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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.

Ivo Terek
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1 Answers1

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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.

N. S.
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