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Let $(R,m)$ be a local Cohen-Macaulay ring and $p$ a prime ideal of $R$. Denote by $\hat{R}$ the completion of $R$ with respect to the $m$-adic topology. Take $q \in \operatorname{Ass}(\hat{R}/p \hat{R})$ and suppose that $q \cap R \neq p$.

Question: Why does that imply that $q$ contains an $\hat{R}/p \hat{R}$-regular element?

My effort: Take $a \in q \cap R - p$, $x = (x_i + m^i)_{i>0} \in \hat{R}$ and suppose $a x \in p \hat{R}$. Then for every $i>0$ we have $a x_i \in p + m^i$. Additionally, since $m=\operatorname{rad}(R)$, $\hat{R}$ is faithfully-flat over $R$ and $p$ is a closed ideal of $R$, i.e. $p = \cap_{i>0} (p+m^i)$ and finally $p \hat{R} \cap R = p$. And that's where i stopped having good ideas. Any hints how to continue?

Motivation: Bruns and Herzog, CMR, proof of Theorem 2.1.15.

Manos
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1 Answers1

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Your question as stated doesn't really make sense: if $q \in \text{Ass}(\hat{R}/p\hat{R})$, then by definition $q$ cannot contain an $\hat{R}/p\hat{R}$-regular element. What Bruns-Herzog is saying is that if some prime $q \in \text{Spec}(\hat{R}/p\hat{R})$ is such that the contraction of $q$ to $R$ strictly contains $p$, then this $q$ contains an $\hat{R}/p\hat{R}$-regular element, hence will not be in $\text{Ass}(\hat{R}/p\hat{R})$. This follows from the flatness of completion: pick $x \in q^c \setminus p$, then the image of $x$ in $R/p$ is nonzero, hence a nonzerodivisor, so under the faithfully flat map $R/p \to \hat{R}/p\hat{R}$, the image of $x$ in $q$ will remain a nonzerodivisor.

zcn
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