prime ideals in regular local rings - MathOverflow

prime ideals in regular local rings

2011-02-07T15:45:03Z

Suppose $R$ is a regular local ring. Let $m$ be the maximal ideal. Then, if the dimension of $R$ is $n$, there is a regular sequence of size $n$, say $x_1,x_2,...,x_n$ s.t. $m=(x_1,x_2,...,x_n)R$. Further, the ideals $(x_{i_1},...,x_{i_j})$ with $i_1,...,i_j\in {1,...,n}$, are prime.

Can we make similar statements about any other kind of prime ideals in a regular local ring $R$? Specifically, do any other prime ideals satisfy the condition: if the ideal is minimally generated by a certain set of generators, then every subset of the generators defines a prime ideal? One example in light of the first paragraph, are the prime ideals generated by a subset of the regular sequence that generates the maximal ideal.

Also, when does a regular sequence define a prime ideal in a regular local ring $R$, and when does a maximal regular sequence define a maximal ideal?

Answer by Sándor Kovács for prime ideals in regular local rings

2011-02-07T18:07:33Z

EDIT: Rephrased the answer in light of Koose's comment.

Claim. Any ideal satisfying the required property has to be a complete intersection.

Proof. Let $\mathfrak p\subset R$ be a prime ideal. Assume that the minimal number of generators for $\mathfrak p$ is $r$ and let $a_1,\dots,a_r\in\mathfrak p$ be a set of generators. Let $I_t=(a_1,\dots,a_t)$ and one has the sequence of ideals:

$$ 0\subsetneq I_1 \subsetneq \dots \subsetneq I_t\subsetneq I_{t+1}\subsetneq \dots\subsetneq I_r=\mathfrak p $$

The containments cannot be equalities, because that would make the corresponding $a_i$ unneeded to generate $\mathfrak p$. If all the $I_t$'s are prime, then $\mathfrak p$ has height at least $r$, but it cannot be more than that, so the claim is proven. $\square$

Example. Take an irreducible projective variety (say the twisted cubic curve) that is not a complete intersection (in projective space). Then the ideal of the affine cone over this in the affine cone over projective space will be a prime, but if you take away one of the generators, that will not be.

Answer by Hailong Dao for prime ideals in regular local rings

2011-02-08T01:49:25Z

As Sandor pointed out, a necessary condition is that the prime ideal $P$ is a complete intersection. Here is a proof that it is also sufficient. It will suffice to prove the following:

Claim: Let $(R,m)$ be a Noetherian local ring and $x\in m$ a regular element on $R$. If $R/(x)$ is a domain, then so is $R$.

Proof: Suppose $ab=0$ in $R$. Then modulo $x$, one of them say $a$, must be $0$. So $a=xa_1$, thus $x(a_1b)=0$. As $x$ is regular, $a_1b=0$, and continuing in this fashion one of $a,b$ must be divisible by arbitrary high power of $x$, so it must be equal to $0$.

As for an example which is not a part of a regular s.o.p, take something like $P=(x^2+y^2+z^2, u^2+v^2+w^2)$ in $\mathbb C[[x,y,z,u,v,w]]$.