16 Dimension and transcendence degree

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1 16 Dimension and transcendence degree 16.A Dimension: desired properties Our goal is to define the dimension of an algebraic variety over k. It should be a welldefinedfunction from the setof isomorphism classes ofalgebraic varieties over k to Z 0 with the following properties: (A) dim(a 0 k )=0; (B) if Z X is closed subvariety, then dim(z) dim(x); (C) if X = r i=1 X i, where the X i are the irreducible components of X, then dim(x)= sup 1 i r {dim(x i )}; (D) if U X is a dense open subvariety, then dim(u)=dim(x); (E) if X A n k is an irreducible affine variety, and f k[x 1,...,x n ] is a nonconstant polynomial such that X V(f ) X, then dim(x V(f ))=dim(x) 1. Example (i) Properties (A) and (E) imply that dim(a n k ) = n for each n Z 0. Indeed, A n 1 k A n k V(x n), so dim(a n k ) 1=dim(An 1 k ), and the claim follows by induction on n. (ii) By (i), Properties (A), (D) and (E) imply that dim(p n k )=dim(u 0)=dim(A n k )=n. (iii) By (i) (resp. (ii)), Properties (A), (B), (E) (resp. and (D)) imply that each affine subvariety X A n k (resp. projective subvariety X Pn k ) is of dimension n. 16.B Dimension: topological definition Definition Let X be a topological space. (i) Its dimension dim(x) is the supremum of the lengths of the chains X 0 X 1 X d X of irreducible closed subpsaces of X. (ii) For each x X, the dimension dim x (X) of X at x is dim x (X) inf x U dim(u), the infimum of the dimensions of the open neighborhoods of x in X. Example (i) If X=A 1 k, then {0} A1 k is a maximal chain of irreducible closed subspaces: the only other closed subspaces are finite disjoint unions of points, which are not irreducible. Thus, dim(a 1 k )=1. (ii) Let X ={p 1,...,p r } with the discrete topology. As dim({p i }) = 0, and {p i } is a chain of irreducible closed subspaces of maximal length, dim(x) = 0. (iii) If X is an affine variety, then we have an inclusion-reversing bijection between irreducible closed subsets of X and prime ideals ofo X (X) (Remark 2.11). Thus, dim(x) 60

2 equals the Krull dimension ofo X (X), i.e., the supremum of the lengths of a the chains p d p d 1 p 0 of prime ideals. Exercise Let X be an topological space. (i) If X 0 X 1 X d =X is a chain of irreducible closed subspaces of maximal length, then dim(x i )=i for 0 i d. (ii) If X is Noetherian, then (C) is satisfied. Lemma Let X be a topological space. Then: (i) dim(x)=sup x X dim x (X); (ii) if{u i } i I is an open cover of X, then dim(x)=sup i I dim(u i ). Proof. Consider (i). Since X is an open neighborhood of each x X, we have dim(x) dim x (X) for each x. Conversely, suppose dim(x) = n 0. If X 0 X 1 X n is a chainofirreducibleclosesubspacesof X,thenforeachx X 0 andeachopenneighborhood x U X,wehaveachainofirreducibleclosedsubspaces X 0 U X 1 U X n U of U.If X i U=X i+1 U,then X i =X i U=X i+1 U=X i+1,sowe have a strictlyincreasing chain of irreducible closed subspaces of U. This is true for each open neighborhood of U, so dim x (X) n. Note that sup i I dim(u i )=supsup i I so (ii) follows from (i). dim x (U i )=supsupdim x (X)=supdim x (X)=dim(X), x U i x U i x X i I 16.C Transcendence degree Definition The dimension dim(x) of an algebraic variety X is the dimension of its underlying topological space. When X is affine, irreducible closed subvarieties correspond to prime ideals, so for any ring A, we let dim(a) denote the supremum of the lengths d of the chains p 0 p 1 p d of prime ideals in A. Proposition This definition satisfies (A), (B), (C). Proof. Properties (A) and (B) are clear. Property (C) is Exercise 16.4(ii). Exercise16.8. An affine variety X is irreducible if and only ifo X (X) is an integral domain. Definition Let X be an irreducible variety. The set of nonempty open subvarieties U Xforms a direct system with respect to containment. The function field k(x) of X is the colimit colim U X O X (U). By the argument of Proposition 5.9, k(x) O X (U) (0) = Frac(O X (U)) for each affine open U X. In particular, k(x) is a field extension of k. 61

3 Lemma Let A be a ring, and let S x A denote the multiplicative subset{x n (1 ax) a A,n Z 0 } Afor each x A. Then dim(a) n if and only if dim(s 1 x A) n 1 for each x A. Proof. For each x Aand each maximal ideal m A, we have m S x : if x m, then x S x ; if x m, then there exists a Asuch that 1 ax m, since x is a unit modulo m. Suppose p 0 p 1 p m is a chain of prime ideals of S 1 x A. Ontheotherhand,ifm Aisamaximalideal,andp maprimeideal,thenp S x = for each x m p: if x n (1 ax) p, then 1 ax p, so 1 ax m, which is a contradiction. Suppose dim(a) n, let x A, and let q 0 q m be a chain of prime ideals of S 1 x A. We have a chain of prime ideals λ 1 q 0 λ 1 q m of A, where λ : A S 1 x A is the localization map. Since λ 1 q m S x =, lest q m = S 1 x A, λ 1 q m is not maximal, so we have a chain λ 1 q 0 λ 1 q m p m+1 of prime ideals of A with p m+1 maximal. It follows that m+1 n, so m n 1, i.e., dim(s 1 x A) n 1. Supposedim(S 1 x A) n 1foreachx A.Letp 0 p m beamaximalchainofprime ideals of A. In particular, p m is a maximal ideal. For each x p m p m 1, p i S x = for 0 i <m, so this chain is of the form λ 1 q 0 λ 1 q m 1 p m for some q i S 1 x A prime. Since dim(s 1 x A) n 1, we must have m 1 n 1, i.e., m n. Theorem Let X be an irreducible affine variety. Then dim(x)=trdeg k (k(x)). 3 Proof. By Lemma 15.6, we know that trdeg k (A)=n<. We wish to prove the following: for each affine k-algebra A, if Ais an integral domain, then trdeg k (Frac(A))=dim(A). We first prove trdeg k (Frac(A)) dim(a). Actually, in order to facilitate an inductive proof, let s prove the following more general claim: if Ais an integral domain, and k A a subring with k a field, and trdeg k (Frac(A))=n<, then trdeg k (Frac(A)) dim(a). Weproceed by induction on n.if trdeg k (Frac(A))=0,then k Frac(A) is an integral ring extension. It follows that the subring A Frac(A) is also integral over k. This implies that Aisafield:if0 α A,thenα satisfiessomemonicpolynomialrelationα r +a r 1 α r 1 + +a 1 α+a 0 =0 with a i k, so 1 α = 1 a 0 (α r 1 +a r 1 α r 2 + +a 1 ) A. 3 See the appendix at the end of this section for a brief review of transcendence degree. 62

4 Thedimensionofafieldis0:theonlychainofprimeidealsinafieldis0.Thus,dim(A)=0. Wenowprovetheinductivestep.Wemayreplacek byitsalgebraicclosureinawithout affectingthetranscendencedegree.letx A.Ifx k,thenx istranscendentaloverk,since k is algebraically closed in A. Thus, trdeg k(x) (Frac(A))=n 1. Note that k(x) S 1 x Awith thenotationoflemma16.10.replacing Aby S 1 x Aandk bythealgebraicclosureofk(x)in S 1 x A, we have dim(s 1 x A) n 1 by induction. If x k, then 0=1 x 1 x S x, so S 1 x A=0, and dim(s 1 x A) n 1. By Lemma 16.10, dim(a) n. We now show trdeg k (Frac(A)) dim(a). By Lemma 15.6, we have an injective, integral ring map k[x 1,...,x n ] A. In particular, trdeg k (Frac(A))=n. By the Going-Up Theorem, each chain of prime ideals of k[x 1,...,x n ] lifts to a chain of prime ideals of A, so dim(a) dim(k[x 1,...,x n ]). Since 0 (x 1 ) (x 1,x 2 ) (x 1,...,x n ) is a chain of prime ideals of k[x 1,...,x n ], we have dim(k[x 1,...,x n ]) n. Corollary Property (D) is satisfied. Proof. It suffices to do so for X irreducible by (C). Each nonempty affine open subvariety U of X is irreducible, hence of (the same) dimension trdeg k (k(u))=trdeg k (k(x)) by Theorem By Lemma 16.5(ii), it follows that dim(x)=dim(u) for each nonempty affine open U X. If U X is a nonaffine, dense, open subvariety, then choose an affine dense open U 0 U. Then the above argument shows that dim(x)=dim(u 0 )=dim(u). Lemma If f : X Y is a surjective morphism of irreducible projective varieties, then dim(x) dim(y). Proof. Let Y 0 Y 1 Y d =Yis a maximal chain of irreducible closed subsets of Y. Ifd =0,thenX,sodim(X) 0.Weproceedbyinductionond.Letf 1 Y d 1 = r i=1 Z i be the irreducible decomposition of f 1 Y d 1. As f is surjective, Y d 1 = r i=1 f (Z i ). Since Y d 1 is irreducible, and f is closed (Theorem 11.2), Y d 1 =f (Z i ) for some i. By Exercise 16.4(i), dim(y d 1 ) = d 1. By induction, dim(z i ) dim(f (Z i )) = dim(y d 1 ) d 1. Since Z i X, and Z i and X are irreducible, it follows that dim(x)>d D Appendix: Transcendence bases Definition Let k Kbe a field extension. (i) We say that α 1,...,α n K are algebraically independent over k if the morphism kϕ : [x 1,...,x n ] Kgiven by x i α i is injective. Otherwise, the α i are algebraically dependent over k: there exists 0 f k[x 1,...,x n ] such that f (α 1,...,α n ) = 0, i.e., there exists a nonzero element f ker(ϕ). 63

5 (ii) We say that A ={α 1,...,α n } Kis a transcendence basis for K over k if A is algebraically independent over k and the morphism k(x 1,...,x n ) Kgiven by x i α i is an algebraic field extension. Lemma Letk Kbe a field extension, A {α 1,...,α n } Kandβ K. Ifβ is algebraic over the subfield k(α 1,...,α n ) Kand algebraically independent over k(α 1,...,α n 1 ), then α n is algebraic over k(α 1,...,α n 1,β). Proof. Let 0 f k[x 1,...,x n,y] such that f (α 1,...,α n,β)=0. Write f (x 1,...,x n,y)= g j (x 1,...,x n 1,y)xn j j for some g j k[x 1,...,x n 1,y]. Since f 0, we have g j0 0 for some j 0. Since β is algebraically independent over k(α 1,...,α n 1 ), g j0 (α 1,...,α n 1,β) 0. Thus, 0 f (α 1,...,α n 1,x n,β) k(α 1,...,α n 1,β)[x n ], but 0 = f (α 1,...,α n,β), so we have a nonzero algebraic dependence relation for α n over k(α 1,...,α n 1,β). Proposition Letk Kbeafieldextension,andletA={α 1,...,α n }andb={β 1,...,β m } besubsetsofk.ifaisalgebraicallyindependentoverk,andaisalgebraicoverk(β 1,...,β m ) K, then n m. Proof. Reindex Aand Bso that A B={α 1,...,α l }, and B={α 1,...,α l,β l+1,...,β m }. If n=l, then A B and n m. If not, then the hypotheses imply that α l+1 is algebraic over k(α 1,...,α l,β l+1,...,β r+1 ), but algebraically independent over k(α 1,...,α l,β l+1,...,β r ) for somel <r <m.bylemma16.15,β r+1 isalgebraicoverl k(α 1,...,α l+1,β l+1,...,ˆβ r,...,β m ). Since A is algebraic over k(β 1,...,β m ) = L[β r ], which is an algebraic extension of L, A is also algebraic over L. Replace Bby{α 1,...,α l+1,β l+1,...,ˆβ r,...,β m } and repeat until we can assume A B, in which case n m. Definition Let k K be a field extension. If K is an algebraic extension of k(α 1,...,α n ), and α 1,...,α n K are algebraically independent over k, then we say that K is of transcendence degree n over k, and write n=trdeg k (K). By Proposition 16.16, if K is an algebraic extension of k(β 1,...,β m ) with β 1,...,β m Kalgebraically independent over k, then m=n, so this definition is independent of our choice of α 1,...,α n. Example If k Kbe a field extension. (i) We have trdeg k (K)=0 if and only if K is algebraic over k. (ii) The function field K=k(x 1,...,x n )=Frac(k[x 1,...,x n ]) is of transcendence degree n over k. 64

6 (iii) Thefield K=k(x)[y]/(y 2 x)=k( x)isalgebraicoverk(x),henceoftranscendence degree 1 over k. Proposition If k K Lare field extensions, then trdeg k (L)=trdeg k (K)+trdeg K (L). Proof. Let α 1,...,α n K be algebraically indepdendent over k, let β 1,...,β m L be algebraically independent over K. Suppose K is an algebraic extension of k(α 1,...,α n ), and Lis algebraic extension of K(β 1,...,β m ). Then trdeg k (K)=n and trdeg K (L)=m. Moreover, Lis an algebraic extension of k(α 1,...,α n,β 1,...,β m ). Finally, note that α 1,...,α n,β 1,...,β m are algebraically independent overk: algebraic dependence relation either containsβ j s, which would provide an algebraic dependence relation for the β j over the K, or it provides an algebraic dependence relation for the α i over k. 65