# Exercises of the Algebraic Geometry course held by Prof. Ugo Bruzzo. Alex Massarenti

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1 Exercises of the Algebraic Geometry course held by Prof. Ugo Bruzzo Alex Massarenti SISSA, VIA BONOMEA 265, TRIESTE, ITALY address:

2 These notes collect a series of solved exercises from the book Algebraic Geometry by R. Hartshorne [Ha]. Chapters and exercises are numbered as in the book. SISSA - Trieste November - December 2011

3 Contents Chapter II - Schemes Sheaves 4 Exercise Exercise Exercise Schemes 7 Exercise Exercise Exercise Exercise Exercise First Properties of Schemes 9 Exercise Exercise Separated and Proper Morphisms 10 Exercise Exercise Exercise Exercise Sheaves of Modules 13 Exercise - Adjunction Formula 13 Exercise 5.1-(b), and a consequence 14 Exercise Bibliography 15 3

4 Chapter II - Schemes 1 - Sheaves Exercise 1.8. For any open subset U X, show that the functor Γ(U, ) from sheaves on X to abelian groups is a left exact functor, i.e. if is an exact sequence of sheaves, then is an exact sequence of groups. 0 F φ F ψ F 0 Γ(U, F ) φ(u) Γ(U, F) ψ(u) Γ(U, F ) PROOF. Since φ is an injective morphism of sheaves, φ(u) is injective for any open subset U X. So it is enough to prove that for any U X. Since is exact, the induced sequence on stalks Im(φ(U)) = ker(ψ(u)) 0 F φ F ψ F 0 F p φ p F p ψ p F p is exact for any p X. Let s Γ(U, F ) be a section of F on U, then ψ p ((φ p )(s p )) = 0 for any p U, that is ψ(φ(s)) p = 0 for any p U. So for any p U there is an open neighborhood U p of p in U such that ψ(φ(s)) Up = 0. So ψ(u)(φ(u)(s)) = 0 and Im(φ(U)) ker(ψ(u)). Now take v ker(ψ(u)), then for any p U there exits s p F p such that φ p (s p ) = v p. Thus the are an open covering {U i } of U and sections s i F (U i ) such that φ(s i ) = v Ui. Now for any i, j, and since φ in injective we get φ(s i Ui U j ) = v Ui U j = φ(s j Ui U j ), s i Ui U j = s j Ui U j for any i, j. Since F is a sheaf there is a section s F (U) such that s Ui = s i for any i. Now, from φ(s Ui ) = v Ui for any i we get φ(u)(s) = v and ker(ψ(u)) Im(φ(U)). Remark. The functor Γ(U, ) need not to be exact, if ψ : F F is surjective the maps on sections ψ(u) : F(U) F (U) need not to be surjective. So it make sense to consider its right derived functors and to define the cohomology of a sheaf. If X is a topological space the category of sheaves of abelian groups on X has enough injectives, that is any sheaf F on X admits an injective resolution 0 F I 0 I

5 1 - SHEAVES 5 of injective sheaves I j. The cohomology groups of F are defined as H i (X, F) := h i (Γ(X, I )). Since any two resolution are homotopy equivalent the definition does not depend on the one we choose. Exercise Skyscraper Sheaves. Let X be a topological space, let p be a point, and let A be an abelian group. Define a sheaf i p (A) as follows: i p (A)(U) = A if p U, 0 otherwise. Verify that the stalk of i p (A) is A at every point q {p}, and 0 elsewhere. Show that this sheaf could also be described as i (A), where A denotes the constant sheaf A on the closed subspace {p}, and i : {p} X is the inclusion. PROOF. If U X is an open subset and q U {p} then p U. In fact if p U c then {p} U c and {p} U =. Take a point q {p}, then any open subset U X containing q has to contain p. So i p (U) = A for any open subset U containing q. The stalk i p (A) q is the direct limit i p (A) q = lim q U i p (U) = lim A = A. q U Now take q / {p}, then there is a closed subset C containing p such that q / C, so V = C c is an open subset containing q such that p / U. Since i p (V) = 0 any section s of i p (A) on V is zero and considering the stalk on q we have s q = 0 for any s q i p (A) q. The direct image sheaf is defined as i (A)(U) := A(i 1 (U)). If p U then i 1 (U) = {p} and i (A)(U) = A. If p / U then i 1 (U) = and i p (A)( ) = 0. Remark. Let C be a smooth projective curve over a field k, and p C be a point. The ideal sheaf I p is the sheaf of regular function on X vanishing at p, it is the invertible sheaf O C ( p). We have an exact sequence 0 O C ( p) O C O p 0. The structure sheaf O p of the point p is a skyscraper sheaf. Its stalk is isomorphic to the base field k on p and zero elsewhere. Exercise Some Examples of Sheaves on Varieties. Let X be a variety over an algebraically closed field k. Let O X be the sheaf of regular functions on X. (a) Let Y be a closed subset of X. For each open set U X, let I Y (U) be the ideal in the ring O X (U) consisting of those regular functions which vanish at all points of Y U. Show that the presheaf U I Y (U) is a sheaf. It is called the sheaf of ideals I Y of Y, and it is a subsheaf of the sheaf of rings O X. PROOF. Let U X be an open subset, and let {U i } be an open cover of U. Consider a collection of regular functions f i I Y (U i ) such that f i Ui U j = f j Ui U j for any i, j. In particular f i O X (U i ), and since O X is a sheaf there exists a regular function f O X (U) such that f Ui = f i for any i. Let y Y U be a point, then y Y U i for some i. Since on U i by construction f = f i, we have f (y) = f i (y) = 0. That is f I Y (U). (b) If Y is a subvariety, then the quotient O X /I Y is isomorphic to i (O Y ), where i : Y X is the inclusion, and O Y is the sheaf of regular functions on Y. PROOF. Restriction of regular functions O X (U) i (O Y )(U) = O Y (Y U), f f Y U, gives a morphism of sheaves O X i O Y. By definition of I Y the sequence 0 I Y O X i O Y

6 6 CHAPTER II - SCHEMES is exact. Let x X be a point. We distinguish two cases. - If x / Y then there is an open neighborhood U x of x such that U x Y =, and the stalk i O Y,x is zero. So the morphism on the stalks O X,x i O Y,x is trivially surjective. - If x Y and f y i O Y,x there exists an open neighborhood U x of x in X and a section f i O Y (U x ) representing f y. We can assume U x to be affine, then the inclusion U x Y U x corresponds to a surjection between the coordinate rings A(U x ) A(U x Y). So there exists a section s O X (U x ) restricting to f and again the morphism on the stalks O X,x i O Y,x is surjective. We conclude that the sequence is exact, and i O Y = OX /I Y. 0 I Y O X i O Y 0 (c) Now let X = P 1, and let Y be the union of two distinct points p, q X. Then there is an exact sequence of sheaves 0 I Y O X i O p i O q 0. Show however that the induced map on global sections in not surjective. This show that the global section functor Γ(X, ) is not right exact. PROOF. In this case X = P 1 is a complete variety over an algebraically closed field k, then regular functions on X are constant, that is Γ(P 1, O P 1) = k. On the other hand i O p i O q is a skyscraper sheaf supported on Y = {p, q}, so Γ(Y, i O p i O q ) = k k. Clearly a surjection k k k does not exist. (d) Again let X = P 1, and let O be the sheaf of regular functions. Let K the constant sheaf on X associated to the function field K of X. Show that there is a natural injection O K. Show that the quotient sheaf O/K is isomorphic to the direct sum of sheaves p X i p (I p ) where I p is the group K/O p, and i p (I p ) denotes the skyscraper sheaf given by I p at the point p. PROOF. A regular function on an open subset U X is a rational function f : U k such that on an open covering {U i } of U the restricted functions f Ui are regular on U i. Then f defines a section in K(U). In this way we get an injective morphism O K. For any rational function f : U k we can consider its image in the quotient K/O p for any p U. So we have a morphism K p X i p (I p ) whose kernel clearly contains O. To conclude we have to prove that the sequence O K i p (I p ) p X is exact. Thus K/O = p X i p (I p ). On the stalk at q X the sequence is O q K q ( p X i p (I p )) q. Now it is enough to observe that K q = K, ( p X i p (I p )) q = K/Oq, and the sequence 0 O q K K/O q 0. is exact. (e) Finally show that in the case of (d) the sequence is exact. 0 Γ(X, O X ) Γ(X, K) Γ(X, K/O X ) 0

7 2 - SCHEMES 7 PROOF. The functor Γ is left exact, so it is enough to prove that Γ(X, K) Γ(X, K/O X ) is surjective. Since K/O X = p X i p (I p ) we have to prove that given a rational function f K and a point p there exists a rational function g K such that g O q for any q = p and g f O p. We can write f as a ratio of polynomials f = P(z) Q(z) = (z a i) (z b j ) = z h (z a i ) (z c j ), and assume p = 0 A 1 P 1. If h 0 then f is regular in p, and g = constant will work. If h > 0 write (z a i ) = i α i z i, (z c j ) = i β i z i and choose g = h i=0 γ i z h, where the γ i are defined recursively as γ 0 = α 0 β 0, γ i = α i i 1 β 0 choices g O q for any q = p and g f O p. Remark. With a bit more technology (e) can be solved easily as follows. The exact sequence of sheaves 0 O X K K/O X 0 yields the following exact sequence in cohomology j=0 c jβ i j 0 Γ(X, O X ) Γ(X, K) Γ(X, K/O X ) H 1 (X, O X )... Now, it is enough to observe that, by Serre duality, on P 1 we have being ω X = OP 1( 2) the canonical sheaf of P 1. H 1 (X, O X ) = H 0 (X, ω X ) = H 0 (X, O X ( 2)) = 0, 2 - Schemes for i > 0. With these Exercise 2.5. Describe Spec(Z), and show that it is a final object for the category of schemes, i.e. each scheme X admits a unique morphism to Spec(Z). PROOF. The ideals of Z are of the form I = (n), and prime ideals are of the form P = (p) such that p Z is a prime integer. Furthermore there is a generic point corresponding to the ideal (0). The closed subsets of Spec(Z) are of the form D(n) = {(p) p n}. The functors Spec and Γ are adjoint. If X is a scheme and A is a ring, there is a bijective correspondence between morphisms X Spec(A) and morphisms of rings A Γ(X, O X ). Recall that we consider morphisms of rings with identity, so there is an unique morphism Z Γ(X, O X ). Exercise 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 giving a point x X and an inclusion map k(x) K. PROOF. Assume to have a morphism ( f, f ) : Spec(K) X. Immediately we get a point x = f (Spec(K)) X. Let Spec(A) be an open affine subset of X containing x, then x Spec(A) corresponds to a prime ideal p. The morphism Spec(K) Spec(A) induces a morphism of rings α : A K whose kernel is p, finally α induces an inclusion A p /pa p = k(x) K.

8 8 CHAPTER II - SCHEMES Fix a point x X and assume to have a morphism k(x) K. The topological space Spec(K) has a unique point, given a point x X we get a continuous map f x : Spec(K) X, Spec(K) x. The sheaf f x (O Spec(K) ) is a skyscraper sheaf whose stalk in x X is isomorphic to K. So to give a morphism of sheaves f x : O X f x (O Spec(K) ) is equivalent to giving for any open subset U X containing x a natural morphism O X (U) K. We take this morphism to be the composition O x (U) O X,x k(x) K. Exercise 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[ɛ]/ɛ 2 be the ring of dual numbers over k. Show that to give a k-morphism of Spec k[ɛ]/ɛ 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 D = Spec k[ɛ]/ɛ 2. Suppose to have a morphism ( f, f ) : D X. We take x X to be the image via f of the unique point in D. The morphism induces an inclusion k(x) k. On the other hand, since f is a k-morphism, the inclusion has to be compatible with the structure morphisms over k. So we have a chain of inclusions k k(x) k, and k(x) = k. The induced morphism on the stalks f x : O X,x k[ɛ]/ɛ 2, maps the maximal ideal m x to the maximal ideal (ɛ), so f x(m 2 x) (ɛ 2 ). Then we get a k-morphism m x /m 2 x k[ɛ]/ɛ 2 k. Now, suppose to have a point x X with k(x) = k and a k-linear morphism L : m x /m 2 x k. We consider the continuous map f : D X mapping D to x X. Consider now the morphisms ev : O X,x k(x) = k, and φ : O X,x k[ɛ]/ɛ 2, g ev(g) + L(g ev(g))ɛ. Note that g ev(g) m x, and that φ is a k-linear ring homomorphism. Now for any open subset U X containing x we consider the composition O X (U) O X,x k[ɛ]/ɛ 2. This gives a morphism of sheaves f : O X f O D. Remark. The meaning of the previous exercise is that to give a morphism of schemes Spec(k[ɛ]/ɛ 2 ) X is equivalent to giving a k-rational point x X and a tangent direction to X at x. Suppose it is given a moduli problem for a certain class of schemes, and let π : X D by a family of these scheme parametrized by D. Let X 0 be the central fiber of this family. Suppose the moduli problem is represented by a scheme M and let x 0 M be the point corresponding to X 0. Then to give morphism from D to M sending D to x 0 is equivalent to give a tangent direction of M at x 0. Naively speaking that s why the scheme D is closely related to the concept of infinitesimal deformation and tangent space to a moduli space.

9 3 - FIRST PROPERTIES OF SCHEMES 9 Exercise Let X be a scheme, let f Γ(X, O X ), and define X f to be the subset of points x X such that the stalk f x of f at x is not contained in the maximal ideal m x of the local ring O x. (a) If U = Spec(B) is an open affine subscheme of X, and if f B = Γ(U, O X U ) is the restriction of f, show that U X f = D( f ). Conclude that X f is an open subset of X. PROOF. Let x U = Spec(B) be the point associated to the prime ideal p B. The maximal ideal of O x is pb p. We have x U X f f / p x D( f ). Since for any affine open subset U X the intersection U X f is open we conclude that X f is open. Exercise In this exercise, we compare some properties of a ring homomorphism to the induced morphism of the spectra of the 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. PROOF. We have, f n = 0 for some non negative integer n f N = p A p f p for any prime ideal p A p / D( f ) for any prime ideal p A. (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. PROOF. Take a point p Spec(A), the stalk ( f O Spec(B) ) p is B A A p. If φ is injective clearly the induced morphism (O Spec(A) ) p ( f O Spec(B) ) p is injective for any p Spec(A), that is f : O X f O Y is injective. Conversely if f : O X f O Y is injective, then the induced morphism on the global sections Γ(X, O X ) Γ(Y, f O Y ) is injective, but this morphism is exactly φ : A B. Let U be the complement of f (Y) in X. The open subset U is covered by subsets of the form D( f ), with f φ 1 (p) for any p Spec(B). Then φ( f ) p for any p Spec(B), so φ( f ) is nilpotent. Since φ is injective f is nilpotent and D( f ) =. (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. PROOF. The morphism φ induces a bijection between the prime ideals of B = A/Ker(φ) and the prime ideals of A containing Ker(φ). The preimage of the open subset D( f ) Spec(A) is D( f + Ker(φ)) Spec(A/Ker(φ)), so principal open subsets of Spec(A/Ker(φ)) are open in the image with respect the induced topology. So f : Y X is continuous and open, therefore it is an homeomorphism onto its image. Finally if φ is surjective, then the induce morphism on the stalks A p B A A p is surjective. Exercise First Properties of Schemes (a) If f : X Y is a morphism, and y Y a point, show that sp(x y ) is homeomorphic to f 1 (y) with the induced topology.

10 10 CHAPTER II - SCHEMES PROOF. Let y Y be a point, and let k(y) be the residue field of y. We have an induced morphism Spec k(y) Y. The fibre of the morphism f over y Y is the scheme X y = X Y Spec k(y). The continuous map induce by the morphism of scheme Spec k(y) Y simply maps g : Spec(k(y)) y. The underlying topological space of X y is sp(x y ) = {(x, ξ) f (x) = g(ξ) = y} = f 1 (y) clearly the projection sp(x y ) f 1 (y), (x, ξ) x is an homeomorphism. (b) Let X = Spec k[s, t]/(s t 2 ), let Y = Spec k[s], and let f : X Y be the morphism defined by sending s s. If y Y is a point a k with a = 0, show that the fiber X y consists of two points, with residue field k. If y Y corresponds to 0 k show that the fiber X y is a nonreduced one-point scheme. If η is the generic point of Y, show that X η is a one-point scheme, whose residue field is an extension of degree two of the residue field of η. (Assume k algebraically closed). PROOF. The fibre is the spectrum of the tensor product k[s, t]/(s t 2 ) k[s] k[s]/(s a) = k[s, t]/(s t 2, s a). The fibre is the zero dimensional subscheme cut out on the parabola s = t 2 by the line s = a. If a = 0 this intersection consists of two distinct points with residue field k. On the other hand, if a a = 0, the line s = 0 is tangent to the parabola at the origin. So the ring of the fibre is k[t]/(t 2 ), that is X 0 is a nonreduced double point whose residue field is an extension of degree 2 of the residue field of the generic point η Y. Exercise (a) Closed immersions are stable under base extension: if f : Y X is a closed immersion, and if X X is any morphism, then f : Y X X X is also a closed immersion. PROOF. Since both X and X can be covered by open affine subsets, we can assume X = Spec(A) and X = Spec(B) to be affine. Since f is a closed immersion Y is an affine subscheme of X, and we can write Y = Spec(A/I). Let φ : A A/I be the morphism induced by f. Consider f : Spec(B A (A/I)) Spec(B). Now B A (A/I) = B/ φ(i), so f is a closed immersion. 4 - Separated and Proper Morphisms Exercise 4.2. Let S be a scheme, let X be a reduced scheme over S, and let Y be a separated scheme over S. Let f and g be two S-morphisms of X to Y which agree on an open dense subset of X. Show that f = g. Give examples to show that this result fails if either (a) X is nonreduced, or (b) Y is nonseparated. PROOF. Let α : X Y S Y, x ( f (x), g(x)) be the diagonal morphism of f, g. Let U X be the open subset on which f U = g U. Consider the fiber product W = X Y S Y Y and the following diagram U U W Y X α Y S Y

11 4 - SEPARATED AND PROPER MORPHISMS 11 Note that since f and g agree on U we have α(u) (Y), so U W is an open subset also in the fiber product. The morphism : Y Y S Y is a closed immersion, since Y is separated over S, furthermore closed immersion are stable under base change, so : W X is a closed immersion as well. The injection U X factors through W which is a closed subset of X, and since U is dense we get sp(w) = sp(x). Let V = Spec(A) X be an open affine subset. The morphism V : W V V is a closed immersion. So W V is a closed subscheme homeomorphic to V and determined by an ideal I of A. The morphism Spec(A/I) Spec(A) is injective, so I N. On the other hand I = N = 0 since A is reduced. So W V = V for any V X open affine, and this implies W = X as schemes. Now it is enough to recall that W = {(x, y) α(x) = ( f (x), g(x)) = (y) = (y, y)} = {x f (x) = g(y)}. Since W = X we get f = g. (a) Consider X = Y = Spec(k[x, y]/(x 2, xy)) the affine line with the origin counted twice. The scheme X is nonreduced, it has nilpotents at the origin. The identity Id : X Y, x x, y y, and the morphism f : X Y, x 0, y y mapping X on the affine line, agree on a dense open subset of X but they differs at the origin. Indeed f kill the nilpotent elements. (b) Take X = A 1 and Y = A 1 0 1,0 2 be the affine line with two origins. Consider f 1 : X Y, mapping x x for any x = 0 and and f 2 : X Y, mapping x x for any x = 0 and Clearly f 1, f 2 coincides on X \ {0} but f 1 = f 2. Exercise 4.3. Let X be a separated scheme over an affine scheme S. Let U, V be open affine subsets of X. Then U V is also affine. Give an example to show that this fails if X is not separated. PROOF. Consider the fiber product W = U V α U S V X X S X Now, X is separated over S so is a closed immersion. Closed immersions are stable under base change, then α is a closed immersion as well. Furthermore U S V is affine, being U, V, S affine, and U V is a closed subscheme of the affine scheme U S V. Then U W has to be affine. Take X = A 2 0 1,0 2 the affine plane with doubled origin, U a copy of the affine plane, and V the other copy of A 2. Then U V = A 2 \ {0}, and this is not affine. Suppose A 2 \ {0} to be affine, then A 2 \ {0} = Spec(A). In this case A is the ring of regular functions A = Γ(A 2 \ {0}, O A 2 \{0} ). On the other hand the origin has codimension two in A 2 and A 2 is a normal scheme. So any regular function on A 2 \ {0} extends to a regular function on A 2, that is A = Γ(A 2, O A 2). So A 2 \ {0} = A 2, a contradiction. Exercise 4.4. Let f : X Y be a morphism of separated schemes of finite type over a noetherian scheme S. Let Z be a closed subscheme of X which is proper over S. Show that f (Z) is closed in Y, and that f (Z) with its image subscheme structure is proper over S.

12 12 CHAPTER II - SCHEMES PROOF. Consider the commutative diagram Z f Z Y α S β Now, Z is proper over S, that is α is a proper morphism. So β f Z = α is proper, furthermore β is separated. We conclude that f Z is proper by Corollary 4.8. Since any proper morphism is closed f (Z) is closed in Y. To prove that f (Z) is closed over S we will show that f (Z) S is of finite type, separated, and universally closed. Since f (Z) is a closed subscheme of Y, and Y is of finite type over S we have that f (Z) is of finite type over S. Since f (Z) Y is a closed immersion, and closed immersions are stable under base change, f (Z) S f (Z) Y S Y is a closed immersion. Then f (Z) S is separated. Let V be a scheme over S. Consider the following situation V S Z g V S f (Z) γ V Z f (Z) S f β Let R V S f (Z) be a closed subset. Then g 1 (R) is closed in V S Z. Furthermore (γ g)(g 1 (R)) is closed in V being Z S universally closed. Finally, being g surjective, we get (γ g g 1 )(R) = γ(r). Then γ(r) is closed in V. Exercise 4.9. Show that the composition of projective morphisms is projective. PROOF. Consider the composition of two projective morphisms X Y Z i f π 1 j g π 2 P n Y P m Z where i, j are closed immersions, and π 1, π 2 are projections. Let Seg : P n P m P nm+n+m be the Segre embedding. Considering the following diagram i X P n Y P n P m Z f π 1 Id j j Y P m Z P nm+n+m Z Seg g Z π 2 π

13 5 - SHEAVES OF MODULES 13 it is clear that g f factors as π (Seg (Id j) i), and since Seg is a closed immersion, we conclude that g f is projective. Remark. If X Spec(k) is a scheme over a field k, the structure morphism is projective if it factors through a closed immersion X P n k. We recover the usual notion of projective scheme. As the projective of a scheme over a field is related to the existence of an ample line bundle on X, the projectivity of a morphism f : X Y is related to the existence of a relatively ample line bundle on X, that is a line bundle on X whose restriction to the fibers of f is ample. 5 - Sheaves of Modules Let X be a scheme, and let F be a sheaf of O X -modules. For any open subset U X, F(U) is an O X (U)-module, and we can consider the sheaves associated to the presheaves U T r F(U), S r F(U), r F(U). These sheaves are respectively the tensor algebra, the symmetric algebra, and the exterior algebra of F. If F is locally free of rank n then T r F, S r F, r F are locally free of rank n r, ( n+r 1 n 1 ) and (n r ) respectively. Consider an exact sequence 0 F F F 0 of O X -modules. For any r there is a finite filtration r F = F 0 F 1... F r F r+1 = 0 with quotients for each p. In particular F p /F p+1 = p F r p F n F = n F n F. This formula is very useful in a number of contests. As instance we can derive the adjunction formula. Exercise - Adjunction Formula. Let Y be a smooth subvariety of a smooth variety X. Consider the exact sequence Taking exterior powers we get 0 T Y T X N Y/X 0. n TX m n m = TY NX/Y O X. Then, on the canonical sheaves, we have n m ω Y = ωx NˇX/Y O X, where n = dim(x) and m = dim(y). In particular for a degree d hypersurface Y P n the canonical sheaf is given by ω Y = OX (d n 1), being ω P n = OP n( n 1).

14 14 CHAPTER II - SCHEMES Exercise 5.1-(b), and a consequence. Let (X, O X ) be a locally ringed space, and let E be a locally free O X -module of finite rank. For any O X -module F, Hom OX (E, F) = Eˇ OX F. PROOF. For any U X open subset we have to define a morphism that is a morphism (Hom OX (E, O X ) F)(U) (Hom OX (E, F))(U), Note that a section s F(U) yields a morphism We define our morphism as follows Hom(E U, O X U ) OX (U) F(U) Hom(E U, F U ). φ s : O X U F U, f s f. Hom(E U, O X U ) OX (U) F(U) Hom(E U, F U ), (ψ, s) φ s ψ. If E is locally free, then the stalk E x is locally free, and the above morphism if clearly an isomorphism at the level of stalks. In particular if E = F = L is a line bundle on X, then Lˇ OX L = Hom OX (L, L) = O X. The line bundle on a locally ringed space (X, O X ) form a group called the Picard group of X and denoted by Pic(X). Exercise Let (X, O X ) be a scheme. There is a one-to-one correspondence between isomorphism classes of locally free sheaves of rank r on X and isomorphism classes of rank r vector bundle on X. PROOF. Let π : E X be a vector bundle on X. For any U X open subset, consider the O X (U)-module Γ(U, E) := {σ : U E π σ = Id U }. The correspondence U Γ(U, E) defined a presheaf E on X, and since E is a vector bundle E is indeed a sheaf. Let {U i } be an open cover of X on which E trivializes. Then E Ui = Ui k r, on the other hand the sheaf of sections of U i k r is isomorphic to O r X U i. We conclude that E is locally free of rank r. Now, let F be a locally free sheaf of rank r. There exists on open covering {U i } of X, and isomorphisms φ i : O r X U i F Ui. These induces on U i,j isomorphisms φ i,j : O r X U i,j O r X U i,j represented by an r r matrix with entries in O X (U i,j ). Since φ i,j is invertible, the matrix A x is invertible for any x U i,j. Then we get a morphism g i,j : U i,j GL(k r ). The morphisms {g i,j } satisfies cocycle conditions, so we can construct the vector bundle F associated to the collection {g i,j }. To conclude it is enough to observe that the assignments E E and F F are inverse to each other.

15 Bibliography [Ha] R. Hartshorne, Algebraic Geometry, Springer. 15

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