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I am working through a proof somewhere, and I want to use this:

Let $(R,\mathfrak m)$ be a local ring (Noetherian commutative) and let $M$ be an $R$-module. If $\mathfrak p$ is an associated prime of $M$, then there exists $\hat{\mathfrak p}$, an associated prime of $M \otimes_R \hat{R}$ such that $\hat{\mathfrak p} \cap R = \mathfrak p$, where $\hat{R}$ denotes the $\mathfrak m$-adic completion of $R$.

Is this the case?

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Let $R\to S$ be a faithfully flat morphism, and $M$ an $R$-module. Let $\mathfrak p$ be an associated prime of $M$. Then there is $P$, an associated prime of $M\otimes_RS$, such that $\mathfrak p=P\cap R$.

There is an injective map $R/\mathfrak p\to M$. By tensoring with $S$ we get an injective map $S/\mathfrak pS\to M\otimes_RS$. Let $P$ be an associated prime of the $S$-module $S/\mathfrak pS$. Then $P$ is an associated prime of $M\otimes_RS$ and $P\cap R=\mathfrak p$: if $a\in R$ and $a\notin\mathfrak p$, then $R/\mathfrak p\stackrel{a\cdot}\to R/\mathfrak p$ is injective, so $a$ is a non-zerodivisor on $S/\mathfrak pS$.

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  • $\begingroup$ Due to faithfully flatness $p \subset P \cap R,$ but can you kindly explain why the other containment ? $\endgroup$ – user371231 Nov 22 '18 at 7:56
  • $\begingroup$ Actually $\mathfrak p\subseteq P\cap R$ since $\mathfrak pS\subseteq P$. The other inclusion is proved in the last part of the answer. $\endgroup$ – user26857 Nov 22 '18 at 8:34
  • $\begingroup$ Truly..I missed it. Thanks. $\endgroup$ – user371231 Nov 22 '18 at 9:29

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