Two sided ideals are maximal right ideal iff they are maximal left ideal. Let $R$ be a ring with unity and $I$ be a two-sided ideal in $R$. Then $I$ is a maximal right ideal if and only if it is a maximal left ideal.
Would anyone give me an idea to prove the statement? Thanks.
 A: Suppose $R/I$ only has trivial right ideals. Then it has only trivial left ideals. For if $x$ is a nonzero member of $R/I$, $x(R/I)=R/I$, and $x$ is right invertible, say by element $y$ similarly $y$ is right invertible, say by element $z$, but it is any easy exercise to prove $x=z$, so $x$ is a unit and $R/I$ is a division ring, and therefore only has trivial left and right ideals.
By a symmetric argument, the words left and right can be interchanged.
A: Hint: Never mind. As rschweib points out in the comments below, this doesn't appear to lead anywhere.  The only somewhat dubious advantage to be gained from trying it would appear to be in showing you that you're better off trying something else. Apologies to anyone whom I sent off on a wild goose chase.
If $\ J\ $ is a one-sided ideal in $\ R\ $ containing $\ I\ $, let $\ j\ $ be an arbirary member of $\ J\ $, and consider the set $\ K=\left\{\,x \in J \mid\,j\,x\,j \in J\,\right\} $.


*

*Is $\ j \in K\ $?

*Can you determine whether $\ K\ $ is any sort of ideal ?

*Can you determine what the intersection $\ K\cap I\ $ is ?

