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

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.

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

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