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 $ere\neq 0$ 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.