# Regular local ring and a prime ideal generated by a regular sequence up to radical

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 $x_1,\dots,x_i$?

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I've given this a fair bit of thought, and the best I can do is give you an upvote. – Alex Becker Jul 7 '12 at 8:21
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. – messi Jul 7 '12 at 10:55
– YACP Apr 28 at 14:26

## 1 Answer

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.

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Thank you, i just noticed the answer right now. – messi Aug 2 '12 at 5:08
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