The Stacks project

Lemma 10.157.2. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. The following are equivalent:

  1. $M$ has no embedded associated prime, and

  2. $M$ has property $(S_1)$.

Proof. Let $\mathfrak p$ be an embedded associated prime of $M$. Then there exists another associated prime $\mathfrak q$ of $M$ such that $\mathfrak p \supset \mathfrak q$. In particular this implies that $\dim (\text{Supp}(M_{\mathfrak p})) \geq 1$ (since $\mathfrak q$ is in the support as well). On the other hand $\mathfrak pR_{\mathfrak p}$ is associated to $M_{\mathfrak p}$ (Lemma 10.63.15) and hence $\text{depth}(M_{\mathfrak p}) = 0$ (see Lemma 10.63.18). In other words $(S_1)$ does not hold. Conversely, if $(S_1)$ does not hold then there exists a prime $\mathfrak p$ such that $\dim (\text{Supp}(M_{\mathfrak p})) \geq 1$ and $\text{depth}(M_{\mathfrak p}) = 0$. Since $\text{depth}(M_{\mathfrak p}) = 0$, we see that $\mathfrak p \in \text{Ass}(M)$ by the two Lemmas 10.63.15 and 10.63.18. Since $\dim (\text{Supp}(M_{\mathfrak p})) \geq 1$, there is a prime $\mathfrak q \in \text{Supp}(M)$ with $\mathfrak q \subset \mathfrak p$, $\mathfrak q \not= \mathfrak p$. We can take such a $\mathfrak q$ that is minimal in $\text{Supp}(M)$. Then by Proposition 10.63.6 we have $\mathfrak q \in \text{Ass}(M)$ and hence $\mathfrak p$ is an embedded associated prime. $\square$

Comments (4)

Comment #1164 by Hu Fei on

''Conversely, if does not'' should add ''hold''

Comment #7078 by Yuto Masamura on

I suggest a proof of the omitted part in the last statement. I think we also need Proposition 10.63.6 in addition to the lemmas cited above.

Since , we see that by the two lemmas (Lemmas 10.63.15, 10.63.18). Since , there is a prime with . We can take such a that is minimal in . Then by Proposition 10.63.6 we have and hence is an embedded associated prime.

There are also:

  • 4 comment(s) on Section 10.157: Serre's criterion for normality

Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 031Q. Beware of the difference between the letter 'O' and the digit '0'.