Is Noetherianity necessary for quasi-regularity to imply regularity? Theorem 16.3 In Matsumura (CRT) reads as follows:

Let $A$ be a Noetherian ring, $M \neq 0$ an $A$-module, and $a_1,\dots,a_n \in A$; set $I=(a_1,\dots,a_n)A$. Under the condition 
  (*) each of $M, M/a_1 M, \dots, M/(a_1,\cdots,a_{n-1})M$ is $I$-adically separated, if $a_1,\dots,a_n$ is $M$-quasi-regular, then it is an $M$-sequence.

After studying the proof, it seems to me that the Noetherianity assumption on $A$ is not needed, and so the more general statement holds, with $A$ being any commutative ring. Is that true? 
 A: You are correct that Noetherianness is not needed: under the condition ($*$), i.e. each of $M/, M/a_1M, ..., M/(a_1,...,a_{n-1})M$ are $I$-adically separated, it is indeed true that $a_1, ..., a_n$ $M$-quasi-regular implies $a_1,...,a_n$ $M$-regular, for any ring $A$ (commutative with $1$, as always). The proof, as you observed, does not use any Noetherian hypotheses. It is worth including two conditions which guarantee ($*$) (these are stated slightly differently in the text):
i) $A$ Noetherian, $I \subseteq \text{rad}(A)$, $M$ finite (here $\text{rad}(A)$ is the Jacobson radical)
ii) $A$ and $M$ are $\mathbb{N}_0$ graded, $a_1, ..., a_n$ homogeneous elements of positive degree
In the text, Matsumura mentions that the conclusion of the theorem can fail if Noetherianness is dropped from i) (even if $M = A$ is local and $I \subseteq \text{rad}(A)$), and he cites an example of Dieudonne, which can be found in this very short (and excellent) paper. I would guess that this is the reason Matsumura included the Noetherian hypothesis in the statement, for ease of presentation. 
