Let $R$ be a regular local ring of dimension $n$ and let $P$ be a height $i$ prime ideal of $R$, where $1< i\leq n-1$. Can we find elements $x_1,\dots,x_i$ such that $P$ is the only minimal prime containing $x_1,\dots,x_i$?

  • $\begingroup$ I also thought about it, i had asked a similar question before for which i got a counterexample, so i thought there should be a counterexample for this too, but i cant find it, so now i am not even sure if this is true or false, perhaps it should be false, but i dont know. Thanks Alex for trying. $\endgroup$
    – messi
    Jul 7, 2012 at 10:55
  • $\begingroup$ Related to Height one prime ideal of arithmetical rank greater than 1. $\endgroup$
    – user26857
    Apr 28, 2013 at 14:26

1 Answer 1


Since $P$ has height $i$, the elements $x_1,...,x_i$ must be a regular sequence. Thus what you are asking is whether $V(P)$ is a set-theoretic complete intersection.

This is a notoriously difficult question in general. For example, it is not known for curves in $A_\mathbb C^3$.

In general, the answer is NO. The simplest example is perhaps $R=\mathbb C[x_{ij}]_{{1\leq i\leq 3},{1\leq j\leq 2}}$ and $P$ is generated by the $2$ by $2$ minors. To prove that $P$ is not generated up to radical by $2$ elements one has to show that the local cohomology module $H_P^3(R)$ is nonzero (basic properties of local cohomology dictates that $H_I^n(R)=0$ if $n$ is bigger than the number of elements that generate $I$ up to radical. That is because local cohomology can be computed with Cech complex on these generators).

Even then, the cleanest way to show $H_P^3(R)\neq 0$ involves a topological argument (the non-vanishing is not true in characteristic $p>0$ by the way).

If you want to know more, the key words are: set-theoretic complete intersection, analytic spread, local cohomology.

  • $\begingroup$ Let me mention that in the above example $\operatorname{height}P=2$. $\endgroup$
    – user26857
    Sep 14, 2016 at 23:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .