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The question is motivated by the following multiple-choice problem:

If $R$ is a ring with the property that $r=r^2$ for each $r\in R$, which of the following must be true?

I. $r+r=0$ for each $r\in R$.
II. $(r+t)^2=r^2+t^2$ for each $r,t\in R$.
III. $R$ is commutative.

Here are my questions:

  • What theorems do I need to solve the problem above?
  • Why is a ring with the property that $r=r^2$ for each $r\in R$ so special? Is there a name for such rings?
  • 1
    It is a theorem of Jacobson that a ring for which $ \rm x^n=x $ for all $\rm x \in R$ must be commutative. Even more generally, if there exists an integer $\rm n(x)$ for each $\rm x\in R$ such that $\rm x^{n(x)}-x \in Z(R),$ then $\rm R$ must be commutative. This general result is much more difficult to prove than your special case though.2011-10-11

1 Answers 1

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Trivially, II holds, since $r^2+t^2 = r+t = (r+t)^2$ by the given condition.

I is a bit harder to see, but it is also true: $r+r = (r+r)^2 = (r+r)(r+r) = r^2+r^2+r^2+r^2 = r+r+r+r,$ so cancelling we get $r+r=0$.

And from this you get III as well: for any $a$ and $b$, $a+b = (a+b)^2 = a^2 + ab+ba+b^2 = a+ab+ba+b.$ Cancelling you get $ab+ba=0$. Since $ab+ab=0$ as well, we conclude that $ba=-ab=ab$, so $R$ is commutative.

No theorems, just some manipulations.

  • 7
    Arturo's proof is obviously correct but I remember a "cleaner trick" for proving $r+r=0$ for all $r\in R$; here it is: if $r\in R$, then $r=r^2=(-r)^2=-r$! The idea is simple: in a Boolean ring, an element is determined by its square.2011-10-11