This question is very related to this one: generators of a prime ideal in a noetherian ring.

Let $\mathfrak{p}$ be a prime ideal in a Noetherian ring and let $k$ be its height. Further suppose that $f_{1},\dots, f_{k} \in \mathfrak{p}$ generate the maximal ideal in the localization $R_{\mathfrak{p}}$, more precisely $f_{1}, \dots, f_{k}$ generate $\mathfrak{p} R_{\mathfrak{p}}$ in $R_{\mathfrak{p}}$. This situation appears for example in the Jacobian criterium in local analytic geometry (see for example the book by DeJong/Pfister).

My question: Is it possible to conclude that $f_{1},\dots, f_{k}$ generate $\mathfrak{p}$?


No - unless I'm misunderstanding your question, here's a simple counterexample. Let $R = \mathbb Z$, and $\mathfrak p = (2)$. Then the element $f = 10$ generates the ideal $(2) \cdot \mathbb Z_{(2)}$, since 5 is invertible there, but clearly doesn't generate $\mathfrak p$ in $\mathbb Z$.

  • $\begingroup$ Thanks - nice and simple example! $\endgroup$ – Sebastian Jun 17 '13 at 14:14
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    $\begingroup$ But what holds (and what I actually needed) is that one can find an element $s \in R\setminus \mathfrak{p}$ so that $s \cdot \mathfrak{p} \subset \left< f_{1},\dots, f_{k} \right>$. $\endgroup$ – Sebastian Jun 19 '13 at 11:05

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