Loading web-font TeX/Math/Italic

The Stacks project

Lemma 10.66.9. Let R be a ring. Let M be an R-module. Let \mathfrak p be a prime ideal of R which is finitely generated. Then

\mathfrak p \in \text{Ass}(M) \Leftrightarrow \mathfrak p \in \text{WeakAss}(M).

In particular, if R is Noetherian, then \text{Ass}(M) = \text{WeakAss}(M).

Proof. Write \mathfrak p = (g_1, \ldots , g_ n) for some g_ i \in R. It is enough the prove the implication β€œ\Leftarrow ” as the other implication holds in general, see Lemma 10.66.6. Assume \mathfrak p \in \text{WeakAss}(M). By Lemma 10.66.2 there exists an element m \in M_{\mathfrak p} such that I = \{ x \in R_{\mathfrak p} \mid xm = 0\} has radical \mathfrak pR_{\mathfrak p}. Hence for each i there exists a smallest e_ i > 0 such that g_ i^{e_ i}m = 0 in M_{\mathfrak p}. If e_ i > 1 for some i, then we can replace m by g_ i^{e_ i - 1} m \not= 0 and decrease \sum e_ i. Hence we may assume that the annihilator of m \in M_{\mathfrak p} is (g_1, \ldots , g_ n)R_{\mathfrak p} = \mathfrak p R_{\mathfrak p}. By Lemma 10.63.15 we see that \mathfrak p \in \text{Ass}(M). \square


Comments (0)

There are also:

  • 3 comment(s) on Section 10.66: Weakly associated primes

Your email address will not be published.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.