Coherent Sheaves on X Form an Abelian Category.”

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 61

7. Coherent sheaves The purpose of this section is to make sense of the statement that “Coherent sheaves on X form an abelian category.”

7.1. Sheaves of modules. For any ring R, the category Mod-R of all R-modules and subcategory mod-R of ﬁnitely generated R-modules are abelian categories. This means in particular that a homomorphism f : M N have kernel ker f, cokernel coker f and image (using the basic property that the cokernel! of ker f M is isomorphic to the kernel of N coker f): ! ! im(f)=M/ker f =ker(N coker f) ! The basic example of a quasi-coherent sheaf over a variety or scheme X isa(X)-module. It is coherent if it is ﬁnitely generated. One very important example is for a grade ring R. Then the quasi-coherent sheaf over X = Proj R corresponding to Rn is denoted oX (n). For more general varieties and schemes X, a quasi-coherent sheaf is a sheaf which is locally a module over (U, oX ):

Deﬁnition 7.1.1. Suppose that (X, oX ) is any ringed space. Then a sheaf of oX -modules over X is a sheaf F of additive groups so that (1) For every open subset U X, F (U)isa(U, o )-module. ✓ X (2) If V U, the restriction map F (U) F (V ) is morphism of (U, oX )-modules. (F (V ⇢) becomes a module over (U) by! the ring homomorphism (U) (V ).) ! This looks like a perfectly good deﬁnition. We need an explicit example to see what is wrong with it and why a “coherence” condition is needed. The basic good example is the ane case X = Spec(R). Let M be any R-module. Then the corresponding sheaf is denoted M and is given on basic open sets by

M(X )=M R f f ⌦R f So, at each point x =[p], the stalk if M R . f ⌦R p Proposition 7.1.2. Every element of M R has the form ⌦R p x 1 := x f ⌦ f where x M,f / p. 2 2 Proof. By deﬁnition, elements of M R are sums x r /f .Butr /f = s / f for ⌦R p i ⌦ i i i i i j some si R. So, 2 P Q si xi ri/fi = ⌦ fj X P Q ⇤ Corollary 7.1.3. For any open U in Spec(R), a section of M over U is given by a section [p]=x/f of the stalk bundle so that x, f are locally constant. I.e., for every point [p], there is a nbh V U so that x, f can be chosen to be constantf on V . ✓ Exercise 7.1.4. Show that, if (U, M)=0 then (V,M) = 0 for all open subsets V of U.

f f 62 KIYOSHI IGUSA BRANDEIS UNIVERSITY

7.2. Deﬁnition.

Deﬁnition 7.2.1. A quasi-coherent sheaf over a scheme X is an oX -module F so that there is an open covering of X by ane open sets U = Spec(R ) so that F U = M for ↵ ↵ | ↵ ↵ some R↵-module M↵. f The following theorem show that the category of quasi-coherent sheaves on X is a gen- eralization of the category of sheaves of the form M on X = Spec(R). Theorem 7.2.2. Every quasi-coherent sheaf F over X = Spec(R) is isomorphic to M f where M = F (X). f This follows from the following lemma.

Lemma 7.2.3. For f R not nilpotent consider Xf = Spec(Rf ). 2 n (a) If s F (X), s Xf =0then n s.t. f s =0 (b) If t 2F (X ) then| n, s F9(X) so that f nt = s X . 2 f 9 9 2 | f Proof that Lemma implies Theorem. Let M = F (X). By definition of oX -module, F (Xf ) is an Rf -module. So, we have: : M = F (X) R F (X ) f ⌦R f ! f given by (x a)=res(x)a. It suces to show that this is an isomorphism for all f. (a) implies⌦ that is a monomorphism: m m m Take any s/f Mf in the kernel of .Then(s/f )=(s)/f =0implies k so k 2 k k n k 9 that f (s)=0=(f s). So, f s Xf = 0. By (a) n so that f (f s)=0inM.Then m | 9 s/f =0inMf . So, is a monomorphism. (b) implies that is onto: Take any t F (X ). By (b) there are n, s M so that t = s/f n X . So, is onto. 2 f 2 | f ⇤ Proof of Lemma. (a) We are given s F (X), X = Spec(R) so that s X = 0. 2 | f Since X is quasi-compact, by definition of quasi-coherent, we have X = Xi, Xi = Spec(R ) so that F X = M . We may assume that X = X . Recall that X X = X . i | i i i gi f \S gi fgi So, Xf = Xfgi and f (X , M )=(M ) S fgi i i f So, s Xf =0i↵s Xfgi =0 (Mi)f for each i. Let yi = s Xi.Thenyi (Xi)f =0implies n | | 2 n | n | f i yi = 0. Let n = max ni.Thenf s restrictedf to Xi is f yi = 0. Since X = Xi this implies that f ns =0inF (X). This proves (a). ni S (b) Let t F (Xf ). Since F (Xfgi )=(Mi)f , t Xfgi = xi/f . Let n = max ni.Then t X = y /f2n. In other words, f nt X = y M|. The restrictions of y ,y to X X may | fgi i | fgi i 2 i i j i\ j not agree. But their restriction to Xi Xj Xf are equal since yi,yj are both restrictions n \ \ mij mij of f t F (Xf ). By (a) this implies that, for some mij, f yi = f yj in F (Xi Xj). 2 m \ Let m = max mij.Thenf yi Mi = F (Xi) agree on Xi Xj. By definition of a sheaf m2 m+n \ !s F (X) so that s Xi = f yi for each i.Sincef t F (Xf ) restricts to the same elements9 2 in X X ,wemusthave| s X = f m+nt. 2 f \ i | f ⇤ Corollary 7.2.4. An oX -module F is quasi-coherent if and only if, for every ane open subset U of X, F U = M where M =(U, o ). | ⇠ X f LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 63

7.3. Coherent sheaves over projective space. We take the simplest case X = P (⌦) = [x, y]:x, y not both zero 1 { } Recall that this is a union of two ane open subsets: X = X X x [ y where 1 Xx = [x, y]:x =0 = A˚ Spec(k[s]) { 6 } ⇡ Xy = [x, y]:y =0 Spec(k[t]) { 6 }⇡ with intersection 1 Xx Xy = [x, y]:xy =0 Spec(k[t, t ]) \ { 6 }⇡ We ignore the fact that the variety has more points than the scheme. (Recall that the scheme is the quotient space of the variety given by identifying x, y if x = y and taking the push-forward of the structure sheaf.)

Proposition 7.3.1. A quasi-coherent sheaf on P1 is given uniquely be the following data. (1) A k[s]-module M0 (2) A k[t]-module M1 1 1 (3) An isomorphism M0 k[s, s ] ⇠= M1 k[t ,t] where the rings are identiﬁed with st =1. ⌦ ⌦

Proof. By the theorem, any quasi-coherent sheaf F restricted to the ane open sets Xx,Xy is given by F X = M and F X = M . | x 0 | y 1 On Xx Xy the restrictions of M0 and M1 agree i↵(3). ⇤ \ f f For Pn, there are more ane sets and the isomorphisms need to be compatible. As an example, take n = 2. (The wordingf suggestsf how to generalize to any n.)

Proposition 7.3.2. A quasi-coherent sheaf on P2 is given uniquely be the following data. (1) An Ri-module Mi for i =0, 1, 2 where R0 = k[T1,T2], R1 = k[T0,T2], R2 = k[T0,T1].) (2) An Rij-module Mij for i = j 0, 1, 2 where Rij = k[T0,T1,T2]/(1 TiTj). 6 2{ } (3) One R012-module M012 where R012 = k[T0,T1,T2]/(1 T0T1T2). so that for any inclusion I J of two nonempty subsets of 0, 1, 2 we have isomorphisms ⇢ { } (7.1) M R = M I ⌦RI J ⇠ J Proof. P2 is a union of three ane open sets X0,X1,X2 and all intersections are also ane Xi Xj = Xij, XI XJ = XI J . Furthermore, XJ = Spec(RJ ) for all nonempty J 0, 1, . \ \ [ ✓{ } By the theorem, any quasi-coherent sheaf on XJ has the form MJ for some RJ -module MJ . The isomorphisms (7.1) are exactly what is needed to paste these together. ⇤ f Deﬁnition 7.3.3. An oX -module F is called a coherent sheaf if X has a covering by ane open sets U = Spec(R ) so that F U = M for some R -module M . ↵ ↵ | ↵ ⇠ ↵ ↵ ↵ It can be shown that, for projective space, these ane open sets can be taken to be the f open set Xi. So, the quasi-coherent sheaf over Pn given above is coherent if and only if each Mi is a ﬁnitely generated Ri module. (And thus each MJ is a f.g. RJ -module. One example is the structure sheaf oX which is given by the ring itself:

MJ = RJ for all = J 0, 1, ,n . This is an example of a locally free sheaf. ;6 ✓{ ··· } 64 KIYOSHI IGUSA BRANDEIS UNIVERSITY

Deﬁnition 7.3.4. A locally free sheaf is a quasi-coherent sheaf on a scheme X so that, X is the union of U↵ = Spec(R↵) and M↵ = F (U↵)isafreeR↵-module for each ↵.Wesay r that F is free of rank r if M↵ ⇠= R↵ for each ↵.

For example, the structure sheaf oX is always locally free of rank 1. Another example of a locally free sheaf of rank 1 on Pn is the “twisting sheaf” oX (1). I will do the example of P1, then give the general deﬁnition, then return to the special case. 1 First, note that R0,R1 are subrings of R01 since R01 = k[t, t ] is the set of all Laurent polynomials a b m 1 m n + n 1 + + yt + zt t t ··· R1 = k[t] is the subring where the negative powers of t do not occur and R0 = k[s]with s =1/t is the subring where positive powers of t do not occur. 1 In terms of the vector space basis for k[t, t ]wehave:

R1 3 2 1 2 3 t ,t ,t , 1 ,t,t ,t , ··· ··· R0 z }| {

The intersection “1” represents| k. It also{z represents} a global section of oX :

(Pn,oX )=k

Proposition 7.3.5. For any quasi-coherent sheaf F on Pn we have: (P ,F) Hom (o ,F) n ⇠= oX X

Deﬁnition 7.3.6. If M = MJ , N = NJ are quasi-coherent sheaves on Pn then an o -homomorphism M N {is a compatible} { collection} of R -homomorphism M N . X ! J J ! J For example, on P1, a morphism is given by a commutative diagram:

M0 / M01 o M1

✏ ✏ ✏ N0 / N01 o N1

Example 7.3.7. On P1, for any integer n, we have the locally free rank one sheaf oX (n) n given by M0 = t R0,M1 = R1,M01 = R01. For example oX (1):

M1 3 2 1 2 3 t ,t ,t , 1,t,t ,t , ··· ··· M0 z }| { 2 has two sections: 1,t span (P1|,oX (1)){z = k . On} the other hand oX ( 1) has no (nonzero) sections: (P ,o ( 1)) = 0. 1 X M1 3 2 1 2 3 t ,t ,t , 1,t,t ,t , ··· ··· M0 z }| {

The locally free sheaves oX (|n) are{z called}twisting sheaves because tensor product with these sheaves “twist” a quasi-coherent sheaf. The tensor product of two quasi-coherent LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 65

sheaves F, G is given locally by (F G)(U↵)=F (U↵) R↵ G(U↵). For any integer n, and any quasi-coherent sheaf F =(M ⌦ M M ) on P⌦, 0 ! 01 1 1 F o (n)=(M n M M ) ⌦ X 0 !t 01 1 Exercise 7.3.8. Show that o (n) o (m)=o (n + m). X ⌦ X X Theorem 7.3.9. If F is coherent then, for some n, F oX (n) has enough global sections to generate F over every point. ⌦ By Proposition 7.3.5, this is equivalent to saying that there is an epimorphism om F o (n) X ⇣ ⌦ X 7.4. General twisting sheaves. Suppose S = Sn is a graded ring and M = Mn is a graded module. This means M is an S-modules and SnMm Mn+m.Then,X = Proj(S) L ⇢ L has a quasi-coherent sheaf M given as follows. For any homogeneous f S , recall that X = P Proj(S):f/P = Spec(S ) 2 d f { 2 2 } (f) where f a S = S a S (f) f n 2 f | 2 nd ⇢ I.e., S(f) is the set of all elements of Sf of degree 0. Then x M =(X , M)= M x M (f) f f n 2 f | 2 nd ⇢ I.e., M is the set of all elements of M = M S of degree 0. (f) ff ⌦ f When M = S, M = oX is the structure sheaf of X = Proj(S). To get the twisting sheaf oX (r), we need to use M = S(r)whichisthemoduleS shifted down in degree by r: f Mn = Sn+r In other words, M is the set of all a/f n S of degree r (a S ). (f) 2 f 2 r+nd We will now show that, in the special case S = k[s, t](X = P1), the two deﬁnitions of the twisting sheaf oX (1) agree:

(7.2) oX (r)=S(r)

We take X = P1 = Proj(k[s, t]), M = S(2) as an example. Then g x M = x S (s) sn | 2 n+2 n a b o Since Sn+2 is spanned by monomials of the form s t where a + b = n + 2, we see that M(t) has basis s2,st,t2,t3/s, t4/s2, . Similarly M has basis ,s3/t, s2,st,t2. When both ··· (s) ··· s, t are inverted then we get all of Sn+2. Thus the images of M0 = M(t) and M1 = M(s) in M01 = S(2) are (the spans of):

M1 4 2 3 1 2 2 1 3 2 4 s t ,s t ,s ,st,t,s t ,s t , ··· ··· M0 z }| { Note that the overlap is the| set of monomials{z of} degree 2. Since there are 3 of them, this is isomorphic to oX (2) in the earlier description. The fact that there are 3 linearly independent sections is actually very easy to see: 3 2 2 (P1,oX (2)) = S(2) = k = span(s ,st,t ). 66 KIYOSHI IGUSA BRANDEIS UNIVERSITY

For oX (1) over P2 = Proj(k[x, y, z]) we have the following analogous description where we recall that, in Mi =(Xi,oX (1)), the variables other than xi are inverted.

M12 = M(yz) M02 = M(xz)

2 2 1 2 1 2 2 2 xy z y z x z x yz

2 2 1 1 2 2 x y z xy z z x yz x y z

3 2 2 1 1 2 2 3 x y x y x y x y x y

M01 = M(xy)

3 1 1 2 1 1 2 1 1 3 1 x y z x z xyz y z x y z

The modules Mi are given intersections. For example, M2 = M(z) = M02 M12 is the inﬁnite triangular region with z at the top. \ The intersection of all the regions is 3 Mij =(P2,oX (1)) = S(1) = span(x, y, z)=k oX (2) is similar with\ 6 things in the middle with entries all monomials of degree 2.