Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 39 (1998), No. 2, 249–254.

A Remark on Koszul Complexes

Winfried Bruns Udo Vetter

Universität Osnabrück, Fachbereich Mathematik/Informatik, D-49064 Osnabrück e-mail: Winfried.Bruns@mathematik.uni-osnabrueck.de Universität Oldenburg, Fachbereich Mathematik, D-26111 Oldenburg e-mail: vetter@mathematik.uni-oldenburg.de

Let R be a commutative ring and M an R-module. A linear form φ: M → R induces the Koszul complex

K(φ): ⋯ ⟶ ⋀^{i}M ⟶^{∂_{φ}} ⋀^{i−1}M ⟶ ⋯ ⟶ ⋀^{2}M ⟶ M ⟶ R ⟶ 0

where ∂_{φ}(m_{1} ∧ … ∧ m_{i}) = ∑_{j}(−1)^{j−1}φ(m_{j})m_{1} ∧ ⋯ m̂_{j} ⋯ ∧ m_{i} for all m_{1}, …, m_{i} ∈ M. When M is a finite free R-module, then the vanishing of the homology of K(φ) is determined by the grade g_{φ} of the ideal Im φ: one has H_{j}(K(φ)) = 0 for j = n−g_{φ}+1, …, n and H_{n−g_{φ}}(K(φ)) ≠ 0. Giving a linear form on a finite free R-module of rank n is equivalent to specifying a sequence a_{1}, …, a_{n} of elements of R.

If M is not a free R-module, then the picture changes drastically. In this note we investigate the simplest case of a non-free module, namely M = F/Rx where x is a nonzero element of the finite free R-module F. We define the linear map ψ: R → F by the assignment 1 ↦ x, so that M = F/ Im ψ. In plain terms, we now consider two sequences of elements of R, namely the sequence (x_{i}) of the components of x, and the sequence (y_{i}) = (φ(ē_{i})) where ē_{i} is the residue class of the ith element in a basis of F. These two sequences are connected by the equation ∑_{i} x_{i}y_{i} = 0.

It is of course not surprising that one can expect completely satisfactory results only when the ideals ∑ Ry_{i} = Im φ and ∑ Rx_{i} = Im ψ^{∗} have “almost” their maximal grade (the superscript ^{∗} denotes the R-dual). The grade g_{ψ^{∗}} of Im ψ^{∗} determines the homological properties of the exterior powers of M.

This note was motivated by a paper of Boffi [1], and its main application certainly is a simple and characteristic free approach to Boffi’s results.

For simplicity we assume in the following that R is noetherian. Using the general theory of grade (for example, see Bruns and Herzog [3], Chapter 9) one can often replace R by an arbitrary commutative ring.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1998 Heldermann Verlag

Let F be a free R-module of rank n, and ψ: R → F an R-homomorphism with cokernel M, as above. Furthermore we have a linear form φ on F such that φψ = 0; the induced linear form on M is denoted by φ̄ (it was φ above). For every non-negative integer i the map ⋀^{i}F → ⋀^{i+1}F, x ↦ x ∧ ψ(1), induces an exact sequence

⋀^{i}M ⟶^{ψ_{i+1}} ⋀^{i+1}F \underrightarrow{⋀^{i+1}π} ⋀^{i+1}M ⟶ 0,

where π: F → M denotes the natural projection.

It is crucial in the following to compare the Koszul complexes K(φ) of φ and K(φ̄) of φ̄. Since obviously

∂φ̄ ∘ (⋀^{i}π) = (⋀^{i−1}π) ∘ ∂_{φ} and ∂_{φ} ∘ ψ_{i} = ψ_{i−1} ∘ ∂φ̄,

we obtain an exact sequence of complexes

K(φ̄)[−1] ⟶^{(ψ_{i})} K(φ) \underrightarrow{(⋀^{i}π)} K(φ̄) ⟶ 0.

Proposition 1. For all i with n − g_{φ} + 1 ≤ i < g_{ψ^{∗}} one has

H_{i+1}(K(φ̄)) ≅ H_{i−1}(K(φ̄)).

Proof. For all j the module ⋀^{j}M has rank \binom{n−1}{j}, so Ker ψ_{j+1} = t(⋀^{j}M) (where t(N) denotes the torsion submodule of the R-module N).

Since i < g_{ψ^{∗}}, the modules ⋀^{j}M are torsionfree for j = 0, …, i; see Lebelt [4], Satz 0. Since this special case of Lebelt’s result is very easy, we include the argument: the sequence

0 ⟶ R ⟶^{ψ(1)∧} F ⟶^{ψ(1)∧} ⋯ ⟶^{ψ(1)∧} ⋀^{j}F ⟶ ⋀^{j}M ⟶ 0

is exact because of g_{ψ^{∗}} > j; it is in fact an augmented “tail” of the dual of the Koszul complex K(ψ^{∗}). An easy depth count yields depth ⋀^{j}M_{𝔭} ≥ min(1, depth R_{𝔭}) for all prime ideals of R, and therefore ⋀^{j}M is torsionfree.

Thus the diagram

0  0  0  0
↑  ↑  ↑  ↑
⋀^{i+2}M  ⟶ ⋀^{i+1}M ⟶^{∂φ̄} ⋀^{i}M ⟶ ⋀^{i−1}M
↑  ↑  ↑_{⋀^{i}π}  ↑
⋀^{i+2}F  ⟶ ⋀^{i+1}F ⟶^{∂_{φ}} ⋀^{i}F ⟶ ⋀^{i−1}F
↑  ↑  ↑^{ψ_{i}}  ↑
⋀^{i+1}M/t(⋀^{i+1}M) ⟶ ⋀^{i}M ⟶^{∂φ̄} ⋀^{i−1}M ⟶ ⋀^{i−2}M
↑  ↑  ↑  ↑
0  0  0  0

has exact columns (the maps in the lower left corner are the induced ones). Since n − g_{φ} + 1 ≤ i, we have H_{i+1}(K(φ)) = H_{i}(K(φ)) = 0. The assertion now follows from the long exact sequence in homology. □

Now we discuss the case in which g_{φ} has its maximal value n, or equivalently, K(φ) is acyclic.

Proposition 2. With the notation from above, suppose that g_{ψ^{∗}} = n. Then Im φ ⊂ Im ψ^{∗}, and for all i with n − g_{φ} ≤ i ≤ n one has

H_{i}(K(φ̄)) = \{ \begin{matrix}R/ Im ψ^{∗} & i ≡ n mod 2, \\ 0 & i ≢ n mod 2. \end{matrix}

In particular, if g_{ψ^{∗}} = n = g_{φ}, then n is even and Im φ = Im ψ^{∗}.

Proof. That H_{n}(K(φ̄)) = R/ Im ψ^{∗} is a consequence of ∂φ̄(⋀^{n}M) ⊂ t(⋀^{n−1}M) = 0 and ⋀^{n}M = R/ Im ψ^{∗}. The last equality implies Im φ ⊂ Im ψ^{∗}. Furthermore one has g_{φ} ≥ 1 if and only if H_{n−1}(K(φ̄)) = 0: if g_{φ} ≥ 1, then H_{n}(K(φ)) = 0, so ∂φ̄: ⋀^{n−1}M → ⋀^{n−2}M is injective. Conversely, if g_{φ} = 0, then there is an a ∈ R, a ≠ 0, with a Im φ = 0. Since ⋀^{n−1}M has rank 1, one has a⋀^{n−1}M ≠ 0; but ∂φ̄(a⋀^{n−1}M) = 0, so H_{n−1}(K(φ̄)) ≠ 0. The remaining assertions follow from Proposition 1. □

We add two

Remarks.

(1) By the Koszul complex, associated to the map φ̄: M → R, one often understands the complex

K^{∗}(φ̄): 0 ⟶ R^{∗} ⟶^{dφ̄} M^{∗} ⟶^{dφ̄} (⋀^{2}M)^{∗} ⟶ … ⟶^{dφ̄} (⋀^{n−1}M)^{∗}

where dφ̄ sends the element α ∈ (⋀^{i}M)^{∗} to φ̄ ∧ α. (Here we use that ⨁_{i}(⋀^{i}M)^{∗} is equipped with a multiplication ∧ which makes it an exterior algebra; for example, see Bruns [2], Theorem 1.4.) Choose an orientation χ on F and set

μ( ⋀^{n−1} π(x)) = χ(x ∧ ψ(1)),

for all x ∈ ⋀^{n−1}F, and

(μ^{i}(u))(v) = μ(u ∧ v)

for all u ∈ ⋀^{i}M, v ∈ ⋀^{n−1−i}M, 0 ≤ i ≤ n − 1, μ^{n} = 0; clearly μ^{i}: ⋀^{i}M → (⋀^{n−1−i}M)^{∗} is a homomorphism, and dφ̄ ∘ μ^{i} = (−1)^{n}μ^{i−1} ∘ ∂φ̄. If g_{ψ^{∗}} = n, then μ^{i} is an isomorphism for i = 0, …, n − 2 and μ^{n−1} is injective (see Bruns [2], Theorem 2.4). Thus, in this case, the homology of K^{∗}(φ̄) is completely determined by the homology of K(φ̄) (and vice versa).

(2) Suppose that x_{1}, …, x_{n} is a regular sequence in R, and consider the first syzygy N of the ideal in R generated by x_{1}, …, x_{n}. We claim that there is a y ∈ N, y ≠ 0, such that (N/Ry)_{𝔭} is free for every prime ideal 𝔭 ⊉ ∑ Rx_{i} if and only if n is even. To prove this set F = R^{n} and φ(e_{j}) = x_{j} where e_{1}, …, e_{n} are the elements of the canonical basis of F. For the ‘only if’ part consider the map ψ: R → F = R^{n} given by ψ(1) = y. The condition for y means that Im ψ^{∗} has grade n = grade φ. Thus, by Proposition 2, n must be even. If n is even, then y = ∑_{j=0}^{n−1}(−1)^{j}x_{n−j}e_{j+1} ∈ Ker φ = N satisfies the condition in the claim. (This construction is closely related to the construction, given by Trautmann in [5], 4.2.4, of a reflexive coherent analytic sheaf on ℂ^{n} with rank n−2 which is locally free on ℂ^{n} \{0} but not on ℂ^{n}.)

The subsequent example has been treated by Boffi in [1]. It was the impulse to our considerations.

Example. Let k be a field, and x_{1}, …, x_{m}, y_{1}, …, y_{p} indeterminates over k. Set \underline{x} = x_{1}, …, x_{m}, \underline{y} = y_{1}, …, y_{p}, R = k[\underline{x}, \underline{y}], and

A = ∑_{{1≤i≤m \atop 1≤j≤p}} a_{ij}x_{i}y_{j}, a_{ij} ∈ k.

Put F = R^{m} ⊕ R^{p}, and define the R-homomorphism ψ: R → F by ψ(1) = (\underline{x}, \underline{y}). Denote by d_{1}, …, d_{m} and e_{1}, …, e_{p} the canonical bases of R^{m} and R^{p}. The linear form φ on F is given by φ(d_{i}) = \frac{∂A}{∂x_{i}} , φ(e_{j}) = −\frac{∂A}{∂y_{j}}. Clearly φψ = 0. As above let φ̄ be the induced linear form on the cokernel of ψ.

Observe that

(φ(d_{1}), …, φ(e_{p})) = ψ(1) ( \begin{matrix}0 & −(a_{ij}) \\ (a_{ij})^{t} & 0 \end{matrix} )

where the upper t means ‘transpose’.

From Proposition 2 we draw:

Proposition 3. Set r = rank(a_{ij}). Then H_{m+p−λ}(K(φ̄)) ≠ 0 for λ = 2r, …, m + p. Furthermore, for 0 ≤ λ ≤ 2r one has

H_{m+p−λ}(K(φ̄)) = \{ \begin{matrix}k & for λ ≡ 0 mod 2 \\ 0 & for λ ≢ 0 mod 2. \end{matrix}

Proof. Since grade Im φ = 2r, the second statement is an immediate consequence of Proposition 2. To prove the first one, observe that H_{λ}(K(φ̄)) = 0 implies Ker ∂_{φ} ⊂ Im ∂_{φ}+Im ψ_{λ}. Consider the systems of linear equations

∑_{i=1}^{m} a_{ij}a_{i} = 0, j = 1, …, p,  and  ∑_{j=1}^{p} a_{ij}b_{j} = 0, i = 1, …, m,

over k. There are linearly independent elements a^{(1)}, …, a^{(m−r)} in k^{m} solving the first system and linearly independent elements b^{(1)}, …, b^{(p−r)} in k^{p} solving the second one. Set v_{i} = (a^{(i)}, 0) ∈ k^{m+p}, i = 1, …, m − r, and w_{j} = (0, b^{(j)}) ∈ k^{m+p}, j = 1, …, p − r. Obviously φ(v_{i}) = φ(w_{j}) = 0. So if 0 ≤ ρ ≤ m − r, 0 ≤ σ ≤ p − r, and v_{i_{1}}, …, v_{i_{ρ}}, w_{j_{1}}, …, w_{j_{σ}} are distinct, then v_{i_{1}} ∧ … ∧ v_{i_{ρ}} ∧ w_{j_{1}} ∧ … ∧ w_{j_{σ}} lies in Ker ∂_{φ} but not in Im ∂_{φ} + Im ψ_{ρ+σ}. Consequently H_{ρ+σ}(K(φ̄)) ≠ 0. □

In particular Proposition 3 implies the main results of Boffi ([1], Theorem 7 and Propositions 8–10) and answers some questions asked there:

Proposition 4. The following conditions are equivalent:

(1) m = p and the matrix (a_{ij}) is invertible.

(2)  H_{i}(K(φ̄)) = \{ \begin{matrix}k & if i ≤ 2m is even \\ 0 & otherwise. \end{matrix}

(3) H_{1}(K(φ̄)) = 0.

Also in the case in which g_{ψ^{∗}} and g_{φ} are “submaximal” one gets a satisfactory result on the behaviour of K(φ̄).

Proposition 5. With the notation introduced above Proposition 1 suppose that g_{ψ^{∗}} = g_{φ} = dim R = n − 1. Then all homology modules H_{i}(K(φ̄)), 1 ≤ i ≤ n, have the same (finite) length. In particular, the following conditions are equivalent:

(1) H_{i}(K(φ̄)) = 0 for i = 1, …, n;

(2) H_{j}(Kφ̄)) = 0 for some j, 1 ≤ j ≤ n.

Proof. First observe that ⋀^{i}M is torsionfree for i = 0, …, n − 2. Corresponding to the diagram in the proof of Proposition 1 we obtain an exact sequence of complexes

0 ⟶ K(φ̄)[−1]′ ⟶^{(ψ′_{i})} K(φ) \underrightarrow{(⋀^{i}π)} K(φ̄) ⟶ 0

where K(φ̄)[−1]′ is the complex K(φ̄)[−1] except that the map ∂φ̄: ⋀^{n−1}M → ⋀^{n−2}M is replaced by the induced map ⋀^{n−1}M/t(⋀^{n−1}M) → ⋀^{n−2}M; correspondingly one chooses ψ′_{i} = ψ_{i} for i < n and replaces ψ_{n} by the induced map ψ′_{n}. From the long exact sequence in homology we then cut the exact piece

0 ⟶ H_{2}(K(φ̄)) ⟶ R/ Im φ ⟶ H_{1}(K(φ)) ⟶ H_{1}(K(φ̄)) ⟶ 0.

Since ℓ(H_{1}(K(φ)) = ℓ(H_{0}(K(φ)) = ℓ(R/ Im φ) < ∞ (ℓ means ‘length’), it follows that ℓ(H_{2}(K(φ̄)) = ℓ(H_{1}(K(φ̄)). The remaining assertions result from Proposition 1. □

Example. Let k be a field of characteristic 0, P the polynomial ring over k in the indeterminates X_{1}, …, X_{n}, graded by deg X_{i} = m_{i} > 0, i = 1, …, n. Choose a nonconstant homogeneous polynomial f ∈ P, and set R = P/(f). Let F = R^{n} and define the R-homomorphism ψ: R → F via ψ(1) = (∂f/∂X_{1}, …, ∂f/∂X_{n}) mod f · F. Then the cokernel of ψ is the R-module D_{k}(R) of Kähler differentials of R over k. The linear form φ on F maps the ith unit vector to m_{i}x_{i} where x_{i} denotes the residue of X_{i} in R; so φ̄ is the linear form on D_{k}(R) induced by the Euler derivation. Clearly φψ = 0. A direct computation yields H_{1}(K(φ̄)) = 0.

Suppose now that g_{ψ^{∗}} = n − 1; this means that (the affine variety with coordinate ring) R has an isolated singularity at 0. By Proposition 5, it results the well-known fact that the Koszul complex of φ̄ is exact.

The same is true for K(ψ̄^{∗}); by ψ̄^{∗} we of course mean the linear form induced by ψ^{∗} on the cokernel of φ^{∗}: R → F^{∗}. In fact, there are only two possibilities for H_{n}(K(ψ̄^{∗})), namely H_{n}(K(ψ̄^{∗})) = 0 and H_{n}(K(ψ̄^{∗})) = k; but the second case is actually impossible since it would imply that H_{1}(K(φ)) = 0.

References

[1] Boffi, G.: Bilinear Forms and (Hyper-)Determinants. Adv. in Math. 123 (1996), 91–103.

[2] Bruns, W.: The Buchsbaum-Eisenbud Structure Theorems and Alternating Syzygies. Comm. Algebra 15 (1987), 873–925.

[3] Bruns, W.; Herzog, J.: Cohen–Macaulay rings. Cambridge University Press 1993.

[4] Lebelt, K.: Torsion äußerer Potenzen von Moduln der homologischen Dimension 1. Math. Ann. 211 (1974), 183–197.

[5] Trautmann, G.: Darstellung von Vektorraumbündeln über ℂ^{n}\{0}. Arch. Math. XXIV (1973), 303–313.

Received June 11, 1996