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Tag 07ZA

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

Lemma 15.8.4. Let $R$ be a ring. Let $M$ be a finite $R$-module.

  1. If $M$ can be generated by $n$ elements, then $\text{Fit}_n(M) = R$.
  2. Given a second finite $R$-module $M'$ we have $$ \text{Fit}_l(M \oplus M') = \sum\nolimits_{k + k' = l} \text{Fit}_k(M)\text{Fit}_{k'}(M') $$
  3. If $R \to R'$ is a ring map, then $\text{Fit}_k(M \otimes_R R')$ is the ideal of $R'$ generated by the image of $\text{Fit}_k(M)$.
  4. If $M$ is of finite presentation, then $\text{Fit}_k(M)$ is a finitely generated ideal.
  5. If $M \to M'$ is a surjection, then $\text{Fit}_k(M) \subset \text{Fit}_k(M')$.
  6. We have $\text{Fit}_0(M) \subset \text{Ann}_R(M)$.
  7. We have $V(\text{Fit}_0(M)) = \text{Supp}(M)$.
  8. Add more here.

Proof. Part (1) follows from the fact that $I_0(A) = R$ in Lemma 15.8.1.

Part (2) follows form the corresponding statement in Lemma 15.8.1.

Part (3) follows from the fact that $\otimes_R R'$ is right exact, so the base change of a presentation of $M$ is a presentation of $M \otimes_R R'$.

Proof of (4). Let $R^{\oplus m} \xrightarrow{A} R^{\oplus n} \to M \to 0$ be a presentation. Then $\text{Fit}_k(M)$ is the ideal generated by the $n - k \times n - k$ minors of the matrix $A$.

Part (5) is immediate from the definition.

Proof of (6). Choose a presentation of $M$ with matrix $A$ as in Lemma 15.8.2. Let $J' \subset J$ be a subset of cardinality $n$. It suffices to show that $f = \det(a_{ij})_{i = 1, \ldots, n, j \in J'}$ annihilates $M$. This is clear because the cokernel of $$ R^{\oplus n} \xrightarrow{A' = (a_{ij})_{i = 1, \ldots, n, j \in J'}} R^{\oplus n} \to M \to 0 $$ is killed by $n$ as there is a matrix $B$ with $A' B = f1_{n \times n}$.

Proof of (7). Choose a presentation of $M$ with matrix $A$ as in Lemma 15.8.2. By Nakayama's lemma (Algebra, Lemma 10.19.1) we have $$ M_\mathfrak p \not = 0 \Leftrightarrow M \otimes_R \kappa(\mathfrak p) \not = 0 \Leftrightarrow \text{rank}(\text{image }A\text{ in }\kappa(\mathfrak p)) \leq n $$ Clearly $\text{Fit}_0(M)$ exactly cuts out the set of primes with this property. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 1402–1423 (see updates for more information).

    \begin{lemma}
    \label{lemma-fitting-ideal-basics}
    Let $R$ be a ring. Let $M$ be a finite $R$-module.
    \begin{enumerate}
    \item If $M$ can be generated by $n$ elements, then
    $\text{Fit}_n(M) = R$.
    \item Given a second finite $R$-module $M'$ we have
    $$
    \text{Fit}_l(M \oplus M') =
    \sum\nolimits_{k + k' = l} \text{Fit}_k(M)\text{Fit}_{k'}(M')
    $$
    \item If $R \to R'$ is a ring map, then $\text{Fit}_k(M \otimes_R R')$
    is the ideal of $R'$ generated by the image of $\text{Fit}_k(M)$.
    \item If $M$ is of finite presentation, then $\text{Fit}_k(M)$
    is a finitely generated ideal.
    \item If $M \to M'$ is a surjection, then
    $\text{Fit}_k(M) \subset \text{Fit}_k(M')$.
    \item We have $\text{Fit}_0(M) \subset \text{Ann}_R(M)$.
    \item We have $V(\text{Fit}_0(M)) = \text{Supp}(M)$.
    \item Add more here.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Part (1) follows from the fact that $I_0(A) = R$ in
    Lemma \ref{lemma-ideals-generated-by-minors}.
    
    \medskip\noindent
    Part (2) follows form the corresponding statement in
    Lemma \ref{lemma-ideals-generated-by-minors}.
    
    \medskip\noindent
    Part (3) follows from the fact that $\otimes_R R'$ is right exact,
    so the base change of a presentation of $M$ is a presentation of
    $M \otimes_R R'$.
    
    \medskip\noindent
    Proof of (4). Let $R^{\oplus m} \xrightarrow{A} R^{\oplus n} \to M \to 0$
    be a presentation. Then $\text{Fit}_k(M)$ is the ideal generated by the
    $n - k \times n - k$ minors of the matrix $A$.
    
    \medskip\noindent
    Part (5) is immediate from the definition.
    
    \medskip\noindent
    Proof of (6). Choose a presentation of $M$ with matrix $A$
    as in Lemma \ref{lemma-fitting-ideal}.
    Let $J' \subset J$ be a subset of cardinality $n$.
    It suffices to show that
    $f = \det(a_{ij})_{i = 1, \ldots, n, j \in J'}$
    annihilates $M$.
    This is clear because the cokernel of
    $$
    R^{\oplus n} \xrightarrow{A' = (a_{ij})_{i = 1, \ldots, n, j \in J'}}
    R^{\oplus n} \to M \to 0
    $$
    is killed by $n$ as there is a matrix $B$ with $A' B = f1_{n \times n}$.
    
    \medskip\noindent
    Proof of (7). Choose a presentation of $M$ with matrix $A$
    as in Lemma \ref{lemma-fitting-ideal}.
    By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK})
    we have
    $$
    M_\mathfrak p \not = 0
    \Leftrightarrow
    M \otimes_R \kappa(\mathfrak p) \not = 0
    \Leftrightarrow
    \text{rank}(\text{image }A\text{ in }\kappa(\mathfrak p)) \leq n
    $$
    Clearly $\text{Fit}_0(M)$ exactly cuts out the set of primes
    with this property.
    \end{proof}

    Comments (2)

    Comment #2062 by Kestutis Cesnavicius on June 11, 2016 a 3:38 pm UTC

    On the left hand side of the equation in (2) the subscript $k$ should be $l$.

    Comment #2092 by Johan (site) on June 16, 2016 a 3:17 pm UTC

    Thanks! Fixed here.

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