If a non-trivial ring $R$ has a unique maximal left ideal $J$ , then $J$ is two-sided and is also the unique maximal right ideal in $R$.
I can prove that it is two sided, but I can't prove that it is unique.
Let $r \in R$. Then $Jr$ is a left ideal. If $Jr = R$, then $jr = 1$ for some $j \in J$. Note that $rj \in J$ since J is a left ideal, so $rj \neq 1$. Using a lemma, $1 - rj$ is not left invertible, which is to say that $R(1-rj) \neq R$. But then R(1-rj) is contained in a maximal left ideal, i.e. $1-rj \in R(1-rj) \subseteq J$, so $1= rj +(1-rj) \in J$, which gives a contradiction. Hence $Jr \neq R$. This implies that $Jr$ is contained in $J$. Since the choice of $r$ is arbitrary, $jr \in J$ for all $j \in J$ and $r \in R$. Hence, $J$ is a two sided ideal.
Can someone tell me how I can prove uniqueness?