Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.