Descent is a technique which allows construction of a global object from local data.

Transcription

1 Descent Étale topology Descent s a technque whch allows constructon of a global object from local data. Example 1. Take X = S 1 and Y = S 1. Consder the two-sheeted coverng map φ: X Y z z 2. Ths wraps the crcle around tself two tmes. Every pont y Y has a small open neghborhood U y Y such that φ 1 (U y ) s a dsjont unon of two open neghborhoods n X, each of whch s mapped by φ homeomorphcally onto U y. Infact, we can make explct calculatons n ths setup. Let y be an arbtrary pont n Y and U y = {e θ a < θ < b} be an open neghborhood of y, such that b a π. Then φ 1 (U y ) = {e θ a 2 < θ < b 2 } {eθ π + a 2 < θ < π + b 2 }. Ths double cover s also defned n algebrac geometry. Example 2. Take X = A 1 k \ {0} = Spec(k[s, s 1 ]) and Y = A 1 k \ {0} = Spec(k[t, t 1 ]), where k s an algebracally closed feld whose characterstc s not equal to 2. Consder the map φ: X Y z z 2. The nduced map φ # on the coordnate rngs s gven by φ # : k[y ] = k[t, t 1 ] k[x] = k[s, s 1 ] t s 2. Unfortunately, the Zarsk-open sets n Y are gven by complements of a fnte number of ponts, whch makes the topology too coarse. Ths prevents φ from beng locally trval n the Zarsk topology as ther nverse mages are stll connected. However, there s a remedy for ths stuaton usng the étale topology. We wll show that φ s an étale map. For any pont x X, we have ÔX,x = ( k[s, s 1 ] (s x) )ˆ whch s further somorphc to k[[α]] by the Cohen structure theorem for complete Noetheran local rngs. Smlarly, we get that ÔY,φ(x) = ( k[t, t 1 )ˆ ] (t x 2 ) = k[[β]]. Now the nduced map ˆφ# on the complete local rngs s gven by ˆφ # : Ô Y,φ(x) = k[[α]] Ô X,x = k[[β]] α βu 1

2 where u s a unt n k[[β]]. Ths shows that φ s étale, snce ˆφ # : Ô Y,φ(x) ÔX,x s an somorphsm x X. Now perform a base change by an étale surjectve morphsm to obtan X Y X X φ X and locally n the étale topology, φ s a trval cover. We observe that φ k[x Y X] = k[s 1, s 1 1 ] k[t,t 1 ] k[s 2, s 1 2 ] φ Y = k[s 1, s 1 1, s 2, s 1 2 ] (s 2 1 s2 2 ) = k [ s 1 /s 2, s 2, s 1 2 (s 1 /s 2 ) 2 1 = k [s 1/s 2 ] (s 1 /s 2 ) 2 1 k k[s 2, s 1 2 ]. ] The correspondng dagram for the map of global sectons s gven as follows: k[s 1 /s 2 ]/ ( (s 1 /s 2 ) 2 1 ) k k[s 2, s 1 2 ] k[s 2, s 1 φ # k[s 1, s 1 1 ] φ # φ # 2 ] k[t, t 1 ] We wll loosely defne the étale topology on Y. Let Ét(Y ) denote the category of all étale morphsms from a scheme to Y. Its objects can be nformally thought of as étale open subsets of Y. The ntersecton of two objects U 1 Y and U 2 Y corresponds to U 1 Y U 2 Y and ther unon corresponds to U 1 U 2 Y. An étale cover of Y can be defned as {u λ : Y λ Y } such that the set of all λ s fnte, each Y λ s affne and each u λ s an étale surjectve morphsm. Descent allows us to glue n the étale topology. Lemma 2.1. Let M be an A-module and f : A B be a fathfully flat rng homomorphsm. Defne two maps e 0 and e 1 as follows: e 0 : B B A B b b 1 e 1 : B B A B b 1 b 2

3 Then the complex s exact. 0 A A M = M f 1 M B A M (e0 e 1 ) 1 M B A B A M (1) Proof. Snce f : A B s fathfully flat, the complex (1) s exact ff the tensored complex B A (1) s exact. It thus suffces to prove that B A (1) s exact. We apply the functor B A to (1) and obtan 0 B A M δ B A B A M 1 B (e 0 e 1 ) 1 M B A B A B A M b m b 1 m We observe that each of the maps e 0 and e 1 admts a secton µ: B A B B, gven by µ (b b ) = bb. Let b m be an arbtrary element of ker(δ). Then we have ( ) δ b m = 0 b 1 m = 0 ( ) (µ 1 M ) b 1 m = (µ 1 M ) (0) b m = 0. Ths shows that the frst map δ s njectve. For convenence, we let γ denote the second map 1 B (e 0 e 1 ) 1 M. It s not hard to see that m(δ) ker(γ). To show the other ncluson we pck an arbtrary element b b m n ker(γ). Ths means that ( 1B (e 0 e 1 ) ( ) ) 1 M b b m = 0 b b 1 m = b 1 b m ( ) ( ) (µ 1 B 1 M ) b b 1 m = (µ 1 B 1 M ) b 1 b m b b 1 m = ( ) δ b b m = b b m b b m. 3

4 We thus conclude that m(δ) = ker(γ) whch justfes the exactness of the complex B A (1). The complex (1) s a specal case of the Amtsur complex of B over A. Defnton 2.1. Gven a rng homomorphsm f : A B, we defne the Amtsur complex of B over A as follows: 0 A f B d0 B 2 d 1 d 1 B d B (+1) d +1 dr B (r+1) where the maps d : B B (+1) are gven by d (b 1 b ) = ( 1) j (b 1 b j 1 b j+1 b ) j=0 Lemma 2.2. Let M be any A-module and f : A B be a fathfully flat homomorphsm of rngs. Then the Amtsur complex of B over A s exact and remans so after tensorng wth M. Proof. Refer to proposton 2.18 of Lectures on Étale Cohomology, James Mlne. Monomorphsms and epmorphsms n the category of schemes Let φ: X Y be a morphsm of schemes. Defnton 2.2. φ s a monomorphsm f φ : Hom(T, X) Hom(T, Y ) s njectve for all schemes T. Unfortunately the correct defnton for an epmorphsm doesn t carry over by replacng the words monomorphsm wth epmorphsm and njectve wth surjectve n Defnton (2.2). Example 3. The ncluson map : R C nduces a map on spectra.e. # : Spec(C) Spec(R). The morphsm # s an epmorphsm but # : Hom (Spec(R), Spec(C)) Hom (Spec(R), Spec(R)) s not surjectve because Hom (Spec(R), Spec(C)) =, whereas Hom (Spec(R), Spec(R)) = {Id Spec(R) }. Defnton 3.1. φ s an epmorphsm f φ : Hom(Y, T ) Hom(X, T ) s njectve for all schemes T. We wll later prove that f φ s fathfully flat then φ s a effectve epmorphsm. Cauton: In general, φ s surjectve φ s an epmorphsm and φ s an epmorphsm φ s surjectve. The next two examples wll llustrate ths strange behavor of morphsms n the category of schemes. We proceed to gve an example of surjectve morphsm whch s not an epmorphsm. 4

5 ( ) k[x] Example 4. Consder the morphsm π # : Spec(k) Spec nduced by the natural quotent map π : k[x] x 2 k. Note that π # s surjectve (nfact ( ) k[x] homeomorphsm) snce both Spec and Spec(k) are one-pont topologcal x 2 spaces. We wll prove that π # s not an epmorphsm by showng that ( ( ) ) k[x] π # : Hom Spec, Spec (k[x]) Hom (Spec(k), Spec (k[x])) x 2 s not one-to-one. We get a commutatve dagram as follows: ψ 1 k[x] k[x]/x 2 ψ 2 x 2 θ π k where the maps are ψ 1 : k[x] k[x] x 2 n a x ā 0 + ā 1 x =0 ψ 2 : k[x] k[x] x n a x ā 0 =0 θ : k[x] k[x] x = k n a x ā 0 =0 π : k[x] k x 2 ā 0 + ā 1 x ā 0. = k k[x] x 2 Note that π # (ψ # 1 ) = θ# = π # (ψ # 2 ) and ψ# 1 ψ# 2, whch shows that π# s not njectve. The next example shows that epmorphsms are not always surjectve. 5

6 Example 5. Consder the morphsm ψ # : nduced by the rng homomorphsm ψ : C[x] n Z C Spec(C) Spec(C[x]) = A 1 C n Z f(x) {f(n)} n Z. Clearly ψ # s not surjectve as the non-nteger ponts n A 1 C are not mapped. To show that ψ # s an epmorphsm, we need to check that ψ # : Hom ( A 1 C, T ) ( ) Hom Spec(C), T s one-to-one for all schemes T. The njectve property be checked locally on T, hence we can assume that T s affne, say T = Spec(R). Now t amounts to showng that the map on the affne coordnate rngs ( ) Hom (R, C[x]) Hom f ψ f n Z R, n Z C s njectve for an arbtrary rng R. Let s suppose that there exst f, g Hom (R, C[x]), such that ψ f = ψ g. Ths means that for any r R, we have f(r)(n) = g(r)(n) n Z. Snce f(r) and g(r) are polynomals over C whch agree on all nteger values, we must have f(r) = g(r). Now t follows that f = g as the element r was arbtrarly chosen. Hence our map s njectve. The notons of monomorphsms and epmorphsms can also be defned n a general category C. Defnton 5.1. f : X Y s a monomorphsm n C f f : Hom C (T, X) Hom C (T, Y ) s njectve for all objects T n C. Defnton 5.2. f : X Y s an epmorphsm n C f f : Hom C (Y, T ) Hom C (X, T ) s njectve for all objects T n C. Theorem 5.1. Let f : X Y be a morphsm of schemes, locally of fnte type. Then the followng condtons are equvalent: f s a monomorphsm. f # : O Y,f(x) O X,x s surjectve for every x X. For every y Y, the fber f 1 (y) s ether empty or somorphc to Spec(k(y)). Proof. Refer to EGA IV, Theorem 5.2. Let f : X Y be a morphsm such that f s surjectve as a map of topologcal spaces and f # : O Y f O X s njectve as a homomorphsm of sheaves of rngs. Then f s an epmorphsm. 6

7 Effectve epmorphsm We wll ntroduce the noton of effectve epmorphsm. Let f : X Y be a morphsm of objects n a category C wth fbre products. Then we have a fber product dagram as follows : X Y X X p 2 f X Y f where p 1 and p 2 are projectons onto the frst and second factor respectvely. Defnton 5.3. We wll say that the sequence Hom C (Y, T ) f Hom C (X, T ) p1 Hom C (X Y X, T ) p 2 s exact f f s njectve and Im(f ) = {h p 1 (h) = p 2 (h)}. Lemma 5.3. If C = Set and f : X Y s a surjectve map of sets, then the sequence (5.3) s exact for all objects T n Set. Proof. In Set we have X Y X = {(x 1, x 2 ) X X f(x 1 ) = f(x 2 )}. Note that f f : X Y s surjectve mples that f : Hom(Y, T ) Hom(X, T ) s njectve. We also have Im(f ) {h p 1 (h) = p 2 (h)} snce p 1 f = p 2 f. Now pck h Hom(X, T ) such that h p 1 = h p 2. We defne g : Y T by g(y) = h(x), where x s any element n the fber f 1 (y) and t s possble snce f s surjectve. If x 1, x 2 f 1 (y) then we have (x 1, x 2 ) X Y X whch further mples that h(x 1 ) = h(x 2 ). Ths shows that g s well-defned and h = g f, thus provng {h p 1 (h) = p 2 (h)} Im(f ). Defnton 5.4. Let C be any category wth fbre products and let X and Y be objects n C. We say that f : X Y s a effectve epmorphsm f the sequence Hom C (Y, T ) f Hom C (X, T ) p1 Hom C (X Y X, T ) p 2 s exact for all objects T n C. We dgress nto fpqc (fathfully flat and quas-compact) and fppf (fathfully flat and locally of fnte presentaton) morphsms. Let f : X Y be a morphsm of schemes. If f s fppf then f s unversally open.e. f s open and remans so after base change. If f s fpqc then f s unversally submersve.e. U Y s open ff f 1 (U) s open. 7

8 We gve two examples to show that there are flat and fathfully flat morphsms whch are nether fppf nor fpqc. Example 6. The morphsm φ # : Spec ( C[x] (x λ) ) Spec (C[x]) nduced by the rng homomorphsm φ: C[x] C[x] (x λ) s flat but nether fathfully flat nor locally of fnte presentaton. Example 7. Let s consder the morphsm φ # : Spec ( ) C[x] (x λ) Spec (C[x]) λ C nduced by the rng homomorphsm φ: C[x] λ C C[x] (x λ). The morphsm φ # s flat and surjectve, so t s fathfully flat. Not that φ s not locally of fnte presentaton snce the nclusonc[x] C[x] (x λ) s not of fnte presentaton. In addton t s not quas-compact snce λ C Spec C[x] (x λ) s not quas-compact. Let T be the scheme obtaned by glung { Spec ( C[x] (x λ) along ther )}λ C ntersectons n Spec(C[x]) and for convenence set X = Spec ( ) C[x] (x λ) and Y = C[x]. We wll check that the sequence λ C Hom(Y, T ) φ# Hom(X, T ) p1 Hom(X Y X, T ) p 2 s not exact, thereby showng that φ # s not a effectve epmorphsm. To see ths, consder the map λ SpecC[x] (x λ) T gven by the open ncluson on each factor. The two pullbacks to λ,µ Spec C[x] x λ Spec C[x] Spec C[x] x µ agree and an equalzer of these maps defne a map of sets Y T whch sends the pont (x λ) to ts mage n the non-separated scheme T. However, ths map s not contnuous wth respect to the Zarsk topology, snce the set λ C (x λ) s closed n T (t s the complement of the generc pont whch s an open set of T but ts nverse mage s not closed n Y. s not closed. Theorem 7.1. [Proposton 2.17 of Lectures on Étale Cohomology, James Mlne] An fpqc or an fppf morphsm f : X Y s a effectve epmorphsm. Proof. We have to show that for every scheme T and morphsm h: X T such that h p 1 = h p 2, there exsts a unque morphsm g : Y Y such that g f = h. Case (): X = Spec(B), Y = Spec(A), T = Spec(C) are all affne. The theorem s equvalent to showng that 0 Hom (C, B A B) Hom(C, B) Hom(C, A) s exact. Now ths follows from applyng Hom(C, ) to the exact sequence (follows from lemma (2.1)) 0 A B e0 e 1 B A B. 8

9 Case (): X, Y and T are arbtrary schemes. We wll gve the proof only when X Y s fppf makng use of the followng lemma. Lemma 7.2. Let f : X Y be an fppf morphsm. Gven x X, let y = f(x). Then there exsts affne open neghborhoods U of x and V of y such that morphsm f restrcts to an fppf morphsm f : U V. Proof. Let V be an affne neghborhood of y and let U be an affne neghborhood of x n f 1 (V ). The map f U U V s open and affne, but not necessarly surjectve. To rectfy ths, replace V by an open affne neghborhood of y contaned n the open set (f U )(U) V. Snce f U s an affne morphsm, f 1 (V ) s now an affne open contanng x. Replacng U wth f 1 (V ) we see that f restrcts to an fppf morphsm U V. We frst show the unqueness of g. Suppose that there exst two maps g 1, g 2 : Y T such that g 1 f = g 2 f = h. Snce f : X Y s surjectve as a maps of topologcal spaces, we must have g 1 = g 2 as maps of the underlyng topologcal spaces. We need to check that they also agree as maps of rnged spaces. But ths can be checked locally on Y. Pck y Y, x f 1 (y) and let U be an open affne neghborhood of g 1 (y) = g 2 (y) n T. By the lemma, there affne neghborhoods W x of x and W y of y such that the restrcton of f to a morphsm W x W y s fppf and we have a commutatve dagram of affne schemes. W x f W y g 1 h g 2 U Thus we are reduced to the affne case we conclude that g 1 = g 2 on W y. Snce we can do ths for every pont we see that unqueness follows. We now construct a map g : Y T such that g f = h. As a map of the underlyng topologcal spaces we defne g : Y T as g(y) = h(x), where x s any pont n the fber f 1 (y). Note that g s well-defned and g f = h by constructon. Now for any open set U T, we have g 1 (U) = f(h 1 (U)). Snce h 1 (U) s open by the contnuty of h and f s an open map, we conclude that g 1 (U) s also open, thereby showng that g s contnuous. Now we need to construct a morphsm of rnged spaces g # : O T g O Y such that f # g # = h #. Usng the same trck as before allows us to defne g # locally. The unqueness condton further shows that the local defntons of g # agree on the overlaps. Thus g # s globally defned whch proves the theorem. Descent for modules We recall the hypothess of lemma (2.1). Let M be an A-module, f : A B be a fathfully flat rng homomorphsm and we defne two A-module maps e 0, e 1 as 9

10 follows: e 0 : B B A B b b 1 e 1 : B B A B b 1 b Let M be a B-module. So we get two B A B-modules.e. M A B and B A M comng from e 0 and e 1 respectvely. The scalar multplcaton n M A B s gven by (b 1 b 2 ).(m b) := b 1 m b 2 b and n B A M s gven by (b 1 b 2 ).(b m) := b 1 b b 2 m. We get a B-module somorphsm σ gven by σ : M A B B A M m b b m. In general σ s not a B A B-module map, however f M = B A M, then one can check that the map σ : B A M A B B A B A M b m c c b m turns out to be an somorphsm of B A B-modules. We also get three maps as follows e 1,2 : B A B B A B A B b c b c 1 e 1,3 : B A B B A B A B b c b 1 c e 2,3 : B A B B A B A B b c 1 b c Usng the same constructon as before we extend the scalars further to get three B A B A B-modules.e. M A B A B, B A M A B and B A B A M comng from e 1,2, e 2,3 and e 1,3 respectvely. Let φ: M A B B A M be an somorphsm of B A B-modules. Ths gves rse to three somorphsms of B A B-modules whch are gven as follows: φ 1,2 : M A B A B φ 1 B B A M A B φ 2,3 : B A M A B 1 B φ B A B A M φ 1,3 : M A B A B B A B A M where φ 1,3 leaves the mddle term fxed and apples φ on the outer two terms of M A B A B. Now suppose M = B A B, then we have three canoncal 10

11 somorphsms of B A B-modules whch are gven below: φ 1,2 : M A B A B = B A M A B A B B A M A B = B A B A M A B b m c d c b m d φ 2,3 : B A M A B = B A B A M A B B A B A M = B A B A B A M b c m d b c d m φ 1,3 : M A B A B = B A M A B A B B A B A M = B A B A B A M b m c d b d c m One can check that the somorphsms satsfy the cocycle condton.e. φ 1,3 = φ 2,3 φ 1,2. Theorem 7.3. Let s suppose we are gven f : A B a fathfully flat rng homomorphsm, M a B-module and an somorphsm φ: M A B B A M of B A B-modules whch satsfy the cocycle condton φ 1,3 = φ 2,3 φ 1,2. Then there exsts an A-module M and an somorphsm σ : M B A M compatble wth φ.e. the followng dagram commutes: M A B φ B A M σ 1 B B A M A B 1 B swtch 1 B σ B A B A M M s also unque upto canoncal somorphsm.e. gven another A-module N and an somorphsm τ : M B A N compatble wth φ, then there exsts an unque somorphsm η : M N such that the followng dagram commutes: M σ B A M τ 1 B η B A N The proof of ths theorem wll be gven later. Defnton 7.1. A descent datum (M, φ) for modules wth respect to a fathfully flat homomorphsm f : A B, s gven by an B-module M and an somorphsm of B A B-modules φ: M A B B A M such that the cocycle condton holds. 11

12 The exstence of an A-module M by theorem (7.3) says that the descent datum s effectve. We now construct a category Des(A B) whose objects are the pars (M, φ) along wth ther descent datum. A morphsm (M, φ) (M, φ) of descent data s a morphsm of B-modules ψ : M M such that the dagram M ψ 1 B A B M A B s commutatve. φ φ B A M 1 B ψ B A M Theorem 7.4. Defne a functor A-mod Des(A B) M (B A M, 1 B (swtch)). Ths functor s an equvalence of categores. Proof. It follows from theorem (7.3). Now we gve the proof of theorem (7.3). Proof. Set M = {m M φ(m 1) = 1 m} and M s an A-module. So we have an exact sequence of A-modules as follows : 0 M M µ B A M m φ(m 1) 1 m Lemma 7.5. The map γ : B A M M defned by γ(b m) = bm s an somorphsm of B-modules. Proof. One can check that γ s a homomorphsm of B-modules. Now note that A B s an exact functor on the category of A-modules snce B s a flat A-module. Thus applyng A B to the above exact sequence we get 0 M A B 1 B M A B µ 1 B B A M A B (2) and ths sequence s also exact. By lemma (2.1) we have 0 M α B A M (e 0 e 1 ) 1 M B A B A M (3) s exact. Combnng (2) and (3) we get the followng dagram of B-modules: 0 M A B 1 B M A B γ φ µ 1 B B A M A B φ 2,3 0 M α B A M (e 0 e 1 ) 1 M B A B A M 12

13 and dagram chasng wll show that t commutes. It s not hard to see that the frst square commutes. For the second square, we see that f m b M A B then φ(( 1 B )(m b)) = φ(m b) = b m α(γ(m b)) = α(bm) = 1 bm = b m so the second square commutes. To show commutatvty of the last square, pck m b M A B and we have φ 2,3 ((µ 1 B )(m b)) = φ 2,3 ((φ(m 1) b) (1 m b)) = φ 2,3 (φ(m 1) b) φ 2,3 (1 m b) = φ 2,3 (φ 1,2 (m 1 b)) φ 2,3 (1 m b) = φ 1,3 (m 1 b) φ 2,3 (1 m b) ((e 0 e 1 ) 1 M )(φ(m b)) = φ 1,3 (m 1 b) φ 2,3 (1 m b). So the dagram commutes and fve lemma shows that γ : B A M M s an somorphsm of B-modules. Thus lemma (7.5) proves the frst part of theorem (7.3). We stll need to show the unqueness of M. Lemma 7.6. The sequence 0 Hom A (M, N) Hom B (B A M, B A N) e0 Hom B A B(B A B A M, B A B A N) e 1 s exact and the maps are defned as follows: For θ Hom A (M, N), defne (θ): B A M B A N b m b θ(m) and f ν Hom B (B A M, B A N), defne e 0 (ν): B A B A M B A B A N b c m b ν(c m) e 1 (ν): B A B A M B A B A N b c m c ν(b m) Proof. We apply lemma (2.1) to get the followng exact sequence: 0 N B A N B A B A N (4) Applyng Hom A (M, ) to (4), we obtan another exact sequence 0 Hom A (M, N) Hom A (M, B A N) Hom A (M, B A B A N) (5) 13

14 We use the tensor-hom adjuncton to get the followng somorphsms: Hom A (M, B A N) = Hom B (B A M, B A N) Hom A (M, B A B A N) = Hom B A B(B A B A M, B A B A N) Now usng the somorphsms above, the lemma follows from sequence (5). Note that the second part of the theorem s equvalent to lemma (7.6). Thus we get that M s unque upto canoncal somorphsm. Descent for schemes Gven an fpqc/fppf morphsm φ: X Y, set X = X Y X, wth ts two projectons p 1 and p 2 from X to X. Let X = X Y X Y X, whch comes wth three projectons p 1,2, p 2,3 and p 1,3 from X to X. We also have three projectons q 1, q 2 and q 3 from X to X, wth q = p 1 p,j and q j = p 2 p,j, 1 < j 3. Theorem 7.7. Let E, F be quas-coherent sheaves on Y. Then, for every morphsm h: φ E φ F on X satsfyng p 1h = p 2h on X, there s a unque morphsm g : E F on X such that φ g = h. In other words, the followng sequence of sheaves Hom OY (E, F) Hom OX (φ E, φ F) Hom OX Y X (p 1φ E, p 2φ F) s exact. Let E be a quas-coherent sheaf on X and τ : p 1E p 2E an somorphsm on X such that p 2,3τ p 1,2τ = p 1,3τ on X. Then there exsts a quascoherent sheaf E on Y and an somorphsm λ: φ E E on X such that p 2λ = τ p 1λ on X. Moreover the par (E, λ) s unque upto canoncal somorphsm. Defnton 7.2. Let φ: X Y be an fpqc/fppf morphsm of schemes and let Z be an X-scheme. Then the descent datum for Z/X/Y s an somorphsm of X -schemes θ : Z Y X X Y Z and satsfyng the expected cocycle condton.e. Z Y X Y X θ 1,3 X Y X Y Z θ1,2 θ 2,3 X Y Z Y X 14

15 Theorem 7.8. If φ: X Y s an fpqc/fppf morphsm of schemes, then the descent datum for an affne or quas-affne morphsm Z X s effectve. Cauton: However the descent datum for Z/X/Y need not be effectve for a projectve morphsm Z X. The reason s that the property of a morphsm X Y beng projectve cannot be checked locally. Indeed there are nonprojectve morphsms f : X Y of schemes of fnte type, such that Y has a Zarsk open cover U α where the morphsms f 1 (U α ) U α are not projectve. (We do not gve an example here.) To avod ths dffculty we consder the data of a (quas)-projectve morphsm together wth an ample lne bundle. Lemma 7.9. Let Z be a scheme and L an nvertble sheaf on Z. Gven a global secton s Γ(Z, L), defne Z s = {z Z s z m z L z } and Z s s open n Z. If F s a sheaf of O Z -modules then defne F(n) = F L n for any n Z. The followng condtons are equvalent: There are sectons s Γ(Z, L) wth n > 0 such that Z s form an open affne cover for Z. For any coherent sheaf F, the sheaf F(n) s generated by global sectons for n >> 0. Defnton 7.3. Let Z be a scheme and L an nvertble sheaf on Z. We say that L s ample f t satsfes the equvalent condtons of lemma (7.9) Defnton 7.4. Let f : Z X be a morphsm of schemes and L an nvertble sheaf on Z. We say that L s f-ample f for every open affne U X, the sheaf L f 1 (U) s ample. Defnton 7.5. A descent datum (Z, L) for schemes wth respect to an fpqc/fppf morphsm φ: X Y, s gven by an X-scheme Z wth an nvertble sheaf L and an somorphsm of X -schemes θ : Z Y X X Y Z satsfyng the cocycle condton θ 1,3ψ = θ 1,2ψ θ 2,3ψ where ψ : θ (1 L) L 1 s an somorphsm. Theorem If φ: X Y s an fpqc/fppf morphsm of schemes, then the descent datum for (Z, L) s effectve. 15