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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?

Xam
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Bryan Shih
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    It follows from associativity and the identity axioms. Just as how you may define a group with lefthanded axioms only and it is equivalent to the ordinary definition. – Randall Sep 27 '17 at 03:19
  • Sorry, I tried to work from the axioms but did not get anywhere, do you mind writing it out? – Bryan Shih Sep 27 '17 at 03:21
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    https://math.stackexchange.com/questions/65239/right-identity-and-right-inverse-implies-a-group – Randall Sep 27 '17 at 03:24

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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.

rschwieb
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