Let $\displaystyle R$ be a ring with a proper left ideal $\displaystyle M$ such that, for any $\displaystyle r \in R$, either $\displaystyle r \in M$ or $\displaystyle r$ is a unit.

(a) Prove that if $\displaystyle L$ is a left ideal which contains $\displaystyle M$ then either $\displaystyle L = R$ or $\displaystyle L = M$.

(b) Prove that every proper left ideal of $\displaystyle R$ is contained in $\displaystyle M$.

For (a) we have $\displaystyle M \subseteq L$ so we need to show that we can take an arbitrary $\displaystyle x \in L$ and rearrange it so it looks like an element of $\displaystyle M$. So if we take $\displaystyle x \in L$ then $\displaystyle x = rl$ for some $\displaystyle r \in R$ and $\displaystyle l \in L$ (because L is a left ideal). I'm not too sure how to proceed from here... how can we relate $\displaystyle L$ with $\displaystyle M$?

Thanks,