LECTURE 21 1. The Hilbert-Samuel function This next result clarifies the degree of the polynomial describing the Hilbert function of a graded module. Theorem 1.1. (Hilbert polynomial Let be a finitely generated graded over R = R 0 [R 1 ] where R is Noetherian, R 0 is Artinian, and dim = d. Then there exists a polynomial HP of degree d 1, called the Hilbert polynomial of, such that for all n 0 H (n = HP (n (the zero polynomial will have degree 1 by convention here. Proof. We will use induction on d. The case d = 0 is immediate since in this case is Artinian and so n = 0 for all n 0, hence H (n = 0 for n 0. First let us prove the statement for the case = R/p where p is a graded prime ideal of R. Assume d > 0 and so let x be a homogeneous element of R/P of degree 1 (this is possible because R is standard graded. We have the following short exact sequence: 0 (R/P ( 1 x R/P R/(xR + P 0. Applky the additivity of the length function in each degree n and get H (n H (n 1 = H N (n, where N = R/(xR + P. Since the dimension of N is d 1 and H N is a function of polynomial type of degree d 2 it follows from the previous equality that H polynomial type of degree (d 2 + 1 = d 1. is of For the general case, note that, similar to the ungraded case, there is prime filtration for : 0 = 0 1 i n =, where i+1 / i (R/p i (a i where p i are 1
2 LECTURE 21 graded prime ideals in R and a i Z. The additivity of the length function gives that H (n = n i=i H (R/p i (a i. Note that d is indeed the maximum of all dimensions of R/p i and the Hilbert polynomials have nonnegative leading coefficients since they define a nonnegative function. In conclusion, the case R/p implies the general case. Definition 1.2. Let be a finitely generated graded over R = R 0 [R 1 ] where R is Noetherian, R 0 is Artinian, and dim = d. The Hilbert iterated functions are defined as H,0 : IN Z, where H,0 = H, and H,i = j n H,i 1 (j. We take it as fact that the latter is a function of polynomial type with degree d + i 1 (the reader can try it as an exercise based upon Theorem 1.1. We write HP (n = e (d 1! nd 1 + lower degree terms. By definition the multiplicity of, e(, is e if d > 0 and l( if d = 0. d ( n + i Note that as a consequence H,1 (n = a i, for n >> 0, with a d = e(, i i=0 and the sum is of polynomial type with degree d. Example 1.3. The curve x 2 y 2 y 3 k[x, y] = 0 defines a one-dimensional ring (x 2 y 2 y 3. The ring is not graded, however we still want to find a natural way to define the multiplicity of the curve at the origin. Now we will lead up to the Hilbert Samuel function of a finitely generated module over a semi-local (has finitely many maximal ideals, Noetherian ring R. First, we need some definitions. Definition 1.4. Let m = Jac(R = m ax(r m. Remember that m n = 0. Let I R. Then I is said to be an ideal of definition for R if θ > 0, integer, such that m θ I m. I n Definition 1.5. Let gr I ( = I n+1 over gr I n I(R = I where n+1 In = R for n=0 n=0 n = 0. Let be finitely ( generated over R, and dim = d. The Hilbert-Samuel function is HS (n = l. I n+1 Consider the exact sequence 0 In I n+1 I n+1 I n 0. n=1
( ( ( I n The exactness implies l = l l I n+1 I n I n+1 Hgr I (,1(n. This can be shown through the following string: Hgr I (,1(n = j n Hgr I ((j = j n LECTURE 21 3 l([gr I (] j = j n l ( I j I j+1, which says HS (n = ( = l = HS I n+1 (n. Hence by Theorem 1.1 and above considerations, the Hilbert-Samuel function of is of polynomial type with degree d, where d = dim(gr I (. Let R be semilocal and Noetherian, m = Jac(R. Take I R such that θ > 0 with m θ I m. Let be a finitely generated R-module. We want to show deg HSP,I = dim, where P S,I is the Hilbert-Samuel polynomial of with respect to I. We know HPgr I ( has degree equal to dim gr I ( 1, and hence HSP,I has degree equal to dim gr I (. So it suffices to show dim = dim gr I (. Denote the degree of HSP,I by the number d. This does not depend on I, our ideal of definition. In fact, m nθ+1 I n+1 m n+1 for all n. So HS,m (n HS,I (n HS,m (nθ. This comes from the fact that the maps in m nθ+1 I n+1 m n+1 are surjective. Thus the degree of HSP,m equals that of HSP,I. Proposition 1.6. Let R, I be as above, and,, be finitely generated R-modules such that 0 0 is exact. Then d = max{ d, d }, and the polynomials HSP,I, HSP,I HSP,I have the same leading coefficient. Proof. We can think of = /. Consider implies 0 In + I n I n I n = / I n / = I n + 0 I n +. This is exact. Thus ( ( ( ( ( + I n l = l + l = l + l. I n I n I n I n I n ( We write φ(n = l, so φ(n is of polynomial type, and φ(n 0 for all I n+1 n 0. So d = max{ d, deg φ}. It now suffices to show deg φ = d. The Artin- Rees lemma implies I i (I k = I k+i for some k, and all i 0. We will
4 LECTURE 21 now show the claim that I n+1 I n+1 I n k+1 for some large n and some fixed k. From the lemma result, we get I k+1 I(I I, and I k+2 I 2 (I k I 2. This implies I m+k I m for all m. Thus I n+1 I n k+1 for every n k 1, and the claim is established. Therefore all the maps in I n+1 I n+1 I n k+1, are surjective. So HS,I(n φ(n HS,I(n k which implies d = deg HSP,I = deg φ, so the first part of the proposition holds. For the leading coefficients, LC(φ = LC(HSP,I, but HSP,I HSP,I = HSP,I, so the second part holds. Remark 1.7. Let (R, m be local and Noetherian, and let R be finitely generated. Let s = min{k x 1,..., x k m with l(/(x 1,..., x k < }. Then s = dim. Proof. Similar to the case = R. Theorem 1.8. Let (R, m be local and Noetherian, R be finitely generated, and I an m-primary ideal of R. Let d = dim gr I (. Then dim = d. Proof. Using the interpretation of the dimension as the degree of the Hilbert-Samuel polynomial we can replace I by m. First we show d dim. We proceed by induction on d. Assume first that = R. If d R = 0, then gr m (R is Artinian which implies m n = m n+1. Now, by NAK, we have m n = 0, which gives dim R = 0. Next, let d R > 0, and take P 0 P n a chain R of prime ideals in R. Let x P 1 \ P 0, and let B =. Note that dim(b n 1. P 0 + xr Then the sequence 0 R/P 0 R/P 0 B 0 is exact with the first nontrivial map being multiplication by x. This gives d R/P0 = max( d R/P0, d B, so d B d R/P0, but in fact we have a strict inequality by a leading coefficient argument using the previous theorem. Since R surjects onto R/P 0, we have R/m n R R/P 0 surjecting onto = m n + P 0 m n (R/P 0, so d B < d R/P0 d R. By the induction hypothesis, we have n 1 dim B d B d R 1. This implies d R n. For a general the proof goes through with the help of prime filtrations and Proposition 1.6 as in the proof of the existence of the Hilbert polynomial.
LECTURE 21 5 Next we show dim d and proceed by induction ( on dim. If dim = 0, then is Artinian which implies l( <. Then l are bounded by l( for all m n n since surjects onto m n, so d = 0. Now let dim = s > 0. By the remark ( x 1,..., x s m such that l <. Let i = for each (x 1,..., x s (x 1,..., x i 1( i s. Note( that dim i = dim i = s i. By the second isomorphism theorem, 1 l = l, and since the sequence m n 1 x 1 + m n 0 x + mn m n m n ( m n x 1 + m n 0 ( l x 1 is exact, the length equation is equal to l, and considering x 1 m n ( (m n : x 1 = {m x 1 m m n }, the equation now becomes l ( m n x 1 l. Consider the map ϕ : which acts by multiplication (m n : x 1 x 1 m n by x 1, and then by the appropriate modulo. We have ker ϕ = {m x 1 m = 0} = {m x 1 m m n } = (m n : x 1. But m n 1 (m n : x 1 since x 1 m n 1 m n because x 1 m. Thus surjects onto which implies that the m n 1 ((m n : x 1 ( length equation above is greater than or equal to l l. Therefore m n m n 1 HSP 1 (n HSP (n = HSP (n 1 (the leading terms on the right hand side cancel out, so d 1 d 1. But dim 1 = dim 1. Applying the induction hypothesis, dim 1 = dim 1 geq d 1 d 1, so dim d. Thus we have equality, and the theorem is established. Remark 1.9. The same statement is true for R a semilocal Noetherian ring with m = Jac(R, I an ideal of definition for R, and a finitely generated R-module. Note that d = dim if I m = Jac(R an ideal of definition for. The leading coefficient of the Hilbert-Samuel polynomial is e(gr I(. d! Definition 1.10. We let e(i, := e(gr I ( denote the multiplicity of with respect to I. The multiplicity of is the multiplicity of with respect to m = Jac(R, but most often this is used for the case when R is local. This Theorem allows us to work with the multiplicity of an ideal I primary to the maximal ideal in a local ring R, without necessarily mentioning the associated graded
6 LECTURE 21 ring of R with respect to I. One can do this by thinking of the multiplicity as a certain limit, as described in the previous lecture. Corollary 1.11. lim n d! l(/in n d exists and equals e(i,. Proof. Consider l( I n+1 = e(gr I( n d + lower degree terms. Then, dividing by d! (n + 1 d and letting n, the limit exists. Corollary 1.12. e(i r, = e(i, r d. Proof. Left as a simple exercise.