I have recently started reading section 6 of Matsumura's "Commutative Ring Theory", and I have noticed that some of the results from Theorems 6.1 through 6.5 make the assumption of Noetherianness of the underlying ring while some others don't. I went a bit further and worked out the exact places where Noetherianness is needed, to find the following:
$(1)$ If $A$ is Noetherian and $M$ a non-zero $A$-module then the set of zero divisors of $M$ is contained in the union of all the associated primes of $M$.
Noetherianness needed to conclude by Zorn that family of annihilators containing $ann_A(a)$, for some zero-divisor $a \in A$ of $M$, has a maximal element.
$(2)$ If $M \neq 0$ is a finitely generated $A$-module over a Noetherian ring $A$, then $$Ass_{A_S} (M_S) \subset Ass_A(M) \cup Spec(A_S),$$ where we have viewed $Spec(A_S)$ as a subset of $Spec(A)$.
Need Noetherianness for the result that for any $\mathfrak P \in Ass_{A_S}(M_S)$, if $\mathfrak P = \mathfrak pA_S = ann_{A_S}(x)$ for some $\mathfrak p \in Spec(A): \mathfrak p \cap S = \phi$ and $x \in M$, then $\mathfrak p = ann_A(tx)$ for some $t \in S$. More precisely, need $\mathfrak p = \mathfrak P \cap A$ to be finitely generated to construct such $t \in S$ using the generators of $\mathfrak p$.
$(3)$ If $A$ is Noetherian and $M \neq 0$ a finitely generated $A$-module (hence also Noetherian), then there exists a chain $0 = M_0 \subset M_1 \subset \cdots \subset M_n=M$ of $A$-submodules of $M$ satisfying, for every $j \in \{1, \cdots , n\}$, $M_j / M_{j-1} \cong A/\mathfrak p_j$ for some prime $\mathfrak p_j \vartriangleleft A$.
Noetherianness of $M$ needed to conclude that the constructed increasing chain of submodules of $M$ terminates. Also need Noetherianness of $A$ only to obtain associated primes via the assertion mentioned in $(1)$ above. (We recall that the underlying ring of a Noetherian module need not itself be Noetherian.)
$(4)$ For $A$ and $M$ as in $(3)$, the minimal elements of $Ass(M)$ and $Supp(M)$ coincide.
Need Noetherianness of $M_{\mathfrak p}$ as an $A_{\mathfrak p}$-module to ensure that $Ass_{A_{\mathfrak p}}(M_\mathfrak p) \neq \phi$ and also need Noetherianness of $A$ to conclude via $(2)$ that $Ass_{A_{\mathfrak p}} (M_{\mathfrak p}) = Ass_A(M) \cup Spec(A_{\mathfrak p})$.
Since I have already found examples where $Ass_A (M)$ need not be nonempty (Example of a non-zero module that has no associated primes) or finite (A module with infinitely many associated primes.) in a non-Noetherian setting (it also seems that we can replace the Noetherianness of $A$ by that of $M$, c.f. Existence of associated primes of modules), I was looking for counterexamples when some of the hypotheses are not true, if there any, and what are the minimal hypotheses for which we can still get the concerned result. I would especially like to know the following about the respective assertions above:
$(1)$, $(2)$, $(4)$ Are there a non-Noetherian ring $A$ for which the stated assertions do not hold?
$(3)$ Is the result still true for Noetherian modules $M$ over non-Noetherian rings $A$? If not, what are simple counterexamples for this and also for when $M$ itself is non-Noetherian?
I would really appreciate counterexamples as simple as possible, since I have been struggling to find or construct them myself. Thank you.
Edit 1: I later realized that the ring $A$ of real-valued continuous functions on the real line is itself a counterexample of $(1)$ since for any two distinct $a, b \in \mathbb R$, Urysohn's Lemma gives us continuous functions $g, h: \mathbb R \rightarrow \mathbb R$ satisfying $g(a)=h(b)=1$ and $gh=0$ on $\mathbb R$, whereupon $g \neq 0$ is a zero divisor of $A$, yielding $Ass_A(A) = \phi \subsetneq ZD_A(A)$ (that $Ass_A(A) = \phi$ is a consequnece of an example in the first link above). I am yet to find examples for the others, and would really appreciate some help regarding the same.
