## 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.

- If $M$ can be generated by $n$ elements, then $\text{Fit}_n(M) = R$.
- Given a second finite $R$-module $M'$ we have $$ \text{Fit}_l(M \oplus M') = \sum\nolimits_{k + k' = l} \text{Fit}_k(M)\text{Fit}_{k'}(M') $$
- 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 $\text{Fit}_k(M)$.
- If $M$ is of finite presentation, then $\text{Fit}_k(M)$ is a finitely generated ideal.
- If $M \to M'$ is a surjection, then $\text{Fit}_k(M) \subset \text{Fit}_k(M')$.
- We have $\text{Fit}_0(M) \subset \text{Ann}_R(M)$.
- We have $V(\text{Fit}_0(M)) = \text{Supp}(M)$.
- 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}
```

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