# uniform bound of the number of generators of prime ideals

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 $R$?

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

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:

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

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@ Karl: I know the paper of Boratyński-Eisenbud-Rees, but it consider the bounddedness o generators of m-primary ideal. Thank you so much for the paper of Moh. So question 2 is not true for dimension bigger than two. – Pham Hung Quy Nov 14 '12 at 16:15

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) $R_{\mathfrak p}$.

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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.

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