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:



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> If $A$ is a quotient of a Cohen-Macaulay local ring and satisfies $(S\_n)$ then the completion $\hat{A}$ also satisfies $(S\_n)$?
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I want to use the proposition 2.1.16 from *Cohen-Macaulay Rings*, Bruns-Herzog. 



Serre condition $(S_n)$
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.


