Lemma 10.63.7. Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. Then

** Over a Noetherian ring each nonzero module has an associated prime. **

**Proof.**
If $M = (0)$, then $\text{Ass}(M) = \emptyset $ by definition. If $M \not= 0$, pick any nonzero finitely generated submodule $M' \subset M$, for example a submodule generated by a single nonzero element. By Lemma 10.40.2 we see that $\text{Supp}(M')$ is nonempty. By Proposition 10.63.6 this implies that $\text{Ass}(M')$ is nonempty. By Lemma 10.63.3 this implies $\text{Ass}(M) \not= \emptyset $.
$\square$

## 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Your email address will not be published.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 (1)

Comment #830 by Johan Commelin on

There are also: