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Let $(R,+,\cdot)$ be a ring, and $e \in R$ be an element such that $ea=a$ for all $a\in R$. I'm trying to prove that if $e$ is unique with this property, then $ae=a$ for all $a\in R $.

So far I have $e^2 = e$ (using uniqueness), but I am stuck. I saw a proof of this fact for groups which used the existence of inverses, which we don't have here. I wonder if the result is really true here. Can someone help? Thanks.

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  • $\begingroup$ Why do you need uniqueness to prove $e^2=e$? $\endgroup$ – Daniel Aug 10 '15 at 15:35
  • $\begingroup$ $e^2a = e(ea) = ea = a$. By uniqueness, $e^2= e$. $\endgroup$ – Ivo Terek Aug 10 '15 at 15:37
  • $\begingroup$ Or simply let $a=e$, isn't it? $\endgroup$ – Daniel Aug 10 '15 at 15:37
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    $\begingroup$ What Solid Snake means is that you get this for free by setting $a=e$. $\endgroup$ – N. S. Aug 10 '15 at 15:37
  • $\begingroup$ Oh, crap. So uniqueness must come in somewhere else $\endgroup$ – Ivo Terek Aug 10 '15 at 15:38
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Let $b \in R$. Then
$$(be-b+e)a=a \forall a \in R$$

By the uniqueness you get $$be-b+e=e$$

As $b \in R$ is arbitrary, you are done.

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  • $\begingroup$ Man, I'm feeling so stupid. We solved an exercise about $be - b + e $ like, $10$ minutes ago. Thanks a lot. $\endgroup$ – Ivo Terek Aug 10 '15 at 15:43
  • $\begingroup$ Amazing answer, I think. $\endgroup$ – Daniel Aug 10 '15 at 15:45
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    $\begingroup$ Very nice. You have to use addition because an associative monoid can have a unique left identity and no right identity. $\endgroup$ – DanielWainfleet Aug 10 '15 at 16:49

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