Lemma 10.62.7. Let $R$ be a Noetherian ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of finite $R$-modules. Then $\max \{ \dim (\text{Supp}(M')), \dim (\text{Supp}(M''))\} = \dim (\text{Supp}(M))$.
Proof. If $R$ is local, this follows immediately from Lemmas 10.62.6 and 10.59.10. A more elementary argument, which works also if $R$ is not local, is to use that $\text{Supp}(M')$, $\text{Supp}(M'')$, and $\text{Supp}(M)$ are closed (Lemma 10.40.5) and that $\text{Supp}(M) = \text{Supp}(M') \cup \text{Supp}(M'')$ (Lemma 10.40.9). $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: