Pseudo associated primes and short exact sequences

Question

Let $A$ be a commutative ring, and $$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$$ a short exact sequence of $A$-modules. The following inclusion relation is well-known: $$\operatorname{Ass}M'\subseteq\operatorname{Ass}M\subseteq\operatorname{Ass}M'\cup\operatorname{Ass}M''$$ This relation still holds for "weak" associated primes; $\operatorname{WeakAss}M$ is the set of those primes minimal over some $\operatorname{ann} m\subset A$, where $m\in M$. Now let us define $\operatorname{PseudoAss}(M)$ to be the set of primes of the form $\sqrt{\operatorname{ann}m}$ for some $m\in M$. (Note that $\operatorname{Ass}M\subseteq\operatorname{PseudoAss}M\subseteq\operatorname{WeakAss}M$.) My question is:

Does the inclusion relation hold for pseudo-associated primes?

The first inclusion is clearly true, but the standard techniques used to prove the second inclusion (either for $\operatorname{Ass}$ or $\operatorname{WeakAss}$) don't seem to work.

The reason why I'm interested in $\operatorname{PseudoAss}M$ is because it occurs naturally in primary decomposition of submodules. I find it a little odd that it is pseudo associated primes and not more general weak associated primes that occur naturally in primary decomposition.

Edit: There is a theorem that says if $A$ is Noetherian then $\operatorname{WeakAss}M=\operatorname{Ass}M$ (same reference as above), so any counter example should be over a non-Noetherian ring.

Answer 1

The standard proof can be modified:

Let $P = \sqrt{\mbox{ann }x}$ for $x \in M$. We have two cases. First suppose that for every $r$ such that $rx \in M'$ we have $r^n x = 0$, for some $n$ (i.e. $r \in P$). Then for the image $\overline x \in M''$, we have that $r \overline x = 0$ implies that $r \in \sqrt{\mbox{ann }x}$. $\ $ Hence $\mbox{ann } x \subset \mbox{ann } \overline x \subset \sqrt{\mbox{ann }x}$ implies $\sqrt{\mbox{ann } \overline x} = P$.

Otherwise, take $r \in R$ such that $r x \in M'$ and $r \not \in P$. Let $s \in \sqrt{\mbox{ann } rx}$. Then $s^nrx = 0$ implies that $s^nr \in \sqrt{\mbox{ann }x} = P$. Therefore we get $s^n \in P$ implies $s \in P$. So $\mbox{ann } x \subset \sqrt{\mbox{ann } rx}\subset P$ implies $\sqrt{\mbox{ann } rx} = P$.