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 $n=3$.

  • $\begingroup$ Playing around with $n = 3$ and assuming that $r$ and $s$ commute, I seem to be getting that $r = -s$. I'm not sure if you get something similar for all $n$, but it seems like forcing $r^n = n$ for all $r \in R$ causes you to have very unusual rings where each element is the negative of every other element. There should probably be some caveat that we must have $s \neq 0 \neq r$, but I wasn't paying much attention to that. $\endgroup$ – pjs36 Feb 18 '15 at 2:23
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    $\begingroup$ 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 $\endgroup$ – Alex Wertheim Feb 18 '15 at 3:02
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    $\begingroup$ 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. $\endgroup$ – vesszabo Feb 25 '15 at 17:56
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    $\begingroup$ Duplicate of: math.stackexchange.com/q/831124/15416, which also doesn't have an answer, but more links. $\endgroup$ – Julian Kuelshammer Mar 8 '15 at 9:53
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    $\begingroup$ 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 $\endgroup$ – Christian Lomp Jun 13 '18 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:

  1. $x^n = x$ implies commutativity, a universal algebraic proof?

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