1
$\begingroup$

Let $(R,\mathfrak{m})$ be a Noetherian local ring. Is there any characterization known for sequences $\mathbf{x}=(x_1,...,x_n)$ such that there exists some module $M\neq 0$ where $\mathbf{x}$ is a regular sequence on $M$? By regular I mean $x_i$ is a non-zero-divisor on $M/(x_1,...,x_{i-1})M$ and $M/\mathbf{x}M\neq 0$.

Instinctually I feel as though these sequences should be somewhat special, but I have no rational basis for this. Perhaps even any sequence of elements in $\mathfrak{m}$, assuming $n\leqslant \mathrm{depth}\,R$, could arise as a regular sequence on some module.

  • 0
    Can a nilpotent element be a regular sequence on any module? – 2017-02-13
  • 0
    @Mohan Ah, no, good point. That's certainly at least one condition. – 2017-02-13
  • 1
    At least for $n=1$, we can phrase the following: Any non-nilpotent element $x \in \mathfrak m$ is regular on some module. By the noetherian property, we can pick some integer with $\operatorname{Ann}(x^n)=\operatorname{Ann}(x^{n+1})$ and then $x$ is regular on $R/\operatorname{Ann}(x^n)$. – 2017-02-13

0 Answers 0