Algebraic Geometry. Appendix A. A.1 Affine Algebraic Sets. Basic Properties

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1 Appendix A Algebraic Geometry We develop the aspects of algebraic geometry needed for the study of algebraic groups over C in this book. Although we give self-contained proofs of almost all of the results stated, we do not attempt to give an introduction to the field of algebraic geometry or to give motivating examples. We refer the interested reader to Cox, Little, and O Shea [1992], Harris [1992], Shafarevich [1994], and Zariski and Samuel [1958] for more details. A.1 Affine Algebraic Sets A.1.1 Basic Properties Let V be a finite-dimensional vector space over C. A complex-valued function f on V is a polynomial of degree k if for some basis {e 1,..., e n } of V one has f ( n i=1 x ) ie i = I k a I x I Here for a multi-index I = (i 1,..., i n ) N n we write x I = x i1 1 xin n. This definition is obviously independent of the choice of basis for V. If there exists a multi-index I with I = k and a I 0, then we say that f has degree k. Let P(V ) be the set of all polynomials on V, P k (V ) the polynomials of degree k, and P k (V ) the polynomials of degree k. Then P(V ) is a commutative algebra, relative to pointwise multiplication of functions. It is freely generated as an algebra by the linear coordinate functions x 1,..., x n. A choice of a basis for V thus gives rise to an algebra isomorphism P(V ) = C[x 1,..., x n ], the polynomial ring in n variables. Definition A.1.1. A subset X V is an affine algebraic set if there exist functions f j P(V ) such that X = {v V : f j (v) = 0 for j = 1,..., m}. When X is an affine algebraic set, we define the affine ring O[X] of X to be the functions on X that are restrictions of polynomials on V : O[X] = {f X : f P(V )}. 605

2 606 APPENDIX A. ALGEBRAIC GEOMETRY We call these functions the regular functions on X. Define I X = {f P(V ) : f X = 0}. Then I X is an ideal in P(V ), and O[X] = P(V )/I X as a commutative algebra. Theorem A.1.2 (Hilbert basis theorem). Let I P(V ) be an ideal. Then there is a finite set of polynomials {f 1,..., f d } I so that every g I can be written as g = g 1 f g d f d with g i P(V ). Proof. Let dimv = n. Then P(V ) = C[x 1,..., x n ]. Since a polynomial in x 1,..., x n with coefficients in C can be written uniquely as a polynomial in x n with coefficients that are polynomials in x 1,..., x n 1, there is a ring isomorphism P(V ) = R[x n ], with R = C[x 1,..., x n 1 ]. (A.1) We call an arbitrary commutative ring R Noetherian if every ideal in R is finitely generated. For example, the field C is Noetherian since its only ideals are {0} and C. To prove the theorem, we see from (A.1) that it suffices to prove the following: (N) If a ring R is Noetherian, then the polynomial ring R[x] is Noetherian We first show that the Noetherian property for a ring R is equivalent to the ascending chain condition for ideals in R: (ACC) If I 1 I 2 R is an ascending chain of ideals in R, then there exists an index p so that I j = I p for all j p. Indeed, suppose R satisfies (ACC). Given an ideal I R, we take f 1 I and set I 1 = Rf 1. If I 1 I, we take f 2 I with f 2 I 1 and set I 2 = Rf 1 + Rf 2. Then I 1 I 2 I. Continuing in this way, we obtain an ascending chain I 1, I 2,... of finitely generated ideals in I. By (ACC) there is a finite n with I n = I, and hence I is finitely generated. Conversely, suppose R is Noetherian. Given an ascending chain I 1 I 2... of ideals in R, set I = i I i. Then I is an ideal in R and hence has a finite set of generators f 1,..., f d by the Noetherian property. But there exists an index p so that f j I p for all j. Thus I = I p, so I j = I p for all j p. Hence R satisfies (ACC). We now prove (N). Assume that R is a Noetherian ring and let I R[x] be an ideal. Choose a nonzero polynomial f 1 (x) I of minimum degree and form the ideal I 1 = R[x]f 1 (x) I. If I 1 = I, we are done. If I 1 I, take f 2 (x) as a nonzero polynomial of minimum degree among all elements of I \ I 1 and set I 2 = R[x]f 1 (x)+r[x]f 2 (x) I. If I = I 2 we are done. Otherwise, we continue this process of choosing f j (x) and forming the ideals I j = R[x]f 1 (x)+ +R[x]f j (x). As long as I j I, we can choose f j+1 (x) as a nonzero polynomial of minimum degree among all elements of I \ I j. If I d = I for some d, then f 1 (x),..., f d (x) is a finite set of generators for I and we are done. We assume, for the sake of obtaining a contradiction, that I j I for all j.

3 A.1. AFFINE ALGEBRAIC SETS 607 If f(x) = c m x m + + c 1 x + c 0 R[x] and c m 0, we call c m x m the initial term and c m the initial coefficient of f(x). Let a j R be the initial coefficient of f j (x) and let J R be the ideal generated by the set {a 1, a 2,...} of all initial coefficients. Since R is Noetherian, there is an integer m so that J is generated by a 1,..., a m. In particular, there are elements u j R so that a m+1 = m j=1 u ja j. Let d j be the degree of f j (x). By the choice of f j (x) we have d j d k for all j < k, since f k (x) I \ I j 1 and f j (x) has minimum degree among all nonzero elements in I \ I j 1. Consider g(x) = m j=1 u jf j (x)x dm+1 dj. The initial coefficient of g(x) is a m+1 and g(x) I m. Hence the polynomial h(x) = f m+1 (x) g(x) has strictly smaller degree than f m+1 (x). Also, h(x) 0 since f m+1 (x) I m. But such a polynomial h(x) cannot exist, by the choice of f m+1 (x). This contradiction proves (N). Suppose A C[x 1,..., x n ] is any collection of polynomials. Let X = {x C n : f(x) = 0 for all f A}, and let I X C[x 1,..., x n ] be the set of all polynomials that vanish on X. Then A I X and I X is an ideal. By the Hilbert basis theorem there are polynomials f 1,..., f d that generate I X. Hence X = {x C n : f j (x) = 0 for j = 1,..., d}. Thus it is no loss of generality to require that algebraic sets be defined by a finite number of polynomial equations. Let a X. Then m a = {f O[X] : f(a) = 0} is a maximal ideal in O[X], since f f(a) m a for all f O[X] and hence dimo[x]/m a = 1. We will show that every maximal ideal is of this form. This basic result links algebraic properties of the ring O[X] to geometric properties of the algebraic set X. Theorem A.1.3 (Hilbert Nullstellensatz). Let X be an affine algebraic set. If m is a maximal ideal in O[X], then there is a unique point a X such that m = m a. Proof. Consider first the case X = C n. Define a representation ρ of the algebra A = C[z 1,..., z n ] on the vector space V = C[z 1,..., z n ]/m by multiplication: ρ(f)(g + m) = fg + m for f, g A. The representation (ρ, V ) is irreducible since m is a maximal ideal. Since A is commutative, we have A End A (V ). Hence Schur s lemma (Lemma 4.1.4) implies that ρ(f) is a scalar multiple of the identity. In particular, there exist a i C such that z i a i m for i = 1,..., n. Set a = (a 1,..., a n ) C n. Then it follows that f f(a) m for all f C[z 1,..., z n ]. Hence m = m a.

4 608 APPENDIX A. ALGEBRAIC GEOMETRY Now consider an arbitrary algebraic set X C n and maximal ideal m O[X]. Let m be the inverse image of m under the canonical restriction map f f X from C[z 1,..., z n ] onto O[X]. Then the quotient rings O[X]/m and C[z 1,..., z n ]/m are naturally isomorphic. Since m is a maximal ideal, these rings are fields, and hence m is a maximal ideal in C[z 1,..., z n ]. By the result just proved there exists a C n so that f f(a) m for all f C[z 1,..., z n ]. Hence f X f(a) m for all f. If f I X, then f X = 0, and so we have f(a) m. But m is a proper ideal; thus we conclude that f(a) = 0 for all f I X. Therefore a X and m = m a. If A is an algebra with 1 over C, then Hom(A, C) is the set of all linear maps ϕ : A C such that ϕ(1) = 1 and ϕ(a a ) = ϕ(a )ϕ(a ) for all a, a A (the multiplicative linear functionals on A). When X is an affine algebraic set and A = O[X], then every x X defines a homomorphism ϕ x by evaluation: ϕ x (f) = f(x) for f O[X]. The coordinate functions from an ambient affine space separate the points of X, so the map x ϕ x is injective. Corollary A.1.4. Let X be an affine algebraic set, and let A = O[X]. The map x ϕ x is a bijection between X and Hom(A, C). Proof. Let ϕ Hom(A, C). Then Ker(ϕ) is a maximal ideal of A, so by Theorem A.1.3 there exists x X with Ker(ϕ) = Ker(ϕ x ). Hence 0 = ϕ(f f(x)) = ϕ(f) f(x) for all f A, since ϕ(1) = 1. Thus ϕ = ϕ x. Corollary A.1.5. Let f 1 (x),..., f s (x) and f(x) be polynomials on C n. Assume that f(x) = 0 whenever f i (x) = 0 for i = 1,..., s. Then there exists an integer r and polynomials g i (x) such that s f(x) r = g i (x)f i (x). i=1 (A.2) Proof. Let x = (x 1,..., x n ) C n. Introduce a new variable x 0 and the polynomial f 0 (x 0, x) = 1 x 0 f(x). Take (x 0, x) as coordinates on C n+1, and view f 1,..., f s as polynomials on C n+1 depending only on x. The assumption on {f i } and f implies that {f 0, f 1,..., f s } have no common zeros on C n+1. Hence by Theorem A.1.3 the ideal generated by these functions contains 1. Thus there are polynomials h i (x 0, x) on C n+1 such that 1 = h 0 (x 0, x)(1 x 0 f(x)) + s h i (x 0, x)f i (x). i=1 (A.3) Let r be the maximum degree of x 0 in the polynomialsh i. Substitutingx 0 = 1/f(x) in Equation (A.3) and multiplying by f(x) r, we obtain Equation (A.2), with g i (x) = f(x) r h i (1/f(x), x).

5 A.1. AFFINE ALGEBRAIC SETS 609 A.1.2 Zariski Topology Let V be a finite-dimensional vector space over C and X V an algebraic subset. Definition A.1.6. A subset Y X is Zariski-closed in X if Y is an algebraic subset of V. If 0 f O[X], then the principal open subset of X defined by f is X f = {x X : f(x) 0}. Lemma A.1.7. The Zariski-closed sets of X give X the structure of a topological space. The finite unions of principal open sets X f, for 0 f O[X], are the nonempty open sets in this topology (the Zariski topology). Proof. We must check that finite unions and arbitrary intersections of algebraic sets are still algebraic. Suppose Y 1 is the zero set of polynomials f 1,..., f r and Y 2 is the zero set of polynomials g 1,..., g s. Then Y 1 Y 2 is the zero set of the functions {f i g j : 1 i r, 1 j s} and is thus algebraic. Given an arbitrary family {Y α : α I} of algebraic sets, their intersection is the zero set of a (possibly infinite) collection of polynomials {f 1, f 2,...}. This intersection is still an algebraic set, however, by the Hilbert basis theorem. By definition, the complement of a proper algebraic subset of X is the union of finitely many sets of the form X f. Unless otherwise stated, we will use the term closed set in this appendix to refer to a Zariski-closed set. Let x 1,..., x n be linear coordinate functions on V determined by a basis for V. Notice that a point a = (a 1,..., a n ) V is a closed set, since it is the zero set of the translated coordinate functions {x i a i : 1 i n}. If X V is any set then X will denote the closure of X in the Zariski topology (the smallest closed set containing X). Let V and W be finite-dimensional complex vector spaces. Suppose X V and Y W are algebraic sets and f : X Y. If g is a complex-valued function on Y define f (g) to be the function f (g)(x) = g(f(x)) for x X. We say that f is a regular map if f (g) is in O[X] for all g O[Y ]. In terms of linear coordinates x 1,..., x m for V and y 1,..., y n for W, a regular map is given by the restriction to X of n polynomial functions y i = f i (x 1,..., x m ), for i = 1,..., n. In particular, when Y = C this notion of regular map is consistent with our previous definition. It is clear from the definition that the composition of regular maps is regular. Lemma A.1.8. A regular map f between algebraic sets is continuous in the Zariski topology. Proof. Let Z Y be a closed set, defined by a set of polynomials {g j }, say. Then f 1 (Z) is the zero set of the polynomials {f g j }, and hence is closed. A.1.3 Products of Affine Sets Let V, W be vector spaces and let X V, Y W be affine algebraic sets. Then X Y is an affine algebraic set in the vector space V W. To see this, let

6 610 APPENDIX A. ALGEBRAIC GEOMETRY f 1,..., f m P(V ) be defining functions for X, and let g 1,..., g n P(W) be defining functions for Y. Extend f i and g i to polynomials on V W by setting f i (v, w) = f i (v) and g i (v, w) = g i (w). Then {f 1,..., f m, g 1,..., g n } is a set of defining functions for X Y. By the universal property of tensor products relative to bilinear maps, there is a unique linear map µ : P(V ) P(W) P(V W) such that µ(f f )(v, w) = f (v)f (w). This map clearly preserves multiplication of functions. Lemma A.1.9. The map µ induces an isomorphism of commutative algebras ν : O[X] O[Y ] O[X Y ]. Proof. Since µ is a vector-space isomorphism (see Proposition C.1.4) and the functions in O[X Y ] are the restrictions of polynomials on V W, it is clear that ν is surjective and preserves multiplication. We will show that ν is injective. Suppose 0 f = i f i f i O[X] O[Y ]. Here we may assume that {f i } is linearly independent and f 1 0. Choose g i P(V ) and g i P(W) with f i = g i X and f i = g i Y and set g(v, w) = i g i (v)g i (w). Then ν(f) = g X Y. Choose x 0 X such that f 1 (x 0) 0. Then by the linear independence of the functions {f i }, we have i f i (x 0)f i 0. Hence the function y g(x 0, y) on Y is nonzero, proving that ν(f) 0. A.1.4 Principal Open Sets Assume X is a Zariski-closed subset of a vector space V. Let f O[X] with f 0. Define ψ : X f V C by ψ(x) = (x, 1/f(x)). This map is injective, and we use it to define the structure of an affine algebraic set on the principal open set X f as follows: Assume X is defined by f 1,..., f n P(V ). Choose f P(V ) so that f X = f. Then ψ(x f ) = {(v, t) V C : f i (v) = 0 for all i and f(v)t 1 = 0}. Thus ψ(x f ) is an algebraic set. We define the ring of regular functions on X f by pulling back the regular functions on ψ(x f ): O[X f ] = {ψ (g) : g P(V C)}. On ψ(x f ) the coordinate t has the same restriction as 1/ f. Hence we see that the regular functions on X f are all of the form g(x 1,..., x n, 1/ f) with g(x 1,..., x n, t) a polynomial in n +1 variables. In particular, 1/f is a regular function on X f. Note that 1/f is not the restriction to X f of a polynomial on V unless f is constant.

7 A.1. AFFINE ALGEBRAIC SETS 611 A.1.5 Irreducible Components Let X V be a nonempty closed set. We say that X is reducible if there are nonempty closed subsets X i X, i = 1, 2 such that X = X 1 X 2. We say that X is irreducible if it is not reducible. Lemma A An algebraic set X is irreducible if and only if I X is a prime ideal (O[X] has no zero divisors). Proof. Suppose X is reducible. There are polynomials f i I Xi such that f 1 does not vanish on X 2 and f 2 does not vanish on X 1. Hence f i / I X but f 1 f 2 I X. Thus I X is not a prime ideal. Conversely, if I X is not prime, then there exist f 1 and f 2 in O[X] with f 1 f 2 vanishing on X but f i not vanishing on X. Set X i = {x X : f i (x) = 0}. Then X = X 1 X 2 and X X i. Hence X is reducible. We shall have frequent use for the following density property of irreducible algebraic sets: Lemma A Let X be an irreducible algebraic set. Every nonempty open subset Y of X is dense in X. Furthermore, if Y X and Z X are nonempty open subsets, then Y Z is nonempty. Proof. By assumption, X \Y is a proper closed subset of X. Since X = (X \Y ) Y, the irreducibility of X implies that Y = X, where Y is the closure of Y (in X) in the Zariski topology. Let Y and Z be nonempty open subsets. If Y Z =, then X = Y c Z c and Y c X, Z c X. This contradicts the irreducibility of X. Lemma A If X is any algebraic set, then there exists a finite collection of irreducible closed sets X i such that X = X 1 X r and X i X j for i j. (A.4) Furthermore, such a decomposition (A.4) is unique up to a permutation of the indices and is called an incontractible decomposition of X. The sets X i are called the irreducible components of X. Proof. Suppose the lemma is false for some X. Then there must be a decomposition X = X 1 X 1 into closed sets with X X 1 properly and the lemma false for X 1. Continuing, we get an infinite strictly decreasing chain X X 1 X 2 of closed subsets. But this gives an infinite strictly increasing chain I X I X1 of ideals, which contradicts the ascending chain condition for ideals in P(V ). By a deletion process, any decomposition of X as a finite union of closed subsets can be written in the form (A.4) with no proper containments among the X i. Suppose X = X 1 X s is another incontractible decomposition. Then for each index i X i = X X i = r j=1 X j X i.

8 612 APPENDIX A. ALGEBRAIC GEOMETRY Since X i is irreducible, this decomposition has only one nonempty intersection. Thus there exists an index j such that X i X j. Similarly, for each index j there exists an index k such that X j X k. Hence i = k and X i = X j. So r = s and there is a permutation σ such that X i = X σ(i). Lemma A If X is an irreducible algebraic set then so is X f for any nonzero function f O[X]. Proof. Let X be a Zariski-closed subset of C n. Suppose u, v O[X f ] with u 0 and uv = 0. There are polynomials g, h in n + 1 variables such that u(x) = g(x, 1/f(x)) and v(x) = h(x, 1/f(x)), for x = (x 1,..., x n ) X f. Hence there is an integer k sufficiently large so that f k u and f k v are the restrictions to X f of polynomials ũ and ṽ, respectively. But f(x)ũ(x)ṽ(x) = f(x) 2k+1 u(x)v(x) = 0 for x X f, and obviously f(x)ũ(x)ṽ(x) = 0 if f(x) = 0. Since O[X] is an integral domain and ũ 0, we conclude that ṽ = 0. Thus if x X f, then v(x) = ṽ(x)/f(x) k = 0. This proves that O[X f ] is an integral domain. Lemma A Let V and W be finite-dimensional vector spaces. Suppose X V and Y W are irreducible algebraic sets. Then X Y is an irreducible algebraic set in V W. Proof. We already verified that X Y is algebraic. Suppose there are closed subsets Z 1, Z 2 in V W such that X Y = Z 1 Z 2. For each x X the set {x} Y is irreducible, since any decomposition of it into proper closed sets would give a decomposition of Y into proper closed sets. Hence for each x X either {x} Y Z 1 or else {x} Y Z 2. This induces a decomposition X = X 1 X 2, where X i = {x X : {x} Y Z i }. We claim that each subset X i is closed. Indeed, let Z i be the zero set of the functions {f α } P(V W) and define for y Y. Then X (i) y X (i) y = {x X : f α (x, y) = 0 for all α} is closed in X and X i = y Y X(i) y, which shows that X i is closed. From the irreducibility of X it now follows that either X = X 1 or else X = X 2. Hence either X Y = Z 1 or else X Y = Z 2. Lemma A Suppose f : X Y is a regular map between affine algebraic sets. Suppose X is irreducible. Then f(x) is irreducible. Proof. Suppose g, h O[Y ] and g(f(x))h(f(x)) = 0 for all x X. Then since O[X] is an integral domain, either f g = 0 or f h = 0. Hence either g or h vanishes on f(x). This proves that O[f(X)] is an integral domain, and hence f(x) is irreducible.

9 A.1. AFFINE ALGEBRAIC SETS 613 A.1.6 Transcendence Degree and Dimension Let X be an irreducible affine algebraic set. The algebra O[X] is finitely generated over C and has no zero divisors. The following result (the Noether Normalization Lemma) describes the structure of such algebras: Lemma A Let k be a field and B = k[x 1,..., x n ] a finitely generated commutative algebra over k without zero divisors. Then there exist y 1,..., y r B such that (1) the set {y 1,..., y r } is algebraically independent over k ; (2) every b B is integral over the subring k[y 1,..., y r ]. The integer r is uniquely determined by properties (1) and (2) and is called the transcendence degree of B over k. A set {y 1,..., y r } with properties (1) and (2) is called a transcendence basis for B over k. Proof. If {x 1,..., x n } is algebraically independent over k, then we can set y i = x i. Otherwise, there is a nontrivial relation J a J x J = 0, (A.5) By relabeling the generators x i if necessary, we may assume that x 1 occurs in (A.5) with a nonzero coefficient. Fix an integer d with d > j k for all J = (j 1,..., j n ) such that a J 0. Set M = (1, d, d 2,..., d n 1 ) N n and introduce the variables y 2 = x 2 x d 1, Then the monomials x J become y 3 = x 3 x d2 1,..., y n = x n x dn 1 1. (A.6) x J = x J M 1 + k<j M f k (y 2,..., y n )x k 1, where J M = j 1 + j 2 d + + j n d n 1 and f k is a polynomial. By the choice of d we see that distinct indices J with a J 0 give distinct values of J M. Thus relation (A.5) can be written as J a J x J M 1 + f(x 1, y 2,..., y n ) = 0, (A.7) with f a polynomial whose degree in x 1 is less than the degree of the first summation in (A.7). Dividing (A.7) by the coefficient a J 0 for which J M is the largest, we see that x 1 satisfies a monic polynomial with coefficients in k[y 2,..., y n ]. Hence by (A.6) x 2,..., x n also satisfy monic polynomials with coefficients in k[y 2,..., y n ]. Thus B is integral over k[y 2,..., y n ]. If {y 2,..., y n } is algebraically independent, then we are done. Otherwise, we repeat the procedure above until we arrive at an algebraically independent set. The number r is the cardinality of any maximal algebraically independent set in B (see Lang [1971, Chap. X, 1, Theorem 1]); hence it is uniquely determined by B.

10 614 APPENDIX A. ALGEBRAIC GEOMETRY Definition A Let X be an affine algebraic set. When X is irreducible, dim X is the transcendence degree of the algebra O[X]. If X is reducible, then dimx is the maximum of the dimensions of the irreducible components of X. With this notion of dimension available, we can obtain a very useful ascending chain property for algebraic sets. Theorem A (1) Let M, N be irreducible affine algebraic sets, such that M N. Then dimm dimn. Furthermore, if dimm = dimn then M = N. (2) Let X 1 X 2 be an increasing chain of irreducible affine algebraic subsets of an algebraic set X. Then there exists an index p so that X j = X p for j p. Proof. (1): The Noether normalization lemma implies that dim M dim N. Suppose that the dimensions are equal. To show M = N, it suffices to show that the restriction homomorphism σ : O[N] O[M] is injective (cf. Corollary A.1.4). Suppose u Ker(σ). Take a transcendence basis S = {f 1,..., f r } for O[M], where r = dimm. Since M, the map σ is surjective. Thus there exist f i O[N] such that σ( f i ) = f i for i = 1,..., r. Clearly, S = { f 1,..., f r } is algebraically independent. Since dimn = dimm = r, it follows that S is a transcendence basis for O[N]. Thus by the Noether normalization lemma there are b i O[ f 1,..., f r ] such that u n + b n 1 u n b 1 u + b 0 = 0. Choose n minimal. Applying the homomorphism σ, we get σ(b 0 ) = 0. But this implies that b 0 = 0, by the algebraic independence of S. Thus n = 1 and u = 0. (2): By (1) we have dimx j dimx j+1 dimx. Hence there exists an index p such that dimx j = dimx p for j p. Then X j = X p for j p by the last part of (1). A.1.7 Exercises 1. Prove that the ring of polynomials k[x] in one variable is integrally closed in its quotient field k(x) for any coefficient field k. (HINT: Use the unique factorization property of k[x].) 2. Let f : C C be a bijective map that is not a polynomial. Show that f is continuous in the Zariski topology but that it is not a regular map. 3. For any subset X C n let I X be the ideal of all polynomials vanishing on X. For any ideal J P(C n ) let V (J) = {x C n : f(x) = 0 for all f J} be the zero set of J. (a) Show that the Zariski closure of X is V (I X ). (b) If I, J are ideals in P(C n ), show that V (IJ) = V (I J).

11 A.2. MAPS OF ALGEBRAIC SETS Suppose f is an irreducible polynomial (this means that f cannot be factored as a product of polynomials of strictly smaller degree). Show that the zero set of f is irreducible. 5. Let f i (t) = g i (t)/h i (t) be rational functions on C n for i = 1,..., r. Let X C r be the Zariski closure of the set {[f 1 (t),..., f r (t)] : t C n and h 1 (t) h r (t) 0}. Prove that V is irreducible. (HINT: Show that the ideal I X is prime.) 6. Let X = {(x 1, x 2 ) C 2 : x 1 x 2 = 0}. (a) Show that the irreducible components of X are X 1 = {(z, 0) : z C} and X 2 = {(0, z) : z C 2 } and hence that X is reducible. (b) Show that X is connected (in the Zariski topology). (HINT: Suppose X = U V with U and V Zariski-open, U V =, and U. Use (a) to argue that U X i = X i for i = 1, 2 and hence V =.) 7. (a) Show that the Zariski-closed subsets of C are precisely the finite sets. (b) Show that the Zariski topology on C C is not the product of the Zariski topologies on each factor. A.2 Maps of Algebraic Sets A.2.1 Rational Maps Let A be a commutative ring with 1 and without zero divisors. Then A is embedded in its quotient field Quot(A). The elements of this field are the formal expressions f = g/h, where g, h A and h 0, with the usual algebraic operations on fractions. For example, let X be an irreducible algebraic set. The algebra A = O[X] has no zero divisors, so it has a quotient field. We denote this field by Rat(X) and call it the field of rational functions on X. We may view the elements of Rat(X) as functions, as follows: If f Rat(X), then we say that f is defined at a point x X if there exist g, h O[X] with f = g/h and h(x) 0. In this case we set f(x) = g(x)/h(x). The domain D f of f is the subset of X at which f is defined. It is a dense open subset of X, since it contains the principal open set X h. A map f from X to an algebraic set Y is called rational if f (ϕ) is a rational function on X for all ϕ O[Y ]. Suppose Y C n and y i is the restriction to Y of the linear coordinate function x i on C n. Set f i = f (y i ). Then f is rational if and only if f i Rat(X) for i = 1,..., n. The domain of a rational map f is defined as D f = D f (ϕ). ϕ O[Y ] By Lemma A.1.11 we know that D f = n i=1 D f (y i) is a dense open subset of X.

12 616 APPENDIX A. ALGEBRAIC GEOMETRY Lemma A.2.1. Suppose X is irreducible and f : X D f = X then f is a regular map. Y is a rational map. If Proof. Let ϕ O[Y ]. Let I O[X] be the set of all functions h such that the function x h(x)ϕ(f(x)) is regular on X. Then I is a nonzero ideal in O[X], since f is a rational map. Suppose I is a proper ideal. Then it is contained in some maximal ideal. In this case Theorem A.1.3 implies that all h I vanish at some point x 0 X. Thus x 0 / D f, a contradiction. Hence 1 I and f (ϕ) O[X] for all ϕ O[Y ]. A.2.2 Extensions of Homomorphisms The points of an affine algebraic set X correspond to the homomorphisms of the algebra O[X] into the base field C. We will use this correspondence to obtain the key results concerning regular maps. The translation from geometric to algebraic language leads us to two main problems: When do homomorphisms of an algebra extend to homomorphisms of a larger algebra? If an extension exists, when is it unique? We shall solve these algebraic problems in this section and give the geometric applications in later sections. We will need some properties of ring extensions. Let A B be commutative rings with 1, and assume that B has no zero divisors. An element b B is said to be integral over A if b satisfies a monic polynomial b n + a 1 b n a n 1 b + a n = 0 (A.8) with coefficients a i A. Lemma A.2.2. An element b B is integral over A if and only if there exists a finitely generated A-submodule C B such that b C C. Proof. Let b satisfy (A.8). Then A[b] = A 1 + A b + + A b n 1 is a finitelygenerated A-submodule, so we may take C = A[b]. Conversely, suppose C exists as stated and is generated by nonzero elements {x 1,..., x n } as an A-module. Since bx i C, there are elements a ij A so that bx i n j=1 a ij x j = 0 for i = 1,..., n. Since B has no zero divisors, this system of equations implies that det[bδ ij a ij ] = 0. This determinant is a monic polynomial in b, with coefficients in A. Hence b is integral over A. Let A be an algebra with 1 over C. Given 0 a A, we set Hom(A, C) a = {ϕ Hom(A, C) : ϕ(a) 0}. For example, if A = O[X] for some affine algebraic set X, then Hom(A, C) is naturally identified with X, by Corollary A.1.4, and Hom(A, C) a corresponds to the principal open set X a.

13 A.2. MAPS OF ALGEBRAIC SETS 617 Theorem A.2.3. Let B be a commutative algebra over C with 1 and no zero divisors. Suppose that A B is a subalgebra such that B = A[b 1,..., b n ] for some elements b i B. Then given 0 b B, there exists 0 a A such that every ϕ Hom(A, C) a extends to ψ Hom(B, C) b. If B is integral over A and b = 1, then one may take a = 1. Proof. We start with the case B = A[u] for some element u B. Let b = f(u), where f(x) = a n X n + + a 0 with a i A. For g(x) A[X], denote by g ϕ (X) O[X] the polynomial obtained by applying ϕ to the coefficients of g(x). If u is transcendental over A, then for any λ C we can define an extension ψ of ϕ by ψ(g(u)) = g ϕ (λ) for g A[X]. (A.9) This extension satisfies ψ(b) = f ϕ (λ). Since b 0, the polynomial f(x) 0. Take for a A any nonzero coefficient of f(x). If we assume that ϕ(a) 0, then f ϕ (X) 0 and any λ C with f ϕ (λ) 0 will serve to define the desired extension of ϕ. Now assume that u is algebraic over A. Then b is also algebraic over A, so there are nonzero polynomials p(x) = a m X m + + a 0 and q(x) = c n X n + + c 0 (where a i and c i in A), whose degrees are minimal and which satisfy p(u) = 0 and q(b) = 0. We have a m 0 and c 0 0 (otherwise, the polynomial q(x)/x would annihilate b). We will prove that the element a = a m c 0 has the desired property. Note that if u is integral over A and b = 1 then a = 1. We first observe that for any g(x) A[X] there is an integer d 0 and polynomials h(x), r(x) A[X] that satisfy deg r(x) < m and (a m ) d g(x) = p(x)h(x) + r(x) (by the Euclidean division algorithm). In particular, if g(u) = 0 then r(x) = 0, since r(u) = 0 and deg r(x) < m. Thus (a m ) d g(x) is divisible by p(x) in A[X]. Suppose ϕ Hom(A, C) a and λ is a root of p ϕ (X). We have ϕ(a m ) d g ϕ (λ) = 0. But ϕ(a m ) = ϕ(a)/ϕ(c 0 ) 0. Hence g ϕ (λ) = 0, and so formula (A.9) determines an extension ψ Hom(B, C) of ϕ. For this extension 0 = ψ(q(b)) = q ϕ (ψ(b)). However, 0 is not a root of q ϕ (X) since ϕ(c 0 ) 0. Thus ψ(b) 0, and the theorem is proved when B has a single generator over A. Let n 1. We assume the theorem is true for all algebras A and all algebras B with n 1 generators over A. We shall show that it is true for the case of n generators. Let B = A[b 1,..., b n ] and let 0 b B be given. Set à = A[b 1,..., b n 1 ] and u = b n. Then B = Ã[u]. By the proof just given there exists ã à so that every complex homomorphism ϕ of à satisfying ϕ(ã) 0 extends to a complex homomorphism ψ of B with ψ(b) 0. Now we invoke the induction hypothesis, with B replaced by à and b replaced by ã. This gives a nonzero element a A so that every complex homomorphism ϕ of A satisfying ϕ(a) 0 extends to a

14 618 APPENDIX A. ALGEBRAIC GEOMETRY complex homomorphism ϕ of à with ϕ(ã) 0. We complete the induction step by combining these two extension processes. Corollary A.2.4. Let B be a finitely generated commutative algebra over C having no zero divisors. Given 0 b B, there exists ψ Hom(B, C) such that ψ(b) 0. Proof. In Theorem A.2.3 take A = C and ϕ(λ) = λ for λ C. Next we consider the uniqueness of the extensions in Theorem A.2.3. Let A B be a subalgebra, and identify Quot(A) with the subfield of Quot(B) generated by A. For example, if A = O[X] for an irreducible variety X, and B = O[X f ] for some nonzero f A, then B = A[b] Quot(A), where b = 1/f. In this example, every ψ Hom(B, C) such that ψ(b) 0 is given by evaluation at a point x X f, and hence ψ is uniquely determined by its restriction to A. The unique restriction property in this example characterizes the general case in which B Quot(A), as follows: Theorem A.2.5. Let B be a finitely generated algebra over C with no zero divisors. Let A B be a finitely generated subalgebra. Assume that there exists a nonzero element b B so that every element of Hom(B, C) b is uniquely determined by its restriction to A. Then B Quot(A). Proof. As in Theorem A.2.3, it suffices to consider the case B = A[u] for some element u B. Let b = f(u), where f(x) = a n X n + + a 0 with a i A. We first show that if u is not algebraic over A, then the unique extension property does not hold. Indeed, in this case every λ C determines an extension ψ Hom(B, C) of ϕ Hom(A, C) by (A.9). The condition ψ(b) 0 implies that f ϕ (X) 0, so there exist infinitely many choices of λ such that ψ(b) 0, as in the proof of Theorem A.2.3. Hence ψ is not uniquely determined by ϕ. Now assume that u is algebraic over A. Let p(x) and q(x) be as in the proof of Theorem A.2.3. We shall prove that m = deg(p) = 1. This will imply the theorem, since then u = a 0 /a 1. Suppose to the contrary that m > 1. Set K = Quot(A) and define ( r(x) = a m X + a ) m m 2 m 1 = a m X m + a m 1X m 1 + d j X j ma m in K[X]. Then p(x) r(x), since p(x) is irreducible over A. Hence there exists some j 0 m 2 such that a j0 d j0. Set a = (a m ) m 1 (a j0 d j0 )c 0. Then 0 a A, and so by Corollary A.2.4 there exists ϕ Hom(A, C) such that ϕ(a) 0. Set α = ϕ(a m ) 0 and β = ϕ(a m 1 )/(mα). Then r ϕ (X) = α(x + β) m, while the condition ϕ(a j0 ) ϕ(d j0 ) implies that j=0 p ϕ (X) α(x + β) m. (A.10)

15 A.2. MAPS OF ALGEBRAIC SETS 619 We claim that if m 2, then p ϕ (X) must have at least two distinct roots. For if not, then p ϕ (X) would be the mth power of a linear polynomial and hence would have to be α(x + β) m, contradicting (A.10). As in the proof of Theorem A.2.3, each root λ of p ϕ (X) determines an extension ψ Hom(B, C) of ϕ by (A.9). In particular, ψ(u) = λ, so distinct roots of p ϕ (X) determine distinct extensions of ϕ. Also, 0 = ψ(q(b)) = q ϕ (ψ(b)). But 0 is not a root of q ϕ (X), since ϕ(c 0 ) 0. Thus ψ(b) 0. So we conclude that if m 2, then there exists ψ 1, ψ 2 Hom(B, C) with ψ 1 (u) ψ 2 (u), ψ i (b) 0, but ψ 1 = ψ 2 on A, contradicting the unique restriction assumption. A.2.3 Image of a Dominant Map Let X and Y be affine algebraic sets and let f : X Y be a regular map. Definition A.2.6. The map f is dominant if f(x) is dense in Y. The definition of dominance is equivalent to the property that f : O[Y ] O[X] is injective. Dominant maps have the following remarkable property (which does not hold for smooth maps of differentiable manifolds): Theorem A.2.7. Assume that X and Y are irreducible affine algebraic sets and f : X Y is a dominant map. Let M X be a nonempty open set. Then f(m) contains a nonempty open subset of Y. Proof. Set B = O[X] and A = f (O[Y ]). Since X is irreducible, it follows that B has no zero divisors. We may assume that M = X b for some 0 b B. Let a A be as in Theorem A.2.3, and let a = f (ā), where ā O[Y ]. We claim that f(x b ) Y ā. (A.11) To prove this, view the points of Y as the homomorphisms from A to C. For y Y ā, the corresponding homomorphism ϕ satisfies ϕ(a) = ā(y) 0. Hence by Theorem A.2.3, there is an extension ψ of ϕ to B such that ψ(b) 0. In geometric language, this means that ψ is given by evaluation at a point x X b. The extension property means that g(f(x)) = g(y) for all g O[Y ]. Since f is dominant, this implies that f(x) = y, proving (A.11). Theorem A.2.8. Let f : X Y be a regular map between affine algebraic sets. Then f(x) contains an open subset of f(x). Proof. Let X 1,..., X r be the irreducible components of X. Then f(x) = f(x 1 ) f(x r ). The theorem now follows from Theorem A.2.7.

16 620 APPENDIX A. ALGEBRAIC GEOMETRY A.2.4 Factorization of a Regular Map Let M, N, and P be irreducible affine varieties. Suppose we have regular maps f : M N and h : M P. If there is a map g that satisfies the commutative diagram N f M h P g (A.12) then h is constant on the fibers of f. Indeed, f(m) = f(m ) implies that h(m) = g(f(m)) = h(m ). Furthermore, if f is surjective, then any such map g is uniquely determined by f and h. We now weaken the fiber and surjectivity conditions with the aim of obtaining a rational map g that satifies the diagram (A.12) in the sense of rational maps. Theorem A.2.9. Assume that f and h are dominant and that there is a nonempty open subset U of M so that f(m) = f(m ) implies h(m) = h(m ) for m, m U. Then there exists a rational map g : N P such that h = g f. Proof. Consider first the case in which M = P and h is the identity map. We may assume that U = M b for some 0 b O[M]. We claim that the conditions of Theorem A.2.5 are satisfied by A = f (O[N]) and B = O[M]. Indeed, every homomorphism ψ : B C such that ψ(b) 0 is given by evaluation at some x M b (cf. Corollary A.1.4). Hence ψ(f p) = p(f(x)) for p O[N]. Since f(m) is dense in N and f is injective on M b, it follows that x is uniquely determined by the homomorphism p p(f(x)). Let x i, for i = 1,..., k, be the coordinate functions defined by some embedding M C k. Applying Theorem A.2.5, we obtain q i Rat(N) so that f (q i ) = x i where defined. The rational map g defined by g(n) = (q 1 (n),..., q k (n)) for n N then satisfies g f(m) = m for m in a dense subset of M, as required. We now reduce the general case to the one just treated. Let F : M N P with F(m) = (f(m), h(m)), and take the projection maps π 1 : N P N and π 2 : N P P. Let L = F(M) and let p i = π i L for i = 1, 2. Then we have the commutative diagram F M L p 1 f N The set L is irreducible and p 1 is a dominant map that is injective on an open set. Hence there is a rational map k : N L such that k p 1 is the identity map on a dense open set. We take g = p 2 k to obtain the desired map.

17 A.3. TANGENT SPACES 621 A.2.5 Exercises 1. Let X and Y be affine algebraic sets and ϕ : X Y a regular map. (a) Show that the graph Γ ϕ = {(x, ϕ(x)) : x X} of ϕ is closed in X Y. (b) Show that the projection map π : Γ ϕ X onto the first coordinate is an isomorphism of affine algebraic sets. 2. Let ϕ : C 2 C 2 be the map ϕ(x, y) = (xy, y). (a) Show that ϕ has dense image but is not surjective. (b) Show that x is not integral over C[xy, y]. 3. Let X = {(x, y) C 2 : y 2 = x 3 }. (a) Show that every element of O[X] can be written uniquely in the form P(x) + Q(x)y, where P and Q are polynomials. (b) Let ϕ : C X be the map ϕ(t) = (t 2, t 3 ). Show that ϕ is bijective and regular, but ϕ 1 is not regular. A.3 Tangent Spaces A.3.1 Tangent Space and Differentials of Maps Let X C n be an algebraic set. Let O[X] denote the linear maps from O[X] (viewed as a C vector space) to C. Definition A.3.1. A tangent vector at a point x X is an element v O[X] such that v(fg) = v(f)g(x) + f(x)v(g) (A.13) for all f, g O[X]. The set of all tangent vectors at x the tangent space T(X) x of X at x. Clearly T(X) x is a linear subspace of O[X]. We observe that if v is a tangent vector at x, then v(1) = v(1 1) = 2v(1). Hence v(1) = 0, so v vanishes on the space of constant functions on X. We now give an alternate description of the tangent space in terms of maximal ideals. Let m x O[X] be the maximal ideal of all functions that vanish at x. Then f f(x) m x for any f O[X], and v(f) = v(f f(x)). Hence v is determined by its restriction to m x. However, by (A.13) we see that v(m 2 x) = 0, so v naturally defines an element ṽ (m x /m 2 x ). Conversely, given any ṽ (m x /m 2 x ), we define a linear functional v on O[X] by v(f) = ṽ(f f(x)). To verify that v satisfies (A.13), we observe that if f, g O[X], then (f f(x))(g g(x)) m 2 x. Hence 0 = v(fg) f(x)v(g) g(x)v(f), since v(c) = 0 for c C. This shows that there is a natural isomorphism T(X) x = (mx /m 2 x). (A.14)

18 622 APPENDIX A. ALGEBRAIC GEOMETRY Let X and Y be algebraic sets and ϕ : X Y a regular map. Then the induced map ϕ : O[Y ] O[X] is an algebra homomorphism. If v T(X) x then the linear functional f v(ϕ (f)), for f O[Y ], is a tangent vector at y = ϕ(x). Indeed, for f, g O[Y ] we have v(ϕ (fg)) = v(ϕ (f)ϕ (g)) = v(ϕ (f))g(y) + f(y)v(ϕ (g)). We denote this element of T(Y ) y by dϕ x (v). At each point x X we thus have a linear map dϕ x : T(X) x T(Y ) ϕ(x) that we call the differential of ϕ at x. Examples 1. If X = C n, then O[X] = P(C n ) = C[x 1,..., x n ]. For u C n and f P(C n ) define D u f(x) = ( / t)f(x + tu) t=0 (the directional derivative of f along the line t x +tu, t C). The linear functional v(f) = D u f(a) is a tangent vector at a C n. If u = (u 1,..., u n ), then we can express v in terms of the partial derivative operators / x i : v = n i=1 u i ( / x i ) a, where we write ( / x i ) a for the tangent vector f ( f/ x i )(a) with a C n. We claim that T(C n ) a has basis ( / x 1 ) a,..., ( / x n ) a and hence has dimension n. Indeed, let a have coordinates (a 1,..., a n ). Then it is clear by the Taylor expansion of a polynomial that the ideal m a is generated by the independent linear functions x i a i, for 1 i n. Thus v T(C n ) a can be uniquely expressed as v(f) = D u f(a), where u = (u 1,..., u n ) C n and u i = v(x i a i ). When convenient, we will identify T(C n ) a with C n by the map v u. 2. Suppose X C n is an algebraic set and a X. Since O[X] = P(C n )/I X, for every v T(X) a there exists ṽ T(C n ) a with ṽ(i X ) = 0 and ṽ(f) = v(f + I X ). Conversely, any ṽ with this property induces a tangent vector v to X at a, with v(f + I X ) = ṽ(f). Hence T(X) a = {ṽ T(C n ) a : ṽ(i X ) = 0}. Let {f 1,..., f r } be a generating set of polynomials for the ideal I X. The defining equation for a derivation shows that ṽ(i X ) = 0 if and only if ṽ(f i ) = 0 for i = 1,..., r. Hence if we set u j = ṽ(x j a j ) then ṽ T(X) a if and only if n j=1 u j f i (a) x j = 0 for i = 1,..., r. (A.15) This is a set of r linear equations for u = (u 1,..., u n ) C n. In particular, we see that dimt(x) a = n rank(j(a)), where J(a) is the r n matrix [ f i (a)/ x j ] of partial derivatives. 3. Let X be an irreducible affine algebraic set. Define m(x) = min a X dimt(x) a.

19 A.3. TANGENT SPACES 623 Let X 0 = {a X : dimt(x) a = m(x)}. The points of X 0 are called smooth. These are the points at which the matrix J(a) defined above has maximum rank d = n m(x). Because this condition can be described by the nonvanishing of some d d minor of J(a), it follows that X 0 is Zariski-dense in X when X is irreducible. If X 0 = X then X is said to be smooth. 4. If X is a reducible algebraic set with irreducible components X i, then we say that X is smooth if each X i is smooth. We define m(x) = max i m(x i ) in this case. A.3.2 Vector Fields Recall that a derivation of an algebra A is a linear map D : A A such that D(ab) = D(a)b + ad(b). If A is commutative and D and D are derivations of A, then any linear combination of D and D with coefficients in A is a derivation, and the commutator [D, D ] = DD D D is a derivation, as we check by the obvious calculation. Thus the derivations of a commutative algebra A form a Lie algebra Der(A), which is also an A-module (see Section 4.1.1). In the case A = O[X], where X is an affine algebraic set, a derivation of A is usually called a vector field on X. We denote by Vect(X) the Lie algebra of all vector fields on X. Given L Vect(X) and x X, we define L x f = (Lf)(x) for f O[X]. Then L x T(X) x, by the definition of tangent vector. Conversely, if we have a correspondence x L x T(X) x such that the functions x L x (f) are regular for every f O[X], then L is a vector field on X. Example Let 0 f O[X] and consider a vector field L on the principal open set X f. Recall that O[X f ] is generated by the restrictions to X f of functions in O[X] together with f 1. However, since L is a derivation, we have L(f)f 1 +fl(f 1 ) = L(ff 1 ) = L(1) = 0, so that L(f 1 ) = f 2 L(f). Hence L is completely determined by its action on O[X]. For example, if X = C n, then A.3.3 L = n i=1 ϕ i Dimension x i with ϕ i = L(x i ) C[x 1,..., x n, 1/f]. (A.16) We now show that the notion of dimension defined geometrically via the tangent space coincides with the algebraic definition in terms of transcendence degree. Theorem A.3.2. Let X be an algebraic set. Then m(x) = dimx. Proof. If X has irreducible components X i, then m(x) = max(m(x i )), so we may assume X C n is irreducible. Set K = Rat(X). We prove first that dimx = dim K Der(K). (A.17) For this, we apply the Noether normalization lemma to B = O[X] to obtain a transcendence basis {u 1,..., u d }, where d = dimx. Let A = C[u 1,..., u d ] and

20 624 APPENDIX A. ALGEBRAIC GEOMETRY k = Quot(A) K. A derivation D of k is determined by an arbitrary choice of d rational functions Du i k, since {u 1,..., u d } is algebraically independent. On the other hand, we claim that every derivation D of k extends uniquely to a derivation of K. To establish this, we use the fact that the field K is a finite extension of k, since B is algebraic over A. Hence the theorem of the primitive element (Lang [1971, Ch VII, 6, Theorem 14]) furnishes an element b K so that K = k(b). Let f(x) k[x] be the minimal polynomial for b. Then f (b) 0, so we may define Db = f D (b)/f (b) K, where for any polynomial g(x) = i a ix i k[x] we let g D (X) = i D(a i)x i. If D can be extended to a derivation of K, it must act by D(g(b)) = g D (b) + g (b)db for all g(x) k[x]. (A.18) To prove that such an extension exists, we must verify that the right side of (A.18) does not depend on the representation of an element of K as g(b). But if g(b) = h(b) for some h(x) k[x], then the polynomial ϕ(x) = g(x) h(x) is of the form ψ(x)f(x) for some ψ(x) k[x]. Hence ϕ (b) = ψ(b)f (b) and ϕ D (b) = ψ(b)f D (b) since f(b) = 0. It follows that g (b)db h (b)db = ψ(b)f (b)db = ψ(b)f D (b) = g D (b) + h D (b) by the definition of Db. Thus (A.18) is unambiguous and defines the desired extension. This completes the proof of (A.17). Let f 1,..., f r be a set of generators for I X. A derivation D of K is uniquely determined by the n functions ξ j = Dx j K, where x j are the linear coordinate functions from C n. It must also satisfy D(f i ) = 0 for i = 1,..., r. Thus by the chain rule we have n f i ξ j = 0, for i = 1,..., r. (A.19) x j j=1 By definition of m(x), system (A.19) has rank n m(x) over K. Hence the solution space to (A.19) has dimension m(x) over K. Combining this fact with (A.17) completes the proof. A.3.4 Differential Criterion for Dominance We begin with a general criterion for a map to be dominant. Proposition A.3.3. Let X and Y be affine algebraic sets and ψ : X Y a regular map. Assume Y is irreducible and dim Y = m. Suppose there exists an algebraically independent set {u 1,..., u m } O[Y ] such that the set {ψ u 1,..., ψ u m } O[X] is also algebraically independent. Then ψ(x) is dense in Y.

21 A.3. TANGENT SPACES 625 Proof. Suppose there exists 0 f O[Y ] such that ψ (f) = 0. Since dim Y = m, f is algebraic over the field C(u 1,..., u m ). Thus there are rational functions a j C(x 1,..., x m ) so that d j=0 a j(u 1,..., u m )f j = 0. We choose the functions a j with d as small as possible (d 1 since f 0). Take 0 γ C[u 1,..., u m ] such that γa j (u 1,..., u m ) C[u 1,..., u m ]. Then d j=0 ψ (γ)a j (ψ u 1,..., ψ u m )ψ (f j ) = 0. Since ψ (f j ) = ψ (f) j = 0 for j 1, we obtain the relation ψ (γ)a 0 (ψ u 1,..., ψ u m ) = 0. (A.20) But we are given that ψ : C[u 1,..., u m ] C[ψ u 1,..., ψ u m ] is an isomorphism, so we have ψ (γ) 0. Thus (A.20) implies that a 0 (ψ u 1,..., ψ u m ) = 0, and hence a 0 (u 1,..., u m ) = 0. This contradicts the definition of d. We conclude that ψ : O[Y ] O[X] is injective. Since Y is irreducible, this implies that ψ(x) is dense in Y. Let X and Y be irreducible affine algebraic sets and let ψ : X Y be a regular map. We have the following criterion for ψ to be dominant: Theorem A.3.4. Suppose there exists a smooth point p X such that ψ(p) is a smooth point of Y and dψ p : T(X) p T(Y ) ψ(p) is bijective. Then ψ(x) is dense in Y. We first prove the following lemma, whose statement and proof is similar to the corresponding result for C manifolds (the local triviality of the tangent bundle): Lemma A.3.5. Let X C n be closed and irreducible and let p X be a smooth point of X. Then there exists a open subset U X with p U and regular maps w j : U C n for j = 1,..., m = dimx such that Proof. Since p is a smooth point, we have T(X) q = m j=1 Cw j(q) for all q U. T(X) p = {v C n : (dϕ) p (v) = 0 for all ϕ I X } by (A.15). Hence there exist ϕ 1,..., ϕ n m I X such that T(X) p = {v C n : (dϕ i ) p (v) = 0 for i = 1,..., n m}. This implies that (dϕ 1 ) p ( dϕ (n m) )p 0, so there is a open subset U 1 of C n such that p U 1 and (dϕ 1 ) q ( dϕ (n m) )q 0 for q U 1. (A.21)

22 626 APPENDIX A. ALGEBRAIC GEOMETRY For any point q U 1 X we have T(X) q W q = {v C n : (dϕ i ) q (v) = 0 for i = 1,..., n m}. Now dimt(x) q m since X is irreducible. Since dim W q = m by (A.21), we conclude that T(X) q = {v C n : (dϕ i ) q (v) = 0, i = 1,..., n m} (A.22) for all q U 1 X. Fix a basis {e 1..., e n } for C n such that det[(dϕ i ) p (e j )] 1 i,j n m 0. Let V = n m i=1 Ce i and W = n i=n m+1 Ce i. We then write x C n as [ ] y x = with y V and z W. z For q C n we write the n n matrix J q = [(dϕ i ) q (e j )] 1 i,j n in block form as J q = [ ] Aq B q C q D q with A q of size (n m) (n m). In terms of the decomposition C n = V W we have A q : V V and B q : W V. Since det A p 0 there exists a Zariski-open subset U 2 U 1 with p U 2 and det A q 0 for all q U 2. We define regular maps w j : U 2 C n by [ A 1 q w j (q) = B ] qe n m+j for j = 1,..., m. e n m+j Clearly {w 1 (q),..., w m (q)} is linearly independent. Also, [ Aq A J q w j (q) = 1 q B q e n m+j + B q e n m+j Set U = U 2 X. Then for all q U we have ] [ 0 = ]. (dϕ i ) q (w j (q)) = 0 for i = 1,..., n m. Thus {w 1 (q),..., w m (q)} is a basis for T(X) q by (A.22). Proof of Theorem A.3.4 Let p X satisfy the conditions of the theorem. Take U and w 1,..., w m be as in Lemma A.3.5. Since {dψ p (w 1 (p)),..., dψ p (w m (p))} is a basis for T(Y ) ψ(p), there exist functions u 1,..., u m in O[Y ] such that dψ p (w j (p))u i = δ ij for i, j = 1,..., m.

23 A.4. PROJECTIVE AND QUASIPROJECTIVE SETS 627 Now for q X we can write dψ q (w j (q))u i = d(ψ u i ) q (w j (q)). Thus there is a open subset V U containing p such that det[d(ψ u i ) q (w j (q))] 1 i,j m 0 for q V. This implies that {ψ u 1,..., ψ u m } is algebraically independent, so {u 1,..., u m } is algebraically independent. The theorem follows from Proposition A.3.3. Proposition A.3.6. Let ϕ : X Y be a dominant regular map of irreducible affine algebraic sets. For y Y let F y = ϕ 1 {y}. Then there is a nonempty open set U X such that dimx = dimy +dim F ϕ(x) and dimf ϕ(x) = dimker(dϕ x ) for all x U. Proof. Let d = dimx dimy, S = ϕ O[Y ], and R = O[X]. Set k = Quot(S) and let B Quot(R) be the subalgebra generated by k and R (the rational functions on X with denominators in S \ {0}). Since B has transcendence degree d over k, Lemma A.1.17 furnishes an algebraically independent set {f 1,..., f d } R such that B is integral over k[f 1,..., f d ]. Taking the common denominator of a set of generators of the algebra B, we obtain f = ϕ g S such that R f is integral over S f [f 1,..., f d ], where R f = O[X f ] and S f = ϕ O[Y g ]. By Theorem A.2.7 we can take g so that ϕ(y g ) = X f. Define ψ : X f Y g C d by ψ(x) = (ϕ(x), f 1 (x),..., f d (x)). Then ψ O[Y g C d ] = S f [f 1,..., f d ], and hence O[X f ] is integral over ψ O[Y g C d ]. By Theorem A.2.3 every homomorphism from S f [f 1,..., f d ] to C extends to a homomorphism from R f to C. Hence ψ is surjective. Let π : Y g C d Y g by π(y, z) = y. Then ϕ = ψ π and F y = ψ 1 ({y} C d ). If W is any irreducible component of F y then O[W] is integral over ψ O[{y} C d ], and hence dimw = d. We have dϕ x = dπ ψ(x) dψ x. The integrality property implies that every derivation of Quot(ψ (Y g C d )) extends uniquely to a derivation of Rat(X f ), as in the proof of Theorem A.3.2. Hence dψ x is bijective for x in a nonempty dense open set U by Lemma A.3.5. For such x, Ker(dϕ x ) = Ker(dπ ψ(x) ) has dimension d. A.4 Projective and Quasiprojective Sets A.4.1 Basic Definitions Let V be a complex vector space. The projective space P(V ) associated with V is the set of lines through 0 (one-dimensional subspaces) in V. For x V \ {0}, [x] P(V ) will denote the line through x. The map p : V \ {0} P(V ) given by p(x) = [x] is surjective, and p(x) = p(y) if and only if x = λy for some λ C. We denote P(C n+1 ) by P n and for x = (x 0,..., x n ) C n+1 we call {x i } the homogeneous coordinates of [x]. If f(x 0,..., x n ) is a homogeneous polynomial in n + 1 variables and 0 x C n+1, then f(x) = 0 if and only if f vanishes on line [x]. Hence f defines a subset A f = {[x] P n : f(x) = 0}