Given a ring $A$ and an $A$-module $M$, an element in $A$ is said to be $M$-regular if it's not a zero divisor of $M$.
A sequence $y_1,\dots,y_n$ of elements of $A$ is said to be $M$-regular if the following conditions hold:
(1) $y_1$ is $M$-regular, $y_2$ is $M/y_1M$-regular, ... , $y_n$ is $M/(y_1,\dots,y_{n-1})M$-regular;
(2) $M/(y_1,\dots,y_n)M\neq0$.
Now we consider the special case when $A$ is Noetherian and $M=A$. Let $P$ be a prime ideal of $A$.
And we suppose that there is an $A_P$-regular sequence $y_1,\dots,y_n$ in $PA_P$.
Then Matsumura said in his book Commutative Ring Theory (proof of Theorem 24.5) that "after passing to a smaller neighbourhood of $P$", i.e. after replacing $A$ by $A_a$ with some suitable $a\in A-P$, we can assume that (a) $y_1, ... , y_n$ is an $A$-regular, and (b) $I = (y_1, ... , y_n)A$ is a $P$-primary ideal.
It seems that this fact is easy to check in his word. Actually, (b) has no difficulty indeed. But I don't think (a) is so. I haven't found a quick method to prove it, even in the case when $n=1$, that is, "being $A$-regular" = " being a non-zero divisor of $A$".
However, I have attempted and got proof of my own, which is a bit complex. It involves associated primes. Let me state it here:
We just consider $n=1$. The general case can be proved by induction. Let $y$ be a non-zero divisor of $A_P$. Then I show that there exists an element $a\in A-P$, s.t. $y\in A_a$ and is also a non-zero divisor of $A_a$.
Proof. Suppose $y=r/s$, $s\in A-P$. First since $A$ is Noetherian, we have $$\operatorname{Ass}_{A_P}(A_P)=\{QA_P:Q\in\operatorname{Ass}_A(A), Q\subseteq P\}.$$ It corresponds bijectively to $\{Q\in\operatorname{Ass}_A(A):Q\subseteq P\}$, whose complement, denoted by $\mathcal{F}$, is a finite set (may be empty) since $\operatorname{Ass}_A(A)$ is finite. If $\mathcal{F}=\emptyset$, we take $a=s$. Otherwise, $\mathcal{F}=\{Q_1,\cdots,Q_r\}$, where $Q_i\nsubseteq P$. So there exists $q_i\in Q_i-P$ for each $i$. And we take $a=s\cdot\prod q_i$. Therefore, we have $$\{Q\in\operatorname{Ass}_A(A):Q\subseteq P\}=\{Q\in\operatorname{Ass}_A(A):a\notin Q\},$$ and the latter corresponds to $\operatorname{Ass}_{A_a}(A_a)$.
Now $y$ is certainly in $A_a$. If $y$ is a zero divisor of $A_a$, it must be in some $QA_a\in\operatorname{Ass}_{A_a}(A_a)$. Thus $y\in (QA_a)_P=QA_P$ and $QA_P\in\operatorname{Ass}_{A_P}(A_P)$ by the argument above, contradiction. Q.E.D.
Is my proof right?
Is there a better way to prove this statement? Can anybody offer some help?
Thanks in advance.