Division ring if and only if it has no proper left ideals.

Let $R$ be an associative ring with $1$. Prove that $R$ is a division ring if and only if $R$ has no proper left ideals.

Clearly, if $R$ is a division ring and $I\neq\{0\}$ is a left ideal, then every nonzero $z\in I$ satisfies $1=z^{-1}z\in I$, hence $I=R$.

Now if $z\in R$ is nonzero and $R$ has no proper left ideals, then $I:=\{rz:r\in R\}$ is equal to $R$ and there exists $r\in R$ such that $rz=1$. However, we still need to prove $zr=1$. How can we do this?

Thank you.

$(rz)r=r(zr)=r$, thus $r(1-zr)=0$, $zr=1$.

Remark that the fact that $zr=1$ follow from the fact that $R$ does not have divisors of zero: if $ab=0$ $a,b\neq 0$ , since $Ra=R$, $ra=1$, $r(ab)=(ra)b=b=0$. Contradiction.

You have the side switched, but it's a minor detail.

If $r\ne0$, then $Rr=R$, so you find $z$ with $zr=1$. Next, also $Rz=R$, so there is $s$ with $sz=1$.

Now prove $r=s$.