Part III Computational Methods
11.1 The Hilbert-Serre Theorem
Chapter 11
Hilbert Series and Dimension
An affine algebraAof positive Krull dimension is always infinite-dimensional as a vector space (see Theorem 5.11). The goal of introducing the Hilbert series is to nevertheless measure the size in some way. The trick is to break upAinto finite-dimensional pieces, given by the degrees. The Hilbert series then is the power series whose coefficients are the dimensions of the pieces. So instead of measuring the dimension by a number, we measure its growth as the degrees rise, and encode that information into a power series. In the first section of this chapter, we show the surprising fact that the Hilbert series can always be written as a rational function, and almost all its coefficients are given by a polynomial, the Hilbert polynomial. We also learn how the Hilbert series can be computed algorithmically. In fact, as in the last chapter, algorithms and theory go hand in hand here. In the second section we show that the degree of the Hilbert polynomial is equal to the Krull dimension of A. Apart from being an interesting result in itself, this leads to a new and better algorithm for computing the dimension. The result also plays an important role in Chapter12.
Throughout this chapter,K[x1, . . . , xn] will denote a polynomial ring over a field.
For I ⊆ K[x1, . . . , xn] an ideal, let A := K[x1, . . . , xn]/I, and for d a non-negative integer, set
A≤d:={f+I|f ∈K[x1, . . . , xn], deg(f)≤d}.
Observe that A≤d is a finite-dimensional K-vector space. The function hI: N0→N0 defined by
hI(d) := dimK(A≤d)
is called theHilbert function ofI. The formal power series HI(t) :=
∞
X
d=0
hI(d)td∈Z[[t]]
is called theHilbert seriesof I.
Example 11.2. (a) LetI= (x1, . . . , xn). ThenhI(d) = 1 for alld, so HI(t) =
∞
X
d=0
td= 1 1−t.
(b) Let I = (x1−x22) ⊂ K[x1, x2] and A = K[x1, x2]/I. For d ∈ N0, the residue classes of 1, x1, . . . , xd1, x2, x1x2, . . . , xd−11 x2 form a basis of A≤d, sohI(d) = 2d+ 1. We obtain
HI(t) = 1 +t (1−t)2.
/ Remark 11.3. (a) In the definition of the Hilbert series, we need not worry about convergence issues, sinceHI(t) is defined as an element of the for- mal power series ring overZ. This also applies to the representations of the Hilbert series as “rational functions” in Example11.2: The polyno- mial 1−t is an invertible element of Z[[t]], and its inverse is P∞
d=0td. Of course, these representations could also be interpreted as identities of real or complex functions which are defined for|t|<1.
(b) It is tempting to define the Hilbert function and Hilbert series of an affine algebra A as follows. By choosing generators of A, we obtain a presentation ofA as A∼=K[x1, . . . , xn]/I. Then set hA(d) :=hI(d) and HA(t) :=HI(t). However, these objects will depend on the choice of the generators. For instance, choosing the (rather unusual) generatorsx2and xfor the polynomial ringA=K[x] yields the Hilbert function 2d+ 1 by Example11.2(b). But choosing justxyieldsd+ 1 by Remark11.5below.
So the Hilbert function and Hilbert series arenot invariants of an affine algebra.
(c) Our definition of the A≤d provides an ascending filtration of A, in the sense thatA≤d⊆A≤d+1 for alldandA=S
d∈N0A≤d. In the literature,
11.1 The Hilbert-Serre Theorem 163 Hilbert series are often defined forgraded vector spaces, i.e., vector spaces V which have a direct sum decomposition
V = M
d∈N0
Vd
with Vd finite-dimensional K-vector spaces. A special case of a graded vector space is agraded algebra, where the grading provides a structure of a graded ring. In this setting, the graded Hilbert series is defined as
HVgrad(t) :=
∞
X
d=0
dimK(Vd)td∈Z[[t]].
But this is strongly related to our definition of the Hilbert series. In fact, a grading can be turned into an ascending filtration by setting V≤d :=
Ld
i=0Vi. Then HVgrad(t) andHV(t) :=P∞
d=0dimK(V≤d)td are obviously connected by
HVgrad(t) = (1−t)HV(t).
Exercise12.3studies the Hilbert series of a graded module over a graded
ring. /
We now calculate the Hilbert series of a principal ideal. As we will see later, this has much more importance than just providing a further example.
Proposition 11.4 (Hilbert series of a principal ideal). Let I = (f) ⊆ K[x1, . . . , xn] be a principal ideal. Then
HI(t) = 1−tdeg(f)
(1−t)n+1 if f 6= 0 and
HI(t) = 1
(1−t)n+1 if f = 0.
Proof. We start with the case f = 0. Since the Hilbert function and Hilbert series of the zero-ideal depend on the number n of indeterminates, we will write them in this proof ashn(d) and Hn(t), respectively. We use induction onn, starting withn= 0. We haveh0(d) = 1 for all d, soH0(t) = 1−t1 . For n >0, we use the direct sum decomposition
K[x1, . . . , xn]≤d= M
i,j∈N0, i+j=d
K[x1, . . . , xn−1]≤i·xjn. (11.1)
With the induction hypothesis, this implies Hn(t) =Hn−1(t)·X∞
j=0
tj
=Hn−1(t)· 1
1−t = 1 (1−t)n+1.
Now assume thatf 6= 0. For everyd∈N0we have
(K[x1, . . . , xn]/(f))≤d∼=K[x1, . . . , xn]≤d/ f·K[x1, . . . , xn]≤d−deg(f) . Since multiplication withf is injective onK[x1, . . . , xn], we obtain
HI(t) = (1−tdeg(f))·H{0}(t) =1−tdeg(f) (1−t)n+1.
u t
Remark 11.5. We can also determine the Hilbert function h{0}(d) of the zero-ideal. Sinceh{0}(d) equals the number of monomials of degree at mostd, it can be determined combinatorially. Taking a different route, we expand the Hilbert series as a binomial series. This yields
H{0}(t) = (1−t)−n−1=
∞
X
d=0
−n−1 d
(−t)d, so
h{0}(d) = (−1)d
−n−1 d
= d+n
d
= d+n
n
.
In particular, we see that h{0}(d) is given by a polynomial of degreenin d.
This will be generalized in Corollary 11.10. /
Our next goal is to link the topic of Hilbert series to the theory of Gr¨obner bases. For this, we need the concept of atotal degree ordering. By defini- tion, this is a monomial ordering onK[x1, . . . , xn] such that two monomials t, t0 with t ≤t0 satisfy deg(t)≤ deg(t0). The most important example of a total degree ordering is the grevlex ordering. A counter example is the lexi- cographic ordering (ifn >1). Recall that ifI⊆K[x1, . . . , xn] is an ideal, we writeL(I) for the leading ideal, which depends on the choice of the monomial ordering.
Theorem 11.6(Hilbert series of the leading ideal). Suppose
that K[x1, . . . , xn] is equipped with a total degree ordering, and let I ⊆ K[x1, . . . , xn] be an ideal. Then
HI(t) =HL(I)(t).
Proof. Set A := K[x1, . . . , xn]/I. By Theorem 9.9, the normal form map NFG, given by a Gr¨obner basis G of I, induces an injective linear map ϕ: A → K[x1, . . . , xn]. For every d ∈ N0 we have a restriction ϕd: A≤d → K[x1, . . . , xn]. LetVd ⊆K[x1, . . . , xn] be the subspace spanned by all mono- mials t with deg(t) ≤dand t /∈ L(I). Since all f ∈ Vd are in normal form w.r.t. G, we get f = NFG(f) = ϕd(f +I), so Vd ⊆ im(ϕd). On the other hand, Definition 9.6(b) and the hypothesis on the monomial ordering imply im(ϕd)⊆Vd. We conclude that
11.1 The Hilbert-Serre Theorem 165
hI(d) = dim(Vd).
Observe that the definition ofVd only depends onL(I). So two ideals with the same leading ideal have the same Hilbert function and Hilbert series.
SinceL(L(I)) =L(I), the result follows. ut
Exercise11.2shows that the hypothesis on the monomial ordering cannot be dropped from Theorem 11.6.
A polynomial f ∈ K[x1, . . . , xn] is called homogeneous if all monomi- als off have the same degree. So every polynomial can be written uniquely as a sum of homogeneous polynomials of pairwise distinct degrees, its ho- mogeneous parts. An ideal I ⊆ K[x1, . . . , xn] is calledhomogeneous if it is generated by homogeneous polynomials. For example, the leading ideal L(I) of any ideal I is homogeneous. For more on homogeneous ideals, see Exercise11.3.
Lemma 11.7 (Hilbert series of the sum and intersection of ideals). LetI, J
⊆K[x1, . . . , xn]be homogeneous ideals. Then
HI+J(t) +HI∩J(t) =HI(t) +HJ(t).
Proof. Let d be a non-negative integer. For an ideal L ⊆K[x1, . . . , xn] we writeL≤d:={f ∈L|deg(f)≤d}. It follows from the hypothesis thatI+J is generated by homogeneous polynomialsg1, . . . , gm∈ I∪J, so everyf ∈ (I+J)≤d can be written as f =Pm
i=1higi, with hi ∈ K[x1, . . . , xn]. This equation still holds if from eachhi we delete all homogeneous parts of degree
> d−deg(gi). This shows that the mapI≤d→(I+J)≤d/J≤d, f7→f+J≤d is surjective. Its kernel is (I∩J)≤d, so
dimK(I≤d)−dimK((I∩J)≤d) = dimK((I+J)≤d)−dimK(J≤d).
Passing to the dimensions of the quotient spaces inK[x1, . . . , xn]≤dand form-
ing power series yields the result. ut
The reduction step given by Theorem11.6is crucial in the following algo- rithm for computing the Hilbert series of an ideal.
Algorithm 11.8 (Hilbert series of a polynomial ideal).
Input: An ideal I⊆K[x1, . . . , xn], given by generators.
Output: The Hilbert series HI(t).
(1) Choose a total degree ordering “≤” on K[x1, . . . , xn] and compute a Gr¨obner basisGofIw.r.t. “≤”. Letm1, . . . , mrbe the leading monomials of the non-zero elements ofG.
(2) Ifr= 0, returnHI(t) := (1−t)1n+1. (3) Set
J := (m2, . . . , mr) and Je:= (lcm(m1, m2), . . . ,lcm(m1, mr)).
(4) Compute the Hilbert series HJ(t) and H
Je(t) by a recursive call of the algorithm.(Notice thatJ andJeare generated by monomials, so there is nothing to do when performing Step (1)onJ andJe.)
(5) Return
HI(t) := 1−tdeg(m1)
(1−t)n+1 +HJ(t)−H
Je(t).
Notice that Algorithm 11.8requires just one Gr¨obner basis computation, and we can use the grevlex ordering, which tends to make computations fastest.
Theorem 11.9. Algorithm11.8terminates after finitely many steps and cal- culates HI(t)correctly.
Proof. With each recursive call of the algorithm, the number r decreases strictly. This guarantees termination.
Let Ie:= (m1, . . . , mr) = L(I). By Theorem11.6, we need to show that Steps (2) through (5) calculateH
Ie(t) correctly. We use induction on r.
For r = 0, the Hilbert series in Step (2) is correct by Proposition 11.4.
Assume r > 0. By induction, the Hilbert series of J and Jeare calculated correctly. We claim that
Je=J∩(m1). (11.2)
Clearly every least common multiple ofm1 and an mi (i≥2) lies inJ and in (m1), so Je⊆ J∩(m1). Conversely, takef ∈J ∩(m1). Thenf =g1m1 and f = Pr
i=2gimi with g1, . . . , gr ∈ K[x1, . . . , xn]. For every monomial t ∈ Mon(g1) there exists i ≥ 2 such thattm1 ∈ Mon(gimi), so mi divides tm1. This implies that lcm(m1, mi) divides tm1, so tm1 ∈Je. We conclude that f ∈J, so (11.2) is established.e
SinceIe=J+ (m1), Lemma 11.7and Proposition11.4yield HIe(t) =H(m1)(t) +HJ(t)−H
Je(t) = 1−tdeg(m1)
(1−t)n+1 +HJ(t)−H
Je(t),
completing the proof. ut
A consequence of the correctness of Algorithm 11.8 is that the Hilbert seriesHI(t) can be written as a rational function with (1−t)m+1as denom- inator. Going one step further, we can extract information about the Hilbert function from this. The results are stated in the following corollary.
Corollary 11.10(Hilbert-Serre theorem). Let I ⊆ K[x1, . . . , xn] be an ideal. Then the Hilbert series has the form
HI(t) = a0+a1t+· · ·+aktk
(1−t)n+1 (11.3)
11.2 Hilbert Polynomials and Dimension 167