# Most challenging problem from I. N. Herstein's Algebra Book

1. Apr 7, 2010

### karthikvs88

Let R be a ring in which x3=x for every x in R. Prove that R is a commutative ring.

This is (word-for-word) in Herstein, Topics in Algebra, Ch. 3 sec. 4, problem 19. Apparently, Herstein commented that this one problem generated more mail than the entire remainder of the book.

The proof (due to Derek Holt) I found is as follows:

An element x is called central if xy=yx for all y in R.
Note that the central elements form a subring of R.

1. xy = 0 => yx = 0.
(Proof: yx = (yx)^3 = y (xy)^2 x = 0.)

2. x^2 = x => x central.
(Proof: x(y - xy) = xy - x^2y = xy-xy=0, so (by 1) (y - xy)x = 0,
and yx = xyx.
Similarly, (y - yx)x = 0 => x(y - yx) = 0 => xy = xyx.)

3. x^2 is central for all x in R. (by 2, because (x^2)^2 = x^4 = x^2).

4. If x^2 = nx for an integer n, then x is central.
(Proof: x = x^3 = qx^2, which is central by 3.)

5. x + x^2 is central for all x in R.
(Proof: By 4, because (x + x^2)^2 = 2(x + x^2).)

6. By 3 and 5, x = (x + x^2) - x^2 is central, completing the proof.

This proof can the found by scrolling to the very end of this page:
http://www.math.niu.edu/~rusin/known-math/99/commut_ring [Broken]

I understand all the steps except for 4. How did he conclude that x is central in step 4?

Last edited by a moderator: May 4, 2017
2. Apr 8, 2010

### rasmhop

Suppose x is central and n is a positive integer. Then for any other element y we have xy=yx and therefore:
\begin{align*} (nx)y&=(x+x+\cdots+x)y \\ &= xy+xy+\cdots+xy \\ &= yx + yx +\cdots + yx \\ &= y(x+x+\cdots+x) \\ &= y(nx)\end{align*}
which shows that nx is also central. The same can be shown to hold if n is negative or 0. Thus if x is central, then so is nx. This may be considered well-known by the author of the proof which is probably why he left it out.

In 4 you know that x^2 is central, and therefore qx^2=x is also central (q=n is an integer).