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Hi,

Let R be commutative regular local ring. Is it true, that for every p \in Spec(R), there is a p-primary R-regular sequence? I.e. an R-regular sequence (x) such that the ideal (x) is p-primary.

Regards, David

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Try $p=0$. –  Martin Brandenburg Nov 21 '10 at 23:16
    
@Martin: it is a matter of convention. See the Remark here jstor.org/pss/2160211 –  Hailong Dao Nov 22 '10 at 9:23

1 Answer 1

EDIT: here is a counter-example for the question in general. Let $P \subset R= \mathbb C[[x,y,z,a,b,c]]$ be generated by the $2$ by $2$ minors of the obvious $2$ by $3$ matrix. Then the local cohomology module $H_P^3(R) \neq 0$ (see page 201 of this book), so $P$ can't be a radical of a $2$-generated ideal.

This is a hard question even for small rings. Let $R=\mathbb C[[x,y,z]]$ and $P$ a prime ideal of height $2$. Your question is the same as asking if the curve $X= \text{Spec} (R/P)$ is always a set-theoretic complete intersection. This is widely open even in this case (space curves over complex numbers).

An amazing partial result is obtained by Cowsik-Nori: every curve in affine space over a field of characteristic $p>0$ is a set-theoretic complete intersection! See this paper by Hartshorne for some relevant references.

Of course, there are a lot of papers on this topic to this day, just google the relevant terms in this answer.

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