# Serre condition $(S_n)$

We know that a finitely generated $R$-module $M$ satisfies the $(S_n)$ condition if $$\operatorname{depth} M_p \geq \min(n,\dim M_p)$$ for every $p\in \operatorname{Spec}R$. It's well known that Cohen-Macaulay rings satisfy $(S_n)$ for all $n \geq 0$. Now is the following conclusion true:

If $A$ is a quotient of a Cohen-Macaulay local ring and satisfies $(S_n)$ then the completion $\hat{A}$ also satisfies $(S_n)$?

I want to use the proposition 2.1.16 from Cohen-Macaulay Rings, Bruns-Herzog.

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I don't see where the definition of fiber is used in your question. Also, if $A$ is not local, does $\hat{A}$ mean completion with respect to Jacobson radical of $A$? – Mahdi Majidi-Zolbanin Feb 26 '12 at 23:01
I'm trying to provide the conditions of proposition 2.1.16 in (cohen macaulay rings by Bruns-Herzog) – Stella Feb 26 '12 at 23:17
What is $(A/p)_p$ in the def. of $F$ ? Shouldn't it be $F=B\otimes_A k(p)$ where $k(p) = A_p/pA_p = \text{Quot}(A/p)$ ? – Ralph Feb 27 '12 at 2:42
$k(p)=A_p/pA_p=(A/p)_p. (A/p)_p means (A/p)_{p/p}$. – Stella Feb 27 '12 at 5:40
This is an exercise in Matsumura's Commutative ring theory, page 185, Exercise 23.2. – Mahdi Majidi-Zolbanin Feb 27 '12 at 15:24

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.
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.