Lemma 15.8.6 (07ZC)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.8: Fitting ideals
Lemma 15.8.6 (cite)

Lemma 15.8.6. 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

$\text{Fit}_ k(M) \not\subset \mathfrak p$,
$\dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p) \leq k$,
$M_\mathfrak p$ can be generated by $k$ elements over $R_\mathfrak p$, and
$M_ f$ 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 May 17, 2020 at 05:46

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.

Comment #5314 by Johan on June 17, 2020 at 10:50

Thanks and fixed here.

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