Math 203A - Solution Set 4

Math 203A - Solution Set 4 Problem 1. Let X and Y be prevarieties with affine open covers {U i } and {V j }, respectively. (i) Construct the product prevariety X Y by glueing the affine varieties U i V j together. (ii) Show that there are projection morphisms π X : X Y X and π Y : X Y Y satisfying the usual universal property for products: given morphisms f : Z X and g : Z Y from any prevariety Z, there is a unique morphism h : Z X Y such that f = π X h and g = π Y h. (iii) Show that if X and Y are varieties then X Y is also a variety. Answer: (i) We have already seen that products of affine algebraic sets are well defined. Thus, by restriction, it follows that products of quasi-affine sets are well defined. Let U ij = U i U j and V ij = V i V j. We glue the products of affine algebraic sets U i V k and U j V l along the products of quasiaffine sets U ij V kl. It is clear that the cocycle condition on triple overlaps for the gluing maps is satisfied. (ii) It is clear that the projections π X and π Y are morphisms. Assume f, g, Z are given. Set theoretically, we define h(z) = (f(z), g(z)) : Z X Y. We need to check that h is a morphism. Clearly h is continuous. We check that if φ is a regular function on some open set W X Y then h φ is regular on h 1 W. Let Z ij be the preimages of U i V j under the morphism h. Clearly, Z ij is open, hence it is a prevariety, hence it can be covered by affines Zij l. Since the restriction h : Wij l U i V j is a morphism of affine sets, it must correspond to a polynomial map. Thus h φ is clearly regular on h 1 (W ) Zij l e.g. it gives rise to a section of the sheaf O Z (h 1 (W ) Zij l ). These sections agree on overlaps and give a regular section of O X (h 1 (W )) which must be equal to h φ. This proves the claim. It follows from the universal property that X Y is unique up to isomorphism. (iii) If h 1, h 2 : Z X Y, we obtain morphisms We have f 1, f 2 : Z X, g 1, g 2 : Z Y. {z Z : h 1 (z) = h 2 (z)} = {z Z : f 1 (z) = f 2 (z)} {z Z : g 1 (z) = g 2 (z)}. The latter two sets are closed since X and Y are varieties, hence their intersection is also closed as well, showing that X Y is a variety. Problem 2. Let X be a prevariety. Consider pairs (U, f) where U is an open subset of X and f O X (U) a regular function on U. We call two such pairs (U, f) and (U, f ) equivalent if there is an open subset V in X with V U U such that f V = f V. 1

2 (i) Show that this defines an equivalence relation. (ii) Show that the set of all such pairs modulo this equivalence relation is a field. It is called the field of rational functions on X and denoted K(X). (iii) If X is an affine variety, show that K(X) is just the field of rational functions as defined in class. Answer: (i) Checking that we have defined an equivalence relation is straightforward. For instance, to verify transitivity, let (U, f), (U, f ) and (U, f ) such that f V = f V and f W = f W for some open sets V U U and W U U. We conclude f T = f T where T = V W U U. Clearly, T is nonemtpy since V, W are open dense in the irreducible set X. (ii) It is easy to see that addition and multiplication pass through the equivalence relation. We will check that non-zero elements have inverses. Let (f, U) be a non-zero element. In particular Z(f) U. Consider the function (1/f, U \ Z(f)) which is regular on the nonempty set U \ Z(f). This is the inverse of (f, U). (iii) If f is a rational function as defined in class, we take U = Domain (f) which is open in X. We obtain an equivalence class (U, f) as in the problem. Conversely, if (f, U) is an equivalence class as in the problem set, then f is regular on U hence by definition, it must be a rational function on X. This assignment (f, U) f passes through the equivalence relation. Indeed, if (f, U ) is another representative of the equivalence class, then f = f on some open set V. Write f = g/h and f = g /h. We show that g/h and g /h define the same element of the function field. Indeed, g h = g h on V = gh g h = 0 on V = gh g h = 0 on X = g h = g as elements of K(X). h For the third implication, we used that V is dense in X. Problem 3. (i) Show that every isomorphism f : A 1 A 1 is of the form f(x) = ax + b. (ii) Show that every isomorphism f : P 1 P 1 is of the form f(x) = ax+b cx+d for some a, b, c, d k, where x is an affine coordinate on A 1 P1. (iii) Given three distinct points P 1, P 2, P 3 P 1 and three distinct points Q 1, Q 2, Q 3 P 1, show that there is a unique isomorphism f : P 1 P 1 such that f(p i ) = Q i for i = 1, 2, 3. Answer: (i) Clearly, a morphism f : A 1 A 1 must be a polynomial map. If f has degree d, the equation f(x) = 0 will have d roots. Therefore, d = 1 hence f(x) = ax + b.

(ii) Let f : P 1 P 1. If f( ) =, then f maps A 1 A 1 and must have the form f(x) = ax + b. Otherwise, if f( ) = λ, let g(x) = 1 x λ. Then g f is an isomorphism which maps to itself hence it must have the form g f = cx + d. This shows f(x) = λ + 1 cx + d = ax + b cx + d for a = cλ, b = dλ + 1. (iii) By transitivity, it suffices to assume P 1 = 0, P 2 = 1, P 3 =. Also, let Q 0 = λ, Q 1 = µ, Q 2 = ν. Assume that none of the greek letters is. We require We may take b d = λ, a + b c + d = µ, a c = ν. a = ν(µ λ), b = λ(ν µ), c = µ λ, d = ν µ. If one of the greek letters is, say ν =, and µ, λ 0, we may apply the transformation x 1 x to reduce to the case we have already studied. If by contrast, λ = 0, then the automorphism f(x) = µx sends the P s to the Q s. If µ = 0, then we may take f(x) = λ(x 1). To prove uniqueness, assume that two morphisms f 1, f 2 have been constructed. The inverse f = f 1 f2 1 has 3 fixed points P 1, P 2, P 3. We may assume that these fixed points are 0, 1,. Since and we conclude f(x) = ax + b cx + d f(0) = 0, f(1) = 1, f( ) = b = 0, a + b = c + d, c = 0 = f = id = f 1 = f 2. Problem 4. Let X P 2 be a projective variety. A morphism f : P 1 X is a polynomial map f([x : y]) = (f 0 ([x : y]), f 1 ([x : y]), f 2 ([x : y])), where f 0, f 1, f 2 are homogeneous polynomials of the same degree, such that f(p 1 ) X. Prove the following facts about lines and conics in projective plane: (i) For any line L P 2, there is a bijective morphism f : P 1 L. (ii) For any irreducible conic C P 2, there is a bijective morphism f : P 1 C. You may wish to change coordinates so that your conic has a convenient expression. 3

4 (iii) Consider the elliptic curve E λ P 2 : y 2 z = x(x z)(x λz). Show that there are no nonconstant morphisms P 1 E λ P 2. Therefore, elliptic curves are not rational curves. Answer: (i) Let L be the line αx+βy+γz = 0. We may assume that γ 0, eventually relabeling the coordinates if necessary. Set f : P 1 L, [x : y] [x : y : αγ x βγ ] y, and g : L P 1 [x : y : z] [x : y]. It is easy to see that both f and g are well defined inverse isomorphisms. (ii) Changing coordinates, we may assume the conic C is given by the equation Set It is easy to check that xz = y 2. f : P 1 C, [s : t] [s 2 : st : t 2 ]. g : C P 1, g([x : y : z]) = is an inverse morphism of f. (iii) Assume that F : P 1 E λ P 2 is a morphism with where { [x : y] if (x, y) (0, 0) [y : z] if (y, z) (0, 0) F ([x : y]) = [f 0 ([x : y]) : f 1 ([x : y]) : f 2 ([x : y])], f 2 1 f 2 = f 0 (f 0 f 1 )(f 0 λf 1 ). If f 2 0, it follows from the equation above that f 0 = 0, hence F is the constant [0 : 1 : 0]. Let us assume f 2 0. Define ( f0 (t, 1) f(t) = f 2 (t, 1), f ) 1(t, 1). f 2 (t, 1) It is clear that f defines a rational map f : A 1 which must be constant by the previous pset. This implies that f 0 (t, 1)/f 2 (t, 1) and f 1 (t, 1)/f 2 (t, 1) are constants. It follows that all points [t : 1] take the same value under F. Similarly, F sends all [1 : t] to the same value, and therefore F must be constant. E λ

Problem 5. Part A. Show that a line and an irreducible conic in P 2 cannot intersect in 3 points. Part B. (i) Four points in P 2 are said to be in general position if no three are collinear (i.e. lie on a projective line in the projective plane). Show that if p 1,..., p 4 are points in general position, there exists a linear change of coordinates with T : P 2 P 2 T ([1 : 0 : 0]) = p 1, T ([0 : 1 : 0]) = p 2, T ([0 : 0 : 1]) = p 3, T ([1 : 1 : 1]) = p 4. (ii) Given five distinct points in P 2, no three of which are collinear, show that there is an unique irreducible projective conic passing though all five points. You may want to use part (i) to assume that four of the points are [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1]. (iii) Deduce that two distinct irreducible conics in P 2 cannot intersect in 5 points. (We will see later that they intersect in exactly 4 points counted with multiplicity.) Answer: Part A. Changing coordinates, we may assume that the conic C is xz = y 2. Let L be the line αx + βy + γz = 0. If β = 0, it is clear that C and L intersect at the points If β 0, then which gives [ γ : ± αγ : α]. y = α β x γ β z xz = 1 β 2 (αx + γz)2. If x = 0, then y = z = 0 which is not allowed. Therefore, we may assume x 0. Dividing by x 2 we obtain the quadratic equation in z x : γ 2 ( z ) ( ) 2 2αγ z + β 2 x β 2 1 x + α2 β 2 = 0. Letting λ 1, λ 2 be the solutions of this equation, we see that the intersection points are [1 : αβ γβ λ i : λ i ]. Part B. (i) Let p i = [a i : b i : c i ] for 1 i 4. Define A = αa 1 αb 1 αc 1 βa 2 βb 2 βc 2, γa 3 γb 3 γc 3 where α, β, γ will be specified later. In fact, we will require that α, β, γ solve the system αa 1 + βb 1 + γc 1 = a 4, 5

6 αa 2 + βb 2 + γc 2 = b 4, αa 3 + βb 3 + γc 3 = c 4. A solution exists since the matrix of coefficients B = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 is invertible. Indeed, the rows of B are independent. Otherwise, a nontrivial linear relation between the rows would give a line on which the points p 1, p 2, p 3 lie. Thus B is invertible. Now, the system above has the solution α β γ = B 1 a 4 b 4 c 4 Note that the same argument shows that A is invertible. Let S : P 2 P 2 be the linear transformation defined by A. Then S is invertible. A direct computation shows S([1 : 0 : 0]) = p 1, S([0 : 1 : 0]) = p 2, S([0 : 0 : 1]) = p 3, S([1 : 1 : 1]) = p 4. The proof is completed letting T be the inverse of S. (ii) After a linear change of coordinates, we may assume that the five points are [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1] and [u : v : w]. The equation f(x, y, z) of any conic passing through the first three points can t contain x 2, y 2, z 2, so. f(x, y, z) = ayz + bxz + cxy. The remaining two points impose the conditions Letting a + b + c = avw + buw + cuv = 0. A = ( 1 1 1 vw uw uv we see that (a, b, c) must be in the null space of A. The rank of A is 2 (the rank cannot be 1 since then the rows would be proportional, hence u = v = w which is not allowed). Therefore, the null space of this matrix is one dimensional, hence the conic passing through the 5 points is unique. The conic cannot be reducible since then it would be union of two lines. One of the lines would have to contain 3 points but that contradicts the general position assumption. (iii) By Part A, a conic and a line cannot intersect in 3 points. Therefore, any 5 points on an irreducible conic are in general position. Now, the claim is evident by (ii). Problem 6. We will make the space of all lines in P n into a projective variety. We define a set-theoretic map φ : {lines in P n } P N ),

7 with ( ) n + 1 N = 1 2 as follows. For every line L P n, choose two distinct points P = (a 0... a n ) and Q = (b 0... b n ) on L and define φ(l) to be the point in P N whose homogeneous coordinates are the maximal minors of the matrix ( ) a0 :... : a n in any fixed order. Show that: b 0 :... : b n (i) The map φ is well-defined and injective. The map φ is called the Plucker embedding. (ii) The image of φ is a projective variety that has a finite cover by affine spaces A 2(n 1). You may want to recall the Gaussian algorithm which brings almost any matrix as above into the form ( 1 0 a 2... a ) n 0 1 b 2.... b n (iii) Show that G(1, 1) is a point, G(1, 2) = P 2, and G(1, 3) is the zero locus of a quadratic equation in P 5. Answer: (i) Let e 0,..., e n be the standard basis of the vector space V = k n+1. A line L corresponds to a 2-dimensional subspace in k n+1, also denoted by L. We claim that the map φ can be described as follows. Picking a basis v, w for the subspace, φ(l) = [v w] P(Λ 2 V ) = P N. Indeed, if P = (a 0 :... : a n ) and Q = (b 0 :... : b n ) are two distinct points, then we may take Thus v w = i,j v = a i e i, w = b i e i. a i b j e i e j = i<j (a i b j a j b i )e i e j. Therefore, in the basis e i e j, the coordinates are the 2 2-minors of the matrix in (i). This shows that φ is well-defined. To check injectivity, let L be another line corresponding to a 2-dimensional subspace. If L L = 0, then pick a basis v 0, v 1, v 2, v 3 for L L with v 0, v 1 L, v 2, v 3 L, and extend it to a basis v 0,..., v n of V = k n+1. Thus v i v j for i < j is a basis for Λ 2 V. In particular, v 0 v 1 and v 2 v 3 are not multiples, or equivalently φ(l) φ(l ). The same argument applies if L and L have a 1 dimensional intersection.

8 (ii) To show projectivity, we will prove that then ω ω = In particular, if i<j,k<l ω Λ 2 V splits as ω = v w if and only if ω ω = 0. ω ij ω kl e i e j e k e l = ω = ω ij e i e j, i<j<k<l so the image of φ is cut by the quadrics ω ij ω kl ω ik ω jl + ω il ω jk = 0. (ω ij ω kl ω ik ω jl + ω il ω jk )e i e j e k e l We now prove the claim. It is clear that if ω = v w then ω ω = 0. Conversely, we will induct on n, the base case n = 2 being clear. Let us write ω = e 0 η + ω where ω, η do not contain the vector e 0. Thus This implies that hence by induction 0 = ω ω = 2e 0 η ω + ω ω. ω ω = 0 ω = v w, with v, w being in the subspace spanned by e 1,..., e n. Also, we know e 0 η ω = 0 = η v w = 0. This shows that η cannot be independent of v, w hence Collecting terms we find η = av + bw. ω = e 0 (av + bw) + v w = (v + be 0 ) (w + ae 0 ) as claimed. For the last part, note that one of the coordinates of φ(l) must be non-zero. Without loss of generality let us assume it is the coordinate corresponding to e 0 e 1. This means that the first 2 2 minor of the matrix ( ) a0 :... : a n b 0 :... : b n is non-zero. The Gaussian algorithm brings this matrix into the form ( 1 0 a 2... a ) n 0 1 b 2.... b n The association L (a 2,..., a n, b 2,..., b n) A 2(n 1) shows how to cover the image of φ by affine opens isomorphic to A 2(n 1).

(iii) All these statements are particular cases of what we proved in part (i). For instance, to see that G(1, 3) is a quadric in P 5 given by x 0 x 5 x 1 x 4 + x 2 x 3 = 0. 9