Math 145 Solution 5 Problem 1. (Products of affine varieties.) Let X ⊂ An and Y ⊂ Am be affine algebraic sets. (i) Show that X × Y ⊂ An+m is also an affine algebraic set. Answer: (i) Let X ⊂ An and Y ⊂ Am be affine algebraic sets such that X = Z(f1; : : : ; fr);Y = Z(g1; : : : ; gs) where fi 2 k[X1;X2;:::;Xn] and gj 2 k[Y1;Y2;:::;Ym] : We regard fi and gj as polynomials in the variables X1;:::;Xn;Y1;:::;Ym: We claim that n+m X × Y = Z(f1; : : : ; fr; g1; : : : ; gs) ⊂ A : This will show that X × Y is an affine algebraic set. To prove the claim, observe that (x1; : : : ; xn; y1; : : : ; ym) 2 Z(f1; : : : ; fr; g1; : : : ; gs) , fi(x1; : : : ; xn) = 0 and gj(y1; : : : ; ym) = 0 for all i; j , (x1; : : : ; xn) 2 X and (y1; : : : ; ym) 2 Y , (x1; : : : ; xn; y1; : : : ; ym) 2 X × Y: Problem 2. (Intersections with hypersurfaces.) (i) Let H ⊂ Pn be a linear hyperplane in Pn (e.g. H is cut out by one linear homogeneous equation in Pn). Show that the quasiprojective algebraic set PnnH is isomorphic to An. (ii) Show that if X is a degree d hypersurface in Pn then PnnX admits a nonconstant morphism to an affine space. (In fact, its not much harder to show that PnnX is isomorphic to an affine variety). (ii) Conclude that if X is a hypersurface in Pn and Y ⊂ Pn is a projective variety which is not a point, then the intersection of X and Y is non-emtpy. (iv) If X is a degree d hypersurface in Pn and L is a line in Pn not contained in X, show that L and X intersect in at most d points. In fact, they intersect in exactly d points counted with multiplicity. 0 0 n Answer: (i) Write temporarily X0;:::;Xn for the coordinates on P . Assume that the hy- perplane H is cut out by the linear equation ( n ) X 0 H = aiXi = 0 : i=0 Without loss of generality, we may assume an 6= 0. Define n X 0 0 Xn = aiXi;Xi = Xi for 0 ≤ i ≤ n − 1: i=0 n Then it’s easy to see that Xi are new coordinates on P . Indeed, the change of coordi- nates matrix 0 1 0 ::: 0 0 1 B 0 1 ::: 0 0 C C = B C @ 0 0 ::: 1 0 A a0 a1 : : : an−1 an is invertible having determinant equal to an 6= 0. In the new coordinates, H is defined by Xn = 0 : 1 2 The map n n X0 Xn−1 π : P nH ! A ; [X0 : ::: : Xn] ! ;:::; Xn Xn n is well-defined because Xn 6= 0 on P nH. Let n n φ : A ! P nH; (Y1;:::;Yn) ! [Y1 : ::: : Yn : 1] Clearly π and φ are inverses, and therefore π is an isomorphism. (ii) Suppose X = Z(f) where X i0 in f = ai0i1:::in X0 ··· Xn = 0 : i0+···+in=d Let v be the d-fold Veronese embedding n N v : P ! P : It is easy to check that v is injective (this follows from the definitions, we need to consider quotients of the monomials od degree d.) N Let Y0;:::;YN be the coordinates on P . For points in the image of the Veronese embedding, Y0;:::;YN is a list of all degree d monomials i0 in X0 ··· Xn ; in some order. To fix notation, we write Yi0:::in for the coordinate corresponding to the i0 in monomial X0 ··· Xn : 0 N Let H be the hyperplane in P defined by the linear equation with coefficients ai0:::in e.g. X ai0:::in Yi0:::in = 0: Clearly, v(X) ⊂ H0: Then n N 0 N v : P n X ! P n H =∼ A ; where in the last equation we used (ii). Therefore, we obtained a non-constant (in fact injective) morphism from Pn n X and AN . (iii) Suppose the intersection of X and Y is empty, then n Y ⊂ P nX: Let n N v : P nX ! A be the morphism described in (ii) and πi be the projection map from N 1 πi : A ! A to the i-th coordinate. Then n 1 πi ◦ v : P nX ! A : 1 Hence πi ◦ v restricts to a morphism from Y ! A . Because Y is a projective variety, πi ◦ φ must be a constant. This is true for every i, therefore v is a constant morphism on Y . But since v is injective, it follows that Y is a point. 3 (iv) After a change of coordinates, we can assume that L is cut out by the equations X2 = ::: = Xn = 0 : Indeed, pick any two points p and q on L, change coordinates such that p and q become [1 : 0 : ::: : 0] and [0 : 1 : ::: : 0]. It is clear that the line L that p and q determine is given by X2 = ::: = Xn = 0. Recall that X i0 in f = ai0i1:::in X0 ··· Xn : i0+···+in=d Let X i0 i1 F (X0;X1) = f(X0;X1; 0;:::; 0) = ai0i10:::0X0 X1 : i0+i1=d Assuming x = [x0 : x1 : ··· : xn] 2 L \ X then x2 = ::: = xn = 0 and f(x) = 0. Therefore X i0 i1 F (x) = ai0i10:::0x0 x1 = 0 : i0+i1=d Note that if F =∼ 0 it follows that L ⊂ X; which is impossible. Therefore F is a non-zero homogenous polynomial of degree at most d. If ad0:::0 = 0, then F = x0G where G has degree d − 1. It follows by induction that G has at most d − 1 roots and thus F has at most d roots, namely the zeros of F and [0 : 1 : ::: : 0]): n Assume now ad0:::0 6= 0. If x1 = 0, this implies x0 = 0, which is not in P . Therefore we can assume x1 = 1 and x0 is a solution of X i0 ai0i10:::0X0 = 0 i0+i1=d This is a nonzero degree d polynomial, with at most d solutions. Therefore there are at most d points in the intersection L \ X. Problem 3. (Rational varieties.) 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. We say that X is rational if X is birational to Pn for some n. (i) Show that a birational isomorphism f : X / Y induces an isomorphism of the fields of rational functions f ? : K(Y ) ! K(X). The converse is also true but you don’t have to prove it. (ii) Show that Pn × Pm is rational, by constructing an explicit birational isomorphism with Pn+m. Show that if X and Y are rational, then X × Y is rational. Answer: (i) For h 2 K(Y ), define f ?(h) = h ◦ f: This is a rational function on X. This fact is explained for instance in the text, section 4:10. In the terminology used there, f is a dominant map. In short, the idea is as follows. We have seen that h is defined on an open set Dom(h) ⊂ Y . By assumption, f establishes 4 an isomorphism between nonempty open sets U and V in X and Y . Then f ?h is defined on the intersection f −1(V \ Dom(h)): To show this intersection is non-empty, we need to argue that the nonempty open sets V and Dom(h) intersect. But if they do not, the proper closed sets Y n V and Y n Dom(h) cover Y contradicting the fact that Y is irreducible. Therefore f ? : K(Y ) ! K(X) is a well defined field homomorphism. Similarly, the birational inverse morphism g : Y / X induces a field homomor- ? ? ? phism g : K(X) ! K(Y ). We claim that f and g are inverses. Indeed, for h1 2 K(Y ), h2 2 K(X), ∗ ∗ ∗ f g (h2) = f (h2 ◦ g) = h2 ◦ g ◦ f = h2 ◦ id = h2 ; ∗ ∗ ∗ g f (h1) = g (h1 ◦ f) = h1 ◦ f ◦ g = h1 ◦ id = h1 : Therefore K(X) =∼ K(Y ). n m (ii) Let U ⊂ P be the open set where the coordinate x0 6= 0. Similarly, let V ⊂ P be the n+m open set where the coordinate y0 6= 0; and let W be the open set in P where the first coordinate is non-zero Define φ : U × V ! Pn+m as x1 x2 xn y1 y2 yn [x0 : x1 : ::: : xn] × [y0 : y1 : ::: : ym] ! 1 : : : ::: : : : ;::: : : x0 x0 x0 y0 y0 y0 It is easy to check that φ establishes an isomorphism between U ×V and W , with inverse : W ! U × V; [1 : z1 : ::: : zn+m] ! [1 : z1 : ::: : zn] × [1 : zn+1 : ::: : zn+m]: Therefore φ and define birational isomorphisms between Pn × Pm and Pn+m. Finally, if X and Y are birational to Pn andPm, then X × Y is birational to Pn × Pm which in turn is birational to Pn+m. Therefore, X × Y is rational. Problem 4. (Quadrics are rational.) (i) Show that a non-singular irreducible quadric Q in P3 can be written in the form xy = zw after a suitable change of homogeneous coordinates. Combining this result with the Segre em- bedding, conclude that any quadric in P3 is isomorphic to P1 × P1, hence it is rational. (ii) In general, show that any non-singular irreducible quadric Q ⊂ Pn+1 is birational to Pn. Answer: (i) A homogeneous quadratic polynomial f(x; y; z; w) can be written as 2x3 6y 7 f = [x y z w] A 6 7 4z 5 w where A is a symmetric 4 × 4 matrix. Because A is symmetric, it can be diagonalized.