr\implies sr=-rs$$

On the other hand $$-r=(-1)r=(-1)^2r=r$$.

So $$sr=-rs=rs$$ as desired.

I seem to be stumped even with $$n=3$$.

• @pjs36 I think you've probably made a mistake. If $n$ is prime, the stated property certainly holds for the prime field $\mathbb F_n$, for example. I think the solution could involve number theory as one can show that the characteristic of $R$ divides $k^n-k$ for each integer $k\ge1$. My number theory isn't strong enough to go further, however. Feb 18, 2015 at 2:41
• This question has been asked here in various forms, though in full generality I think this is a bit difficult for a homework problem. Perhaps you'll find this link helpful: math.ucla.edu/~ggim/W14-110BH.pdf Feb 18, 2015 at 3:02
• See the answer in math.stackexchange.com/questions/360958/… . I have no the book of Herestein. However, in this special case might exist a simple proof. Please, ask your professor to show the solution of this homework, and share the ideas with us. Feb 25, 2015 at 17:56
• Duplicate of: math.stackexchange.com/q/831124/15416, which also doesn't have an answer, but more links. Mar 8, 2015 at 9:53
• Here is one more link, namely Herstein has a paper entitled "An elementary proof of a theorem of Jacobson", where he proves it on 4 pages: projecteuclid.org/download/pdf_1/euclid.dmj/1077465581 Jun 13, 2018 at 9:41

The proof is found in the literature but is perhaps too long for an answer here on math.SE. We have long suffered from a lack of a canonical answer for this question, so this is an attempt at an approximation to one.

Even more is true:

If for each $$x$$ there exists an integer $$n(x)>1$$ such that $$x^{n(x)}-x$$ is in the center of $$R$$, then $$R$$ is commutative.

There are a few proofs in the literature, ones which are probably too long for a post here.

You should consult your local library to obtain a copy of one of these (probably preferring the newest one you can get):

N. Jacobson, "Structure theory for algebraic algebras of bounded degree", Ann. of Math. 46 (1945), 695–707.

Herstein, I. N. "An elementary proof of a theorem of Jacobson", Duke Mathematical Journal 21.1 (1954): 45-48.

Rogers, Kenneth. "An elementary proof of a theorem of Jacobson", Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg. Vol. 35. No. 3. Springer Berlin/Heidelberg, 1971.

A proof of an even more general nature is also discussed as Theorem 3.2.3 in:

Herstein, Israel Nathan. Noncommutative rings. Vol. 15. American Mathematical Soc., 1994.

There are also useful related questions on this site:

• The easiest proof appears in "Elementary proofs of a theorem of Wedderburn and a theorem of Jacobson" by Takasi Nagahara, Hisao Tominaga. link.springer.com/article/10.1007%2FBF02993501 Feb 16, 2020 at 1:22

The most elementary proofs of Jacobson's Theorem (with a fixed exponent $$n$$) were given by Wamsley, Dolan and Nagahara-Tominaga:

• J. W. Wamsley, On a condition for commutativity of rings, J. Lond. Math. Soc. 2.2 (1971), 331-332.
• S. W. Dolan, A proof of Jacobson's Theorem, Canad. Math. Bull. 19.1 (1976), 59-61.
• T. Nagahara, H. Tominaga. Elementary proofs of a theorem of Wedderburn and a theorem of Jacobson, Abh. Math. Semin. Univ. Hambg. 41 (1974), 72-74.

In contrast to other proofs, they do not use the axiom of choice. Unfortunately, they are still not constructive, in particular not equational.

I have recently found a method how to find an equational proof for every $$n$$: Equational proofs of Jacobson's Theorem. The rough idea is to prove the reduction to prime characteristic and then to $$n=p^k$$ in a constructive way. Then, the (more or less equivalent) proofs by Dolan and Nagahara-Tominaga, which reduce the Theorem to Wedderburn's Theorem, can be made constructive as well. The paper discusses several examples. For example, every ring with $$x^{2023}=x$$ is commutative by an equational proof. This is just an example of infinitely many examples.