Koszul complex

Definition

Let R be a commutative ring and E a free module of finite rank r over R. We write ∧ i E for the i-th exterior power of E. Then, given an R-linear map s : E → R , the Koszul complex associated to s is the chain complex of R-modules:

where the differential d_k is given by: for any e_i in E,

The superscript ⋅ ^ means the term is omitted. (Showing d k ∘ d k + 1 = 0 is straightforward; alternatively, this identity also follows using the #Self-duality of a Koszul complex.)

Note ∧ 1 E = E and d 1 = s . Note also ∧ r E ≃ R ; this isomorphism is not canonical (for example, a choice of a volume form in differential geometry is an example of such an isomorphism.)

If E = R^r (i.e., an ordered basis is chosen), then, giving an R-linear map s: R^r → R amounts to giving a finite sequence s₁, ..., s_r of elements in R (namely, a row vector) and then one sets K ∙ ( s 1 , … , s r ) = K ∙ ( s ) .

If M is a finitely generated R-module, then one sets:

which is again a chain complex with the induced differential ( d ⊗ 1 M ) ( v ⊗ m ) = d ( v ) ⊗ m .

The i-th homology of the Koszul complex

is called the i-th Koszul homology. For example, if E = R^r and s = [ s 1 ⋯ s r ] is a row vector with entries in R, then d 1 ⊗ 1 M is

and so

Similarly,

Koszul complexes in low dimensions

Given a ring R, an element x in R, and an R-module M, the multiplication by x yields a homomorphism of R-modules,

Considering this as a chain complex (by putting them in degree 1 and 0, and adding zeros elsewhere), it is denoted by K ( x , M ) . By construction, the homologies are

the annihilator of x in M. Thus, the Koszul complex and its homology encode fundamental properties of the multiplication by x.

This chain complex K_•(x) is called the Koszul complex of R with respect to x, as in #Definition. The Koszul complex for a pair ( x , y ) ∈ R 2 is

with the matrices d 1 and d 2 given by

Note that d_i is applied on the left. The cycles in degree 1 are then exactly the linear relations on the elements x and y, while the boundaries are the trivial relations. The first Koszul homology H₁(K_•(x, y)) therefore measures exactly the relations mod the trivial relations. With more elements the higher-dimensional Koszul homologies measure the higher-level versions of this.

In the case that the elements x₁, x₂, ..., x_n form a regular sequence, the higher homology modules of the Koszul complex are all zero.

Example

If k is a field and X₁, X₂, ..., X_d are indeterminates and R is the polynomial ring k[X₁, X₂, ..., X_d], the Koszul complex K_•(X_i) on the X_i's forms a concrete free R-resolution of k.

Properties of a Koszul homology

Let E be a finite-rank free module over R, s: E → R an R-linear map and t an element of R. Let K ( s , t ) be the Koszul complex of ( s , t ) : E ⊕ R → R . Let M be a finitely generated module over R.

Using ∧ k ( E ⊕ R ) = ⊕ i = 0 k ∧ k − i E ⊗ ∧ i R = ∧ k E ⊕ ∧ k − 1 E , there is the exact sequence of complexes:

where [-1] signifies the degree shift by -1 and d K ( s ) [ − 1 ] = − d K ( s ) . One notes: for (x, y) in ∧ k E ⊕ ∧ k − 1 E ,

(In the language of homological algebra, the above means that K ( s , t ) is the mapping cone of t : K ( s ) [ − 1 ] → K ( s ) .)

Taking the long exact sequence of homologies, we get:

Here, the connecting homomorphism

is computed as follows. By definition, δ ( [ x ] ) = [ d K ( s , t ) ( y ) ] where y is an element of K ( s , t ) that maps to x. Since K ( s , t ) is a direct sum, we can simply take y to be (0, x). Then the early formula for d K ( s , t ) gives δ ( [ x ] ) = t [ x ] .

The above exact sequence can be used to prove the following.

Proof by induction on r. If r = 1, then H 1 ⁡ ( K ( x 1 ; M ) ) = Ann M ⁡ ( x 1 ) = 0 . Next, assume the assertion is true for r - 1. Then, using the above exact sequence, one sees H i ⁡ ( K ( x 1 , … , x r ; M ) ) = 0 for any i ≥ 2. The vanishing is also valid for i = 1, since x r is a nonzerodivisor on H 0 ⁡ ( K ( x 1 , … , x r − 1 ; M ) ) = M / ( x 1 , … , x r − 1 ) M . ◻

Proof: By the theorem applied to S and S as an S-module, we see K(y₁, ..., y_n) is an S-free resolution of S/(y₁, ..., y_n). So, by definition, the i-th homology of K ( y 1 , … , y n ) ⊗ S M is the right-hand side of the above. On the other hand, K ( y 1 , … , y n ) ⊗ S M = K ( x 1 , … , x n ) ⊗ R M by the definition of the S-module structure on M. ◻

Proof: Let S = R[y₁, ..., y_n]. Turn M into an S-module through the ring homomorphism S → R, y_i → x_i and R an S-module through y_i → 0. By the preceding corollary, H i ⁡ ( K ( x 1 , … , x n ) ⊗ M ) = Tor i S ⁡ ( R , M ) and then

For a Noetherian local ring, the converse of the theorem holds. More generally,

Proof: We only need to show 2. implies 1., the rest being clear. We argue by induction on r. The case r = 1 is already known. Let x' denote x₁, ..., x_r-1. Consider

Since the first x r is surjective, N = x r N with N = H 1 ⁡ ( K ( x ′ ; M ) ) . By Nakayama's lemma, N = 0 and so x' is a regular sequence by the inductive hypothesis. Since the second x r is injective (i.e., is a nonzerodivisor), x 1 , … , x r is a regular sequence. (Note: by Nakayama's lemma, the requirement M / ( x 1 , … , x r ) M ≠ 0 is automatic.) ◻

Tensor products of Koszul complexes

In general, if C, D are chain complexes, then their tensor product C ⊗ D is the chain complex given by

with the differential: for any homogeneous elements x, y,

where |x| is the degree of x.

This construction applies in particular to Koszul complexes. Let E, F be finite-rank free modules, s: E → R, t: F → R R-linear maps. Let K ( s , t ) be the Koszul complex of the linear map ( s , t ) : E ⊗ F → R . Then, as complexes,

To see this, it is more convenient to work with an exterior algebra (as opposed to exterior powers). Define the graded derivation of degree -1

by requiring: for any homogeneous elements x, y in ΛE,

One easily sees that d s ∘ d s = 0 (induction on degree) and that the action of d_s on homogeneous elements agrees with the differentials in #Definition.

Now, we have ∧ ( E ⊕ F ) = ∧ E ⊗ ∧ F as graded R-modules. Also, by the definition of a tensor product mentioned in the beginning,

Since d K ( s ) ⊗ K ( t ) and d K ( s , t ) are derivations of the same type, this implies d K ( s ) ⊗ K ( t ) = d K ( s , t ) .

Note, in particular,

The next proposition shows how the Koszul complex of elements encodes some information about sequences in the ideal generated by them.

Proof: (Easy but omitted for now)

As an application, we can show the depth-sensitivity of a Koszul homology. Given a finitely generated module M over a ring R, by (one) definition, the depth of M with respect to an ideal I is the supremum of the lengths of all regular sequences of elements of I on M. It is denoted by depth ⁡ ( I , M ) . Recall that an M-regular sequence x₁, ..., x_n in an ideal I is maximal if I contains no nonzerodivisor on M / ( x 1 , … , x n ) M .

The Koszul homology gives a very useful characterization of a depth.

Proof: To lighten the notations, we write H(-) for H(K(-)). Let y₁, ..., y_s be a maximal M-regular sequence in the ideal I; we denote this sequence by y _ . First we show, by induction on l , the claim that H i ⁡ ( y _ , x 1 , … , x l ; M ) is Ann M / y _ M ⁡ ( x 1 , … , x l ) if i = l and is zero if i > l . The basic case l = 0 is clear from #Properties of a Koszul homology. From the long exact sequence of Koszul homologies and the inductive hypothesis,

which is Ann M / y _ M ⁡ ( x 1 , … , x l ) . Also, by the same argument, the vanishing holds for i > l . This completes the proof of the claim.

Now, it follows from the claim and the early proposition that H i ⁡ ( x 1 , … , x n ; M ) = 0 for all i > n - s. To conclude n - s = m, it remains to show that it is nonzero if i = n - s. Since y _ is a maximal M-regular sequence in I, the ideal I is contained in the set of all zerodivisors on M / y _ M , the finite union of the associated primes of the module. Thus, by prime avoidance, there is some nonzero v in M / y _ M such that I ⊂ p = Ann R ⁡ ( v ) , which is to say,

Self-duality

There is an approach to a Koszul complex that uses a cochain complex instead of a chain complex. As it turns out, this results essentially in the same complex (the fact known as the self-duality of a Koszul complex).

Let E be a free module of finite rank r over a ring R. Then each element e of E gives rise to the exterior left-multiplication by e:

Since e ∧ e = 0 , we have: l e ∘ l e = 0 ; that is,

is a cochain complex of free modules. This complex, also called a Koszul complex, is a complex used in (Eisenbud 1995). Taking the dual, there is the complex:

Using an isomorphism ∧ k E ≃ ( ∧ r − k E ) ∗ ≃ ∧ r − k ( E ∗ ) , the complex ( ∧ E , l e ) coincides with the Koszul complex in #Definition.

Use

The Koszul complex is essential in defining the joint spectrum of a tuple of commuting bounded linear operators in a Banach space.