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Let $R$ be a commutative ring and let $M\rightarrow N$ be an essential morphism of $R$-modules. Then, $M$ and $N$ have the same associated primes.

Over non-noetherian rings the notion of associated primes does not behave very well, and it is sometimes appropriate to rather consider the so-called weakly associated primes of a module. (A prime ideal of $R$ is weakly associated with an $R$-module $L$ if it is a minimal prime of the annihilator of an element of $L$; see Bourbaki, AC.IV.1 Exer. 17, for basic facts about this notion.)

Unfortunately, if $M\rightarrow N$ is an essential morphism of $R$-modules then $M$ and $N$ need not have the same weakly associated primes. (This happens for example over every non-noetherian valuation ring with maximal ideal of finite type.)

This leads to the following question:

What are examples of (classes of) non-noetherian rings over which weakly associated primes do not change along essential morphisms?

EDIT: A sufficient condition is that weakly associated primes and associated primes coincide for every module. This happens e.g. for one-dimensional valuation rings. So, a sub-question of the above is the following:

What are examples of (classes of) non-noetherian rings over which weakly associated primes coincide with associated primes for every module?

EDIT 2: Neil pointed out, and rightly so, that in the previous edit I wrote some nonsense. Weakly associated primes and associated primes coincide for every module for example if $R$ is a local domain such that every non-zero ideal of $R$ contains a power of the maximal ideal. What I intended to say about one-dimensional valuation rings has in fact nothing to do with valuation rings, but should rather be the following: One-dimensional local domains are an example of a class of not necessarily noetherian rings that have the property asked for in the original question, i.e., weakly associated primes do not change along essential morphisms.

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    $\begingroup$ Well, any zero-dimensional quasilocal ring will do... $\endgroup$
    – Neil Epstein
    Commented Jan 30, 2014 at 3:08
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    $\begingroup$ Regarding your edit: I don't think your statement about one-dimensional valuation rings is true. If $(R,m)$ is a valuation ring with value group $\mathbb{Q}$, and $I$ is the ideal of elements with values greater than the square root of 2, then the $R$-module $R/I$ has no associated primes. But $m$ is obviously a weakly associated prime. $\endgroup$
    – Neil Epstein
    Commented Feb 12, 2014 at 22:10
  • $\begingroup$ Dear @Neil, thanks - I think now I wrote what I wanted to write... $\endgroup$
    – Fred Rohrer
    Commented Feb 12, 2014 at 22:47

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