Koszul complex
In mathematics,[] the Koszul complex was first introduced to define a cohomology theory for Lie algebras, by Jean-Louis Koszul. 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 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
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 is given by: for any in E,
The superscript means the term is omitted.
Note that and. Note also that ; this isomorphism is not canonical
If , then, giving an R-linear map amounts to giving a finite sequence of elements in R 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 and is a row vector with entries in R, then is
and so
Similarly,
Koszul complexes in low dimensions
Given a commutative 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, 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 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
Note that 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 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 form a regular sequence, the higher homology modules of the Koszul complex are all zero.
Example
If k is a field and are indeterminates and R is the polynomial ring, the Koszul complex on the 's forms a concrete free R-resolution of k.Properties of a Koszul homology
Let E be a finite-rank free module over R, let be an R-linear map, and let t be an element of R. Let be the Koszul complex of.Using,
there is the exact sequence of complexes:
where signifies the degree shift by -1 and. One notes: for in,
In the language of homological algebra, the above means that is the mapping cone of.
Taking the long exact sequence of homologies, we obtain:
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. Then the early formula for gives.
The above exact sequence can be used to prove the following.
Proof by induction on r. If , then. Next, assume the assertion is true for r - 1. Then, using the above exact sequence, one sees for any. The vanishing is also valid for, since is a nonzerodivisor on
Proof: By the theorem applied to S and S as an S-module, we see K is an S-free resolution of S/. 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.
Proof: Let S = R. Turn M into an S-module through the ring homomorphism S → R, yi → xi and R an S-module through. By the preceding corollary, and then
For a 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,..., 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, is a regular sequence.
Tensor products of Koszul complexes
In general, if C, D are chain complexes, then their tensor product is the chain complex given bywith 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, and let and be two 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. Define the graded derivation of degree
by requiring: for any homogeneous elements x, y in ΛE,
- when
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.
Proof:
As an application, we can show the depth-sensitivity of a Koszul homology. Given a finitely generated module M over a ring R, by 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.
Proof: To lighten the notations, we write H for H. 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
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.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. Taking the dual, there is the complex:
Using an isomorphism, the complex coincides with the Koszul complex in #Definition.