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I need to show that a domain $R$ with a minimal left ideal is a division ring. Suppose that $I$ is a minimal left ideal, then take $a\in I\setminus \{0\}$ and consider the left ideal generated by $a$, that is, $Ra$. It is a subset of $I$ and by minimality of $I$ we get that $$ Ra=I $$ and $I$ is principal. Take $r\in R$ to be nonzero, then $ra\in I$ and $ra\neq 0$ since $r\neq 0, a\neq0$. Take $r' \in R$ and suppose that $$ ra=r'a $$ then $(r-r')a=0$ and hence $r-r'=0$, so that $r=r'$. I am not sure how to proceed from here.

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2 Answers 2

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Take any $x\in R, x\neq 0 ,$ then $R(xa) =Ra $ hence there exists $ v\in R $ such that $a=vxa $ and hence $$0=a-vxa =a(1-vx),$$ and since $R$ is domain we get $$1=vx.$$

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I think you didn't quite get the right angle, but you started on the right foot.

With your $Ra$, notice that $Ra=Ra^2$ is and think about what that entails.

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