Math 203, Solution Set 4.
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1 Math 203, Solution Set 4. Problem 1. Let V be a finite dimensional vector space and let ω Λ 2 V be such that ω ω = 0. Show that ω = v w for some vectors v, w V. Answer: 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 where ω, η do not contain the vector e 0. Thus This implies that hence by induction ω = e 0 η + ω 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 as claimed. η = av + bw. ω = e 0 (av + bw) + v w = (v + be 0 ) (w + ae 0 ) Problem 2. The set X of degree d homgeneous polynomials in n + 1 variables can be identified with a projective space P N, by recording the coefficients in some order. What is N? Show that the set of reducible polynomials form a closed subset of X. Answer: The space V d of degree d polynomials in n + 1 variables has dimension ( ) n+d d. Consider the morphism φ k : P(V k ) P(V d k ) P(V d ), (f, g) f g. Clearly, φ k is a morphism. This can be seen by writing which shows f = a I z I, g = b J z J = f g = K φ k (a I, b J ) = (c K ), c K = 1 I+J=K ( I+J=K a I b J. a I b J ) z K
2 This is a morphism. By the main theorem of projective geometry, Y k = Image φ k is closed. The reducible polynomials are given as This set is therefore also closed in X. Y = d 1 k=1 Y k. Problem 3. Show that P 1 A 1 and P 2 \ {x} are neither affine nor projective. Answer: Write t for the coordinate on A 1. We claim that the regular functions on P 1 A 1 are polynomials f(t). This will show that P 1 A 1 is not projective, because projective varieties only have constants as regular functions. It also shows X = P 1 A 1 is not affine since it were affine, its coordinate ring would be A(X) = k[t] = A(A 1 ). Then X A 1, but this is clearly impossible for dimension reasons. Indeed, let U, V be two affine opens covering P 1. We have U A 1, V A 1 with coordinates z, w and w = 1 z over overlaps. Let φ be a regular function on P1 A 1. Then φ is regular on U A 1 A 2 so φ = p(z, t) for some polynomial p. Similarly, φ is regular on V A 1 A 2 so φ = q(w, t) for some polynomial q. Over (U V ) A 1 we must have p(z, t) = q( 1 z, t). The powers of z on the left have nonnegative exponents, while the powers of z on the right have nonpositive exponents, so the exponents must be 0. Thus, p(z, t) = q( 1, t) = f(t) = φ = f(t) z for some polynomial f. For Y = P 2 \ {x}, we claim that all regular functions are constant. This will show that Y cannot be affine because A(Y ) = k = A(point) = Y point 2
3 which is clearly impossible for dimension reasons. Indeed, if φ is regular on P 2 \ {x}, then consider the restriction of φ to U = A 2 \ {0}. This extends to a regular function on A 2 by the removable singularity theorem. Thus φ extends to a regular function on P 2, showing then that φ must be constant. To see Y is not projective, assume otherwise. Let L = {l = 0} be a line in P 2 through x. Then, Z = L\{x} is closed in Y so it must be projective. This is not true since Z = L\{x} A 1. Z admits nonconstant regular functions, so it cannot be projective. This contradiction shows that Y is not projective. Problem 4. (Joins.) Let G(1, n) be the Grassmannian of lines in P n as in the previous homework. Show that: (i) The set {(L, P ) : P L} G(1, n) P n is closed. (ii) If Z G(1, n) is any closed subset then the union of all lines L P n such that L Z is closed in P n. (iii) Let X, Y P n be disjoint projective varieties. Then the union of all lines in P n intersecting X and Y is a closed subset of P n. It is called the join J(X, Y ) of X and Y. Answer: (i) We let J = {(P, L) : P L} P n G(1, n). We will think of lines L in terms of their Plucker coordinates z ij = a i b j a j b i where a, b are two points on L with a = [a 0 :... : a n ], b = [b 0 :... : b n ]. In fact, it will be useful to form the vectors a = a i e i, b = b i e i. Similarly, a point P P n will have an associated vector p = p i e i. Now, if P L, then p = sa + tb hence p a b = 0. Then ( ) ( ) ( ) ( ) pi e i ai e i bi e i = pi e i z jk e j e k j<k = (p i z jk p j z ik + p k z ij )e i e j e k. i<j<k 3
4 The conclusion is that J is defined by the equations p i z jk p j z ik + p k z ij = 0 which are bihomogeneous in the variables. Thus, J is projective. (ii) Let p : J G(1, n), q : J P n be the natural projections. Then, for any Z closed in G(1, n), the preimage p 1 (Z) is also closed. Thus q(p 1 (Z)) is closed by the main theorem of projective varieties. This set consists in points P lying on lines L such that L Z, hence it can be identified with the union of all lines in Z. (iii) We let A be the set of lines intersecting X and B be the set of lines intersecting Y. We show A and B are closed in G(1, n), hence so is Z = A B. The join J(X, Y ) is simply the union of lines contained in Z hence it must be closed in P n by item (ii). It suffices to prove A is closed in G(1, n). Indeed, we can think of A as the projection under p of the set {(P, L) : P L} X G(1, n)) = J q 1 (X). Hence A = p(j q 1 (X)) which is closed because p is closed and q is continuous. Problem 5. (Rational varieties.) The definition of birational isomorphisms given in class extends to the projective category. Two projective varieties X and Y are birational if there are rational maps f : X Y, g : Y X. which are rational inverses to each other. Just as in the affine case, a birational isomorphism f : X Y induces an isomorphism of the fields of rational functions f : K(Y ) K(X). (i) Explain that if X is rational, then K(X) = k(t 1,..., t n ). (ii) Show that P n P m is rational, by constructing an explicit birational isomorphism with P n+m. Show that if X and Y are rational, then X Y is rational. (iii) Show that P 2 is not isomorphic to P 1 P 1. 4
5 (iv) The group of automorphisms of the field of fractions in two variables k(x, y) is called the Cremona group. Explain that the elements of the Cremona group correspond to birational self-isomorphisms of P 2. Explain that the Cremona involution (x, y) (x 1, y 1 ) extends to an automorphism of k(x, y). What is the corresponding birational involution of P 2? Where is this birational automorphism regular? (v) More generally, show that GL 2 (Z) is a subgroup of the Cremona group. Answer: (i) Clearly, A n is birational to P n, hence K(P n ) = K(A n ) = k(t 1,..., t n ). Thus X is rational iff K(X) = k(t 1,..., t n ). (ii) Let U P n be the open set where the coordinate x 0 0. Similarly, let V P m be the open set where the coordinate y 0 0, and let W be the open set in P n+m where the first coordinate is non-zero. Define φ : U V P n+m as [x 0 : x 1 :... : x n ] [y 0 : y 1 :... : y m ] [ 1 : x 1 x 0 : x 2 x 0 :... : x n x 0 : y 1 y 0 : y 2 y 0,... : y n y 0 It is easy to check that φ establishes an isomorphism between U V and W, with inverse ψ : W U V, [1 : z 1 :... : z n+m ] [1 : z 1 :... : z n ] [1 : z n+1 :... : z n+m ]. Therefore φ and ψ define birational isomorphisms between P n P m and P n+m. Finally, if X and Y are birational to P n andp m, then X Y is birational to P n P m which in turn is birational to P n+m. Therefore, X Y is rational. (iii) Two closed subsets {a} P 1 and {b} P 1 in P 1 P 1 have nonemtpy intersection. This is false in P 2 by problem 1(iii). Hence P 2 and P 1 P 1 cannot be isomorphic. (iv) We explained in (i) that K(P 2 ) = k(x, y) hence any automorphism of k(x, y) corresponds to an automorphism of K(P 2 ) which in turn gives a birational isomorphism of P 2. The involution corresponds to the birational map (x, y) (x 1, y 1 ) f[x : y : z] = [x 1 : y 1 : z 1 ]. This map is regular on P 2 \ {[1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1]}. Indeed, to show that the map is regular at the points where (x, y) (0, 0), we rewrite it in the form [ z f[x : y : z] = x : z ] y : 1. 5 ].
6 (v) The automorphism (x, y) (x a y b, x c y d ) where belongs to the Cremona group. Its inverse is ( ) a b GL 2 (Z) c d (x, y) (x a y b, x c y d ) where ( a b ) is the inverse of the matrix above. c, d Problem 6. (Quadrics are rational.) Show that any non-degenerate irreducible quadric Q P n+1 is birational to P n. Answer: All nondegenerate quadrics are isomorphic as seen in class. We can assume that the quadric Q is defined by x 0 x 1 x 2 2 x x 2 n = 0. Pick the point p = [1 : 0 :... : 0] Q, and let H be the hyperplane X 0 = 0. The line passing through p = [1 : 0 :... : 0] and q = [x 0 :... : x n ] is [r + sx 0 : sx 1 :... : sx n ]. This line intersets the hyperplane H when r + sx 0 = 0. Therefore the line intersects H at [0 : sx 1 :... : sx n ] = [0 : x 1 :... : x n ]. This may be undefined when x 1 = x 2 = = x n = 0, i.e. when q = p. We obtain a morphism f : Q \ {p} H, [x 0 : x 1 :... : x n ] [0 : x 1 :... : x n ]. The rational inverse of f is given by [ g : H x 2 Q, [x 1 : x 2 :... : x n ] = 2 + x 2 n x 1 ] : x 1 : x 2 :... : x n. This may be undefined at the points where x 1 = 0. Since f has a rational inverse, Q is birational to H. 6
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