The Stacks project

Lemma 15.8.7. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$. Let $\mathfrak p \subset R$ be a prime ideal. The following are equivalent

  1. $\text{Fit}_ k(M) \not\subset \mathfrak p$,

  2. $\dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p) \leq k$,

  3. $M_\mathfrak p$ can be generated by $k$ elements over $R_\mathfrak p$, and

  4. $M_ f$ can be generated by $k$ elements over $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$.

Proof. By Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $M_ f$ can be generated by $k$ elements over $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$ if $M \otimes _ R \kappa (\mathfrak p)$ can be generated by $k$ elements. Hence (2), (3), and (4) are equivalent. Using Lemma 15.8.4 part (3) this reduces the problem to the case where $R$ is a field and $\mathfrak p = (0)$. In this case the result follows from Example 15.8.5. $\square$


Comments (2)

Comment #5107 by typo_bot on

In (3), element' should beelements'. In (2) it would be good to parenthesize the argument of dim, even though one could argue that no confusion is possible and that this is a matter of taste.

There are also:

  • 8 comment(s) on Section 15.8: Fitting ideals

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Your email address will not be published.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 07ZC. Beware of the difference between the letter 'O' and the digit '0'.