Every proper ideal contained in a maximal ideal?

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.

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Well, if commutativity is not used in the proof... – anon Nov 29 '12 at 5:45
You should be careful when talking about ideals in noncommutative rings: every proper left resp. right resp. two-sided ideal is contained in a maximal left resp. right resp. two-sided ideal. – Qiaochu Yuan Nov 29 '12 at 6:05

1 Answer

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.

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