# Show that $I$ is an ideal

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?

• Suppose no, then there exist $r\in R$ and $x\in I$ such that $rx\notin I$. Consider $\langle I,rx\rangle$... – MrSelberg Apr 10 '16 at 0:56
• Then we have that $I\subseteq \langle I, rx\rangle$, right? To find a contradiction we use the fact that $I$ is the only maximum right ideal of $R$, or not? @MrSelberg – Mary Star Apr 10 '16 at 9:55

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.
• Why is $ir$ an idempotent element?  Also, could you explain to me your last paragraph? – Mary Star Apr 10 '16 at 13:44
• @MaryStar if you know what an idempotent element is then I should not have to explain why $ir$ is idempotent. If you ask a more specific question about the last paragraph I may respond. – rschwieb Apr 10 '16 at 14:54
• @MaryStar There might be another way, but it seems like this is the simplest path. An element $e$ is called idempotent if $e^2=e$. – rschwieb Apr 11 '16 at 2:43
• Ah ok... What does the expression $R=irR\oplus(1-ir)R$ ? Which operation is this? – Mary Star Apr 11 '16 at 21:42