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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.

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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
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up vote 2 down vote accepted

This is true for a Noetherian module. Please see page 109 in the following notes. http://www.math.lsa.umich.edu/~hochster/614F10/614.pdf

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