# 12 Hilbert polynomials

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1 12 Hilbert polynomials 12.1 Calibration Let X P n be a (not necessarily irreducible) closed algebraic subset. In this section, we ll look at a device which measures the way X sits inside P n. Throughout this section, let S = k[x 0,, X n ] be the homogeneous coordinate ring of P n, and let S(X) be the homogeneous coordinate ring of X. Then S(X) is a graded module over the graded ring S. Define the Hilbert function of X by h X (l) = dim S(X) l, the dimension of the l th graded piece of the coordinate ring of X. Recall that this is isomorphic to S l /I(X) l. Example This means that h P n(l) = ( n+l l ). We saw this before when were examining the Veronese embedding. h X (l) = h P n(l) dim k I(X) l Example line... Suppose that X consists of three distinct points in P 2. They either do or don t live on a If the points are collinear, then there is a linear relation in I(X), and h X (1) = 2 1 = 1. If the points are not collinear, then there is not a linear relation in I(X), and h X (1) = 2 0 = 2. Having said that, we have: Claim If X consists of three distinct points in P 2, then h X (2) = 3. Proof For each point P i, let L i be a linear form which vanishes on P i, but not on the others. Then the product L i L j is a quadratic function which vanishes on P i and P j, but not the other. This gives a surjective map from S(P 2 ) 2 to the space of functions on X, so that h X (3) = ( ) 3 = 3. In fact, for all l 3, h X (l) = 3, no matter what position the points are. Two things worth pointing out: For small l, h X (l) depends on the arrangment of X. 69 M672: Algebraic geometry

2 For sufficiently large l, this difference is erased. Here is another example. Suppose X is a hypersurface in P n ; say I(X) = (F), with F k[x 0,, X n ] of degree d. Then I(X) m is the polynomials of degree m which are divisible by F. This means that multiplication by F gives an isomorphism between S m d and I(X) m, so that h X (m) = h P n(m) h P n(m d) Assume l d; then ( ) ( ) n + l n + l d = l l d For instance, if n = 2, then = dl d(d 3). 2 There s a regularity there which is independent of the curve. The goal of this chapter is to generalize these remarks Algebra Numerical polynomials See homework. The point is that a function h : N Z is called a numerical polynomial if there s some P Q[z] such that, for l 0, h(l) = P(l) Hilbert polynomials of graded modules Let S be a graded noetherian ring. A S-module M is graded if it comes equipped with a decomposition M = M d such that S d M e M d+e. If l N, the twist of M by l is the same module with the grading shifted: The annihilator of M is If M is graded, this is a homogeneous ideal in S. M(l) d = M d+l. Ann(M) = {x S : x M = 0}. 70 M672: Algebraic geometry

3 Lemma Let M be a finitely generated graded module over S. There exists a filtration 0 M 0 M 1 M r = M by graded submodules such that for each i, M i /M i 1 = (S/pi )(l i ) where p i S is a homogeneous prime ideal and l Z. Proof The proof is just like the existence of prime ideal factorizations in noetherian rings. Given any M, let F(M) be the set of graded submodules of M which admit such a filtration. It s nonempty, since (0) is in this class. Let M M be a maximal element; it exists, since M is a noetherian module. Consider M = M/M. If M = 0, we re done. Otherwise, consider the collection of ideals which are annihilators of homogeneous elements, I = {I m = Ann(m) : m M {0} homogeneous }. Then each I m is a proper homogeneous ideal. By noetherianness, we can take a maximal element I m of I. Claim I m is prime. Suppose a, b S, ab I m, b I m ; we ll show a I m. By taking homogeneous components, we may assume a and b are homogeneous. Consider bm M. Since b I m, bm = 0. Since I m bm, by maximality I m = I bm ; but then abm = 0, so a I bm = I m. So, I m is a homogeneous prime ideal, call it p. Let deg(m) = l. Then the module N M generated by m is isomorphic to (S/p)( l). Lift this: N M N = (S/p)( l) M Then M N, N /M = (S/p)( l); N has a suitable filtration, which contradicts the maximality of M. Therefore, M = M. The prime ideals {p 1,, p r } which show up should be thought of as the elementary divisors of the S-module M. Among them, we distinguish the minimal ones; these are the minimal primes of M. Lemma Let p S be a homogeneous prime ideal. Then p Ann(M) if and only if p p i for one of the minimal primes p i of M. 71 M672: Algebraic geometry

4 Proof p Ann(M) if and only if p annihilates some M i /M i 1. But Ann((S/p i )(l)) = p i. Recall that the localization of S at a prime ideal p is S p = { x s : deg x = deg s, s p}. Lemma Let p be a minimal prime of M, and choose any filtration as above. Then the number of i for which p i = p is the length of M p as S p -module, where S p is the localization. Proof that Choose some filtration. For each i with p i = p, there exists a p i which is a unit in S p, so M i p/m i 1 p = S p S (S/p i ) = (0). (Use the fact that 1 1 = (a 1 a) 1 = a 1 a = 0.) On the other hand, if p = p i, then M i p/m i 1 p = S p (S/p) = (S/p). Therefore, M p is an S p -module of the advertised length. Definition The multiplicity of M at p is µ p (M), the length of M p over S p Hilbert-Serre theorem Attached to a graded module is the Hilbert function, h M (l) = dim k M l. Theorem [Hilbert-Serre] Let S = k[x 0,, x n ] with the standard grading, and let M be a finitely generated graded S-module. Then there is a unique polynomial P M (z) Q[z] such that, for l 0, h M (l) = P M (l). Moreover, deg P M = dim Z P n(ann(m)). Remark We assign deg 0 = 1, and dim = M672: Algebraic geometry

5 Proof A homomorphism of graded modules necessarily preserves the grading. Therefore, any short exact sequence of graded modules yields the equality 0 M 1 M 2 M 3 0 h M 1(l) + h M 2(l) = h M 3(l) for each l. After choosing a filtration as above, it suffices to prove the theorem for (S/p)(l). Since h M(l) (d) = h M (d + l), it suffices to prove the theorem for M = (S/p). We distinguish two cases, the first of which is the base case and the second is an inductive step, by induction on dim Z(Ann(M)). Suppose p = (X 0,, X n ). Then h M (l) = 0 for l > 0, and thus for l 0. Moreover, Z(p) =, so that dim Z(p) = deg h M = 1. Otherwise, suppose X i p. Then remember that deg x i = 1 we have an exact sequence 0 M( 1) x i M M = M/x i M 0, and h M (l) = h M (l) h M (l 1) = ( h M )(l 1). Now, Z(Ann(M )) = Z(p) Z(x i ). The hyperplane section Z(x i ) doesn t contain Z(p) (by hypothesis), so that dim Z(Ann(M )) = dim Z(p) 1. By induction, h M is a numerical polynomial, represented by a polynomial P M of degree dim Z(Ann(M )). Since ( h M ) is a numerical polynomial of degree dim Z(Ann(M)) 1, h M is a numerical polynomial of degree dim Z(Ann(M)). The only missing ingredient is the lemma promised in the homework: Lemma Let f : Z Z be any function. Suppose that there exists a numerical polynomial q such that ( f )(n) = q(n) for all n 0. Then there exists a numerical polynomial p such that f (n) = p(n) for all n 0. Proof By the homework, there exists a polynomial Q(z) = r j=0 c j( z j ) such that, for n 0, Q(n) = q(n). Let P(z) = r j=0 ( ) z c j. j + 1 Then for n 0, (P) = ( f ) = Q(n) = q(n), so that (P f )(n) = 0 for n 0. Therefore, (P f )(n) = a for some a Z and all n 0, so that f (n) = P(n) a is a numerical polynomial. 73 M672: Algebraic geometry

6 12.4 Geometry of the Hilbert polynomial Definition Suppose Y P n. The Hilbert polynomial P Y of Y is the Hilbert polynomial of its homogeneous coordinate ring S(Y). Note that deg P Y = dim Y. The degree of Y is (dim Y)! times the leading coefficient of P Y. Remark Since P Y is a numerical polynomial, it has a binomial representation P Y (z) = r ) z c j(. (1) j j=0 with c j Z; then deg Y = c r. Lemma If Y P n is nonempty, then deg(y) N. Proof Since Y =, P Y is a nonzero polynomial of degree r = dim Y. Then deg Y = c r Z, as above. It s positive since there are functions on Y of arbtirarily large degree, so that h Y (l) > 0 infinitely often. Lemma Suppose Y P n, Y = Y 1 Y 2, dim Y 1 = dim Y 2 = r, dim(y 1 Y 2 ) < r. Then deg Y = deg Y 1 + deg Y 2. Proof Let I j = I(Y j ), and let I = I 1 I 2 = I(Y). Note that I(Y 1 Y 2 ) = (I 1 + I 2 ). There is an exact sequence 0 S/I S/I 1 S/I 2 S/(I 1 + I 2 ) 0 Since dim(y 1 Y 2 ) < r, deg P S/(I1 +I 2 ) < r, and the leading coefficient of S/I is the sum of the leading coefficients of those of S/I 1 and S/I 2. Lemma deg P n = 1 Proof Use the earlier calculation of P P n(z). Lemma If H P n is a hypersurface whose ideal is generated by a homogeneous polynomial of degree d, then deg H = d. 74 M672: Algebraic geometry

7 Proof Suppose I(H) = (F), F homogeneous of degree d. There s a diagram of graded modules 0 S( d) F S S(H) = S/(F) 0 Then h H (l) = h P n(l) h P n(l d) h H (z) = h P n(z) h P n(z d) ( ) ( ) z + n z d + n = n n d = (n 1)! zn 1 + and the degree of H is d. Remark The degree depends not just on the variety, but on the way it sits in the ambient projective space. Do an example of the d-uple embedding? 12.5 Intersection theory Let X P n be a projective variety, not necessarily irreducible, of pure dimension r. Let H P n = Z(F) be some hypersurface choose F reduced. By the principal ideal theorem, X H = Z 1 Z m, where dim Z j = r 1. Each Z j corresponds to a homogeneous prime ideal p j = I X (Z j ). Let S = k[x 0,, X n ]. Then we have homogeneous coordinate rings S(X) = S/I(X) S(H) = S/I(H) = S/(F) S(X H) = S/ (I(X) + I(H)) Note that, as S-module, Ann(S(X H)) = I(X) + I(H), and the minimal primes of this module are simply the p 1,, p m. Definition The intersection multiplicity of X and H along Z j is i(x, H; Z j ) = µ p j (S/(I(X) + I(H))). 75 M672: Algebraic geometry

8 Example Consider the elliptic curve E = Z P (Y 2 Z X 3 + XZ 2 ) P 2, and assume we re not in characteristic three. Let H be the hyperplane H = Z(Y). Then we compute intersection multiplicities by looking at the module k[x, Y, Z] M = (Y 2 Z X 3 + XZ 2, Y) k[x, Y, Z] = (X 3 XZ 2, Y) k[x, Z] = X(X Z)(X + Z) = k[x, Z] (X 3 XZ 2 ) Note that as a k-module, this has dimension three. There are three maximal ideals of k[x, Z] which contain the ideal (X 3 XZ 2 ), namely, (X), (X Z), (X + Z). These are the three (minimal) prime ideals associated with this module. Let, say, p = (X Z). If we localize k[x, Y] at p, then we invert (in particular) X and X + Z, so that M p = k[x, Z]/(X Z) has length one as M p -module. The same is true for the other ideals. Example Same elliptic curve, but now consider the hyperplane W = Z(X Z). Then the module in question is k[x, Y, Z] N = (Y 2 Z X 3 + XZ 2, X Z) k[x, Y] = (Y 2 X X 3 + X 3 ) k[x, Y] = (Y 2 X). The minimal primes of this module are (Y) and (X), corresponding to the (projective) points [1, 0, 1] and [0, 1, 0] (the point at infinity ), respectively. The multiplicities at these points are two and one, respectively. Theorem With all notation as above (especially, X H), s i(x, H; Z j ) deg(z j ) = (deg X)(deg H). j=1 76 M672: Algebraic geometry

9 Proof Suppose deg H = d, H = Z(F), F reduced. Let M = S/(I(X) + I(H)). As before, we have an exact sequence 0 S(X)( d) F S(X) M 0 so that P M (z) = P X (z) P X (z d). Consider a (maximal) filtration M 0 M 1 M q = M, with quotients M i /M i 1 = (S/q i )(l i ). Then P M (z) = P (S/qi )(l i )(z). Suppose Z(q i ) is a projective variety of dimension r i and degree d i ; then its Hilbert polynomial has degree r i. Only the minimal primes contribute to the leading coefficient of the Hilbert polynomial of M, since the rest have degree less than r 1. Twisting doesn t affect the leading coefficient of the Hilbert polynomial, so that the leading coefficient is multiply by (r 1)!, then as desired. lcoeff P M (z) = lcoeff q j=1:q j minimal P (S/qi )(z) = µ pj (M) lcoeff P (S/pj )(z) j = i(x, H; Z j ) lcoeff P (S/qi )(z) j deg(x H) = i(x, H; Z j ) deg(z j ) j 77 M672: Algebraic geometry

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