Show that $I$ is an ideal

Question

Let $R$ be a ring and $I\subseteq R$ the only maximal right ideal of $R$. I want to show that $I$ is an ideal.

To show that $I$ is an ideal, we have to show that $I$ is a left ideal, right?

How could we show that?

Could you give me some hints?

Answer 1

First off, $rI$ is certainly some right ideal, and we hope it is contained in $I$ for al $r$ so that it's a left ideal.

So you're looking for a reason that $rI\neq R$. If to the contrary $rI=R$, then $ri=1$ for some $i$. Now $ir$ is an idempotent element, so $R=irR\oplus(1-ir)R$ (just as you have for any idempotent element.)

Considering there is only one maximal right ideal, the summands have to be a trivial pair. If $irR=\{0\}$, then $r=rir=0$ contradicts $rI=R$. If $(1-ir)R=\{0\}$, then $ir=1$ and $i$ is a unit, contradicting the properness of $I$. So by contradiction, we do have after all that $rI\subseteq I$ for all r.