Proof of brauer's lemma, $eRe$ being a division ring. On page 1 of this article, the author proves the following claim: > Brauer's Lemma: Let $K$ be a minimal left ideal of a ring $R$, with $K^2 \not= 0$. Then $K=Re$ where $e^2=e \in R$, and $eRe$ is a division ring. **What I do not understand** is why $eRe$ is a division ring. They showed for $b\not=0$ _in_ $eRe$ exists $(ere)b=e$ (where $e$ is the identity in this ring). This shows left invertibility, but not right. What am I missing?

As suggested, then $ere\
eq 0$ has a left inverse too, which consequently you will show is equal to $b$, so that $ere$ and $b$ are mutually inverse.

Another way to do it is to show that $End_R(Re, Re)\cong eRe$, and if you're familiar with Schur's lemma, that makes it obvious $eRe$ is a division ring too.