Let $R$ be commutative regular local ring. Is it true, that for every $\mathfrak p \in \mathrm{Spec}(R)$, there is a $\mathfrak p$-primary $R$-regular sequence? (I.e. an $R$-regular sequence $\bf x$ such that the ideal $({\bf x})$ is $\mathfrak p$-primary.)
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$\begingroup$ Try $p=0$. $\endgroup$– Martin BrandenburgCommented Nov 21, 2010 at 23:16
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$\begingroup$ @Martin: it is a matter of convention. See the Remark here jstor.org/pss/2160211 $\endgroup$– Hailong DaoCommented Nov 22, 2010 at 9:23
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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.
answered Nov 22, 2010 at 2:46
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$\begingroup$ While I suppose "this book" refers to Twenty-Four Hours of Local Cohomology, the link failed to show it. $\endgroup$ Commented Feb 18, 2019 at 22:15