Question

We know that a finitely generated R-module M satisfies. the $S_n$ condition if $depthM_p \geq min(n,dimM_p)$ for every $p\in SpecR$. It's well known that every cohen-macaulay rings satisfies $(S_n)$ for all $n \geq 0$,now is the following conclusion true? note that: Let $\phi: A \to B $ be a homomorphism of Noetherian rings. we say $F=B\otimes_{A} (A/p)_p $ is fiber ring of $\phi$ over $p\in Spec A$. If A be a quotient of a cohen-macaulay local ring and satisfies $(S_n)$ then the completion $\hat{A}$ also satisfies $(S_n)$.

I wanna use the proposition 2.1.16 from " Cohen-macaulay rings-Bruns-Herzog"

Answer 1

This is too long for a comment, so I am writing it here. It reduces the problem to the case where $A$ can be assumed to be Cohen-Macaulay. But there is still an exercise remaining for you to do! All theorem and page numbers refer to Matsumura's Commutative ring theory.

Here are steps to do exercise 23.2 (p. 185). I am assuming all rings are local.

1. By Theorem 23.9 (p. 184) it suffices to show that all fibers of $A\rightarrow\hat{A}$ satisfy ($S_n$).
2. By assumption $A$ is a quotient of a Cohen-Macaulay ring, say $A=R/I$, with $R$ Cohen-Macaulay.
3. Let $\mathfrak{p}\in\mathrm{Spec}\ A$. This means $\mathfrak{p}=\mathfrak{p}^\prime/I$, for some $\mathfrak{p}^\prime\in\mathrm{Spec}\ R$. The fiber of $A\rightarrow\hat{A}$ at $\mathfrak{p}$ coincides with the fiber of $R\rightarrow R^\prime$ at $\mathfrak{p}^\prime$. (This is explained at the bottom of page 184. To use the explanation given there we should also note that $\hat{R}/I\hat{R}=(R/I)^\hat{\ }=\hat{A}$. This is Theorem 8.11, p. 61.)
4. So now we are down to showing that all fibers of $R\rightarrow\hat{R}$ satisfy ($S_n$). It suffices to show that all fibers of $R\rightarrow\hat{R}$ are Cohen-Macaulay. This is Exercise 23.1, p. 185 again. Maybe you can spend some time thinking about this one.