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The question is as follows.

Let $ R $ Be a ring such that $ x^3 = x $ for every $ x \in R $. Prove that $ R $ is commutative

I proved it as follows.

$1.$ Since $ a^3 = a $ for every a $ \in R$, $ a^3 = a^2\cdot a = a \cdot a^2 = a $. So the square of any element in R is an identity of the ring.
$2.$ Let a , b $ \in R $. Then $ (ab)^3 = ab = ab \cdot (ba)^2 = abbaba = ab^2aba = a^2ba = ba.$ So $ab = ba $ for all $ a, b \in R $

Is my proof is correct? I'm not sure I can say that the square of any element in R is an identity of the ring if $ a^3 = a^2\cdot a = a \cdot a^2 = a $ for every a $ \in R$

rschwieb
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alryosha
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    $a^3=a$ implies $a^2=1_R$ only for units $a\in R$, i.e., elements $a\in R$ which have a multiplicative inverse. Also, you don't necessarily have $1_R$ in $R$ (the usual definition of rings include rngs as well). – Prasun Biswas Mar 29 '18 at 01:45
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    If you would like to contribute your attempted solution as a solution you can add it to the duplicate. Also next time use a better title. – rschwieb Mar 29 '18 at 02:36
  • @Prasun Biswas Could you elaborate more why $ a^3=a $ implies $ a^2=1_R $ only for units $a \in R? $ Doesn't $ a^3=a $ imply $ a^2=1_R $ even if R is a ring with identity? – alryosha Mar 29 '18 at 03:00
  • @alryosha Haven't you tried with the ring of $2\times 2$-matrices? Why should $A^3=A$ imply that $A^2=I_2$ for a singular matrix $A$? – Dietrich Burde Mar 29 '18 at 09:44

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