The length of any saturated prime ideal chains between $P$ and $Q$ are the same.

Question

Let $M$ be a finitely generated Cohen-Macaulay $(R,\mathfrak{m})$-module, and $P,Q\in\mathrm{Supp}_{R}(M)$ such that $P\subset Q$. Prove that if the length of any saturated prime ideal chains between $P'\in\mathrm{Min}_{R}(M)$ and $\mathfrak{m}$ are the same, then the length of any saturated prime ideal chains between $P$ and $Q$ are the same.

Answer 1

Take $M$-regular sequence $x=(f_{1},f_{2},...,f_{s})$ in $P$ whence $depth(P,M)=depth_{R}(M/xM)$ or $P\subseteq Zdv(M)=\cup_{P'\in Ass_{R}(M/xM)} P'$. By Prime Avoidance, $P\subseteq P'\in Ass_{R}(M/xM)$.

+$dim(R/P')=dim_{R}(M/xM)=dim_{R}(M)-s$

+$dim(R/P)=dim_{R}(M)-depth_{R_{P}}(M_{P})=dim_{R}(M)-depth(P,M)=dim_{R}(M)-s$ because $M$ is CM then $depth(P,M)=depth_{R_{P}}(M_{P})$

So, $dim(R/P')=dim(R/P)$ then $P'=P$.

Localize $M/xM$ at $Q$ we have done.

One thing which is probably more simple. When $P\in P'$, $P'\in Ass_{R}(M/xM)=Min_{R}(M/xM)$ we can deduce that $P\in Ass_{R}(M/xM)$ because $M/xM$ CM and $P\in Supp_{R}(M)$.