uniform bound of the number of generators of prime ideals - MathOverflow

uniform bound of the number of generators of prime ideals

2012-11-13T02:40:22Z

These questions are inspired from the well known fact (by Sally et. al.) as follows:

Theorem 1. Let $(R, \mathfrak{m})$ be a Noetherian local ring of dimension one. Then the minimal number of generators of ideals of $R$ is bounded above by a constant i.e. there exists a positive integer $C$ such that $\ell (I/\mathfrak{m}I) \leq C$ for all ideal $I$.

In higher dimension, it is easy to see Theorem 1 is not true.

Question 2. Let $(R, \mathfrak{m})$ be a Noetherian local domain of dimension two. Does the exist a positive integer $C$ such that $\ell (\mathfrak{p}/\mathfrak{mp}) \leq C$ for all prime ideal $\mathfrak{p}$ of $R$?

Question 3. Let $(R, \mathfrak{m})$ be a Noetherian local domain of dimension two. Does the exist a positive integer $C$ such that $\ell (\mathfrak{p}R_{\mathfrak{p}}/\mathfrak{p}^2R_{\mathfrak{p}}) \leq C$ for all prime ideal $\mathfrak{p}$ of $R$?

Answer by Qing Liu for uniform bound of the number of generators of prime ideals

2012-11-14T00:08:54Z

The answer to Question 3 is yes if $R$ is excellent (it is enough that the normalization of $R$ is finite over $R$). Indeed the normal locus of $\mathrm{Spec}(R)$ is then open, so there are only finitely many $\mathfrak p$ of height $1$ with non-normal $R_{\mathfrak p}$. The other prime ideals are either $0, \mathfrak m$ or height $1$ with normal (hence regular) $R_{\mathfrak p}$.

Answer by Pham Hung Quy for uniform bound of the number of generators of prime ideals

2012-11-14T03:29:18Z

Here is answer for question 3 in arbitrary dimension with a certain restriction on $R$.

We assume $R$ is a image of a regular local ring $(S, \mathfrak{n})$ (we always have this assumption by passing the completion). $R = S/ I$ for some ideal $I$ of $S$. Recall that the embedded dimension of $R$ is $\mu(\mathfrak{m}) = \ell (\mathfrak{m}/\mathfrak{m}^2)$, we denote it by $C$. It is not difficult to see that we can assume that the embedded dimension of $S$ is equal to $\mu(\mathfrak{m})$. So $S$ is a regular local ring of dimension $C$.

Now let $\mathfrak{p}$ be a prime ideal of $R$. Let $\mathfrak{q} \in Spec (S)$ such that $\mathfrak{p} = \mathfrak{q}/I$. It is well known that $S_{\mathfrak{q}}$ is also a regular local ring of (embedded) dimension $\leq C$. Moreover $R_{\mathfrak{p}} = S_{\mathfrak{q}}/I S_{\mathfrak{q}}$. So the embedded dimension of $R_{\mathfrak{p}}$ is less than or equal to the embedded dimension of $S_{\mathfrak{q}}$. Thus $\mu (\mathfrak{p}R_{\mathfrak{p}}) = \ell (\mathfrak{p}R_{\mathfrak{p}}/ \mathfrak{p}^2R_{\mathfrak{p}}) \leq C$.

Edit:

I have just read from Sally's book (page 52) the following result:

Theorem: Let $(R, \mathfrak{m})$ be a NOetherian local ring. Then $\dim R \leq 2$ iff there is a uniform bound of the number of generators of all ideals which do not have $\mathfrak{m}$ as a associated prime.

Thus, Question 3 has an affirmative answer.

Answer by Karl Schwede for uniform bound of the number of generators of prime ideals

2012-11-14T14:59:08Z

The following paper seems to indicate that there is no such bound in question 2.

T. T. Moh, On the unboundedness of generators of prime ideals in powerseries rings of three variables. J. Math. Soc. Japan Volume 26, Number 4 (1974), 722-734. http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jmsj/1240435038

He seems to construct a sequence of prime ideals $P_n$ in $k[[x,y,z]]$ with $n$ generators.

On the other hand, in a positive result, in this paper:

M. Boratyński, D. Eisenbud, D. Rees, On the number of generators of ideals in local Cohen-Macaulay rings. J. Algebra 57 (1979), no. 1, 77–81. http://www.sciencedirect.com/science/article/pii/0021869379902096

There they show some bounds for 2 -dimensional Cohen-Macaulay rings.