TITLE: Every proper ideal contained in a maximal ideal?
QUESTION [7 upvotes]: This is a true in a commutative ring with $1$, but does it also hold in a noncommutative ring with $1$?   The proof in my book is just an application of Zorn's lemma, but the commutativity of the ring is not used anywhere.

REPLY [6 votes]: The theorem is known as the Krull's Theorem and stated in its complete form it says:
Let $R$ be a ring with identity, and let $I$ be a (left, right, two-sided) ideal of $R$ that is distinct from $R$. Then there exists a maximal (left, right, two-sided) ideal of $R$ containing $I$.
The proof is similar to the standard proof given when $R$ is commutative.