Sheaves of Modules. Chi-yu Cheng. This is a supplement for the study of Hartshorne s Algebraic Geometry

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1 Sheaves of Modules Chi-yu Cheng This is a supplement for the study of Hartshorne s Algebraic Geometry Currently a graduate student at the University of Washington 1

2 1 Preliminary Observations Observation 1: Let F be a presheaf of abelian groups and G be a presheaf of rings on a topological space X. If F is a presheaf of G -modules, then the sheafification F + is a sheave of G + -modules. Observation 2: Sheafification commutes with finite product. i.e. (F G ) + = F + G +. Observation 3: The push-forward (resp. inverse image) of a quasi-coherent sheaf of some affine scheme to another affine scheme is again quasi-coherent. Observation 4: The following three conditions are equivalent without invoking proposition F is quasi-coherent. 2. For all open subset U X, F U is quasi-coherent. 3. There is an open cover α V α of X such that F Vα is quasi-coherent. and similarly for coherent sheaves. Next I collect some technical facts about sheaves of O X modules. Observation 5: For any sheaf of O X modules F, there is an isomorphism of sheaves of O X -modules: H om OX (O X, F ) = F. Observation 6: If F, G, H are sheaves of O X modules, then there is a morphism of sheaves of O X modules H om OX (F, G ) OX G H om OX (F, G OX H ) defined in the following manner: Given an open subset U X and a pair (ϕ U, h u ) where ϕ U H om OX U (F U, G U ) and h u H (U), associate this pair to a map F (V ) G (V ) OX(V ) H (V ) for every open subset V U by sending any element f v F (V ) to the image of ϕ u (V )(f v ) h u V in G OX F (V ). As ϕ u is a morphism of sheaves, one checks easily that we obtain an element in H om OX U (F U, G U OX U H U ). Hence we obtain a morphism of presheaves of O X modules and hence a morphism of sheaves of O X modules H om OX (F, G ) OX G H om OX (F, G OX H ). This morphism is an isomorphism if either F or H is finite locally free! A special case for this is that if C is locally free, then for any O X -module F, H om OX (C, F ) = C ˇ OX F. Observation 7: If f : X Y is a morphism of locally ringed spaces, then f O Y O X. There are two ways to prove this. The first one is to proceed by concrete construction of the functorf. The other one uses the abstract definition of f as the left adjoint of f. We will proceed in the second. Let F be an arbitrary sheaf of O X modules, then by (7.4.5) in [?] and the adjunction formula, we have Hom OX (f O Y, F ) Hom OY (O Y, f F ) Γ(Y, f F ) = Γ(X, F ) Hom OX (O X, F ). 2

3 As these isomorphisms are functorial in F, the result follows from Yoneda lemma. Observation 8: It is noteworthy that in parallel to the concept of a homogeneous ideal of a graded ring, there is also the notion of a homogeneous submodule of a graded module and that a submodule is homogeneous if and only if it is generated by homogeneous elements. The main purpose of this observation arises from proposition 5.11 (c). Suppose S is a noetherian graded ring and M is a finitely generated graded S module, then for any homogeneous element f S +, the S (f) module M (f) is finitely generated. To prove this, let us consider the homogeneous submodule N generated by the set: {m M m is homogenous, m f l M (f)for some l N.} As M is noetherian, N is finitely generated, say by {m 1,, m n } and note that each m i is homogeneous and there is an l i N such that mi f M l i (f). I claim that { m1 f,, mn l 1 f } generate ln M (f) as an S (f) module. Let m f M l (f), then m M so that there are homogeneous s i such that i s im i = m where each s i m i has degree l deg f. Dividing f l on both sides we have m f = l i simi f. Then by construction of N there is an l l i N such that mi f M l i (f) Hence we may split the simi f as si l f where ɛ ɛ i i + l i = l. This implies that si f S ɛ i (f). Hence M (f) is finitely m i f l i generated as an S (f) -module! Observation 9: Here we give ourselves the liberty to discuss twisted sheaf on a projective space P n k. Let us motivate our discussion by the following observation, albeit somewhat oblique. Suppose (X i, U ij, ϕ ij ) is a gluing data of schemes. Namely, for each i, j, U ij X i is an open sub-scheme that is isomorphic to U ji X j via the isomorphism ϕ ij that satisfies the co-cycle conditions. Let ι i : X i X be the standard open immersion. Then to give a sheaf of O X -modules, it is sufficient to give a sheaf F i of O Xi -modules for each i, and an isomorphism a ij : ι i F i Uij ι j F j Uji for each pair (i, j) satisfying the co-cycle conditions. But on U ji we have ι j = ι i ϕ ji so to give an isomorphism a ij : ι i F i Uij ι j F j Uji = ι i ϕ ji F j is the same as giving an isomorphism b ij : F i Uij ϕ ji F j of O Uij -modules such that ι b ij satisfies the co-cycle conditions. A little calculation shows that b ij has to satisfy b ij = ϕ ki b kj b ik on U ij U ik. Now let us consider P 2 k that is obtained by gluing D(x 0) = Spec k[x 1/0, x 2/0 ], D(x 1 ) = Spec k[x 0/1, x 2/1 ] and D(x 2 ) = Spec k[x 0/2, x 1/2 ] along principal open sets. exact, we have isomorphisms ϕ 10, ϕ 20 and ϕ 12 corresponding to the ring isomorphisms To be k[x 1/0, x 2/0 ] x1/0 k[x 0/1, x 2/1 ] x0/1 where x 1/0 1/x 0/1 and x 2/0 x 2/1 1/x 0/1. k[x 1/0, x 2/0 ] x2/0 k[x 0/2, x 1/2 ] x0/2 where x 1/0 x 1/2 1/x 0/2 and x 2/0 1/x 0/2. k[x 0/2, x 1/2 ] x1/2 k[x 0/1, x 2/1 ] x2/1 where x 0/2 x 0/1 1/x 2/1 and x 1/2 1/x 2/1. To apply the above discussion to P 2 k, note that we need to give D(x i)-modules F i and isomorphisms b ij : F i Uij ϕ ij F j Uji of O Uij -modules where U ij = D(x i x j ). To define the twisted sheaf, say O X (l), one takes F 0 = k[x1/0, x 1/2 ], F 1 = k[x0/1, x 2/1 ] and F 2 = k[x0/2, x 1/2 ]. Next, we need to define isomorphism b ij : F i Uij ϕ ji F j Uji of O Uij -modules. For example, we need an isomorphism b 01 : k[x1/0, x 2/0 ] x1/0 ϕ 10 k[x0/1, x 2/1 ] x0/1. This is equivalent 3

4 to give a k[x 1/0, x 2/0 ] x1/0 -modules isomorphism k[x 1/0, x 2/0 ] x1/0 k[x 0/1, x 2/1 ] x0/1 where the k[x 1/0, x 2/0 ] x1/0 -module structure for k[x 0/1, x 0/2 ] x0/1 is induced by ϕ 10. In the case of O X (l), b 01 : k[x 1/0, x 2/0 ] x1/0 k[x 0/1, x 2/1 ] x0/1 is multiplication by (x 0/1 ) l b 02 : k[x 1/0, x 2/0 ] x2/0 k[x 0/2, x 1/2 ] x0/2 is multiplication by (x 0/2 ) l and b 21 : k[x 0/2, x 1/2 ] x1/2 k[x 0/1, x 2/1 ] x2/1 is multiplication by (x 2/1 ) l. It can be checked that these {b ij } satisfies the required condition. 2 Pull-back and Push-forward Are Adjoints This section is devoted to the very important and useful fact that the inverse image sheaf construction is the left adjoint of the push-forward functor. Let us start by recalling the constructions. In general, if π : X Y is a continuous map of topological spaces, then we have the inverse image π G for any sheaf G on Y. The first step is to assign each open subset U X to the direct limit direct limit says there is a map x lim V π(u) lim V π(u) lim V π(u) G (V ). Then note that if U U, the universal property of G (V ) lim V π(u ) G (V ). The map is rather concrete: if G (V ), represented by some element y G (V ), then x is mapped to y s image in lim V π(u ) image sheaf π G is defined to be the sheafification of π G pre. G (V ). In this way we obtain a presheaf on X denoted by π G pre. Then the inverse Example 2.1. π G pre in general is not a sheaf. Consider X = Spek k Spec k, Y = Spec k and π : X Y is given by the diagonal ring map k k k. Then π G pre (X), π G pre (Spec k ) and π G pre ( Spec k) are all k and the restriction maps are all identity maps on k. But we can choose two distinct elements from π G pre (Spec k ) and π G pre ( Spec k). Then there is no way we can glue a single element in π G pre (X) by the identity map on k. π is actually a covariant functor: if φ : F G is a morphism of sheaves on Y, then if U X and V 1 V 2 are two open subsets of Y containing π(u), the universal property of direct limit ensures us that is a unique arrow filling in the dotted arrow in the commutative diagram lim F (V ) V π(u) F (V 1 ) F (V 2 ) lim G (V ) V π(u) G (V 1 ) G (V 2 ) It is easily checked that the collection of dotted arrows forms a morphism between presheaves π F pre and π G pre. Then as sheafification is also a functor, we obtain π as a functor from the category of sheaves on Y to the category of sheaves on X. 4

5 Notation If F is a sheaf, then F will be the presheaf obtained by the forgetful functor that forgets the sheaf structure. For any topological space X, Hom X pre(f, F ) denotes the set of morphisms of pre-sheaves between F and F. Lemma With the same notations, if G is a sheaf on Y, then there is a morphism of presheaves G π π G pre. Proof. Let V 1 V 2 be two open subsets of Y, then the fact that V i π(π (V i )) gives two arrows G (V i ) π π G pre (V i ) = π G pre (π (V i )). Then the defining property of π G pre (π (V 2 )) implies that the diagram commutes. G (V 1 ) π G pre (π (V 1 )) G (V 2 ) π G pre (π (V 2 )) Lemma If ψ : π G pre F is a morphism of presheaves on X, then there is a corresponding morphism α(ψ) : G π F of sheaves on Y. Proof. Let V be an open subset of Y. Then the fact that V π(π (V )) implies there is an arrow G (V ) π G pre (π (V )). So that G (V ) maps to π F (V ) = F (π (V )) by the diagram ψ(π (V )) π G pre (π (V )) F (π )(V ) α(ψ)(v ) G (V ) V 1 V 2 be two open subsets of Y. Now if V 1 V 2 are two open subsets of Y, then lemma implies the following diagram commutes G (V 1 ) α(ψ)(v 1) π G pre (π (V 1 )) F (π (V 1 )) G (V 2 ) α(ψ)(v 2) π G pre (π (V 2 )) F (π (V 2 )) Hence we obtain a morphism of sheaves α(ψ) : G π F on Y. Lemma If ϕ : G π F is a morphism of sheaves on Y, then there is a corresponding morphism of presheaves β(ϕ) : π G pre F on X. Proof. Given an open subset U X, we want to define an arrow β(ϕ)(u) : π G pre (U) F (U). This can be achieved by the following commutative diagram G (V 1 ) G (V 2 ) π G pre (U) ϕ(v 1) ϕ(v 2) F (U) F (π (V 1 )) F (π (V 2 )) 5

6 where Y V 1 V 2 π(u). Note that F (π (V i )) maps to F (U) as π (V i ) U. Then by universal property of direct limit, the dotted arrow can be fitted in by an arrow which we call β(ϕ)(u). Again it is easy to check that these β(ϕ)(u) induce a morphism of presheaves β(ϕ) : π G pre F. Proposition 2.2. The assignments in lemma and are inverse to each other. Therefore, we obtain a bijection α : Hom X pre(π G pre, F ) Hom Y (G, π F ). Proof. We first show that βα(ψ) = ψ : π G pre F. Let U X be an open subset and suppose y π 1 G pre (U) is represented by some element x G (V ) where V π(u) is an open subset of Y. Then we have the inclusions π (V ) π π(u) U. Denote the equivalence class of x in π G pre (U) by [x] and that of x in π G pre (π (V )) by [x]. By lemma and 2.1.3, we have the following commutative diagram G (V ) π G pre (U) π G pre (π (V )) α(ψ)(v ) β(α(ψ))(u). ψ(π (V )) F (π (V )) F (U) Therefore, β(α(ψ))(u)([x]) = (α(ψ)(v )(x)) U = (ψ(π (V ))([x] )) U = ψ(u)([x]). Now we prove that αβ(ϕ) = ϕ : G π F. Let V Y be an open subset and x G (V ) be an element. Denote the equivalence class of x in π G pre (π (V )) by [x]. By lemma 2.1.2, the diagram π G pre (π (V )) F (π )(V ) G (V ) β(ϕ)(π (V )) αβ(ϕ)(v ) commutes. Moreover, by lemma 2.1.3, the following diagram commutes. Hence we have G (V ) π G pre (π (V )) ϕ(v ) β(ϕ)(π (V )) F (π (V )) F (π (V )) αβ(ϕ)(v )(x) = β(ϕ)(π (V ))([x]) = ϕ(v )(x). The last step is to accept this important fact: Proposition 2.3. The sheafification functor is the left adjoint of the forgetful functor. The corollary is the result: 6

7 Corollary 2.4. Let π : X Y be a continuous map of topological spaces, F a sheaf on X and G a sheaf on Y. Then there is a natural bijection Hom X (π G, F ) Hom Y (G, π F ). Proof. Hom X (π G, F ) Hom X pre(π G pre, F ) Hom Y (G, π F ). Now we turn to sheaves of O X -modules where X is a locally ringed space. We begin with a property that I believe to be true. The proof will be updated once I have luxurious time. Proposition 2.5. Let H be a presheaf of rings on X. Then sheafification is a functor from the category of presheaves of H modules to the the category of sheaves of H + -modules on X. Moreover, sheafification is a left adjoint of forgetful functor from the category of sheaves of H + modules to the category of presheaves of H -modules. Namely, if G is a presheaf of H -modules and F is a sheaf of H + -modules, there is a natural bijection Hom H pre(g, F ) Hom H +(G +, F ) In all the discussion that follows, π : X Y is a morphism of locally ringed spaces, F is a sheaf of O X -modules and G is a sheaf of O Y -modules. The push-forward π F is a π O X -module. But π : O Y π O X is part of the data so π F is automatically a sheaf of O Y -modules. As for inverse image sheaf π G, it is a sheaf of π O X -modules (π G pre is a presheaf of π O pre Y -modules). By the adjoint property of inverse image stated in corollary 2.4, π : O Y π O X gives π O Y O X, hence it makes sense to write π G π O Y O X. Definition 2.6. If π : X Y is a morphism of locally ringed spaces and G is a sheaf of O Y moduels. Then we define the pull back π G := π G π O Y O X. As was expected, pull back and push-forward should still be adjoint pairs. We break the proof down to smaller steps. Notation Hom π O pre Y (π G, F ) is the set of all morphisms of pre-sheaves of π O pre Y -modules. Lemma In lemma 2.1.2, α induces a map ᾱ : Hom π O pre Y (π G, F ) Hom OY (G, π F ) and β induces the inverse for ᾱ. Proof. Omitted. Proposition 2.7. There is a natural bijection Hom OX (π G, F ) Hom OX (G, π F ). Proof. As scalar extension is a left adjoint for scalar restriction, we have the natural bijection Hom OX (π 1 G π O Y O X, F ) Hom π O Y (π G, F ). By left adjoint property of sheafification and the previous lemma, we have another two bijections Hom π O Y (π G, F ) Hom π O pre Y (π G pre, F ) Hom OY (G, π F ). 7

8 We now explore further about some nice properties of pull-back. Notation Let f : U V be a morphism of locally ringed spaces. Denote to be the natural bijection specified in 2.7. Θ f (N, M) : Hom OV (f N, M) Hom OU (N, f M) Proposition 2.8. Let id U : U U be the identity map on a locally ringed space. Then there is an isomoprhism ɛ U (M) : id U M M for any sheaf of O U-modules. Moreover, this isomorphism is functorial in M. In other words: ɛ U is a natural isomorphism between id U and the identity functor. Proof. Given an open subset A U, there is an isomorphism M(A) id U Mpre (A). Note that id U Mpre is a sheaf, we do not need to sheafify so id U Mpre = id U M. Similarly, id U M O U O pre U is a sheaf, we do not need to sheafify. Hence M(A) id UM(A) in an obvious way. It is also note worthy that the inverse of this isomorphism coincide with the isomorphism obtained by Yoneda lemma applied to this natural bijection Θ idu (M, M) : Hom OU (M, M) Hom OU (id UM, M). Funtoriality can be checked using definition of id U or purely abstractly using the fact that pull back is the left adjoint of push forward as I did in summer notes for Vistoli s note. Proposition 2.9. Given two morphisms U f V g W, there is a natural isomorphism α f,g : f g (gf). Proof. Given an open subset A W and a sheaf P of O W -modules, a section in 3 Exercises Exercise 5.3: We start with a useful but simple observation. Let F and G be sheaves on a topological space. If for every base open set T, there is a map ϕ T : F (T ) G (T ) such that if T is another base open set and T T T is any base open set, the diagram ϕ T F (T ) G (T ) ρ F T ρ G T ϕ T F (T ) G (T ) commutes, then there is a unique ϕ : F G such that ϕ(t ) = ϕ T. Moreover, if each ϕ T is an isomorphism, so is ϕ. Given an A module homomorphism ϕ hom A (M, Γ(X, F )). Define α : hom A (M, Γ(X, F )) hom OX ( M, F ) in the following manner. First we pick an element f A. Then the homomorphism ϕ : M Γ(X, F ) gives rise to a homomorphism of A f modules ϕ f : M f Γ(X, F ) f. As F (D(f)) is an A f module, the restriction map ρ : Γ(X, F ) F (D(f)) induces a homomorphism of A f modules ρ f : Γ(X, F ) f F (D(f)). Then by composing ρ f ϕ f we obtain a map M(D(f)) = M f F (D(f)). In order to obtain a 8

9 morphism of sheaves of O X modules, we need to check that for any other g A, the following diagram commutes: ρ f ϕ f M f Γ(D(f), F ) ρ fg ϕ fg M fg Γ(D(fg), F ) where the left downward arrow is the localization map and the right downward arrow is the restriction map. This is tautological. In this way we obtain α(ϕ) hom OX ( M, F ) with α(ϕ)(d(f)) = ρ f ϕ f. In particular, if we take f = 1, then α(ϕ)(x) = ϕ. Conversely, we easily define β : hom OX ( M, F ) hom A (M, Γ(X, F )) by taking global sections. Namely, if ψ hom OX ( M, F ), then β(ψ) = ψ(x) : M Γ(X, F ). Then it is easy to see that βα(ϕ) = α(ϕ)(x) = ϕ. Next, we show that α β(ψ) = ψ for any ψ : M F. It is sufficient to prove the equality on an arbitrary D(g). For this, let m g be some function in k M(D(g)) = M g, then ψ(d(g))( m g ) = 1 k g ψ(d(g))( m k 1 D(g)) = 1 g ψ(x)(m) k D(g). On the other hand, α β(ψ)(d(g))( m g ) = α(ψ(x))(d(g))( m k g ) = ρ k g ψ(x) g ( m g ) = 1 k g ψ(x)(m) k D(g). Hence α β(ψ)(d(g)) = ψ(d(g)) for any g A hence α β(ψ) = ψ. Exercise 7: (a) We start with some algebraic comments. If M is a finitely generated module over a noetherian ring A, then if M x = 0 for some prime ideal x A, we have M f = 0 for some f / x. Let us prove this by starting with a set {m 1,, m s } M of generator of M over A. Since M x = 0, for each i, there exists an f i / x such that f i m i = 0. Take f = i f i, then it is easy to see that m i = 0 in M f. But it is also clear that m i generates M f as an A f -module. Hence we get the result. Next, we assume that M is still a finitely generated module over a noetherian ring A and M x is a free A x -module for some prime ideal x A. By clearing denominators, we can assume that {m 1,, m r } M is a basis for M x. Let N M be the sub A-module spanned by it. Our assumption implies M x /N x = (M/N) x = 0. Hence by earlier comment we get an f A such that (M/N) f = 0. This means M f = N f. Now it is tempting to think that {m 1,, m r } is a basis for M f as an A f -module. This set certainly generates M f as we just saw, but it might not be linearly independent over A f. To resolve this, one needs to find a smaller open set inside A f. Again the set {m 1,, m r } induces an exact sequence of A-modules 0 ker ψ A r ψ N 0. This induces another exact sequence 0 ker ψ x = 0 A r x N x 0 ker ψ x = 0 because N x is freely generated by {m 1,, m r } by assumption. As A r is noetherian, ker ψ is finitely generated. So by earlier comment ker ψ g = 0 for some g A. Hence we have an induced exact sequence 0 ker ψ g = 0 A r g N g 0 which means N g A r g is free. Now take A fg A f, one combines two results together and concludes that M fg is free as an A fg -module. Now it is easy to solve this problem. We may assume that X = Spec A and F = M for some finitely generated A-module M. Then M x = M x is free implies that there is a principal open set D(g) Spec A such that M D(g) = M g is free. (b) Easy consequence of (a). (c) There is a natural morphisms of pre-sheaves of O X -modules: β pre : F OX Hom OX (F, O X ) defined by the following: if U X is an open subset, t F (U) is a section over U and 9

10 φ : F U O U is a morphism of O U -modules. Then β pre (U)(t φ) = φ(u)(t). If F is locally free and U X is an open set over which F is free, then β pre (U) is an isomorphism. This means the induced map β : F OX Hom OX (F, O X ) of sheaves is an isomorphism. The hard part is the converse. Suppose U = Spec A X is an open affine sub-scheme. By assumption there exists two finitely generated A-modules M, N such that M F U and Ñ G U. By shrinking to smaller affine, we assume that F U G U O U. Hence we have M A N A. By (a), it is sufficient to show that M p is a free p-module for all p U. Localizing the isomorphism at p, we have M p N p A p. Passing this isomorphism to the residue field κ(p), we see that M p κ(p) and N p κ(p) are one dimensional vector spaces so by Nakayama, M p and N p are generated by one element. Let the generators be m p and n p respectively. Then it is easy to see that m p n p is a generator for M p N p. But it is free of rank 1 so m p n p must be a basis for M p N p. Consequently, m p is torsion free in M p. For if there is a k A such that km p = 0, we would have km p n p = k (m p n p ) = 0. This means M p is a free A p -module generated by m p. Exercise 8: (a). Suppose x X is such that F x Ox k(x) = n, then I will construct an open neighborhood V of x such that F y Oy k(y) has at most dimension n y V. The first step I take would be to pick an affine neighborhood Spec A = U of x so that 1. F x Ox k(x) is spanned by f 1,x, f n,x where each f i F (U). 2. F U = M where M is a finitely generated A module with the generators {m1,, m s }. By Nakayama s lemma, we know that f 1,x,, f n,x generate M x as an A x module. By the definition of stalks, for each m i, there is some g i A such that m i D(gi) = n j=1 a ij D(gi)f j D(gi) for a ij in the section of F over some open neighborhood containing D(g i ). Then consider g = g 1 g s. Then M g can be generated by {f 1 D(g),, f n D(g) } as an A g module. Therefore, for every point y D(g), M y can be generated by f 1,y,, f n,y so that dim k(y) F y Oy k(y) has dimension at most n. What I have prove is that the set {x ϕ(x) < n} is open. (b). The result follows at once as ϕ in this case is locally constant. (c). Again I choose an affine open neighborhood Spec A = U of x such that the above two properties hold. Then from what I have done in (a) I can select a g A such that M g is generated by f i D(g). Clearly { f 1,y,, f n,y } spans the vector space F y Oy k(y). As ϕ is constant, we have that {f 1,y,, f n,y } linearly independent for each y D(g). I claim that M g is a free A g module with basis {f 1 D(g),, f n D(g) }. If n i=1 a if i D(g) = 0, then n i=1 a i,yf i,y = 0 for all y D(g), a i has to lie inside every prime ideal in A g. As X is reduced, we conclude that a i = 0. Hence F is free on D(g). This argument works for all x X so F is locally free. Exercise 5.9: (a) In general, for any graded module M and degree zero homogenous element m M 0, m determines naturally a global section m Γ(X, M). Now if d Z is any integer, M(d)0 = M d so any element m M d determines a global section m Γ(X, M(d)) Γ(X, M(d)). Summing over all d Z we get a graded homomorphism α : M Γ ( M). Here I discuss some properties that will be used later (exercise (c) for example). S is a graded ring that is finitely generated by S 1 over S 0. Lemma Let d Z be an integer. For any quasi-coherent sheaf F on X = Proj S, there is a natural map α(d) : Γ(X, F (d)) Γ(X, Γ (F )(d)). Let β : Γ (F ) F be the isomorphism given by proposition 5.15 and let β (d) be β id OX(d). Then β (d)(x) is just α(d). 10

11 Proof. The map α(d) is just the d-th piece of the map α constructed in (a) where M = Γ (F ) here. Let m Γ(X, F (d)) To prove m := α(d)(m) = β (d)(x)(m), it is sufficient to prove α(d)(m) Xf = β (d)(x)(m) Xf for all f S 1 where we view f as a global section of O X (1). For the right hand side, β (d)(x)(m) Xf = β (d)(x f )(m Xf ). Since F (d)(x f ) = F (X f ) S (f) (d), we may write m Xf = m (ɛ/f deg ɛ ) f d = m (ɛ/f deg ɛ ) f d. Hence we may even assume m Xf = m f d for some m F (X f ). This means the section f d m extends to the section m Γ(X, F (d)). So by construction of β, β (d)(x f )(m f d ) = m f d /f d f d = m/f d f d = m Γ(X f, Γ (F )(d)). This proves the lemma. Corollary 3.1. With the same X and F as in the lemma, the natural map α : Γ (F ) Γ ( Γ (F )) given by (a) is Γ (β ) and hence is an isomorphism. Proof. α = d Z α(d) = d Z β (d)(x) = Γ (β ). (b) Let X = Proj S. By proposition 7.4 I, there is a filtration by graded sub-modules with quotients 0 = M 0 M 1 M r = M M i /M i (S/p i )(l i ) for some homogeneous prime ideal p i S. Then for each i and d 0, the sequence 0 M i d M id (S/p i )(l i + d) 0 induces the following commutative diagram with exact rows 0 M i d M id S/p i (l i + d) 0 0 Γ(X, M i (d)) Γ(X, Mi (d)) Γ(X, S/p i (l i + d)) When i = 1, the mapm i d Γ(X, M i (d)) is an isomorphism. Suppose for any integral graded S, finitely generated over a noetherian ring A in degree one, the map S d Γ(Proj S, S(n)) is an isomorphism for all large d, then there is a d 0 0 such that S/p 0 (l 0 +d) Γ(X, S/p i (l i + d)) is an isomorphism for all d d 0. Then M 1d γ(x, M1 (d)) is an isomorphism for all d d 0. This process proceeds inductively so we get M d Γ(X, M(d)) is an isomorphism for all d large enough. Thus we have reduced to show that if S is an graded integral domain finitely generated over a noetherian ring A in degree 1, then S d Γ(Proj S, S(d)) is an isomorphism for all d large enough. For this, let x 0,, x r S 1 be the generators. Let S = n 0 Γ(X, O X (n)). Then we have inclusions S S S xi. Since S is an integral domain, S S xi so S S is an inclusion. That S is a subring of S xi is slightly more subtle. Take one element t S say t S n for some n 0. Then t is just a r + 1-tuple (t 0,, t r ) where t i S xi (n) and t i = t j in S xix j (n). For each i let ι i : S xi S x0 x r be the natural inclusion. We will identify S as i Im ι i S x0 x r. These t i all 11

12 agree on D + (x 0 x r ) so they represents the same element in S x0 x r (n). Hence any ι i (t i ) is in i Im ι i. If each ι i (t i ) = 0 for all i, we have t i = 0 as ι i is an inclusion. This means S i S xi in S x0 x r. S is an integral domain now follows from the fact that it is a subring of S xi. Let us first show S is integral over S. Let s S. Then Hartshorne showed that there is an integer N such that for every q, S N (s ) q S N. In particular, x N 0 (s ) q S N. Hence (s ) q (1/x 0 ) N S. Then we have the inclusions S S[s ] (1/x 0 ) N S K(S) where K(S) is the fraction field. But (1/x 0 ) N S is a finite S-sub-module of K(S) so that s has to be integral over S. Whence S is integral over S, S is contained in the integral closure S of S. But K(S ) = K(S) and S is a finitely generated k-algebra so the finiteness of integral closure ensures that S is a finite S-module. Then S is noetherian as S is. Whence S is finitely generated over S. Let s 1,, s l be the generators. Pick an integer N such that S N s i S N for all i. Let M = N + i deg s i. Then I claim S M = S M. Pick a homogenous α S M, then α = i α is i for some α i S. I may assume each α i is homogeneous and deg α i + deg s i = deg α. Then deg α i + deg s i = deg α M = N + i deg s i so that deg α i N. But then α i s i S for all i. Consequently α S. (c) I think a more precise definition is needed for the relation. We say two graded S-module M 1 M 2 if and only if there are an integer d and an isomorphism of graded S-modules f : (M 1 ) d (M 2 ) d. Suppose M is a graded S-module and M M for some finitely generated S-module, we will show that M M. Then M will be a coherent sheaf on X. If s S is of degree 1, there is an induced map f s : M (s) M (s) of S (s)-modules. To define this f s, take a fraction m/s deg m M (s) and raise the power of m by multiplying with f until s ɛ m M d. Then set f s (s ɛ m/s deg m+ɛ ) = s ɛ f(m)/s deg m+ɛ. This is a well defined map of S (s) -modules. f s is in fact an isomorphism. These f s is compatible with localization so we get an isomorphism M M. Since M is coherent, so is M. Hence we get a functor mapping from the category of quasi-finitely generated graded S-modules, modulo the equivalence relation (abbreviated as QFGSM) to Coh (X). Conversely, if F is a coherent sheaf on X, Γ (F ) is a graded S-module M. As in the proof for theorem 5.19 II, there is a finitely generated sub-module M M such that the inclusion M M induces and isomorphism M M. Consider the following commutative diagram M M Γ ( M ) Γ ( M) Part (b) says there is an integer d such that the left vertical arrow is an isomorphism in degree greater than d. Corollary 3.1 says the right vertical arrow is an isomorphism. Hence M d = M d. Therefore Γ (F ) is quasi-finitely generated. Proposition 5.15 ensures that Γ 12.

13 is isomorphic to id Coh (X). Part (b) says Γ is isomorphic to id QFGSM. Exercise 5.14 (a) That X is normal implies the local ring of X at every point is an integral domain. Hence the irreducible components of X are disjoint from each other. But X is connected so X is actually irreducible. X is also reduced so by proposition 3.1 II, X is integral. I will also show that the homogenous coordinate ring S is an integral domain. There are two possible ways to do this. First, let me show that I = Γ (I X ) is radical. For this, let s k[x 0,, x r ] be an element such that s n I for some n N. We want to show s I. We may assume s is a homogeneous element. Then for each i, s n deg s n /xi I (xi). Since X = Proj S is integral, I (xi) is a prime ideal for each i. Hence s/x deg s i I (xi). Then there is an integer m i such that x mi i s I. Since I is saturated by exercise 5.10 (a), we see that s I. Next, let me show that I is a prime ideal. Let a, b k[x 0,, x r ] be two homogeneous polynomials and a b I. We then have V (I) = (V (I) V (a)) (V (I) V (b)). Since V (I) is irreducible as a topological space, we may assume V (I) = V (I) V (a) = V (I + a). Then since k is algebraically closed, exercise 2.3 (d) I implies I = I = I + a. Hence a I. This shows I is a prime ideal so S = k[x 0,, x r ]/I is an integral domain. For the other way to see that S is a domain, let cl (X) be the set of closed point in X. Then cl (X) is a projective variety. Let J = I(cl (X)) be its ideal consisting of polynomials vanishing on X. Then t(cl (X)) = Proj k[x 0,, x r ]/J and Proj S has the same topological sub-space of P r k. Since both Proj S and t(cl (X)) are reduced closed sub-schemes, we see that Proj S = t(cl (X)). By exercise 5.10 (b), I = J. But J is a prime ideal so J = J. Hence I is a prime ideal so that S is an integral domain. We have already seen that S is integral over S. To show that S is the integral closure for S, it is sufficient to show S is integrally closed. For this, we noted in exercise 5.9 (b) that S = n 0 S (xi)(n). Hence it will be sufficient to show n 0 S (xi) is integrally closed for each i. But n 0 S (xi) = n S (xi)x n i S (xi)[t]. Since X is normal, S (xi) is integrally closed. Then so is S (xi)[t]. Hence S is integrally closed. (b) Take M = S. (c) Let d be the integer such that S d = S d. Let X = Proj S (d) X. Then by (a), n Γ(X, O X (n)) is integrally closed. But n Γ(X, O X (n)) n Γ(X, O X (dn)) by exercise Since both S 0 and S 0 = k, by (b) we have nγ(x, O X (dn)) = S (d). This shows S (d) is integrally closed. (d) Projective normality implies normality as we have seen earlier. Moreover, we have S = S. So the map Γ(P r, O p r(n)) Γ(X, O X (n)) is isomorphic to the quotient map k[x 0,, x r ](n) S(n). Because Γ(P r, O P r) Γ(X, O X ) is surjective, X is connected. Then by (a), S is integrally closed. So it is sufficient to show S = S. Since the map Γ(P r, O P r(n)) Γ(X, O X (n)) factors through β n : S(n) Γ(X, O X (n)), we see that β n is also surjective. Hence S = S. Exercise 16: (a). First note that if F is a presheaf on X, then for any open set U X, (F U ) + = F + U. Now it is clear that T r (F ) pre U = T r (F U ) pre so that T r (F ) U = T r (F U ). By this we can 13

14 assume that F is free of rank n. Then it is easy to see that T r (OX n )pre (U) OX nr. This isomorphism sends the basis e U i e U j of T r (OX n )pre (U) to that e U ij of Onr X (U) is compatible with restriction maps and hence defines an isomorphism of sheaves of O X -modules. Similar arguments apply to S r (F ) and r (F ). (b). We start with a classical result that if M is a free A-module of rank n, then there is an isomorphism r M HomA (n r n M, M). Establishing such an isomorphism is a lof of fun! Now let U X be an open subset over which F is free. Then we have an isomorphism ( r F ) U H om OX U ( n r F U, n F U ). As U runs through every open set over which F is free, these isomorphisms glue together to an isomorphism r F H om OX ( n r F, n F ). Then apply exercise 5.1 (b). (d). Set up: If 0 F α β F F 0 is an exact sequence of locally free sheaves of finite rank on a scheme X, then for every integer r there is a filtration r F = F 0 F 1 F r 0 of locally free sheaves with quotients F i /F i+1 i (F ) r i F. To achieve this, note firs that ker r (β) : r F r F can be chosen to be F 1. This is because over a trivializing affine open U = Spec A for F and F, β splits so r β splits on U. This implies r β is surjective on U. Since trivializing open affine sub-scheme of X covers X, we see that r β is surjective. Hence r F / ker r β = F 0 /F 1 r F. Moreover, r α factors through ker r β. Next, let us construct F 2. On a trivializing open affine sub-scheme U = Spec A X for F and F, we have exact sequence of the form 0 A n α A m+n β A m 0. Then convince yourself that on this U, ker r β = F 1 = r i=1 i F r i F. Choose e 1,, e n F (U) and f 1, f m F (U) as basis over Γ(U, O X ). Then e 1,, e n, f 1, f n F (U) is a basis where f i is a lift for f i. Then on this U one defines F 1 F r F by the projection r i=1 i A n r i A m A n r A m. One checks that this projection map does not depend on the choice of e i, f i and f i s. Hence we obtain a surjective map β 1 : F 1 1 F r F. Let F 2 be ker β 1. Then F 1 /F 2 F r F. 14

15 Convince yourself that on this U, F 2 = r i=2 i F r i F. Moreover, 3 F F 1 factors through F 2. Continuing, we arrive at β r : F r r F F with the property that 1. On U, ker β r = r F and 2. r F F r factors through ker β r F r. Hence F r = r F as globally F r r F and locally they agree. Exercise 5.17: (c). We need this basic fact for quasi-coherent sheaves. Lemma Let Y be a scheme and F be a quasi-coherent sheaf on Y. Then for any nested affine open sub-schemes V U in Y, the restrictions F (U) F (V ) and O Y (U) O Y (V ) induces an isomorphism F (V ) F (U) OY (U) O Y (V ) of O Y (U)-modules. If F is a quasi-coherent algebra, the above isomorphism is an isomorphism of O Y (V )-algebras. Proof. Cover V by principal affine open sub-scheme D(f) for f O Y (U) and argue locally. Let U be an affine open sub-scheme of Y. Then for any f O Y (U), D( f) Spec A (U) is a principal affine open sub-scheme of Spec A (U). Here f is the image of f in A (U) under the ring map O Y A that gives A the structure of O Y -algebras. If U and V be two affine open sub-scheme of Y. Then there is an open sub-scheme T UV = f D( f) Spec A (U) where the union is taken over all simultaneously principal affine open D(f) U V. Suppose D(f) = D(g) for some g O Y (V ). Then there is an isomorphism of rings ϕ fg : A (D( f)) = A (D(ḡ)). If D(f 1 ) = D(g 1 ) and D(f 2 ) = D(g 2 ) for some f i O Y (U) and g i O Y (V ). The commutativity of the following diagram ensures that the maps ϕ fg : Spec A (D( f)) = Spec A (D(ḡ)) glues to a map ϕ UV = T UV T V U. D( f i ) D(ḡ i ). D( f 1 f 2 ) D( g 1 g 2 ) The vertical maps are the restriction maps. Let us check that if U, V, W are three affine open sub-schemes of Y, these ϕ satisfies 1. ϕ UV (T UV T UW ) = T V U T V W and 2. ϕ UW = ϕ V W ϕ UV on T UV T UW. 15

16 For this, it would be convenient to show first that T UV T V U is the union of D( h) where D( h) is a simultaneously principal affine open sub-scheme in U V W. Of course T UV T UW = ff D( ff ) where f O Y (U), f O Y (U) and D(f) = D(g) U V and D(f ) = D(g ) U W for some g O Y (V ) and g O Y (W ). Let us verify that D( ff ) is a simultaneously principal open affine sub-scheme for U V W. It is already principal in U. Note that f D(f ) =f D(g ) f D(f) =f D(g). Hence there exist an x O Y (V ) and an x O Y (W ) such that f =x /(g ) k f =x/g k Then D(ff ) = D(gx) is principal in V and D(ff ) = D(g x ) is principal in W. I proved that T UV T UW is contained in the union of D( h) where D(h) is a simultaneously principal affine open in U V W. For the converse, use the stupid identity D( h) = D( hh). Hence I have shown that T UV T UW = h D( h) where the union is taken over all simultaneously principal affine open sub-scheme D(h) U V W. Then it is immediate by our construction of ϕ that they satisfy the co-cycle conditions. Hence we obtained a scheme X by glueing these Spec A (U). To construct a map π : X Y, simply take advantage of the following commutative diagram O Y (U) A (U) O Y (D(f)) A (D(f)) O Y (D(g)) A (D(g)) O Y (V ) A (V ) where the vertical maps are restrictions and horizontal maps are just the structure morphism for A as a O Y -algebra. Let me show that if U Y is an affine open sub-scheme, π (U) is precisely Spec A (U). For this, if x Spec A (V ) and π(x) U V, then pick a simultaneously principal affine open sub-scheme D(f) = D(g) for U V. Then the lemma about quasi-coherent sheaves imply that the diagram D(ḡ) Spec A (V ) D(g) V is cartesian. Hence x D(ḡ) = D( f) Spec A (U). We see that π (U) is precisely Spec A (U). By the lemma again and what we just proved, if V U are two affine open- 16

17 schemes of Y, then the following diagram Spec A (V ) Spec A (V ) X V U Y is cartesian. Here I also give a functorial approach. The goal is to construct a Y -scheme f : X Y such that if u : W Y is another Y -scheme, then there is a bijection Hom Y (W, X) Hom OY (A, u O W ). This property is universal: if W is another Y -scheme satisfies the above bijection, then there is a Y -isomorphism between W and W. Step 1: When Y = Spec A is an affine scheme. Let Spec C Spec A be an open affine sub-scheme. I already showed that the square Spec A (C) Spec C γ Spec A (A) Spec A where γ : Spec A (C) Spec A (A) is induced by the restriction A (A) A (C), is cartesian. But there is a purely functorial approach to prove this. Let u : W Spec C be a scheme. By some adjunction formulas, we have Hom Spec C (W, Spec A (C)) Hom C (A (C), Γ(W, O W )) Hom OSpec C (i A, u O W ) Hom OSpec A (A, i u O W ) Hom A (A (A), Γ(W, O W )) Hom Spec A (W, Spec A (A)) where i : Spec C Spec A is the open inclusion. This bijection is functorial in W. Replacing W by Spec A (C), we see that the identity map goes to γ. If φ : W Spec A (C) is a Spec C- morphism, consider the commutative diagram Hom Spec C (W, Spec A (C)) Hom Spec A (W, Spec A (A)). Hom Spec C (Spec A (C), Spec A (C)) Hom Spec A (Spec A (C), Spec A (A)) This diagram implies that the bijection Hom Spec C (W, Spec A (C)) Hom Spec A (W, Spec A (A)) is given by φ γ φ. This proves that the square is cartesian. It is clear that Spec A (A) Spec A satisfies the universal property. This proves (c) in the case that Y is affine. Step 2: General case. Note that the assignment w : W Y Hom OY (A, w O W ) is a contravariant functor F from scheme to sets. This question is equivalent to prove that the functor F is representable by a Y -scheme. To do this, we will need to prove that F is a zariski sheaf and then cover F by open sub-functors that are representable. That is, we need to find a set I such that for each i I, 1. If w : W Y is a Y -scheme and ξ F(W ), then there is a an open cover i I W i such that ξ i = ξ Wi F i (W i ), 17

18 2. There is a sub-functor F i F, 3. F i is representable, 4. The natural injection F i F is representable by an open immersion. That F is a zariski sheaf follows from the local property of maps between sheaves. To find an open representable cover for F, let us cover Y by all affine open sub-schemes Y i. For each i, define the sub-functor F i by F i (W ) = Hom OY (A, w O W ) if w : W Y factors through Y i and empty otherwise. Then condition 1 and 2 are easily satisfied. For representability of F i, note that F i (W ) = Hom OY (A, w O W ) = Hom OY (A, ι i w i O W ) Hom OYi (A Yi, w i O W ) Hom Y (W, Spec A (Y i )) where ι i : Y i Y is the inclusion and ι i w i = w. The harder part is to show that F i F is representable by an open immersion. For this, suppose t : T Y is a Y -scheme and T F is given by an element ɛ F(T ) we want to show that there is a bijection T (W ) F(W ) F i (W ) T (W ) Y (W ) Y i (W ) functorial in W. This proves that F i F is represented by T Y Y i T. Let (φ, a) T (W ) F(W ) F i (W ). If w : W Y does not factor through Y i, then both sets are empty. If w factors through Y i, then w = ι i w i. Let ψ(w ) : T (W ) F(W ) F i (W ) T (W ) Y (W ) Y i (W ) be defined by if w does not factor through Y i and (φ, a) (φ, w i ) if w factors through Y i. To see that ψ(w ) is surjective, if (φ, w i ) T (W ) Y (W ) Y i (W ), this means we have the following commutative diagram W Y i w i φ T t. ι i Denote φ : O T φ O W be the morphism given by φ, then we have a morphism t φ ɛ : A t φ O W = ι i w i O W. Then it is clear that (φ, t φ ɛ) T (W ) F(W ) F i (W ) and that ψ(w )(φ, t φ ɛ) = (φ, w i ). To see that ψ(w ) is injective, suppose ψ(w )(φ, a) = ψ(w )(ϕ, b). Then the definition of ψ(w ) gives us (φ, w i ) = (ϕ, w i ) hence we have φ = ϕ. Now (φ, a) T (W ) F(W ) F i (W ) implies a = t φ ɛ. Again (ϕ, b) = (φ, b) T (W ) F(W ) F i (W ) implies b = t φ ɛ so a = b. Hence ψ(w ) is injective. By representability criterion, this functor F is represented by a scheme α : X Y obtained by gluing Spec A (Y i ) together. Now if Y i is some affine open sub-scheme of Y, then since X F F i is represented by X Y Y i = α (Y i ) and by F i as well. This implies Spec A (Y i ) α (Y i ). Let us set up the notations. Let X i = Spec A (Y i ), α i : X i Y i that corresponds to the ring map Γ(Y i, O Yi ) A (Y i ), ι i : Y i Y be the inclusion ξ i Hom OY (A, ι i α i O Xi ) be the universal family given by the identity map id A (Yi) and U ij = αi (Y i Y j ). Then the goal is to show that the isomorphism ϕ ij : U ij U ji, obtained by stack projects procedures, agrees with the isomorphism obtained from the fact that both U ij and U ji satisfies the universal property for the pair (A Yi Y j, Y i Y j ). We start with the Y following commutative diagram δ i j U ij X i Y i Y ι j j ι j This diagram yields the α i Uij α i ι j i Y i Y j Y j. ι i 18

19 following relations Hom OY (A, ι i α i δj i O Uij ) Hom OYi (A Yi, ι i j α i Uij O U ij ) Hom Yi (U ij, X i ) Hom Yi (U ij, U ij ) Hom OYi Y j (A Yi Y j, α i Uij O U ij ) β ξi (U ij) Hom OYj (A Yj, ι j i α i Uij O U ij ) Hom Yj (U ij, X j ) Hom Yj (U ij, U ji ) As one can check, the universal family ξ Uij Hom OY (A, ι i α i δj i O Uij ) goes down to a map in Hom OYi (A Yi, ι i j α i Uij O U ij ) that corresponds to the restriction A (Y i ) O Uij (U ij ), which agrees with the identity map in Hom Yi (U ij, U ij )! Therefore, the isomorphism ϕ ij : U ij U ji obtained by stack project s procedures agrees with the one obtained from the fact that both U ij and U ji satisfies the universal property for the pair (A Yi Y j, Y i Y j ). At this point I do not how to write properly for these isomorphisms to satisfy the co-cycle conditions. I would rather say that the verification is saved by the elegant functorial machineries. In stacks project that the isomorphisms ϕ ij satisfy the co-cycle conditions was proved using purely functorial languages. Now that we proved that the isomorphism one would naturally get agrees with stack project s, we can confirm that these isomorphisms glue. After these, we obtained a Y -scheme α : X Y with the following data: There are open immersion ψ i : X i X such that ψ i = ψ j ϕ ij on U ij. α ψi(x i) = α i ψ i. One checks that ψ i (X i ) = α (Y i ). For the assertion that if Y i Y j are two affine sub-schemes of Y, then α(y i ) α(y j ) corresponds to A (Y j ) A (Y j ), note that ϕ ij obtained from the diagram above does correspond to the restriction map. Finally, one checks that if we begin with just an affine cover, {Y i } i I, the the glued Y -scheme α : X Y will be Y -isomorphic to X. The universal families ξ i glue to a universal family ξ F(X) that corresponds to the identity map on X. To see this, we need the following observation. If t : T Y is a scheme satisfying the universal property and there is a Y -morphism of schemes S s f t T Y. Then if V Y is an open sub-scheme, we have a commutative diagram Hom OY (A, s O S ) Hom Y (S, T ) Hom OV (A V, s O S V ) Hom V (s (V ), t (V )). 19

20 where Hom OY (A, s O S ) Hom OV (A V, s O S V ) and Hom Y (S, T ) Hom V (s (V ), t (V )) are given by restrictions. For this, observe that we have a commutative diagram Hom OY (A, s O S ) Hom Y (S, T ) Hom OY (A, s ι s (V ) O s (V )) Hom Y (S (V ), T ) Hom OV (A V, s O S V ) Hom V (s (V ), t (V )). where ι s (V ) : S V is the inclusion. The upper square commutes because of the functoriality as part of the definition of the universal property. For the lower square, first note that s ι s (V ) = ι V s s (V ) where ι V : V Y is the inclusion. Then Hom OY (A, s ι s (V ) O s (V )) Hom OV (A V, s s (V ) O s (V )) = Hom OV (A V, s O S V ). Now define Hom OV (A V, s O S V ) Hom V (s (V ), t (V )) so that the lower square commutes. Finally the composition of the two left down arrows is the restriction to V. As a consequence, we have the following commutative diagram Hom OYi (A Yi, α i O Xi ) Hom Yi (Spec A (Y i ), Spec A (Y i )) Hom OY (A, α O X ) Hom Y (Spec A, Spec A ). So the map x : Spec A Spec A corresponding to ξ goes to identity on Spec A (Y i ). This holds for all i so x is the identity. Exercise 5:18 (a) I want to mention that the condition that for any open affine Spec A U i U j, the transition maps ψ induces A-algebra linear automorphism A[x 1,, x n ] ensures that for any point x U i U j, the isomorphisms ψ x A n are linear and is given by GL(O(U i U j )) as x runs through U i U j. Since there is not much content in this exercise I think it is more instructive to compute Spec S(O(1)) over P 1 k = Proj k[x 0, x 1 ]. O(1) over D + (x 0 ) is generated over k[x 1/0 ] by x0 1 and O(1) over D +(x 1 ) is generated by x1 1 over k[x 0/1]. Hence Spec S(O(1)) is obtained by glueing Spec k[x 10, y] and Spec k[x 01, z] along Spec k[x 1/0, y] x1/0 Spec k[x 0/1, z] x0/1 via the map k[x 0/1, z] x0/1 k[x 1/0, y] x1/0 of k[x 0/1 ] x0/1 -algebras where x 0/1 1/x 1/0, z x 1/0 y. (b) Before we delve into the abstract proof, I think it is instructive to continue the example above. Let X = Spec S(O(1)) be a geometric vector bundle over Y = P 1 k. Let us show that 20

21 S (X/Y ) = O(). For this, we know that S (X/Y ) is generated by the k[x 0/1 ]-algebra homomorphism ϕ z : k[x 0/1, z] k[x 0/1 ] that sends z to 1. The restriction of this homomorphism to k[x 0/1 ]x 0/1 sends y first to x 0/1 z, then to x 0/1 and to 1/x 1/0. Hence the transition map for S (X/Y ) agrees with the transition map for O(). This will be even clearer if we try to compute the global sections for S (X/Y ). Let s : Y X be a global section. Then locally on D + (x 1/0 ), s corresponds to a map s y : k[x 1/0, y] k[x 1/0 ] of k[x 1/0 ]-algebras. Similarly, we have another map s z : [x 0/1, z] k[x 0/1 ] of k[x 0/1 ]-algebras. They must agree on k[x 1/0, y] x1/0 k[x 0/1, z] x0/1. Suppose f(x 1/0 ) = s y(y) and g(x 0/1 ) = s z(z). Then the commutative diagram k[x 1/0, y] x1/0 k[x 0/1, z] x0/1 s y s z k[x 1/0 ] x1/0 k[x 0/1 ] x0/1 yields the following relation f(1/x 0/1 ) = g(x 0/1 ) x 0/1 by looking at how y travels from the top left to the bottom right. This is a relation that we see when we compute that there is no non-zero global sections of O()! Let us attack this exercise. I list some undergraduate algebra properties that will be used. Lemma Let B be an A-algebra and M, N be A-modules. Then the zero map 0 : M N pulls back to zero map 0 id B : M A B N A B. If f, g : M N are two maps of A-modules, then f id B + g id B = (f + g) id B. If α A is an element and f : M N is a map of A-modules, then (α f) id B = α B (f id B ). Let f : Y X be a geometric vector bundle of rank n. Note that in this case f is necessarily an affine morphism. It is apparent that S (X/Y ) is a sheaf of sets. I will give it a structure of sheaf of abelain groups. For this, fix a cover of Y given by trivializing affine open sub-schemes {U i } and fix isomorphisms f (U i ) A n U i. A zero section 0 S (X/Y ) can be defined by setting it to be the zero map O Ui [x 1,, x n ] O Y (U i ) of O Ui algebras where x i 0. The fact that f is affine implies both O f U i and O Ui are quasi-coherent O Ui -algebras. Hence the map on global section determines a map of sheaves of O Ui -algebras on U i. By the lemma the zero section I defined on each affine U i f (U i ) glues to a global section 0 : Y X. If s 1 and s 2 are two global sections Y X, I can add them locally on each affine U i. By the lemma the addition of s 1 and s 2 defined on each U i glues to a global section s 1 + s 2 : Y X. It is apparent now S (X/Y ) has a structure of sheaf of abelian groups, if we fix a trivializing cover and the trivialization isomorphisms. To give S (X/Y ) a O Y -module, pick an element α Γ(Y, O Y ) and a section s : Y X. We can multiply s by α locally on each U i. Then the lemma glues these multiplication together to a global section α s : Y X. Hence we obtain an O Y -module structure on S (X/Y ). Apparently 21

22 S (X/Y ) Ui OU n i. This construction does not depend on the choice of the cover {U i } nor does it depend on the choice of trivialization f (U i ) A n U i. To see this, we may first fix a cover {U i } and choose different trivialization ψ i : f (U i ) A n U i. Then ψ i ψ i is given by a matrix I ij GL n (O Y (U i )). Let the coordinate of A n U i be given by (x 1,, x n ). Then define by setting x k : O Y (U i )[x 1,, x n ] O Y (U i ) x l δ kl. Then {x k ψ i} S (f (U i )/U i ) forms a O Y (U i )-basis and {x k ψ i } is another basis. Then x k ψ i = l I lk x l ψ i. We can define S (f (U i )/U i ) S (f (U i )/U i ) by setting x k ψ i l I lk x l ψ i which we see is an identity map. This shows different trivialization does not affect the O Y - structure on S (X/Y ). Next, we may fix trivialization and refine cover. Combine these two we get the result. (c). It is clear in the construction in (b) that we look at maps from E to O Y locally. And if one tries to compute the transition function for S (X/Y ), they agree with the transition functions for E. (d). The bijection would be E Spec S(E ) Y with inverse f : X Y S (X/Y ). 4 An Interesting Adjoint Pair A consequence that one may deduce from exercise 5.3 is the following. If X = Spec A is a scheme. Let O X -Mod be the category of sheaves of O X -modules and QCoh(X) be the category of quasi-coherent sheaves on X. Then the functor Q : O X -Mod QCoh(X) that takes a sheaf F to Γ(X, F ) is the right adjoint to the forgetful functor f : QCoh(X) OX -Mod. Indeed, if F is an O X -module and M is a quasi-coherent module for some A-module M, then Hom OX (f( M), F ) Hom A (M, Γ(X, F )) Hom QCoh(X) ( M, Γ(X, F ) = Q(F )). A natural problem that arises is to generalize the construction of Q to an arbitrary scheme. Currently, there is a construction if one imposes X to be quasi-compact and quasi-separated. We abbreviate this condition on a scheme as qcqs. Let X be qcqs and F a sheaf of O X -modules. Fix a finite cover for X consisting of affine open sub-schemes N i=1 U i. For any i, j, U i U j is quasi-compact, so we may also fix a finite cover for U i U j, say nij k=1 U ijk. We already constructed 22