HARTSHORNE EXERCISES


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1 HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing up Y at O = (0, 0) is nonsingular. (b) We define a node (also called ordinary double point) to be a double point (i.e., a point of multiplicity 2) of a plane curve with distance tangent directions. If P is a node on a plane curve Y, show that ϕ 1 (P ) consists of two distinct nonsingular points on the blownup curve Ỹ. We say that blowing up P resolves the singularity at P. (c) Let P Y be the tacnode x 2 = x 4 + y 4. If ϕ : Ỹ Y is the blowing up at P, show that ϕ 1 (P ) is a node. Using (b) we see that the tacnode can be resolved by two successive blowingsup. (d) Let Y be the plane curve y 3 = x 5, which has a higher order cusp at O. Show that O is a triple point; that blowing up O gives rise to a double point (what kind?) and that one further blowing up resolves the singularity. Proof. (a) We first work with the cusp. Let [t : u] be homogeneous coordinates of P 1. The variety Ỹ is covered by two open affine sets, those where either t or u doesn t vanish. First, consider t 0. Then we may assume that t = 1 and treat u as an affine coordinate, so that we obtain an affine variety in A 3. This variety is given by the polynomial equations: x u 2 x 2 x 2 u 4 = 0 xu y = 0 Therefore, the Jacobian of Ỹ at any point (x, y, u) of this affine piece is ( ) 1 2x 2xu 4 0 2u 4x 2 u 3 u 1 x This matrix has less than full rank if and only if 1 2x 2xu 4 = 2u 4x 2 u 3 = 0. Hence, to show that each point is nonsingular, we must show that the system of equations (1) x u 2 x 2 x 2 u 4 = 0 (2) 1 2x 2xu 4 = 0 (3) 2u 4x 2 u 3 = 0 has no solution. Notice that if u = 0 or x = 0 or char k = 2, then the system has no solution. So we can assume that neither u nor x are 0 and that the characteristic is not Date: Spring
2 2. Also, notice that the 2 (1) x (2) shows that x = 2u 2. This reduces us to the following system of equations: (4) 1 4u 2 4u 6 = 0 (5) 1 + 8u 6 = 0 2 (4) + (5) shows us that 3 = 8u 2 = 4x. Plugging this into (5) gives the equality = 0 which is only true if the characteristic of k is 7 or 13. If the characteristic of k is 7, the triple (6, 6u, u) is a singular point for any value of u k satisfying u 2 = 3. If the characteristic of k is 13, the triple (4, 4u, u) is a singular point for any value of u k satisfying u 2 = 2. In fact, the point (6, 6u) is singular on Y if char k = 7 and the point (4, 4u) is singular on Y if char k = 13, so it makes sense that they remain singular under the blowup. If u 0, we may assume that u = 1 and treat t as an affine coordinate. The resulting variety in A 3 is given by the equations: yt 3 1 y 2 t 4 y 2 = 0 x yt = 0 This leads to the Jacobian (0 ) t 3 2yt 4 2y 3yt 2 4y 2 t 3 1 t y The only point we need to check on this affine piece of Ỹ is the point t = 0, whose Jacobian is: ( ) 0 2y y If t = 0, then x = 0 and y satisfies the equation y = 0. Hence the point (0, y, 0) is singular on Ỹ if and only if 2y = 0 = y If char k = 2, then we have the singular points (0, 1, 0). In summary, Ỹ is nonsingular if and only if the characteristic of the field is not 2, 7, or 13. Proceeding as above, it can be seen that the blowup of the node at O is nonsingular no matter the characteristic of the field. This follows from the fact that the only singularity of the node is the origin for all fields k, and the singularity is a double point, so it is resolved by the blowup (part (b) below). (b) Suppose P = (a, b). If Y is defined by the polynomial f(x, y), then Y is isomorphic to the polynomial defined by f(x + a, y + b) under the map x x a, y y b, so we may assume that P = (0, 0). We can also rotate Y to assume that the tangent directions at the origin are not vertical. Decompose f(x, y) into its homogeneous components by f(x, y) = f 2 (x, y) + f 3 (x, y) f d (x, y), where f 2 (x, y) = (ax + by)(a x + b y), with [a : b] [a : b ] P 1 because P must have distinct tangent directions. Also, since the tangent directions are not vertical, 2
3 we have b 0 b. As in part (a), let s first consider the affine piece of the blowup where t 0. This is given by the equations x 2 f(x, xu) = (a + bu)(a + b u) + x 2 (f 3 (x, xu) f d (x, xu)) = 0 xu y = 0 Then varphi 1 ((0, 0)) are points of the form (0, 0, u) where (a + bu)(a + b u) = 0. Since both b and b are nonzero, this equation has two distinct solutions given by u = ab 1 and u = a b 1. Notice the Jacobian at (0, 0, u) is then given by ( x 3 f 3 (x, xu) (0,0,u) 0 ) ab + a b + 2bb u u 1 x which has rank 2 because ab + a b + 2bb u 0 for both solutions u (this follows from our assumption that [a : b] [a : b ] P 1 ). (c) First suppose P = (0, 0). Checking the affine pieces of the blowup reveals that the only point in ϕ 1 (P ) is (x, y, t) = (0, 0, 0) when u 0. This affine piece in A 3 is defined by the equations t 2 y 2 y 2 t 4 = 0 x yt = 0 We can isomorphically project this onto the y, tplane, to obtain the plane curve defined by f(y, t) = t 2 y 2 y 2 t 4 = (t y)(t + y) y 2 t 4. Notice then that the point (0, 0) is a node so long as char k 2, otherwise it is not a node. The result is not true for any other P (0, 0), for no other point is singular on Y when char k 2. (d) O is a triple point because the smallest nonzero homogeneous component of f(x, y) = y 3 x 5 has degree 3. Blowing up at O gives a curve with an affine open piece isomorphic to the cusp u 3 = x 2, where the preimage of O is the point (0, 0). The point (0, 0) is a double point on the cusp defined by f(x, u) = x 2 u 3, but it is not ordinary because x 2 has the same linear factors. On a previous homework assignment, we showed that blowing up the cusp u 3 = x 2 at the origin resolves the singularity, so one further blowing up of (0, 0) will resolve the singularity of Y. Hartshorne, Exercise I.6.3. Show by example that the result of (6.8) is false if either (a) dim X 2, or (b) Y is not projective. Proof. (a) Let X = A 2, P = (0, 0), and Y = P 1 with ϕ : X P Y (x, y) [x : y] and suppose that ϕ extended to ϕ : X Y where ϕ((0, 0)) = [a : b] P 1. Then, consider any point [c : d] [a : b] P 1. This point is closed (defined by the homogeneous polynomial equation dx = cy). However, ϕ 1 ([c : d]) = {(x, y) dx = cy} (0, 0) is not closed, so the map ϕ is not continuous. 3
4 (b) Let X = P 1, P = [0 : 1], and Y = A 1 with ϕ : X P Y [x : y] y x and suppose that ϕ extended to ϕ : X Y. A morphisms of varieties into A 1 is a regular function, so that ϕ is a regular function of P 1, so by a previous homework, ϕ is constant. This contradicts the definition of ϕ, so no such extension exists. Hartshorne, Exercise I.6.7. Let P 1,..., P r, Q 1,..., Q s be distinct points of A 1. If A 1 {P 1,..., P r } is isomorphic to A 1 {Q 1,..., Q s }, show that r = s. Is the converse true? Proof. Let U = A 1 {P 1,..., P r } and V = A 1 {Q 1,..., Q s }, and view U and V as open subsets of P 1. Let ϕ : U V be an isomorphism, and i : V P 1 be inclusion. Then the morphism i ϕ : U P 1 can be uniquely extended (Hartshorne I.6.8) to a map ϕ : P 1 P 1. I claim that ϕ is an isomorphism. If so, The result follows because then ϕ must map the set {P 1,..., P r, } bijectively to the set {Q 1,..., Q s, }. To see that ϕ is an isomorphism, we will exhibit its inverse. Namely, begin with ϕ 1 : V U and complete to ϕ 1 : P 1 P 1 as above. Notice that ϕ ϕ 1 restricts to the identity on V, so it must be the unique extension of the identity morphism on V, which is the identity on P 1. It follows that ϕ 1 = ϕ 1. The converse is true only for 0 r 2. For r = 0, A 1 = A 1. For r = 1, A 1 \ P = A 1 \ Q via the map x x P + Q. For r = 2, A 1 \ {P 1, P 2 } = A 1 \ {Q 1, Q 2 } via the map x (Q 1 Q 2 )x Q 2 P 1 + Q 1 P 2 P 2 P 1 To see that the converse is false for r > 3, we use without proof the result that an automorphism of P 1 is determined by where it sends three points (Hartshorne, exercise I.6.6). Let P 1,..., P r, Q 1,..., Q r be distinct points such that no automorphism of P 1 bijectively maps the set {P 1,..., P r, } to the set {Q 1,..., Q r, }. If A 1 \ {P 1,..., P r, } = A 1 \ {Q 1,..., Q r, }, as shown above, there must be an automorphism of P 1 which maps {P 1,..., P r, } to the set {Q 1,..., Q r, } bijectively, a contradiction. Hartshorne, Exercise II.2.3. Reduced Schemes. A scheme (X, O X ) is reduced if for every open set U X, the ring O X (U) has no (nonzero) nilpotent elements. (a) Show that (X, O X ) is reduced if and only if for every P X, the local ring O X,P has no (nonzero) nilpotent elements. (b) Let (X, O X ) be a scheme. Let (O X ) red be the sheaf associated to the presheaf U O X (U) red, where for any ring A, we denote by A red the quotient of A by its ideal of nilpotent elements. Show that (X, (O X ) red ) is a scheme. We call it the reduced scheme associated to X, and denote it by X red. Show that there is a morphism of schemes X red X, which is a homeomorphism on the underlying topological spaces. (c) Let f : X Y be a morphism of schemes, and assume that X is reduced. Show that there is a unique morphism g : X Y red such that f is obtained by composing g with the natural map Y red Y. 4
5 Proof. (a) Suppose (X, O X ) is reduced, P X, and s, U O X,P for some open U X containing P and some s O X (U). Suppose that s, U n = s n, U = 0 O X,P for some n N. This means there is some open V U containing P such that (s n ) V = (s V ) n = 0 O X (V ). Since (X, O X ) is reduced O X (V ) has no nilpotent elements, so s V = 0. It follows that s, U = s V, V = 0 so that the local ring O X,P has no nilpotent elements. Next, suppose that for every P X, O X,P has no nilpotent elements. For any U X, let s O X (U), and suppose that s n = 0 for some n N. For each P U, let s P = s, U O X,P. By our hypothesis, it follows that s P = 0 for all P U, that is, for each P there exists open V P U containing P such that s VP = 0. The V P cover U, so by the sheaf axiom, it follows that s = 0 O X (U), and thus, (X, O X ) is reduced. (b) First, assume that X = Spec A is affine. We will show that X red = Spec Ared. The canonical projection A A red induces a map of schemes ϕ : Spec A red Spec A. If N is the nilradical of A, then since all prime ideals in A contain N, there is a onetoone correspondence between prime ideals p A and prime ideals p A red. The map ϕ is a homeomorphism sending p to p. Now, ϕ # U : O Spec A(U) ϕ O Spec Ared (U) must vanish on nilpotent elements of O Spec A (U) because Spec A red is a reduced scheme (its localizations are of the form (A red ) p = (Ap ) red, so they are reduced, so by part (a), Spec A red is reduced). We thus have a map of presheaves (O Spec A (U)) red ϕ O Spec Ared (U) which by the universal property extends to a map of sheaves (O Spec A ) red ϕ O Spec Ared. To see that this is an isomorphism, we will see that the induced map on stalks is an isomorphism. The induced map on stalks is the middle map in the following sequence of compositions: (6) (A p ) red ((O Spec A ) red ) p (ϕ O Spec Ared ) p (A red ) p ( a ) a a b b, D(b) b, D ( b ) a b Here a A, b A \ p, D(b) = {q Spec A b / q}, D(b) = {q Spec A red b / q}, and for any commutative ring R, prime ideal p R, and element r R s p, we view r as s a function on D(s) via: D(s) p D(s) p r s (6) is the canonical isomorphism between (A p ) red and (A red ) p, so that the map of stalks is also an isomorphism. We have thus shown that as affine schemes, (Spec A) red = Spec A red. We have also exhibited the desired morphism of affine schemes Spec A red Spec A which is just the map induced by the projection A A red. Now, if X is any scheme, let x X, and choose some open affine neighborhood of x, U = Spec A. This means that (U, O X U ) = (Spec A, O Spec A ) as affine schemes. Thus, (U, (O X U ) red ) = (Spec A red, O Spec Ared ) as affine schemes. For any open V U, (O X U ) red and (O X ) red U are both the sheaf associated to the presheaf V O X (V ), they are isomorphic. It follows that (Spec A red, O Spec Ared ) is an open affine set of x X red so that X red is indeed a scheme. 5 R p
6 To obtain the desired map X red X, let ϕ : X X be the identity. Then, for any open U X, define ϕ # U : O X(U) (O X ) red (U) to be the composition of the quotient map O X (U) ((O X (U)) red with the natural map from the presheaf (O X (U)) red to its associated sheaf (O X ) red (U). (c) The natural map (ϕ, ϕ # ) from Y red to Y is the identity on topological spaces. Thus we can define g(x) = f(x). Next, for any open U Y, consider the map f # U : O Y (U) O X (f 1 (U)). Since X is reduced, f # U must factor through (O Y (U)) red. This gives a map of presheaves (O Y (U)) red O X (f 1 (U)). By the universal property of sheafification, there is a unique map of sheaves g # U : (O Y ) red (U) O X (g 1 (U)). The pair (g, g # ) is the desired morphism of schemes X Y red. Hartshorne, Exercise II.2.7. Let X be a scheme. For any x X, let O x be the local ring at x, and m x its maximal ideal. We define the residue field of x on X to be the field k(x) = O x /m x. Now let K be any field. Show that to give a morphism of Spec K to X it is equivalent to give a point x X and an inclusion map k(x) K. Proof. Let Y = Spec K, and suppose (f, f # ) : Y X is a morphism of schemes, that is f : Y X is a continuous map and f # : O X f O Y is a morphism of sheaves of rings on X. As a topological space Y consists of one point (the zero ideal in K), so we take x X to be the image of this point. Now let U be any open set in X containing x, so that we have a map f # U : O X(U) O Y (Y ) = K. Since f # is a morphism of sheaves, the maps f # U are compatible with restrictions, so that the map f x # : O x K s, U f # U (s) is welldefined, where s O X (U). Being a map of schemes, we require this induced morphism to be a local homomorphism of local rings. In this case, this condition means that ker f x # = m x so that f x # induces an inclusion k(x) K. Next, suppose we fix a point x X and an inclusion i : k(x) K. Define a continuous map f : Y X by f((0)) = x. For any open U X not containing X, f 1 (U) = so define f # U : O X(U) O Y ( ) = 0 to be the trivial ring map. If U contains x, then define : O X(U) O Y (Y ) = K as follows: f # U f # U : O X(U) K s i( s, U + m x ) In other words, f # U is the composition of maps O X(U) O x k(x) K. To see that this defines a map of sheaves, we must verify that for any V U, the diagram: O X (U) O Y (f 1 (U)) O X (V ) O Y (f 1 (V )) 6
7 commutes. If V does not contain x, then the diagram commutes. If V contains X, then the diagram becomes O X (U) O X (V ) K The commutativity of this diagram follows from that of O X (U) O X (V ) O x By definition, the map induced on stalks by f # is O x k(x) K, whose kernel is m x, so it is a local map of local rings, and thus (f, f # ) is a morphism of schemes. Hartshorne, Exercise II.2.8. Let X be a scheme. For any point x X, we define the Zariski tangent space T x to X at x to be the dual of the k(x)vector space m x /m 2 x. Now assume that X is a scheme over a field k, and let k[ε]/ε 2 be the ring of dual numbers over k. Show that to give a kmorphism of Spec k[ε]/ε 2 to X is equivalent to giving a point x X, rational over k (i.e., such that k(x) = k), and an element of T x. Proof. Let Y = Spec k[ε]/ε 2, and suppose (f, f # ) is a kmorphism from Y to X. This means the following diagram of schemes commutes: Y f Spec k X As a topological space, Y consists of one point (the ideal (ε)), so let x be the image of this point, and consider the induced commutative diagram of stalks: k[ε]/ε 2 f # x O x Each arrow above is a local homomorphism of local rings. In particular, the maps from k must be injections, and (f # x ) 1 ((ε)) = m x. If k[ε]/ε 2 k is the map obtained by dividing by (ε), then the map O x k[ε]/ε 2 k vanishes on m x so that we have a map k(x) = O x /m x k. This discussion results in the following commutative diagram of fields: k k f # x k(x) k 7
8 where the map k k is an isomorphism. This shows that k(x) = k so that x is rational. Notice also that f x # restricts to a kalgebra map α : m x (ε) = k. This map must vanish on m 2 x because ε 2 = 0, so α induces an element in T x. Next, suppose we are given a point x X such that k(x) = k and a klinear map α : m x /m 2 x k. To construct a morphism (f, f # ) from Y to X, first let f((ε)) = x. Next, for any open U X not containing x, define f # U : O X(U) O Y ( ) = 0 to be the zero map. Now suppose that U contains x. The kscheme structure on X gives O X (U) the structure of a kalgebra. Also, since x is a rational point, given any a O X (U), we can canonically decompose a into a = a(x) + (a a(x)), where a(x) = a + m x k(x) = k. Notice that a a(x), U m x O x. Now, define f # U : O X(U) O Y (f 1 (U)) = k[ε]/ε 2 by: O X (U) O x O x /m 2 x k[ε]/ε 2 a a, U a, U + m 2 x a(x) + α( a a(x), U + m 2 x)ε 2 That this is a map of rings follows from the klinearity of α and the fact that (a a(x)) 2, U m 2 x. f # is a map of sheaves because the natural maps O X (U) O x and O X (V ) O x commute with restriction for all open V U containing x. Hartshorne, Exercise II Let k = F p be the finite field with p elements. Describe Spec k[x]. What are the residue fields of its points? How many points are there with a given residue field? Proof. k[x] is a PID, so its nonzero prime ideals are maximal and are in one to one correspondence with monic, irreducible polynomials in k[x]. Let f(x) be a monic, irreducible polynomial of degree n in k[x], and let m = (f). Since quotienting and localizing commute, we can either view the residue field of f as k[x] m /(m k[x]), or equivalently as K = (k[x]/m) (0) = k[x]/m. K/k is the splitting field of f, and thus has degree equal n. By uniqueness of finite fields, it follows that K = F p n, so we see that the residue fields of the points of Spec k[x] are all of the form F p n for some n. For fixed n, to determine how many points there are with residue field F p n, we must determine how many monic, irreducible polynomials of degree n exist in k[x]. For this, we reference without proof a result of Gauss, which states that the number of monic, irreducible polynomials of degree n in k[x] is 1 ( n ) µ p d n d d n where µ is the Möbius function given by 0 if p 2 m for some prime p µ(m) = 1 if m = 1 or m is the product of an even number of distinct primes 1 if m is the product of an odd number of distinct primes Hartshorne, Exercise II In this exercise, we compare some properties of a ring homomorphism to the induced morphism of the spectra of rings. (a) Let A be a ring, X = Spec A, and f A. Show that f is nilpotent if and only if D(f) is empty. 8
9 (b) Let ϕ : A B be a homomorphism of rings, and let f : Y = Spec B X = Spec A be the induced morphism of affine schemes. Show that ϕ is injective if and only if the map of sheaves f # : O X f O Y is injective. Show furthermore in that case f is dominant, i.e., f(y ) is dense in X. (c) With the same notation, show that if ϕ is surjective, then f is a homeomorphism of Y onto a closed subset of X, and f # : O X f O Y is surjective. (d) Prove the converse to (c), namely, if f : Y X is a homeomorphism onto a closed subset, and f # : O X f O Y is surjective, then ϕ is surjective. Proof. (a) This follows from the fact that any nilpotent element is contained in all prime ideals. (b) f # : O X f O Y is an injective map of sheaves if and only if f # U : O X(U) f O Y (U) is injective for all open U X. Take U = X to obtain a map Γ(X, O X ) = A Γ(Y, O Y ) = B. This is ϕ, so it is injective. Next, suppose that ϕ is injective, so we assume A B is a subring of B. To show that f # is an injective map of sheaves, we show it is injective on all stalks. Let p X, and notice that the stalk of O X at p is the localization A p. The stalk of f O Y at p can be calculated as follows: (f O Y ) p = lim p U O Y (f 1 (U)) = lim p D(g) O Y (f 1 (D(g))) = lim B g = S 1 B g A\p where S is the multiplicative set A \ p B. The second equality above results from the fact that sets of the form D(g) form a base for the topology on X, and the third equality results from the identification of f 1 (D(g)) = D(ϕ(g)) with D(g), viewing A as a subring of B. The map induces by ϕ on stalks is then inclusion: f p # : A p S 1 B a a a a which is clearly injective. To show that f is dominant, consider any nonempty open set U X. Then O X (U) is nonzero, so by injectivity it follows that O Y (f 1 (U)) is nonzero, so that f 1 (U) is nonzero. Hence, f(y ) meets every nonempty open set U X, and thus f(y ) is dense in X, so that f is dominant. (c) We show that f is injective and closed, so that it is a homeomorphism onto its image. Suppose f(p) = f(p ) so that ϕ 1 (p) = ϕ 1 (p ). For any b p, by surjectivity choose a with ϕ(a) = b. Then we have b p if and only if ϕ(a) p if and only if a ϕ 1 (p) = ϕ 1 (p ) if and only if ϕ(a) p if and only if b p. So f is injective. Next, consider any ideal I B and the closed set it defines V (I). Since ϕ is surjective, we have the equivalence I p if and only if ϕ 1 (I) ϕ 1 (p), so that f(v (I)) = V (ϕ 1 (I)) so that f is a closed map. Now, since ϕ is surjective, we have A/ ker ϕ = B so that the prime ideals of B are precisely the prime ideals of A containing ker ϕ. It follows that f(spec B) = V (ker ϕ) is closed. 9
10 Finally, to see that the map of sheaves is surjective, again we check on stalks. Similar to part (b), here we have (f O Y ) p = lim p U O Y (f 1 (U)) = lim p D(g) O Y (f 1 (D(g))) = lim B ϕ(g) = S 1 B g A\p where now S is the multiplicative set ϕ(a\p) B because ϕ is not necessarily injective. The map on stalks is then f # p : A p S 1 B a ϕ(a) a ϕ(a ) whose surjectivity follows from that of ϕ. (d) Let A = A/ ker ϕ and let X = Spec A. We can think of ϕ as the composition of the surjective map π : A A and the injective map i : A B. It follows that f is the composition of g : Y X induced by i and h : X X induced by π. It is sufficient to show that g is an isomorphism of schemes, so that A = B from which it follows that ϕ(a) = B. Since π is surjective, it follows from part (c) that h is a homeomorphism onto a closed subset of X and that the induced map on sheaves h # : O X h O X is surjective. Similarly, since i is injective, it follows from part (b) that g is dominant and that the induced map on sheaves g # : O X g O Y is injective. Since f = h g is a homeomorphism onto a closed set, as is h, it follows that g must also be a homeomorphism onto a closed set. Since g is dominant, it follows that g is a homeomorphism. Next, we know that g # is injective on sheaves, so we are finished if we show it is also surjective. By hypothesis, f # : O X f O Y is surjective as is h #, and since f # = h g # h # it follows that h g # is also surjective. Finally, we show that the surjectivity of h g # implies that of g #. Take any p X. We have a commutative diagram on stalks h g # (h O X ) h(p) h(p) (f O Y ) h(p) (O X ) p g # h(p) (g O X ) p where the vertical maps are actually inclusions. Explicitly, the left vertical map is given by the inclusion lim O X (h 1 (V )) lim O X (U) p h 1 (V ) p U Where V ranges over open sets in X containing h(p) and U ranges over open sets in X containing p. It turns out that this map along with the right vertical map in the diagram are surjections, which, with the commutativity of the above diagram, proves that g # is surjective. To see that the above map is actually surjective, notice that it is enough to show that the directed system of the limit on the left hand side includes all basic open sets D(a ) X which contain p, some a A. Since π is surjective, we can find find 10
11 a A with π(a) = a. Then h 1 (D(a)) = D(a ). Hence the limit on the left hand side actually recovers the full stalk, and the map is surjective. Hartshorne, Exercise II.4.7. Schemes over R. For any scheme X 0 over R, let X = X 0 R C. Let α : C C be complex conjugation, and let σ : X X be the automorphism obtained by keeping X 0 fixed and applying α to C. Then X is a scheme over C, and σ is a semilinear automorphism, in the sense that we have a commutative diagram X σ X Since σ 2 = id, we call σ an involution. Spec C α Spec C (a) Now let X be a separated scheme of finite type over C, let σ be a semilinear involution on X, and assume that for any two points x 1, x 2 X, there is an open affine subset containing both of them. (This last condition is satisfied for example if X is quasiprojective.) Show that there is a unique separated scheme X 0 of finite type over R, such that X 0 R C = X, and such that this isomorphism identifies the given involution of X with the one on X 0 R C described above. For the following statements, X 0 will denote a separated scheme of finite type over R, and X, σ will denote the corresponding scheme with involution over C. (b) Show that X 0 is affine if and only if X is. (c) If X 0, Y 0 are two such schemes over R, then to give a morphism f 0 : X 0 Y 0 is equivalent to giving a morphism f : X Y which commutes with the involutions, i.e., f σ X = σ Y f. (d) If X = A 1 C, then X 0 = A 1 R. (e) If X = P 1 C, then either X 0 = P 1 R, or X 0 is isomorphic to the conic in P 2 R given by the homogeneous equation x x x 2 2 = 0. Proof. (a) The case for X = Spec A is treated in part (b), where X 0 = Spec A σ with A σ the subring of fixed elements under σ : A A. (b) If X 0 = Spec A, then X = Spec A R C is affine. If X = Spec A, then X 0 = Spec A σ, where A σ is the subring of fixed elements under σ : A A. To see this, note that A σ R C = A under the map a 1 + b i a + bi for a, b A σ. This map is surjective because for any a A we can write: ( ) ( ) a + σ (a) a σ (a) a = + i 2 2i (c) By the universal property of the fiber product, given a map of schemes over R, f 0 : X 0 Y 0, we get map of schemes over C, f : X Y, satisfying the following commutative 11
12 diagram: X 0 X Spec C f 0 Y 0 f Y id Spec C Spec R Now σ X is a map X X, where X is considered a scheme over X 0, and similarly for σ Y. It follows that the above diagram commutes when we replace f with σ Y f σ X. By uniqueness, it follows that σ Y f = f σ X. For the reverse direction, first suppose X = Spec A and Y = Spec B are affine. Then, the induced map f : B A restricts to a ring homomorphism f 0 : B σ A σ because f commutes with σ. Then f 0 : X 0 Y 0 is a morphism of schemes. (d) This follows from (b) and the observation that C[t] σ = R[t]. (e) Hartshorne, Exercise II.5.7. Let X be a noetherian scheme, and let F be a coherent sheaf. (a) If the stalk F x is a free O x module for some point x X, then there is a neighborhood U of x such that F U is free. (b) F is locally free if and only if its stalks F x are free O x modules for all x X. (c) F is invertible (i.e., locally free of rank 1) if and only if there is a coherent sheaf G such that F G = O X. Proof. (a) We can assume that X = Spec A, and that F = M for some finitely generated A module M = x 1,..., x k. Let x = p X, so by hypothesis M p = A n p. Let m 1,..., m n generate M p freely. Consider the equations n a ij x i m j b ij and let b = b ij. Then we have a surjective map of A b modules j=1 f : A n b M b (a 1,..., a n ) a i m i The kernel of this map is finitely generated because A n b is noetherian. For each generator (a 1,..., a n ), there exists some s / p such that s (a 1,..., a n ) = 0. Let g be the product of b and all such s. I claim that m 1,..., m n generate M g freely over A g. This follows from the fact that the localization of f at g is an isomorphism. (b) The reverse direction follows from (a). For the forward direction, take any x X and find an open affine neighborhood U containing x such that F U = O n X U (we know there must be finitely many direct summands because F is coherent). Now we have n n F x = lim F U (V ) = x V U lim O n X U(V ) = x V U 12 lim x V U i=1 O X (V ) = i=1 O x
13 where the last equality is justified by the following isomorphism: n n O x lim O X (V ) i=1 x V U i=1 n ( s 1, V 1,..., s n, V n ) (s 1,..., s n ), V i Notice that the rank of F x is equal to the rank of F at x. (c) Suppose F is invertible and let G = F ˇ = H om OX (F, O X ). Define a map of sheaves ϕ U : F (U) OX (U) Hom OX U (F U, O X U ) O X (U) by s, U f f U (s), U. By the sheaf axiom, it is enough to show that this induces an isomorphism on an open cover of X. Choose an open cover U i such that F Ui is free on U i of rank 1. On such a U, the above map has inverse given by s, U 1, U f s where (f s ) V : O X (V ) O X (V ) is defined by t, V s V t, V. To see that Fˇ is coherent... For the reverse direction, I first claim that Pic(Spec R) is trivial for R a commutative local ring, that is, if M and N are finitely generated Rmodules such that M R N = R, then M = N = R. To see this, let k be the residue field of R and notice that by extension of scalars we have (M R k) k (N R k) = k. It follows that both M R k and N R k are 1 dimensional vector spaces over k. We now need to invoke the following form of Nakayama s lemma: if M is a finitely generated Rmodule, and J is the Jacobson radical of R, and the images of m 1,..., m n M generate the quotient M/JM as an Rmodule, then the m 1,..., m n also generate M as an Rmodule. In our case, J = m R, and M/mM = M R k is generated as an Rmodule by one element x + mm. Nakayama s lemma then says that x M generates M as an Rmodule, so that M = R. The same follows for N. Back to our problem, if we assume that F G = O X, then taking stalks at any point x X we have F x Ox G x = Ox. By what we have just show, it follows that F x = Ox, so that F is invertible by part (b). Hartshorne, Exercise II.5.8. Again let X be a noetherian scheme, and F a coherent sheaf on X. We will consider the function ϕ(x) = dim k(x) F x Ox k(x), where k(x) = O x /m x is the residue field at the point x. Use Nakayama s lemma to prove the following results. (a) The function ϕ is uppersemicontinuous, i.e., for any n Z, the set {x X ϕ(x) n} is closed. (b) If F is locally free, and X is connected, then ϕ is a constant function. (c) Conversely, if X is reduced, and ϕ is constant, then F is locally free. Proof. (a) We will show that {x X ϕ(x) < n} is open. First, find an open affine neighborhood of x, U = Spec A, such that F U = M for some finitely generated A module M, and let m 1,..., m n be generators of M over A. Next, if x = p Spec A, 13 i=1
14 let [ x1 ] [ xk ],..., 1 1 be a k(p)basis of M p /pm p. Nakayama s lemma then states that (1) x 1 1,..., x k 1 is a minimal generating set of M p over A p. Write m i k 1 = a ij x j b ij 1 j=1 and let b = b ij and V = D(b). For and q D(b), (1) is still a generating set of M q over A q. By Nakayama s lemma again, it follows that ϕ(q) k < n so that p D(b) {x X ϕ(x) < n}, and thus {x X ϕ(x) < n} is open. (b) If F is locally free, then by part (b) of the previous exercise, its stalks F x are free O x  modules for all x X, and the rank of F x is equal to the rank of F at x. Thus for any open set U on which F is free, the value of ϕ on U is the rank of F on U, which must be finite because F is coherent. For any n, by the definition of locally free, both the set on which F has rank n and the set on which F has rank different from n are open. Since X is connected, it follows that there is some n such that ϕ(x) = n for all x X. (c) We may assume X = Spec A and F = M for some finitely generated Amodule M. Let p X, and suppose that ϕ(p) = n, so that by Nakayama s lemma there is a set m 1,..., m n M generating M p over A p (by clearing denominators we may assume the m i M). If x 1,..., x k generate M over A, write x i = n j=1 a ij b ij m j Let b = b ij. The same equation above then shows that m 1,..., m n generate M b over A b. Also, for any q D(b), the m i form a k(q)basis for M q /qm q because they are a generating set of nelements, and ϕ is constant. We now show that the generation of M b over A b is free. Suppose that we have a dependence relation n i=1 a i s i m i = 0 This gives a dependence relation in M q /qm q for each q D(b), which by linear independence, gives that a i q. It follows that a i N (A b ) = (N (A)) b = 0, because localization commutes with quotients, and A is reduced. This shows that the m i generate M b freely because coherent sheaves on an affine noetherian scheme are determined by their global sections (Corollary 5.5 in Hartshorne). 14 over A b. This is enough to show that M D(b) = O X n U
15 Hartshorne, Exercise II Vector Bundles. Let Y be a scheme. A (geometric) vector bundle of rank n over Y is a scheme X and a morphism f : X Y, together with additional data consisting of an open covering {U i } of Y, and isomorphisms ψ i : f 1 (U i ) A n U i, such that for any i, j, and for any open affine subset V = Spec A U i U j, the automorphism ψ = ψ j ψ 1 i of A n V = Spec A[x 1,..., x n ] is given by a linear automorphism θ of A[x 1,..., x n ], i.e., θ(a) = a for any a A, and θ(x i ) = a ij x j for suitable a ij A. An isomorphism g : (X, f, {U i }, {ψ i }) (X, f, {U i}, {ψ i}) of one vector bundle of rank n to another one is an isomorphism X X of underlying schemes, such that f = f g, and such that x, f together with the covering of Y consisting of all the U i and U i, and the isomorphisms ψ i and ψ i g, is also a vector bundle structure on X. (a) Let E be a locally free sheaf of rank n on a scheme Y. Let S(E ) be the symmetric algebra on E, and let X = Spec S(E ), with projection morphism f : X Y. For each open affine subset U Y for which E U is free, choose a basis of E, and let ψ : f 1 (U) A n U be the isomorphism resulting from the identification of S(E (U)) with O(U)[x 1,..., x n ]. Then (X, f, {U}, {ψ}) is a vector bundle of rank n over Y, which (up to isomorphism) does not depend on the bases of E U chosen. We call it the geometric vector bundle associated to E, and denote it by V(E ). (b) For any morphisms f : X Y, a section of f over an open set U Y is a morphism s : U X such that f s = id U. It is clear how to restrict sections to smaller open sets, or how to glue them together, so we see that the presheaf U {set of sections of f overu} is a sheaf of sets on Y, which we denote by S (X/Y ). Show that if f : X Y is a vector bundle of rank n, then the sheaf of sections S (X/Y ) has a natural structure of O Y module, which makes it a locally free O Y module of rank n. (c) Again let E be a locally free sheaf of rank n on Y, let X = V(E ), and let S = S (X/Y ) be the sheaf of sections of X over Y. Show that S = E, as follows. Given a section s Γ(V, E ) over any open set V, we think of s as an element of Hom(E V, O V ). So s determines an O V algebra homomorphism S(E V ) O V. This determines a morphism of spectra V = Spec O V Spec S(E V ) = f 1 (V ), which is a section of X/Y. Show that this construction gives an isomorphism of E to S. (d) Summing up, show that we have established a onetoone correspondence between isomorphism classes of locally free sheaves of rank n on Y, and isomorphism classes of vector bundles of rank n over Y. 15
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