Chap 11: Dimension Theory
MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 67
11. Dimension theory The dimension of a Noetherian local ring can be defined in several ways. The main theorem is that all of these descriptions give the same number. So far we have the Krull dimension dim A which is the length of the maximal chain of prime ideals. 11.1. Poincar´eseries and Hilbert functions. Suppose that m is a maximal ideal of A and Gr(A)= A = A/m m/m2 m2/m3 n ⊕ ⊕ ⊕··· is the associated graded ring. If A is Noetherian then each An = n n+1 m /m is a finite dimensional vector space over the field k = A0 = A/m and we have the Poincar´eseries
∞ n PGr(A)(t)= t dimk An n=0 11.1.1. length function. In greater generality we consider a graded ring
∞ A = A = A A A n 0 ⊕ 1 ⊕ 2 ⊕··· n=0 so that A is Noetherian and A0 is Artinian. Suppose that
∞ M = M = M M M n 0 ⊕ 1 ⊕ 2 ⊕··· n=0 is a finitely generated graded module over A. Then each Mn is a finitely generated A0-module. Since A0 is Artinian, each Mn has finite length (Mn) and we can define the Poincar´eseries of M to be
∞ n PM (t)= (Mn)t n 0 − Recall that the length of an A0-module N is defined to be the largest integer m = (N) so that there is a filtration
N = N0 N1 N2 Nm =0 ··· by A0-submodules Nn. It can be shown that any two maximal fil- trations have the same length. The subquotients Nn/Nn+1 are called composition factors of N. They are simple modules which are unique up to isomorphism and permutation of indices. Given this property it is easy to see that the length function satisfies the property that, for any short exact sequence of f.g. A0-modules
0 M M M 0 → → → → 68 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA
we have (M)=(M )+(M ). Any function on f.g. A0-modules with this property is called an additive function. Some trivial observations about the Poincar´eseries:
(1) 0 PM (1) in general. (2) P ≤(1) = ≤∞(M ) < iff M = 0 for all but finitely many n. M n ∞ n (We are given that (Mn) < for all n 0.) ∞ ≥ (3) PM (1) = 0 iff M =0. 11.1.2. rational functions. Theorem 11.1. The Poincar´eseries of M is a rational function of the form f(t) PM (t)= s (1 tki ) i=1 − where f(t) is a polynomial in t with coefficients in .) Z Proof. The proof is by induction on s, the number of homogeneous generators xi of the ideal A+. If s = 0 then A = A0 and each Mn is an A-submodule of M. Since M is f.g. this implies that Mn = 0 for all but finitely many n. So n PM (t)=f(t)= (Mn)t is a polynomial in t with coefficients in Z. Now suppose that s 1 and consider the last generator xs Ak . ≥ ∈ s Multiplication by xs gives a graded map M M of degree ks.LetK be the kernel and L the cokernel of this map.→ Then, for each n 0we get an exact sequence: ≥
xs 0 Kn ks Mn ks Mn Ln 0 → − → − −→ → →
where Kn ks = Mn ks =0forn (Kn ks ) (Mn ks )+(Mn) (Ln)=0 − − − − Multiply by tn n n n n (Kn ks )t (Mn ks )t + (Mn)t (Ln)t =0 − − − − Sum over all n tks P (t) tks P (t)+P (t) P (t)=0 K − M M − L So: ks PL(t) t PK (t) PM (t)= − 1 tks − Since xs annihilates K and L, these are graded modules over the graded ring A = A/(x ) whose positive grade ideal has s 1 generators s − x1, x2, , xs 1, the images of xi A. So, PL(t) and PK (t) are rational ··· − s 1∈ k functions with denominator − (1 t i ). The theorem follows. i=1 − MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 69 There is one special case when we can easily calculate f(t) by induc- tion. This is the case in which K = 0, i.e., x z = 0 for all z = 0 in M. s In this case we say that xs is M-regular. Then PM/xsM (t) PM (t)= 1 tks − Next, suppose that xs 1 is M/xsM-regular. Then − PM/xs 1xsM (t) − PM/xsM (t)= 1 tks 1 − − Definition 11.2. We say that (any sequence of elements of A) xk, ,xs is an M-regular sequence if this continues in the analogous way··· up to xk. When M = A, we call this a regular sequence for A. By induction we would get: PM/xkxk+1 xsM (t) PM (t)= s ··· (1 tki ) i=k − Example 11.3. Take A = M = K[x , ,x ] graded in the usual way 1 ··· d so that x1, ,xd span A1 = M1. Then x1, ,xd is an M-regular sequence and···M/x x M = K is 1-dimensional··· and 1 ··· d 1 n + d 1 P (t)= = − tn A (1 t)d d 1 − − Proof. (of the second equality) Start with 1 n 2 3 (1 t)− = t =1+t + t + t + − ··· and differentiate d 1 times to get − d n d+1 (d 1)!(1 t)− = n(n 1) (n d +2)t − − − − ··· − d n n d+1 n + d 1 n (1 t)− = t − = − t − d 1 d 1 − − ∞ nd 1 nd 2 = − − + tn (d 1)! − (d 3)! ··· n=0 − − 11.1.3. dimension using PM (t). Definition 11.4. Let d = d(M) be the smallest positive integer so that d lim(1 t) PM (t) < t 1 → − ∞ Proposition 11.5. d(M) s for all f.g. graded A-modules M. ≤ 70 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Proof. 1 t 1 1 − k = k 1 1 t 1+t + + t − → k as t 1. So, (1 t)s−P (t)convergesas··· t 1. → − M → Lemma 11.6. If x A is M-regular then d(M)=d(M/xM)+1. ∈ k Proposition 11.7. If A has an M-regular sequence of length r then d(M) r. ≥ In Example 11.3, d = s = r. So, d(K[x1, ,xd]) = d. The idea is that d(M) is the number of generators of A ···which act as independent variables on M. 11.1.4. Hilbert function. The Hilbert function gives a formula for (Mn) for large n. Theorem 11.8. Suppose that ki =1for i =1, 2, ,s. Then (Mn) is a polynomial of degree d(M) 1 for large n. ··· − Proof. By Theorem 11.1 we have: f(t) (M )tn = n (1 t)s − where f(t) is a polynomial. By definition of d = d(M) we have f(t)= s d (1 t) − g(t) where g(1) = 0. So, − g(t) n + d 1 (M )tn = = − tng(t) n (1 t)d d 1 − − m Suppose that g(t)= a tk = a + a t + + a tm. Then k=0 k 0 1 ··· m m n n + d 1 n+k (M )t = a − t n k d 1 k=0 − When n m we get ≥ m n k + d 1 (M )= a − − n k d 1 k=0 − (When n Definition 11.9. The Hilbert function of M is the polynomial func- tion hM (n)=(Mn) for sufficiently large n. 11.1.5. Samuel function. For each n consider the finite sum: g(n)=(M0)+(M1)+ + (Mn 1). · − Lemma 11.10. g(n) is a polynomial of degree d(M) for sufficiently large n. Proof. For m>m0 we have d 1 (Mm)=hM (m)=c0 + c1m + + cd 1m − ··· − Therefore n 1 − g(n)= hM (m)+const m=0 n 1 k for n>m0. But, anyone who has taught Calculus knows that m−=0 m is a polynomial in n of degree k + 1. The lemma follows. Suppose that A is a Noetherian local ring with unique maximal ideal m and quotient field K = A/m.LetM be any f.g.A-module. For any n, M/mnM is a f.g. module over the Artin local ring A/mn. Therefore it has a finite length: n 2 n 1 n (M/m M) = dim (M/mM mM/m M m − M/m M) K ⊕ ⊕···⊕ Proposition 11.11. For sufficiently large n, (M/mnM) is a polyno- mial in n of degree s = dim (m/m2). ≤ K Proof. Take the associated graded ring Gr(A)=A/m m/m2 m2/m3 ⊕ ⊕ ⊕··· 2 2 3 has positive ideal Gr(A)+ = m/m m /m generated by s ele- ments of degree 1. The associated Gr⊕ (A)-module⊕··· Gr(M)=M/mM mM/m2M ⊕ ⊕··· is a f.g.Gr(A)-module with d(Gr(M)) s by Proposition 11.5. So, (M/mnM) is a polynomial of degree d≤(Gr(M)) s for sufficiently ≤ large n by the lemma above. Definition 11.12. The Samuel function for M is defined to be the polynomial function given for large n by: M n χm (n)=(M/m M) deg χM (n)=d(Gr(M)) dim (m/m2) m ≤ K 72 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 11.2. Noetherian local rings. Suppose that A is a Noetherian lo- cal ring with maximal ideal m. We will prove the main theorem of dimension theory which says: d(A) = dim A = δ(A) Definition 11.13. If A is a Noetherian ring δ(A) is defined to be the smallest number of elements which can generate a proper ideal q so that A/q is Artinian. When A is Noetherian local, this is equivalent to saying that q is m-primary. If M is a f.g. A-module we define δ(M)to be δ(A/ Ann(M)). Once we review the definition of d(A) we will see that we have already proven that d(A) δ(A) ≤ 11.2.1. definition of d(M). Suppose that s = δ(A) and q is an m- primary ideal generated by s elements. Then Gr (A)=A/q q/q2 q2/q3 q ⊕ ⊕ ⊕··· is a graded Noetherian local ring whose positive ideal Grq(A)+ = q/q2 q2/q3 is generated by s homogeneous elements in degree ⊕ ⊕··· 1. Suppose that M is a f.g. A-module and (Mn) is a stable q-filtration. Then Gr(M)=M/M M /M M /M 1 ⊕ 1 2 ⊕ 2 3 ⊕··· is a f.g. Grq(A)-module. This implies that the Poincar´eseries ∞ f(t) P (t)= (M /M )tn = M n n+1 (1 t)s n=0 − where f(t) Z[t]. If we reduce the fraction we get (1 t)d in the denominator∈ where d = d(M). So, clearly d s = δ(A). Replacing− A with A/ Ann M we get ≤ d(M) δ(M) ≤ (in particular d(A) δ(A)) with one particular definition of d(M). Next recall that ≤ h(n)=(Mn/Mn+1) is a polynomial in n of degree d(M) 1 for large n. This was called the Hilbert function of M. As a formal− consequence of this theorem we concluded that, for large n, g(n)=(M/Mn) MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 73 is a polynomial in n of degree d(M). (The theorem was stated only in n the case Mn = m M but the same proof works for any filtration.) In n the particular case Mn = q M we have the Samuel function M n χq (n)=(M/q M) 11.2.2. d(M) is well defined. We will show that d(M) is independent of the choice of q and the choice of the filtration (Mn). Proposition 11.14. Given any m-primary ideal q and any stable q- M n filtration (Mn), the polynomials g(n)=(M/Mn) and χq (n)=(M/q M) have the same degree and leading coefficient. Furthermore, this leading coefficient is a positive rational number. Proof. (M ) being q-stable means n s.t. n ∃ 0 qn+n0 M M = qnM qnM ⊆ n+n0 n0 ⊆ This implies that (M/qn+n0 ) (M/M ) (M/qnM) ≥ n+n0 ≥ χM (n + n ) g(n + n ) χM (n) q 0 ≥ 0 ≥ q M M But χq (n+n0) and χq (n) are polynomials in n of the same degree and the same leading coefficient. So, g(n+n0)=(M/Mn+n0 ) and therefore g(n) also has the same degree and leading coefficient. Positivity is obvious. M Proposition 11.15. The degree of the polynomial χq (n) is indepen- dent of the choice of the m-primary ideal q. Proof. If q is m-primary then ma q m for some a. This implies mna qn mn which implies ⊆ ⊆ ⊆ ⊆ (M/mnaM) (M/qnM) (M/mnM) ≥ ≥ χM (na) χM (n) χM (n) m ≥ q ≥ m M But it is easy to see that χm (na) is a polynomial in n with the same M d degree, say d,asχm (n) (with leading coefficient multiplied by a ). So, M χq (n) is a polynomial of degree d in n. Theorem 11.16. If A is a Noetherian local ring and M is a finitely generated A-module then d(M) δ(M). In particular ≤ d(A) δ(A). ≤ 74 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 11.2.3. Krull dimension. The plan is to prove that δ(A) d(A) dim A δ(A) ≥ ≥ ≥ showing that all three are equal. We discussed the definitions of the first two terms and proved the first inequality. Now we prove the second. The proof will be by induction on d(A) (we have not yet proven that dim A is finite). Lemma 11.17. If 0 M M M 0 is a short exact sequence of A-modules then → → → → d(M) = max(d(M ),d(M )) Furthermore, if M ∼= M then d(M) >d(M ). Proof. For each n we have the short exact sequence: n n n 0 M m M m M m M 0 → ∩ → → → n By the Artin-Rees Lemma, (M m M) is a stable m-filtration of M . This exact sequence gives an exact∩ sequence of quotient modules M M M 0 0 → M mnM → mnM → mnM → ∩ Since length is additive, this gives M M M = + mnM M mnM mnM ∩ M M χm (n)=g(n)+χm (n) M By Proposition 11.14, the polynomial g(n) has degree d(M )andχm (n) has degree d(M ). Since both of these polynomials have positive lead- ing terms, the leading terms cannot cancel and their sum has degree equal to the maximum of the two degrees. In the special case when M = M , we use the fact that g(n) will ∼ M have the same degree and the same leading coefficient as χm (n) causing M χm (n) to have smaller degree. Theorem 11.18. If A is a Noetherian local ring with Krull dimension dim A then d(A) dim A ≥ Proof. By induction on d(A). If d(A) = 0 then mn = 0 for some n making A Artinian with dim A = 0. So, suppose d = d(A) > 0 and the theorem holds for B with d(B) We are trying to show that dim A d. So, suppose by contradiction that dim A>d. Then there exists≤ a tower of prime ideals in A of length d +1: p0 p1 p2 pd+1 ··· Choose any x p p .Letx be the image of x in the integral domain ∈ 1 − 0 A = A/p0. Then x is A-regular (not a unit and not a zero divisor). Therefore we have an exact sequence of A-modules x 0 A/p A/p A/(x) 0 → 0 −→ 0 → → By the lemma, d = d(A) d(A) >d(A/(x)) d(A/p ) ≥ ≥ 1 By induction on d we have d(A/p ) dim A/p d 1 ≥ 1 ≥ since p1 p2 pd+1 gives a tower of prime ideals of length d in ··· A/p1. This is a contradiction proving the theorem. 11.2.4. the fundamental theorem of dimension theory. Theorem 11.19. If A is a Noetherian local ring then dim A δ(A) ≥ Proof. Let d = dim A. Claim 1 x , ,x m so that ∃ 1 ··· d ∈ dim A/(x , ,x) d i 1 ··· i ≤ − In particular, dim A/(x1, ,xd)=0 (x1, ,xd) is m-primary δ(A) d. So, it suffices··· to prove this⇐⇒ claim. ··· ⇒We prove≤ Claim 1 by induction on d.Ifd = 0 the statement is clearly true. So, suppose that d>0. Claim 2 If d = dim A>0 then x m so that ∃ 1 ∈ dim(A/(x1)) (If m = pi then by prime avoidance we would have m pi for some i contradicting the assumption that dim A>0.) ⊆ 76 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Let x m so that x/p for any i. Then dim(A/(x )) q0 q1 qd ··· in A all of which contain (x1). But x1 is not contained in any min- imal prime. So, q0 is not minimal. So, the tower can be extended contradicting the hypothesis that dim A = d. This proves Claim 2 which implies Claim 1 which proves the theorem. Corollary 11.20 (Fundamental theorem of dimension theory). If A is a Noetherian local ring then dim A = d(A)=δ(A) In particular, dim A is finite. Corollary 11.21. If A is a Noetherian local ring with maximal ideal m and k = A/m then dim A dim m/m2 ≤ k Corollary 11.22. In any Noetherian ring, every prime ideal has finite height. In particular, the prime ideals satisfy the DCC. MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 77 11.3. regular local rings. Suppose that A is a Noetherian local ring of dimension d. Then there exist d elements x1, ,xd in A which generate an m-primary ideal q in A. Such a collection··· of elements is called a system of parameters for A. Definition 11.23. A system of parameters for A is called regular if it generates the maximal ideal m.IfA has a regular system of parameters it is called a regular local ring. 11.3.1. first characterization of regular local rings. Proposition 11.24. Suppose that A is a Noetherian local ring with dim A = d. Then the following are equivalent. (1) Grm(A) = k[t1, ,td]. ∼2 ··· (2) dimk m/m = d. (3) m is generated by d elements. (4) A is a regular local ring. Proof. Certainly (1) (2) and (3) (4) by definition of regular local ring. Nakayama’s⇒ lemma implies⇔ that (2) (3). To show that ⇔ (3) (1) suppose that x1, ,xd generate m, i.e., they form a regular system⇒ of parameters for A···. Then we get an epimorphism of graded rings ϕ : k[t]=k[t , ,t ] Gr (A) 1 ··· d → m 2 sending ti to the image xi of xi in m/m . If this homomorphism is not an isomorphism then it has a nonzero element in its kernel, say f(t). But k[t] is an integral domain. So, f(t) is not a zero divisor. And it cannot be a unit, being in the kernel of ϕ. So f is regular and thus d(k[t , ,t ]/(f)) = d(k[t , ,t ]) 1=d 1 1 ··· d 1 ··· d − − But Gr (A) is a quotient of k[t , ,t ]/(f). So, m 1 ··· d d = d(Gr (A)) d(k[t , ,t ]/(f)) = d 1 m ≤ 1 ··· d − which is a contradiction. So, ϕ must be an isomorphism as claimed. Corollary 11.25. A regular local ring is an integral domain. Proof. By Krull’s theorem 10.26, mn = 0. Therefore the following lemma applies. Lemma 11.26. Suppose that a is an ideal in A so that an =0. Suppose that Gra(A) is an integral domain. Then A is an integral domain. 78 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Proof. Suppose that x, y are nonzero elements of A. Since an =0, x an, x/an+1 for some n 0 and similarly y am, y/am+1 for∈ some m∈ 0. Then the images≥ of these elements∈ x a∈n/an+1, y am/am+1 ≥are nonzero in Gr (A) and therefore their product∈ ∈ a x y = xy an+m/an+m+1 ∈ is nonzero (since Gra(A) is an integral domain). This implies that xy / an+m+1 and, in particular, xy =0. ∈ Corollary 11.27. A ring A is a regular local ring of dimension 1 iff it is a discrete valuation ring. Proposition 11.28. Let A be a Noetherian local ring. Then A is a Noetherian local ring with the same dimension and A is regular iff A is regular. 11.3.2. Koszul complex. We will use the Koszul complex to prove part of the following well known theorem. Theorem 11.29 (Serre). Suppose that A is a Noetherian local ring. Then the following are equivalent. (1) A is regular. (2) gl dim A< (3) gl dim A =∞ dim A. Suppose that x1, ,xd A. Then the Koszul complex K (x1, ,xd) is the chain complex··· of f.g.∈ free A modules given by: ∗ ··· d (n) Kn(x)=freeA-module generated by ei1i2 in = A ··· ∼ where 1 i1 We went over the basic definitions in class. The global dimension of a ring is the supremum of the projective dimensions of all f.g. modules. This is defined in terms of projective modules. Definition 11.31. A module P is projective if it satisfies the prop- erty that for any epimorphism of modules p : X Y and any homo- morphism f : P Y there exists a morphism f : P X so that → → f = p f. ◦ X f ∃ p ef P Y n If we take X = A and Y = P with f = idP , we get the following characterization of projective modules. Proposition 11.32. A module P is projective iffit is isomorphic to a direct summand of a free module. Every f.g. projective module is isomorphic to a direct summand of An for some positive integer n. Definition 11.33. The projective dimension pd dim M of any A- module M is defined to be the smallest integer n so that there exists an exact sequence: 0 Pn Pn 1 P1 P0 M 0 → → − →···→ → −→ → where the Pi are all projective modules. In class I “rotated” the augmentation map : P0 M to get the following diagram: → P : 0 Pn Pn 1 P1 P0 0 ∗ − ··· M : 0 M 0 ··· The exactness of the original sequence is equivalent to the statement that is a quasi-isomorphism of chain complexes. Definition 11.34. A quasi-isomorphism of chain complexes is de- fined to be a chain map which induces an isomorphism in homology in all degrees. In the case at hand, the projective chain complex P is exact except ∗ in degree 0 where the map P1 P0 is not onto. So, H0(P ) = M and → ∗ ∼ : P0 M gives a quasi-isomorphism P M. → ∗ Definition 11.35. The global dimension of any ring A is defined to be the supremum of pf dim M for all f.g. A-modules M. 80 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Lemma 11.30 (proved below) implies that pr dimAk d. We need to strengthen this to an equality and then we need another≤ lemma: Lemma 11.36. If A is a local ring with maximal ideal m and k = A/m then gl dim A = pr dim k. Proof. By definition of global dimension we have gl dim A pr dim k. So, we need to show the reverse inequality. Suppose that pr dim≥ M = n. Then there is a projective complex 0 Pn Pn 1 P1 P0 0 → → − →···→ → → A which is quasi-isomorphic to M.Torn (M,k) is the kernel of the map- ping Pn k Pn 1 k ⊗ → − ⊗ But this mapping is 0 by Nakayama (assuming the projective complex A P is minimal). So, Torn (M,k) = 0. But ∗ A A Torn (k, M) ∼= Torn (M,k) So, this implies that pr dim k n. Therefore, ≥ pr dim M pr dim k ≤ for all f.g. A-modules M. The lemma follows. We need another lemma to do the induction step for the proof that the Koszul complex is exact. Lemma 11.37. If x , ,x is a regular system of parameters for 1 ··· d a regular local ring A then A/(x1, ,xn) is a regular local ring of dimension d n for n =1, 2, ,d.··· − ··· Proof. By induction it suffices to do the case n = 1. In that case, x1 is a regular element since A is an integral domain. So, dim A/(x1)=d 1. But the images x , , x of x , ,x generate the maximal ideal− of 2 ··· d 2 ··· d A/(x1). So, A/(x1) is regular. Proof of Lemma 11.30. The proof is by induction on the length of the regular sequence. The induction hypothesis is: Claim: For all 1 n d, the Koszul complex K (x1, ,xn) is ≤ ≤ ∗ ··· quasi-isomorphic to A/(x1, ,xn). For n = d this is the statement··· that we want to prove. Start with n = 1. Since x1 is regular, multiplication by x1 is a monomorphism and we have an exact sequence: 0 A x1 A A/(x ) 0 → −→ → 1 → MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 81 When we “rotate” the augmentation map this becomes a quasi-isomorphism K (x1) A/(x1). So, the Claim holds for n =1. ∗The case n = 2 shows the idea of the induction step. In this case the Koszul complex K (x1,x2) looks like: ∗ x2 Ae1 x1 − Ae12 A ⊕ x1 x2 Ae2 This complex contains two copies of the smaller complex K (x1) one is shifted and the boundary map from one copy to the other∗ is multi- plication by x2. This means we have a short exact sequence of chain complexes: 0 K (x1) K (x1,x2) ΣK (x1) 0 → ∗ → ∗ → ∗ → We know by induction that K (x1) A/(x1). So, we can compare this to the complex: ∗ 0 A/(x1) X ΣA/(x1) 0 → → ∗ → → where X is the chain complex: ∗ x2 X : A/(x1) A/(x1) ∗ −→ This is the Koszul complex K (x2) for the regular local ring A/(x1). So, ∗ it is quasi-isomorphic to A/(x1,x2). We have the following commuting diagram where the rows are exact and the first and third vertical maps are quasi-isomorphisms by induction. So, the middle vertical arrow is also a quasi-isomorphism by the 5-lemma. 0 K (x1) K (x1,x2) ΣK (x1) 0 ∗ ∗ ∗ 0 A/(x1) X ΣA/(x1) 0 ∗ Therefore K (x1,x2) X A/(x1,x2). ∗ ∗ The next steps are similar. 11.3.3. multiplicity. I gave a very brief summary of another use for the Koszul complex. Namely, in the equation for the leading terms of the Samuel function A n χq (n)=(A/q ) 82 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA Theorem 11.38. Suppose that A is a Noetherian local ring of di- mension d and x1, ,xd is a system of parameters for A. Let q = ··· A (x1, ,xd). Then the leading term of the Samuel function χq (n) is equal··· to χ(K (x1, ,xd)) d ∗ ··· t d! where χ(K (x)) is the Euler characteristic of the Koszul complex: ∗ d i χ(K (x)) = ( 1) (Hi(K (x))) ∗ − ∗ i=0 For example in the case of a regular local ring the leading term is 1 td d! and k if i =0 Hi(K(x)) = 0 otherwise