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 Smodule 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 Smodule 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 HilbertSerre theorem Attached to a graded module is the Hilbert function, h M (l) = dim k M l. Theorem [HilbertSerre] Let S = k[x 0,, x n ] with the standard grading, and let M be a finitely generated graded Smodule. 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 duple 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 Smodule, 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 kmodule, 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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