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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?

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  • $\begingroup$ Not really worth much but certainly if $R/I$ is regular, then $I$ satisfies the condition you pointed out (because then the generators of $I$ can be made to be part of a regular sequence defining the maximal ideal). $\endgroup$
    – Karl Schwede
    Commented Feb 7, 2011 at 17:59
  • $\begingroup$ @Karl: Thanks for your response. Does this really give us a new example? Since, $R$ is regular, $R/I$ is regular if and only if $I$ is generated by a regular system of parameters. This takes us back to the family I mention. Am I missing something here? $\endgroup$
    – Koose Muniswamy
    Commented Feb 7, 2011 at 20:13
  • $\begingroup$ No no, not a new example at all. Just a different way of phrasing your example. $\endgroup$
    – Karl Schwede
    Commented Feb 8, 2011 at 3:58
  • $\begingroup$ @Karl: Yes, definitely a more compact (better) way of putting it. $\endgroup$
    – Koose Muniswamy
    Commented Feb 8, 2011 at 4:06

2 Answers 2

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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 $\mathbb C[[x,y,z,u,v,w]]$.

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  • $\begingroup$ @Hailong: Thanks for the answer. Just to make it completely explicit. If $P$ is a complete intersection in a regular local ring (hence Cohen Macaulay), there is a regular sequence that generates it. We can reduce to case where $P$ is $2$ generator using induction. Then, using your claim and the fact that regular sequences are permutable in a Noetherian local ring, we get the result. Does this sound OK? $\endgroup$
    – Koose Muniswamy
    Commented Feb 8, 2011 at 4:14
  • $\begingroup$ @Koose: I am not sure how to reduce to 2-generator case. What I had in mind is pretty simple: if $P=(x_1,\cdots,x_n)$, then my claim shows that $(x_2,\cdots, x_{n})$ is prime. Also, since one can permute reg sequence, all subsets of min. generators of $P$ generate prime ideals. $\endgroup$
    – Hailong Dao
    Commented Feb 8, 2011 at 19:55
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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.

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  • $\begingroup$ @Sandor: Thanks for the answer. I am not familiar with all the terminology here, but this only seems to show that the particular ideal does not satisfy the desired condition. I don't see how it shows there are no other prime ideals satisfying the condition in a regular local ring apart from the family mentioned in the question. $\endgroup$
    – Koose Muniswamy
    Commented Feb 7, 2011 at 20:16
  • $\begingroup$ @Koose: I suppose I misunderstood your question. However, the same idea shows that any ideal having this property has to be a complete intersection (that is, it is generated by as many elements as its height). $\endgroup$
    – Sándor Kovács
    Commented Feb 7, 2011 at 21:45
  • $\begingroup$ @Sandor: Thanks for your clarification. I do realize this, but, I was looking for sufficient conditions for an ideal to satisfy the above property. $\endgroup$
    – Koose Muniswamy
    Commented Feb 8, 2011 at 0:01

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