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Tag 080Z

Chapter 15: More on Algebra > Section 15.8: Fitting ideals

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 10.19.1. 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 10.14.5 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. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 1540–1546 (see updates for more information).

    \begin{lemma}
    \label{lemma-principal-fitting-ideal}
    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.
    \end{lemma}
    
    \begin{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 \ref{algebra-lemma-NAK}. 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 \ref{algebra-lemma-matrix-right-inverse} 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.
    \end{proof}

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