MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

share|cite|improve this question
Try $p=0$. – Martin Brandenburg Nov 21 '10 at 23:16
@Martin: it is a matter of convention. See the Remark here – Hailong Dao Nov 22 '10 at 9:23

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.