
Lemma 15.8.7. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $r \geq 0$. The following are equivalent

1. $M$ is finite locally free of rank $r$ (Algebra, Definition 10.77.1),

2. $\text{Fit}_{r - 1}(M) = 0$ and $\text{Fit}_ r(M) = R$, and

3. $\text{Fit}_ k(M) = 0$ for $k < r$ and $\text{Fit}_ k(M) = R$ for $k \geq r$.

Proof. It is immediate that (2) is equivalent to (3) because the Fitting ideals form an increasing sequence of ideals. Since the formation of $\text{Fit}_ k(M)$ commutes with base change (Lemma 15.8.4) we see that (1) implies (2) by Example 15.8.5 and glueing results (Algebra, Section 10.22). Conversely, assume (2). By Lemma 15.8.6 we may assume that $M$ is generated by $r$ elements. Thus a presentation $\bigoplus _{j \in J} R \to R^{\oplus r} \to M \to 0$. But now the assumption that $\text{Fit}_{r - 1}(M) = 0$ implies that all entries of the matrix of the map $\bigoplus _{j \in J} R \to R^{\oplus r}$ are zero. Thus $M$ is free. $\square$

Comment #1423 by Kestutis Cesnavicius on

In (1), "locally free of rank $k$" ---> "locally free of rank $r$".

There are also:

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

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).