The Stacks project

Lemma 10.63.16. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Via the canonical injection $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ we have

  1. $\text{Ass}_ R(S^{-1}M) = \text{Ass}_{S^{-1}R}(S^{-1}M)$,

  2. $\text{Ass}_ R(M) \cap \mathop{\mathrm{Spec}}(S^{-1}R) \subset \text{Ass}_ R(S^{-1}M)$, and

  3. if $R$ is Noetherian this inclusion is an equality.

Proof. For $m \in S^{-1}M$, let $I \subset R$ and $J \subset S^{-1}R$ be the annihilators of $m$. Then $I$ is the inverse image of $J$ by the map $R \to S^{-1}R$ and $J = S^{-1}I$. The equality in (1) follows by the description of the map $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ in Lemma 10.17.5. For $m \in M$, let $I \subset R$ be the annihilator of $m$ in $R$ and let $J \subset S^{-1}R$ be the annihilator of $m/1 \in S^{-1}M$. We have $J = S^{-1}I$ which implies (2). The equality in the Noetherian case follows from Lemma 10.63.15 since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset $ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$. $\square$


Comments (4)

Comment #8249 by Et on

I don't think the first statement of the proof about the annhilator of m is true. It does hold however if the annhilator is a prime ideal which doesn't intersect S, which is what you need anyway.

Comment #9034 by on

Regarding "the displayed inclusion and equality in the Noetherian case follows from Lemma 10.63.15": I think the displayed inclusion is independent of the invoked lemma (one observes that the -annihilator of in is the -annihilator of in if ), and it is only the converse inclusion what requires the mentioned lemma.

There are also:

  • 16 comment(s) on Section 10.63: Associated primes

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Your email address will not be published.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 05BZ. Beware of the difference between the letter 'O' and the digit '0'.