MA 252 notes: Commutative algebra

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1 MA 252 notes: Commutative algebra (Distilled from [Atiyah-MacDonald]) Dan Abramovich Brown University April 1, 2017 Abramovich MA 252 notes: Commutative algebra 1 / 21

2 The Poincaré series of a graded module Recall that an additive function λ : A 0 -mod Z is a function such that λ(m ) + λ(m ) = λ(m) whenever 0 M M M 0 is exact. The key case is A 0 = k a field and λ = dim k, but the flexibility will be useful. Say A = A n is a graded ring, M = M n a graded module. Then M n is an A 0 module. We assume A noetherian, generated by homogeneous x 1,..., x s of degrees k 1,..., k s over A 0. The key case is k i = 1. Assuming M is ginitely generated, then each M n is a finitely generated A 0 -module. Definition The Poincaré series of M is P λ (M, t) := λ(m n )t n. Abramovich MA 252 notes: Commutative algebra 2 / 21

3 Rationality Theorem P λ (M, t) = f (t)/ (1 t k i ), with f (t) Z[t], in particular it is a rational function. Apply induction on the number of generators s of A over A 0. We have an exact sequence 0 K M xs M L 0. x This breaks up as 0 K n M s n Mn+ks L n+ks 0, giving λ(k n ) λ(m n ) + λ(m n+ks ) λ(l n+ks ) = 0. Each of K and L are A/(x s )-modules, so induction applies. To balance degrees we multiply by t ks and get t ks P(K, s) t ks P(M, s) + P(M, s) P(L, t) = g(t) Q[t]. Abramovich MA 252 notes: Commutative algebra 3 / 21

4 Dimension, act one. Define d(m) to be the order of pole of P(M, t) at t = 1. Corollary If each k i = 1 we have that λ(m n ) is eventually a polynomial Q[n] of degree d 1. (Read the proof in the book. Relies on Taylor expansion of 1/(1 t) d.) This is called the Hilbert polynomial. The actual function λ(m n ) is the Hilbert function. Proposition If x A k not a zero divisor in M then d(m/xm) = d(m) 1. Follow the proof of the theorem! K = 0 and L = M/xM. If A = k[x 1,... x s ] and λ = dim then P(A, t) = 1/(1 t) s, so d = dimension. Abramovich MA 252 notes: Commutative algebra 4 / 21

5 Hilbert-Samuel functions (A, m) noetherian local ring, q an m-primary having s generators, M finitely generated, (M n ) stable q-filtration. Proposition (i) M/M n of finite length. (ii) l(m/m n ) is eventually a polynomial g(n), of degree s. (ii) deg(g(t)) and the leading coefficient do not depend on the chosen filtration, only on M, q. Gr(A) = q n /q n+1, Gr 0 (A) = A/q Artin local, Gr(M) = M n /M n+1 finitely generated Gr(A)-module. Gr n (M) = M n /M n+1 noetherian A/q module, so finite length. So l(m/m n ) = n 1 r=0 l(m r /M r+1 ) is finite, giving (i). Abramovich MA 252 notes: Commutative algebra 5 / 21

6 Hilbert-Samuel functions - continued (i) M/M n of finite length. (ii) l(m/m n ) is eventually a polynomial g(n), of degree s. (ii) deg(g(n)) and the leading coefficient do not depend on the chosen filtration, only on M, q. If x 1,..., x s generate q the ring Gr(A) is generated by x 1,..., x s. So l(m r /M r+1 ) is eventually a polynomial f (n) Q[n] of degree s 1. As l n+1 l n = f (n) we have l n is eventually a polynomial g(n), of degree s, giving (ii). Two stable filtrations (M n ), (M n) are of bounded difference so g(n + n 0 ) g (n), g (n + n 0 ) g(n). So lim(g(n)/g (n)) = 1, so they have the same degree and leading term, giving (iii). Abramovich MA 252 notes: Commutative algebra 6 / 21

7 Dimension, Act 2 Notation: χ M q (n) := l(m/q n M) for large n. χ q (n) := χ A q (n). Reformulate: Corollary l(a/q n ) is a polynimial χ q (n) for large n, of degree s. Proposition deg χ q (n) = deg χ m (n), in particular independent on q. m q m r so m n q n m rn so χ m (n) χ q (n) χ m (nr) for large n, so the degrees agree. Notation d(a) := deg χ q (n). Note: d(a) = d(gr(a)). Abramovich MA 252 notes: Commutative algebra 7 / 21

8 Inequalities For (A, m) noetherian local define δ(a) = minimum number of generators of a m-primary ideal. Since s d(a) we have Proposition δ(a) d(a). Proposition If x A is not a zero divisor for elements of M and M = M/xM then deg χ M q < deg χ M q. In particular if x is not a zero divisor then d(a/(x)) d(a) 1. Writing N = xm we have N M. Writing N n = N q n M we have exact sequence 0 N/N n M/q n M M /q n M 0. So l(n/n n ) χ M q (n) + χ M q (n) = 0 for large n. Now (N n ) is a stable q-filtration by Artin-Rees, so l(n/n n ) is eventually a polynomial with the same leading term as χ M q (n), so χ M q (n) has lower degree. Abramovich MA 252 notes: Commutative algebra 8 / 21

9 Inequalities - continued Keeping (A, m) noetherian local, we have Proposition d(a) dim A. We apply induction on d = d(a). If d = 0 then l(a/m n ) is eventually constant so m n = m n+1. By Nakayama m n = 0 so A Artinian and dim A = 0. If d > 0 consider a chain p 0 p r a chain of primes. Pick x 1 p 1 but x 1 p 0, and set A = A/p 0, an integral domain, with x 0 the image of x. We therefore have d(a /(x )) < d(a ). Also we have A/m n A /m n so l(a/m n ) l(a /m n ) so d(a) d(a ). So d(a /(x )) d(a) 1 = d 1. We have the primes p 1 /(p 0, x) p r /(p 0, x), so by induction r 1 dim A /(x ) d(a /(x )) d 1, hence r d. Abramovich MA 252 notes: Commutative algebra 9 / 21

10 Height of a prime Corollary If A noetherian local then dim A finite. Moreover, define the height ht(p) of p to be the length of chain p = p r p 0. Corollary Let A be noetherian. Then ht(p) is finite. In particular primes in A satisfy the descending chain condition. Indeed we can pass to the noetherian local ring A p. (The depth of a prime ideal is not necessarily finite.) Abramovich MA 252 notes: Commutative algebra 10 / 21

11 Dimension, Act 3 Proposition Let (A, m) be noetherian local. There is an m-primary q generated by dim A elements. This implies Theorem (The dimension theorem) If A noetherian local, then dim A = δ(a) = d(a). Abramovich MA 252 notes: Commutative algebra 11 / 21

12 Existence of good primary ideal - proof Let (A, m) be noetherian local. There is an m-primary q generated by d = dim A elements. One constructe (x 1,..., x d ) inductively so that if p (x 1,..., x i ) then ht(p) i. Take p j the minimal primes containing (x 1,..., x i ), having height i 1. We have i 1 < dim A = ht(m). So m p j and so m p j. Take x i m p j. If q contains (x 1,..., x i ), it contains a minimal prime p of (x 1,..., x i ). If p = p j then q p since x i p. So ht(q) i. If p is another, then ht(p) i so ht(q) i anyway. It follows that any prime containing (x 1,..., x d ) as height at least d, but since ht(m) = d we have that (x 1,..., x d ) is m-primary. Abramovich MA 252 notes: Commutative algebra 12 / 21

13 Consequences One gets immediately that dim(k[x 1,..., x m ]) (x1,...,x m) = m. Corollary Let (A, m) be noetherian local. Then dim A dim A/m (m/m 2 ). Corollary Let A be noetherian. Every prime containing (x 1,..., x r ) has height r. Indeed, the ideal (x 1,..., x r ) becomes m-primary in A p. Abramovich MA 252 notes: Commutative algebra 13 / 21

14 Krull s theorem Theorem (Krulls Hauptidealsatz) A noetherian, x A not a unit and not a zero divisor. Every minimal prime of (x) has height 1. The height is 1 by a previous result. Every element of a prime of height 0 is a zero divisor by a result on primary decompositions. So the height is 1. Corollary If A noetherian local and x m not a zero divisor, then dim A/(x) = dim A 1. We have seen that d := dim A/(x) dim A 1. If ( x 1,..., x d ) is an m primary ideal then (x 1,..., x d, x) is m primary, giving d + 1 dim A. Abramovich MA 252 notes: Commutative algebra 14 / 21

15 More consequences Corollary Let Â be the completion of (A, m). Then dim A = dim Â. Indeed A/m n Â/ ˆm n so χ m (A) = χˆm (Â). We say (x 1,... x d ) a system of parameters if they generate a primary ideal with d = dim A. Proposition Assume (x 1,... x d ) a system of parameters with primary q. Assume f (t 1,..., t d ) A[t 1,..., t d ] s, and f (x 1,..., x d ) q s+1. Then f m[t 1,..., t d ]. First consider α : A/q[t 1,..., t d ] Gr q (A) sending t i x i + q. This is surjective since t i generate q. It follows that f mod q is in Kerα. (... continued) Abramovich MA 252 notes: Commutative algebra 15 / 21

16 Parameters continued Assume (x 1,... x d ) a system of parameters with primary q. Assume f (t 1,..., t d ) A[t 1,..., t d ] s, and f (x 1,..., x d ) q s+1. Then f m[t 1,..., t d ]. f is in Ker(α : A/q[t 1,..., t d ] Gr q (A)). Assume by contradiction a coefficient of f is a unit. Then f can t be a zero divisor by an old statement you did in homework. dim(gr q (A)) dim(a/q[t 1,..., t d ]/(f ) = dim(a/q[t 1,..., t d ] 1 = d 1 But by the dimension theorem dim(gr q (A)) = d. Abramovich MA 252 notes: Commutative algebra 16 / 21

17 Parameters with coefficient field A coefficient field for (A, m) is a field k A such that k A/m an isomorphism. So k[[x]] has a coefficient field by Z p does not. Corollary Assume A has a coefficient field k and (x 1,... x d ) parameters. Then x 1,... x d are algebraically independent. Assume f (x 1,..., x d ) = 0 where f has coefficients in k. Write f = f s + h.o.t. with f s 0. The proposition applies of f s, so it has its coefficients on m, and since the coefficients are in k they are 0, contradiction! Abramovich MA 252 notes: Commutative algebra 17 / 21

18 Regular local rings Theorem Let (A, m) noetherial local, dim A = d, k = A/m. The following are equivalent: (i) Gr m (A) k[t 1,..., t d ], (ii) dim k m/m 2 = d (iii) m can be generated by d elements. (i) implies (ii) by definition of grading. Elements x i generating m/m 2 generate m by Nakayama so (ii) implies (iii), and (iii) implies (i) since as in the proof of the proposition k[t 1,..., t d ] Gr m (A) an isomorphism. Such rings are called regular local rings. For instance k[x 1,..., x d ] m0 regular, since its gradation is k[x 1,..., x d ]. Abramovich MA 252 notes: Commutative algebra 18 / 21

19 Domains, regular local rings and completions Lemma Let a A such that a n = 0, and assume Gr a A a domain. Then A is a domain. This evidently implies that a regualr local ring is a domain. Assume x, y A nonzero. Let r maximal so x a r and s maxial for y a s. So x Gr r (A) and ȳ Gr s (A) are nonzero, so xȳ = (xy) Gr r=s (A) nonzero, so xy 0. Proposition For A noetherian, (A, m) regular if and only if (Â, ˆm) regular. Indeed Â noetherian of the same dimension and with isomorphic graded ring! So k x 1,..., x d is regular. Abramovich MA 252 notes: Commutative algebra 19 / 21

20 Dimensions of varieties Assume k algebraically closed. The dimension dim V of an affine variety V with coordinate ring A(V ) = k[x 1,..., x n ]/p is the transcendence degree of the fraction field k(v ) of A(V ). Theorem For any V and m A(V ) we have dim V = dim(a(v ) m ). Lemma Assume B A domains, B integrally closed, A integral over B. Let m A maximal and n = B m. Then n maximal and dim A m = dim B n. We have seen by going up that n is maximal, and any chain of primes m q 1 q d in A restricts to distinct primes in B, giving dim A m dim B n. The opposite follows by going down. Abramovich MA 252 notes: Commutative algebra 20 / 21

21 Dimensions of varieties - conclusion For any V and m A(V ) we have dim V = dim(a(v ) m ). By a previous corollary we have that dim V dim(a(v ) m ), since a system of parameters is algebraically independent. By Noether normalization there is a polynomial ring and integral extension B = k[x 1,..., x d ] A(V ), with d = dim V. Since B integrally closed the lemma applies so we need to show d = dim B n. By weak nullstellensatz we may as well assume n is the ideal of the origin, and the local ring has dimension d. Abramovich MA 252 notes: Commutative algebra 21 / 21

Homework 2 - Math 603 Fall 05 Solutions

Homework 2 - Math 603 Fall 05 Solutions 1. (a): In the notation of Atiyah-Macdonald, Prop. 5.17, we have B n j=1 Av j. Since A is Noetherian, this implies that B is f.g. as an A-module. (b): By Noether