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 $R$?
 A: The following paper seems to indicate that there is no such bound in question 2.  


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*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:


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*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.
A: 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) $R_{\mathfrak p}$.
A: 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. 
