Math 203A, Solution Set 6.


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1 Math 203A, Solution Set 6. Problem 1. (Finite maps.) Let f 0,..., f m be homogeneous polynomials of degree d > 0 without common zeros on X P n. Show that gives a finite morphism onto its image. f : X P m, f(x) = [f 0 (x) :... : f m (x)] Answer: Let v : P n P N be the Veronese embedding. Let F = v f. Since v is an isomorphism onto image, f finite F finite. Let l 0,..., l m be the linear polynomials corresponding to f 0,..., f m under Veronese. Then F (x) = [l 0 (x) :... : l m (x)]. Without loss of generality, we may assume l 0,..., l m are independent. For otherwise, assuming l 0,..., l k are the independent ones and setting F (x) = [l 0 (x) :... : l k (x)] we have F = A F for an injective linear map A : C k C l. Since A is an isomorphism onto its image, it suffices to prove that F is finite. After a change of coordinates, we may assume l 0 = x 0,..., l k = x k. The map F becomes the composition of projections away from a point, which we have seen is finite. Problem 2. (Semicontinuity of fiber dimensions.) Assume that f : X Y is a surjective morphism of projective varieties. Show that Y k = {y Y : dim f 1 (y) k} is closed. Answer: Let n = dim X, m = dim Y. We will use induction on m, the case m = 0 being clear. Let d = n m 0. When k d there is nothing to prove since Y k = Y by the theorem of dimension of fibers. Asssume k > d. Let U be an open subset over which dim f 1 (y) = d. The existence of U is proven in the theorem of dimension of fibers. Let Z = Y \ U is a closed subset of Y, so dim Z < dim Y = dim Z m 1. Note that Y k Z for k > d. We wish to show that Y k is closed in Z, so then it will be closed in Y as well. To this end, consider the restriction f : W Z, where W = f 1 (Z). We clearly have Y k = {y Y : dim f 1 (y) k} = {y Z : dim f 1 (y) k}. We use induction on the dimension of the base to conclude. Indeed, dim Z m 1, and working with each irreducible component of Z one at a time, we may assume Z is irreducible. In this case however the domain W = f 1 (Z) may have components 1
2 2 W 1,..., W r. Since Z is closed, W i are closed in X hence projective in X. Let f i : W i Z be the restricted morphism. Then Y k = Y k ( f i ) where the sets of the right are calculated with respect to the morphisms f i. By the induction hypothesis applied to f i, it follows that Y k ( f i ) are closed in the image of f i which in turn is closed in Z by projective hypothesis, hence Y k is closed in Z as well. Problem 3. (Criterion for irreducibility.) Assume that f : X Y is a surjective morphism of projective algebraic sets such that Y is irreducible and all fibers of f are irreducible of the same dimension. Show that X is irreducible as well. Answer: Write n for the common dimension of the fibers. irreducible components of X. Then Write X 1,..., X r for the f(x i ) = Y and f(x i ) are closed in Y since f is a morphism of projective sets, hence a closed map. Since Y is irreducible, there exists i such that f(x i ) = Y. Assume that X 1,..., X s are chosen so that f(x 1 ) =... = f(x s ) = Y but f(x j ) Y for j > s. Construct U 1,..., U s nonempty open sets in Y such that y U i, 1 i s = dim(f Xi ) 1 (y) = n i = dim X i dim Y. In fact, even for j > s we can define U j = Y \ f(x j ) and for y U j we have (f Xj ) 1 (y) =. Write U = r i=1u i, which is open and nonempty. For y U, f 1 (y) is irreducible and nonempty, and is covered by X 1,..., X r so it must exist i 0 such that f 1 (y) X i0. It is clear from the choice of i 0 that the entire fiber over y can be computed in X i0 that (f Xi0 ) 1 (y) = f 1 (y) so so f Xi0 : X i0 Y must be surjective by the definition of U i0, and i 0 s. Furthermore n = n i0 is the common dimension of the fibers since the fiber dimension can be calculated at y and (f Xi0 ) 1 (y) = f 1 (y). If z Y, then (f Xi0 ) 1 (z) f 1 (z)
3 3 and the left hand side is at least of dimension n i0 = dim X i0 dim Y by the theorem on dimension of fibers. But f 1 (z) is irreducible and n = n i0 dimensional, so must have equality. Thus f 1 (z) X i0 for all z Y. This shows that there are no components in X other than X i0, so X is irreducible. Problem 4. (Intersections in projective space.) Let X and Y be two subvarieties of P n. Show that if dim X + dim Y n, then X Y is not empty. Answer: Let H 1, H 2 be two disjoint linear subspaces of dimension n in P 2n+1. We write [x 0 : x 1 :... : x n : y 0 : y 1 :... : y n ] for the homogeneous coordinates in P 2n+1. Without loss of generality, we may assume H 1 is given by the equations y 0 = y 1 =... = y n = 0, while H 2 is given by x 0 =... = x n+1 = 0. We regard X H 1 = P n P 2n+1, Y H 2 = P n P 2n+1 as subvarieties of P 2n+1. We form the join J(X, Y ) in P 2n+1. We first prove that J(X, Y ) has dimension dim X + dim Y + 1. Indeed, any point P in J(X, Y ) lies on a line L which intersects both X and Y in two points Q and R. The map f : J(X, Y ) X Y, P (Q, R) is a welldefined morphism since given P, then Q and R are uniquely defined. Indeed, if P J(X, Y ) has coordinates [p 0 :... : p 2n+1 ] then Q = [p 0 :... : p n : 0... : p n ] and R = [0 :... : 0 : p n+1 :... : p 2n+1 ], as claimed. The fibers of f are lines QR, hence they are 1 dimensional. Thus dim J(X, Y ) = dim(x Y ) + 1 = dim X + dim Y + 1 n + 1. Even stronger, by the previous problem, J(X, Y ) is irreducible since it fibers over the irreducible set X Y with equidimensional fibers. Next, let K i be the hyperplane x i y i = 0 for 0 i n. We claim that X Y = J(X, Y ) K 0 K 1... K n. Indeed, any point P J(X, Y ) lies on a line QR with Q X, R Y, hence P = αq + βr = [αq : βr],
4 4 where q and r are the homogeneous coordinates of Q and R in P n. The requirement that P means that 0 i n αq i = βr i hence Q = R. This means P = Q = R X Y, proving the above equality. Finally, Intersecting with a hyperplane either keeps the same dimension or cuts the dimension down by 1, hence K i dim (J(X, Y ) K 0... K n ) 0 = X Y. Problem 5. (Lines on hypersurfaces.) (i) Let d > 2n 3. Show that a general degree d hypersurface in P n contains no lines. (ii) Any cubic surface in P 3 contains at least one line. (iii) Let f be a degree 4 homogeneous polynomial in 4 variables and let Z f be the quartic surface f = 0 in P 3. Show that there is a single polynomial Φ in the coefficients of f which vanishes if and only if the quartic surface Z f P 3 contains a line. Answer: (i) We think of a hypersurface X = Z(f) as a point in projective space P N for N = ( ) n+d d 1, by means of the coefficients ai of its defining equation f = a I X I. We form the incidence correspondence and we let be the two projections. J = {(L, X) : L X} G(1, n) P N p : J G(1, n), q : J P N We claim that the fibers of p have dimension N (d + 1). Indeed, fix a line L and study p 1 (L). Without loss of generality, we may assume L is given by the equations x 0 =... = x n 2 = 0. If X p 1 (L) is given by the polynomial the requirement L X means f = 0, f(0 :... : 0 : s : t) = 0
5 for all s, t. In particular, the d+1 coefficients of s i t d 1 for 0 i d must vanish: a 0...0,i,d i = 0, while the other coefficients are arbitrary. Thus p 1 (L) has codimension d + 1 in P N, as claimed. Also the fibers of p are irreducible so J is irreducible as well by Problem 2. With this understood, we conclude by looking at the fibers of p that dim J = dim G(1, n) + N (d + 1) = (2n 2) + N (d + 1) < N. Therefore, the morphism q is not surjective. In particular, the image q(j) is a proper subvariety of P N. For hypersurfaces X belonging to the complement P n \ q(j), the preimage q 1 (X) is therefore empty, or in other words, for there are no lines lying on such hypersurfaces. (ii) We have d = n = 3, so that N = 19. In this case, the above computation shows dim J = N = 19. It suffices to show q is surjective onto P N, since then for each cubic X there should be an element in q 1 ([X]) in J, thus giving a line L X. The image of q is closed and irreducible in P N. 5 If q is not surjective, the image is of dimension N 1 or lower. All fibers of q will have dimension at least N (N 1) = 1. We construct a cubic containing only finitely many lines. For example, we can take X = {x 3 + y 3 + z 3 + w 3 = 0}. By symmetry, we may search for lines of the form x = az + bw, y = cz + dw and substituting we find This gives (az + bw) 3 + (cz + dw) 3 + z 3 + w 3 = 0. a 3 + c = b 3 + d = 0, a 2 b + c 2 d = 0, ab 2 + cd 2 = 0. We claim that there are finitely many solutions for a, b, c, d. If a = 0 then it is easy to conclude that d = 0 and b, c have to satisfy b 3 = c 3 = 1, and the solution set is finite. Assume now that neither a, b, c, d is zero. Then a 2 b = c 2 d, ab 2 = cd 2 = (a2 b) 2 (ab 2 ) = d) 2 (c2 (cd 2 ) which contradicts a 3 + c 3 = 1. (iii) In this case, we have d = 4, n = 3, N = 34. Let J = {(L, X) : L X}. = a 3 = c 3 In this case, the above computation show that dim J = N 1. We claim that the image q(j) is a codimension 1 subvariety of P N. We complete the proof letting Φ be a polynomial cutting out q(j).
6 6 To prove q(j) is of dimension 33, assume otherwise, namely that the dimension is 32 or lower. By the theorem of dimension of fibers, for all [X] q(j), the fiber q 1 ([X]) has dimension at least = 1. In other words all quartics that contain at least one line in fact contain infinitely many lines. One example is the quartic x 4 + y 4 + z 4 + w 4 = 0. By symmetry, we may search for lines of the form x = az + bw, y = cz + dw and substituting we find (az + bw) 4 + (cz + dw) 4 + z 4 + w 4 = 0. This gives a 4 + c = b 4 + d = 0, a 2 b 2 + c 2 d 2 = 0, a 3 b + c 3 d = 0, ab 3 + cd 3 = 0. We claim that there are finitely many solutions for a, b, c, d. If a = 0 then it is easy to conclude that d = 0 and b, c have to satisfy b 4 = c 4 = 1, and the solution set is finite. Assume now that neither a, b, c, d is zero. Then a 3 b = c 3 d, ab 3 = cd 3 = (a/b) 2 = (c/d) 2 and in addition (ab) 2 = (cd) 2 so multiplying we find a 4 = c 4 which contradicts a 4 + c 4 = 1.
k k would be reducible. But the zero locus of f in A n+1
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