TITLE: Domain with a minimal left ideal is a division ring.
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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 $$1=vx.$$