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.
$\square$

## 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Your email address will not be published.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.

## Comments (2)

Comment #5510 by Manuel Hoff on

Comment #5706 by Johan on

There are also: