## 15.8 Fitting ideals

The Fitting ideals of a finite module are the ideals determined by the construction of Lemma 15.8.2.

Lemma 15.8.1. Let $R$ be a ring. Let $A$ be an $n \times m$ matrix with coefficients in $R$. Let $I_ r(A)$ be the ideal generated by the $r \times r$-minors of $A$ with the convention that $I_0(A) = R$ and $I_ r(A) = 0$ if $r > \min (n, m)$. Then

$I_0(A) \supset I_1(A) \supset I_2(A) \supset \ldots $,

if $B$ is an $(n + n') \times m$ matrix, and $A$ is the first $n$ rows of $B$, then $I_{r + n'}(B) \subset I_ r(A)$,

if $C$ is an $n \times n$ matrix then $I_ r(CA) \subset I_ r(A)$.

If $A$ is a block matrix

\[ \left( \begin{matrix} A_1
& 0
\\ 0
& A_2
\end{matrix} \right) \]

then $I_ r(A) = \sum _{r_1 + r_2 = r} I_{r_1}(A_1) I_{r_2}(A_2)$.

Add more here.

**Proof.**
Omitted. (Hint: Use that a determinant can be computed by expanding along a column or a row.)
$\square$

Lemma 15.8.2. Let $R$ be a ring. Let $M$ be a finite $R$-module. Choose a presentation

\[ \bigoplus \nolimits _{j \in J} R \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0. \]

of $M$. Let $A = (a_{ij})_{i = 1, \ldots , n, j \in J}$ be the matrix of the map $\bigoplus _{j \in J} R \to R^{\oplus n}$. The ideal $\text{Fit}_ k(M)$ generated by the $(n - k) \times (n - k)$ minors of $A$ is independent of the choice of the presentation.

**Proof.**
Let $K \subset R^{\oplus n}$ be the kernel of the surjection $R^{\oplus n} \to M$. Pick $z_1, \ldots , z_{n - k} \in K$ and write $z_ j = (z_{1j}, \ldots , z_{nj})$. Another description of the ideal $\text{Fit}_ k(M)$ is that it is the ideal generated by the $(n - k) \times (n - k)$ minors of all the matrices $(z_{ij})$ we obtain in this way.

Suppose we change the surjection into the surjection $R^{\oplus n + n'} \to M$ with kernel $K'$ where we use the original map on the first $n$ standard basis elements of $R^{\oplus n + n'}$ and $0$ on the last $n'$ basis vectors. Then the corresponding ideals are the same. Namely, if $z_1, \ldots , z_{n - k} \in K$ as above, let $z'_ j = (z_{1j}, \ldots , z_{nj}, 0, \ldots , 0) \in K'$ for $j = 1, \ldots , n - k$ and $z'_{n + j'} = (0, \ldots , 0, 1, 0, \ldots , 0) \in K'$. Then we see that the ideal of $(n - k) \times (n - k)$ minors of $(z_{ij})$ agrees with the ideal of $(n + n' - k) \times (n + n' - k)$ minors of $(z'_{ij})$. This gives one of the inclusions. Conversely, given $z'_1, \ldots , z'_{n + n' - k}$ in $K'$ we can project these to $R^{\oplus n}$ to get $z_1, \ldots , z_{n + n' - k}$ in $K$. By Lemma 15.8.1 we see that the ideal generated by the $(n + n' - k) \times (n + n' - k)$ minors of $(z'_{ij})$ is contained in the ideal generated by the $(n - k) \times (n - k)$ minors of $(z_{ij})$. This gives the other inclusion.

Let $R^{\oplus m} \to M$ be another surjection with kernel $L$. By Schanuel's lemma (Algebra, Lemma 10.108.1) and the results of the previous paragraph, we may assume $m = n$ and that there is an isomorphism $R^{\oplus n} \to R^{\oplus m}$ commuting with the surjections to $M$. Let $C = (c_{li})$ be the (invertible) matrix of this map (it is a square matrix as $n = m$). Then given $z'_1, \ldots , z'_{n - k} \in L$ as above we can find $z_1, \ldots , z_{n - k} \in K$ with $z_1' = Cz_1, \ldots , z'_{n - k} = Cz_{n - k}$. By Lemma 15.8.1 we get one of the inclusions. By symmetry we get the other.
$\square$

Definition 15.8.3. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$. The *$k$th Fitting ideal* of $M$ is the ideal $\text{Fit}_ k(M)$ constructed in Lemma 15.8.2. Set $\text{Fit}_{-1}(M) = 0$.

Since the Fitting ideals are the ideals of minors of a big matrix (numbered in reverse ordering from the ordering in Lemma 15.8.1) we see that

\[ 0 = \text{Fit}_{-1}(M) \subset \text{Fit}_0(M) \subset \text{Fit}_1(M) \subset \ldots \subset \text{Fit}_ t(M) = R \]

for some $t \gg 0$. Here are some basic properties of 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 $f$ 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

\[ M_\mathfrak p \not= 0 \Leftrightarrow M \otimes _ R \kappa (\mathfrak p) \not= 0 \Leftrightarrow \text{rank}(\text{image }A\text{ in }\kappa (\mathfrak p)) < n \]

Clearly $\text{Fit}_0(M)$ exactly cuts out the set of primes with this property.
$\square$

Example 15.8.5. Let $R$ be a ring. The Fitting ideals of the finite free module $M = R^{\oplus n}$ are $\text{Fit}_ k(M) = 0$ for $k < n$ and $\text{Fit}_ k(M) = R$ for $k \geq n$.

Lemma 15.8.6. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$. Let $\mathfrak p \subset R$ be a prime ideal. The following are equivalent

$\text{Fit}_ k(M) \not\subset \mathfrak p$,

$\dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p) \leq k$,

$M_\mathfrak p$ can be generated by $k$ element over $R_\mathfrak p$, and

$M_ f$ can be generated by $k$ elements over $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$.

**Proof.**
By Nakayama's lemma (Algebra, Lemma 10.19.1) we see that $M_ f$ can be generated by $k$ elements over $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$ if $M \otimes _ R \kappa (\mathfrak p)$ can be generated by $k$ elements. Hence (2), (3), and (4) are equivalent. Using Lemma 15.8.4 part (3) this reduces the problem to the case where $R$ is a field and $\mathfrak p = (0)$. In this case the result follows from Example 15.8.5.
$\square$

Lemma 15.8.7. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $r \geq 0$. The following are equivalent

$M$ is finite locally free of rank $r$ (Algebra, Definition 10.77.1),

$\text{Fit}_{r - 1}(M) = 0$ and $\text{Fit}_ r(M) = R$, and

$\text{Fit}_ k(M) = 0$ 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 10.22). 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.
$\square$

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 10.14.6 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.
$\square$

Lemma 15.8.9. Let $R$ be a ring. Let $M$ be a finitely presented $R$-module. Let $k \geq 0$. Assume that $\text{Fit}_ k(M) = (f)$ for some nonzerodivisor $f \in R$ and $\text{Fit}_{k - 1}(M) = 0$. Then $M$ has projective dimension $\leq 1$.

**Proof.**
Choose a presentation

\[ R^{\oplus m} \xrightarrow {A} R^{\oplus n} \to M \to 0 \]

We claim the image of $A$ is finite locally free of rank $n - k$. If the claim holds, then the lemma is true by definition of projective dimension. To prove the claim we may replace $R$ by the localization at a prime, see Algebra, Lemma 10.77.2. This reduces us the the case discussed in the next paragraph.

Assume $R$ is local. Set $M' = \{ x \in M \mid fx = 0\} $. By Lemma 15.8.8 we can choose $x_1, \ldots , x_ k \in M$ which generate $M/M'$. Then $x_1, \ldots , x_ k$ generate $M_ f = (M/M')_ f$. Hence, if there is a relation $\sum a_ ix_ i = 0$ in $M$, then we see that $a_1, \ldots , a_ k$ map to zero in $R_ f$ since otherwise $\text{Fit}_{k - 1}(M) R_ f = \text{Fit}_{k - 1}(M_ f)$ would be nonzero. Since $f$ is a nonzerodivisor, we conclude $a_1 = \ldots = a_ k = 0$. Thus $M \cong R^{\oplus k} \oplus M'$. After a change of basis in our presentation above, we may assume the first $n - k$ basis vectors of $R^{\oplus n}$ map into the summand $M'$ of $M$ and the last $k$-basis vectors of $R^{\oplus n}$ map to basis elements of the summand $R^{\oplus k}$ of $M$. Having done so, the last $k$ rows of the matrix $A$ vanish. In this way we see that, replacing $M$ by $M'$, $k$ by $0$, $n$ by $n - k$, and $A$ by the submatrix where we delete the last $k$ rows, we reduce to the case discussed in the next paragraph.

Assume $R$ is local, $k = 0$, and $M$ annihilated by $f$. Now the $0$th Fitting ideal of $M$ is $(f)$ and is generated by the $n \times n$ minors of the matrix $A$ of size $n \times m$. (This in particular implies $m \geq n$.) Since $R$ is local, some $n \times n$ minor of $A$ is $uf$ for a unit $u \in R$. After renumbering we may assume this minor is the first one. Moreover, we know all other $n \times n$ minors of $A$ are divisible by $f$. Write $A = (A_1 A_2)$ in block form where $A_1$ is an $n \times n$ matrix and $A_2$ is an $n \times (m - n)$ matrix. By Algebra, Lemma 10.14.6 applied to the transpose of $A$ (!) we find there exists an $n \times n$ matrix $B$ such that

\[ BA = B(A_1 A_2) = f \left( \begin{matrix} u 1_{n \times n}
& C
\end{matrix} \right) \]

for some $n \times (m - n)$ matrix $C$ with coefficients in $R$. Then we first conclude $BA_1 = fu 1_{n \times n}$. Thus

\[ BA_2 = fC = u^{-1}fuC = u^{-1}BA_1C \]

Since the determinant of $B$ is a nonzerodivisor we conclude that $A_2 = u^{-1}A_1C$. Therefore the image of $A$ is equal to the image of $A_1$ which is isomorphic to $R^{\oplus n}$ because the determinant of $A_1$ is a nonzerodivisor. This finishes the proof.
$\square$

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