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Tag 07ZD

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

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 10.23). 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. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 1510–1521 (see updates for more information).

    \begin{lemma}
    \label{lemma-fitting-ideal-finite-locally-free}
    Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $r \geq 0$.
    The following are equivalent
    \begin{enumerate}
    \item $M$ is finite locally free of rank $r$
    (Algebra, Definition \ref{algebra-definition-locally-free}),
    \item $\text{Fit}_{r - 1}(M) = 0$ and $\text{Fit}_r(M) = R$, and
    \item $\text{Fit}_k(M) = 0$ for $k < r$ and $\text{Fit}_k(M) = R$
    for $k \geq r$.
    \end{enumerate}
    \end{lemma}
    
    \begin{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 \ref{lemma-fitting-ideal-basics}) we see that (1) implies (2) by
    Example \ref{example-fitting-free}
    and glueing results (Algebra, Section \ref{algebra-section-more-glueing}).
    Conversely, assume (2). By
    Lemma \ref{lemma-fitting-ideal-generate-locally} 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.
    \end{proof}

    Comments (2)

    Comment #1423 by Kestutis Cesnavicius on April 17, 2015 a 6:26 pm UTC

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

    Comment #1436 by Johan (site) on April 22, 2015 a 1:18 pm UTC

    Oops! Thanks, see here.

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