# If there is a unique left identity, then it is also a right identity

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.

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

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.
• Man, I'm feeling so stupid. We solved an exercise about $be - b + e$ like, $10$ minutes ago. Thanks a lot. – Ivo Terek Aug 10 '15 at 15:43