Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I am sorry, this may not be a good question here. I am looking a good reference about when the ideal $I$ of a given commutative ring $R$ (local or may not be local) with identity is a projective module.

share|cite|improve this question

Here are some partial results if you strengthen the hypotheses on the ring:

Definition: Let $R$ be a domain, $Q=\operatorname{Frac}(R)$. An ideal $I$ is invertible if there are elements $a_1,\dots,a_n\in I$ and $q_1,\dots,q_n \in Q$ such that $q_iI\subset R$ for all $i$, and $1=\sum_{i=1}^n q_ia_i$.

Theorem: If $R$ is a domain, then a nonzero ideal $I$ is projective iff it is invertible.

If we move from domains to UFD's, then:

Theorem: If $R$ is a UFD, then a nonzero ideal $I$ is projective iff it is principal.

A reference for the above theorems is Rotman's "An Introduction to Homological Algebra", 2nd. edition, pp. 167-168.

share|cite|improve this answer
Thank you Bruno. – Rajnish Apr 15 '12 at 18:18
See also Dedekind domains and hereditary rings for (commutative) rings having the property that all ideals are projective. – t.b. Apr 15 '12 at 18:41
@Rajnish: you're welcome :) – lentic catachresis Apr 15 '12 at 20:18

For rings in general (with identity), the ring is called right hereditary if all right ideals are projective, and right semihereditary if all finitely generated right ideals are projective. There are a lot of other interesting equivalences and properties that these rings have, which you can find here.

A commutative hereditary domain is the same thing as a Dedekind domain, and a commutative semihereditary domain is the same thing as a Prüfer domain.

For examples that aren't domains (usually), semisimple rings are hereditary (on both sides) and von Neumann regular rings are semihereditary (on both sides).

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.