HILBERT FUNCTION AND HILBERT
SERIES. PURE RESOLUTION ∗
This exposition uses essentially the approach to this subject given in the
book of W. Bruns and J. Herzog “Cohen-Macaulay Rings”, Cambridge University Press 1993, 4.1.
• Let M be the Z−module of all polynomials P ∈ Q [X] such that P (z) ∈
Z for all z ∈ Z. M is a free Z−module with basis
(X + i) (X + i − 1) . . . (X − 1) X
=
i!
If P ∈ M is a polynomial of degree d, then
P =
d
X
µ
d−i
(−1)
ed−i
i=0
µ
¶
X +i
, i ∈ N.
i
¶
X +i
.
i
(1)
Definition 1. We say that a numerical function f : Z → Z is of polynomial
type (of degree d ) if there exists a polynomial P ∈ Q [X] (of degree d ) such
that f (n) = P (n) for all n À 0. The zero polynomial has degree −1.
Proposition 2. Let f : Z → Z be a function and d > 0 on integer. The
following conditions are equivalent:
a) f is of polynomial type of degree d.
b) ∆ f (n) = f (n + 1) − f (n) is of polynomial type of degree d − 1.
Proof. a) ⇒ b) This is easy.
b) ⇒ a) We show this by induction on d. For d = 0 the implication is
obvious. Let d > 0 and assume that ∆f (n) = f (n + 1) − f (n) is of polynomial
type of degree d−1. Then there exist an integer no and a polynomial P ∈ Q[X]
of degree d − 1 such that f (n + 1) − f (n) = P (n) for all n ≥ no. Then
f (n + 1) = F (no ) +
n
X
P (i) , for n ≥ no ,
i=no
∗ This lecture was held by Nicolae Radu:
University of Bucharest, Faculty of Mathematics, Romania
e-mail: [email protected]
1
2
Hilbert Function and Hilbert Series. Pure Resolution
and this last sum is a polynomial function in n of degree d.
Corollary 3. A function f : Z → Z is of polynomial type of degree d ≥ 0 if
and only if ∆d f = ∆(∆d−1 f ) is a constant 6= 0 for n À 0.
∞
£
¤
P
Lemma 4. Let F =
ai X i ∈ Z [[X]] X −1 and f : Z → Z , f (i) = ai ,for
i=s
all i ∈ Z. Then
(i)
n
(1 − X) F =
∞
P
∆n f (i − n)X i .
i=s
(ii) For d > 0, the following conditions are equivalent:
a) There exists a polynomial P ∈ Q[X] of degree d − 1 such that P (n) =
f (i), n À 0.
¤
£
Q
b) F = (1−X)
, where Q ∈ Z X, X −1 and Q(1) = 0.
d
Proof. i) This follows easily by induction on n
ii) a) ⇒ b) According to i),
d
(1 − X) F =
∞
X
£
¤
∆d f (i − d)X i = Q ∈ Z X, X −1 ,
i=s
because ∆d f (m) = 0, for m À 0. Suppose that Q(1) = 0. Then
0
=
Q(1) =
=
d−1
∞
X
∆d f (i − d) =
i=s
∆
∞
X
(∆d−1 f (i + 1 − d) − ∆d−1 f (i − d)) =
i=s
(m) ,
for m À 0 and according to Corollary 1.3, ∆d−1 (m) is a constant 6= 0,a
contradiction.
b) ⇒ a) The relation
d
(1 − X) F =
∞
X
∆d f (i − d)X i = Q
(4.1)
i=s
shows that ∆d f (m) = 0 for m À 0, thus ∆d−1 f (m + i) = ∆d−1 f (m) , for
m À 0. The equality
(1 − X)
d−1
F =
∞
X
i=s
f (i − d + 1)X i =
Q
1−X
shows that ∆d−1 (m) 6= 0 for m À 0, thus f is of polynomial type of degree
d − 1.
3
Hilbert Function and Hilbert Series. Pure Resolution
Lemma 5. Under the assumption of Lemma 1.4, ii), if
¶
µ
d−1
X
X +i
d−1−i
,
P =
(−1)
ed−1−i
i
i=0
then
ei =
Q(i) (1)
i!
(5.1)
for i = 0, ..., d − 1.
Proof. Let
D =Q−
d−1
i
X
(−1)
i=0
Then
H=
£
¤
i
Q(i) (1) (1 − X) ∈ Z X, X −1 .
i!
d−1
i
X
(−1)
d−i
i!
i=0
Q(i) (1)
(1 − X)
+
D
d
(1 − X)
.
Since D(i) (1) = 0 for i = 0, 1, ..., d − 1, it follows that D is divisible by
£
¤
d
D
−1
(1 − X) . So (1−X)
and for large n
d ∈ Z X, X
d−1
i
X
(−1)
i=0
i!
Qi (1)
(1 − X)
d−i
=
X
P (n) X n ,
(5.2)
n≥0
hence the equality holds for all n ∈ Z , because the coefficients of both series
are polynomial functions in n.
Since
¶
∞ µ
X
1
j−d−i−1
=
Xj
d−i
d
−
i
−
1
(1 − X)
j=0
we have
d−1
i
X
(−1)
i=0
i!
Q(i) (1)
(1 − X)
d−i
!
Ãd−1
∞
X
X (−1) Q(i) (1) µj − d − i − 1¶
=
Xj
i!
d
−
i
−
1
j=0
i=0
Ãd−1
µ
¶!
∞
X
X (−1)d−1−i
j+i
(d−1−i)
=
Q
(1)
Xj.
i!
i
j=0
i=0
Now (5.2) and (5.3) imply (5.1).
(5.3)
4
Hilbert Function and Hilbert Series. Pure Resolution
Let R =
∞
L
i=s
Ri be a graded ring such that R0 is artinian local ring and R
is finitely generated over R0 by elements of degree 1.
∞
L
Let M =
Mn by a finitely generated graded R-module. Then all Mn
n=t
have finite length, l(Mn ), the numerical function: H(M, ·) : Z → Z with
H(M, n)P= l(Mn ) for all n ∈ Z is called the Hilbert function of M and
HM =
H(M, n)X n is called the Hilbert series of M.
n∈Z
Theorem 6. (Hilbert) Let M be a finitely generated graded R-module of dimension d. Then H(M, n) is of polynomial type of degree d − 1.
Proof. Induction on the dimension d of M . If d = 0 M is artinian, thus
H(M, n) = 0 for n large. For d > 0, let P ∈ Ass(M ), then P is a graded Rmodule. There exists a graded submodule N ⊂ M, such that N ≈ (R/P )(a).
Since M is Noetherian there exist a chain 0 = N0 ⊂ N1 ⊂ . . . ⊂ Nn = M of
graded submodules of M such that for each i we have Ni+1 /Ni ≈ (R/Pi )(ai ),
where Pi is a graded prime ideal of R. Since the length of modules is additive
on short exact sequences, it follows that
H(M, n) =
n
X
H((R/Pi (ai )), n).
(6.1)
i=1
Assuming that the assertion of theorem is proved for M = R/P, where P
is a graded ideal with R of dim R/P = d, then from (6.1) the assertion follows
for M, since the polynomials H((R/Pi (ai )), n) are zero or have positive leading
coefficients, the degree of the sum is the maximum of dim(R/Pi ) − 1 and the
maximum of dim(R/Pi ) is equal with d = dim M.
Let be M = R/P , where P is a graded prime ideal of R and dim(R/P ) > 0.
Then we may choose a homogeneous element x ∈ R/P , x 6= 0 of degree 1.
The exact sequence
0 → (R/P )(−1) → R/P → R/(x, P ) → 0
gives the relation
∆H(R/P, n) = H(R/P, n + 1) − H(R/P, n) = H(R/(x, P ), n + 1).
As dim R/(x, P ) = d−1, the induction hypothesis implies that ∆H(R/P, n)
is of polynomial type of degree d-2, then Proposition 2 implies that H(R/P, n)
is a polynomial type of degree d-1.
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Hilbert Function and Hilbert Series. Pure Resolution
Corollary 7. Let M 6= 0 be a finitely generated graded R- module of dimension d. Then
QM
HM (X) =
,
(1 − X)d
where QM ∈ Z[X, X −1 ] with QM (1) 6= 0 and the coefficient ei of PM in
representation (1) equals Q(i) (1) /i!.
Definition 8. Let M be a graded R-module of dimension d. The unique
polynomial PM such that H(M, n) = PM (n) for n À 0 is called the Hilbert
polynomial of M and the multiplicity of M is defined to be

e0
if d > 0
e(M ) =
l(M ) if d = 0 ,
where e0 is the leading coefficient of PM in representation (1).
Remark 9. The higher iterated Hilbert functions Hi (M, n), i ∈ N of a finitely
generated graded R-module M are defined recursively as follows:
X
H0 (M, n) = H(M, n) and Hi (M, .n) =
Hi−1 (M, j) for i > 0.
j≤n
Since
∆Hi (M, n) = Hi (M, n + 1) − Hi (M, n) = Hi−1 (M, n + 1), for i > 0,
it follows from Proposition 1.3 and Theorem 6 that Hi (M, n) is of polynomial
type of degree d + i − 1, where d = dim M. In the representation of Hi (M, n)
as in (1), for n large, a0 = e(M ).
Let A be a local noetherian ring with the maximal ideal N and let I be an
∞
L
N -primary ideal. The graded ring R =
I i /I I+1 is noetherian, R0 = R/I
i=0
is a local artinian ring and R is generated over R0 by the elements of degree
∞
L
1. If E is a finitely generated A- module, then grI (E) =
I i E/I I+1 E is a
i=0
finitely generated graded R-module, H1 (E, n) = l(E/I n E) and the associated
polynomial is called the Hilbert-Samuel polynomial of E with respect to I.
Proposition 10. Let M be a finitely generated graded R-module of finite projective dimension and let
M
X
0→
R(−j)βpj → ... →
R(−j)β0j → M → 0
j
j
6
Hilbert Function and Hilbert Series. Pure Resolution
be a graded free resolution of M. Then HM = SM HR , where
SM =
p X
X
βαj X j .
α=0 j=1
In particular, for R = k[X1 , ..., Xn ], the polynomial ring over the field k , we
have
SM
HM =
.
(1 − X)n
Proof. The Hilbert function is additive on short exact sequences, so
HM =
p
X
i=0
(−1)i
X
βij HR(−j)
j
and HR(−j) = X j HR . If R = k[X1 , ..., Xn ] then HR =
1
(1−X)n .
Corollary 11. Let R = k[X1 , ..., Xn ] , where k is a field and M be a finitely
generated graded R-module of dimension d. Then:
(i) SM = (1 − X)n−d QM .
(i)
(ii) n − d = inf{i | SM (1) 6= 0}.
(n−d−i)
(iii) ei = (−1)n−d SM
(1)/(n − d + i)! .
Proof. It follows from Proposition 10 and Corollary 7.
Definition 12. Let R = k[X1 , ..., Xn ] , where k is a field and I ⊂ R a graded
ideal. We say that R/I has a pure resolution of type (d1 , ..., dp ) if its minimal
free graded resolution has the form
0 → R(−dp )βp → ... → R(−d1 )β1 → R → R/I → 0 .
Then d1 < d2 < ... < dp .
Theorem 13. Let R = k[X1 , ..., Xn ] be a polynomial ring over a field k and I
be an ideal of R such that R/I has a pure resolution of type (d1 , ..., dp ) where
p = dim R − dim(R/I) (for example if R/I is Cohen-Macaulay). Then:
Q
(i) βi = (−1)i+1
dj /(dj − di ).
j6=i
(ii) e(R/I) = (1/p!)
p
Q
i=1
di .
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Hilbert Function and Hilbert Series. Pure Resolution
Proof. From Proposition 10 and Corollary 11, ii) we obtain the system of
linear equations
p
P
(−1)i βi = −1
i=1
(j)
SR/I (1) = 0, j = 0, 1, ..., p − 1 ,
because d0 = 0, β0 = 1. Applying elementary row operations which do not
affect the solution of this system, we obtain the system:
p
P
(−1)i βi = −1
i=1
p
P
i=1
p
P
i=1
(−1)i βi di = 0
(−1)i βi d2i = 0
...................................
p
P
i=1
(−1)i βi dp−1
= 0,
i
whose solution is stated in (i).
(ii) By Corollary 1.11,iii,
(p)
e(R/I) = (−1)p SR/I (1)/p! =
=
1
p!
p
Q
i=1
di (
P
p−1
Q
¡ ¢
P
(−1)p+1 βi dpi =
(di −dj )
j=1
Q
(di −dj ) ).
j6=i
It remains to show that the sum in this expression equals 1. For this, take the
rational complex function
p−1
Q
f (z) =
(z − j)
j=1
p
Q
(z − dj )
j=1
which has simple poles at worst in the points d1 , ..., dp and the sum of all
residues of this function at all points including ∞ is zero. Then
p
P
i=1
p−1
Q
(di −j)
( Q (di −dj ) ) = −Res∞ f (z) = Res0 f (1/z)
=
z2
j=1
j6=i
= Res0
Ã
1
z
p−1
Q
j=1
(1 − jz)
p
Q
j=1
!
−1
(1 − dj z)
= 1.
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Hilbert Function and Hilbert Series. Pure Resolution