Lemma 15.8.8. Let $R$ be a local ring. Let $M$ be a finite $R$-module. Let $k \geq 0$. Assume that $\text{Fit}_ k(M) = (f)$ for some $f \in R$. Let $M'$ be the quotient of $M$ by $\{ x \in M \mid fx = 0\}$. Then $M'$ can be generated by $k$ elements.

Proof. Choose generators $x_1, \ldots , x_ n \in M$ corresponding to the surjection $R^{\oplus n} \to M$. Since $R$ is local if a set of elements $E \subset (f)$ generates $(f)$, then some $e \in E$ generates $(f)$, see Algebra, Lemma 10.20.1. Hence we may pick $z_1, \ldots , z_{n - k}$ in the kernel of $R^{\oplus n} \to M$ such that some $(n - k) \times (n - k)$ minor of the $n \times (n - k)$ matrix $A = (z_{ij})$ generates $(f)$. After renumbering the $x_ i$ we may assume the first minor $\det (z_{ij})_{1 \leq i, j \leq n - k}$ generates $(f)$, i.e., $\det (z_{ij})_{1 \leq i, j \leq n - k} = uf$ for some unit $u \in R$. Every other minor is a multiple of $f$. By Algebra, Lemma 10.15.6 there exists a $n - k \times n - k$ matrix $B$ such that

$AB = f \left( \begin{matrix} u 1_{n - k \times n - k} \\ C \end{matrix} \right)$

for some matrix $C$ with coefficients in $R$. This implies that for every $i \leq n - k$ the element $y_ i = ux_ i + \sum _ j c_{ji}x_ j$ is annihilated by $f$. Since $M/\sum Ry_ i$ is generated by the images of $x_{n - k + 1}, \ldots , x_ n$ we win. $\square$

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