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This question already has an answer here:

I am stuck in a question for Algebraic Structures:

I need to prove the following:

Let $R$ be a ring. If $a^2=a$ for any $a\in R$, then $a+a=0$ for any $a\in R$.

(of course $+$, $\cdot$ and $0$ regarding $R$)

P.S. This is similar to a different question I asked (if $a \cdot a = 0$ then $a + a = 0$) but I wasn't able to apply the ideas from there. In that question we assumed that $R$ is a unity ring.

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marked as duplicate by Stella Biderman, Namaste, C. Falcon, Daniel W. Farlow, user223391 Mar 21 '17 at 17:47

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  • $\begingroup$ Not true in the integers modulo $4$. $1*1=1$, but $1+1=2$ $\endgroup$ – Mastrem Mar 20 '17 at 20:12
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    $\begingroup$ I think what is meant is: if $R$ is a ring such that $a^2=a$ for all $a$, then $a+a=0$ for all $a\in A$. Is this true? $\endgroup$ – Aweygan Mar 20 '17 at 20:13
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    $\begingroup$ The statement is unclear to me. Who is $a$ in $a+a=0$ ? Is it all those that satisfy $a*a=a$ ? Is it, if for any $a$ one has $a*a=a$ then for any $a$ one had $a+a=0$ ? $\endgroup$ – James Well Mar 20 '17 at 20:16
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    $\begingroup$ Oh, I am sorry! Yes, it is different (but with almost the same proof). $\endgroup$ – Dietrich Burde Mar 20 '17 at 20:18
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    $\begingroup$ @DietrichBurde and whoever else voted to close, I would recommend you retract your close votes, so no one else makes the same mistake. $\endgroup$ – Aweygan Mar 20 '17 at 20:26
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Let $a\in R$ be given. Then $$a+a=(a+a)^2=a\cdot a+a\cdot a+a\cdot a+a\cdot a=a+a+a+a.$$ Adding $-(a+a)$ to both sides yields the result.

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