The Stacks project

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 $\text{Fit}_0(M \oplus M') = \text{Fit}_0(M)\text{Fit}_0(M')$ and more generally

    \[ \text{Fit}_ l(M \oplus M') = \sum \nolimits _{k + k' = l} \text{Fit}_ k(M)\text{Fit}_{k'}(M') \]

  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. If $I$ is an ideal of $R$, then $\text{Fit}_0(R/I) = I$.

  9. 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 10.20.1) 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:

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

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07ZA. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.8.4 as pdf