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Let $(A,\mathfrak{m})$ be a Noetherian local ring, $M\neq0$ a finite $A$-module. Suppose $$d=\min\{{\operatorname{depth}A,\operatorname{depth}M\}}\geq1.$$ Then does there always exists $a_1,\ldots,a_d\in\mathfrak{m}$ which is both an $A$-sequence and an $M$-sequence?

I came up with a very simple proof, but I'm a little surprised because I don't remember it being mentioned in any book. Can anyone confirm if this is true?

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up vote 2 down vote accepted

One easily reduces to the case $d=1$ by induction.

Let $\mathrm{Ass}(A)$, $\mathrm{Ass}(M)$ be the set of associated primes of $A$ and of $M$ respectively. Then $\mathfrak m$ is not contained is the union of $\mathfrak p\in \mathrm{Ass}(A)$ because $\mathrm{depth}(A)=1$. Same for $M$. By the prime avoidance lemma, there exists $$a\in \mathfrak m\setminus \cup_{\mathfrak p\in \mathrm{Ass}(A)\cup \mathrm{Ass}(M)} \mathfrak p.$$ Then $a$ is both $A$-regular and $M$-regular.

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That's also how I proved it. Thanks for the confirmation! – ashpool Nov 27 '11 at 15:45

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