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.

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.

• Why is "$R/I$ only has trivial right ideals implies that it has the only trivial left ideals" true? – bozcan Jan 22 '19 at 13:22
• @bozcan Literally everything I wrote is dedicated to proving that. – rschwieb Jan 22 '19 at 13:57
• If there is any further question about that, there are numerous posts to read about it, such as math.stackexchange.com/questions/1151319/… – rschwieb Jan 22 '19 at 14:23

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 ?
• Is there a typo? Because as written, $J=K$ and the hint doesn’t seem to go anywhere. – rschwieb Jan 22 '19 at 10:20
• Not a typo. A blunder, rather, in my supposed proof that $\ K\$ had to be two-sided. The idea was to find a description of $\ J\$ which it was not too difficult to show to be two-sided. If such a thing is possible, it appears to be much more difficult than I had initially hoped, and $\ K\$ certainly doesn't qualify. – lonza leggiera Jan 22 '19 at 21:33