TITLE: If there is a unique left identity, then it is also a right identity
QUESTION [15 upvotes]: 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.

REPLY [24 votes]: 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.