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.)

$\begingroup$ Try $p=0$. $\endgroup$– Martin BrandenburgNov 21, 2010 at 23:16

$\begingroup$ @Martin: it is a matter of convention. See the Remark here jstor.org/pss/2160211 $\endgroup$– Hailong DaoNov 22, 2010 at 9:23
1 Answer
EDIT: here is a counterexample 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 settheoretic complete intersection. This is widely open even in this case (space curves over complex numbers).
An amazing partial result is obtained by CowsikNori: every curve in affine space over a field of characteristic $p>0$ is a settheoretic 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.

$\begingroup$ While I suppose "this book" refers to TwentyFour Hours of Local Cohomology, the link failed to show it. $\endgroup$ Feb 18, 2019 at 22:15