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The Stacks project

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

  1. I_0(A) \supset I_1(A) \supset I_2(A) \supset \ldots ,

  2. 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),

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

  4. 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).

  5. 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 10.109.1) 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 kth 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.

  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

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.

Example 15.8.6. Let R be a ring and let I, J\subset R be ideals. Then \text{Fit}_0(R/I) = I, \text{Fit}_0(R/I \oplus R/J) = IJ, \text{Fit}_1(R/I\oplus R/J) = I+J.

Lemma 15.8.7. 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

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

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

  3. M_\mathfrak p can be generated by k elements over R_\mathfrak p, and

  4. 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 10.20.1) 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 15.8.8. Let R be a ring. Let M be a finite R-module. Let r \geq 0. The following are equivalent

  1. M is finite locally free of rank r (Algebra, Definition 10.78.1),

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

  3. \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 10.23). Conversely, assume (2). By Lemma 15.8.7 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 15.8.9. 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 10.20.1. 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 10.15.6 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 15.8.10. 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

  1. M has projective dimension \leq 1,

  2. M' = \mathop{\mathrm{Ker}}(f : M \to M) is the f-power torsion submodule of M,

  3. M' has projective dimension \leq 1,

  4. M/M' is finite locally free of rank k, and

  5. M \cong M/M' \oplus M'.

Proof. Choose a presentation

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

for some matrix A with coefficients in R.

We first prove the lemma when R is local. Set M' = \{ x \in M \mid fx = 0\} as in the statement. By Lemma 15.8.9 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 0th 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 10.15.6 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. Hence M has projective dimension \leq 1.

We return to the case of a general ring R. By the local case we see that M/M' is a finite locally free module of rank k, see Algebra, Lemma 10.78.2. Hence the extension 0 \to M' \to M \to M/M' \to 0 splits. It follows that M' is a finitely presented module. Choose a short exact sequence 0 \to K \to R^{\oplus a} \to M' \to 0. Then K is a finite R-module, see Algebra, Lemma 10.5.3. By the local case we see that K_\mathfrak p \cong R_\mathfrak p^{\oplus a} for all primes. Hence by Algebra, Lemma 10.78.2 again we see that K is finite locally free of rank a. It follows that M' has projective dimension \leq 1 and the lemma is proved. \square


Comments (8)

Comment #3040 by SE user on

Lemma 15.8.4, Proof of (6): "killed by " or "killed by "?

Comment #3385 by shanbei on

In the third line of proof of (7) in Lemma 07ZA, perhaps you meant the rank of image is less than n instead of \leq?

Comment #7652 by algori on

In lemma 0F7M, part (4), isn't it supposed to be "locally free of rank " rather than of rank ?

Comment #8719 by Bjorn Poonen on

Maybe add some easy but useful special cases to Lemma 07ZA:

One could add a new item "If is an ideal of , then ."

In (2), after "we have" and before the formula one could insert " and more generally" (I think that that is the most commonly used case of (2)).

In the proof of (4), it would be good to put in parentheses, twice.


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