Koszul Complexes (version 22 Feb 2007)

8 Pages·2007·0.118 MB·English

by  Adam Boocher

Koszul Complexes
Adam Boocher
Spring 2007
1 Mapping Cone Construction
We begin this talk by discussing how to obtain a free resolution inductively by adding regular
elements to the ring R/I. For simplicity, at the moment, we assume that R is a polynomial ring in
some number of variables and that I is a homogeneous ideal. We begin with a lemma.
Lemma: Let I be an R-ideal, f a homogeneous polynomial of degree d, then the following is
a graded exact sequence.
0 → R(−d)/(I : f) → R/I → R/(I,f) → 0.
Note that if f is a regular element on R/I then I : f = I so that we actually have a graded
exact sequence
0 → R(−d)/I → R/I → R/(I,f) → 0.
Let G be a free resolution for the module R/I. Also suppose that G is a free resolution for
. .
R(−d)/I (which can be obtained just by shifting the grading). Thus we have a pair of resolutions
// // // // //
··· F F F R(−d)/I 0
2 1 0
·f ·f ·f ·f
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
// // // // //
··· G G G R/I 0
2 1 0
where the vertical maps are just multiplication by f. If f is a regular element of degree d on R/I
we can of course complete this diagram a bit by using the exact sequence from the lemma. To
arrive at
0
(cid:15)(cid:15)
··· //F ψ2 //F ψ1 //F ψ0//R(−d)/I //0
2 1 0
φ2 φ1 φ0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
··· //G ξ2 //G ξ1 //G ξ0 //R/I //0
2 1 0
π
(cid:15)(cid:15)
R/(I,f)
(cid:15)(cid:15)
0
1
∼
Note that the map α = πξ is a map from G = R → R/(I,f) a first step toward a free resolution
0 0 0
of R/(I,f).
Claim: The kernel of α is generated by the images of φ and ξ .
0 0 1
Proof. Chasing diagrams it is clear that φ (F ) and ξ (G ) are mapped to 0 by α . The other
0 0 1 1 0
direction involves similar chasing, which ends by showing that if α(x) = 0 then we can write
x = ξ (z)+φ (y) for z,y in the appropriate sets.
1 0
Thus we can define a map from F ⊕G → R/(I,f) and get a start to our free resolution. We
0 1
can continue this process in general to obtain a chain complex
··· //Fi−1⊕Gi ∂i //Fi−2⊕Gi−1 //··· //F0⊕G1 //R/(I,f) //0.
Where the map ∂ is given by the matrix
i
(cid:20) (cid:21)
ψ 0
i−1
(−1)i−1φ ξ
i−1 i
It is easy to check that this is indeed a chain complex as ∂2 = 0. Further, it is also straightforward
to check in the same way that this sequence is exact. This is because f was a regular element on
R/I. If this were not the case, then we would not have an exact sequence but would still have a
complex. This is called the Koszul Complex. Now we compute an example.
Let I = (f ,...,f ) be a complete intersection with f ,...,f forming a regular sequence with
1 n 1 n
f of degree d . Let I = 0. Then we have a free resolution for R/I easily
i i 0 0
// // //
0 R R/I 0.
0
Then from our previous work we know that we have a free resolution for R/(f ) from the Koszul
1
complex by
0
(cid:15)(cid:15)
// // //
0 R(−d ) R(−d )/I 0
1 1 0
(cid:15)(cid:15) (cid:15)(cid:15)
// // //
0 R R/I 0
0
(cid:15)(cid:15)
R/(f )
1
(cid:15)(cid:15)
0
Tracing diagonals we get the resolution,
// // // //
0 R(−d ) R R/I 0.
1
2
Of course, this is no surprise and isn’t even that impressive since it is simply the vertical column.
But now we replace this sequence with the bottom row in the previous diagram to compute a free
resolution of R/(f ,f ).
1 2
0
(cid:15)(cid:15)
// // // //
0 R(−d −d ) R(−d ) R(−d )/(f ) 0
1 2 2 2 1
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
// // // //
0 R(−d ) R R/(f ) 0
1 1
(cid:15)(cid:15)
R/(f ,f )
1 2
(cid:15)(cid:15)
0
Yielding the resolution
// // // // //
0 R(−d −d ) R(−d )⊕R(−d ) R R/(f ,f ) 0
1 2 1 2 1 2
Ignoring the grading for a moment we can simply look at the betti numbers of the resolution as we
add f to our ideal.
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// // // // //
0 R R2 R R/(f1,f2) 0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
// // // // //
0 R R2 R R/(f1,f2) 0
We see that the next set of Betti numbers will be 1,3,3,1 and in general it is easy to see the
following fact
Proposition If f ,...,f is a regular sequence then ignoring the grading, the free resolution
1 n
for R/(f ,...,f ) is given by
1 n
// // // // // //
0 Rb0 Rb1 ··· Rbn R/(f1,...,fn) 0
where b is given by (cid:0)n(cid:1).
i i
Example: This gives a concrete free resolution for a field k when considered a quotient
k[x ,...,x ]/(x ,...,x ).
1 n 1 n
Before presenting our next example, we state a lemma concerning Tor.
Lemma: Suppose F is a finitely generated graded S = k[x ,...,x ] module. If Tor (F,k) = 0
1 n 1
then F is free.
Proof. Note that F ⊗ k = F/IF where I = (x ,...,x ), a maximal ideal of S. Thus F ⊗ k is
1 n
naturally an A/I = k-module, or a k-vector space. Choose homogeneous elements v ,...,v in F
1 r
so that their images in F/IF form a basis for this vector space. Let L be a free module with basis
e ,...,e . Let
1 r
L → F
3
be the graded homomorphism sending e 7→ v for i = 1,...,n. We will show that this is an
i i
isomorphism. Let C be the cokernel, so
L → F → C → 0
is exact. Tensoring with k yields an exact sequence
L⊗k → F ⊗k → C ⊗k → 0.
Since by construction, the map L⊗k → F ⊗k is surjective, it follows that C ⊗k = 0. We now
prove a Nakayama like lemma to show that C = 0.
Lemma: Let N be a graded module over A. Let I = (x ,...,x ). Then if N/IN = 0 then
1 n
N = 0. Note that N ⊗k = N/IN).
Proof. The proof is clear. Just take an element f of minimum degree in N. Then since f+IN = 0,
we must have f ∈ IN but each element of I has degree > 0 so this is impossible.
Thus if E is the kernel of our map L → F then we now have an exact sequence
0 → E → L → F → 0
we must show that E = 0. But this is where we get to use Tor! The Tor exact sequence gives us
0 = Tor (F,k) → E ⊗k → L⊗k → F ⊗k → 0.
1
By construction, L⊗k is isomorphic to F ⊗k, so E⊗k = 0 and by the lemma, again, we see that
E = 0.
Example: (Hilbert Syzygy Theorem) Let F be a finitely generated module over A. Then A
has projective dimension less than n. In other words, if
0 → K → F → ··· → F → F → 0 (1)
n−1 0
is exact with F free, then K is free.
i
Proof. Since the Koszul complex is a free resolution for k of length n, tensoring with F we see that
Tor (k,F) = 0. Tensoring (1) with k we see that 0 =Tor (F,k) =Tor (K,k) so K is free.
n+1 n+1 1
2 The Koszul Complex
In this section we follow Matsumura pp. 127-130 and show that we can in general construct a
complex for any sequence of elements in a ring and that exactness is a more general result.
Given a ring R and x ,...,x ∈ R, define the Koszul complex K as follows: set K = A, and
1 n • 0
K = 0 if p > n and for 1 ≤ p ≤ n define
p
M
K = R·e ∧···∧e
p i1 ip
to be the free module of rank (cid:0)n(cid:1) with basis {e ∧···∧e : 1 ≤ i < ··· < i ≤ n} (A standard
p i1 ip 1 p
construction)
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The differential map ∂ : K → K is defined by
p p−1
p
X
∂(e ∧···∧e ) = (−1)r−1x e ∧···∧e ∧···∧e
i1 ip ir i1 cir ip
r=1
(for p = 1, set ∂(e ) = x ). This is no doubt a horrendous expression, but nonetheless one that the
i i
reader has probably encountered before. In words it is simple, remove one of the e at a time and
ij
multiply by x and then sum while alternating signs.
ij
For example,
∂(e ∧e ∧e ) = x (e ∧e )−x (e ∧e )+x (e ∧e ).
1 2 3 1 2 3 2 1 3 3 1 2
It is clear that ∂2 = 0 as is checked in any basic topology course. This is called the Koszul
complex and is written K (x ,...,x ) or K (x).
• 1 n •
Let’s do some examples now, if n = 1 then the the complex K (x) is given by
•
// // // x // //
··· 0 0 A A 0.
If n = 2 then K (x,y) is the complex
•
2 3
x
4 5 h i
−y x y
// // // // //
··· 0 A A2 A 0
Finally, for n = 3, K (x,y,z) is the complex
•
··· //0 //A ψ3 //A3 ψ2 //A3 ψ1 //A //0
where
   
x 0 −z −y
(cid:2) (cid:3)
ψ3 = −y, ψ2 = −z 0 x , ψ3 = x y z
z y x 0
Note that one has to be careful about choosing the order of the bases for these matrices in order
for everything to make sense. It is perhaps better not to worry about these things and just use the
definition using wedge products.
We now give a third definition of the Koszul complex which is given by forming the tensor
product of two complexes.
Definition: Let K and L be two complexes. Then we form the complex K ⊗L by
• •
M
(K ⊗L) = K ⊗L
n p q
p+q=n
and the boundary map given by
∂(x⊗y) = ∂x⊗y+(−1)px⊗∂y
for x ∈ K and y ∈ L .
p q
Notice that if K = L are the complex 0 → R → R → 0 then their tensor product is
0 → R → R2 → R → 0.
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If we tensor this again with K we obtain
0 → R → R3 → R3 → 0.
and can define the Koszul complex in the following way: Let K 1 ≤ i ≤ n be the complex
i
0 → R → R → 0 where the map m : R → R is given by multiplication by x . Then the Koszul
i i
N
complexisgivenbyK = K thereadercanverifythatthemapsgiveninthisconstructionmatch
i
up with those in the first two examples. The second definition in terms of wedge products is the
one that Koszul proposed.
Definition: For an R-module M we can define K (x,M) as the complex K (x)⊗ M. More-
• • R
over, if C is an arbitrary complex of R modules then we define C (x) = C ⊗K (x).
• • • •
Remark: Note that since K (x ,...,x ) = K (x ) ⊗ ··· ⊗ K (x ) and tensor products of
• 1 n • 1 • n
complexes are commutative, we have that the Koszul complex is invariant under permutation of
x ,...,x .
1 n
The Koszul complex has homology groups H (K(x,M)), which we abbreviate by H (x,M). It
p p
is easy to see that
∼
H (x,M) = M/xM = M/(x M +···x M)
0 1 n
and
∼
H (x,M) = {m ∈ M : x m = 0 for all i}.
n i
Theorem: Let C be a complex of R modules and x ∈ R. Then we get an exact sequence of
complexes
0 → C → C(x) → C0 → 0
where C0 is the complex obtained by shifting the degrees in C up by 1. (That is C0 = C and
p+1 p
they share the same differential.) The homology long sequence obtained by this has the form
// // // (−1)p−1x // //
··· H (C) H (C(x)) H (C) H (C) ···.
p p p−1 p−1
Finallly, we have x·H (C(x)) = 0 for all p.
p
Proof. Since K(x) is given by 0 → R → R → 0, the complex C(x) is given by
// // // // //
··· Cn⊕Cn−1 ··· C1⊕C0 C0 0.
The boundary map for tensor products of complexes gives that
∂(ξ,η) = (∂ξ+(−1)p−1xη,∂η).
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The exactness follows easily by considering the following diagram
0 0 0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
// // // // //
··· C ··· C C 0
n 1 0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
// // // // //
··· Cn⊕Cn−1 ··· C1⊕C0 C0 0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
// // // // //
··· Cn−1 ··· C0 0 0
(cid:15)(cid:15) (cid:15)(cid:15)
0 0 .
We always have a long homology sequence. What remains is to show that the induced map
from H (C) → H (C) is given by (−1)p−1x. So suppose we have an element η ∈ C = C0
p−1 p−1 p−1 p
that is in the kernel of ∂. Then in C(x) we have ∂(0,η) = ((−1)p−1xη,0) so in C (the top line)
n−1
we see that η is sent to (−1)p−1xη.
Finally, if d(ξ,η) = 0 then ∂(η) = 0 and ∂ξ = (−1)pxη, so that we have
x(ξ,η) = ∂(0,(−1)pξ) ∈ ∂C (x),
p+1
and therefore x·H (C(x)) = 0.
p
Remark: Applying this theorem to K(x,M) we see that the ideal (x) = (x ,...,x ) annihi-
1 n
lates the homology groups.
Theorem(i) Let R be a ring, M an R-module, and x ,...,x a regular sequence for M. Then
1 n
H (x,M) = 0 for p > 0 and H (x,M) = M/xM.
p 0
(ii) Suppose that one of the following two conditions holds:
(α) (R,m) is a local ring, x ,...,x ∈ m and M is a finite R-module.
1 n
(β) R is an N-graded ring, M is an N-graded R-module, and x ,...,x are homogeneous elements
1 n
of degree > 0. Then the converse of (i) holds in the following strong form: if H (x,M) = 0 and
1
M 6= 0 then x ...,x is a regular sequence for M.
1 n
Proof. (i) We induct on n. When n = 1 we have H (x,M) = {m ∈ M : xm = 0} = 0, so that the
1
first step is done. When n > 1 and p > 1 the previous theorem gives an exact sequence
0 = H (x ,...,x ,M) → H (x ,...,x ,M) → H (x ,...,x ,M) = 0.
p 1 n−1 p 1 n p−1 1 n−1
So that H (x ,...,x ,M = 0. For p = 1, we have
p 1 n
0 → H (x,M) → H (x ,...,x ,M) = M/(x ,...,x )M → M/(x ,...,x )M
1 0 1 n−1 1 n−1 1 n−1
where the last map is just multiplication by ±x . Since x is a regular element, This shows that
n n
H = 0.
1
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(ii) Assuming either (α) or (β) if M 6= 0 then M = M/(x ,...,x )M 6= 0. By hypothesis and
i 1 i
by the previous theorem, we have
H (x ,...,x ,M) ±xn //H (x ,...,x ,M) //H (x,M) = 0;
1 1 n−1 1 1 n−1 1
but by the hypotheses, these homology groups are finite R modules or N graded modules, so by
NAK we have H (x ,...,x ,M) = 0. Thus by induction we know that x ,...,x is exact. Then
1 1 n−1 1 n
from the exact sequence
0 → H (x,M) → H (x ,...,x ,M) = M/(x ,...,x )M → M/(x ,...,x )M,
1 0 1 n−1 1 n−1 1 n−1
since the last map is multiplication by x and exactness implies injectivity, we know that x is a
n n
regular element as well.
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