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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 $f$ 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. $\square$


Comments (2)

Comment #2062 by Kestutis Cesnavicius on

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

There are also:

  • 4 comment(s) on Section 15.8: Fitting ideals

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