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Suppose $R$ is a Noetherian ring and $P$ is a prime ideal. If the number of generators of $PR_P$ as an ideal in $R_P$ is $n$, can we say anything about the number of generators of $P$ as an ideal of $R$?

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This looks like a homework problem. You're welcome to ask those there, but you ought to review meta.math.stackexchange.com/questions/1803/… . –  Dimitrije Kostic Nov 21 '11 at 4:28
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@Dimitrije Kostic: What made you think this is a HW problem? It is not, and from the several google searches I did, I don't think it has a straightforward answer either. –  Zuben Nov 21 '11 at 15:53
    
The homework problems that show up around here tend to be tersely stated, as your question is. Sorry if I was mistaken. –  Dimitrije Kostic Nov 21 '11 at 16:08
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You can deduce that there exists $f \in R \setminus P$ such that the ideal $PR_f$ is generated by $n$ elements, as ideal of $R_f$. But it might be the case that you cannot choose $f = 1$, i.e. $P$ may not be generated by $n$ elements as ideal of $R$. For example, take $R = \mathbb{Z}[\sqrt{-5}]$ and $P = (2, 1+ \sqrt{-5})$: $P$ is not principal but $P R_P$ is principal. –  Andrea Nov 22 '11 at 22:59

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Certainly $P$ has at least $n$ generators, as the images of the generators in $R_p$ generate $PR_p$. As you can see from Andrea's comment, the we cannot give an upper bound in general. This is because the number of generators is not really a "nice" property of ideals. A closely related property is the height of an ideal, which for a prime ideal $P$ is the supremum of the lengths of chains of prime ideals contained in $P$, and for a general ideal is the infimum of the heights of prime ideals containing it. Since we have a bijection between the ideals of $R$ contained in $P$ and the ideals of $R_P$, the height of $P$ is the same as the height of $PR_P$. While this is not quite what you're looking for, it's close. By the generalized principal ideal theorem, the height of an ideal in a Noetherian ring is at most the number of generators. In some cases we can say more, especially for prime ideals. For example, prime ideals of height $1$ in a UFD have exactly one generator. In a regular local ring, the height of the maximal ideal is equal to its number of generators.

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