Lemma 10.63.3. Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{Ass}(M') \subset \text{Ass}(M)$ and $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$. Also $\text{Ass}(M' \oplus M'') = \text{Ass}(M') \cup \text{Ass}(M'')$.

Proof. If $m' \in M'$, then the annihilator of $m'$ viewed as an element of $M'$ is the same as the annihilator of $m'$ viewed as an element of $M$. Hence the inclusion $\text{Ass}(M') \subset \text{Ass}(M)$. Let $m \in M$ be an element whose annihilator is a prime ideal $\mathfrak p$. If there exists a $g \in R$, $g \not\in \mathfrak p$ such that $m' = gm \in M'$ then the annihilator of $m'$ is $\mathfrak p$. If there does not exist a $g \in R$, $g \not\in \mathfrak p$ such that $gm \in M'$, then the annilator of the image $m'' \in M''$ of $m$ is $\mathfrak p$. This proves the inclusion $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$. We omit the proof of the final statement. $\square$

Comment #5510 by Manuel Hoff on

I would suggest to add the following immediate consequence to the Lemma: If $M$ and $M'$ are two $R$-modules, then $\text{Ass}(M \oplus M') = \text{Ass}(M) \cup \text{Ass}(M')$.

There are also:

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

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).