Koszul complex

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In mathematics, the Koszul complex was first introduced to define a cohomology theory for Lie algebras, by Jean-Louis Koszul (see Lie algebra cohomology). It turned out to be a useful general construction in homological algebra. As a tool, its homology can be used to tell when a set of elements of a (local) ring is an M-regular sequence, and hence it can be used to prove basic facts about the depth of a module or ideal which is an algebraic notion of dimension that is related to but different from the geometric notion of Krull dimension. Moreover, in certain circumstances, the complex is the complex of syzygies, that is, it tells you the relations between generators of a module, the relations between these relations, and so forth.

Definition[edit]

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

where the differential dk is given by: for any ei in E,

The superscript means the term is omitted. (Showing is straightforward; alternatively, this identity also follows using the #Self-duality of a Koszul complex.)

Note and . Note also ; 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 = Rr (i.e., an ordered basis is chosen), then, giving an R-linear map s: RrR amounts to giving a finite sequence s1, ..., sr of elements in R (namely, a row vector) and then one sets

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

,

which is again a chain complex with the induced differential .

The i-th homology of the Koszul complex

is called the i-th Koszul homology. For example, if E = Rr and is a row vector with entries in R, then is

and so

Similarly,

Koszul complexes in low dimensions[edit]

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 . 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 is

with the matrices and given by

and

Note that di 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, ..., xn form a regular sequence, the higher homology modules of the Koszul complex are all zero.

Example[edit]

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

Properties of a Koszul homology[edit]

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

Using , there is the exact sequence of complexes:

where [-1] signifies the degree shift by -1 and . One notes:[1] for (x, y) in ,

(In the language of homological algebra, the above means that is the mapping cone of .)

Taking the long exact sequence of homologies, we get:

Here, the connecting homomorphism

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

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

Theorem — Let R be a ring, M a finitely generated module over R. If a sequence x1, x2, ..., xr of elements of R is a regular sequence on M, then

for all i ≥ 1. In particular, when M = R, this is to say

is exact; i.e., is an R-free resolution of .

Proof by induction on r. If r = 1, then . Next, assume the assertion is true for r - 1. Then, using the above exact sequence, one sees for any i ≥ 2. The vanishing is also valid for i = 1, since is a nonzerodivisor on

Corollary[2] — Let R, M be as above and x1, x2, ..., xn a sequence of elements of R. Suppose there are a ring S, an S-regular sequence y1, y2, ..., yn in S and a ring homomorphism SR that maps yi to xi. (For example, one can take S = Z[y1, ..., yn].) Then

where Tor denotes the Tor functor and M is an S-module through SR.

Proof: By the theorem applied to S and S as an S-module, we see K(y1, ..., yn) is an S-free resolution of S/(y1, ..., yn). So, by definition, the i-th homology of is the right-hand side of the above. On the other hand, by the definition of the S-module structure on M.

Corollary[3] — Let R, M be as above and x1, x2, ..., xn a sequence of elements of R. Then both the ideal I = (x1, ..., xn) and the annihilator of M annihilate

for all i.

Proof: Let S = R[y1, ..., yn]. Turn M into an S-module through the ring homomorphism SR, yixi and R an S-module through yi → 0. By the preceding corollary, and then

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

Theorem — Let R be a Noetherian ring and M a nonzero finitely generated module over R . If x1, x2, ..., xr are elements of the Jacobson radical of R, then the following are equivalent:

  1. The sequence is a regular sequence on M,
  2. ,
  3. for all i ≥ 1.

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, ..., xr-1. Consider

Since the first is surjective, with . By Nakayama's lemma, and so x' is a regular sequence by the inductive hypothesis. Since the second is injective (i.e., is a nonzerodivisor), is a regular sequence. (Note: by Nakayama's lemma, the requirement is automatic.)

Tensor products of Koszul complexes[edit]

In general, if C, D are chain complexes, then their tensor product CD 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: ER, t: FR R-linear maps. Let be the Koszul complex of the linear map . 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,

  • when

One easily sees that (induction on degree) and that the action of ds on homogeneous elements agrees with the differentials in #Definition.

Now, we have as graded R-modules. Also, by the definition of a tensor product mentioned in the beginning,

Since and are derivations of the same type, this implies

Note, in particular,

.

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

Proposition — Let R be a ring and I = (x1, ..., xn) an ideal generated by some n-elements. Then, for any R-module M and any elements y1, ..., yr in I,

where is viewed as a complex with zero differential. (In fact, the decomposition holds on the chain-level).

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 . Recall that an M-regular sequence x1, ..., xn in an ideal I is maximal if I contains no nonzerodivisor on .

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

Theorem (depth-sensitivity) — Let R be a Noetherian ring, x1, ..., xn elements of R and I = (x1, ..., xn) the ideal generated by them. For a finitely generated module M over R, if, for some integer m,

for all i > m,

while

then every maximal M-regular sequence in I has length n - m (in particular, they all have the same length). As a consequence,

.

Proof: To lighten the notations, we write H(-) for H(K(-)). Let y1, ..., ys be a maximal M-regular sequence in the ideal I; we denote this sequence by . First we show, by induction on , the claim that is if and is zero if . The basic case is clear from #Properties of a Koszul homology. From the long exact sequence of Koszul homologies and the inductive hypothesis,

,

which is Also, by the same argument, the vanishing holds for . This completes the proof of the claim.

Now, it follows from the claim and the early proposition that for all i > n - s. To conclude n - s = m, it remains to show that it is nonzero if i = n - s. Since is a maximal M-regular sequence in I, the ideal I is contained in the set of all zerodivisors on , the finite union of the associated primes of the module. Thus, by prime avoidance, there is some nonzero v in such that , which is to say,

Self-duality[edit]

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 , we have: ; 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 , the complex coincides with the Koszul complex in #Definition.

Use[edit]

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

See also[edit]

Notes[edit]

  1. ^ Indeed, by linearity, we can assume where . Then
    ,
    which is .
  2. ^ Eisenbud, Exercise 17.10.
  3. ^ Serre, Ch IV, A § 2, Proposition 4.

References[edit]