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

Is it true that if $M\neq0$ is Noetherian then $\mbox{Ass}(M)\neq\emptyset$ and $|\mbox{Ass}(M)|<\infty$ ? Many books (e.g. Eisenbud, Atiyah) seem to require the ring to be Noetherian, as well, but I have a convincing proof which requires only the module to be Noetherian (based on proofs from Injective Modules, by Sharpe & Vamos). But it's hard to ignore Eisenbud and Atiyah. Does anyone have any opinion on this?

p.s. Here by associated primes I mean prime annihilators of elements of the module, not those primes related to primary decomposition of the zero submodule. I understand that they coincide if the ring is Noetherian.

share|cite|improve this question
Exercise:If $M$ is Noetherian, then $R/\mathrm{ann}(M)$ is a Noetherian ring! Then we may assume $R$ is Noetherian. – wxu Jul 7 '11 at 1:34
up vote 2 down vote accepted

This is true for a Noetherian module. Please see page 109 in the following notes.

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.