MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton. Timothy J. Ford April 4, 2016

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1 MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton Timothy J. Ford April 4, 2016 FLORIDA ATLANTIC UNIVERSITY, BOCA RATON, FLORIDA address:

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3 Contents List of Figures Disclaimer iii v Chapter 1. Affine Algebraic Sets Algebraic Preliminaries Affine Space and Algebraic Sets The Ideal of a Set of Points The Hilbert Basis Theorem Irreducible Components of an Algebraic Set Algebraic Subsets of A 2 (k) Hilbert s Nullstellensatz Modules, Finiteness Conditions Integral Elements Field Extensions 17 Chapter 2. Affine Varieties Coordinate Rings Polynomial Maps Coordinate Changes Rational Functions and Local Rings Discrete Valuation Rings Forms Direct Products of Rings Operations on Ideals Ideals with a Finite Number of Zeros Quotient Modules and Exact Sequences Free Modules 33 Chapter 3. Local Properties of Plane Curves Multiple Points and Tangent Lines Multiplicities and Local Rings Intersection Numbers 39 Chapter 4. Projective Varieties Projective Space Projective Algebraic Sets Affine and Projective Varieties Multiprojective Space 51 Chapter 5. Projective Plane Curves 53 i

4 ii CONTENTS Chapter 6. Varieties, Morphisms and Rational Maps The Zariski Topology Varieties Morphisms of Varieties Products and Graphs Algebraic Function Fields and Dimension of Varieties Rational Maps 60 Chapter 7. Resolution of Singularities Rational Maps of Curves 65 Bibliography 67

5 List of Figures 1.1 y = x(x 2 1) y 2 = x(x 2 1) y 2 = x 2 (x + 1) Real Quadric Cone. x 2 + y 2 z Real Quadric Hyperboloid of One Sheet. x 2 + y 2 z Real Quadric Paraboloid. x 2 + y 2 z Real Quadric Hyperbolic Paraboloid. x 2 y 2 z Real Quadric Hyperboloid of Two Sheets. z 2 x 2 y Mapping V (y 2 x) onto A Mapping A 1 onto y 2 = x The real plane curve 3x 2 y y 3 + (x 2 + y 2 ) 2 = The real plane curve 4x 2 y 2 (x 2 + y 2 ) 3 = iii

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7 Disclaimer These notes are for classroom use only. They are based on the classic Algebraic Curves by Fulton [3]. All references to material in these notes should cite the original text, not these notes. It is very unfortunate that this book is out of print. v

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9 CHAPTER 1 Affine Algebraic Sets 1.1. Algebraic Preliminaries By a ring we always mean commutative ring with multiplicative identity 1. In a ring the elements 0 and 1 are different except in the degenerate case where the ring has only one element. A homomorphism of rings always maps 1 to 1. For any ring R, the ring of polynomials with coefficients in R is denoted R[x]. The ring of polynomials in n variables over R is written R[x 1,...,x n ]. If k is a field, the quotient field of k[x 1,...,x n ] is the field of rational functions in n variables over k, written k(x 1,...,x n ) Problems. PROBLEM 1.1. Let R be a domain. (1) If f, g are forms of degree r, s respectively in R[x 1,...,x n ], show that f g is a form of degree r + s. (2) Show that any factor of a form in R[x 1,...,x n ] is also a form. PROBLEM 1.2. Let R be a UFD, K the quotient field of R. Show that every element z of K may be written z = a/b, where a,b R have no common factors; this representation is unique up to units of R. PROBLEM 1.3. Let R be a PID. Let P be a non-zero, proper, prime ideal in R. (1) Show that P is generated by an irreducible element. (2) Show that P is maximal. PROBLEM 1.4. Let k be an infinite field, f k[x 1,...,x n ]. Suppose f (a 1,...,a n ) = 0 for all a 1,...,a n k. Show that f = 0. (Hint: Write f = f i x i n, f i k[x 1,...,x n 1 ]. Use induction on n, and the fact that f (a 1,...,a n 1,x n ) has only a finite number of roots if any f i (a 1,...,a n 1 ) 0.) PROBLEM 1.5. Let k be any field. Show that there are an infinite number of irreducible monic polynomials in k[x]. (Hint: Suppose f 1,..., f n were all of them, and factor f 1 f n + 1 into irreducible factors.) PROBLEM 1.6. Show that any algebraically closed field is infinite. (Hint: The irreducible monic polynomials are x a, a k.) PROBLEM 1.7. Let k be a field, f k[x 1,...,x n ], a 1,...,a n k. (1) Show that f can be written as where λ (i) k for each i. f = λ (i) (x 1 a 1 ) i1 (x n a n ) i n, 1

10 2 1. AFFINE ALGEBRAIC SETS (2) If f (a 1,...,a n ) = 0, show that f can be written as f = n i=1 (x i a i )g i for some (not unique) g i k[x 1,...,x n ] Affine Space and Algebraic Sets Let k be any field. Affine n-space, denoted A n (k), A n k or just An, is the set k n = {(a 1,...,a n ) a i k}. Take f k[x 1,...,x n ]. A point (a 1,...,a n ) A n is a zero of f if f (a 1,...,a n ) = 0. The hypersurface defined by f is V ( f ) = {P A n f (P) = 0}. If n = 2 and f k[x,y], V ( f ) is an affine plane curve. EXAMPLE 1.1. Some examples of real plane curves. Let k = R. 1. For any polynomial f (x) k[x], consider y f (x) k[x,y]. Then V (y f (x)) is an algebraic curve, usually called the graph of f (x). For example, say f (x) = x(x 2 1). Then V (y f (x)) is a non-singular rational cubic (as we will see later). See Figure 1.1. FIGURE 1.1. y = x(x 2 1) 2. V (y 2 x(x 2 1)) is a non-singular elliptic cubic (as we will see later). See Figure 1.2. FIGURE 1.2. y 2 = x(x 2 1) 3. V (y 2 x 2 (x + 1)) is a singular cubic (as we will see later). See Figure 1.3.

11 1.2. AFFINE SPACE AND ALGEBRAIC SETS 3 FIGURE 1.3. y 2 = x 2 (x + 1) FIGURE 1.4. Real Quadric Cone. x 2 + y 2 z 2 EXAMPLE 1.2. Examples of quadric surfaces over R are shown in Figures If S k[x 1,...,x n ], the set of zeros of S is V (S) = {P A n f (P) = 0 for all f S} = f SV ( f ) If X A n, we say X is an algebraic set in case X = V (S) for some S k[x 1,...,x n ]. LEMMA 1.3. If I is the ideal generated by S, then V (S) = V (I). PROOF. Clearly V (I) V (S). Say f I. Then f = m i=1 g i f i where f i S, g i k[x 1,...,x n ]. Given any P V (S), So V (S) V (I). f (P) = g i (P) f i (P) = g i (P) 0 = 0

12 4 1. AFFINE ALGEBRAIC SETS FIGURE 1.5. Real Quadric Hyperboloid of One Sheet. x 2 + y 2 z 2 1 FIGURE 1.6. Real Quadric Paraboloid. x 2 + y 2 z 1 LEMMA 1.4. If {I α } is any collection of ideals, then V ( I α ) = V (I α ). This proves that the intersection of any collection of algebraic sets is an algebraic set. PROOF. Take P V ( I α ). Show P V (I α ) for each α. Given f α I α, f α is also in I α. So f α (P) = 0.

13 1.2. AFFINE SPACE AND ALGEBRAIC SETS 5 FIGURE 1.7. Real Quadric Hyperbolic Paraboloid. x 2 y 2 z FIGURE 1.8. Real Quadric Hyperboloid of Two Sheets. z 2 x 2 y 2 1 Take P V (I α ). Given any f I α, f is in some I α. So f (P) = 0 and P V ( I α ). LEMMA 1.5. If I J, then V (I) V (J).

14 6 1. AFFINE ALGEBRAIC SETS PROOF. If P V (J), then f (P) = 0 for all f J. But I J, hence g(p) = 0 for all g I. Therefore P V (I). LEMMA 1.6. For all f,g k[x 1,...,x n ], V ( f g) = V ( f ) V (g). For any two ideals I, J in k[x 1,...,x n ], V (I) V (J) = V ({ f g f I,g J}) = V (IJ). This proves that the union of finitely many algebraic sets is an algebraic set. PROOF. A point P is in V ( f g) if and only if f (P)g(P) = 0 which is true if and only if f (P) = 0 or g(p) = 0 which is true if and only if P is in V ( f ) V (g). For the second part, V (IJ) = V ( f g) f I g J ( ) = V ( f ) V (g) f I g J = f I (V ( f ) V (J)) = V (I) V (J). LEMMA V (0) = A n 2. V (1) = /0 3. V (x 1 a 1,...,x n a n ) = {(a 1,...,a n )} for any a i k. This proves any finite subset of A n is an algebraic set. PROOF. Obvious Problems. PROBLEM 1.8. Show that the algebraic subsets of A 1 (k) are just the finite subsets, together with A 1 (k) itself. PROBLEM 1.9. Show that the following are algebraic sets. (1) Show that { (t,t 2,t 3 ) A 3 (k) t k } is an algebraic set. (2) Show that { (cost,sint) A 2 (R) t R } is an algebraic set. (3) Show that the set of points in A 2 (R) whose polar coordinates (r,θ) satisfy the equation r = sinθ is an algebraic set. PROBLEM Let f be a non-constant polynomial in k[x 1,...,x n ], k an algebraically closed field. Show that A n (k) V ( f ) is infinite if n 1, and V ( f ) is infinite if n 2. Conclude that the complement of any proper algebraic set is infinite. PROBLEM Let V A n and W A m be algebraic sets. Show that V W = {(a 1,...,a n,b 1,...,b m ) (a 1,...,a n ) V,(b 1,...,b m ) W} is an algebraic set in A n+m. It is called the product of V and W.

15 1.3. THE IDEAL OF A SET OF POINTS The Ideal of a Set of Points If X A n k, we define the ideal of X to be I(X) = f k[x 1,...,x n ] f (p) = 0 for all p X. Check that so I(X) is an ideal in k[x 1,...,x n ]. (α f + βg)(p) = α(p) f (p) + β(p)g(p) LEMMA 1.8. Let X, Y be algebraic sets. (1) If X Y, then I(X) I(Y ). (2) I(X Y ) = I(X) I(Y ) = α(p) 0 + β(p) 0 = 0 PROOF. (1) If f I(Y ), then f (p) = 0 for each p Y. Thus f (p) = 0 for each p X and f I(X). (2) Note that f I(X Y ) if and only if f I(X) and f I(Y ). LEMMA I(/0) = k[x 1,...,x n ] 2. I(A n ) = (0) if k is an infinite field. 3. I({(a 1,...,a n )}) = (x 1 a 1,...,x n a n ) for a i k. PROOF. 1. I(/0) = k[x 1,...,x n ] by default. 2. Follows from Problem Follows from Problem 1.7. LEMMA I(V (S)) S for each S k[x 1,...,x n ]. 2. V (I(X)) X for each X A n. PROOF. 1. Take f S and p V (S). Then f (p) = 0 so f I(V (S)). 2. Take p X. For each f I(X) it follows that f (p) = 0. So p V (I(X)). LEMMA V (I(V (S))) = V (S) for each S k[x 1,...,x n ]. 2. I(V (I(X))) = I(X) for each X A n. PROOF. 1. Since I(V (S)) S, it follows V (I(V (S))) V (S). Take p V (S) and f I(V (S)). Then f (p) = 0 hence p V (I(V (S))). 2. Since V (I(X)) X, it follows I(V (I(X))) I(X). Take f I(X). For each p V (I(X)) f (p) = 0. Thus f I(V (I(X))). If R is a commutative ring and I is an ideal in R, the radical of I, written RadI is RadI = {a R a n I for some n > 0} It is proved in Problem 1.14 that RadI is an ideal containing I. If I = RadI we say that I is a radical ideal. LEMMA For each X A n, I(X) is a radical ideal. PROOF. If f n I(X), then for each p X 0 = f n (p) = f (p) n hence f (p) = 0. Therefore f I(X). So I(X) = RadI(X). EXAMPLE The twisted cubic curve C in A 3 is C = {(t,t 2,t 3 ) t k} = V (x 2 y,x 3 z)

16 8 1. AFFINE ALGEBRAIC SETS EXAMPLE A nodal cubic curve in A 2 is V (y 2 x 2 (x + 1)). EXAMPLE A cuspidal cubic curve in A 2 is V (y 2 x 3 ). EXAMPLE A quadric surface in A 3 with a node is V (z 2 x 2 y 2 ). EXAMPLE Let k be algebraically closed, chark 2. A plane conic curve C is V ( f ) where f k[x,y] is a polynomial of degree 2. Every conic curve C in A 2 is one of the following types, up to a linear change of variables: A. If f factors into two linear factors, then C is reducible and is one of the following three types: 1. double line, x 2 = 0 2. parallel lines, x(x + 1) = 0 3. intersecting lines, xy = 0 B. If f is irreducible, then C is one of 1. C has one point at infinity, y = x 2 2. C has two points at infinity, xy = c, c 0 PROOF. Start with the general polynomial of degree two f k[x,y], f = ax 2 + bxy + cy 2 + dx + ey + f If f is not irreducible, C = V ( f ) is reducible and of one of the types in A. Assume f is irreducible. If the quadratic form has only one variable, that is, looks like ax 2 or cy 2, then it is easy to see we are in case B.1. Otherwise, at least two of a,b,c are non-zero. Since k is algebraically closed, the quadratic form in f factors into a product of linear forms: ax 2 + bxy + cy 2 = (αx + βy)(γx + δy) If this quadratic form is not a square, a linear change of variables transforms it into a sum of squares. In either case, after a change of variables we can assume f = gx 2 + hy 2 + dx + ey + f If g = 0 or h = 0 we are in case B.1. Otherwise complete the square in the two variables, After a change of variables, f = g(x + i) 2 + h(y + j) 2 + k f = x 2 + y 2 + k where k 0 if f is irreducible. This equation is of type B Problems. PROBLEM Let V, W be algebraic sets in A n. Show that V = W if and only if I(V ) = I(W). PROBLEM Prove the following. (1) Let V be an algebraic set in A n, p A n a point not in V. Show that there is a polynomial f k[x 1,...,x n ] such that f (q) = 0 for all q V but f (p) = 1. (Hint: I(V ) I(V {p})) (2) Let {p 1,..., p r } be a finite set of points in A n. Show that there are polynomials f 1,..., f r in k[x 1,...,x n ] such that { 0 if i j f i (p j ) = 1 if = j

17 1.4. THE HILBERT BASIS THEOREM 9 (3) Let V be an algebraic set in A n. Let p 1, p 2 be points in A n which are not in V. Show that there is a polynomial f k[x 1,...,x n ] such that f I(V ), f (p 1 ) 0, and f (p 2 ) 0. (Hint: Find f i I(V ) so that f i (p i ) 0. Then take f to be f 1, f 2, or f 1 + f 2 ) PROBLEM Let I be an ideal in a commutative ring R. If a n I, b m I, show that (a + b) n+m I. Show that RadI is an ideal, in fact a radical ideal. Show that any prime ideal in R is a radical ideal. PROBLEM Show that for any ideal I in k[x 1,...,x n ], V (I) = V (RadI), and RadI I(V (I)). PROBLEM Show that I = (x 1 a 1,...,x n a n ) k[x 1,...,x n ] is a maximal ideal, and that the natural homomorphism from k to k[x 1,...,x n ]/I is an isomorphism The Hilbert Basis Theorem LEMMA The following are equivalent for a commutative ring R. (1) Every ideal of R is finitely generated. (2) R has the ascending chain condition (ACC) on ideals. (3) Every non-empty set of ideals S = {I α } has a maximal element. PROOF. (1) implies (2). Let I 0 I 1 I 2... be a chain of ideals in R. Let I = i=0. Then I is an ideal, hence I is finitely generated. Say I = (r 1,...,r m ). There is n such that {r 1,...,r m } I n. Therefore I = I n = I n+1 =... (2) implies (3). Pick I 0 S. If I 0 is not maximal, then there exists I 1 S such that I 0 I 1. Iterate to get an ascending chain of ideals in R, each an element of S. I 0 I 1 I 2... By the ACC, this chain becomes constant. For some n, I n = I n+1 =... Therefore I n is a maximal element of S. (3) implies (1). Let I be an ideal in R. Let S = {J I J is a finitely generated ideal in R} Since (0) S, S /0. Let M S be a maximal element. Then M is finitely generated. Let x 1,...,x m be a generating set for M. If y I, then M y = M + Ry = (x 1,...,x m,y) is finitely generated and M M y I. Since M is maximal, M = M y. Hence y M and M = I. This proves I is finitely generated. A commutative ring R is noetherian if R satisfies the ACC on ideals. By Lemma 1.18 R is noetherian if and only if ideals are finitely generated. THEOREM 1.19 (Hilbert s Basis Theorem). If R is noetherian, then R[x 1,...,x n ] is noetherian. PROOF. By induction on n, it suffices to show R[x] is noetherian. Let I be an ideal in R[x]. For n 0 let J n be the ideal in R generated by the leading coefficients of polynomials of degree n in I. If deg f = n and f I and f has leading coefficient a n, then a n J n. Moreover f = a n x n +... x f = a n x n

18 10 1. AFFINE ALGEBRAIC SETS shows a n J n+1 and the ideals form a chain J 0 J 1 J 2... Because R has ACC, there exists N 0 such that J i J N for all i 0. Since each J i is finitely generated, choose m sufficiently large and a set of polynomials S = { f n1,..., f nm } N n=0 I such that the leading coefficients of f n1,..., f nm generate J n. Let I denote the ideal generated by S in R[x]. Then I I. Suppose G I I. Choose G such that the degree of G is minimal. The proof is reduced to two cases. Case 1: degg N. Then the leading coefficient of G is in J N. So there exist g 1,...,g m in R such that the leading coefficient of G is equal to the leading coefficient of m i=1 g i f Ni. Let H = G m i=1 g i f Ni. Then H has degree less than degg. But H I I contradicts the choice of G, so G I. Case 2: degg = n < N. As before, we can use f n1,..., f nm to lower the degree of G. THEOREM The following hold for A n k where k is a field. (1) Every algebraic set in A n is the intersection of a finite number of hypersurfaces. (2) The descending chain condition (DCC) holds for algebraic sets. (3) A non-empty collection of algebraic sets has a minimal member. PROOF. By Theorem 1.19, k[x 1,...,x n ] is noetherian for any field k. In particular k[x 1,...,x n ] has the three properties of Lemma (1) Say X = V (I). Let f 1,..., f m be a set of generators for I. Then by Lemma 1.3 m X = V (I) = V ({ f 1,..., f m }) = V ( f i ). i=1 (2) Suppose there is a descending chain of algebraic sets Then by Lemma 1.8 the ideals X 0 X 1 X 2... I(X 0 ) I(X 1 ) I(X 2 )... form an ascending chain of ideals in k[x 1,...,x n ]. This chain becomes constant by ACC, so there is some N such that I(X N ) = I(X N+1 ) =... By Problem 1.12, X N = X N+1 =... (3) Let S = {X α } be a family of algebraic sets. Then T = {I(X α )} is a family of ideals in k[x 1,...,x n ]. Then T has a maximal member, say I(X α ). By Lemma 1.8, X α is a minimal member of S. LEMMA The homomorphic image of a noetherian ring is noetherian. PROOF. Let ϕ : R S be a homomorphism of commutative rings. Assume ϕ is surjective and R is noetherian. If K = kerϕ, then R/K is isomorphic to S. Let π : R R/K be the natural map. Every ideal J of R/K is the image under π of an ideal I in R that contains K. A set of generators for I maps to a set of generators for J Irreducible Components of an Algebraic Set An algebraic set X A n is called reducible if X = X 1 X 2 where X 1, X 2 are algebraic sets and X X 1, X X 2. Otherwise X is said to be irreducible. An irreducible algebraic subset of A n is also called an affine variety. PROPOSITION The algebraic set X is irreducible if and only if I(X) is a prime ideal.

19 1.5. IRREDUCIBLE COMPONENTS OF AN ALGEBRAIC SET 11 PROOF. Assume X is reducible. Say X = X 1 X 2 and X X 1, X X 2. Then by Problem 1.12, I(X) I(X i ), i = 1,2. For each i pick f i I(X i ) I(X). So for each point p X we have { 0 f 2 (p) if p X 1 ( f 1 f 2 )(p) = f 1 (p) 0 if p X 2 So f 1 f 2 I(X) and this shows I(X) is not a prime ideal. Conversely assume I(X) is not a prime ideal. Assume f g I(X) where f, g each are not in I(X). Then I(X) I(X) + R f. So X = V (I(X)) V (I(X)) V ( f ) = X V ( f ) Similarly X X V (g). Now f g I(X) implies Intersect both sides with X. V ( f ) V (g) = V ( f g) X (V ( f ) X) (V (g) X) = X This proves that X is reducible. THEOREM Let V A n be an algebraic set. Then there are unique irreducible algebraic sets V 1,...,V m such that V = V 1 V m and V i V j for all i j. The decomposition of an algebraic set which is provided by Theorem 1.23 is called the decomposition of V into its irreducible components. PROOF OF THEOREM (Existence part) Consider the collection of algebraic sets S where S consists of all algebraic sets V such that V cannot be written as a union of a finite number of irreducible sets. Suppose S is non-empty. By Theorem 1.20, S has a minimal member, call it V. Then V is itself not irreducible, so we can write V = V 1 V 2 where each V i is a proper subset of V. By minimality of V, it follows that each V i is not in S. Therefore each V i can be decomposed into irreducibles. However this means V = V 1 V 2 can also be decomposed into irreducibles, which is a contradiction. (Uniqueness part) Say V = V 1 V m. Throw away any V i such that V i V j and i j. Say V = W 1 W p. is another such decomposition. Then V 1 = (V 1 W 1 ) (V 1 W p ) Since V 1 is irreducible, V 1 = V 1 W i for some i. Therefore V 1 W i. Likewise W i V j for some j. This implies V 1 W i V j It follows that V 1 = W i. By a finite induction argument, we are done Problems. PROBLEM Prove the following in the complex affine plane A 2 C. (1) Show that V (y x 2 ) is irreducible. In fact, I(V (y x 2 )) = (y x 2 ). (2) Decompose V (y 4 x 2,y 4 x 2 y 2 + xy 2 x 3 ) into irreducible components. PROBLEM If k is an infinite field, show that A n k is irreducible.

20 12 1. AFFINE ALGEBRAIC SETS 1.6. Algebraic Subsets of A 2 (k) PROPOSITION Let f,g k[x,y] with no common factors. Then V ( f,g) = V ( f ) V (g) is a finite set of points. PROOF. If f and g have no common factors in k[x,y], then they have no common factors in k(x)[y]. Set R = k[x]. Then R is a UFD. Factor f g into irreducibles in R[y]. The irreducible factors in R[y] remain irreducible in k(x)[y]. In k(x)[y], f and g are relatively prime, hence generate the unit ideal. There exist p,q k(x)[y] such that p f + qg = 1. Multiply by some h k[x] to clear the denominators: h = f (hp) + g(hq) Now h k[x] has only finitely many zeros. But for each (α,β) V ( f ) V(g) we have h(α) = 0. There are only finitely many values for α. Similarly there are only finitely many values for β. This proves that V ( f, g) is finite. COROLLARY If f k[x,y] is irreducible and V ( f ) is infinite, then I(V ( f )) = ( f ) and V ( f ) is irreducible. PROOF. Let g I(V ( f )). Then V (g) V ( f ). So V (g) V ( f ) = V ( f ) is infinite. Proposition 1.24 implies f and g have a factor in common. Since f is irreducible, f divides g. This shows I(V ( f )) = ( f ). Since f is irreducible and k[x,y] is a UFD, the principal ideal ( f ) is prime. The last claim follows from Proposition are COROLLARY Say k is an infinite field. The irreducible algebraic subsets of A 2 k (1) A 2 (2) /0 (3) singleton sets {p} (4) irreducible plane curves V ( f ) where f is irreducible and V ( f ) is infinite. PROOF. Let V A 2 be irreducible. Then I(V ) is a prime ideal. If I(V ) = (0), then V = A 2. If V = /0, then I(V ) = k[x,y]. If V is finite and non-empty, then it follows from Lemma 1.6 and Lemma 1.9 that V is a singleton set V = {p}. Otherwise assume V is infinite and I(V ) contains a non-constant polynomial f. Since I(V ) is prime, an irreducible factor of f is in I(V ). Without loss of generality assume f is irreducible. If g I(V ) ( f ) we have I(V ) (g, f ). Therefore V V ( f ) V(g). This is a contradiction to Proposition We conclude that I(V ) = ( f ). COROLLARY Say k is algebraically closed and f k[x,y]. Let f = f n 1 1 f n r r be the unique factorization of f into irreducibles. Then V ( f ) = V ( f 1 ) V ( f r ) is the decomposition of V ( f ) into irreducible components and I(V ( f )) = ( f 1 f 2 f r ). PROOF. Clearly V ( f n i i ) = V ( f i ). Therefore V ( f ) = V ( f n 1 1 ) V ( f n r r ) = V ( f 1 ) V ( f r ). Since k is algebraically closed, Problem 1.10 and Corollary 1.25 imply I(V ( f i )) = ( f i ) and V ( f i ) is irreducible. Since f 1,..., f r are distinct irreducible polynomials, it follows that

21 1.7. HILBERT S NULLSTELLENSATZ 13 V ( f 1 ),...,V ( f r ) are distinct irreducible algebraic sets. Now I(V ( f 1 ) V ( f r )) = I(V ( f 1 )) I(V ( f r )) = ( f 1 ) ( f r ) = ( f 1 f r ) The first equality comes from Lemma 1.8 and the third equality follows since k[x,y] is a UFD Problems. PROBLEM Let k = R. (1) Show that I(V (x 2 + y 2 + 1)) = (1). (2) Show that every algebraic subset of A 2 R is equal to V ( f ) for some f R[x,y] Hilbert s Nullstellensatz Throughout this section k is an algebraically closed field. THEOREM 1.28 (Weak Nullstellensatz). If I is a proper ideal in k[x 1,...,x n ], then V (I) /0. PROOF. Take any maximal ideal M I. Let L = k[x 1,...,x n ]/M. Considering the composite homomorphism of fields (1.1) k k[x 1,...,x n ] L we view L/k as an extension of fields. Since L is a finite extension field of k and k is algebraically closed the next lemma implies k = L. LEMMA Say L/k is an extension of fields such that L is finitely generated as a k-algebra. If k is algebraically closed, then L = k. The proof of Lemma 1.29 will come later. Return to the proof of Theorem The composite (1.1) is an isomorphism. So there are a 1,...,a n k such that a i + M = x i + M for i = 1,...,n. That is, x i a i M for each i. Hence J = (x 1 a 1,...,x n a n ) M. But J is maximal by Problem 1.16, so J = M. Therefore V (M) = V (J) = {(a 1,...,a n )}. THEOREM 1.30 (Hilbert s Nullstellensatz). Let I be an ideal in k[x 1,...,x n ], k algebraically closed. Then I(V (I)) = Rad I. There is a one-to-one correspondence between algebraic sets in A n and radical ideals in k[x 1,...,x n ]. Under this correspondence prime ideals correspond to irreducible sets and maximal ideals correspond to points. PROOF. We already know from Problem 1.15 that RadI I(V (I)). Let g I(V (I)). We will show that g N I for some N. Let f 1,..., f r be a generating set for I. Let J denote the ideal in k[x 1,...,x n,x n+1 ] generated by f 1,..., f r,x n+1 g 1. J = ( f 1,..., f r,x n+1 g 1) k[x 1,...,x n,x n+1 ] Then V (J) A n+1 is empty, since for each point p V (I) = V ( f 1,..., f r ) we have G(p) = 0. By Weak Nullstellensatz (Theorem 1.28), J = (1). Thus (1.2) 1 = A 1 f A r f r + B(x n+1 g 1).

22 14 1. AFFINE ALGEBRAIC SETS for some A 1,...,A r,b in k[x 1,...,x n,x n+1 ]. Let N be the highest power of x n+1 that occurs in all of the A 1,...,A r,b. Set y = 1/x n+1 and multiply both sides of (1.2) by y N to get an equation in k[x 1,...,x n,y] (1.3) y N = C 1 f 1 + +C r f r + D(g y). where D = By N 1 and C i = A i y N. Now consider the evaluation homomorphism ϕ : k[x 1,...,x n,y] k[x 1,...,x n ] defined by ϕ(y) = g. From (1.3) we see that ϕ(y N ) = g N I. COROLLARY Say k is algebraically closed and f k[x 1,...,x n ]. Factor f uniquely into irreducibles f = f e 1 1 f e r r. Then V ( f ) = V ( f 1 ) V ( f r ) and I(V ( f )) = ( f 1 f r ). There exists a one-to-one correspondence between irreducible polynomials f and irreducible hypersurfaces in A n. COROLLARY Say k is algebraically closed and I is an ideal in k[x 1,...,x n ]. Then V (I) is a finite set if and only if k[x 1,...,x n ]/I is a finite dimensional vector space over k. If this occurs, the number of points in V (I) is less than or equal to the k-dimension of k[x 1,...,x n ]/I. PROOF. Assume d = dim k k[x 1,...,x n ]/I is finite. Let {p 1,..., p r } be any finite set of points in V (I). We prove r d. By Problem 1.13 there exist f 1,..., f r in k[x 1,...,x n ] such that { 0 if i j f i (p j ) = 1 if = j Consider the natural homomorphism η : k[x 1,...,x n ] k[x 1,...,x n ]/I. Suppose there exist λ 1,...,λ r k such that λ i η( f i ) = 0. That is, λ i f i I. Therefore for each j, 0 = λ i f i (p j ) = λ j. This proves the set {η( f 1 ),...,η( f r )} is linearly independent, hence r d. Conversely assume V (I) = {p 1,..., p r }. Say p i = (a i1,...,a in ). For j = 1,...,n, set f j = r i=1 (x j a i j ) Then f j is a monic polynomial of degree r in k[x j ]. Viewing each f j as a polynomial in k[x 1,...,x n ] we see that f j (p i ) = 0 for each pair i, j. In particular, for each j, f j I(V (I)). Since I(V (I)) = RadI, for some N > 0 we have f j N I for all j. We can take N large enough to work for all j. Since f j N is monic of degree rn in k[x j ], we can write (1.4) f N j = x rn j + α j1 x rn 1 j + α j2 x rn 2 j +... N Using bars to signify I-residues, f j = 0. From (1.4), x rn j is a k-linear combination of x rn 1 j,..., x j, 1. Let T j denote the k-subspace of k[x 1,...,x n ]/I spanned by S j = { x m j 0 m < rn}. Inductively assume s rn and that x s j T j. Since x s+1 j = x j x s j = x j u, where u T j, it is easy to check that x s+1 j T j. So T j contains all residues of the form x s j. Let T denote the k-subspace of k[x 1,...,x n ]/I spanned by m1 m S = { x 1 x n n 0 m i < rn} Consider an arbitrary monomial v = x e 1 1 xe n n. Check that v T. This proves k[x 1,...,x n ]/I = T is finite dimensional over k.

23 1.8. MODULES, FINITENESS CONDITIONS Problems. PROBLEM Assume k = C. (1) Decompose V (x 2 + y 2 1,x 2 z 2 1) A 3 into irreducible components. (2) Let V = {(t,t 2,t 3 ) t k}. Find I(V ) and prove that V is irreducible. We call V the twisted cubic curve in A 3. PROBLEM Let R be a UFD. (1) Show that a monic polynomial of degree two or three in R[x] is irreducible if and only if it has no root in R. (2) Show that x 2 a R[x] is irreducible if and only if a is not a square in R. PROBLEM Show that V (y 2 x(x 1)(x λ)) A 2 is an irreducible curve for any algebraically closed field k and any λ k. PROBLEM Assume k = C. Let I = (y 2 x 2,y 2 + x 2 ) k[x,y]. Find V (I) A 2 and dim k k[x,y]/i. PROBLEM Let k be an algebraically closed field, V = V (I) A n. Prove that there is a natural one-to-one correspondence between algebraic subsets of V and radical ideals in k[x 1,...,x n ]/I. Under this correspondence prime ideals correspond to irreducible sets and maximal ideals correspond to points. This completes the proof of Theorem Modules, Finiteness Conditions Say R and S are rings and ϕ : R S is a homomorphism of rings. We say S is modulefinite over R in case S is finitely generated as an R-module. In this case, if v 1,...,v n are a set of generators, we write S = Rv i. We say S is ring-finite over R (or is a finitely generated R-algebra) in case there exist v 1...v n in S and the evaluation homomorphism ϕ : R[x 1,...,x n ] S x i v i is surjective. In this case, we write S = R[v 1,...,v n ]. If K and L are fields and K L, we say L is a finitely generated field extension of K in case there exist v 1,...,v n in L and the smallest subfield of L containing K and v 1,...,v n is L itself. We write L = K(v 1,...,v n ) Problems. PROBLEM Let K be a field and L = K(x) the field of rational functions in one variable over K. Show that L is a finitely generated field extension of K. Show that L is not ring-finite over K. (Hint: If L were ring-finite over K, there would be an element b K[x] such that for all z L, b n z K[x] for some n; but let z = 1/c, where c doesn t divide b (Problem 1.5).) PROBLEM Let R be a subring of S and S a subring of T. Show that all of the finiteness conditions are transitive. (1) If S = Rv i and T = Sw j, then T = Rv i w j. (2) If S = R[v 1,...,v n ] and T = S[w 1,...,w m ], then T = R[v 1,...,v n,w 1,...,w m ]. (3) Say K, L and M are fields. If L = K(v 1,...,v n ) and M = L(w 1,...,w m ), then M = K(v 1,...,v n,w 1,...,w m ).

24 16 1. AFFINE ALGEBRAIC SETS 1.9. Integral Elements Let S/R be an extension of rings. If v S we say v is integral over R in case there exists f R[x] such that f is monic and f (v) = 0. If R and S are fields, we usually say v is algebraic over R. PROPOSITION Let S/R be an extension of rings and v S. The following are equivalent. (1) v is integral over R. (2) R[v] is module-finite over R. (3) There is a subring T of S containing R[v] and T is module-finite over R. PROOF. (1) implies (2). Say there exist r n 1,...,r 0 in R such that v n + r n 1 v n r 1 v + r 0 = 0 Then v n = r n 1 v n 1 r 1 v r 0 is in R1 + Rv + + Rv n 1. Inductively suppose s n and v s is in R1 + Rv + + Rv n 1. Then v s+1 = vv s = ρ 0 v + ρ 1 v ρ n 2 v n 1 + ρ n 1 v n = ρ 0 v + ρ 1 v ρ n 2 v n 1 ρ n 1 (r n 1 v n r 1 v + r 0 ) is in R1 + Rv + + Rv n 1. (2) implies (3). Take T = R[v]. (3) implies (1). Say T = Rw Rw n. We show every element of T is integral over R. Left-multiplication by v defines an R-module endomorphism of T. ϕ : T T x vx By [2, Exercise 5.4.4] or [1, Proposition 2.4, page 21], ϕ satisfies an equation over R In particular ϕ(1) = v and ϕ m + a m 1 ϕ m a 1 ϕ + a 0 = 0 v m + a m 1 v m a 1 v + a 0 = 0 so v is integral over R. COROLLARY The set of elements in S integral over R is a subring of S containing R, called the integral closure of R in S. PROOF. Let a and b be elements of S that are integral over R. We need to show that a ± b, ab are integral over R. Since b is integral over R, clearly b is integral over R[a]. Therefore R[a, b] is module-finite over R[a]. By Problem 1.26, R[a, b] is module-finite over R. Since a ± b, ab are in R[a,b], they are integral over R. We say that the ring S is integral over the ring R in case every element of S is integral over R.

25 1.10. FIELD EXTENSIONS Problems. PROBLEM Suppose S is ring-finite over R. Show that S is module-finite over R if and only if S is integral over R. PROBLEM Let K be a subfield of a field L. (1) Show that the set of elements of L that are algebraic over K is a subfield of L containing K. (Hint: if v n + a 1 v n a n = 0, and a n 0, then v(v n ) = a n.) (2) Suppose L is module-finite over K, and K R L, R a subring of L. Show that R is a field Field Extensions Suppose L/K is an extension of fields and that L is generated as a field by v L. Then L = K(v). Let ϕ : K[x] L be the homomorphism that maps x to v. Since K[x] is a PID, the kernel of ϕ is principal, kerϕ = ( f ) for some polynomial f. Since K[v] L is an integral domain, ( f ) is a prime ideal. The diagram looks like this. There are two cases. onto K[x] K[x]/( f ) ϕ x v = L K[v] subring CASE 1. Say f = 0. Then k[x] = k[v] and L = K(v) is isomorphic to K(x). By Problem 1.25 L is not ring-finite over K (or module-finite). So v is not algebraic over K. We say v is transcendental over K. CASE 2. Say f 0. Then K[v] is a domain, so f is an irreducible polynomial and ( f ) is a maximal ideal in K[x]. Therefore K[x]/( f ) = K[v] is a field. Since f (v) = 0, v is algebraic over K and K[v] is finite dimensional over K. We also have K[v] = K(v) = L. PROPOSITION (Zariski) If L is ring-finite over K then L is finite dimensional as a K-vector space and hence L is algebraic over K. PROOF. Suppose that L = k[v 1,...,v n ]. If n = 1, then by Case 2 we are done. Assume inductively that n 2 and that the conclusion is true if n 1 elements generate L (as a ring). Let K 1 = K(v 1 ). Then L = K 1 [v 2,...,v n ]. By induction L/K 1 is finite dimensional. If K 1 = K(v 1 ) is finite dimensional over K, then we are done by transitivity. So assume K(v 1 ) is not finite over K. That is, assume v 1 is transcendental over K and argue until a contradiction is reached. Each v i is algebraic over K(v 1 ), so for some finite set {a i j } K(v 1 ) we have (1.5) v n i i + a i1 v n i 1 i + = 0 for each i. Find a K[v 1 ] which is a multiple of all of the denominators of all of the elements in {a i j }. This is possible because K[v 1 ] is a UFD. Multiply (1.5) by a n i to get (1.6) (av i ) n i + aa i1 (av i ) n i 1 + = 0

26 18 1. AFFINE ALGEBRAIC SETS which is a polynomial equation for av i over the ring K[v 1 ]. Thus av i is integral over the ring K[v 1 ] for each i. Given any z L = K[v 1,...,v n ] we write (1.7) z = λ j v e j 1 1 v e jn n j where the coefficients λ j are in K. Pick N to be the maximum of the degrees of the monomials occurring in (1.7), N = max j {e j1 + + e jn }. Multiply (1.7) by a N to get (1.8) a N z = λ j a f j (av 1 ) e j 1 (av n ) e jn j where f j 0 depends on j. Then (1.8) shows that a N z is integral over K[v 1 ]. In particular, if z K(v 1 ), then there exists N such that a N z is integral over K[v 1 ]. We are close to the contradiction. Find g K[v 1 ] such that g is irreducible and g does not divide a. This is possible by Problem 1.5. Now take z = 1/g K(v 1 ) and by the above, find N such that a N z is integral over K[v 1 ]. Then there is a polynomial equation (1.9) (a N z) r + α 1 (a N z) r 1 + = 0 with each α i K[v 1 ]. Substituting z = 1/g and multiplying by g r yields (a N /g) r + α 1 (a N /g) r = 0 (a N ) r + gα 1 (a N ) r = 0 Since g is irreducible in the polynomial ring K[v 1 ], this implies g divides a, which is a contradiction. PROOF OF LEMMA Let L/k be an extension of fields where k is algebraically closed and L is ring-finite over k. By Proposition 1.35, L is finite dimensional over k. By Proposition 1.33 every element v of L is algebraic over k, hence v k. This shows L = k as desired Problems. PROBLEM Let R be an integral domain with quotient field K, and let L be a finite algebraic extension of K. (1) For any v in L, show that there is a non-zero a R such that av is integral over R. (2) Show that there is a basis v 1,...,v n for L over K (as a vector space) such that each v i is integral over R.

27 CHAPTER 2 Affine Varieties Let k be an algebraically closed field. An affine variety is an irreducible algebraic subset V A n Coordinate Rings Let V be an affine variety in A n. The affine coordinate ring of V is Γ(V ) = k[x 1,...,x n ]/I(V ) Since I(V ) is a prime ideal, Γ(V ) is an integral domain. We can view Γ(V ) as the ring of functions f : V k such that f is the restriction of a polynomial function on A n. If f and g are polynomial functions on A n and f and g define the same function on V, then f g I(V ) and f = ḡ in Γ(V ). Say V is a variety and W V A n. If W is an algebraic subset of V, then I(W) I(V ). There is a one-to-one correspondence between algebraic subsets W V and radical ideals I(W) I(V ). Also there is a one-to-one correspondence between radical ideals I(W) I(V ) and radical ideals I(W) Γ(V ). There is a commutative diagram with exact rows. 0 I(V ) k[x 1,...,x n ] k[x 1,...,x n ]/I(V ) 0 = 0 I(W) k[x 1,...,x n ] k[x 1,...,x n ]/I(W) 0 The kernel of ρ is I(W) and the third column gives rise to the surjective homomorphism of coordinate rings 0 I(W) Γ(V ) Γ(W) 0 So Γ(W) = Γ(V )/I(W). The map ρ is the restriction map. It restricts a polynomial function on V to a polynomial function on W. EXAMPLE 2.1. This is an example of a plane curve which is a finite covering of the affine line. Say f (x,y) k[x,y] is monic in the variable y, (2.1) f (x,y) = y n + p 1 (x)y n p n 1 (x)y + p n (x) Let V = V ( f ) A 2. Let ϕ be defined by the composition of the other two homomorphisms ϕ k[x] Γ(V ) = k[x,y]/( f ) ρ ρ k[x, y] 19

28 20 2. AFFINE VARIETIES Given any polynomial g(x) k[x], it is clear that f does not divide g. Therefore ϕ : k[x] Γ(V ) is an extension of rings. This extension is module-finite because y generates Γ(V ) over k[x] and y is integral because y satisfies the monic polynomial (2.1). As a module over k[x], Γ(V ) is generated by 1,y,...,y n 1. The map ϕ on coordinate rings works in this way. Consider a point P = (a,b) in V A 2, and a polynomial g(x) in k[x]. Then ϕ(g) simply acts on the x-coordinate of P. Equivalently, if π : A 2 A 1 is the projection onto the x-axis, then ϕ(g(a,b)) = g(π(a,b)). Under π the curve V is being projected onto the affine line. EXAMPLE 2.2. Using the notation of Example 2.1, let f = y 2 x k[x,y] and V = V ( f ). Then and ϕ : k[x] k[y] is defined by x y 2. x y 2 = Γ(V ) k[x,y]/(y 2 x) k[y] = π FIGURE 2.1. Mapping V (y 2 x) onto A Problems. PROBLEM 2.1. Let V A n be a variety. Show that there is a natural one-to-one correspondence between subvarieties of V and prime ideals of Γ(V ). Show that under this correspondence points of V correspond to maximal ideals of Γ(V ). PROBLEM 2.2. Let V A n be a variety. Show that the following are equivalent: (1) V is a point. (2) Γ(V ) = k. (3) dim k Γ(V ) <.

29 2.2. POLYNOMIAL MAPS Polynomial Maps Given T 1,...,T m k[y 1,...,y n ], let T = (T 1,...,T m ). Then p T (p) defines a polynomial map T A n A m p (T 1 (p),...,t m (p)) The associated k-algebra homomorphism is defined by mapping x i to T i for each i T k[x 1...,,x m ] k[y 1,...,y n ] x i T i For any f k[x 1,...,x m ], note that T ( f ) is the composition T followed by f : T ( f ) = f T. Any k-algebra homomorphism k[x 1...,x m ] k[y 1,...,y n ] is completely determined by the images of x 1...,x m. So there is a one-to-one correspondence between polynomial maps T : A n A m and k-algebra homomorphisms T : k[x 1...,x m ] k[y 1,...,y n ]. Let V A n and W A m be algebraic varieties. Say ϕ : V W is a function such that ϕ = T V for some polynomial map T. ϕ V W A n T A m Given f I(W) and v V we have ( T ( f ))(v) = f (T (v)) = 0, since T (v) = ϕ(v) W. Therefore T (I(W)) I(V ) and ϕ induces a k-algebra homomorphism ϕ : Γ(W) Γ(V ). The following diagram commutes and has exact rows. 0 I(W) k[x 1,...,x m ] Γ(W) 0 T 0 I(V ) k[y 1,...,y n ] Γ(V ) 0 Conversely, let α : Γ(W) Γ(V ) be a k-algebra homomorphism. We lift α to a k- algebra homomorphism T : k[x 1...,x m ] k[y 1,...,y n ] in the following way. For each i = 1,...,m choose T i in k[y 1,...,y n ] such that T i + I(V ) = α(x i + I(W)). Define T by x i T i. The diagram I(W) k[x 1,...,x m ] Γ(W) T I(V ) k[y 1,...,y n ] Γ(V ) commutes. Set T = (T 1,...,T m ). Then T : A n A m is a polynomial function. If f I(W) and v V, then f (T (v)) = T ( f )(v) = 0 since T ( f ) I(V ). If we set ϕ = T V, then ϕ(v ) ϕ α

30 22 2. AFFINE VARIETIES W. As a function on V, ϕ does not depend on the choices for the T i. The diagram ϕ V W commutes. We have proved the following. A n T A m PROPOSITION 2.3. There is a one-to-one correspondence between k-algebra homomorphisms α : Γ(W) Γ(V ) and polynomial maps T : V W. A polynomial map ϕ : V W is an isomorphism if there exists a polynomial map ψ : V W such that ϕ and ψ are inverses. In this case we say V and W are isomorphic and write V = W. By Proposition 2.3, V = W if and only if Γ(W) = Γ(V ). The composition of polynomial maps is a polynomial map. This is pretty clear, isn t it? EXAMPLE 2.4. Say n m. We have the projection pr : A n A m defined by pr(a 1,...,a n ) = (a 1,...,a m ). The associated homomorphism on coordinate rings is pr: k[x 1,...,x m ] k[y 1,...,y n ] where pr(x i ) = y i. Note pr is a monomorphism. EXAMPLE 2.5. An example of a closed immersion. Let V A n. The natural epimorphism k[x 1,...,x n ] Γ(V ) induces a one-to-one polynomial map V A n which is the restriction to V of the identity function on A n. EXAMPLE 2.6. Consider ϕ : A 1 A 2 defined by ϕ(t) = (t 2,t 3 ). The image of ϕ is the variety W = V (y 2 x 3 ). So ϕ is a parametrization of the cuspidal cubic curve W as shown in Figure 2.2. Check that ϕ is one-to-one. Assume ϕ(a) = ϕ(b). Then a 2 = b 2 and a 3 = b 3. So a = ζ 2 b and a = ζ 3 b where ζ2 2 = 1 and ζ 3 3 = 1. This implies a = b. Check that ϕ is onto. Let (a,b) satisfy a 3 = b 2. Let t be any solution to t 2 = a. Then ϕ(t) = (t 2,t 3 ) = (a,a 3/2 ) = (a,±b). Either ϕ(t) = (a,b) or ϕ( t) = (a,b). However, ϕ is ϕ FIGURE 2.2. Mapping A 1 onto y 2 = x 3 not an isomorphism. As k[x,y] k[t] is induced by x t 2 and y t 3, the diagram k[x,y] Γ(W) = k[x,y]/(y 2 x 3 ) T k[t] = ϕ k[t]

31 2.3. COORDINATE CHANGES 23 commutes. The image of ϕ is k[t 2,t 3 ] which is a proper subring of k[t]. EXAMPLE 2.7. Consider ϕ : A 1 A 2 defined by ϕ(t) = (t 2 1,t(t 2 1)). It is easy to prove that the image of ϕ is the variety W = V (y 2 x 2 (x + 1)). Since ϕ(1) = ϕ( 1) we see that ϕ is not one-to-one. The curve W is called a nodal cubic curve W, the real points on W are shown in Figure 1.3 on page 3. EXAMPLE 2.8. If T : A n A m is a polynomial map and V is an algebraic set, is W = T (V ) an algebraic set? Consider this example. Take for T the projection pr : A 2 A 1 onto the first coordinate pr(x,y) = (x). For V take the curve V (xy 1). Then T (V ) is the subset W = A 1 0. A proper algebraic subset of A 1 is a finite set of points (Problem 1.8). So, the answer is no Problems. PROBLEM 2.3. If ϕ : V W is a surjective polynomial map, and X W is an algebraic subset, show that ϕ 1 (X) is an algebraic subset of V. If ϕ 1 (X) is irreducible, show that X is irreducible. This gives a useful test for irreducibility Coordinate Changes If T = (T 1,...,T m ) is a polynomial map from A n to A m, then there is an induced map on polynomial rings k[x 1,...,x m ] T k[y 1,...,y n ] where T ( f ) = f (T 1,...,T m ). Let V A m be an algebraic set with ideal I = I(V ). The set of polynomials T (I) defines an algebraic set in A n. One can check that as a set V ( T (I)) is equal to T 1 (V ). If V = V ( f ) is a hypersurface in A m, then T 1 (V ) = V ( T ( f )) is a hypersurface in A n, unless T ( f ) is a unit. An affine change of coordinates is a polynomial map T : A n A n, T = (T 1,...,T n ) where each T i is a linear polynomial (i.e. degt i = 1) and such that T is a bijection. If T i = a i1 x a in x n + a i0, then we can write T as the composition of a linear map with a translation, T = T T, where T i = a i1 x a in x n and T = x i + a i0. A translation is always a bijection. Since T is invertible, T is invertible and T is an isomorphism of A n with itself Problems. PROBLEM 2.4. A set V A n is called a linear subvariety of A n in case V =V ( f 1,..., f r ) for some polynomials f i of degree 1. (1) Show that if T is an affine change of coordinates on A n, then the inverse image of V under T, T 1 (V ), is also a linear subvariety of A n. (2) If V /0, show that there is an affine change of coordinates T of A n such that T 1 (V ) = V (x m+1,...,x n ). (Hint: use induction on r.) So V is a variety. (3) Show that the m which appears in part (2) is independent of the choice of T. It is called the dimension of V. V is then isomorphic (as a variety) to A m. (Hint: Suppose there were an affine change of coordinates T such that T 1 (V (x m+1,...,x n )) = V (x s+1,...,x n ), m < s; show that T m+1,...,t n would be dependent.) PROBLEM 2.5. Let P = (a 1,...,a n ), Q = (b 1,...,b n ) be distinct points of A n. The line through P and Q is defined to be {(a 1 +t(b 1 a 1 ),...,a n +t(b n a n )) t k}

32 24 2. AFFINE VARIETIES (1) Show that if L is the line through P and Q, and T is an affine change of coordinates, then T (L) is the line through T (P) and T (Q). (2) Show that a line is a linear subvariety of dimension one, and that a linear subvariety of dimension one is the line through any two of its points. (3) Show that in A 2 a line is the same thing as a hyperplane (that is, a hypersurface V ( f ) where deg f = 1). (4) Let P, P be points in A 2, L 1, L 2 two distinct lines through P, and L 1, L 2 distinct lines through P. Show that there is an affine change of coordinates T of A 2 such that T (P) = P and T (L i ) = L i, i = 1, Rational Functions and Local Rings Let V be an irreducible variety in A n k. So the affine coordinate ring, Γ(V ), is an integral domain. The quotient field of Γ(V ) is called the field of rational functions on V and is denoted K(V ). If f K(V ) and P V, we say f is defined at P if there exist a,b Γ(V ) such that f = a/b and b(p) 0. In this case, the value of f at P is f (P) = a(p)/b(p). Check that f (P) does not depend on the choices of a and b hence f is a function from the domain U = {P V f is defined at P} to the field k. EXAMPLE 2.9. Consider the polynomial wx yz in k[w,x,y,z] which one can check is irreducible. The quadric hypersurface V = V (wx yz) in A 4 is a variety and Γ(V ) = k[w,x,y,z]/(wx yz) is an integral domain. Denote residues modulo (wx yz) by putting a bar over the function name. So w, x,ȳ, z are the residues of w,x,y,z in Γ(V ). In Γ(V ) we have w x = ȳ z, so that Γ(V ) is an example of an integral domain that is not a UFD. Consider the function f = x/ȳ = z/ w K(V ). Then f is defined at a point P = (α,β,γ,δ) V if α 0 or γ 0. Let V be a variety and P a point in V. The local ring of V at P is Clearly O P (V ) is a ring and O P (V ) = { f K(V ) f is defined at P}. k Γ(V ) O P (V ) K(V ). The pole set of f is {P V f is not defined at P}. PROPOSITION Let V be a variety in A n and f a rational function on V. (1) The pole set of f is an algebraic subset of V. (2) Γ(V ) = O P (V ) P V PROOF. Let f K(V ) and J = {g Γ(V ) g f Γ(V )}. Clearly J is an ideal in Γ(V ). Let η : k[x 1,...,x n ] Γ(V ) be the natural homomorphism. Lift J to an ideal J = η 1 ( J) in k[x 1,...,x n ] containing I(V ). Then V (J) is an algebraic subset of V (Lemma 1.5). Let P V (J). Given a representation f = ā/ b, it follows that b f = ā Γ(V ), so b J. Lift b to get b J. Then b(p) = 0, which implies f is not defined at P. Conversely if P V (J), then there exists b J such that b(p) 0 and b f Γ(V ) so f is defined at P. This proves the pole set of f is V (J). Now let f P V O P (V ). Then f is defined at every P in V, hence the pole set of f is the empty set. Set J = {g k[x 1,...,x n ] ḡ f Γ(V )}. By part (1), V (J) is the pole set of f. Theorem 1.28 says that J is the unit ideal. If 1 J, then f Γ(V ).

33 2.5. DISCRETE VALUATION RINGS 25 Fix P V and consider O P (V ). Evaluation at P defines a homomorphism and the kernel is denoted m P (V ): f f (P) 0 m P (V ) O P (V ) k 0 So m P (V ) is a maximal ideal and f m P (V ) if and only if f (P) is zero. If f (P) 0, then f is a unit in O P (V ). So m P (V ) is the set of all non-units. Every proper ideal of O P (V ) is contained in m P (V ), hence O P (V ) is a local ring with maximal ideal m P (V ). In the same way, we have the evaluation at P homomorphism on Γ(V ): f f (P) 0 M P (V ) Γ(V ) k 0 where M P (V ) is the maximal ideal consisting of those f such that f (P) = 0. It is now obvious that O P (V ) is the localization of Γ(V ) at the maximal ideal M P (V ). Since Γ(V ) is noetherian and every ideal of O P (V ) is the localization of an ideal of Γ(V ), it follows that O P (V ) is noetherian. EXAMPLE Let V = V (wx yz) be the quadric hypersurface of Example 2.9. Consider the function f = x/ȳ = z/ w K(V ). Then f is defined at a point P = (α,β,γ,δ) V if α 0 or γ 0. The pole set of f is the subvariety W = {(w,x,y,z) w = y = 0}. Then I(W) = (w,y) and modulo wx yz, the ideal I = ( w,ȳ) is a prime ideal since Γ(V )/( w,ȳ) = k[x,z] is an integral domain Problems. PROBLEM 2.6. Let ϕ : V W be a polynomial map of affine varieties, ϕ : Γ(W) Γ(V ) the induced map on coordinate rings. Suppose P V, ϕ(p) = Q. Show that ϕ extends uniquely to a ring homomorphism (also written ϕ) from O Q (W) to O P (V ). (Note that ϕ may not extend to all of K(W).) Show that ϕ(m Q (W)) m P (V ). PROBLEM 2.7. Let T : A n A n be an affine change of coordinates. If P A n and T (P) = Q, show that T : O Q (A n ) O P (A n ) is an isomorphism. If V A n is a variety and P T 1 (V ), show that T induces an isomorphism from O Q (V ) to O P (T 1 (V )) Discrete Valuation Rings PROPOSITION Let R be a domain which is not a field. The following are equivalent. (1) R is noetherian, local, and the maximal ideal is principal. (2) There is an irreducible element t in R such that every non-zero element z in R may be written uniquely in the form z = ut n, u R, n 0. PROOF. (1) implies (2): Let M be the maximal ideal, M = Rt. For the uniqueness claim, assume ut n = vt m, n m. Then ut n m = v. So t n m is a unit, hence n = m and u = v. For the existence claim, assume z is not a unit. Then z M. So z = z 1 t for some z 1 R. If z 1 is a unit, we are done. Else z 1 = z 2 t, z 2 R. Iterate to get z = z 1 t = z 2 t 2 =... The chain of ideals (z) (z 1 ) (z 2 )... becomes constant because R is noetherian. Thus for some n, (z n ) = (z n+1 ). Hence there exists a unit v such that z n = z n+1 v. If z n is not a unit, then z n = z n+1 t implies z n+1 v = z n+1 t which implies t = v is a unit. This is a contradiction. Therefore z n is a unit and z = z n t n.