Processing math: 100%

The Stacks project

Lemma 38.14.1. Let R be a local ring with maximal ideal \mathfrak m. Let R \to S be a ring map. Let N be an S-module. Assume

  1. N is projective as an R-module, and

  2. S/\mathfrak mS is Noetherian and N/\mathfrak mN is a finite S/\mathfrak mS-module.

Then for any prime \mathfrak q \subset S which is an associated prime of N \otimes _ R \kappa (\mathfrak p) where \mathfrak p = R \cap \mathfrak q we have \mathfrak q + \mathfrak m S \not= S.

Proof. Note that the hypotheses of Lemmas 38.7.1 and 38.7.6 are satisfied. We will use the conclusions of these lemmas without further mention. Let \Sigma \subset S be the multiplicative set of elements which are not zerodivisors on N/\mathfrak mN. The map N \to \Sigma ^{-1}N is R-universally injective. Hence we see that any \mathfrak q \subset S which is an associated prime of N \otimes _ R \kappa (\mathfrak p) is also an associated prime of \Sigma ^{-1}N \otimes _ R \kappa (\mathfrak p). Clearly this implies that \mathfrak q corresponds to a prime of \Sigma ^{-1}S. Thus \mathfrak q \subset \mathfrak q' where \mathfrak q' corresponds to an associated prime of N/\mathfrak mN and we win. \square


Comments (0)


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.