3
$\begingroup$

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 $\mathfrak{p}$?

$\endgroup$
2
$\begingroup$

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 $\mathbb Z$.

$\endgroup$
  • $\begingroup$ Thanks - nice and simple example! $\endgroup$ – Sebastian Jun 17 '13 at 14:14
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.