# Problem of the Week #208 - May 24, 2016

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#### Euge

##### MHB Global Moderator
Staff member
The following problem is a nice exercise that both the undergrad and grad members here can enjoy:

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Let $R$ be a ring such that for all $r\in R$, $r^3 = r$. Prove $R$ is commutative.

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#### Euge

##### MHB Global Moderator
Staff member
Surprisingly, no one answered this week's problem! You can read my solution below.

First, I'll show that the squares in $R$ commute with every element of $R$. Let $r,s\in R$. Let $t = r^2$. Then $t^2 = r^4 = r^3r = rr = t$, and so $$(st - tst)^2 = stst - st^2st - tstst + tst^2st = stst - stst - tstst + tstst = 0$$ and $$(ts - tst)^2 = tsts - tstst - tst^2s + tst^2st = tsts - tstst - tsts + tstst = 0$$ Therefore, $0 = (st - tst)^3 = st - tst$ and $0 = (ts - tst)^3 = ts - tst$, whence $st = tst = ts$.

Now given $a,b\in R$, we have
$$ab = a^3b = a^2(ab) = aba^2 = ab^3a^2 = (ab)b^2a^2 = b^2aba^2 = b(ba)^2a = (ba)^2ba = (ba)^3 = ba$$
Hence, $R$ is commutative.

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