The Stacks project

Lemma 10.62.6. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Then $d(M) = \dim (\text{Supp}(M))$ where $d(M)$ is as in Definition 10.59.8.

Proof. Let $M_ i, \mathfrak p_ i$ be as in Lemma 10.62.1. By Lemma 10.59.10 we obtain the equality $d(M) = \max \{ d(R/\mathfrak p_ i) \} $. By Proposition 10.60.9 we have $d(R/\mathfrak p_ i) = \dim (R/\mathfrak p_ i)$. Trivially $\dim (R/\mathfrak p_ i) = \dim V(\mathfrak p_ i)$. Since all minimal primes of $\text{Supp}(M)$ occur among the $\mathfrak p_ i$ (Lemma 10.62.5) we win. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 10.62: Support and dimension of modules

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Your email address will not be published.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 00L8. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.62.6 as pdf