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 Answer
$\begingroup$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.
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