Math 256B Notes

James McIvor

April 16, 2012

Abstract These are my notes for Martin Olsson’s Math 256B: Algebraic Ge- ometry course, taught in the Spring of 2012, to be updated continually throughout the semester. In a few places, I have added an example or ﬂeshed out a few details myself. I apologize for any inaccuracies this may have introduced. Please email any corrections, questions, or suggestions to mcivor at math.berkeley.edu.

Contents

0 Plan for the course 4

1 Wednesday, January 18: Diﬀerentials 4 1.1 An Algebraic Description of the Cotangent Space ...... 4 1.2 Relative Spec ...... 5 1.3 Dual Numbers ...... 6

2 Friday, January 20th: More on Differentials 7 2.1 Relationship with Leibniz Rule ...... 7 2.2 Relation with the Diagonal ...... 8 2.3 KählerDifferentials ...... 10

3 Monday, Jan. 23rd: Globalizing Differentials 12 3.1 Definition of the Sheaf of Differentials ...... 12 3.2 Relating the Global and Affine Constructions ...... 13 3.3 Relation with Infinitesimal Thickenings ...... 15

4 Wednesday, Jan. 25th: Exact Sequences Involving the Cotangent Sheaf 16 1 4.1 Functoriality of ΩX/Y ...... 16 4.2 First Exact Sequence ...... 18 4.3 Second Exact Sequence ...... 20

n 5 Friday, Jan. 27th: Diﬀerentials on P 22 5.1 Preliminaries ...... 22 5.2 A Concrete Example ...... 22 1 5.3 Functorial Characterization of ΩX/Y ...... 23

1 1 6 Monday, January 30th: Proof of the Computation of Ω n 24 PS /S 7 Wednesday, February 1st: Conclusion of the Computation 1 of Ω n 28 PS /S 7.1 Review of Last Lecture’s Construction ...... 28 7.2 Conclusion of the Proof ...... 29 7.3 An Example ...... 31 7.4 Concluding Remarks ...... 31

8 Friday, February 3rd: Curves 32 8.1 Separating Points and Tangent Vectors ...... 32 8.2 Motivating the Terminology ...... 32 8.3 A Preparatory Lemma ...... 33

9 Monday, February 6th: Proof of the Closed Embedding Criteria 34 9.1 Proof of Proposition 8.1 ...... 34 9.2 Towards Cohomology ...... 36

10 Wednesday, Feb. 8th: Cohomology of Coherent Sheaves on a Curve 37 10.1 Black Box: Cohomology of Coherent Sheaves on Curves . . 37 10.2 Further Comments; Euler Characteristic ...... 39 10.3 Statement of Riemann-Roch; Reduction to the More Fa- miliar Version ...... 40

11 Friday, Feb. 10th: Proof and Consequences of Riemann- Roch 41 11.1 Preparatory Remarks ...... 41 11.2 Proof of the Riemann-Roch Theorem ...... 42 11.3 Consequences ...... 43

12 Monday, Feb. 13th: Applying Riemann-Roch to the Study of Curves 44 n 12.1 Using Riemann-Roch to Detect Closed Immersions to P .. 44 12.2 Examples ...... 45

13 Wednesday, Feb. 15th: “Where is the Equation?” 46

14 Friday, Feb. 17th: Ramiﬁcation and Degree of a Mor- phism 49

15 Wednesday, Feb. 22nd: Hyperelliptic Curves 51 15.1 Canonical Embeddings ...... 51 15.2 Hyperelliptic Curves ...... 52

16 Friday, Feb. 24th: Riemann-Hurwitz Formula 53 16.1 Preparatory Lemmata ...... 53 16.2 Riemann-Hurwitz Formula ...... 55

2 17 Monday, Feb. 27th: Homological Algebra I 56 17.1 Motivation ...... 56 17.1.1 Algebraic topology ...... 56 17.1.2 Algebraic geometry – a.k.a. the Black Box . . . . . 56 17.2 Complexes, cohomology and snakes ...... 57 17.3 Homotopy ...... 59

18 Wednesday, Feb. 29th: Homological Algebra II 61 18.1 Additive and abelian categories ...... 61 18.2 δ-functors ...... 63 18.3 Left- and right-exact functors ...... 65 18.4 Universality of cohomology ...... 66

19 Friday, March 2nd: Homological Algebra III 67 19.1 Derived functors – a construction proposal ...... 67 19.1.1 Derived functors – resolution of technical issues . . . 69 19.2 Derived functors – conclusion ...... 71 19.3 Two useful lemmas ...... 72

20 Monday, March 5th: Categories of Sheaves Have Enough Injectives 72

21 Wednesday, March 7th: Flasque Sheaves 75

22 Friday, March 9th: Grothendieck’s Vanishing Theorem 77 22.1 Statement and Preliminary Comments ...... 77 22.2 Idea of the Proof; Some lemmas ...... 78

23 Monday, March 12th: More Lemmas 79

24 Wednesday, March 14th: Proof of the Vanishing Theorem 80 24.1 Base step ...... 80 24.2 Reduction to X irreducible ...... 81 24.3 Reduction to the case where F = ZU ...... 81 24.4 Conclusion of the Proof ...... 83

25 Friday, March 16th: Spectral Sequences - Introduction 83 25.1 Motivating Examples ...... 83 25.2 Deﬁnition ...... 84 25.3 Spectral Sequence of a Filtered Complex ...... 85

26 Monday, March 19th: Spectral Sequence of a Double Com- plex 86 26.1 Double Complex and Total Complex ...... 86 26.2 Application: Derived Functors on the Category of Complexes 88

27 Wednesday, March 21st: Spectral Sequence of an Open Cover 89

28 Friday, March 23rd: Cechˆ Cohomology Agrees with De- rived Functor Cohomology in Good Cases 92

3 29 Monday, April 2nd: Serre’s Aﬃneness Criterion 96

30 Wednesday, April 4th: Cohomology of Projective Space 99

31 Friday, April 6th: Conclusion of Proof and Applications 101

32 Monday, April 9th: Finite Generation of Cohomology 103

33 Wednesday, April 11th: The Hilbert Polynomial 106

0 Plan for the course

1. Diﬀerentials 2. Curve Theory (Assuming Cohomology) 3. Build Cohomology Theory 4. Surfaces (Time Permitting) Professor Olsson’s oﬃce hours are Wednesdays 9-11AM in 879 Evans Hall.

1 Wednesday, January 18: Diﬀerentials

1.1 An Algebraic Description of the Cotangent Space We want to construct the relative cotangent sheaf associated to a morphism f : X → Y . The motivation is as follows. A differential, or dually, a tangent vector, should be something like the data of a point in a scheme, together with an “infinitesimal direction vector” at that point. Algebraically, this can be described by taking the morphism Spec k → Spec k[]/(2) and mapping it into our scheme. These two schemes have the same topological space, and the nilpotency in k[]/(2) is meant to capture our intuition of the “infinitesimal”. More generally, we can take any “infinitesimal thickening”, i.e., a closed immersion T0 ,→ T defined by a square-zero ideal sheaf J ⊂ OT and map that into X. Of course, we want this notion to be relative, so we should really map this thickening into the morphism f, which amounts to filling in the dotted arrow in following diagram

T0 / X _ >

T / Y which should be thought of as representing the following picture in the case when Y = Spec k picture omitted/pending 2 Example 1.1. Take Y = Spec k, T = Spec k and T0 = Spec k[]/( ). Consider the set of morphisms ﬁlling in following diagram:

4 x∈X Spec k / X or, equivalently, k o OX,x 8 O O

7→0 z Spec k[]/(2) / Spec k k[]/(2) o k

Commutativity of the square forces the composite k → k to be the identity, so the maximal ideal m ⊂ OX,x maps to zero in k, hence along the dotted arrow, m gets mapped into (), giving a new diagram

a k o O /m2 O X,xO 7→0 b y k[]/(2) o k

2 2 But k ⊕ m/m =∼ OX,x/m via the map (α, β) 7→ b(α) + β. More- over, k ⊕ m/m2 can be given the structure of a k-algebra by deﬁning the multiplication (α, β) · (α0, β0) = (αα0, αβ0 + α0β). Also, as k-algebras, k[]/(2) =∼ k ⊕ k · , so in fact we are looking for dotted arrows ﬁlling in the following diagram of k-algebras:

a k o k ⊕ m/m2 O O 7→0 b y k[]/(2) o k Such arrows are determined by the map m/m2 → k. Thus the set of arrows filling in diagram 1 is in bijection with (m/m2)∨. This justifies the definition of the tangent space of X over k at x as (m/m2)∨. Greater generality can be achieved by working with X as a scheme over some arbitrary scheme Y , and also by replacing the ring of dual numbers k[]/(2) by a more general construction. Before doing so, we need the notion of relative Spec.

1.2 Relative Spec Given a scheme X and a quasicoherent sheaf of algebras A on X, we show how to construct a scheme Spec (A). Loosely speaking, the given sheaf X A includes both the data of a ring over each aﬃne open of X, and the gluing data for these rings. We take Spec of each ring A(U) for opens U of X, and glue them according to the speciﬁcations of the sheaf A. Alternatively, Spec (A) is the representing object for the functor X F : (Schemes/X)op → Set

f ∗ which sends T → X to the set of OT -algebra morphisms f A → OT . Two things need to be checked, namely that this functor is representable at all,

5 and that the scheme obtained by gluing described above actually is the representing object. Here’s why it’s locally representable: If X and T are aﬃne, say X = Spec B, and T = Spec R, with A = A˜ ∗ ∼ for some B-algebra A, then f A = (A⊗B R) , by deﬁnition of the pullback sheaf. By the equivalence of categories between quasicoherent sheaves of ∗ OT -modules and R-modules, the set of OT -algebra maps f A → OT is in natural bijection with the set of R-algebra maps A ⊗B R → R, so we are looking for dotted maps in the following diagram

A ⊗B R / R O w; ww ww ww ww R w But the vertical map on the left ﬁts into a ﬁbered square as follows

A / A ⊗B R / R O O w; ww ww ww ww B / R w

Moreover, to give a map of R-algebras A ⊗B R → R is equivalent, by the universal property of the ﬁbered square, to giving a map A → R along the top: ( A / A ⊗B R / R O O w; ww ww ww ww B / R w ∗ This shows that in this case, the set of maps of OT -algebras f A → OT is in bijection with the maps of R-algebras A → R, and hence that Spec A represents the functor F . The remaining details of representability are omitted. In fact, the relative Spec functor deﬁnes an equivalence of categories

Spec X s {aﬃne morphisms Y → X} {qcoh O -algebras} 3 X

f where the map in the other direction sends the aﬃne morphism Y → X to f∗OY .

1.3 Dual Numbers In the example of 1.1 we took a scheme X over k and mapped the “in- ﬁnitesimal thickening” Spec k → Spec k[]/(2) into X → Spec k. In fact this thickening can be globalized as follows. If X is a scheme and F a quasicoherent sheaf of OX -algebras, deﬁne a quasicoherent sheaf of OX - algebras by OX [F] = OX ⊕ F

6 with the multiplicative structure given open-by-open by the rule

(a, b) · (a0, b0) = (aa0, ab0 + a0b) just like before. We then have a diagram of OX -algebras

pr1 OX o OX ⊕ F O u: uu Id uu uu uu OX where pr1 denotes projection onto the ﬁrst factor and Id is the identity map. We apply the relative Spec functor to this diagram, and deﬁne the scheme X[F] to be Spec OX [F], which by the construction comes with X a diagram

X / X[F] zz Id zz zz z| z X

f Example 1.2. Consider the aﬃne case, where B → A is a ring map, and M an A-module. Here we are thinking of X in the above discussion as the scheme Spec A over the scheme Spec B, and the dual numbers are just A[M] =∼ A ⊕ M, with multiplication deﬁned as above. Thus the diagram of OX -algebras above, when working over B, looks like

A o A[M] O = O zz Id zz zz zz A o B A natural question to ask is: what maps along the diagonal, A → A[M], make this diagram commute? We take up this question in the next lecture.

2 Friday, January 20th: More on Diﬀerentials

Given a ring map B → A and A-module M, we form A[M] as last time, with the “dual multiplication rule”. In the case A = M = k, we recover the initial example: A[M] =∼ k[]/(2).

2.1 Relationship with Leibniz Rule The module A[M] comes with maps of A-algebras

σ0 π A / A[M] / A whose composition is the identity. We would like to study sections of the map A[M] → A. First we need a deﬁnition.

7 Definition 2.1. If B → A is a ring map and M is an A-module, a B- linear derivation of A into M is a map ∂ : A → M satisfying ∂(b) = 0 for all b in B, and ∂(aa0) = a∂(a0) + a0∂(a) for al a, a0 in A. The second property is referred to as the Leibniz rule. Note also that by ∂(b), we really mean map b into A using the given ring map, and then apply ∂. The set of B-linear derivations of A into M is denoted DerB (A, M), or often just Der(M) if the ring map B → A is clear from context. Note also that Der(M) itself has the structure of an A-module (even though its elements are not A-linear maps!): if a is in A and ∂ is in Der(M), define (a · ∂)(x) = a · ∂(x), the dot on the right being the action of A on M. ∞ 1 For an example, let B = R, A = C (R), and M = Ω (R), the A- module of smooth 1-forms on R. Then take ∂ = d, the usual differential operator. The conditions in this case say simply that d kills the constants and satisfies the usual product rule for differentiation. Proposition 2.1. The set of A-algebra maps φ: A → A[M] such that π ◦ φ = Id is in bijection with the set DerB (A, M).

Proof. The bijection is deﬁned by sending a derivation ∂ to the map A → A[M] given by a 7→ a + ∂(a). The resulting map is a section of π since it is the identity on the ﬁrst component; it is A-linear since it is additive in both factors A and M, and for a, a0 in A,

aa0 7→ aa0 + ∂(aa0) = aa0 + a∂(a0) + a0∂(a) = (a + ∂(a))(a0 + ∂(a0)), the last equality as a result of the deﬁnition of the multiplication in A[M]. The map in the other direction sends an A-linear map φ: A → A[M] to ∂(a) = φ(a) − a. Note that the image of ∂ actually lies in M since π ◦ φ = Id. Moreover, ∂ is a derivation since, writing φ(a) = a + m for some m in M, we have

∂(aa0) = φ(aa0) − aa0 = φ(a)φ(a0) − aa0 = (a + m)(a0 + m0) − aa0 = am0 + a0m = a(φ(a0) − a0) + a0(φ(a) − a) = a∂(a0) + a0∂(a)

2.2 Relation with the Diagonal Now, to study these maps φ from a slightly diﬀerent point of view, we think of M above as varying over all A-modules, and construct a universal derivation. To begin with, take Spec of the maps above, giving a diagram

Spec A π / Spec A[M] L LLL LL σ0 Id LL LL& Spec A Now, putting the diagram over B, we are looking for φ such that the following diagram commutes:

8 Id

φ % Spec A π / Spec A[M] / Spec A L LLL LL σ0 Id LL LL& Spec A / Spec B

Let ∆: Spec A → Spec A ⊗B A denote the diagonal morphism, corresponding to multiplication A ⊗B A → A. Then to give a morphism φ above is equivalent to giving a map φ such that

Spec A π / Spec A[M] MM MM MMM MM φ MM σ0⊗φ MM ∆ MMM MMM M& pr2 M& σ0 Spec A ⊗B A / Spec A

pr1 % Spec A / Spec B commutes, by the universal property of the ﬁber product. If the appearance of the diagonal is surprising, you should think of it in the following way. The composite

X → X ×Y X → X can be interpretted as expressing X as a retract of the diagonal, sort of like a tubular neighborhood of X in X ×Y X. We think of the image of a point x in X in X ×Y X as “two nearby points”, which then map back down along the second map to the original point in X. In any case, the above situation leads to the following algebraic problem. We are interested in φ: A⊗B A → A[M] (note that this is not exactly the same φ as above) such that the following commutes:

A[M] V d π zz φ zz zz z} z A o A ⊗B A σ0 O a7→a⊗1

Let I be the kernel of the multiplication A ⊗B A → A. Then the composite φ π I → A ⊗B A → A[M] → A is zero, by commutativity of the triangle above; hence the image of I lies in M ⊂ A[M], and there is an induced dotted map

9 A[M] o M V d g z JJ π z JJ φ zz JJ zz JJ z} z J A o A ⊗B A o I σ0 O a7→a⊗1

A 2 2 Now since M = 0 in A[M], I ⊂ A ⊗B A maps to 0 in A[M] under 2 any such φ, hence the desired φ must factor through A ⊗B A/I . Thus 2 A ⊗B A/I is the universal object of the form A[M], as M varies. In fact, 2 2 2 A ⊗B A =∼ A[I/I ], since the map A[I/I ] → A ⊗B A/I sending a + γ to (a ⊗ 1) + γ ﬁts into the following map of exact sequences

π 0 / I/I2 / A[I/I2] / A / 0

∼= ∼=

mult 0 / I/I2 / A ⊗ A/I2 / A / 0 so it is an isomorphism by the ﬁve lemma. Summarizing our results, we have established Theorem 2.1. The following four sets are in natural bijection:

1. DerB (A, M) 2. {A-algebra maps φ: A → A[M] | π ◦ φ = Id} 3. {A-algebra maps | A[I/I2] / A[M] } KKK% { w A w 2 4. HomA(I/I ,M) The bijection between 3 and 4 is given by sending φ to its restriction to the second factor, I/I2. We can express the equivalence of 1 and 4 as saying that the functor

Der: ModA → ModA as described above is represented by I/I2.

2.3 Kähler Differentials We have seen that every A-linear map A → A[M] corresponds to a derivation ∂ : A → M; the map A → A[I/I2] is associated to the universal derivation d: A → I/I2, defined by

d(a) = a ⊗ 1 − 1 ⊗ a

The fact that this lies in I is implied by the following:

Lemma 2.1. The kernel I of the multiplication map A ⊗B A → A is generated by elements of the form 1 ⊗ a − a ⊗ 1.

10 Proof. Clearly these elements are in I. Conversely, let Σx ⊗ y be any element of I. Then Σxy = 0, so Σ(xy ⊗ 1) = 0, giving Σx ⊗ y = Σx(1 ⊗ y − y ⊗ 1).

Theorem 2.2. The map d deﬁned above is a derivation, and is universal in the following sense: for any B-linear derivation ∂ : A → M, there is a unique A-linear map φ: I/I2 → M such that the diagram

d A / I/I2 B BB BB φ ∂ B BB! M commutes.

Proof. The universal property follows from the preceding discussion of I/I2. It remains only to check that d is actually a derivation. For b in B, we have d(b) = 1 ⊗ b − b ⊗ 1 = b ⊗ 1 − b ⊗ 1 = 0 using the fact that the tensor product is taken over B. For the Leibniz rule, ﬁrst note that I is generated by elements of the form 1 ⊗ a − a ⊗ 1. Then let a and a0 be elements of A, and compute

d(aa0) − a d(a0) − a0 d(a) = aa0 ⊗ 1 − 1 ⊗ aa0 − a(a0 ⊗ 1 − 1 ⊗ a0) − a0(a ⊗ 1 − 1 ⊗ a) = aa0 ⊗ 1 − 1 ⊗ aa0 − aa0 ⊗ 1 + a ⊗ a0 − a0a ⊗ 1 + a0 ⊗ a = −1 ⊗ aa0 − aa0 ⊗ 1 + a ⊗ a0 + a0 ⊗ a = (a ⊗ 1 − 1 ⊗ a)(a0 ⊗ 1 − 1 ⊗ a0), and this is zero in I/I2.

Remark. The above computation used the A-module structure on A⊗B A given by a · (x ⊗ y) = ax ⊗ y. But there is also the action given by a · (x ⊗ y) = x ⊗ ay. Both of these actions, of course, restrict to I, but in fact when working modulo I2, these two restrictions agree. This follows from Lemma 1: if a is an element of A, then the difference of the two actions of a on I can be described as multiplication by (1 ⊗ a − a ⊗ 1) (using the usual multiplication for the tensor product of rings). But the lemma shows that this is in I, and hence its action on I is zero modulo I2. Professor Olsson used this in his computation that d satisfies the Leibniz rule. Definition 2.2. The A-module I/I2 is called the module of (relative) Kählerdifferentials of B → A, and denoted ΩA/B . Note that it comes together with the derivation d: A → ΩA/B .

In light of the previous discussion, one thinks of ΩA/B as being the algebraic analogue of the cotangent bundle. Here is an example to further support this terminology.

11 Example 2.1. With notation as above, let B = k, and A = k[x, y]/(y2 − x3 − 5). Then A dx ⊕ A dy Ω =∼ , A/B 2y dy − 3x2 dx as you might expect from calculus. The universal derivaton d: A → ΩA/B ∂f ∂f is given by the usual diﬀerentiation formula: df = ∂x dx + ∂y dy; the quotient ensures that this is well deﬁned when we work with polynomials modulo y2 − x3 − 5. The universal property is expressed by the following diagram:

d A / ΩA/B @ @@ @@ ∂ @ φ @@ | M

The map φ:ΩA/B → M guaranteed by the theorem sends dx and dy to ∂(x) and ∂(y), respectively.

3 Monday, Jan. 23rd: Globalizing Diﬀerentials

Today we will globalize the construuctions of last week, obtaining the 1 sheaf ΩX/Y of relative differentials associated to a morphism of schemes X → Y . We could do this by gluing together the modules ΩA/BB of last time, but instead we give a global definition, and then check that it reduces to the original definition in the affine case. Recall that for a ring map B → I, letting I be the kernel of A ⊗B A → 2 A, we defined ΩA/B = I/I . To relate this to today’s construction, it will be helpful to bear in mind the isomorphism 2 ∼ I/I = I ⊗A⊗B A A which emphasizes the fact that this object is a pullback along the diagonal.

3.1 Deﬁnition of the Sheaf of Diﬀerentials

Let f : X → Y be a morphism of schemes. It induces a map ∆X/Y : X → X ×Y X, from the following diagram:

X Id ∆X/Y % X ×Y X / X Id f f ) X / Y

# The map ∆X/Y comes with a map of sheaves ∆X/Y : OX×Y X → (∆X/Y )∗OX on X ×Y X, which we will just abbreviate by ∆. Although really a map of sheaves of rings, we will regard it as a map of sheaves of OX×X -modules. Let J be its kernel, again regarded as an OX×X -module.

Deﬁnition 3.1. The sheaf of relative diﬀerentials of X → Y is the OX - 1 ∗ module ΩX/Y = ∆ J .

12 Remark. Note that ∆X/Y is not necessarily a closed immersion, so a priori J is just a sheaf of OX×Y X -modules. Now from the diagram

∆X/Y X / X ×Y X HH HH HH pr1 pr2 Id HH HH$ Ø X −1 we get maps of sheaves δi : pri OX → OX×Y X for i = 1, 2. Applying −1 the functor ∆X/Y to these morphisms gives maps

−1 −1 −1 ∆ δi −1 ∆ pri OX −→ ∆ OX×Y X , −1 −1 ∼ but by the commutativity of the diagram of schemes, we have ∆ pri OX = −1 −1 OX . Thus we have maps of sheaves on X, ∆ δi : OX → ∆ OX×Y X . −1 Both maps (for i = 1, 2) agree when followed by the map ∆ OX×Y X → −1 OX , so their diﬀerence maps into the kernel of ∆ OX×Y X → OX , giving

−1 −1 −1 ∆ δ1 − ∆ δ2 : OX → ker(∆ OX×Y X → OX )

Recall that J was deﬁned as the kernel of OX×Y X → ∆∗OX . Applying ∆−1, we have

−1 −1 ∆ J = ∆ ker(OX×Y X → ∆∗OX ) Using the left-exactness of ∆−1, we can push it across the kernel and then −1 apply the adjunction between ∆ and ∆∗ to get

−1 −1 ∆ J = ker(∆ OX×Y X → OX ) Putting this together we now have a map

−1 −1 −1 ∆ δ1 − ∆ δ2 : OX → ∆ J

−1 ∗ 1 Now follow this with the map ∆ J → ∆ J = ΩX/Y . We have now, at −1 −1 great length, constructed the universal derivation d = ∆ δ1−∆ δ2 : OX → 1 1 ΩX/Y , which is the global version of the map dA → ΩA/B in the aﬃne case.

3.2 Relating the Global and Affine Constructions We now verify that our new definition of the cotangent sheaf agrees with 1 the affine definition locally, in the process showing further that ΩX/Y is quasicoherent. 1 Proposition 3.1. 1. ΩX/Y is quasicoherent. 2. If f : X → Y is locally of finite type and Y is locally Noetherian, 1 then ΩX/Y is coherent.

13 Proof. Pick x in X, let y = f(x), take an open aﬃne neighborhood Spec B of y, and choose an open Spec A of X which lies in the preimage of Spec B and contains x. This gives a commutative diagram

open Spec A / X

open Spec B / Y mapping the inclusion Spec A → X into the ﬁber product gives a commutative diagram

∆A/B Spec A / Spec A ×Spec B Spec A

open

∆X/Y X / X ×Y X

Let J = ker(A ⊗B A → A). Our goal is to show that

1 ∼ 2 ∼ ΩX/Y = (J/J ) , Spec A

1 which will prove simultaneously that ΩX/Y is quasicoherent, and that it agrees with our old deﬁnition in the aﬃne case. To do so we make the following computation 1 ∼ −1 ΩX/Y = ∆X/Y J ⊗∆−1O OX (1) Spec A X×X Spec A −1 =∼ ∆ J ⊗ OX (2) (∆−1O ) Spec A X×X Spec A Spec A

∼ −1 ∼ = ker(∆ OX×Y X → OX ) ⊗∆−1(A⊗ A)∼ A (3) Spec A B −1 ∼ ∼ ∼ ∼ ∆ ker (A ⊗ A) → A ⊗ −1 ∼ A (4) = A/B B ∆ (A⊗B A) −1 ∼ ∼ ∼ ∆ J ⊗ −1 ∼ A (5) = ∆ (A⊗B A) =∼ ∆∗(J ∼) (6) ∼ ∼ = J ⊗A⊗B A A (7) ∼ =∼ J/J 2 (8)

The justifications for the steps are as follows: 1 (1) Definition of ΩX/Y (and definition of pullback sheaf) (2) Tensor product commutes with restriction (3) Definition of J (4) Left-exactness of ∆−1 and commutativity of the square immediately above this computation, and for the subscript on the ⊗, the fact that ∆−1 commutes with restriction

14 ∼ ∼ (5) Most importantly, the fact that ker (A⊗B A) → A is quasicoherent, which follows from the fact that in the affine case, the diagonal is a closed immersion (note that J , on the other hand, may not be quasicoherent, precisely because in general, the diagonal map need not be a closed immersion) Thanks to Chang-Yeon for pointing this out to me (6) Definition of pullback (7) Pullback corresponds to tensor product over affines (8) An observation made at the very beginning of this lecture This proves part 1 of the theorm. Part 2 was an exercise on HW 1.

Note: At this point Prof. Olsson made some remark about “the key fact here is a certain property of the diagonal map ...” I didn’t write this down, but if anyone remembers what he said here, please let me know. Corollary 3.1. If B → A is a ring map and f is an element of A, then the unique map 1 1 ΩA/B ⊗A Af → ΩA0/B is an isomorphism.

Proof. This follows immediately from quasicoherence, but here’s an al- ternate proof. Check that the localization of d: A → ΩA/B gives an

Af -linear derivation df : Af → ΩAf /B . Precomposing df with the localization map A → Af gives a derivation A → ΩAf /B so by the universal property of ΩAf /B there is a unique A-linear map ΩA/B → ΩAf /B which commutes with the localiztion map, d, and df . Now one checks that this map ΩA/B → ΩAf /B satiﬁes the universal property of the localization ΩA/B → ΩA/B ⊗ Af . This also was assigned on HW 1, so I will not ﬁll in the details.

3.3 Relation with Infinitesimal Thickenings We motivated our discussion of differentials with talk of mapping infinitesimal thickenings into a morphism X → Y . Having now made our explicit definition of the cotangent sheaf, it’s time we related it back to that initial discussion. The following theorem says loosely that if there is a way to do this at all, the number of ways to do so is controlled by the cotangent sheaf. First fix some notation. Let T0 → T be a closed immersion defined by a square-zero quasicoherent (follows from closed immersion) ideal sheaf J . Although J is a sheaf on T , the fact that J 2 = 0 means the action of OT on J factors through that of OT0 , so in fact we may regard J as a sheaf on T0 (recall that their topologial spaces are the same). Let x0 be a point of X, and let S be the set of morphisms filling in the following diagram x0 / T0 > X

i f y T / Y

15 Theorem 3.1. Either S = ∅ or else there is a simply transitive action of ∗ 1 HomO (x Ω , J ) on S. T0 0 X/Y Intuitively, this says that if we look at a point x in our scheme X, although there is no preferred choice of tangent direction at X, once one is chosen, we can obtain all the other by a rotation. In other words, the difference of two tangent directions is given by a vector in the cotangent space. We illustrate the theorem with a concrete example. Example 3.1. Let k be an algebraically closed field and X/k a finite-type k-scheme, with X regular, connected, and of dimension 1. Thus if x is a closed point, the local ring OX,x is a DVR. 1 Proposition 3.2. In the situation above, ΩX/k is locally free of rank 1.

1 Proof. We know that ΩX/Y is coherent by part 2 of Proposition 2. We 1 ﬁrst show that for any closed point x of X, the k-vector space ΩX/k(x) = 1 1 ΩX/k,x mxΩX/k,x (equality since k is algebraically closed) is one-dimensional. 1 For this we calculate the dimension of the dual vector space Homk(ΩX/Y (x), k), which by the pevious theorem is equal to the set of dotted arrows

x Spec k /8 X

Spec k[]/2 / Spec k

2 and we calculated in Example 1 of Lecture 1 that this is equal to Homk(mx/mx, k), which is one-dimensional since OX,x is a DVR. Now, applying “geometric 1 Nakayama” (see Vakil 14.7D), we can lift a generator of ΩX/k to some 1 aﬃne neighborhood of x. Thus ΩX,k is locally free of rank 1.

4 Wednesday, Jan. 25th: Exact Sequences In- volving the Cotangent Sheaf

Before proving two sequences which give a further geometric perspective 1 on all we’ve done so far, it would be nice if the construction of ΩX/Y were functorial in some appropriate sense. Since this sheaf is associated to a morphism X → Y , we should consider a “morphism of morphisms”, i.e., a commutative diagram.

1 4.1 Functoriality of ΩX/Y Theorem 4.1. Given a commutative diagram

g X0 / X

Y 0 / Y

∗ 1 1 0 there is an associated morphism g ΩX/Y → ΩX0/Y 0 of sheaves on X .

16 Proof. In the following diagram, the two outer “trapezoids” are the commutative square given to us, and the upper left square is the usual ﬁber square 0 0 0 X ×Y 0 X / X / X

X0 / Y 0 AA AA AA AA X / Y 0 0 We get a map g × g : X ×Y 0 X → X ×Y X by the universal property of X ×Y X, which also commutes with the relevant diagonal maps, giving us the square

∆X0/Y 0 0 0 0 X / X ×Y 0 X

g g×g X / X ×Y X ∆X/Y From the deﬁnition of J we have a sequence

JX/Y → OX×Y X → (∆X/Y )∗OX

0 0 of sheaves on X ×Y X, and an analogous sequence of sheaves on X ×Y 0 X . Applying (g × g)∗ to the ﬁrst sequence gives

∗ ∗ ∗ (g × g) J → (g × g) O 0 0 → (g × g) (∆ ) O X/Y X ×Y 0 X X/Y ∗ X

0 0 on X ×Y 0 X . Now commutativity of the square above shows that (g×g)◦ ∗ ∗ ∆X0/Y 0 = ∆X/Y ◦ g, so (g × g) (∆X/Y )∗OX = (∆X0/Y 0 )∗ g OX . But the pullback of the structure sheaf along any morphism is always the structure ∗ ∗ sheaf, so g O ∼ O 0 , and also (g × g) O ∼ O 0 0 . This gives X = X X×Y X = X ×Y 0 X 0 0 us a diagram of sheaves on X ×Y 0 X :

∗ ∗ O 0 0 (g × g) JX/Y / X ×Y 0 X / (g × g) (∆X/Y )∗OX

∼= 0 0 O 0 0 0 0 0 JX /Y / X ×Y 0 X / (∆X /Y )∗OX

Even after applying (g×g)∗, the composite along the top is still zero, hence ∗ ∼= the map (g × g) J → O 0 0 → O 0 0 → (∆ 0 0 ) O 0 is X/Y X ×Y 0 X X ×Y 0 X X /Y ∗ X ∗ zero, so it factors through JX0/Y 0 , giving us a map (g × g) JX/Y → ∗ JX0/Y 0 . Now if we apply ∆X0/Y 0 to this morphism we get a map

∗ 1 ∗ ∗ ∗ ∗ ∗ 1 g ΩX/Y = g ∆X/Y JX/Y = ∆X0/Y 0 (g×g) JX/Y → ∆X0/Y 0 JX0/Y 0 = ΩX0/Y 0 which is what we were after.

17 In the aﬃne case, this construction goes as follows. We are given a diagram Spec A0 / Spec A

Spec B0 / Spec B which yields maps of rings

0 0 0 0 / A ⊗ 0 A / IO BO AO α β

I / A ⊗B A / A

Here α is just the restriction of β to I; commutativity of the square on the right ensures that it factors through I0. 0 0 0 0 0 0 First we form the A ⊗B A -module I ⊗A⊗B A A ⊗B A , which is ∗ the analogue of the pullback (g × g) J . Now we have an A ⊗B A-linear map α: I → I0, and by the change of base adjunction

0 0 0 0 Hom 0 0 I ⊗ A ⊗ 0 A ,I ∼ Hom (I,I ) A ⊗B0 A A⊗B A B = A⊗B A

0 0 0 0 0 0 0 this gives a map I ⊗A⊗B A A ⊗B A → I of A ⊗B A -modules, which ∗ corresponds to the map (g × g) JX/Y → JX0/Y 0 in the above. Finally we 0 apply the functor − ⊗ 0 0 A to this map, which gives a map A ⊗B0 A

0 0 0 0 0 I ⊗ A ⊗ 0 A ⊗ 0 0 A → I ⊗ 0 0 A A⊗B A B A ⊗B0 A A ⊗B0 A

0 0 1 But I ⊗ 0 0 A = Ω 0 0 by deﬁnition, and by “changing base all over A ⊗B0 A A /B the place,”

0 0 0 0 0 I ⊗ A ⊗ 0 A ⊗ 0 0 A = I ⊗ A = I ⊗ A ⊗ A A⊗B A B A ⊗B0 A A⊗B A A⊗B A A

1 0 and this is ΩA/B ⊗A A , so we have produced the desired map. If you work through the construction, you ﬁnd that this map is just da ⊗ a0 7→ a0 · d(g(a)), where g is the given map A → A0. In particular, this map is surjective if g is.

4.2 First Exact Sequence The following construction is what Vakil calls the “relative cotangent exact sequence” (Thm 23.2.24). f g Theorem 4.2. Given morphisms of schemes X → Y → Z, there is an exact sequence of sheaves on X:

∗ 1 1 1 f ΩX/Y → ΩX/Z → ΩY /Z

18 Proof. The result essentially follows from the following diagram

∆X/Z

∆X/Y $ X / X ×Y X / X ×Z X H HH HH HH f HH H# ∆Y /Z Y / Y ×Z Y in which the square is cartesian - it is a magic square (see Vakil 2.3R and take X1 = X2 = X). Let’s see what this says in the aﬃne case. Then the given morphisms correspond to ring maps A ← B ← C, and the diagram above looks like

A o A ⊗B A o o A ⊗C A Gc G O O GG GG GG GG B o B ⊗C B

We have a right exact sequence

ker(B ⊗C B → B) → B ⊗C B → B → 0 to which we apply the right exact functor − ⊗B⊗C B A ⊗C A, giving us A⊗C A ⊗B⊗C B ker(B⊗C B → B) → A⊗C A → A⊗C A ⊗B⊗C B B → 0 Here we have changed base at the middle term: A ⊗C A ⊗B⊗ B B ⊗C C B =∼ A ⊗C A. This is where the magic square comes in: it says that A ⊗B A, which sits at the top left of that square, satisﬁes the universal property of the tensor product of B and A ⊗C A over the base B ⊗C B, so our sequence becomes A ⊗C A ⊗B⊗C B ker(B ⊗C B → B) → A ⊗C A → A ⊗B A → 0 Now the term on the left is already what we want; the other two terms sit in their own sequences

AAO O

A ⊗ A ⊗ ker(B ⊗ B → B) / A ⊗ A / A ⊗ A / C OC OB 0

ker A ⊗C A → A ker A ⊗B A → A

There is an induced map ker A ⊗C A → A → ker A ⊗B A → A . We would like to also ﬁnd a map A ⊗C A ⊗ ker(B ⊗ B → B) → ker A ⊗C A → A . For this we have to show that the composite A ⊗C A ⊗ ker(B ⊗ B → B) → A ⊗C A → A

19 is zero. But this is clear if you remember that we just changed the base on the middle term. So the map on the left is just Id ⊗i, where i: ker(B ⊗ B → B) → B ⊗ B is the inclusion. Then we follow this with multiplcation, mapping us into A. But the magic square above show that anything in the kernel of B ⊗ B → B maps to 0 in A. So the leftmost map factors through the kernel of the center column, and we obtain our sequence A⊗C A ⊗B⊗C B ker(B⊗C B → B) → ker A⊗C A → A → ker A⊗B A → A → 0 which by deﬁnition is

∗ 1 1 1 f ΩC/B → ΩC/A → ΩB/A → 0

4.3 Second Exact Sequence The next Theorem describes how differentials behave under a closed immersion. Of course, since the construction is relative, we should look at a closed immersion X → Y over some base scheme Z. Theorem 4.3. Let i X / Y @@ @@ @@ @@ Z be a diagram of schemes, where i is a closed immersion defined by the ideal d 1 sheaf I ⊂ OY . Then the map I → OY → ΩY /Z induces an OX -linear ∗ ∗ 1 map i: i I → i ΩY /Z which fits into an exact sequence

∗ d ∗ 1 1 i I → i ΩY /Z → ΩX/Z → 0

Proof. (This is the corrected version, actually given in the following lecture) We’ll prove it only in the aﬃne case. In this setting, we have a diagram

Spec A/I i / Spec A KKK KK f KKK KK% Spec B and we wish to exhibit an exact sequence

2 d 1 1 I/I → ΩA/B ⊗A A/I → ΩA/I/B → 0 We claim that the sequence can be found buried in the belly of this beast:

20 I ⊗A⊗B A A ker(A ⊗B A → A) VVVV a⊗f7→a⊗f−f⊗a VVVV VVVV+ I ⊗ A ⊕ A ⊗ I / ker(A ⊗B A → A/I) / / ker(A/I ⊗B A/I → A/I) UUUU UUUU Σ UUUU UU* I 0

Here Σ is the map deﬁned by a ⊗ f + a0 ⊗ f 0 7→ af + a0f 0, extended by linearity. I haven’t worked through this yet. I’ll come back to it. In case I forget, the argument in Matsumura is nice and short. Read that instead.

Example 4.1 (Jacobian Matrix Calculation). Let B → A be a ﬁnite type 1 ring map, with A and B Noetherian. We compute ΩA/B as follows. First write A = B[x1, . . . , xn]/(f1, . . . , fs) Then we have morphisms

n Spec A / AB

Spec B

We have computed already that Ω1 =∼ B dx ⊕ · · · ⊕ B dx , and B[xi]/B = 1 n pulling this back to Spec A is the same as tensoring up with A, giving

1 ∼ ΩB[xi]/B = A dx1 ⊕ · · · ⊕ A dxn A

1 Now by the previous proposition, this surjects onto ΩA/B with kernel 2 I/I , where I = (f1, . . . , fs). Furthermore, there is an obvious surjection s 2 s A → I/I , sending the i-th basis element of A to fi. Since precomposing with a surjection does not change the cokernel, the sequence

s 1 A → A dx1 ⊕ · · · ⊕ A dxn → ΩA/B

1 s expresses ΩA/B as the cokernel of the map A → A dx1 ⊕ · · · ⊕ A dxn. But consider the image of a basis element ei under this map. It maps 2 2 ﬁrst to fi mod I . Recall that the map I/I sends the class of f to P ∂f s df = dxi. Thus in terms of the bases {ei} for A and { dxj } for ∂xi ∂fi A dx1 ⊕ · · · ⊕ A dxn, this map is represented by the matrix , the ∂xj i,j 1 familiar Jacobian matrix. This is the key to computing ΩA/B in many applications.

21 5 Friday, Jan. 27th: Differentials on Pn Remark. The first problem on HW 1 should have convinced you that at least in algebraic geometry, inseparable field extensions are not patholo- gies, despite what your Galois Theory course may have led you to believe. One common situation in which they arise is the following. Say we have a finite type k-scheme X, where k is algebraically closed, and a morphism X → Y whose fibers are curves. Then one is led to consider the map of fields k(x) → k(y), where k(y) is typically not a perfect field. It may be, for example, of the form k(t1, . . . , tn), such as appeared on the assignment.

5.1 Preliminaries

n Anyway, today will will compute the cotangent sheaf of PS . There are many more concrete proofs (e.g., in Hartshorne), but we will take the n functorial approach. Recall that PS represents the functor op Schemes/S → Set

n+1 sending T → S to the set of all surjections OT → L, where L is a locally free sheaf of rank one on T , up to isomorphism. In particular, n plugging PS itself in to this functor, we get an isomorphism

n n h n ( ) ∼ F ( ), PS PS = PS and the image of Id n under this isomorphism is the surjection PS

n+1 O n → O n (1) PS PS where O n (−1) is the so-called “twisting sheaf”. A HW problem from last PS n term computed the abelian group structure of Pic(PS ) under tensor product. Applying − ⊗O n O n (−1) (which is right exact) to this surjection PS PS gives another surjection

n+1 O n (−1) → O n PS PS n The following theorem asserts that the cotangent sheaf of PS is the kernel of this surjection. 1 Theorem 5.1. Ω n is locally free of rank n, and there is an exact PS /S sequence 1 n+1 0 → Ω n → O n (−1) → O n → 0 PS /S PS PS

5.2 A Concrete Example To warm up, let’s take S = Spec k, with k algebraically closed. We have seen that cotangent vectors can be regarded as dotted maps ﬁlling in the following diagram / n Spec k 8 Pk

Spec k[]/2 / Spec k

22 Since k is algebraically closed, the map along the top is given by choos- n ing a (closed) point [a0 : ... : an] of Pk . 2 n To give a map from k[]/ into Pk is to give a point [a0 + b0 : ... : an + bn] which agrees with the ﬁrst point modulo (since the vertical map on the left is reduction mod . So since the ais have already been chosen, ﬁlling in the diagram entail choosing the bis. But we could rescale our coordinates by an element of k[]/2 and possibly still get the same map. Since (ai + bi)(x + y) = aix + (bix + aiy), we are forced to take x = 1 in order to get the diagram to commute. So the map is determined by choosing the bis, and the ambiguity in doing so can be expressed as the corkernel of the map sending y 7→ (a0y, . . . , any). Thus we conclude, loosely speaking, that the (co)tangent space is the cokernel of a map k → kn+1. To relate this to the theorem, take the dual of the sequence which appears there:

n+1 1 ∨ 0 → O n → O n (1) → Ω n → 0 PS PS PS n where the ﬁnal term is by deﬁnition the tangent sheaf of PS over S. This is the coordinate-free expression of our computation above. Compare this with the topological desription of complex projective ∗ space. There we have a C -bundle

n+1 C \{0}

f n CP

n+1 n+1 and if p is a nonzero vector in C , the tangent space to C at p is just n+1 n n C , whereas the tangent space of CP at f(p) is isomorphic to C , and n+1 we think of this as being the tangent space of C module the tangent space of the ﬁber over f(p), which is just a line:

0 → T −1 (p) → T n+1 (p) → T n (f(p)) → 0 f f(p) C \{0} CP

n+1 n 0 → C → C → C → 0

1 5.3 Functorial Characterization of ΩX/Y

1 Before we tackle the proof, let’s ask what functor ΩX/Y represents. We’ve basically already done this. Given X → Y , a morphism of schemes, deﬁne

TX/Y : QCoh(X) → Set

1 by sending F to HomOX (ΩX/Y , F). Given a quasicoherent sheaf F, we saw the other day how to construct the following diagram

Id / X < X z s0 z zz zz zz X[F] / Y

23 Recall that, by analogy with the usual dual numbers, the scheme X[F] comes equipped with morphisms of sheaves OX → OX ⊕ F → OX . In particular, there is a distinguished arrow s0 ﬁlling in the diagram, though of course there may be others. Since we have a distinguished choice of arrow, Theorem 3.1 says that the set of all such arrows is isomorphic 1 to the set of morphisms ΩX/Y , F). In other words, the content of that 1 theorem is really that ΩX/Y represents this functor TX/Y .

6 Monday, January 30th: Proof of the Computa- 1 tion of Ω n PS /S Today we aim to prove Theorem 5.1. We will do this in three steps, the second of which is the most involved. Remember that we have to show 1 that Ω n fits into an exact sequence PS /S 1 n+1 0 → Ω n → O n (−1) → O n/S → 0. PS /S PS PS Following the discussion of last time, all that remains is to check that 1 Ω n is the kernel of the map on the right. Let K be this kernel. The PS /S following is an outline of the argument: n 1. For any quasicoherent sheaf F on PS , n+1 coker Hom(O n , F) → Hom(O n (−1) , F) =∼ Hom(K, F) PS /S PS /S n 2. For any quasicoherent sheaf F on PS , n+1 ∼ 1 coker Hom(O n/S , F) → Hom(O n/S (−1) , F) = Hom(Ω n , F) PS PS PS /S 1 1 3. By 1 and 2, Hom(K, F) =∼ Hom(Ω n , F), and therefore K =∼ Ω n PS /S PS /S Step 3 is basically just an application of Yoneda: once we’ve shown 1 that Hom(K, F) =∼ Hom(Ω n , F), then taking global sections gives PS /S ∼ 1 1 ∼ HomO n K, F = HomO n Ω n , F , which implies Ω n = K, fin- P P PS /S PS /S ishing the proof. Now we prove the first two steps.

Proof of Step 1. We already have an exact sequence n+1 0 → K → O n (−1) → O n → 0 PS PS /S We claim that application of the functor Hom gives another exact n sequence of sheaves on PS : n+1 0 → Hom(O n , F) → Hom(O n (−1) , F) → Hom(K, F) → 0 PS PS The exactness follows from the following lemma Lemma 6.1. Let 0 → K → E → E 0 → 0 be an exact sequence of quasicoherent sheaves on X, where E 0 is locally free of ﬁnite rank. Then for all F in QCoh(X), the sequence of sheaves on X 0 → Hom(E 0, F) → Hom(E, F) → Hom(K, F) → 0 is exact in the category of OX -modules.

24 Proof of Lemma. Exactness can be checked on stalks. Since E 0 is locally free of rank r for some r, every point of X has a neighborhood U on which E 0 ∼ Or . Therefore replacing X by U we may assume E 0 ∼ Or , U = U = X 0 generated by basis elements ei in Γ(X, E ), i = 1, . . . , r. Then deﬁne a section by lifting each ei to one of its preimages in Γ(X, E), giving a splitting E =∼ K ⊕ E 0 Then Hom(E, F) =∼ Hom(K ⊕ E 0, F) =∼ Hom(K, F) ⊕ Hom(E 0, F) which implies the exactness of the sequence in the statement of the lemma.

n+1 0 Now, taking E = O n (−1) and E = O n (which is locally free of PS PS rank one), we have proved step 1.

Step 2 is the heart of the proof, and we will not ﬁnish it today. Before we begin, here are a few facts which we will be useful to bear in mind. n Our approach involves the fact that PS represents the functor op F : Schemes/S → Set which sends an S-scheme T → S to the set of surjections of OT -modules n+1 OT → L where L is a line bundle on T , up to some equivalence relation that we won’t worry about here. But recall the way in which this F is actually a (contravariant) functor: given a morphism

g T 0 / T @@ @@ @@ @ S

n+1 π of S-schemes, the induced morphism F (f) sends the surjection OT → L ∗ n+1 g π ∗ 0 in F (T ) to the surjection OT 0 −→ g L in F (T ). Why is g∗π still a surjection? Because g∗ is right exact, being a left ∗ adjoint to the pushforward g∗. Why is g L a line bundle? Because pullback sends vector bundles to vector bundles: this can be checked locally, and if T 0 = Spec A, T = Spec B, and L =∼ (B∼)n, we have ∗ ∼ ∼ n ∼ ∼ n ∼ ∼ ∼ n ∼ g L = (B ) ⊗(B∼) A = (B ⊗B A) = (A ) as B -modules. Now we beign the proof.

n Proof of Step 2. Let U be any open of PS , and F any quasicoherent sheaf n 1 1 on . Then Hom Ω n , F (U) = HomO Ω n , F , and by Theo- PS PS U PS U U rem 3.1, this set is equal to the set of dotted maps ﬁlling in the following diagram: π U / n = PS

U[F] / S

25 Notice that in applying the theorem, we have used the fact that this set is non-empty: we have a distinguished “retract” r : U[F] → U, which when n composed with the inclusion U → PS , ﬁlls in the diagram. Now we consider such diagrams from the point of view of the functor F n n which is represented by PS . As noted above, a map T → PS corresponds n+1 to a surjection of OT -modules OT → L, with L a line bundle on T . So n n+1 the inclusion U → PS corresponds to a section OU LU , and since the n n inclusion commutes with the identity map PS → PS , this line bundle LU is actually just O n (1) , but we’ll continue to denote it by LU for now. PS U We also have our distinguished dotted arrow, the retract r (followed n n+1 ∼ by inclusion into PS ), which corresponds to a surjection OU[F] → L . ∼ ∗ Notice that OU[F] = r OU since pullback sends the structure sheaf to the ∗ n+1 ∼ structure sheaf. So our second surjection really looks like r OU → L . But the fact that the square above commutes imposes some constraints on these two surjections. Namely, in light of our review of the functoriality of F , we must have a commuting square

∗ ∗ n+1 r π ∗ r OU / / r LU

n+1 OU / / LU

Now we are asking whether there are other maps which fill in the square of schemes above, or equivalently, fill in this square of sheaves, and the functoriality therefore requires us to look for not just any surjection r∗On+1 → ∼ ∗ L , but specifically a surjection onto r LU . So what can be said about ∗ this r LU ? Firstly, it fits into an exact sequence of OU -modules

∗ 0 → LU ⊗ F → r LU → LU → 0

∗ The reason is that r LU can be rewritten as follows:

r∗L = O ⊗ L = L ⊗ (O ⊕ F ) = L ⊕ (L ⊗ F), U U[F] OU U U OU U U U U OU and this ﬁts into the (split) exact sequence above. But the map on the right is just the right hand vertical arrow in our square of sheaves two paragraphs up, so we can ﬁt the two diagrams together, giving

∗ 0 / LU ⊗ F / r LU / LU / 0 O O r∗π

n+1 n+1 ∼ ∗ n+1 n+1 OU ⊕ F = r OU / OU

∗ All the maps here are OU -linear, and in addition, r π is OU[F]-linear. We ask: what other OU[F]-linear maps are there that fill in this diagram? Observe first that by commutativity of the square any two such must ∗ agree when composed with the map r LU → LU killing F. Thus by the exactness of the horizontal sequence their difference factors through n+1 LU ⊗OU F, i.e., is a map OU[F] → LU ⊗F. So we can produce the desired

26 ∗ n+1 maps by adding to r π an OU[F]-linear map φ: OU[F] → LU ⊗ F. This associates to each element of

n+1 HomOU[F] OU[F], LU ⊗OU F a dotted arrow ﬁlling in our diagram. But by changing base to the category of OU -modules, this Hom set is in natural bijection with

n+1 HomOU OU , LU ⊗OU F , where LU ⊗OU F is regarded simply as an OU -module. Thus given any n+1 ∗ element of this set, we have produced a surjection OU[F] → r LU of sheaves on U[F]. Now recall that the line bundle LU is actually OU (1), and that maps 1 ﬁlling in the diagram correspond to elements of HomO Ω n , F , so U PS U we have produced a map

n+1 1 HomO O , OU (1) ⊗O F → HomO Ω n , F U U U U PS U But with a little calculation, we can rewrite the domain as follows:

n+1 HomOU OU , OU (1) ⊗OU F ∼ n+1 = HomOU OU (−1) ⊗OU OU (1), OU (1) ⊗OU F ∼ n+1 = HomOU OU (−1) , HomOU OU (1), OU (1) ⊗OU F ∼ n+1 = HomOU OU (−1) , F

The ﬁnal isomorphism being because OU (1) is locally isomorphic to OU . For more detail, see the lemma below. Summarizing, we now have a map

n+1 1 HomO OU (−1) , F → HomO Ω n , F U U PS U n for every open U ⊆ PS , which is compatible with restriction maps by construction. Thus it determines a morphism of sheaf-Homs

n+1 1 HomO n O n (−1) , F → HomO n Ω n , F P P P PS The rest of the proof of Part 2 will be ﬁnished in the next lecture.

Here’s a lemma which was referred to above, and which gets used in the next lecture also, but was neither explicitly stated nor proved in lecture. Lemma 6.2. Let L be a line bundle on X, and F a quasicoherent sheaf on X. Then there is an isomorphism of sheaves

∼ σ : Hom(L, L ⊗OX F) → F Proof. Choose a covering of X on which L is trivial and F is the sheaf associated to some module, i.e., consisting of open aﬃnes U = Spec A, on which we have isomorphisms

L ∼ O ∼ A and F ∼ F, U = U = e U = e

27 for some A-module F . Then we have, by the equivalence of categories QCoh(U) =∼ ModA, ∼ Hom(L, L ⊗OX F)(U) = HomA(A, A ⊗A F )

using which σ is given explicitly by the following: if φ ∈ HomA(A, A⊗AF ), and φ(1) = 1 ⊗ f (we can always write it like this since the tensor product is taken over A), we define σ(φ) = f. We need to check that this definition of σ agrees on the intersection W of two affines U and V . But in this case, we may restrict further if necessary to ensure that (F ) ∼ (F ) . U W = V W Moreover, the isomorphisms L ∼ O and L ∼ O , when restricted U = U V = V to W , differ by multiplication by a unit in OW . Since this unit appears in both sides of the equation defining σ (specifically, we multiply the 1 in both sides by this unit), it may be cancelled, so that the equations defining σ on U and on V are the same when restricted down to W . Thus the definition of σ on open affines glues to give a morphism of sheaves, which is an isomorphism of sheaves since it is so locally: HomA(A, A ⊗A F ) =∼ HomA(A, F ) =∼ F .

7 Wednesday, February 1st: Conclusion of the 1 Computation of Ω n PS /S 7.1 Review of Last Lecture’s Construction By the end of the last lecture, we claimed to have produced, for each F n in QCoh(PS ), a morphism of sheaves n+1 1 γ : Hom(O n (−1) , F) → Hom(Ω n , F). PS PS /S Let’s ﬁrst clarify the construction of this map γ. Consider an aﬃne n ∼ open Spec R ⊆ PS , with a sheaf F = Fe, for an R-module F . We are interested in dotted arrows / n Spec R : PS

Id Spec R[F ] / S

r Spec R

n n+1 π The inclusion Spec R → PS corresponds to a surjection OR → L, where L is an invertible sheaf on Spec R. 1 We observed that in fact ∗ ∼ L = O(1), and that r L = L ⊕ (F ⊗OR L) as OR-modules. This implies that there is an exact sequence

∗ 0 → F ⊗OR L → r L → L → 0,

1 In the following, we will abbreviate OSpec R by OR, and similarly for R[F ].

28 in which the map r∗L =∼ L ⊕ (F ⊗ L) → L is projection onto the ﬁrst factor; we ﬁt this into the following diagram involving π and a potential lifting πe:

∗ 0 / F ⊗O L / r L / L / 0 R O O π πe

n+1 n+1 OR[F ] / / OR

n+1 If ei are basis elements of OR , then these lift to basis elements of OR[F ] [Reason: elements of F ⊆ R[F ] are nilpotent, hence contained in the nilradical, hence the radical of R[F ]. Thus Nakayama says we can lift generators of R[F ]/F = R to generators of R[F ]]. Abusing notation and denoting these lifts also by ei, the commutativity of the square means that for any πe ﬁlling in the diagram, we must have πe(ei) = (π(ei), αi) for some αi in L ⊗ F . Thus πe is determined by these αi, so the set of maps ﬁlling in the diagram is in bijection with n-tuples of sections of L ⊗ F, which is in turn in bijection with Hom (L−1)n+1, F,according to the following computation:

(L ⊗ F)n+1 =∼ Hom(On+1, L ⊗ F) =∼ Hom (L−1)n+1 ⊗ L, L ⊗ F =∼ Hom (L−1)n+1, F, where we used Lemma 6.2 at the last step. This construction is compatible with restriction, so this construction of πe gives a map of sheaves n+1 n n Hom(OP (−1) , F) → {maps U → P ﬁlling in the diagram above}. −1 ∼ n (We used L = OP (−1) here). Now the sheaf on the right is a priori just n a sheaf of sets, but one can check that there is an OP module structure 1 on it, in such a way that it is isomorphic as sheaves to Hom(Ω n , F). P Now we consider this map

n+1 1 γ : Hom(O n (−1) , F) → Hom(Ω n , F). PS PS /S

7.2 Conclusion of the Proof To ﬁnish the proof of Step 2, it suﬃces to check that γ is surjective and 1 that its kernel is Hom(O n , F). This shows that Hom(Ω n , F) is the P PS /S n+1 cokernel of Hom(O n , F) → Hom(O n (−1) , F) because sheaves of P PS n OP -modules form an abelian category, so every surjection (here, γ) is the cokernel of its kernel. n For surjectivity, we start with an open U ⊆ P , and consider a surjection π0 in the following diagram:

0 π 0 OU[F]n+1 / / L

n+1 OU / / L

29 Remember that our construction began with the “canonical surjection n+1 n+1 ∗ OU → OU (1) and lifted it to a surjection OU[F] → r L. Thus a map π0 as above comes from our construction if and only if there exists an isomorphism ρ: L0 → r∗L over L, i.e., a diagram

∼= L0 / r∗L @ @@ || @ || @@ || @ |} } | L which must furthermore ﬁt into the following:

φ+π0

0 ρ $ n+1 π 0 ∗ OU[F] / / L / r L = L ⊕ (L ⊗ F) qq qqq qqq qqq n+1 qx q OU / / L We exhibit this ρ locally: first find an open affine on which all three of L, L0, and r∗L are trivial. Choose a basis for L, lift it via the surjections L → L and r∗L in such a way that the triangle above commutes (this can always be done, since locally, all three are isomorphic to some ring R). Then our choice of bases for L0 and r∗L determines ρ locally. This proves surjectivity. n+1 n To determine the kernel, we ask: when do two maps φ1, φ2 : OP (−1) → n+1 ∗ F determine the same π : OU[F] → r L?. In other words, for which φ1, φ2 is there an isomorphism σ : r∗L → r∗L filling in the following diagram

r∗L π+φ2 7 K oo KK ooo KK n+1 o K% O σ U[F] 9 L O ss OOO ss π+φ1 OO' ss r∗L

× Any isomorphism σ is given by multiplication by some unit in OU[F]; the requirement that it be an isomorphism over L forces this unit to map to 1 in OU after collapsing F. In other words, such units have the form 1 + x ∈ O , where x ∈ F; in more otherer words, the sections of F U[F] U × × inject into the kernel of the map of sheaves of groups OU[F] → OU . Now for any section x of F , we consider the maps U

n+1 π+φ2 OU[F] / LU ⊕ (L ⊗ F) L LLL LL ·(1+x) π+φ1 LL LL& L ⊕ (L ⊗ F)

30 Commutativity requires that for each basis element ei of OU[F],

(π(ei), φ1(ei)) = (π(ei), π(ei) · x + φ2(ei)).

Thus the two maps agree if φ1(ei) − φ2(ei) = π(ei) · x. So the kernel of γ consists of these sections x of F; that is, ker γ =∼ F =∼ Hom(OU , F). n+1 We regard this latter as a subsheaf of Hom(Ou(−1) , F) by applying n+1 the functor Hom(−, F) to the surjection OU (−1) → OU , giving an n+1 inclusion Hom(O, F) ,→ Hom(OU (−1) , F). Since our choice of U was arbitrary and the constructions are compatible with restriction, the n conclusions hold with U replaced by PS , concluding the proof.

7.3 An Example

1 We’d like to compute Ω 1 for a ﬁeld k. We know it’s a line bundle, and Pk/k 1 we know what the line bundles on P are. But which line bundle is it? Theorem 5.1 gives us the exact sequence

1 2 0 → Ω 1 → O 1 (−1) → O 1 /k → 0. Pk/k Pk Pk Now since O is free, the sequence splits, so the middle term is isomorphic to the direct sum of the outer two. Hence the second exterior power of the middle term is isomorphic to the tensor product of the outer two terms. This means that

2 2 1 1 Λ O(−1) =∼ Ω 1 ⊗ O =∼ Ω 1 , Pk/k Pk/k but Λ2O(−1)2 =∼ O(−2), so we have calculated

1 Ω 1 =∼ O(−2) Pk/k 1 1 1 ∼ More concretely, if we look on the aﬃne open A ⊂ P , we have Ω 1 = 1 P1 Ω ∼ O 1 , generated by, say, dx. Similarly on the other chart ∼ 1 = A A = A 1 D(y): there Ω 1 is generated by dy. How are the two related? At ∞ ∈ P D(y) \ D(x), the local ring O 1 is a DVR with uniformizer x = 1/y, P ,∞ giving dx = −1/y2 dy. Thus the transition map for Ω1 is multiplication by x2 = 1/y2. The transition map for O(1) is multiplication by y. Thus 1 Ω n ∼ O(−2). P =

7.4 Concluding Remarks Before moving on to the study of curves, let’s review the techniques we have available for calculating diﬀerentials on a scheme: 1 1. Try to use the calculation of Ω n P 2. Use the closed immersion exact sequence. 3. Use the relative cotangent exact sequence. These ideas lead naturally into curve theory, where one of our main n 1 tasks will be to understand embeddings of curves into P using Ω .

31 8 Friday, February 3rd: Curves

Now we begin our study of curves by embedding them into projective space. An equivalent approach which we exploit is to study line bundles (and sections thereof) on a curve (especially Ω1).

8.1 Separating Points and Tangent Vectors Let k be an algebraically closed field, and X a proper k-scheme. Given a n+1 n surjection OX → L, when is the corresponding map φ: X → Pk a closed immersion? First we fix some notation. Let V = kn+1, with a fixed choice n+1 ∼ of basis {ei}. We write OX = V ⊗k OX , where in the tensor product we are really tensoring with the constant sheaf associated to V . n+1 Let L be a line bundle on X, and OX → L a surjection. For p a point of X, recall our notation L(p) = Lp/mpLp. For any vector s in V , we obtain an element sp of L(p) by the composite map

v7→v⊗1 V −→ Γ(X,V ⊗k OX ) → Γ(X, L) → Lp → L(p)

Recall also that by the Nullstellensatz, the residue field k(p) at any closed point p of X is isomorphic to k (remember we assume k is algebraically closed). With these observations in place, we answer the question above, viz. when do surjections onto line bundles give embeddings into projective space, by the following. Proposition 8.1. If X/k is proper over an algebraically closed field, and n n+1 π φ: X → P the map associated to the surjection OX → L, then φ is a closed immersion if and only if the following two conditions hold 1. (Separating Points) For every pair of distinct point p, q in X, there is an element s of V such that sp (as defined above) is nonzero in L(p), whilst sq is zero in L(q). 2. (Separating Tangent Vectors) For every closed point p of X, the map

2 ker V → L(p) → mpLp/mpLp

is surjective.

8.2 Motivating the Terminology We’ll prove this in the next lecture. First we motivate the conditions. As a warmup, write V = kx0 ⊕ ... ⊕ kxn, and consider the xi as coordinates n on P . If we take, for example, s = x0, then what is the set of closed points p in X such that s goes to zero in L(p)? n First recall that we can write the element φ(p) ∈ Pk as [a0 : ... : an] as follows. Choose a basis element for L(p), i.e., an isomorphism L(p) →σ k. Then we get a map n+1 σ k = V → Lp → k which sends each basis element xi to some ai in k. Of course, it is not well-deﬁned, since we had to choose a basis for L(p); but any other basis diﬀers from our chosen one by a nonzero element of k, so our coordinates

32 ai are well-deﬁned up to nonzero scalars, giving a usual point of projective space (in the classical sense). Thus we see that the set of closed points of X for which x0 maps to zero is just the preimage under φ of the set n {[0 : a1 : ... : an]} ⊂ Pk . More generally, if s is any linear form in the xi (i.e., an element of V ), then the points of X for which s maps to zero in L(p) are the preimage n of the zero locus of s in Pk . The zero locus of such a linear form is n usually called a hyperplane in Pk . So what condition 1 says is that for any p, q ∈ X, there exists a hyperplane passing through p but not q; succinctly, the map φ is injective on points. Of course, this does not guarantee that φ is a closed immersion. For an example where this fails, let Y be the cusp Spec k[x, y]/(y2 − x3) and X its normalization Spec k[t]. Then there is a morphism φ: X → Y corresponding to the ring map x 7→ t2, y 7→ t3. This is a homeomorphism of topological spaces but not a closed immersion: the tangent space at the origin on Y is “too big”. To see why, look at the map on tangent spaces at the origin. On Y , 2 ∼ 2 ∼ we have mO/mO = k · x ⊕ k · y, and on X, we have mO/mO = k · t. But the induced map between them is zero, since the images of x and y are 2 both in mO ⊂ k[t]. In order for this map φ to be a closed immersion, this map would need to be surjective. ∗ ∗ n Condition two says exactly this: the map TP φ(p) → TX (p) is surjective, or dually, the map on tangent spaces is injective.

8.3 A Preparatory Lemma We’ll need the following in our proof next time. n Lemma 8.1. A map φ: X → Pk is a closed immersion if and only if, for every i = 0, . . . , n,

1. The subscheme Xi = {p ∈ X xi 67→ 0 in L(p)} is aﬃne, and

2. the map k[y0, . . . , yn] → Γ(Xi, OXi ) sending yj to π(xj )/π(xi) is surjective. n+1 We explain the second condition: π is our surjection OX = V ⊗k OX → L, which we think of as determining the homogeneous coordinates n+1 of a point p ∈ X. Restricting to Xi, we get a map π : Γ(Xi, O ) → Xi Xi Γ(X , L ). For each p ∈ X , if we restrict to the stalk at p, we ﬁnd i X i i that π (xi) is nonzero (it’s nonzero in L(p), hence nonzero in Lp by Xi Nakayama). But this holds for any p ∈ Xi, therefore π(xi) must have been a unit in Γ(X , L ) (reason: it’s a unit in every stalk, hence not i X i contained in any prime ideal of Γ(Xi, L )). Xi Now since L is a line bundle, we are guaranteed an open cover of X on which it trivializes. The above remarks show that in fact the Xi form such a cover, with the trivialization

OXi → L Xi given by 1 7→ π(xi), since we just saw that π(xi) generates L over Xi. Therefore each global section π(xj ) of L over Xi (some or all of which may

33 be zero), maps back under this isomorphism to an element π(xj )/π(xi).

Then of course since k[y0, . . . , yn] is free, we may deﬁne a map to Γ(Xi, OXi ) by sending each yj to π(xj )/π(xi). This is the map which appears in the second condition of Lemma 8.1, which we now prove.

Proof of Lemma 8.1. Since the condition of being a closed immersion is n aﬃne-local on the target, the map φ: X → Pk is a closed immersion if and only if, for each i = 0, . . . , n, the restriction

−1 n n φ Pk \ V (xi) → Pk \ V (xi) is a closed immersion. By deﬁnition, this morphism is just

y0 yn Xi → Spec k[ ,..., ]. yi yi Moreover, since the target now is affine, this morphism is a closed immersion if and only if Xi also is affine (a closed immersion must be an affine morphism) and the corresponding map of sheaves, which when Xi is affine y0 yn corresponds to a ring map k[ ,..., ] → Γ(Xi, OX ), is surjective. yi yi i

9 Monday, February 6th: Proof of the Closed Embedding Criteria

9.1 Proof of Proposition 8.1 We need yet another lemma: Lemma 9.1. Let f : A → B be a local homomorphism of local Noetherian rings, such that the induced map on residue ﬁelds A/mA → B/mB is an isomorphism, and the map on cotangent spaces

2 2 mA/mA → mB /mB is surjective. Then f itself is surjective.

The intuition behind the lemma is that if, say, A = k[[x1, . . . , xn]] and B = k[[x1, . . . , xn]]/(f1, . . . , fr), then surjectivity on cotangent spaces means that in each equation fi = 0, we can solve for xi in terms of elements of A.

Proof. The ideal mAB ⊂ B is contained in mB since f is a local homomor- 2 2 phism. Since mA/mA → mB /mB is surjective, the images of generators of 2 2 mA/mA generate mB /mB , and we can lift these by Nakayama to generators of mB . This implies that the images of generators of mA generate mB , so we have mAB = mB . Thus by our ﬁrst assumption, A/mA =∼ B/mAB. In particular we have a surjection A → B/mAB, so the image of 1 generates B/mAB as an A-module. Applying Nakayama again, this time to B as an A-module, we can lift the image of 1 in B/mAB to a generator of B as an A-module, which means f is surjective.

34 Proof of Proposition 8.1. For the “if” direction, first observe that since n n X is proper over k, X → Pk is also proper. Therefore the image Z ⊂ |Pk | of |X| under φ is a closed set. The same holds with |X| replaced by any closed subset of X and Z replaced by φ(V ) (note closed immersions are proper and compositions of proper morphisms are still proper), so φ is a closed map. Next, condition (1) implies that the map φ of topological spaces is injective on closed points. We claim that this implies φ is actually injective. I’ll come back to this at some point. Now a closed map that is also injective is a homeomorphism onto its image. Therefore (check!) φ is quasi-finite. But a proper and quasifinite map is in fact finite (check!) and a finite map is always affine, so Xi = −1 φ (Ui) is affine. It remains to show that if p ∈ X(k), then the map

2 2 m n m n m P ,φ(p)/ P ,φ(p) → X,p/mX,p is surjective. [Recall our notation: X(k) is the set of k-valued points, i.e., maps Spec k → X, which correspond to closed points of X since k is algebraically closed. So we’re saying the above should be surjective for all n closed points p]. This will suﬃce, because the condition that OP → φ∗OX be surjective can be checked locally, since the global sections functor is exact on aﬃnes. For notational convenience, we make a linear change of coordinates so n that φ(p) = [0 : ... : 0 : 1] ∈ P . Then

2 x0 xn−1 m n m n ∼ P ,φ(p)/ P ,φ(p) = k · ⊕ ... ⊕ k · , xn xn

x0 xn−1 the linear forms on the open aﬃne Un = Spec k[ ,..., ]. Since xn xn p ∈ Xn, the global section sn = xn maps to a basis of Lp, i.e., gives a trivialization of Lp (the section sn canonically trivializes L over Xn). So 2 ∼ 2 mX,p/mX,p = mpLp/mpLp, and this isomorphism of OX -modules ﬁts into the following diagram

∼ 2 = x0 xn−1 m n /m n / k · ⊕ ... ⊕ k · P ,φ(p) P ,φ(p) xn xn

∼ 2 = 2 mX,p/mX,p / mpLp/mpLp

The map on the left is the one we want to show is surjective, and the map on the right is the map of condition (2). This proves surjectivity of the map of sheaves locally at closed points. It is left as an exercise to check at the non-closed points (possible hint: use ﬁber products somehow). Note: we hand-waved the other implication (closed immersion implies separates points and tangent vectors) in the statement of the Proposition in class last time. I’ll try to come back and write it up clearly at some point.

35 9.2 Towards Cohomology As before, consider a scheme X, proper over an algebraically closed ﬁeld, of dimension one, normal, and connected. Such a curve has only two types of points: a generic point, and some closed points X(k), at each of which the local ring is a DVR.

Remark. If P is a closed point of X, then IP , the (quasicoherent) ideal sheaf associated to the embedding P → X, is invertible. In general, if L −1 is an invertible sheaf on X, we may write L(P ) = L ⊗OX IP . n+1 Now, when we are given a rank one surjection OX → L, how do we check whether conditions (1) and (2) of Proposition 8.1 hold? First we consider condition (1). Suppose given two distinct (closed) points P and Q of X. We hope to produce a section s ∈ V such that s(P ) = 0 but s(Q) 6= 0. Consider the case when V = Γ(X, L). Then the s we desire is, in particular, an element of the kernel of the map of k-vector spaces

Γ(X, L) = V → LP ⊗OX,P k(P ), and this kernel is just Γ(X, L(−P )). To see this, note that the map in queston is obtained by tensoring the exact sequence 0 → IP → OX → k(P ) → 0 with L, and then taking global sections (which is left-exact on the left, so ∼ respects kernels). Then note that Γ(X, L ⊗OX k(P )) = LP ⊗OX,P k(P ) since k(P ) is a skyscraper sheaf 2. In order to determine whether we can ﬁnd an s such that, also, s(Q) 6= 0, we ask whether the map

Γ(X, L(−P )) → LQ ⊗ k(Q)

is surjective. Here we regard LQ ⊗ k(Q) =∼ LQ/mQLQ simply as a one- dimensional vector space; but we may also regard LQ⊗k(Q) as a skyscraper sheaf at Q, which ﬁts into the following exact sequence of sheaves of OX - modules: 0 → L(−P − Q) → L(−P ) → LQ ⊗ k(Q) → 0 Now the global sections functor is not right-exact, so the map

Γ(X, L(−P )) → Γ(X, LQ ⊗ k(Q)) = LQ ⊗ k(Q)

is not necessarily surjective. But since LQ ⊗ k(Q) is a one-dimensional k-vector space, there are only two possibilities: the map is zero, or else it’s surjective. Condition (1) holds exactly when the map is not zero. Similar considerations apply to condition (2) by taking P = Q in the above. So we are led to the following general problem. Given the short exact sequence

0 → L(−P − Q) → L(−P ) → LQ ⊗ k(Q) → 0,

when is the sequence

0 → Γ(X, L(−P − Q)) → Γ(X, L(−P )) → Γ(X, LQ ⊗ k(Q))

2This means that the sections over any open containing P are all the same, so taking the limit does nothing.

36 obtained by applying the global sections functor also exact? This is the question of cohomology. Once we have deﬁned cohomology, we will see that there is a long exact sequence

0 / Γ(X, L(−P − Q)) / Γ(X, L(−P )) / Γ(X, LQ ⊗ k(Q)) dddddd ddddddd dr dddddd H1(X, L(−P − Q)) / H1(X, L(−P )) / ... and thus the vanishing of H1(X, L(−P − Q)) for all P and Q (not neces- n sarily distinct) is suﬃcient to ensure that the associated map X → P is a closed immersion.

10 Wednesday, Feb. 8th: Cohomology of Coher- ent Sheaves on a Curve

Before proving Riemann-Roch, we introduce, without proof of its existence, the cohomology theory we need, together with its relevant properties. The ﬁrst section is a reproduction of a handout Professor Olsson distributed in class. The second and third contain my notes on his comments about the handout.

10.1 Black Box: Cohomology of Coherent Sheaves on Curves For the results on curves that we will prove in this class, we will assume the existence of a cohomology theory satisfying the following properties. The existence of such a cohomology theory will be a consequence of the general development of cohomology which we will study later. Fix an algebraically closed ﬁeld k. In the following, a curve means a proper normal scheme X/k of dimension 1. If X is a curve and F is a coherent sheaf on X, we write H0(X,F ) for Γ(X,F ). As part of our black box we will include the following result: (0) For every coherent sheaf F on a curve X, H0(X,F ) is ﬁnite dimensional over k. We assume given for every curve X a functor

H1(X, −) : (coherent sheaves on X) → (ﬁnite dimensional k-vector spaces) and for every short exact sequence of coherent sheaves on X

E : 0 → F 0 → F → F 00 → 0 a map 00 1 0 δE : Γ(X,F ) → H (X,F ) such that the following conditions hold. (i) The functor H1(X, −) is k-linear in the following sense. If a ∈ k and F is a coherent sheaf, and if ma : F → F denotes the multiplication by a map on F (which is an endomorphism in the category of coherent sheaves on X) then the induced map

1 1 1 H (ma): H (X,F ) → H (X,F )

37 is multiplication by a on the vector space H1(X,F ). (ii) For every short exact sequence E as above, the sequence

0 / H0(X,F 0) / H0(X,F ) / H0(X,F 00) h δ hhh E hhhh hhhh hhhh ht hhh H1(X,F 0) / H1(X,F ) / H1(X,F 00) / 0 is exact.

(iii) The maps δE is functorial in the exact sequence E in the following sense. If 0 00 0 / F1 / F1 / F1 / 0 (9)

a b c 0 00 0 / F2 / F2 / F2 / 0 is a commutative diagram of coherent sheaves on a curve X with exact rows, then the diagram

δ 0 00 E1 1 0 H (X,F1 ) / H (X,F1)

H0(c) H1(a) δ 0 00 E2 1 0 H (X,F2 ) / H (X,F2) commutes, where we write E1 (resp. E2) for the top (resp. bottom) short exact sequence in (9). (iv) If F is a coherent sheaf on a curve X whose support is a ﬁnite set of points, then H1(X,F ) = 0. 1 1 (v) If X is a curve then there is an isomorphism tr : H (X, ΩX ) → k such that for any locally free sheaf E on X with dual E∨ the pairing

0 1 ∨ 1 1 1 tr H (X,E) × H (X,E ⊗ ΩX ) / H (X, ΩX ) / k is perfect. Remark. Axiom (i) implies that if F and G are coherent sheaves on X and if z : F → G is the zero map then the induced map H1(z): 1 1 H (X,F ) → H (X,G) is also the zero map. Indeed z = m0 ◦ z, where 1 1 1 m0 : G → G is multiplication by 0, and therefore H (z) = H (m0)◦H (z). 1 1 Since H (m0) is multiplication by 0 on H (X,G) by (i) this implies that H1(z) = 0. Remark. Axiom (ii) implies that H1(X, −) commutes with ﬁnite direct sums. Indeed to show this it suﬃces by induction to consider two coherent sheaves F and G. In this case we have a split exact sequence 0 → F → F ⊕ G → G → 0. Since the projection map F ⊕ G → G admits a section, then induced map H0(X,F ⊕ G) → H0(X,G) is surjective, whence (ii) gives an exact sequence 0 → H1(X,F ) → H1(X,F ⊕ G) → H1(X,G) → 0

38 split by the standard inclusion G,→ F ⊕ G. Since finite direct sums are isomorphic to finite direct products in both the category of coherent sheaves on X as well as the category of finite dimensional vector spaces, it follows that H1(X, −) also commutes with finite products. This has many consequences. For example, if F is a coherent sheaves and σ : F ⊕ F → F is the summation map, then the induced map

H1(σ) H1(X,F ) ⊕ H1(X,F ) ' H1(X,F ⊕ F ) / H1(X,F ) is the summation map on the vector space H1(X,F ). Similarly if f, g : F → G are two maps of coherent sheaves, then the two maps

H1(f + g),H1(f) + H1(g): H1(X,F ) → H1(X,G) are equal.

10.2 Further Comments; Euler Characteristic Axiom (v) is known as Serre duality. We explain how to obtain the ﬁrst map, leaving the existence of the trace map tr as part of the axiom. For a perfect pairing, we would like to associate to each e ∈ H0(X,E) an 1 ∨ 1 operator he, −i on H (X,E ⊗ ΩX ), in such a way that this isomorphism 0 1 ∨ 1 identiﬁes H (X,E) with the dual space of H (X,E ⊗ΩX ). First observe 0 that for each e ∈ H (X,E) = Γ(X,E), we get a map OX → E, which we tensor with E∨ to obtain maps of sheaves

∼ e⊗1 ∨ = ∨ ∨ E / OX ⊗OX E / E ⊗ E

/ Ox where the vertical map is just evaluation. The dotted composite we now tensor with Ω1 to give a map E∨ ⊗ Ω1 → Ω1, to which we apply the functor H1(X−). So for each such global section e, we have a map 1 ∨ 1 1 1 H (X,E ⊗ ΩX ) → H (X, Ω ), and this deﬁnes the pairing. Note also that the existence of the trace map in the axiom should be expected if you are familiar with Hodge theory on complex manifolds. We will state Riemann-Roch using Euler characteristic, and then red- erive the more familiar form. The theorem is about calculating the dimensions of the k-vector spaces H0(X, L) for various line bundles L. This can be quite hard, and it turns out to be easier to look instead at the Euler characteristic χ deﬁned by, for F ∈ Coh(X),

0 1 χ(X, F) = dimk H (X, F) − dimk H (X, F)

One useful fact is that χ is additive on short exact sequences. More precisely, we have the following: Lemma 10.1. If the sequence 0 → F 0 → F → F 00 → 0 is exact, then χ(X, F) = χ(X, F 0) + χ(X, F 00).

39 Proof. From the given sequence we get a long exact sequence

0 0 0 q8 LL O qqq LLL& I q I o7 1 3O OO oooo OO' 0 / H0(F 0) / H0(F) / H0(F 00) / H1(F 0) / H1(F) / H1(F 00) / 0 K : KK% tt I t q8 3 LL qqq LLL 0 q & 0

Here the Ik are the kernels and images of the appropriate maps. This gives us four short exact sequences involving the cohomology groups and the Ik. Each short exact sequence relates the dimensions of three of the vector spaces. If you combine the four equations appropriately, you get the result. Note: This diagram diﬀers slightly from the one used in class, though the argument is essentially the same. This more symmetric form of the diagram was pointed out to me by Alex Kruckman.

Remark. Let’s briefly recall the relationship between divisors and line bundles. On a curve X, we can define a divisor as a formal sum of closed P points, D = nP · P . To such an object, we associate a line bundle, denoted by either OX (D) or L(D), as follows. Each closed point P may be regarded as a closed subscheme of X, and thus has an associated ideal sheaf IP [Recall further that a closed point is the same as a map # i: Spec k → X, which induces a map i : OX → i∗OSpec k, where the latter is most usefully regarded as the skyscraper sheaf of k(P ) = k at the # point P . The ideal sheaf IP is defined to be the kernel of i ]. Then we set X O −nP L(D) = L( nP · P ) = IP P where here the superscript means tensor power. Conversely, every invertible sheaf on X is of this form. Suppose given a line bundle L on X. Take an open subset U ⊂ X on which we have a trivialization σ : O →∼ L . Let s = σ(1), a section of L over U. U U The complement of U is then a finite set of points {P1,...,Pr} (possibly empty, if L itself happened to be trivial). Since X is a curve, the local ring at each Pi is a DVR. We want to use the valuations in each of these DVRs to produce the integers nPi . Now we proved last semester that there is an inclusion L ,→ K , where K is the constant sheaf associated to the function field k(X). So we take s, and map it into K (U) =∼ k(X). Then at each Pi we have a valuation νP on k(X), and we define nP = νP (s). i P i i These nPi , for each Pi, thus define a divisor D = nPi · Pi on X. Of course, we still have to check that L =∼ L(D). For this, you can refer to the notes from last semester.

10.3 Statement of Riemann-Roch; Reduction to the More Fa- miliar Version

i i From now on, we will use the notation h (X, L) = dimk H (X, L).

40 Theorem 10.1 (Riemann-Roch Theorem). If X is a curve, and L a line bundle on X, then

χ(X, L) =∼ deg L + χ(X, OX ).

We will prove this next lecture. First notice that

1 1 ∨ 1 ∨ 1 H (X, L) = H X, (L ⊗ ΩX ) ⊗ ΩX ,

0 ∨ 1 1 and by Serre Duality, this is dual to H (X, L ⊗ ΩX ). Hence h (X, L) = 0 ∨ 1 h (X, L ⊗ΩX ). This allows us to rewrite the left hand side of the equation 0 0 ∨ 1 in the theorem as h (X, L) − h (X, L ⊗ ΩX ). Moreover, applying the 1 1 0 1 same trick to H (X, OX ), we get h (X, OX ) = h (X, ΩX ), which is by 0 deﬁnition the genus, g, of X. Finally, we have h (X, OX ) = 1, by either of two arguments presented at the beginning of the next lecture. Thus we can rewrite Riemann-Roch as

0 0 ∨ 1 h (X, L) − h (X, L ⊗ ΩX ) = deg L + 1 − g.

Remark. If H0(X, L) 6= 0, then deg L ≥ 0. For if s is a nonzero global section, then as an element of the fraction ﬁeld, its valuation at each point P is non-negative (i.e., it’s in OX,P ). Loosely, it’s everywhere regular, so has no poles. In fact, you can calculate the degree of L explicitly by looking at the exact sequence

0 → OX → L → L/OX → 0,

where the term on the right is supported on a ﬁnite set of points.

11 Friday, Feb. 10th: Proof and Consequences of Riemann-Roch

11.1 Preparatory Remarks

0 0 Remark. Since H (X, OX ) is a k-algebra, h (X, OX ) is at least one. Here are two diﬀerent reasons why it’s equal to one.

0 1 1 Proof 1. Let f ∈ H (X, OX ). Then f defines a map X → Ak ,→ Pk. 1 1 Since Pk is proper, f(X) is a closed subset of Pk, hence a finite set of 1 points in Ak. Since X is connected, so is f(X), which must therefore 1 3 1 consist of a single point α ∈ Ak. So the map to Ak defined by f agrees 0 with the map defined by α ∈ k ⊂ H (X, OX ). Therefore the global sections f and α differ by nilpotence, and since X is reduced, they are 0 the same global section. Hence H (X, OX ) = k.

3 1 Recall that the map to Ak deﬁned by f sends x to the residue of f in k(x) for each x ∈ X. For closed points, k(x) = k, so in our case, we have that f maps to the same α ∈ k for 1 every closed point, and we think of this α as living in Ak (i.e., as corresponding to the ideal (t − α) ⊂ k[t].

41 0 Proof 2. Since H (X, OX ) is finite-dimensional over k, it’s a finitely generated integral (commutative) algebra over k. This implies it’s a finitely generated field extension of k: to find the inverse to a nonzero α, use the n n−1 monic polynomial cancelling α. That is, if α + bn−1α + ... + b0 = 0, n−1 n−1 then α(α + bn−1α + ... + b1) = −b0, which is nonzero since it’s a domain, so α is a unit. Then since k is algebraically closed, there are no nontrivial finitely 0 generated extensions, hence H (X, OX ) = k.

Remark. Recall from last time that if L is an invertible sheaf on X, and there is a nonzero global section s ∈ H0(X, L), then deg L ≥ 0. Also, if deg L = 0, then L =∼ OX (so s ∈ k). To see this, note that the section s defines a map OX → L. For over each open U, we can define O(U) → L by sending 1 (i.e., the restriction of 1 ∈ H0(X, O )) to s . U X U This obviously commutes with restriction maps, so it defines a morphism of sheaves. Of course, there’s no reason for this to be an isomorphism, but it is injective since s is nonzero over every open U (using that X is integral). We are interested in the cokernel of this map s. To measure the failure of s to be an isomorphism, we begin with an open U on which L is trivial; s itself furnishes a trivialization. The complement of U is a finite point set {P1,...,Pr}. The cokernel is trivial over U; looking at germs near

Pi, LPi is a free rank one OX,Pi -module, and the cokernel of s locally has ni the form LP /m LP . Each ni is non-negative since s is a global section. i Pi i Thus the cokernel of s is the sheaf

O ni coker(s) = LP /m LP i Pi i {P1,...,Pr } where each factor in the tensor product is a skyscraper sheaf at the appropriate Pi. The global sections of this sheaf forms a ﬁnite-dimensional P k-vector space, whose dimension is the degree of L, hence deg L = ni. It follows, then, that if deg L = 0, each ni is zero, and the cokernel is trivial, so s deﬁnes an isomorphism L =∼ OX .

11.2 Proof of the Riemann-Roch Theorem Proof. We will prove the relation

χ(X, L) = deg L + χ(X, OX ) by induction on the degree of L. For deg L = 0, it’s a consequence of the above remarks, since we just saw that in this case, L =∼ OX . For the inductive step, we prove that the equation holds for L if and −1 only if it holds for L(P ) = L ⊗ IP . To see this, tensor the sequence

0 → IP → OX → k(P ) → 0 with L(P ), giving 0 → L → L(P ) → k(P ) → 0

42 (note that the term on the right is a skyscraper sheaf, so tensoring with L(P ) does nothing). Since the Euler characteristic is additive on exact sequences, this yields χ(X, L(P ) = χ(X, L) + 1 (note that h1(X, k(P ) = 0 by one of our axioms for cohomology). Since deg L(P ) = deg L + 1, this proves the equivalence. This proves the theorem, for we have seen above that any line bundle can be obtained from OX by adding or subtracting a ﬁnite number of points

11.3 Consequences

1 Corollary 11.1. deg Ωk = 2g − 2. 1 This comes from setting L = Ωk in the statement of Theorem 10.1. It suggests that the study of curves should divide into three cases: g = 0, 1 g = 1, or g ≥ 2, according as the degree of Ωk is negative, zero, or positive, respectively. Corollary 11.2. If deg L > 2g − 2,

h0(X, L) = deg L + 1 − g

Corollary 11.3. If deg L > 2g −1, then L is generated by global sections. 0 Remark. We could also express this by saying that the map H (X, L)⊗k OX → L is surjective. What map is this? Let f : X → Spec k be the ∗ structure morphism. Since f is a left adjoint to f∗, there is a natural ∗ 0 map f f∗L → L. But f∗ is just H (X, L) sitting over the unique point −1 0 of Spec k. This pulls back to f H (X, L) ⊗ −1 O , and since f f OSpec k X 0 is actually an open map here, and H (X, L) and OSpec k are constant sheaves f −1 just pulls them back to the corresponding constant sheaves −1 0 0 on X, i.e., f H (X, L) ⊗ −1 O ∼ H (X, L) ⊗ O . f OSpec k X = k X

Proof of Corollary 11.3. We prove that the cokernel of the map H0(X, L) → L described in the previous remark is zero. Since the support of the cokernel (and indeed any sheaf) is closed, it’s enough to check that the cokernel is zero at closed points. Thus we have to show that for any closed point P , there is a global section f whose image in L ⊗ k(P ) =∼ LP /mP LP is nonzero (we are using Nakayama here). This is equivalent to ﬁnding a global section f which is in Γ(X, L) but not in the subspace Γ(X, L(−P )). This we can verify simply by comparing dimensions. Our degree assumption ensures that the h1 term in Riemann-Roch vanishes for both sheaves (Corollary 11.2), so we have h0(X, L) = deg L+1−g and h0(X, L(−P )) = deg L−1+1−g = deg L−g

This Γ(X, L(−P )) is a proper subspace of Γ(X, L), so we can ﬁnd the desired global section f.

Remark. (In response to a question) Recall that L(−P ) has stalks equal to L away from P , and equal to mP ⊗ L at P .

43 Of course, it is not really necessary to require that deg L be greater than 2g − 1, for if L has any positive degree, then we can take a high tensor power to obtain a sheaf whose degree is at least 2g − 1. 0 Definition 11.1. A line bundle L on X is very ample if the map π : H (X, L)⊗k n L → L defines a closed immersion into Pk . L is ample if there exists an 0 ⊗n ⊗n n > 0 such that π : H (X, L ) ⊗k L → L defines a closed immersion. We will prove next time that if deg L > 0, then L is ample.

12 Monday, Feb. 13th: Applying Riemann-Roch to the Study of Curves

First we introduce some new terminology. Recall that we have constructed 0 a map H (X, L) ⊗k OX → L of sheaves, for a line budle L on a curve X. We call the set of points at whose stalks this map is not surjective the base locus of L, and if it surjective everywhere (i.e., L is generated by global sections), we say L is base point-free. In this language, we saw last time that if the genus of X is g and deg L > 2g −1, then L is base point-free, in which case we get a morphism 0 of schemes X → PH (X, L).

n 12.1 Using Riemann-Roch to Detect Closed Immersions to P n+1 n Any rank one quotient of OX corresponds to some morphism X → Pk . When L is generated by global sections, we get a rank one quotient of 0 the form H (X, L) ⊗k OX → L, with corresponding morphism X → 0 PH (X, L). When is this a closed immersion? By our earlier criteria, L must separate points and tangent vectors. We now rephrase these conditions in such terms as allow us to apply Riemann-Roch. The condition of separating points is equivalent to requiring that for any two distinct closed poitns P and Q in X, there is a section s in H0(X, L(−P )) \ H0(X, L(−P − Q)). The condition of separating tangent vectors is equivalent to requiring that for any closed point P , there is a section s in H0(X, L(−P )) \ H0(X, L(−2P )). We can guarantee the satisfaction of both conditions at once by requiring that, for any P and Q in X(k)4, not necessarily distinct, there is a section in H0(X, L(−P )) \ H0(X, L(−P − Q)). This can be checked with a simple dimension count, and Riemann-Roch gives the following criterion: Proposition 12.1. If deg L > 2g, then L is very ample.

Proof. For any closed points P,Q (not necessarily distinct), our assumption on the degree of L ensures that h1 is zero for both L(−P ) and L(−P − Q), so Riemann-Roch computes the following:

h0(X, L(−P )) = (deg L − 1) + 1 − g h0(X, L(−P − Q)) = (deg L − 2) + 1 − g,

4Recall that this is the set of k-valued points of X, namely the closed points.

44 This means the latter is a proper subspace of the former, and we can ﬁnd the section we desire, as described in the preceding discussion.

We also have the following: Corollary 12.1. If deg L > 0, L is ample.

12.2 Examples Genus zero Let X be a curve with g = 0. We will show that X is isomorphic to 1 −1 Pk. For any P ∈ X(k), let, as usual, OX (P ) = IP , where IP is the ideal sheaf of P,→ X. It has degree 1 > 2g − 2 = −2, so

0 h (X, OX (P )) = deg L + 1 − g = 2

0 ∼ 2 By choosing an isomorphism H (X, OX (P )) −→ k , we get a closed immersion (by Proposition 12.1, since deg L = 1 > 2g = 0) i: X → 1 Pk. We claim this is actually an isomorphism of schemes. Since it’s a closed immersion, i is a closed map. Therefore its image is a closed 1 set, hence all of P . Since it’s an immersion, it’s injective, and since proper also, it’s a homeomorphism (cf. a related statement in point-set topology - a continuous bijection from a compact space to a Hausdorﬀ space is automatically a homeomorphism.) Now we check the map of sheaves is an isomorphism. Since i is a closed immersion, it induces a surjection of sheaves, and hence a surjection of stalks O 1 → O P ,P X,P (here we identify X with its image, namely P 1, and thus speak of the same P on both sides). The map i is a surjective morphism of schemes, hence in particular a dominant rational map, and so it induces a map of 1 funtion ﬁelds k(P ) → k(X). This map is not the zero map, so it must be injective. Thus we obtain the following diagram

O 1 / / P ,P OX,P

1 k(P ) / k(X)

where the top arrow is surjective and the other three are injective. This forces the top arrow to be an isomorphism, like we wanted. This shows 1 that X is isomorphic to Pk.

Genus one Now let X be a genus one curve5 over k. Since 2g = 2, we need a line bundle of degree at least three to embed in projective space. Pick a point P ∈ X(k) and consider OX (3P ). After choosing a basis for the three- 0 2 dimensional space H (X, OX (3P )), we get a closed immersion i: X → Pk. 2 We will show next time that i identiﬁes X with a cubic curve in Pk, which can be given in characteristics diﬀerent from 2 or 3 by an equation of the form y2 = x3 + Ax + B.

5This is not the same as saying that X is an elliptic curve - these are given by a genus one curve together with a chosen point.

45 Remark. We chose a degree 3 line bundle to satisfy the hypothesis of the theorem, but this is not necessarily the best we can do. For an example, 1 0 1 take X = Pk, and the line bundle OX (2). Then h (Px, O(2)), and we get 1 2 an embedding Pk → Pk. Exercise: ﬁnd the equation of this embedding (hint: it’s the Veronese map).

13 Wednesday, Feb. 15th: “Where is the Equa- tion?”

Today we prove: Theorem 13.1. Any curve of genus one (over k) is isomorphic to a cubic 2 in Pk. Recall that a homogeneous polynomial F of degree d in the variables, n say, x0, . . . , xn, deﬁnes a closed subscheme of Pk , which we denote V (F ). n We ask, given a morphism g : X → Pk corresponding to a rank one quo- n+1 tient OX → L, how can we detect whether g factors through V (F )? To answer this, let the generating global sections from the surjection n+1 n n OX → L be s0, . . . , sn ∈ Γ(X, L). Recall that the identity map P → P n+1 n gives a surjection O n → OP (1), with corresponding global sections P n x0, . . . , xn, which we think of as the homogeneous coordinates on Pk . Now n+1 n if we have a morphism g : O n → OP (1), we can pull back the surjection n+1 n P n+1 ∗ O n → O n (1) over to the surjection O → g O n (1) (recall that P P Pk X P the pullback of the structure sheaf is again the structure sheaf). We will abuse notation slightly and refer to the global sections of this pulled-back surjection also as x0, . . . , xn. The surjections ﬁt into a diagram

(s ,...,s ) n+1 0 n OX / / L PP PPP PPP ∼= (x0,...,xn)PP PPP( ( ∗ n g OP (1) Here the vertical map is an isomorphism because both surjections corre- n spond to the same morphism X → Pk .

Proposition 13.1. The map g factors through V (F ) if and only if F (s0, . . . , sn) is zero in Γ(X, L⊗d).

n Coarse Proof. The expression of a closed point P of P in terms of homogeneous coordinates [x0 : ... : xn] is obtained by taking each of the global n ∼ sections x0, . . . , xn of OP (1) and evaluating them in O(1)P /mP O(1)P = k. A point P is in V (F ) if and only if these evaluations satisfy F = 0, or put another way, if and only if F (x0, . . . , xn), regarded as an element of the sheaf O(1)⊗d, maps to zero at P . This makes sense, since locally, ⊗d O(1) =∼ O(1) via the multiplication map a0 ⊗ ... ⊗ an 7→ a1 ··· an (of n course, they’re both isomorphic to OP locally). So the evaluation of F at P is obtained by taking F , restricting it to some open containing P on which O(1) is trivial, multiplying the sections occuring in F 6, and then

6 It doesn’t make sense to multiply, say, x0x1 over an arbitrary open since O(1) may not be a ring over any open - it only is once we trivialize the line bundle.

46 passing to O(1)P /mP O(1)P . This gives an element of k, which we require to be zero in order for P to be in V (F ). To ask whether a point of X maps into V (F ), note that the image of a point P ∈ X under g is given by evaluating the sections s0, . . . , sn at P . So the discussion in the previous paragraph applies to (the images of) points of X, simply replacing xi by si.

Let’s look at some examples. Example 13.1. Let X have genus zero, and pick a point P ∈ X(k). The 0 0 space H (X, OX ) is a proper subspace of H (X, OX (P )). We’ve already 0 0 computed dim H (X, OX (P )) = 2, so we can write H (X, OX (P )) as the span of two global sections 1 and x. Here 1 spans the constant sections, 0 namely H (X, OX ), while x necessarily has a pole at P , or else it would 0 be in H (X, OX ). We’ve seen last time that a basis of global sections of OX (P ) deter- 1 1 mines a map X → P (an isomorphism, in fact). Identifying X with P via this isomorphism, one thinks of the global sections x and 1 as de- homogenizations of the global sections (i.e., homogeneous coordinates) x 1 and y on P . Example 13.2. Let X have genus one, and ﬁx a point P ∈ X(k). The in- 2 0 vertible sheaf L = O(3P ) deﬁnes an embedding X → Pk, since h (X, L) = 0 3 > 2g = 2. We have a chain of subspaces (writing H (X, OX ) = 0 H (OX ), etc., for short)

H0(O) ⊆ H0(O(P )) ⊆ H0(O(2P )) ⊆ H0(O(3P )) ⊆ H0(O(4P )) ⊆ H0(O(5P )) ⊆ H0(O(6P )) ⊆ ...

Riemann-Roch allows us to compute the dimensions: 1,1,2,3,4,5,6,. . . . Pick a basis 1, x for H0(O(2P )). Then x has a pole of order two at P , or else the previous space would have been two-dimensional. Similarly, we have a basis 1, x, y for H0(O(3P )), where y has a triple pole at P . Then x2 extends this to a basis for H0(O(4P )), and xy extends it further to a basis for H0(O(5P )). But in H0(O(6P )), we have the seven elements 1, x, x2, x3, xy, y, y2, all of which have poles or order at most 6 at P . Since this space is six-dimensional, there must be a dependence relation amongst them, and this relations must involve both x3 and y2, or else one of these two would live in a previous space, but they both have poles of order exactly six. Thus, after rescaling the coordinates if necessary, we obtain a relation of the form

q(x, y) = y2 + bxy + cy − x3 − ex2 − fx − g = 0

2 This implies that X ⊆ Y = V (q(x, y)) ⊆ P . Now one argues that since X is a curve, q(x, y) is irreducible, and Y is one-dimensional, X must be homeomorphic to Y . To check the map X → Y is an isomorphism, it’s

47 enough to look at the stalks. For the stalk at P , we have a diagram

OY,P / / OX,P

∼ k(Y ) = / k(X) where the vertical maps are inclusions, and the bottom arrow is an isomorphism since Y is integral (we mentioned above it’s irreducible. Check it’s reduced). Thus the map on stalks is an isomorphism at P . In general, given X and L, with corresponding embedding X,→ 0 PH (X, L), it possible to express

M ⊗n X =∼ Proj Γ(X, L ). n≥0

We also have n M n Pk = Proj S Γ(X, L), n≥0 where SnV denotes the nth symmetric power. Then the embedding X,→ n n Pk is given by a map of graded rings given in degree n by S Γ(X, L) → Γ(X, L⊗n). The equations defining X come from the kernel of this map of graded rings. Going back to Example 2, the equation q(x, y) = 0 defines an open 2 2 subset U of X inside the affine chart Ax,y of Pk (the point P is in the 2 complement of U). We can regard this inclusion Y ⊂ A as being given by the coordinate functions x and y (this is basically the Spec −Γ adjunction for morphisms to affine targets). But the coordinate function x also defines 2 1 2 a map A → A . Precomposing this with the inclusion U ⊆ A and then 1 1 1 following with the inclusion A → P gives a map U → P . We have seen 1 that such a map extends to a morphism X → P . Now we ask: what is the line bundle and corresponding global sections 1 2 1 associated to this map X → P ? Since we mapped A to A by “forget- 1 ting” the y-coordinate, it is perhaps not surprising that the map to P comes from the space H0(X, O(2P )), which had as basis only 1 and x.

Remark. It is no accident that the degree of the line bundle O 1 (1), P when pulled back to X, is two. This is also the degree of

k(x)[y]/(y2 − (x − α)(x − β)(x − γ)).

1 Note also that, letting f be the map X → P , we have

∗ O eP −1 X deg f O 1 (1) = deg I −1 = deg I = − f (∞) = e P f (∞) P

P f(P )=∞

Here eP denotes the ramiﬁcation index of f at P . We will discuss degree and ramiﬁcation in the next lecture.

48 14 Friday, Feb. 17th: Ramiﬁcation and Degree of a Morphism

Definition 14.1. Let f : X → Y be a nonconstant morphism of curves. The degree of f is the degree of the field extension [k(Y ): k(X)]. If P is a closed point of X, with f(P ) = Q, the ramification index eP of f at P is defined to be the ramification index of the induced map of DVRs OY,Q → OX,P . This isn’t much of a definition until we have defined ramification indices for maps of DVRs. Let πP , πQ uniformizers of OX,P and OY,Q, eP respectively. Then πQ maps to a unit times πP , where eP ≥ 1 since this is a local homomorphism. Then eP is the degree of this map of DVRs. We say f is unramified at P if eP = 1, and ramified at P if eP > 1, in which case we also call Q a branch point of f. The degree of the morphism f is, intuitively, the cardinality of the preimage of any point Q ∈ Y , but we have to be careful at the branch points. Proposition 14.1. For any closed point Q of Y , X eP = deg f. P ∈f −1(Q)

Proof. Let = f∗OX . Then X = Spec . The idea of the proof is to A Y A first prove that A is a locally free sheaf of finite rank on Y . Then we can compute the rank at any point of Y . Over the generic point, we show the rank is [k(X): k(Y )] = deg f, while over a closed point Q, we will show P the rank is P ∈f −1(Q) eP , giving the desired equality. First let’s prove A is locally free of finite rank. Since f is proper and quasifinite, f is finite, so we will have finite rank if we can show A is locally free. For this, it is enough to check at closed points Q ∈ Y (k), since the generic point is a localization of any closed point. First we observe that AQ is a flat OY,Q-module. Flatness is equivalent to the map I ⊗ AQ → A being injective for any ideal I of OY,Q. Since OY,Q is a DVR, it’s enough to check that the action of the uniformizer πQ on AQ −1 is nonzero. But over an open U, A (U) = OX (f (U)), and the action −1 of πQ is given by mapping πQ into OX (f (U)) and multiplying. But −1 OX (f (U)) is a domain, so this action is nonzero. Now the stalks of A (which are stalks of OX ), are finitely generated over the stalks of OY since f is finite. But flat and finitely generated impies free, so A has free stalks. We conclude that A is locally free by the following.

Lemma 14.1. If F is a coherent sheaf on Y such that FQ is a free OQ- module for all Q, then F is locally free.

r 0 0 Proof. At each Q we have an isomorphism OY,Q → FQ. Let f1, . . . , fr ∈ FQ be the image of the generators under this isomorphism, and f1, . . . , fr their representatives on some neighborhood U containing Q. Then these f deﬁne a map of sheaves on U (in fact, of O -modules) σ : Or → F i U U U U which is an isomorphism at the stalk at Q. Let K and G be the kernel and cokernel, respectively, of σU .

49 r Since OU and F are coherent, so are K and G. Thus their supports are closed subsets of U not containing Q. In particular, there is an open neighborhood V ⊆ U containing Q such that KP and GP are both zero for all P ∈ V . Since isomorphisms can be checked at stalks, this means that σ is an isomorphism over V , and hence F is free. U V V Returning to the proof, we have now that A is locally free, and want to compute its rank at both a closed point and the generic point η of Y . First, letting ξ be the generic point of X, we note that

−1 Aη = lim OX (f (U)) = lim OX (V ) = k(X), η∈U ξ∈V the equality in the middle coming from the fact that for any open V of X, we can find U ⊆ Y such that f −1(U) ⊆ V (“preimages of opens in Y are cofinal amongst opens in X”). Namely, if V = X \{Pi}, set U = Y \{f(Pi)}. We use this to compute the rank of A at η: it’s [k(X): k(Y )], i.e., deg f. Now let Q be a closed point of Y . We will compute the rank of A at Q. This is by definition the rank of AQ as an OY,Q-module, which is also equal to the dimension of AQ ⊗OY,Q k(Q) as a k(Q)-vector space. First we claim that ∼ AQ ⊗OY,Q k(Q) = Γ(X ×Y Spec k(Q), OX×Spec k(Q)) To see this, we can replace X and Y by affines Spec R and Spec S, respectively, where Q is contained in Spec S and f(Spec R) = Spec S since f is affine. The fiber product in question corresponds to the diagram

R ⊗ k(Q) o k(Q) S O O

R o S which we can expand into a double ﬁbered diagram

R ⊗ k(Q) o k(Q) S O O

R ⊗ O o O S O Y,Q Y,QO

f # R o S ∼ Thus the claim follows from the observation that in fact AQ = R⊗S OY,Q, since

−1 R ⊗S OY,Q = lim R ⊗S Sg = lim Rf #g = lim OX (f (D(g))), Q∈D(g) Q∈D(g) Q∈D(g) which is (f∗OX )Q = AQ. We must therefore compute the dimension over k(Q) of the global sections of X ×Y k(Q). Since f is ﬁnite, X ×Y Spec k(Q) is a disjoint

50 −1 union of points: f (Q) = {P1,...,Ps}, and therefore it is an aﬃne scheme:

s s ∼ a ∼ Y X×Y Spec k(Q) = Spec OX,Pi ⊗OY,Q k(Q) = Spec OX,Pi ⊗OY,Q k(Q). 1=1 i=1

∼ ei Now, OX,Pi ⊗OY,Q k(Q) = OX,Pi /πi , since the OY,Q-algebra structure ei on OX,Pi is given by sending the uniformizer of OY,Q to πi , where πi is the uniformizer of OX,Pi , and ei the ramiﬁcation index of f at Pi. Thus we conclude that

s ∼ Y ei AQ ⊗OY,Q k(Q) = OX,Pi /πi , i=1

s X which has dimension ei. This is therefore the rank of A at Q, and i=1 equating it with the rank at the generic point (deg f) gives the result.

Example 14.1. If X is a curve of genus 1, and P a closed point, then 0 1 h (X, OX (2P )) = 2, so we get a map f : X → P . On the complement of 1 # P , we have a map to Ak = Spec k[t] given by f : k[t] → Γ(X \ P, OX ). # The section f (t) extends to a global section of Ox(2P ) with a pole of order 1 or 2 at P . Thus deg f ≤ 2. But the degree cannot be one, or else f would be an isomorphism, which is impossible since X has genus 1. So 1 the divisor 2P deﬁnes a degree two map to P .

15 Wednesday, Feb. 22nd: Hyperelliptic Curves

So far in our study of curves, we have seen that genus one curves are 0 isomorphic to P while genus one curves are cut out by cubic equations 2 in P . Today we study curves of genus greater than one. We will see that they split into two camps: those that are “canonically embedded” in some n P , and hyperelliptic curves.

15.1 Canonical Embeddings

1 For a curve X of genus g ≥ 2, we have deg ΩX > 0, so we can ask whether 1 the line bundle ΩX deﬁnes a morphism, or better yet, an embedding, into n 0 1 g−1 P . Since h (X, ΩX ) = g, we will get a morphism to P ; in the case that it’s an embedding, we will call this the canonical embedding. 1 To get a morphism at all, we need ΩX to be base point free, which you recall means that there is no point of X at which all global sections 1 0 1 of ΩX vanish. Since the dimension h (X, ΩX (−P )) is always one less 0 1 than or equal to h (X, ΩX ), it is equivalent to require that that for all 0 1 0 1 P ∈ X(k), h (X, ΩX (−P )) < h (X, ΩX ), i.e., there is some global section not vanishing at P . 1 If furthermore we ask that this morphism be an embedding, ΩX must separate points and tangent vectors, which we have seen earlier is equiva- 0 1 lent to requiring that for all P,Q ∈ X(k), not necessarily distinct h (X, ΩX (−P − 0 1 Q)) < h (X, ΩX (−P )). Thus we obtain easily the following criterion.

51 1 g−1 Lemma 15.1. ΩX deﬁnes a closed embedding X,→ P if and only if, 0 1 for all closed points P,Q of X (not necessarily distinct), h (X, OX (−P − Q)) = g − 2.

0 1 Proof. This follows from the above discussion: the dimension of H (X, OX ) is g. Being base point free is equivalent to requiring that the dimension 1 drop by one when we subtract a point from OX ; separating points and tangent vectors happens exactly when the dimension drops one further after subtraction of another point.

When is the condition of the lemma satisﬁed? We apply Riemann- 1 Roch to the line bundle ΩX (−P − Q):

0 1 0 h (X, ΩX (−P − Q)) − h (X, OX (P + Q)) = g − 3.

0 So we get a closed embedding exactly when, for all P,Q in X, h (X, OX (P + 0 Q)) = 1. Of course, there is always an inclusion k ,→ H (X, OX (P + Q)), so we are really asking that there be no nonconstant functions with simple poles at one or both of P and Q, for any choice of P and Q. But the case when there is such a function is also interesting, for such functions give 1 degree two morphisms to P . We now turn our attention to this latter situation.

15.2 Hyperelliptic Curves Definition 15.1. A curve X is hyperelliptic if there exists a degree two 1 map X → P . With this terminology, we summarize our results thus far by saying that any curve of genus g ≥ 2 can either be (canonically) embedded in g−1 P or else is hyperelliptic. We now ask for a classification of hyperelliptic curves. This will be accomplished by explicitly writing down an equation for such curves. 1 So suppose X → P is a degree two map. Then we have a separated 1 degree two, hence Galois, field extension k(P ) ,→ k(X). This gives a Z/2 1 action on k(X), which lifts to an action of Z/2 on X over P . For example, 2 if X were defined by an equation of the form y = (x − a1) ··· (x − an), then the action would be given by sending y to −y. How can we obtain the Z/2 action on X from that on k(X)? In general, 1 since the map f : X → P is affine, A = f∗OX is a locally free sheaf of 1 O 1 -algebras of rank two. For an open affine U of , we have maps P P

k(X) o A (U) O O

1 o O 1 (U) k(P ) P

1 The maps O 1 (U) ,→ k( ) ,→ k(X) are inclusions, and hence identify P P O 1 (U) with a subring of k(X). We claim that (U) is the integral P A closure of O 1 (U) in k(X). Since the map f is ﬁnite, (U) is a ﬁnite P A O 1 (U)-module, and hence (U) is integral over O 1 (U). This shows P A P

52 1 1 that we can recover X over P from the inclusion of ﬁelds k(P ) ,→ k(X) (since X = Spec A ). By the functoriality of this description of A , we get a /2-action on over O 1 . This breaks into eigenspaces; we write Z A P A A = A+ ⊕ A−, where the generator σ of Z/2 acts trivially on A+ and by −1 on A−. Both “eigensheaves” are locally free of rank one, for this can be checked locally, say at the generic point, where it’s true because 1 k(X)/k(P ) is Galois. Also, O 1 injects into , and in fact this is an isomorphism (check), so P A+ 1 there is some line bundle L on (so L ∼ O 1 (d)) such that ∼ O 1 ⊕L P = P A = P as O 1 -modules. How does the algebra structure of look under this P A isomorphism? We already know how to multiply two elements of O 1 , P or an element of O 1 and an element of L, so the algebra structure on P ⊗2 O 1 ⊕ L is determined by a map ρ: L → O 1 . P P 1 1 1 Pick an open Spec k[t] = ⊂ ; then O 1 ∼ k[t]. Since Pic is A P A = g A ∼ ∼ ⊗2 ∼ 2 ∼ trivial, L 1 = O 1 = kg[y]. The map ρ 1 : O 1 = k][y ] → kg[t] = O 1 A A A P A sends y2 to an element g(t) ∈ k[t]. Thus ∼ ∼ 2 A 1 = O 1 ⊕ L 1 = k[t, y]/(y − g(t)), A A A with the Z/2 action given by y 7→ −y. In conclusion, hyperelliptic curves 2 2 2 X are given (in an open aﬃne A ⊂ P ) by equations of the form y = g(t). We will see next lecture how to (almost) obtain the degree of g(t) solely from the genus of X.

16 Friday, Feb. 24th: Riemann-Hurwitz Formula

Today we show how to use a morphism f : X → Y of curves to relate the genus gX of X to the genus gY of Y . The key ingredient in this computation is the familiar exact sequence

∗ 1 1 1 f ΩY /k → ΩX/k → ΩX/Y → 0.

We will use the following terminology: the morphism f is separable if the corresponding ﬁeld extension k(Y ) ,→ k(X) is separable.

16.1 Preparatory Lemmata

∗ 1 α 1 Lemma 16.1. The map f ΩY /k → ΩX/k is injective if and only if f is separable.

Proof. Injectivity of α can be checked at stalks. Both X and Y are integral schemes, so the vertical maps in the following diagram are injective:

f ∗Ω1 / 1 Y x ΩX,x

f ∗Ω1 1 Y η / ΩX,η where η is the generic point of X. Thus the top arrow is injective if and only if the bottom one is, so it suﬃces to check injectivity at the generic

53 point. The inclusions into X and Y of the respective generic points gives a ﬁbered diagram Spec k(X) / X

Spec k(Y ) / Y This implies that Ω1 ∼ Ω1 , and this is zero if and only if f X/Y η = k(X)/k(Y ) is separable, as we checked on a previous HW assignment.

Lemma 16.2. The diagram

deg Pic(X) X / O ZO f ∗ · deg f

degY Pic(Y ) / Z commutes. P Proof. Recall that any line bundle on Y has the form OY ( nP P ), and P the degree of such a line bundle is just nP . For the proof, it is enough to check it on the generators of Pic(Y ), namely those line bundles of the form OY (P ) for P ∈ Y . Since such a line bundle has degree one, we ∗ need to show that deg f = degX (f OY (P )). Let IP be the ideal sheaf of ∗ the inclusion P,→ Y . Then f IP is an ideal sheaf on X, isomorphic to N eQ f(Q)=P IQ . Therefore

∗ ∗ O eQ degX (f OY (P )) = − degX (f IP ) = − degX IQ f(Q)=P X = − −eQ = deg f, f(Q)=P where the last equality is from Proposition 14.1.

Our ﬁnal lemma concerns the local structure of the module of diﬀer- entials.

Lemma 16.3. Let πX be a uniformizer of the local ring OX,P . Then 1 ∼ ΩX,P = OX,P · dπX .

1 Proof. ΩX,P is generated by elements of the form df, for f ∈ OX,P . But df = d(f − f(P )), since d(f(P )) = 0. Since f − f(P ) is in the maximal ideal m of OX,P , we can write it as gπX . Then

df = d(gπX ) = g dπX + πX dg.

1 1 Thus the image of dπX generates ΩX,P /mΩX,P , and therefore dπX gen- 1 erates ΩX,P by Nakayama.

54 16.2 Riemann-Hurwitz Formula We now assume that f is separable, so by Lemma 16.1 the sequence ∗ 1 α 1 1 0 → f ΩY /k → ΩX/k → ΩX/Y → 0. is exact. In particular, α is injective; we want to understand its image. For this we look at the stalk, where the sequence looks like

1 αP 1 1 0 → ΩY,f(P ) ⊗OY,f(P ) OX,P → ΩX,P → ΩX/Y,P → 0.

Let πY be a uniformizer for OY,f(P ) and πX a uniformizer for OX,P . eP We can write the image of πY in OX,P as uπX for some unit u and 1 integer eP ≥ 1. Since ΩY,f(P ) is generated as an OY,P -module by dπY , ∗ 1 ∼ f ΩY,f(P ) = dπY OX,P , so using this isomorphism, αP sends dπY to eP eP eP −1 d(uπX ) = πX du + epuπX dπX . If it happens that the characteristic of k divides eP , the situation is considerably more complicated, and gets a name, but we will not treat this case. Deﬁnition 16.1. The morphism f has wild ramiﬁcation at P if the characteristic of k divides eP .

So assume that the characteristic of k does not divide eP . Then fol- 1 eP 1 eP lowing αP with the quotient map to ΩX,P /m ΩX,P , the term πX du eP −1 in the above vanishes, so the image is generated by πX dπX . By eP −1 Nakayama, then, the image of αP itself is generated by πX dπX . Thus ∗ 1 ∼ eP −1 1 f ΩY,f(P ) = πX ΩX,P . Allowing P to vary, we obtain the main theorem of the lecture. Theorem 16.1. If f is separable and has no wild ramiﬁcation, then ∗ 1 X ∼ 1 f ΩY (eP − 1) · P = ΩX P ∈X Theorem 16.2 (Riemann-Hurwitz). If f is separable and has no wild ramiﬁcation, then X 2gX − 2 = deg f(2gY − 2) + (eP − 1) P ∈X Proof. Take degrees in Theorem 16.1.

1 Example 16.1. Let f : X → P be a degree two map, i.e., X be a hyperelliptic curve. By the previous lecture, we can write the equation 2 for X as y = (x − a1) ··· (x − ar), so f is ramified at the points ak and possibly at ∞. For those points at which f is ramified, its ramification index is two, so using 16.2 we find

2gX − 2 = 2(−2) + r + , where comes from the ramification at ∞: it is 1 if f is ramified there, and 0 if not. By parity, we see that if r is odd, = 1, so f is ramified at ∞ and gX = (r −1)/2. If r is even, f is unramified at ∞ and gX = (r −2)/2. This makes good on the promise made at the end of the previous lecture. Example 16.2. As a reality check, consider the case where r = 3, i.e., X is an elliptic curve. It’s ramified at infinity, and the computation above agrees with our knowledge that gX = 1. Phew!

55 17 Monday, Feb. 27th: Homological Algebra I

Lecture and typesetting by Piotr Achinger

17.1 Motivation 17.1.1 Algebraic topology was the main motivation for the concepts of homological algebra. The term homology itself comes from Poincar´e’sﬁrst attempts on algebraic topology. Given a nice topological space7 X and a commutative ring R, one can associate to it the following sequence of maps between R-modules:

0 → C0(X) −→d C1(X) −→d C2(X) −→d ...

where Ci(X) is the dual of the free R-module spanned by the set of all continuous maps f : [0, 1]n → X (,,singular cubes in X”) and the maps d are dual to the maps taking such ,,singular cubes” to their ,,boundaries”. It is intuitively clear that the boundary of the boundary is zero, i.e., d◦d = 0. We can therefore look at the cohomology groups8 :

ker(d : Ci(X) → Ci+1(X)) Hi(X,R) := . im(d : Ci−1(X) → Ci(X)) These encode important topological information of X, and have many useful properties, to note a few: (a) H0(X,R) is a free R-module spanned by the (path-)connected components of X. (b) The Hi(X,R) are functorial in both X and R – contravariantly in X and covariantly in R (the map Hi(f) induced by f : X → Y is usually denoted by f ∗), (c) given a nice inclusion9 i : Y,→ X, one gets a long cohomology exact sequence

p∗ ∗ → Hi−1(Y,R) −→δ Hi(Z,R) −→ Hi(X,R) −→i Hi(Y,R) −→δ Hi+1(Z,R) → ... (10) where Z = X/Y is X with Y contracted to a point, and p : X → Z is the projection.

17.1.2 Algebraic geometry – a.k.a. the Black Box Our main goal is to deﬁne and study the sheaf cohomology functors Hi(X, F ), where X is a topological space and F a sheaf of abelian groups on X. We already know some of their properties, having used them as a ,,black box” to study algebraic curves: (a) H0(X, F ) = Γ(X, F ),

7e.g. a CW-complex 8If we didn’t take duals here, we would get homology groups – but cohomology serves a better motivation for our purposes. 9e.g. a coﬁbration

56 (b) The H0(X, F ) are functorial in both X and F – contravariantly in X and covariantly in F (contravariance in X is to be understood as follows: if f : X → Y is a continuous map and G is a sheaf on Y , we get a map Hi(Y, G ) → Hi(X, f −1G )). (c) Given a short exact sequence 0 → F → G → H → 0 of sheaves on X, one gets a long cohomology exact sequence

δ δ → Hi−1(X, H ) −→ Hi(X, F ) → Hi(X, G ) → Hi(X, H ) −→ Hi+1(X, F ) → (11) Propaganda 17.1. Our goal, motivated by algebraic topology, is to develop a general notion of a ,,cohomology theory”, encompassing (17.1.1) and (17.1.2) above10 and many other examples in a uniform fashion. Our second goal is to develop a theory which would handle the non-exactness of the functor Γ(X, −) (and other non-exact functors).

17.2 Complexes, cohomology and snakes We start with a rather lengthy definition. I hope most readers will be familiar with most of the notions presented here. Definition 17.2. Let R be a commutative ring. • n n n 1.A complex of R-modules is a pair C = (C , d ) where C (n ∈ Z) are R-modules and dn : Cn → Cn+1 are maps (called differentials) satisfying the condition dn+1 ◦ dn = 0. (In practice, one usually writes d instead of dn, so our condition can be written succintly as d2 = 0). 2.A morphism of complexes f • : C• → D• is a family of maps f n : n n n n n+1 C → D commuting with the differentials, i.e., dD ◦ f = f ◦ n dC . One therefore gets a category Ch(R − mod) of complexes of R-modules. 3. The cohomology groups of a complex C• are the groups Hi(C•) := ker dn/ im dn−1. One easily checks that they are covariant functors Hi : Ch(R − mod) → R − mod. Elements of ker dn are called cycles and denoted by Zn(C) and elements of im dn−1 are called boundaries and denoted by Bi(C). Therefore Hi(C) = Zi(C)/Bi(C). 4. The kernel, image and cokernel of a morphism of complexes are defined in the usual way. One can then talk about subcomplexes, quotient complexes, exact sequences of complexes, complexes of complexes, cohomology complexes of complexes of complexes and so on. Propaganda 17.3. Here is one principle of our approach to homological algebra: to study an object A, associate to it a complex C•(A). This complex will depend on the ,,cohomology theory”. In other words – to construct a cohomology theory H ∗, one first describes a way to attach a complex C•(A) to A and define H i(A) as Hi(C•(A)). Usually C•(A) is not functorial in A (but its cohomology is).

10In fact (again for ,,nice” spaces), (1) is a special case of (2): Hi(X,R) = Hi(X, R) where R is the constant sheaf associated to R.

57 How do we get long exact sequences? We extend our principle as follows: if an object X ﬁts into a ,,short exact sequence” Y,→ X Z, then there should be an exact sequence of complexes 0 → C•(Y ) → C•(X) → C•(Z) → 0. Then we get a long exact sequence from the Snake Lemma (below). This encompasses both long exact sequences (10) and (11). Lemma 17.1 (Snake Lemma11). (a) Given a commutative diagram of R-modules with exact rows

. C D E 0

f g h

0 C0 D0 E0 .

there exists an arrow δ : ker g → coker f such that the sequence

ker f → ker g → ker h −→δ coker f → coker g → coker h

is exact. Here is the whole picture:

ker f ker g ker h

C D E 0

f g h

0 C0 D0 E0

coker f coker g coker h

(b) The arrow δ is functorial, that is, given two diagrams as above and a commutative system of arrows between them, the two δ-arrows keep the system commutative. p (c) Given a short exact sequence of complexes 0 → C• −→i D• −→ E• → 0, there is a long exact sequence of cohomology

p ... → Hi−1(D•) −→δ Hi(C•) −→i∗ Hi(D•) −→∗ Hi(E•) −→δ Hi+1(C•) → ...

58 (d) The arrows δ above are functorial, that is, given two short exact sequences as above and a commutative system of arrows between them, one gets a commutative ladder between the two long cohomology exact sequences.

Proof. You should prove (a) and (b) yourself. For (c), apply (a) to the diagram

. Ci/Bi(C) Di/Bi(D) Ei/Bi(E) 0

i i i dC dD dE

0 Zi+1(C) Zi+1(D) Zi+1(E) .

(it is straightforward to check that the rows are exact) obtaining

Hi(C) Hi(D) Hi(E)

Ci/Bi(C) Di/Bi(D) Ei/Bi(E) 0

i i i dC dD dE

0 Zi+1(C) Zi+1(D) Zi+1(E)

Hi+1(C) Hi+1(D) Hi+1(E)

Then (d) follows from (b) and (c).

17.3 Homotopy In Section 1, we discussed cohomology groups Hi(X,R) of a topological space X. In fact, these are homotopy invariants of X: that two maps f, g : X → Y are homotopic if there exists a continuous map H : X × [0, 1] → Y such that H(x, 0) = f(x) and H(x, 1) = g(x). One proves that homotopic maps induce the same maps on cohomology groups. To prove this, one ﬁrst constructs a family of maps si : Ci(Y ) → Ci−1(X) such

59 that ds + sd = f ∗ − g∗, as in the diagram below.

di−1 di ... Ci−1(Y ) Y Ci(Y ) Y Ci+1(Y ) ...

∗ ∗ ∗ ∗ ∗ ∗ f g si+1 f g si+1 f g

... Ci−1(X) Ci(X) Ci+1(X) ... i−1 i dX dX

Then, the statement follows formally from the equation ds+sd = f ∗−g∗12. Deﬁnition 17.4. Let C•, D• be complexes and let f, g : C• → D• be two maps of complexes. We say that f and g are homotopic (denoted f ∼ g) if there exists a family of maps si : Ci → Di−1 such that i i i−1 i i+1 i f − g = dD ◦ s + s ◦ dC . Here is the diagram for your convenience:

d ... Ci−1 Ci Ci+1 ...

s f g s

... Di−1 Di Di+1 ... d

Lemma 17.2. If f ∼ g, then Hi(f) = Hi(g) for all i.

Proof. Let z ∈ Zi(C). We want to prove that there is an x ∈ Di−1 such that f(z) − g(z) = dx. But f(z) − g(z) = d(s(z)) + s(d(z)) = d(s(z)) + 0 since dz = 0 by assumption. Therefore we can take x = s(z).

Some more vocabulary: Deﬁnition 17.5. The family of maps si is called a homotopy between f and g. A map which is homotopic to zero is called nullhomotopic. We call an f : C• → D• a homotopy equivalence if there exists a g : D• → C• such that f ◦ g is homotopic to idD and g ◦ f is homotopic to idC . Remark. By the Lemma, a homotopy equivalence induces an isomorphism on cohomology groups (such a map is called a quasi-isomorphism). The problem with quasi-isomorphisms is that they are not preserved under additive functors, for example a short exact sequence of sheaves α β 0 → F 0 −→ F −→ F 00 → 0 can be seen as a quasi-isomorphism as in the diagram below:

... 0 F 0 0 0 ...

... 0 F F 00 0 ... β

12For the construction of si, see e.g. A. Hatcher Algebraic Topology, Theorem 2.10

60 This map of complexes will not however remain a quasi-isomorphism after applying Γ(X, −). On the other hand, homotopy equivalences are preserved by additive functors, by the nature of their deﬁnition. This means that homotopy equivalence is a much better behaved notion. In fact, they will play a key role in our construction of ,,cohomology theories” (i.e., derived functors).

18 Wednesday, Feb. 29th: Homological Algebra II

Lecture and typesetting by Piotr Achinger Here is what we know after Lecture 1: • we know what is a complex (of abelian groups or R-modules), what are its cohomology groups, what it means for it to be exact, • we know the Snake Lemma and that a short exact sequence of complexes gives us a long cohomology exact sequence, • we know the notion of homotopy of maps between complexes and that homotopic maps induce the same map on cohomology.

18.1 Additive and abelian categories A model example of a category in which one can make homological considerations is the category R- Mod of (left) modules over a ring R. Observation 18.1. In the category R- Mod there exist: (a) a zero (initial and terminal) object 0. (b) a structure of an abelian group on the sets of morphisms Hom(M,N) satisfying the distributivity laws of left and right composition with repect to addition (i.e. for f : M → N the functions

x 7→ x ◦ f : Hom(N,N 0) → Hom(M,N 0),

x 7→ f ◦ x : Hom(M 0,M) → Hom(M 0,N) are group homomorphisms). (c) direct sums M ⊕ N which are simultaneously products (i.e. we have both the projections pM : M ⊕ N → M, pN : M ⊕ N → N and the inclusions iM : M → M ⊕ N, iN : N → M ⊕ N satisfying pM ◦ iM = idM , pN ◦ iN = idN , iM ◦ pM + iN ◦ pN = idM⊕N ). Deﬁnition 18.2. A category satisfying (a)–(c) above is called additive13. An additive functor is a functor F : C → D between additive categories taking 0 to 0, direct sums to direct sums and such that the induced maps Hom(M,N) → Hom(FM,FN) are group homomorphisms. In an additive category, it makes sense to talk about complexes, but not cohomology. Therefore we need additional properties.

13One may think that ,,additive” is an extra structure on a category. In fact, whenever a category is additive, the group structure on the sets Hom(M,N) is uniquely determined. Therefore ,,additive” is a property of a category

61 Observation 18.3. In the category R- Mod there also exist: (d) kernels, i.e. equalizers of any morphisms with the zero morphism (that is, for any f : M → N there is an object ker f = K together with a map if = i : K → M with the following universal property: given g : M 0 → N such that f ◦ g = 0, there is a unique g¯ : M 0 → K satisfying g = i ◦ g¯), (e) cokernels, i.e. coequalizers of any morphisms with the zero morphism (that is, for any f : M → N there is an object coker f = C together with a map pf = p : N → C with the following universal property: given g : N → N 0 such that g ◦ f = 0, there is a unique g¯ : C → N 0 satisfying g =g ¯ ◦ p),

(f) canonical isomorphims ker(pf ) = coker(if ), that is ,,the cokernel of the kernel is the kernel of the cokernel” (by the universal property of kernel and cokernel, there is a canonical map ker(pf ) → coker(if ) – we require it to be an isomorphism). Deﬁnition 18.4. An additive category satisfying additionally (d)–(f) is called abelian. In an abelian category, the usual notions of cohomology of a complex or an exact sequence make sense. An additive functor F : C → D between two abelian categories is called exact if it takes kernels to kernels and cokernels to cokernels (or equivalently – takes exact sequences to exact sequences). Observation 18.5. In the category R- Mod we have the Snake Lemma, which gives us long cohomology exact sequences for short exact sequences of complexes. At the ﬁrst glance, it is not obvious how the proof of the Snake Lemma should go in any abelian category – recall that our proof used ,,elements” of the objects in the diagram. However, the Snake Lemma and its conclusions hold in any abelian category, thanks to the following result: Theorem 18.1 (Freyd-Mitchell, 1964). Let A be a small14 abelian category. Then A is a full abelian subcategory of the category R- Mod of modules over a certain ring R. More precisely, there exists a ring R and a fully faithful exact additive functor F : A → S − Mod. Remark. In fact, the Freyd-Mitchell theorem allows us to perform diagram chasing in any abelian category – given a diagram (whose objects form a set) we can pass to the smalles abelian subcategory containing the objects of the diagram (which should be small). I will ignore set theoretic issues of this type from now on. Remark. The category Ch(A ) of complexes of objects of an abelian category A is abelian (we take sums, kernels and cokernels ,,levelwise”). The functors

Hi : Ch(A ) → A ,Hi(A•) = ker(di)/im(di−1) are additive, but obviously not exact. Our considerations concerning R- Mod and the Freyd-Mitchell theorem show that 14A category is small if its objects form a set.

62 f g Theorem 18.2. If A is an abelian category and ξ : 0 → A• −→ B• −→ C• → 0 is a short exact sequence of complexes in A , then there exists the long cohomology exact sequence

i−1 δ Hif Hig ... → Hi−1C• −−−→ξ HiA• −−−→ HiB• −−→ HiC• → ...

f g which is natural in the following sense: if ξ0 : 0 → A0• −→ B0• −→ C0• → 0 is another such sequence and we are given a morphism of short exact sequences15 φ : ξ → ξ0, then the diagrams

i δξ HiC• Hi+1A•

φ∗ φ∗ i δξ0 HiC0• Hi+1A0•

commute. Examples 18.6. The following additive categories are abelian: • abelian groups, R-modules, • sheaves and presheaves of abelian groups on a topological space X, sheaves of OX -modules on a ringed space (X, OX ),

• coherent and quasi-coherent OX -modules on a scheme X, • complexes in an abelian category A . and the following are not: • ﬁnitely generated free abelian groups,

• vector bundles (i.e., locally free OX -modules of ﬁnite rank) on a scheme X.

18.2 δ-functors Propaganda 18.7. Recall that our slogan was ,,replace the objects we study by complexes”. One can ﬁnd manifestations of this idea on every level of homological algebra, the most reﬁned of them being the notion of a derived category, developed by J.-L. Verdier in 1960s. Trying to realize our motto in a naive way, imagine that to a given object A of an abelian category A there is assigned a complex X• in some other abelian category B. For simplicity (and better analogy with our motivational examples coming from topology) we will assume that Xi = 0 for i < 0. With this picture there should be an associated cohomology theory, that is, a sequence of functors Hi(A) = Hi(X•) and for every short exact sequence 0 → A → B → C → 0 a sequence of arrows δi : Hi(C) → Hi+1(A) (as if 0 → A → B → C → 0 gave a short exact sequence of complexes), giving a long cohomology exact sequence. The notion realizing the above picture is called a δ-functor.

15This is simply a commutative ,,ladder” between ξ and ξ0.

63 Deﬁnition 18.8. Let A and B be abelian categories. By a (covariant, cohomological) δ-functor T • from A to B we mean the following data • a sequence of additive functors T i : A → B (i ≥ 0), i i i+1 • a sequence of morphisms δξ : T C → T A (i ≥ 0) for any short exact sequence ξ : 0 → A → B → C → 0 in A , satisfying the following conditions f g 1. for any exact sequence ξ : 0 → A −→ B −→ C → 0 in A the sequence

0 i T 0f T 0g δ T if T ig δ 0 → T 0A −−→ T 0B −−→ T 0C −→ξ ... → T iA −−→ T iB −−→ T iC −→ξ T i+1A → ...

is exact in B, 2. for any two exact sequences ξ : 0 → A → B → C → 0 and ξ0 : 0 → A0 → B0 → C0 → 0 and a morphism φ : ξ → ξ0 the diagram

i δξ T iC T i+1A

φ∗ φ∗ i δξ0 T iC0 T i+1A0

commutes Remark. There also exist notions of a contravariant δ-functor and of a homological δ-functor. Propaganda 18.9. What we would also like to have (this condition may not sound very well motivated to a person without background from say algebraic topology) is for this ,,cohomology theory” to be ,,determined by its coefficients” – where by ,,coefficients” we mean the functor T 0. Indeed, the cohomology functors Hi(X,R) satisfy 1. H0(X,R) = R for X connected, 2. if Hi(X) is an ordinary cohomology theory, then H0(pt) determines the whole theory (at least for CW-complexes), 3. a morphism R → R0 of coefficient rings extends uniquely to a ,,morphism of theories”, i.e., a system of natural transformations Hi(−,R) → Hi(−,R0), compatible with the connecting homomorphisms δ in the long exact sequences. Another motivation comes from our Black Box – we want universal functors Hi satisfying the properties we need. To understand, what 3. above should mean in the context of δ- functors, we should introduce the notion of a morphism of δ-functors (,,natural transformation of cohomology theories” – in fact the motivating example for Eilenberg and MacLane’s definition of a natural transformation of functors).

64 Deﬁnition 18.10. Let T •, T 0• be two δ-functors from an abelian category A to an abelian category B.A morphism of δ-functors τ : T • → T 0• is a sequence of natural transformations τ i : T i → T 0i which commute with δi, δ0i, that is, for every short exact sequence ξ : 0 → A → B → C → 0 the diagram i δξ T iC T i+1A

i i+1 τC τA 0i δξ T 0iC T 0i+1A

commutes. We can now say what it means that a δ-functor T • is determined by its coefficients T 0. Definition 18.11. Let T • be a δ-functor from an abelian category A to an abelian category B. We say that T • is universal if for every other δ-functor S• from A to B and a natural transformation σ : T 0 → S0 there is a unique morphism of δ-functors τ : T • → S• with τ 0 = σ. As usual, the universal property implies that whenever a universal δ-functor with a fixed T 0 exists, it is unique.

18.3 Left- and right-exact functors What additive functors F : A → B does there exist a (not necessarily universal) δ-functor T • with T 0 = F ? From the ,,long exact sequence” property it follows that for every short exact sequence 0 → A → B → C → 0 the sequence 0 → FA → FB → FC (with no 0 on the right!) is exact. Deﬁnition 18.12. Let A , B be abelian categories and let F : A → B be an additive functor. We call F left-exact if for any short exact sequence 0 → A → B → C → 0 in A , the sequence

0 → FA → FB → FC

is exact; F is right-exact if for any short exact sequence 0 → A → B → C → 0 in A , the sequence

FA → FB → FC → 0

is exact16. Remark. There also is a notion of a contravariant left- or right-exact functor. We just consider such a functor as a covariant functor A op → B.

16That is, F is left/right exact iﬀ it preserves kernels/cokernels – or equalizers/coequalizers – or (because F is additive) ﬁnite limits/colimits.

65 Examples 18.13. For any object A ∈ A , the functors Hom(A, −) and Hom(−,A) are left-exact. If A ∈ R- Mod, then A ⊗R − is right-exact. If f : X → Y is a map of topological spaces, then f∗ : Sh(X) → Sh(Y ) is left- exact and f −1 : Sh(Y ) → Sh(X) is right exact (where Sh are the categories ∗ of sheaves of abelian groups) and similarly for f∗ and f for sheaves of modules if X and Y are ringed spaces. In particular, Γ : Sh(X) → A b is left-exact. More generally, if F : A → B is left adjoint to G : B → A then F is right exact and G is left exact (Problem 4). A few observations are in order. First, a functor is exact if and only if it is both left- and right-exact. Secondly, a functor extends to a δ-functor only if it is left-exact. Finally, if F : A → B is exact, then T 0 = F , T i = 0 for i > 0 deﬁnes a universal δ-functor. Therefore F can give an interesting universal δ-functor only if it is left-, but not right-exact. Propaganda 18.14. We now change our paradigm a little: we consider (universal) δ-functors as a tool to study not (or not only) objects, but the non-exactness of left-exact functors. That is, we focus on the second goal mentioned in the ﬁrst lecture: to develop a theory which would handle the non-exactness of the functor Γ(X, −) (and other non-exact functors). Our main goal is now the construction of a universal δ-functor extend- ing a given left-exact functor (the construction for right-exact functors is analogous) – we shall call them (left) derived functors.

18.4 Universality of cohomology To make sure that our approach may be right and imagine how the general construction should work, we solve the following ,,toy” problem: since the main prototype of the notion of a δ-functor are the cohomology functors i H : Ch≥(A ) → A (Ch≥(A ) is the category of complexes concentrated in degrees ≥ 0), do they form a universal δ-functor? i Proposition 18.1. The functors H : Ch≥(A ) → A together with the morphisms δ from Theorem 2 form a universal δ-functor.

• Proof (sketch). Let there be given a δ-functor S from Ch≥(A ) to A and a natural transformation τ 0 : H0 → S0. We shall construct the τ i : Hi → Si inductively. Suppose that τi−1 has already been constructed. From the Lemma below it follows that for any complex A• there is an injection A• → I• into an exact complex I•. This gives us a short exact sequence of complexes 0 → A• → I• → B• → 0. Looking at the long cohomology exact sequences gives us the diagram

Hi−1A• Hi−1I• Hi−1B• HiA• HiI• = 0

τ i−1 τ i−1 τ i−1 τ i

Si−1A• Si−1I• Si−1B• SiA• SiI•

By some easy diagram chasing we convince ourselves that this determines τ i : HiA → SiA. We then have to check that τ does not depend on the choice of A• → I•, that it is a natural transformation etc.

66 Lemma 18.1 (Problem 3). For any A• ∈ Ch(A ) there is an injection A• → I• where I• ∈ Ch(A ) and Hi(I•) = 0 for all i. Remark for algebraic topologists. We can look at the above proof in the following way: we embed a given space A into the cone I = CA, which is contractible. Then B = I/A = ΣA is the suspension of A. By the suspension axiom we get Hi(A) = Hi−1(B) for any cohomology theory H•. But on Hi−1 everything is already defined, hence the inductive step. An important observation here is that the proof of Proposition 18.1 essentially shows the following Lemma 18.2 (Effaceable δ-functors are universal. See Problem 5). Let T • : A → B be a δ-functor which is effaceable, that is, for every A ∈ A there exists an injection A,→ J into a T •-acyclic object J (J is called acyclic for T • if T i(J) = 0 for i > 0). Then T • is universal.

19 Friday, March 2nd: Homological Algebra III

Lecture and Typesetting by Piotr Achinger Here is what we know after Lecture 2: • what is an additive or abelian category, what is an additive or exact functor, • what is a δ-functor and what it means for it to be universal, • what is a left- or right-exact functor, • that an eﬀaceable17 δ-functor is universal.

Our goal is to extend a given left-exact functor to a universal δ-functor.

19.1 Derived functors – a construction proposal Let us think about what was crucial in our proof of ,,universality of cohomology”. The key idea was that any object A could be embedded into an object I on which the given functor was ,,exact” (that is, the ,,coordinates” of the δ functor for i > 0 vanished on I). Lemma 19.1 (Effaceable δ-functors are universal). Let T • : A → B be a δ-functor which is effaceable, that is, for every A ∈ A there exists an injection A,→ J into a T •-acyclic object J (J is called acyclic for T • if T i(J) = 0 for i > 0). Then T • is universal. Let A and B be abelian categories and let F : A → B be a left- exact functor. We are looking for a universal δ-functor T • with T 0 = F . Motivated by the above Lemma, we want to find a good class I of objects of A , such that 0. every object A ∈ A injects into an object I(A) ∈ I , 1. objects of I are acyclic for F .

17A δ-functor T • is eﬀaceable if any object injects into an object on which T 1,T 2,... vanish.

67 It is not clear what it should mean for an I ∈ I to be acyclic, since we don’t have a δ-functor yet, but only its zero component T 0 = F . But in any case, the following should hold: whenever 0 → I → M → N → 0 is a short exact sequence, 0 → FI → FM → FN → 0 is also exact (because the next term after FN should be T 1I = 0). This of course holds when 0 → I → M → N → 0 splits, since a split short exact sequence stays split (therefore exact) after applying F . Let us be greedy here and ask for the class of objects I for which all such sequences split. Deﬁnition 19.1. An object I ∈ A is called injective if any short exact sequence 0 → I → M → N → 0 splits. An object P ∈ A is called projective if any short exact sequence 0 → M → N → P → 0 splits (that is, if P is injective in A op). Problem 6. What are the injective and projective objects in the category of abelian groups? The class of projective objects plays the same role as injective objects when one wants to study right-exact functors (or contravariant left-exact functors). Then one has to consider surjections P A and the whole program carries over. With these assumptions in place, set I to be the class of injective objects in A and let us try to construct T i using the principle of ,,induction by dimension shifting” as in the proof of Lemma 19.1. Given A ∈ A , ﬁnd A,→ I, with I ∈ I and let B be the cokernel:

β 0 → A −→α I −→ B → 0.

After applying F , we get

F β 0 → FA −−→F α FI −−→ FB which we want to continue to the right

F β 0 → FA −−→F α FI −−→ FB → T 1A → T 1I → ..., and since we wish that T iI = 0 for i > 0, we should put T 1A = coker F β and T i+1A = T iB for i ≥ 1. Thus to compute say T 2A, we should ﬁnd an short exact sequence γ 0 → B → J −→ C → 0 and compute coker F γ. This seems already cumbersome, but we can make a long exact complex 0 → A → I → J → ... (an ,,injective resolution” of A) by ,,splicing” all the short exact sequences. Construction. Given A ∈ A , construct an injective resolution A →

68 I• of I as follows

0 0

B2 iB2

i 0 A A I0 I1 I2 ... iB1 iB3

B1 B3

0 0 0 0

0 1 i i i+1 (where I = I(A), B = coker iA, I = I(B ) for i > 0, B = coker(Bi → Ii)). Then apply F to I•:

F (I•): F (I0) → F (I1) → F (I2) → ... and define T iA = Hi(F (I•)). By left-exactness of F , since 0 → A → I0 → I1 is exact, 0 → FA → FI0 → FI1 is exact, hence T 0A = FA. Obviously this agrees with our previous approach. Remark. It is convenient to view 0 → A → I• → ... as a quasi- isomorphism A → I• (where A is identified with the complex ... → 0 → A → 0 → ...). This means that in the derived category of A (the category of complexes where quasi-isomorphisms are made to be actual isomorphisms), A and I• are the same. There are several obvious problems here: 2. Why should T iA be well-defined (i.e., independent of the choice of A → I•)? 3. Why should it be functorial in A? 4. And how to define the connecting homomorphisms δ to get a δ- functor? Suppose that (2)-(4) were resolved. Then (1) follows, since we can choose 0 → I → I → 0 to be the injective resolution of I, which gives us T iI = 0 for i > 0. If in additions (0) would hold, then by Lemma 19.1 we have constructed a universal δ-functor.

19.1.1 Derived functors – resolution of technical issues Regarding (2) – T iA are well-deﬁned Note that T i constructed as above will not depend on the choice of 0 → A → I• if any two such resolutions will be homotopic (since homotopic maps 1. remain homotopic after applying any additive functor,

69 2. induce the same map on cohomology). What we need is the following Theorem 19.1. 1. for any two resolutions 0 → A → I• amd 0 → A → J • there exist φ, ψ – morphisms between 0 → A → I• i 0 → A → J • (i.e. dφ = φd oraz dψ = ψd) giving the identity on A:

0 A I0 I1 I2 ...

ψ φ ψ φ ψ φ

0 A J 0 J 1 J 2 ...

2. any two such φ, ψ be the homotopy inverses of each other – i.e. there exist si : Ii → Ii−1 and ti : J i → J i−1 (i ≥ 0, we put I−1 = A = J −1) such that ψφ − id = ds + sd and φψ − id = dt + td:

0 A I0 I1 I2 ...

s ψφ − id s ψφ − id s ψφ − id s

0 A I0 I1 I2 ...

0 A J 0 J 1 J 2 ...

t φψ − id t φψ − id t φψ − id t

0 A J 0 J 1 J 2 ...

Let us try to construct φ inductively – the ﬁrst step looks as follows:

0 A I0 I1 I2 ...

0 A J 0 J 1 J 2 ...

We would like to extend the dotted arrow A → J 0 to I0. This follows from Problem 7. An object I ∈ A is injective iff any morphism A → I defined on A ⊆ B can be extended to B (or, equivalently, iff the functor Hom(−,I) is exact).

Regarding (3) – Functoriality In fact, a more general version of Theorem 19.1 holds: Theorem 19.2. Let f : A0 → A be a map in A and let A → I•, A0 → I0• be resolutions of A and A0, respectively, with Ii injective (the objects I0i do not have to be injective). Then there exists a chain map f¯ : I• → I0•, inducing f on H0, which is unique up to homotopy.

70 Problem 8. Prove Theorem 19.2 and deduce Theorem 19.1. Thus, the exercise above resolves both our issues with (2) and (3).

Regarding (4) – Construction of δi This is called the Horseshoe Lemma: Lemma 19.2. Given a short exact sequence 0 → A → B → C → 0 and • • already chosen injective resolutions A → IA and C → IC one can ﬁnd an • injective resolution B → IB so that one obtains a short exact sequence of • • • complexes 0 → IA → IB → IC → 0 which is 0 → A → B → C → 0 on H0. 0 0 0 Problem 9. Prove the above statement. Hint: set IB = IA ⊕ IC and 0 construct an arrow B → IB . Use the Snake Lemma, obtaining a short 0 0 0 exact sequence 0 → IA/A → IB /B → IC /C → 0. Proceed by induction. Thus, given a short exact sequence 0 → A → B → C → 0, we obtain the morphisms δi : T iC → T i+1A from the long cohomology exact sequence associated to the short exact (why?) sequence of complexes 0 → • • • • • • F (IA) → F (IB ) → F (IC ) → 0 where IA,IB ,IC are as in the Horseshoe Lemma. We need to check that the morphisms δ are well deﬁned and natural, which is also left to the reader.

Regarding (0) – Sufficiently many injective objects The only problem may happen with (0). For example, if A is the category of all finitely generated abelian groups, then A has no nonzero injective objects! Definition 19.2. An abelian category A is said to have sufficiently many injective objects if for any object there exists an injection into an injective one. We say that A has sufficiently many projective objects if for any object there is a surjection from a projective one. Theorem 19.3. For any ring R, the category R- Mod has sufficiently many injective and projective objects.

19.2 Derived functors – conclusion Theorem 19.4. Let A and B be abelian categories and let T 0 : A → B be a left-exact functor. If A has suﬃciently many injective objects, then T 0 extends to a universal δ-functor T • with T 0 = F . Deﬁnition 19.3. We denote the functors T i by RiT 0 and call the right derived functors of T 0. A similar Theorem holds for right-exact functors and projective objects (then one has to use homological δ-functor). The left-derived functors of a right-exact functor T0 are denoted by LiT0. Examples 19.4. Let A ∈ A , then Exti(A, −) := Ri Hom(A, −) and Exti(−,A) = Ri Hom(−,A) (these two coincide, so one can compute Exti(A, B) from an injective resolution of B or a projective resolution of A). For A ∈ R − Mod, → ri(A, B) := Li(A ⊗R B) (here one proves

71 i that → ri(A, B) '→ ri(B,A)). For a topological space X, H (X, −) := RiΓ : Sh(X) → A b are called the sheaf cohomology functors (one checks ﬁrst that the category of abelian sheaves on X has suﬃciently many injective objects). For a continuous map f : X → Y , the derived functors i R f∗ : Sh(X) → Sh(Y ) are called the higher direct image functors. For a group G, the functor of G-invariants (−)G : G − Mod → A b is left-exact and we call its derived functors Hi(G, −) = Ri(−)G the group cohomology G functors. One checks easily that M = Hom(Z,G) where Z has the triv- i i ial Z-action. Therefore H (G, M) = Ext (Z,M) can be computed from a projective resolution of Z, which is useful since we do not have to pick a new resolution whenever M changes, and also because projective objects are generally easier to work with than injective objects.

19.3 Two useful lemmas Lemma 19.3 (Acyclic resolutions suﬃce). Let F : A → B be left-exact, where A has suﬃciently many injective objects. Let A ∈ A and let 0 → A → J 0 → J 1 → ... be an exact sequence with all J n acyclic for F (that is, RiF (J n) = 0 for i > 0 and all n ≥ 0). Then there are natural isomorphisms Hi(F (I•)) ' RiF (A). Lemma 19.4. Let A , B and C be abelian categories, F : A → B and G : B → C be left-exact functors. Suppose that A and B have enough injectives (so that the derived functors RiF , RiG and Ri(G ◦ F ) exist) and that F maps injective objects of A to G-acyclic objects of B (that is, if I ∈ A is injective, then (RiG)(F (A)) = 0 for i > 0). Let A ∈ A . Then (a) if (RiF )(A) = 0 for i > 0, then Ri(G ◦ F )(A) ' (RiG)(F (A)), (b) if (RiG)(F (A)) = 0 for i > 0, then Ri(G ◦ F )(A) ' G(RiF (A)). In general, under the assumptions of the lemma, there is a spectral sequence

pq p q p+q E2 = (R G)(R F (A)) ⇒ R (G ◦ F )(A).

20 Monday, March 5th: Categories of Sheaves Have Enough Injectives

We have seen last week how to construct the right derived functors of an additive left-exact functor, so long as the categories we are interested in have enough injectives. Today we verify that this is the case for the categories we are interested in, namely sheaves of OX -modules ModOX , or perhaps just sheaves of abelian groups Sh(X). The functor whose right derived functors we construct is f∗, where f : X → Y is a morphism of schemes. We will be especially interested in the case where Y = Spec k; in this case f∗ is the just the global sections functor Γ(X, −): Sh(X) → Ab. Remark. In some sense, it won’t really matter what categories between which we regard f∗ as a functor. In fact this is reﬂected in the notation: i R f∗ does not specify Sh(X) or ModOX , etc. Of course, some care needs to be taken. For instance, we will not consider categories of coherent sheaves, because over Spec Z, there aren’t any (except for the zero sheaf):

72 the reason is that injective Z-modules are things like Q and Q/Z, which are not ﬁnitely-generated. The goal of today’s lecture is to establish the following, which holds even for ringed spaces with noncommutative rings:

Theorem 20.1. If (X, OX ) is a ringed space, then ModOX has enough injectives.

Remark. Note that by taking OX = Z, we obtain as a corollary that Sh(X) has enough injectives.

Proof. We ﬁrst show that categories of modules over a ring have enough injectives, and then globalize. The following simple lemma is at the heart of both steps of the proof. Lemma 20.1. If F : A → B is an additive functor which has an exact18 left adjoint L: B → A , then F takes injectives to injectives.

Proof. Let I ∈ A be an injective object. To show that F (I) is an injective object of B we must show there exists a dotted arrow in the following commutative diagram in B

0 / M / N

} FI

By the adjunction, this is equivalent to ﬁlling in the dotted arrow in

0 / LM / LN

| I

which can be done because I is injective. Note that the map LM → LN is still injective because L is left exact.

Now let R be a ring (commutative for simplicity). Then ModR has enough injectives. This follows easily from the following lemma.

Lemma 20.2. Let R be a ring. Then the forgetful functor L: ModR → Ab has a right adjoint F : Ab → ModR, and if M ∈ ModR, and LM ,→ I is a monomorphism in Ab, then the composite M → FL(M) → FI is a monomorphism.

Proof. We deﬁne, for A any abelian group, FA = HomAb(R,A) which has an R-module structure given by (r · f)(x) = f(rx). To see this gives an adjunction we need to establish a natural bijection, for M ∈ ModR, A ∈ Ab, HomAb(LM, A) ' HomR(M, HomAb(R,A))

18We actually only use left-exactness of L, but right exactness comes for free because left- adjoints preserve colimits

73 The map going to the right sends f : M → A to the map m 7→ (r 7→ f(rm)). The map going to the left sends φ: M → HomAb(R,A) to the map m 7→ φm(1), where φm is the image of m under φ. We omit the veriﬁcation that these constructions are inverse and functorial in M and A. Now we must check that given an inclusion LM ,→ I, the map M → FL(M) → FI is also an inclusion. But this map is exactly the map coming from the adjunction HomAb(LM, I) ' HomR(M,FI). We need to check that given φ such that

0 * K / M / FL(M) φ

then actually φ is zero. Applying L and using the adjunction gives

0 ) LK / LM / LM Lφ Id

Thus Lφ = 0, hence φ = 0, since L is faithful (Lφ is the same map as φ, it just ignores the R-module structure). This ﬁnishes the proof.

To complete the proof of the theorem, we now consider OX -modules. Fix x ∈ X, i.e., a morphism j : x ,→ X, and for ease of notation, let

A = ModOX , the category of OX -modules, and B = ModOX,x , the category of modules over the local ring OX,x at x. Then we have an adjoint pair of functors

j−1 ) A i B j∗

Now j∗ is a right adjoint, so if I ∈ B is injective, then so is j∗I, by the lemma19. For the remainder of the discussion, ﬁx F ∈ A , and since B has enough (x) injectives, pick for each x an inclusion Fx → I for some injective OX,x- module I(x). Then we have a map

α Y (x) F −→ jx∗I , x∈X

where each map jx is the inclusion of the point x. A product of injectives is still injective, coarsely speaking, because both products and injectives are deﬁned by maps in. Moreover, α is a monomorphism. We check this at the stalk at an arbitrary point z ∈ X:

αz Y (x) (z) (z) F −−→ j I → j I ∼ I , z x∗ z z∗ = x∈X

19Note that j−1 is exact: in general, for f : X → Y , and F a sheaf on Y , the stalk of f −1F at x ∈ X is the same as the stalk of F at f(x) ∈ Y .

74 −1 where the isomorphism on the right is because jz∗ and jz are an adjoint pair and j−1 is exact. The composite map here is just the chosen inclusion (z) Fz ,→ I . Since this is injective, so is αz. Thus we have take an arbitrary sheaf F of OX -modules, and embedded it into an injective object of A , which shows that A has enough injectives.

21 Wednesday, March 7th: Flasque Sheaves

Having shown that ModOX has enough injectives, we can deﬁne , for F an OX -module

Hi(X, F) = HiΓ(X,I0) → Γ(X,I1) → Γ(X,I2) → · · · ,

where F → I• is an injective resolution of F. However, in practice it is easier to work with flasque sheaves, which we now define, rather than injectives. Definition 21.1. A sheaf F on a topological space is flasque if for every inclusion, U ⊂ V , F(V ) → F(U) is surjective. This notion is useful because, in consideration of whether a sheaf is flasque, it does not matter whether we regard it as a sheaf of O-modules or abelian groups. Moreover, every injective sheaf is flasque. In the proof, we will use the following. Definition 21.2. Let j : U,→ X be an open subset, and F be a sheaf of OU -modules. Then the extension by zero of F to X is defined to be the sheafification of the presheaf ( F(W ) if W ⊂ U W 7→ 0 if not.

It is denoted j!F.

Note that j!F has stalks equal to those of F inside U, and zero else- −120 where. Also, j!F is a left adjoint to j .

Lemma 21.1. Let (X, OX ) be a ringed space. If I ∈ ModOX is injective, then I is ﬂasque.

Proof. Let V ⊂ U be two opens in X, with inclusions i: U,→ X, j : V,→ U. By the adjunction, we have for any sheaf F of OX -modules,

−1 HomOX (i!OU , F) ' HomOU (OU , i F) 'F(U)

−1 Now there is an inclusion of OU -modules j!j OV → OU , which gives a map −1 HomOX (i!OU , F) → HomOX (i!(j!j OV ), F)

by restricting along the above inclusion. But HomOX (i!OU , F) 'F(U), −1 and similarly HomOU (OU , i F) 'F(V ). So we have a natural map F(U) → F(V ). This is for any F, but now take F = I an injective

20 −1 Remember that j is itself a left adjoint to j∗.

75 21 −1 OX -module. Then since i! is exact , the inclusion j!j OV → OU gives −1 an inclusion i!j!j OV → i!OU , and then injectivity implies the map −1 HomOX (i!OU , I) → HomOX (i!(j!j OV ), I) is surjective, i.e., I(V ) → I(U) is surjective.

Our next goal is to show that ﬂasque sheaves are acyclic. For this we need a(nother) lemma.

Lemma 21.2. Let F be a ﬂasque sheaf of OX -modules, and embed F ,→ I into an injective sheaf I, with cokernel G. Then for all opens U of X, I(U) → G(U) is surjective.

Proof. We have an exact sequence 0 → F → I → G → 0. Fix a section g ∈ G(U). By exactness at the stalk, for each point of U, there is a neighborhood V over which we can lift g to some g ∈ I(V ). Suppose V eV we have also some other neighborhood W , and a lifting g of g to eW W I(W ). We would like to glue these to a lift in I(V ∪ W ). Since g eV W ∩V and g both map to g in G(W ∩ V ), the exactness of the eW W ∩V W ∩V above sequence implies that their diﬀerence g − g lies in eV W ∩V eW W ∩V F(W ∩ V )22. Since F is ﬂasque, F(W ) surjects onto F(W ∩ V ), so there is some ∈ F(W ) which maps to g −g in F(W ∩V ). Since eV W ∩V eW W ∩V maps to zero in G(W ), g + ∈ I(W ) is a lift of g to I(W ). Moreover, eW W g − (g + )| = 0 in I(W ∩ V ), so they glue to give a lifting eV W ∩V eW W ∩V of g over W ∪ V . Thus we have shown that we can lift sections of G “one open at a time”. We now apply Zorn’s lemma to the set

{(V, geV | geV lifts g over V }, 0 0 partially ordered by (V, g ) ≤ (V , g 0 ) if V ⊂ V and g 0 = g . Any V V V V V increasing chain of such objects has an upper bound by taking the union of the V . Thus we have a maximal such pair, which must be a lifting of g over all of U: if it were not, we could extend further by the above and violate maximality.23

We can now prove i Proposition 21.1. Let F be a ﬂasque sheaf of OX -modules. Then H (X, F) = 0 for i > 0.

Proof. First we pick an embedding F ,→ I of F into an injective I. Then I is ﬂasque by Lemma 21.1. The quotient G is also ﬂasque, because if V ⊂ U is an inclusion, then we have a diagram I(U) / / G(U)

I(V ) / / G(V )

21This follows from the description of the stalks. 22We are really using the left-exactness of the functor Γ(W ∩ V, −) here. 23Cf. also Hartshorne Ex II.1.16

76 in which the horizontal arrows are surjective by Lemma 21.2. This implies that the vertical arrow on the right is surjective also, hence G is ﬂasque. Now H1(X, F) = 0 since it is the cokernel of the map Γ(X, I) → Γ(X, G), which is surjective by the previous lemma. The short exact sequence 0 → F → I → G → 0 gives rise to a long exact sequence in cohomology, in which the middle terms Hr(X, I) are zero because I is injective. Thus we obtain isomorphisms Hi(X, G) ' Hi+1(X, F). Since G is ﬂasque, the result follows by induction.

22 Friday, March 9th: Grothendieck’s Vanishing Theorem

22.1 Statement and Preliminary Comments Ther next few lectures will be devoted to the proof of the following Theorem 22.1. If X is a Noetherian topological space of dimension n, and F a sheaf of abelian groups on X, then Hi(X, F) = 0 for i > 0.

Our primary application of this theorem will be when (X, OX ) is a ringed space and F an OX -module. In order to do so, we will need the fact that computations of cohomology in the categories of abelian groups give the same result. More precisely, we have a universal δ-functor i {HM (X, −)} from ModOX → Ab (the subscript “M” is for module), and i also a δ-functor {HA(X, −)} from Sh(X) → Ab (“A” is for abelian group), which is deﬁned by regarding a sheaf of abelian groups F ∈ Sh(X) as a 24 sheaf of Z-modules, where Z is the constant sheaf There is also the

forgetful functor : ModOX → Sh(X). Since is exact, the composition i {HA(X, −) ◦ } is a δ-functor. We ask whether it is isomorphic, as δ- i functors, to {HM (X, −)}.

Theorem 22.2. If (X, OX ) is a ringed space and F an OX -module, then i i HA(X, F) ' HM (X, F). Proof. We have an isomorphism

0 0 HM (X, −) ' HA(X, (−))

i which, by the universality of HM , induces a morphism of δ-functors

i i {HM (X, −)} → {HA(X, −) ◦ }.

i This is an isomorphism because {HA(X, −) ◦ } is also universal. This

follows because it is effaceable: for F ∈ ModOX , pick an embedding into an injective OX -module I. This injective is also flasque, and so I is i flasque in Sh(X). Flasque sheaves in Sh(X) are acyclic for HA by applying Lemma 21.1 with OX = Z.

24 Note that the same argument as in the case of ModOX shows that Sh(X) has enough injectives, by taking OX = Z.

77 22.2 Idea of the Proof; Some lemmas The proof of the vanishing theorem begins by reducing first to the case when X is irreducible, and then reducing to the case where F = j! Z for j : U,→ X and open. Then we use induction on the dimension of X. To make the reduction on F, we will need to know how flasque sheaves and cohomology interact with direct limits. Lemma 22.1. On a Noetherian topological space, a direct limit of flasque sheaves is flasque.

Proof. HW exercise.

The next lemma says that “cohomology commutes with direct limits”. More precisely, if (Fα) is a directed system in Sh(X), then for each α we have a map F → lim F . Applying Hi we get maps Hi(X, F ) → α −→ α α Hi(X, lim F ) for each α. By the universal property of direct limit, these −→ α induce a map lim Hi(X, F ) → Hi(X, lim F ) −→ α −→ α Lemma 22.2. If X is a Noetherian topological space, the map lim Hi(X, F ) → −→ α Hi(X, lim F ) is an isomorphism. −→ α Proof. Let A be a directed set, which we can regard as a category. We denote by Sh(X)A the category of functors A → Sh(X). Note that it is an abelian category; kernels, cokernels, etc., are deﬁned “index by index”. It also has arbitrary products, and enough injectives. From the following diagram −→lim Sh(X)A / Sh(X)

Γ Γ AbA / Ab −→lim we obtain two δ-functors Sh(X)A → Ab, namely the right-derived functors of the two different compositions in the square. More precisely, the two δ-functors in question are 1.( F ) 7→ Hi(X, lim F ) α −→ α 2.( F ) 7→ lim Hi(X, F ) α −→ α We will be done if we can show that these are both universal, for then the universality gives a canonical isomorphism between them. We show that they are both effaceable. The reason is that for any sheaf F, there is a functorial embedding F ,→ Fe of F into a flasque sheaf, namely take Fe to be the so-called sheaf of discontinuous sections of F, given by [ Fe(U) = {s: U FP s(p) ∈ Fp for each p}. p∈U

Then the first δ-functor is effaceable since this is functorial, i.e., given an A object (Fα) ∈ Sh(X) , the construction produces a system Feα of flasque sheaves, whose direct limit is flasque by Lemma 22.1, hence Hi(X, lim F ) = −→ α

78 0 for i > 0. The second is eﬀaceable since each Hi(X, Fe) = 0 for i > 0, so lim Hi(X, F ) = 0. −→ α 23 Monday, March 12th: More Lemmas

We stated without proof last time that the functor lim: Sh(X) → Ab is −→ exact. Let’s prove that now. Proposition 23.1. For A a directed set, the functor lim: Sh(X)A → −→ Sh(X) is exact.

Proof. Let 0 00 0 → (Fα) → (Fα) → (Fα ) → 0 be a short exact sequence in Sh(X)A. For each x ∈ X, and each α ∈ A, the sequence 0 00 0 → Fα,x → Fα,x → Fα,x → 0 is exact, and the functor lim: ModA → Mod is exact, so we have −→ OX,x OX,x short exact sequences

0 → lim F 0 → lim F → lim F 00 → 0 −→ α,x −→ α,x −→ α,x for each x ∈ X. Hence we will be done if we can show that lim commutes −→ with stalks, i.e., lim(F ) = lim F −→ α x −→ α,x But stalk is a special case of pullback, so it suﬃces to prove more generally that for f : Y → X, the functor f −1 : Sh(X) → Sh(Y ) commutes with direct limits. For this we use Yoneda: let (Fα) ∈ Sh(X) and G ∈ Sh(Y ). Then

Hom (f −1 lim F , G) = Hom (lim F , f G) Y −→ α X −→ α ∗ = lim Hom (F , f G) −→ X α ∗ = lim Hom (f −1F , G) −→ Y α = Hom (lim F , G), Y −→ α so f −1 lim F = lim F .25 −→ α −→ α The next lemma says that if we have a sheaf F on a closed subset Z ⊂ X, we can compute the cohomology of F directly on Z, or push it forward to X and compute there, and we will get the same result. Lemma 23.1. Let i: Z,→ X be a closed subset; then for any F ∈ Sh(Z),

i i H (Z, F) ' H (X, i∗F)

25The argument is a common one: “colimits commute with left adjoints”. Also note that this isomorphism holds for OX -modules, although our proof only works in Sh(X), because −1 f is not well-deﬁned for OX -modules.

79 26 Proof. First note that since i is a closed immersion, i∗ is exact. Also, i∗ has an exact left-adjoint, so by Lemma 20.1, it takes injectives to injectives. Thus taking an injective resolution F → I• of F in Sh(Z), we • get an injective resolution i∗F → i∗I of i∗F in Sh(X). But for any sheaf G on Z, Γ(X, i∗G) = Γ(Z, G), so applying the global sections functor to these two injective resolutions gives the same sequence of abelian groups, hence the result.

Remarks. 1. To see that i∗ is exact when i is a closed immersion, note that the stalk of i∗F is ( Fx if x ∈ Z (i∗F)x = . 0 if not

Since exactness can be checked at stalks, it is easy to see in this case that i∗ is exact. However, this fails when i is not a closed immersion: consider the open immersion

2 2 i: Y = Ak \{(0, 0)} ,→ Ak = X

Then the stalk (i∗OY )(0,0) is just OX,(0,0). Intuitively, an open neighborhood U of the origin, and the punctured neighborhood U \ {(0, 0)} have the same sections, because the point has codimension two. 2. In applying the lemma, we will frequently use the following fact. If F is a sheaf on a space X, whose support is contained in a closed −1 subset i: Z,→ X, then F = i∗i F. At this point we began the proof of Theorem 22.1, but I’ll consolidate the whole argument into the notes for the next lecture.

24 Wednesday, March 14th: Proof of the Vanish- ing Theorem

We recall ﬁrst the statement: If X is a Noetherian topological space of dimension n, and F a sheaf of abelian groups on X, then Hi(X, F) = 0 for i > 0.

Proof. 24.1 Base step We proceed by induction on n = dim X. For the case n = 0, the theorem says exactly that the global sections functor is exact. But here X is a disjoint union of points, finite by the Noetherian hypothesis. A sheaf on a point is just an abelian group, so a sheaf on X is just a finite product of abelian groups. Thus Γ(X, −) is exact - it sends a finite product of abelian groups to ... that same finite product of abelian groups. Now we fix n > 0, and assume the theorem holds for any space of dimension less than n. Fix a sheaf F of abelian groups on X.

26It’s a right adjoint, so it’s always left exact (even if i is not a closed immersion). For right-exactness in the case of a closed immersion, see the remark following the proof.

80 24.2 Reduction to X irreducible Here we argue that it is enough, under our inductive hypothesis, to assume that X is irreducible. Suppose it were not. Then pick an irreducible closed subset X0 of X, let U be the complement in X0 of the other irreducible components of X, and let Y = X \ U. Then we have inclusions

j i U ,→ X ←-Y

with j open and i closed. Let F = j (F ) and F = i i−1F. We U ! U Y ∗ obtain27 an exact sequence of sheaves on X:

0 → FU → F → FY → 0,

which gives rise to a long exact sequence of cohomology

0 0 0 1 1 H (X, FU ) → H (X, F) → H (X, FY ) → H (X, FU ) → H (X, F) → · · ·

i i −1 By Lemma 23.1, we have H (X, FY ) = H (Y, i F). Also, since FU is supported on X0, we may regard it as a sheaf on X0, and we have i i 0 28 H (X, FU ) = H (X , FU ), again by Lemma 23.1 . So the long exact sequence looks like

i−1 −1 i 0 i i −1 · · · → H (Y, i F) → H (X , FU ) → H (X, F) → H (Y, i F) → · · ·

Since X0 is irreducible, and Y 0 is a closed subset of X with less irreducible components than X, this sequence will prove that Hi(X, F) = 0 for i > n by induction on the number of irreducible components, so long as we can establish this in the case where X is irreducible (i.e., the base step for the induction on the irreducible components).

24.3 Reduction to the case where F = ZU . Now we assume X is irreducible. For j : U,→ X any open subset, we let

as above ZU = j! Z, where Z is the constant sheaf Z on U. We will simply write ZU from now on. Deﬁne a S = F(U), U⊂X

and let A be the set of ﬁnite subsets of S , partially ordered by inclusion. A typical element of A looks like

α = {(U1, f1),..., (Ur, fr)}, fk ∈ F(Uk)

Such a thing deﬁnes a map

r M (fi) ZUi → F i=1

27Cf. Hartshorne, Ex II.1.19(c). 28 i i 0 −1 0 Abusing notation slightly: should be H (X, FU ) = H (X , s∗s FU ), where s: X ,→ X is the inclusion. See remark 2 following Lemma 23.1.

81 For each α as above, we let Fα be the image of this map. Then the inclusions F ,→ F induce a map lim F → F which is an isomorphism. α −→ α It’s injective since the Fα are subsheaves of F. It’s surjective on each open, since if f ∈ F(U), then take α = (U, f); the image of 1 under (U) → F (U) → (lim F )(U) is a preimage for f over U. ZU α −→ α Since Hi(X, lim F ) ' lim Hi(X, F ), it is now enough to prove that −→ α −→ α i when i > n, H (X, Fα) = 0. With α = {(U1, f1),..., (Ur, fr)} as above, 0 and setting α = {(U1, f1),..., (Ur−1, fr−1)}, we have a diagram

0 / Fα0 / Fα / Q / 0 O O O

/ Lr−1 / Lr / / 0 i=1 ZUi i=1 ZUi ZUr 0 where the horizontal rows are exact by construction (here Q is just the cokernel of Fα0 → Fα), and the left two vertical arrows are surjective by deﬁnition. It follows that the rightmost vertical arrow is also surjective. Now from the long exact sequence in cohomology induced by the top sequence here, and using induction on the cardinality of α, it suﬃces to show that Hi(X, Q) = 0 for i > n, where Q is any quotient of a sheaf of the form ZU . For such a sheaf Q, we have an exact sequence

0 → K → ZU → Q → 0 and again using the long exact sequence, we see it will be enough to show i that H (X, K ) = 0 for i > n, where K is any subsheaf of ZU (including ZU itself). So fix some subsheaf K of ZU ; then the stalk Kx at x ∈ X is a principal ideal (dx) ⊂ Z, where we may assume that dx ≥ 0. If all integers dx are zero, then K is the zero sheaf, whose higher cohomology vanishes, so there’s nothing to show. Otherwise, set d = inf{dx}, where the infimum is taken over all x such that dx 6= 0. Then we will have dx = d for at least one point of X. Take some neighborhood V ⊂ U containing this x, and a section s ∈ K (V ) representing d on this neighborhood. The section s defines an injective map ZV ,→ K sending 1 to d. Composing with the inclusion K ,→ ZU gives a map

ZV → K → ZU of sheaves on U, which is multiplication by d at stalks of V , and the zero map elsewhere. Now we claim that in fact KV is isomorphic to ZV . We check this at the stalk at a point y of V . There the inclusions above look like ZV,y ,→ Ky ,→ ZU,y, where ZV,y ' ZU,y ' Z and the composite map is multiplication by d. But this multiplication must factor through the image of Ky ,→ ZU,y, which is just the subgroup generated by dy. So (d) ⊆ (dy); but d is minimal amongst the dx as x varies, therefore dy = d and Ky ' (d) ' ZV,y. In particular, ' . Z V K V

82 This isomorphism extends to an inclusion ZV → KV ,→ K . Let the cokernel of ZV ,→ K be D. Then since ZV ' K over V , D is supported on U \ V , and hence on U \ V , which has lower dimension than X. For if Z0 ⊂ · · · Zr is a chain of irreducible closed subsets in U \ V , then the inclusion Zr ⊂ X extends this chain, since X has been assumed irreducible. Thus the long exact sequence in cohomology coming from

0 → ZV → K → D → 0

i i shows that H (X, ZV ) ' H (X, K ) for i > n, so it will be enough to i prove H (X, ZV ) = 0 for i > n (note that the long exact sequence even n+1 n+1 shows H (X, ZV ) ' H (X, K ), since the cohomology of D vanishes in degrees greater than or equal to n).

24.4 Conclusion of the Proof

i Now we have finished our reduction. To show that H (X, ZV ) = 0 in degrees greater than n, consider the inclusion ZV ,→ Z (the constant sheaf on X), and let Q be the cokernel. Since X is irreducible, Z is flasque, so its cohomology vanishes in positive degree. The long exact sequence coming from 0 → ZV → Z → Q → 0 i i+1 thus shows that H (X, Q) ' H (X, ZV ) for all i > 0. But Q is supported on X \ V , which as above has lower dimension than X since X is irreducible29. So Hi(X, Q) = 0 for i ≥ n by induction30. Thus i H (X, ZV ) = 0 for all i ≥ n + 1, finishing the proof.

25 Friday, March 16th: Spectral Sequences - In- troduction

In the following few lectures, we will state without proof the results about spectral sequences which we will use, along with some common applications. Our aim is to obtain the following two results 1. If X is aﬃne and F a quasicoherent sheaf on X, then the higher cohomology of F vanishes (we saw the vanishing of H1 last semester). 2. If f : X → Y is an aﬃne morphism and F ∈ QCoh(X), then the i higher direct image sheaves R f∗F are zero for i > 0. This general- izes the previous result by taking Y =point.

25.1 Motivating Examples

Examples 25.1. 1. Let X be a topological space and U = {Ui} an open cover. If F ∈ Sh(X), we ask: how can we compute Hs(X, F)

29The V from above was non-empty by construction. If we were proving this for general V , the case when V is empty implies that ZV is the zero sheaf, so there’s nothing to show. 30We’re using Lemma 23.1 and Remark 2 following it again.

83 s in terms of the H (Ui, F )? Consider ﬁrst the case when s = 0. Ui Then by the sheaf property, H0(X, F) is the kernel of the map

Y 0 Y 0 31 H (Ui, F) ⇒ H (Uij , F). i i,j

For the case s = 1, let

Y 1 Y 1 K = ker H (Ui, F) ⇒ H (Uij , F) . i i,j

We also deﬁne groups (the “Cechˇ cohomology groups”)

Q 0 Q 0 ker i,j H (Uij , F) ⇒ i,j,k H (Uijk, F) Hˇ 1(U, F) = Q 0 Q 0 im i H (Ui, F) ⇒ i,j H (Uij , F) Q 0 Q 1 ker i,j,k H (Uijk, F) ⇒ i,j,k,l H (Uijkl, F) Hˇ 2(U, F) = Q 0 Q 0 im ij H (Uij , F) ⇒ i,j,k H (Uijk, F)

At this point you should ask “what actually are all those maps ⇒?” We’ll come to that later. We will see that these groups ﬁt into an exact sequence

0 → Hˇ 1(U, F) → H1(X, F) → Hˇ 2(U, F)

In the case mentioned above, namely when X is aﬃne and F quasi- Q 1 ˇ 1 coherent, the i H (Ui, F) = 0, so K = 0, and we get H (U, F) ' H1(X, F). f g 2. Consider continuous maps X → Y → Z, and a sheaf F on X. We q p would like to relate the composite right derived functors R g∗(R f∗F) n with the right derived functors of the composite R (fg)∗F. This is accomplished by the Leray spectral sequence, which we will see soon.

25.2 Deﬁnition

Definition 25.2. Let A be an abelian category, and a ∈ N.A spectral sequence starting at page a consists of the following data: pq 1. Objects Er of A , for p, q ∈ Z and r ≥ a some integer. The collection pq Er for fixed r is called the “rth page”. pq pq p+r,q−r+1 2 2 2. Morphisms dr : Er → Er such that d = 0 (whenever d makes sense). The maps go “r to the right and r − 1 down”. pq ∗ pq 3. Isomorphisms Er+1 ' H (Er ). Usually we will consider situations where p, q ≥ 0, in which case for each p, q there exists a page r beyond which the differentials d are all zero pq pq (this r will depend on p, q), so Er = Er+1 = .... We denote this common pq value by E∞ . We say then that the spectral sequence converges.

31 As usual, Uij = Ui ∩ Uj , and in the following we will denote triple intersections by U = Ui ∩ Uj ∩ U , etc. Also, we will carelessly write F even where we should actually be ijk k writing F , etc. Ui

84 ∗ n Definition 25.3. Let H = {H }n∈Z be a collection of objects of A . We pq ∗ say that Er converges to H if a) it converges in the sense described n 32 • p p+1 n,n−p above, and b) each H has a filtration F such that F /F ' E∞ . pq ∗ We use the notation Er ⇒ H in this case. pq ∗ In other words, Er ⇒ H means that the nth antidiagonal of E∞ gives the graded pieces of a filtration on Hn. Remark. Note that this information is not always enough to determine Hn. It is, for example when A is the category of k-vector spaces. On the other hand, the two complexes of abelian groups

·2 0 → Z/2 → Z/4 → Z/2 → 0 and 0 → Z/2 → Z/2 × Z/2 → Z/2 → 0

Both have ﬁltrations with quotients Z/2, but are not isomorphic complexes.

25.3 Spectral Sequence of a Filtered Complex Let K• ... 0 → Kr → Kr+1 → ... be a complex in A , with a ﬁltration by subcomplexes F i, so that

0 = F s ⊆ F s−1 ⊆ ... ⊆ F 1 ⊆ F 0 = K•.

We say that the ﬁltration is ﬁnite because the F i are eventually zero, and exhaustive because F 0 = K•. We will use frequently the following result: Theorem 25.1. In the situation above, there is a spectral sequence

pq p+q p • p+q E1 = H (gr K ) ⇒ H (K•).

pq p+q p • p+q A word on notation. When we write E1 = H (gr K ) ⇒ H (K•), we mean that the spectral sequence begins at page 1, with the first page being Hp+q(grp K•). Here grp K• means the pth graded piece of the given filtration on K•, namely F p/F p+1 (note that this, too, is a complex, so we can take cohomology). The notation for the spectral sequence gives no indication of what the differentials are, not even on the first page. To understand what they are, you must study the proof of the theorem. See Weibel’s book, section 5.4, for instance. One thing we can say is that the map Hp+q(grp K•) → Hp+1+q(grp+1 K•) is the boundary map coming from application of the Snake Lemma to the short exact sequence

0 → F p+1/F p+2 → F p/F p+2 → F p/F p+1 → 0.

32For us a ﬁltration will always mean a decreasing ﬁltration, i.e., a sequence of inclusions ... ⊂ F i+1 ⊂ F i ⊂ ...Hn.

85 Example 25.1. The “ﬁltration bˆete”on a complex

K• = K0 → K1 → K2 → ...

is as follows:

F 2 : 0 / 0 / K2 / ...

F 1 : 0 / K1 / K2 / ...

K• : K0 / K1 / K2 / ...

In this case grp K• = F p/F p+1 is the complex with the object Kp in degree p and zeroes elsewhere, which we denote Kp[−p]33 Therefore ( Kp if q = 0 Hp+q(grp K•) = 0 if q 6= 0

Thus the ﬁrst page of our spectral sequence is

0 0 0 0 ...

K0 / K1 / K2 / K3 / ...

which converges by the second page since the differentials are zero there. p0 p • But on the second page the objects are E2 = H (K ). So the Theorem tells us that the objects Hp+q(K•) have a filtration whose pth graded piece is 0 unless q = 0, in which case it is Hp(K•). In other words, Hn is filtered by Hn. Bête,indeed.

26 Monday, March 19th: Spectral Sequence of a Double Complex

26.1 Double Complex and Total Complex Today we discuss another important example of a spectral sequence. Sup- pose we have a double complex I•,•; that is, a collection of objects Ipq

33In general, whenever we write an object as standing for a complex, we mean the complex containing that object in degree zero with zeroes elsewhere. Here we have shifted the degree to make the object sit in degree p.

86 and two diﬀerentials d and δ, such that the following diagram commutes

δ p,q+1 / p+1,q+1, I O I O d d

Ip,q / p+1,q δ I

Then we can form a new complex, the total complex tot(I) of I•,•, deﬁned as follows: n M i,j (tot I) = I i+j=n The diﬀerential of this complex is

D = d + (−1)j δ Iij There are other sign conventions in the literature - other authors take D = d + δ but insist that the squares in the original double complex anticommute; the important point is that in order to get a complex, the square in the following diagram must anticommute

i,j+2 I O d

Ii,j+1 / Ii+1,j+1 O ∓δ O d d

i,j / i+1,j / i+2,j I ±δ I δ I

D2 is the sum of all the paths from the lower left to the three objects on the upper right antidiagonal - these should sum to zero. There is a ﬁltration on tot(I•,•) whose pth term is

p n M ij (F ) = I i+j=n i≥p In other words, the pth term F p of the ﬁltration is that part of the nth antidiagonal of I•,• which lies to the right of the p − 1 column. From this description, we see that the quotient F p/F p+1 is a complex whose objects are precisely the pth colum of I•,•. But Ipq sits in degree n = p+q according to the above deﬁnition, so the complex is graded by q, but with a shift down in degree by p. I.e.,

F p/F p+1 = Ip,•[−p]34

Theorem 25.1 in this situation can be stated as: Theorem 26.1. In the situation above, there is a spectral sequence

pq p+q p,• p+q •,• E1 = H (I [−p]) ⇒ H (tot I ) 34Recall our convention about grading shifts: (C[r])n = Cr+n.

87 Note that Hp+q(Ip,•[−p]) = Hq(Ip,•). Corollary 26.1. Let f : I•,• → J •,• be a morphism of double complexes such that for each p, Ip,• → J p,• is a quasi-isomorphism. Then tot I•,• → tot J •,• is a quasi-isomorphism.

Proof. The morphism f induces a morphism of spectral sequences, as follows. By assumption, the maps Hq(Ip,•) → Hq(J p,•) induced by f are isomorphisms between the first pages of the two spectral sequences. Thus when we take cohomology we get isomorphisms between the 2nd pages, and so on. So isomorphisms between the first pages give rise to isomorphisms pq pq E∞ (I) → E∞ (J) Now it is a general fact that a map of filtered complexes which induces isomorphisms on each graded piece must be an isomorphism of complexes. So the above implies that we get an isomorphism

Hn(tot I•,•) →' Hn(tot J •,•).

26.2 Application: Derived Functors on the Category of Com- plexes The above will be useful for us in the following situation. Let F : A → B be a left exact functor between abelian categories, where A has enough injectives. Let K• be a complex in A with Kp = 0 for p << 0. We want to compute RiFK• ∈ B. First resolve the complex K• by injectives:

... / Kp−1 / Kp / Kp+1 / ...

... / Ip−1,• / Ip,• / Ip+1,• / ...

Here the rows and columns are complexes, and the columns are also resolutions (i.e. exact). In particular, it’s a double complex I•,•. We can thus deﬁne RiFK• = Hi(tot FI•,•) Then of course we have to check that this does not depend on the resolution I•,•. Suppose given two injective resolutions I•,• and J •,• of the complex K•. Then one argues as in the HW a few weeks ago that it is enough to consider the case when we have a morphism I•,• → J •,• which commutes with the inclusions of K• into each double complex. Fixing p, we have injective resolutions Kp → Ip,• and Kp → J p,•. We know that the derived functors of F on objects do not depend on the resolution, i.e., the maps I•,• → J •,• induce isomorphisms Hn(FIp,•) ' Hn(FJ p,•), and hence the conditions of the corollary are met, showing that the derived functors RnF are well-deﬁned on the complex K•.

88 27 Wednesday, March 21st: Spectral Sequence of an Open Cover

Let (X, OX ) be a ringed space and F an OX -module. Fix an open cover U = {Ui}i∈I of X, and assume I is totally ordered. We use the notation

i = (i0, . . . , ik), i0 < i1 < . . . < ik for a multi-index in I, with |i| = k denoting the length (length minus one, really, since our multi-indices start with i0). For such a multi-index, deﬁne Ui = Ui0 ∩ ... ∩ Uik , and let

ji : Ui ,→ X be the inclusion. We deﬁne a complex C •(U, F) of sheaves of abelian groups on X, the Cechˆ complex of F relative to the open cover U, as follows. For each k ≥ 0, set k Y C (U, F) = ji∗F . Ui |i|=k The ﬁrst few terms are thus 0 Y C (U, F) = ji∗F Ui i∈I 1 Y (U, F) = j F C (i1,i2)∗ U ∩U i1 i2 i1

k k+1 dj : C (U, F) → C (U, F) which sends an element (a ) 7→ (a ) i i=i0i1···ik s0s1···scj ···sk s0s1···sk

That is, the component of dj ((ai)) on the s0s1 ··· sk factor is obtained by deleting the jth index sj and taking the appropriate entry of (ai). Then we deﬁne the diﬀerential on C k(U, F) by

k+1 X d = dj j=0 One checks that this defines a complex of sheaves. 1 Example 27.1. Take (X, OX ) to be Pk with its structure sheaf, and 1 U = {U0,U1} the standard open cover of P by affines. Then 0 (U, 1 ) = j ⊕ j C OP 0∗OU0 1∗OU1 1 (U, 1 ) = j C OP 01∗OU01 k (U, 1 ) = 0 for k > 1 C OP 1 Then if (a0, a1) is a local section of C , the two maps dj for j = 0, 1 send 0 1 it to a0 and a1, respectively, so the differential d: C → C maps (a0, a1) to a1 − a0.

89 The reason we care about this Cechˆ complex is the following: Proposition 27.1. There is a quasi-isomorphism of complexes F → C •(U, F). Remark. Recall that when we speak of a sheaf F as a complex, we mean the complex which has F in degree zero and zeroes elsewhere, and with diﬀerential zero. Then to say that this complex is quasi-isomorphic to some other complex K• (with Kp = 0 for p ≤ 0) means that K• must be exact except in degree zero, since the cohomology of the “complex” F is zero in higher degrees. Moreover, the kernel of K0 → K1 must be isomorphic to F because of the isomorphism of cohomology in degree zero. In other words, “F quasi-isomorphic to K•” means “K• is a resolution of F”.

Proof. First note that we have a map Y F → ji∗F Ui i∈I Given by the product of the restriction maps. This map is actually an isomorphism onto the kernel of d0 because of the sheaf axiom for F. It remains to show that the higher cohomology sheaves H q(C •(U, F)) are zero for q > 0. The question is local since we’re working with sheaves, so we can assume X is one of the Uis. In fact, after some fussing around with indices and signs, we may assume that X = U0 for simplicity. Now we deﬁne a homotopy operator

k+1 k hk : C (U, F) → C (U, F)

which sends a local section (ai0···ik+1 ) to (a0i0···ik )i0···ik . One checks that these maps satisfy

dhk + hk+1d = IdC • , that is, they deﬁne a homotopy between the identity map on C •(U, F) and the zero map. This shows that the identity map and the zero map induce the same map on cohomology, so it must be that (C •(U, F), d) is exact (except, of course, in degree zero).

Next we want to apply our results on spectral sequences to Cechˆ complexes. Take an injective resolution F → I • of F. Then the Cechˆ sheaves C (U, I ) are all injective sheaves, and we obtain a double complex

. . .O .O

C 0(U, I 1) / C 1(U, I 1) / ··· O O

C 0(U, I 0) / C 1(U, I 0) / ···

90 The rows are (injective) resolutions of the sheaves I k, by the proposition. We can regard the complex I • as a double complex I •,• with only one column. Since the maps I k → C •(U, I k) are resolutions (i.e., quasi- isomorphisms), Corollary 26.1 from the previous lecture applies, and says that the map of total complexes tot I •,• → tot C •(U, I •) is a quasi- isomorphism. But the total complex of I •,• is just I •. Thus we have a chain of quasi-isomorphisms

F' I • ' tot(C p(U, I q)) In other words, the total complex of the double complex above is a resolution of F, also. Now we take global sections all over the place. Let

q q Y C (U, F) = Γ(X, C (U, F)) = F(Ui). |i|=q

Then the quasi-isomorphisms I k → C •(U, I k) give rise to quasi-isomorphisms Γ(X, I k) → C•(U, I k) of complexes of abelian groups35. Then we have a double complex

. . .O .O

C0(U, I 1) / C1(U, I 1) / ··· O O

C0(U, I 0) / C1(U, I 0) / ···

where here the rows are resolutions of Γ(X, I k). By the same argument as above, we ﬁnd that the total complex tot C•(U, I •) is a resolution of Γ(X, I •), so we can use this total complex to compute the cohomology groups of F on X:

H∗(tot C•(U, I •)) ' (H∗(X, F)). On the other hand, our discussion of the spectral sequence of a double complex showed that there is a spectral sequence

pq q p • p+q • • E1 = H (C (U, I )) ⇒ H (tot C (U, I )) But

q p • q Y • 36 Y q • Y q H (C (U, I )) = H I (Ui) = H (Ui, I ) = H (Ui, F) |i|=p |i|=p |i|=p

35This is only true because we began with resolutions of injective objects I k. In general, if K• and K0• are quasi-isomorphic complexes of injectives, then application of a left exact functor F gives quasi-isomorphic complexes FK• and FK0•, but the fact that both K• and K0• are injective is essential. 36Not sure under what circumstances we can interchange cohomology and products like this.

91 In conclusion, we have constructed a spectral sequence pq Y q p+q E1 = H (Ui, F) ⇒ H (X, F) |i|=p This is the spectral sequence of the open cover U.

28 Friday, March 23rd: Cechˆ Cohomology Agrees with Derived Functor Cohomology in Good Cases

Last time, given an open cover U = {Ui}i∈I of a topological space X, and a sheaf F of abelian groups on X, we constructed a spectral sequence pq Y q p+q E1 = H (Ui, F) ⇒ H (X, F) |i|=p Recall that the “global” Cechˆ complex consisted of groups p Y C (U, F) = F(Ui), |i|=p with differentials defined in the same way as for the sheafy version C • last time. We call the cohomology groups Hp(C•(U, F)) the Cechˆ cohomology of F relative to the open cover U, and denote them by Hˇ p(U, F). Today we will use the above spectral sequence to prove that in good circmstances, the Cechˆ cohomology groups agree with the cohomology groups defined by derived functors of Γ(X, −). Let us write out the first few pages of the “global” spectral sequence from last time. The E1 page looks like

. . . .

Q 1 γ Q 1 i H (Ui, F) / i

C0(U, F) / C1(U, F) / ···

0 0 ··· The second page looks as follows, letting K = ker γ.

. . . .

K XXX ··· VV ······ XXXX ∂ VVVV XXXXX VVVV XXXXX+ VVVVV Hˇ 0(U, F) Hˇ 1(U, F) Hˇ 2(U, F) VV+ ··· XXXX WWW XXXXX WWWWW XXXXX WWWW 0 0XXX+ 0 WW+ ···

92 The map in to Hˇ 0(U, F) is also zero, so this shows that Hˇ 0(U, F) is the only term in our ﬁltration of H0(X, F), so

Hˇ 0(U, F) ' H0(X, F)

To study the ﬁltrations on the higher cohomology groups Hi(X, F), we look now at the third page:

. . . .

ker ∂ ········· R RRR RRR 0 1RRR Hˇ (U, F) Hˇ (U,RFRR) ······ RRR RRR RRR 0 0 0R( 0 Note that now the ﬁrst antidiagonal p + q = 1 stabilizes, as the maps into and out of both ker ∂ and Hˇ 1(U, F) are zero hereafter. So we know that there is a ﬁltration on H1(X, F):

0 = F 2 ⊆ F 1 ⊆ F 0 = H1(X, F) with

0 1 0,1 0,1 F /F ' E∞ = E3 ' ker ∂ 1 2 1,0 1,0 ˇ 1 F /F ' E∞ = E3 ' H (U, F)

Since F 2 = 0, we have

ker ∂ ' F 0/F 1 ' F 0/F 2/F 1/F 2 ' H1(X, F)/Hˇ 1(U, F)

So we get an exact sequence

0 → Hˇ 1(U, F) → H1(X, F) → ker ∂ → 0.

It turns out that the map on the right, H1(X, F) → ker ∂, is compatible with the inclusions

Y 1 ker ∂ ,→ K,→ H (Ui, F) i Thus we obtain a commutative diagram

0 / Hˇ 1(U, F) / H1(X, F) / ker ∂ / 0 GG GG GG K GG GG# Q 1 i H (Ui, F)

This is good because it relates H1(X, F) to Hˇ 1(U, F), but at the expense Q 1 of the “smaller” groups i H (Ui, F). Of course, what we want is a way to

93 1 get a handle on these H (Ui, F). More generally, we would like to handle i the groups H (Ui, F), where i > 0 and the multi-index i has arbitrary length. A first step in this direction is to take the cover U to consist of open affines. The next theorem shows why this is useful. It is also the first of the two desired results mentioned at the beginning of our discussion of spectral sequences. Theorem 28.1. Let X be an affine scheme and F ∈ QCoh(X). Then for all q > 0, Hq(X, F) = 0.

Proof. We proceed by induction on q. The case when q = 1 was established last semester: “the global sections functor is exact on affines”. So now assume that for any affine scheme X and any quasicoherent sheaf F on X, Hi(X, F) = 0 for 1 ≤ i ≤ q. We fix an affine scheme X, with a 37 cover by open affines X = U0 ∪ ... ∪ Ur. We can arrange it so that all possible intersections of the Ui are also affine (by taking them to be basic opens, say). We consider the spectral sequence from before, where by our i inductive assumption, all H (Ui, F) are zero, where i is a multi-index and 0 < i ≤ q. The first page is thus

Q q+1 Q q+1 i H (Ui, F) / i

0 0 0

C0(X, F) / C1(X, F) / ··· We include only two rows of zeroes for simplicity, but of course there will be more for larger q. These rows are zero by our inductive assumption. We now proceed to page two. The ﬁrst row will consist of the groups Hp(C•(U, F)), which are the right derived functors of Γ(X, −), applied to the “Cechˆ sheaves” C •(U, F). These sheaves are quasicoherent, so since X is aﬃne, Γ(X, −) is exact and the higher right derived functors vanish. So the second page is K ······

0 0 0

H0(X, F) 0 0

Q q+1 Here K is just some subgroup of i H (Ui, F). Since this spectral sequence converges to Hp+q(X, F), we have that Hq+1(X, F) ' K,→ Q q+1 i H (Ui, F). It turns out that, by the construction of the spectral sequence (which we did not go into in any detail), this map

q+1 Y q+1 H (X, F) ,→ H (Ui, F) i 37Recall that an affine scheme is quasicompact, so we may take a finite cover here, although we will not use the finiteness.

94 is induced by the restriction maps along Ui ,→ X. We want to use this to show that Hq+1(X, F) = 0, for which we need the following lemma. Lemma 28.1. IF X is any scheme, and α ∈ Hi(X, F) for i > 0. Then S there is an open cover X = j Uj such that the image of α in each i H (X,Uj ) is zero.

Proof. First choose an injective resolution F → I •. Our element α ∈ i i i+1 H (X, F) can be represented by an element αe ∈ ker(Γ(X, I ) → Γ(X, I )). Since the copmlex of sheaves I • is exact, we can ﬁnd a neighborhood U i−1 i of any point in X such that α is in the image of I (U) → I (U). S e U Now choose an open cover j Uj of X by such U. Then we have [αe] = 0 i in H (Uj , F), for each j, as desired.

Using the Lemma, we may replace, i.e., reﬁne, our original cover to get q+1 q+1 some new Uj such that α ∈ H (X, F) restricts to 0 in each H (Uj , F). q+1 Q q+1 But the map H (X, F) ,→ i H (Ui, F) is injective, so α is zero, hence Hq+1(X, F) = 0.

Corollary 28.1. If f : X → Y is an affine morphism, and F a quasico- i herent sheaf on X, then R f∗F = 0 for all i > 0. The proof follows immediately from the theorem and the following lemma. Lemma 28.2. If f : X → Y is a morphism of schemes, and F a sheaf i on X, then the higher pushforward R f∗F is the sheaf associated to the presheaf on Y i i −1 TF : U 7→ H (f (U), F). Proof. My notes for Professor Olsson’s argument are unclear here. They include only the cryptic phrase “sheafification commutes with quotients”. See Hartshorne III.8.1 for a different argument.

Now we make good on the title of today’s lecture r Theorem 28.2. If X is a scheme and U = {Ui}i=1 a cover by open aﬃnes with all possible intersections aﬃne38, then for any quasicoherent sheaf F on X, we have Hn(X, F) ' Hˇ n(U, F) Proof. Here the spectral sequence of the open cover, namely

pq Y q p+q E1 = H (Ui, F) ⇒ H (X, F), |i|=p is especially simple. On the first page, all the rows except the first are zero, by Theorem 28.1. On the second page, the sequence has already converged, and the nonzero stable groups are the Hˇ n(U, F). By the theorem on this spectral sequence, these groups form a one term filtration of Hn(X, F), which proves it.

38This happens, for instance, when X is separated.

95 ∗ 1 39 Example 28.1. We’ll use the above to compute H ( k, O 1 ) . We have P Pk 1 a cover of P by two aﬃnes Spec k[x] and Spec k[y], whose intersection is Spec k[x±], where y corresponds to 1/x on this intersection. The Cechˆ complex is

C0(U, O) = k[x] × k[y] → k[x±] = C1(U, O) where this map is given by (p(x), q(y)) 7→ p(x)−q(1/x). The kernel of this map consists of the constants k '{(c, c)} ⊂ k[x] × k[y]. It is surjective, so the cokernel is zero. Therefore ( n 1 k i = 0 H (Pk, O) ' 0 else

29 Monday, April 2nd: Serre’s Aﬃneness Crite- rion

Today we prove a useful result of Serre that uses cohomology of ideal sheaves to test whether a Noetherian scheme is affine. The idea of the proof introduces the point of view that cohomology groups should be thought of as containing “obstructions” to various things. We begin with a few more examples to review the concepts from last time. Recall that the hypotheses of Theorem 28.2 are automatically sat- isfied if X is a scheme over R which is separated. The reason is that for two open affines U, V ⊂ X, we have a diagram

U ∩ V / U ×R V

∆ X / X ×R X

and when X/R is separated, ∆ is a closed immersion, hence so is the top arrow. But a closed immersion is an affine morphism and U ×R V is affine, hence so is U ∩ V . Example 29.1. Let X be the affine line with doubled origin, 1 ` 1 . Ak Gm Ak This scheme is not separated, but nonetheless we have a cover by two affines, with affine intersection, so the theorem applies. Example 29.2. If X is affine, then we can take the open cover to consist of just X itself, so Theorem 28.2 applies and Hˇ i = Hi. Moreover, the Cechˆ groups are zero in positive degree since our open cover has only one element. This isn’t really an application of the theorem, however, since we used this fact in the proof; but we see another point of view on why affine schemes have no higher cohomology. Now to the main course: Theorem 29.1 (Serre). If X is a Noetherian scheme, the following are equivalent: 1. X is affine.

39 1 Hereafter we’ll drop the subscript Pk from the notation of the structure sheaf.

96 2. Hi(X, F) = 0 for all quasicoherent sheaves F on X and all i > 0. 3. H1(X, I ) = 0 for all quasicoherent ideal sheaves I on X.40

Proof. All that needs to be shown is that 3 implies 1. For this we use the following two facts 1. (Hartshorne II.2.17) X is aﬃne if and only if there are global sections

f1, . . . , fr of OX such that for each i, Xfi is aﬃne and the ideal (f1, . . . , fr) = Γ(X, OX ). 2. If X is a Noetherian scheme, and U an open subset which contains all the closed points of X, then U = X.

Now we produce the sections fi “one point at a time.” Specifically, pick a closed point P ∈ X, and an affine neighborhood U of P . Let Y = X \ U, and give it a scheme structure (the reduced one, say). We have the skyscraper sheaf k(P ) at P , and it fits into a short exact sequence

0 → IY ∪P → IY → k(P ) → 0.

This comes from the fact that Y is a closed subscheme of Y ∪ P , so we have an inclusion IY ∪P ,→ IY . This is an isomorphism away from P ; the cokernel is then just the sheaf k(P ). Taking cohomology gives the long exact sequence

0 → Γ(X, IY ∪P ) → Γ(X, IY ) → k(P ) → 0

where the rightmost term is zero because of our assumption (3) that H1 vanishes for quasicoherent ideal sheaves. Thus the map Γ(X, IY ) → k(P ) is surjective, so we can find a global section f of IY which does not vanish at P (namely a preimage of any nonzero element of k(P )). This means that P ∈ Xf , and moreover, since f vanishes on Y , we have that Xf ⊆ U. Therefore the points where f vanishes are exactly the points where the image of f in OX /IY ' OU vanishes, which is just Uf . So Xf is affine. So we can cover all the closed points of X by these affine sets Xf . Therefore their union is all of X, by fact 2. Again using the Noetherian hypothesis, we see that finitely many such Xf suffice to cover X. Thus S we have produced global sections f1, . . . , fr such that X = Xfi . It remains to check that (f1, . . . , fr) = Γ(X, OX ). The global sections fi define a map r OX → OX P by sending (a0, . . . , ar) to aifi. This map is surjective since we can

check at stalks, and at each point of X, not all fi vanish since the Xfi cover X. So we obtain an exact sequence

r 0 → K → OX → OX → 0

r for some subsheaf K of OX . Looking at the induced long exact sequence in cohomology, we see that if we can show that H1(X, K ) = 0, then we r will have a surjection Γ(X, OX ) → Γ(X, OX ), which will ﬁnish the proof.

40 (In reply to a question) An example of an ideal sheaf which is not quasicoherent: j!I ⊂ OX , where j : U,→ X is any proper open subset, and I ⊂ OU an ideal sheaf on U.

97 The idea is to filter K in such a way that we can apply our assumption 3 to the quotients of the filtration. The filtation

r−2 r−1 r OX ,→ ... OX ,→ OX ,→ OX , where Or−j has local sections of the form (∗,..., ∗, 0,..., 0). This gives rise to a ﬁltration of K :

Kr ,→ ...,→ K2 ,→ K1 ,→ K

r−i where Ki = OX ∩ K . The local sections of Ki/Ki+1 have zeroes in all r−i r−i−1 but one slot, so they are isomorphic to a subsheaf of OX ' OX /OX . More precisely, we have the following diagram

0 0 ker α

0 / Ki+1 / Ki / Ki/Ki+1 / 0

α r−i−1 r−1 0 / OX / OX / OX / 0

r−i−1 r−1 0 / OX /Ki+1 / OX /Ki

0 0

The map on the bottom is injective since the upper left square is cartesian. From the snake lemma, we obtain an exact sequence

r−i−1 r−1 0 → ker α → OX /Ki+1 → OX /Ki → ...

r−i−1 r−1 and since the map OX /Ki+1 → OX /Ki is injective, ker α = 0. This shows that Ki/Ki+1 is a subsheaf of OX . We now use this to show that 1 H (X, Ki) = 0. The long exact sequence induced by 0 → Ki+1 → Ki → Ki/Ki+1 → 0 gives, in degree 1,

1 1 1 ... → H (X, Ki+1) → H (X, Ki) → H (X, Ki/Ki+1) → ...

1 Now since Ki/Ki+1 is an ideal sheaf, H (X, Ki/Ki+1) = 0 by our assump- 1 tion of (3). Assume inductively that H (X, Ki+1) = 0; then it follows that 1 1 H (X, Ki) = 0. So we get that H (X, K ) = 0 by decreasing induction on i. As we mentioned above, this suffices to finish the proof that X is affine.

98 30 Wednesday, April 4th: Cohomology of Pro- jective Space

We begin with a quick warm-up, using the criterion of last time to show 2 that X = Ak \{(0, 0)} is not aﬃne. By 29.1 it will be enough to ﬁnd a quasicoherent sheaf F on X for which H1(X, F) 6= 0. Using the open cover

± U1 = Spec k[x , y] ± U2 = Spec k[x, y ] ± ± U1 ∩ U2 = Spec k[x , y ], we have

1 (a,b)7→a−b H (X, F) = coker F(U1) × F(U2) −→ F(U1 ∩ U2)

−1 −1 Now we can take F = OX . Then x y ∈ OX (U1 ∩ U2) is not in the 1 image of the diﬀerence map, so H (X, OX ) is non-zero (it’s not even ﬁnite- dimensional). Now we turn to the calculation of the cohomology of the sheaves r O r (n). Let A be a ring, S = A[x0, . . . , xr], and set X = PA = Proj S. P L We use the notation Γ∗(OX ) for the graded ring n≥0 Γ(X, OX (n)). ' Theorem 30.1. (a) There is an isomorphism S → Γ∗(OX ) induced by the universal sections s0, . . . , sr ∈ Γ(X, OX (1)). i (b) H (X, OX (n)) = 0 for 0 < i < r and i > r. r 41 (c) H (X, OX (−r − 1)) ' A . (d) For each n, the map

0 r r H (X, OX (n)) × H (X, OX (−n − r − 1)) → H (X, OX (−r − 1)) ' A is a perfect pairing42 Remarks. 1. The theorem allows us to compute all cohomology groups of OX (n) for any n. 0 2. The pairing of (d) is defined as follows. If s ∈ H (X, OX (n)), it defines a map of sheaves OX → OX (n), or equivalently, a map OX (−n) → OX . Tensoring with OX (−r − 1) gives a map OX (−n − r − 1) → OX (−r − 1). Taking rth cohomology gives our map r r H (OX (−n − r − 1)) → H (X, OX (−r − 1)). L 43 i Proof. Let F = n∈ OX (n) . We will compute H (X, F). The stan- Z r dard open cover U of PA satisfies the hypotheses of Theorem 28.2, so we can use the Cechˆ complex to compute this. Moreover, formation of the Cechˆ complex commutes with direct sums, so we have

i M i H (X, F) = H (X, OX (n)), n∈Z 41Not canonically. 42 0 r ∨ This means it induces an isomorphism H (X, OX (n)) ' H (X, OX (−n − r − 1)) . 43 r+1 This sheaf is actually the pushforward of the structure sheaf of X = AA \{O} along r the map X → PA.

99 i so we will be able to recover the cohomology groups H (X, OX (n)) for each n from our calculation. More precisely, the Cechˆ complex looks like

r M M F(Ui) → F(Uij ) → ... i=1 i

Note that for each n, the sections of OX (n) over the open set Ui = D+(xi) are jsut the degree n part of Sxi . Summing over all n gives F(Ui) ' Sxi ,

F(Uij ) ' Sxixj , etc., so we are interested in the following graded complex of graded S-modules

r M M Sxi → Sxixj → ... i=1 i

Now the cohomology groups of this complex are by deﬁnition Hi(C•(X, F))). But because the diﬀerential in the complex is just localization, it is compatible with the grading of the individual groups so the degree n piece of i • i H (C (X, F))) is just H (X, OX (n)). We now use the observations to prove the result, in the following order: (a),(c),(d),(b). We’ve basically already seen (a) earlier. Here’s an example in the case r = 1. then H0 is the kernel of the map

A[x±, y] × A[x, y±] → A[x±, y±], which is isomorphic to A[x, y]. For (c), since the Cechˆ complex vanishes in degree r + 1 and larger, we have

r r Y H (X, F) = coker S → S . x0···xck···xr x0···xr k=0

l0 lr But Sx0···xr is a free A-module with basis {x0 ··· xr | li ∈ Z}, so this l0 lr cokernel is free with basis {x0 ··· xr | li < 0}. Looking at the degree n P piece, we see that since each li < 0, li ≤ −(r + 1), so

r H (X, OX (n)) = 0 whenever n > −r − 1, and for n = −r − 1, the only monomial in the basis −1 −1 r is x0 ··· xr , so H (X, OX (−r −1)) is a free A-module of rank one. This proves (c). Now for (d). First, if n < 0, then OX (n) has no global sections, and r since −n − r − 1 > −r − 1, H (X, OX (−n − r − 1)) = 0, as was observed in the previous paragraph. So there is nothing to show. When n = 0, 0 r H (X, OX ) ' A and H (X, OX (−r − 1)) ' A by (c). For each global section s ∈ A we get a map

r r A ' H (X, OX (−r − 1)) → H (X, OX (−r − 1)) ' A

0 r and this association deﬁnes an isomorphism of H (X, OX ) with H (X, OX (−r− ∨ r 1)) since H (X, OX (−r − 1)) is free of rank one. 0 Now we consider the case when n > 0. Here H (X, OX (n)) is free with m0 mr P r basis {x0 ··· xr | mi ≥ 0, mi = n}, while H (X, OX (−n − r − 1)) has

100 l0 lr P basis {x0 ··· xr | li < 0, li = −n − r − 1}. The pairing acts on these bases by m0 mr l0 lr m0+l0 mr +lr hx0 ··· xr , x0 ··· xr i = x0 ··· xr r m0+l0 mr +lr But in H (X, OX (−r − 1)), x0 ··· xr is zero unless each mi + li is -1, in which case it is 1. So these form dual bases with respect to the pairing, which is therefore perfect. This proves (d). We will prove (b) next time.

31 Friday, April 6th: Conclusion of Proof and Applications

We will prove (b) by induction on r. Note that the statement for i > r is true, regardless of r, simply because the Cechˆ complex is zero after the rth degree. If r = 1, the claim is vacuous. • ˆ Note that (U, F)xi is the Cech complex of F with respect to the C Ui r open cover Ui = {Ui ∩ Uj }j=1. Therefore since Ui is aﬃne, it is an exact complex except in degree 0. This means that for each j, an for all i > 0, i H (X, F)xj = 0. But to say that this localization is zero is to say that i the S-module H (X, F) is annihilated by xj , so to prove (b), it will be enough to show that

i i ·xj : H (X, F) → H (X, F) is injective for each 0 < i < r, 0 ≤ j ≤ r. Now xj ∈ Γ(X, OX (1)) gives a map OX (n − 1) → OX for each n, so it deﬁnes a short exact sequence of graded S-modules

·xj 0 → S(−1) → S → S/(xj ) → 0.

r This gives rise to a short exact sequence of sheaves on P

0 → F(−1) → F → FH → 0,

r−1 r where H = V+(xj ) ' PA ⊂ PA. Taking cohomology gives the long exact sequence

·x 0 j 0 0 H (X, F) / H (X, F) / H (H, FH ) g ggggg ggggg ggggg ggggg 1 gs g 1 1 H (X, F) / H (X, F) / H (H, FH ) hhh hhhh hhhh hhhh hhhh hhhh . hs hhh . . . . .

·xj Hr−1(X, F) / Hr−1(X, F)

101 0 0 First, the map H (X, F) ' S → H (H, FH ) ' S/(xj ) is surjective, so the map ·x H1(X, F) −→j H1(X, F) is injective. Next, we may assume by our induction hypothesis that i H (H, FH ) = 0 for 0 < i < r − 1, so the right hand column consists of zeroes except in the top row. Therefore each map

·x Hi(X, F) −→j Hi(X, F) for i = 2, 3, . . . , r − 1 is injective. This ﬁnishes the proof.

Example 31.1. Let F ∈ k[x0, . . . , xr] be a homogeneous polynomial of degree d, and consider the hypersurface

i r X = V+(F ) ,→ Pk. i r r We want to calculate H (X, OX ). The global section F ∈ Γ(P , OP ) deﬁnes an exact sequence

·F r r 0 → OP (−d) → OP → i∗OX → 0 i i r Recall that since i is a closed immersion we have H (X, OX ) = H (P , i∗OX ). So the long exact sequence can be written as

0 r 0 r 0 H (P , O r (−d)) / H (P , O r ) / H (X, OX ) P P g ggggg ggggg ggggg s ggggg 1 r g 1 r 1 r / r / H (P , OP (−d)) H (P , OP ) H (X, OX ) ggg ggggg ggggg ggggg ggggg . gs gggg . . . . .

r r α r r r r r H (P , OP (−d)) / H (P , OP ) / H (X, OX ) We have already computed the ﬁrst two columns; the left column is zero except in degree r, the middle column is zero except in degrees 0 r r 0 and r. Moreover, in degree zero, H (P , OP ) ' k. Also, we know i H (X, OX ) = 0 in degrees greater than r−1 for dimension reasons. Hence we have

0 H (X, OX ) ' k i H (X, OX ) ' 0 for r 6= 0, r − 1

r−1 The only tricky part is to determine H (X, OX ), which is the kernel r r r r r r of the map α: H (P , OP (−d)) → H (P , OP ). Recall that the global section F deﬁnes a map

r r ·F : OP → OP (d).

102 Twisting by −r − 1 and taking global sections we have

0 r F 0 r r r H (P , OP (−r − 1)) −→ H (P , OP (d − r − 1)) Taking duals gives a map

0 r ∨ F 0 r ∨ r r H (P , OP (d − r − 1)) −→ H (P , OP (−r − 1)) .

r r r r r r Applying Serre duality, we get a map H (P , OP (−d)) → H (P , OP ). Unwinding the deﬁnitions, it turns out that this map is in fact α. So 0 r F r the kernel of α is isomorphic to the kernel of H (P , OP (−r − 1)) −→ 0 r r H (P , OP (d − r − 1)). Example 31.2. Specializing the previous example, let F be the polyno- 2 3 2 3 2 mial zy − (x − axz + bz ); this deﬁnes an elliptic curve E ⊂ Pk. We have

0 h (E, OE ) = 1 ! 1 2 h (E, OE ) = = 1 2

Example 31.3. Recall that the canonical sheaf ωX is the top exterior 1 r power of ΩX ; it’s a line bundle on X. In the case of P , we’ve calculated 1 1 that ω r is the line bundle r (−r − 1). For , we have ω 1 = Ω = P OP P P 1 1 1 1 1 P 2 (−2); so H ( , ω 1 ) = H ( , 1 (−2)). OP P P P OP

32 Monday, April 9th: Finite Generation of Co- homology

Today we discuss the question of whether the cohomology groups of a coherent sheaf are ﬁnitely generated modules over the base ring. The answer is yes in good circumstances. Theorem 32.1. If A is a Noetherian ring, X/A projective, L is a very ample sheaf on X, and F is any coherent sheaf on X, then (a) For all i > 0, Hi(X, F) is a ﬁnitely generated A-module.

(b) There is an integer n0 such that for all i > 0 and n ≥ n0,

Hi(X, F ⊗ L ⊗n) = 0.

r Remark. Recall that L is very ample if there is an immersion i: X → PA ∗ r such that i OP (1) ' L . In the following we will write L as OX (1), in order to use the simpler notation F(n) = F ⊗ L ⊗n. The theorem also admits the following generalization. Theorem 32.2. 1. If A is a Noetherian ring and X is proper over A, then for any coherent sheaf F on X and all i > 0, Hi(X, F) is a ﬁnitely generated A-module. 2. If f : X → Y is a proper morphism of locally Noetherian schemes, i then for any coherent sheaf F on X and all i > 0, R f∗F is a coherent sheaf on Y .

103 Statement 2 is the most general; this is proved by reducing it to statement 1, and then reducing that to the proof of the previous theorem, which as we will now see, involves a reduction to the calculation of the cohomology of projective space. We will need the following technical lemma r Lemma 32.1. Let F be a coherent sheaf on P and let s ∈ Γ(D+(xi)). m Then there is an integer m such that sxi extends to a global section of r r Γ( , F(m)), i.e. there is a section s ∈ Γ( , F(m)) such that s = P e P e D+(xi) m sxi ∈ Γ(D+(xi), F(m)). L Proof. Here’s a sketch of the idea. Given coherent F, set G = n≥0 F(n). By the sheaf property, we have the following two equalizer diagrams, where the vertical arrows are the localizations maps Q Q G(X) / j G(D+(xj )) / j

Q Q G(D+(xi)) / j G(D+(xixj )) / j

One can lift sections up along the vertical maps by multiplying by a suﬃciently high power of xi. Use this and a diagram chase to lift s ∈ G(D+(xi)) to a section of G(X).

r Corollary 32.1. If F is a coherent sheaf on PA, then F admits a surjection E → F, where E is a ﬁnite direct sum of sheaves of the form r OP (q).

Proof. If we have generators s1, . . . , st for F over D+(xi), multiply each one by an appropriate power of xi to get a section of F(m). These sections t deﬁne a map r → F(m), where this m is the maximum of the twists OP involved in lifting the various sj . This map is surjective over D+(xi). r r Tensoring with OP (−m) gives a map of sheaves OP → F which is surjective over D+(xi). Do this oover each D+(xi), and then take the direct sum to prove the corollary.

r Proof of Theorem 32.1. First we reduce to the case X = PA. Let i: X → r PA be the closed immersion. Since i∗ is exact and has an exact left adjoint, it doesn’t change the cohomology. Also, it preserves coherence, since this can be checked locally. If on some open affine, F corresponds to the B/I- module M, then i∗F corresponds to the module MB obtained by base change along the ring map B → B/I, which is still finitely generated i i r since this is a surjective ring map. Thus H (X, F) ' H (P , i∗F), so we r may assume X = P . i r The result is true for i > r: in that case H = 0 since P can be covered by r + 1 affines44. We also know the result in the case that F i r is OX (n), since we have explicitly calculated H (P , OX (n)); we also get 44Note - we cannot argue that Hi = 0 for i > r because of dimension reasons, since A is r arbitrary, so it is difficult to get our hands on the dimension of P .

104 the result when F is a finite direct sum of copies of these since Hi is an additive functor. Now for the inductive step: assume that for some fixed i0, and for all i r coherent G, H (P , G) is finitely generated whenever i > i0. We will use i r this to show that H 0 (P , F) is finitely generated. Now use the lemma L r above to find an E = i OP (qi) and a morphism E → F, and let K be the kernel. Then we have an exact sequence

0 → K → E → F → 0, which gives a long exact sequence in cohomology in which the following terms appear:

i r α i r β i +1 r ... → H 0 (P , E ) → H 0 (P , F) → H 0 (P , K ) → .... i r Note that H 0 (P , E ) is finitely generated, as we observed above, and that i +1 r H 0 (P , K ) is finitely generated by induction. Since this sequence is exact, we can break off the following short exact sequence

i r 0 → im α → H (P , F) → ker β → 0, and both im α and ker β are finitely generated, since kernels and images of finitely generated modules over a Noetherian ring are also finitely generated. This proves (a). For (b), we only provide a sketch; the crux of the argument is the same as (a). We will need the following result, which is useful enough to isolate as a separate lemma.

Lemma 32.2 (Projection Formula). If i:(X, OX ) → (Y, OY ) is a morphism of ringed spaces, F an OX -module, and E a locally free OY -module of ﬁnite rank, then

∗ i∗(F ⊗OX i E ) ' (i∗F) ⊗OY E A similar argument was made for this statement in the special case r when E is a line bundle in an earlier lecture. In our case, taking Y = PA, r and E = OP (1), we have ⊗n ⊗n i∗F(n) = i∗(F ⊗O OX (1) ) = (i∗F) ⊗O r O r (1) = (i∗F)(n), X P P r which shows that we can assume that X = PA. Now, the statement of (b) is true when i > r, as in (a). We also know that for F = i r OX (n), H (X, OX (n) = 0 for 0 < i < r and H (X, OX (n)) is dual to 0 H (X, OX (−n − r − 1)), which shows that the statement holds when F = OX (n); it holds for ﬁnite direct sums of these again by additivity of Hi. For the inductive step ﬁx i0, and assume that for any G and any i > i0, i there is an integer n0 such that H (X, G(n)) = 0 if n ≥ n0. Then one i0 shows that there is an m0 such that H (X, F(m)) = 0 for m ≥ m0, by arguments similar to those of (a).

Note in conclusion that the integer n0 of (b) in principle depends on all of X, A, L , and F, but the essential dependence is on F. In particular, we can always twist F down to force larger values of n0.

105 33 Wednesday, April 11th: The Hilbert Polyno- mial

Fix a projective scheme X over a Noetherian ring A. Let OX (1) be a very ample sheaf. For a coherent sheaf F on X, we deﬁne the Euler characteristic of F to be the integer

X i i χ(F) = (−1) h (X, F). i Note that by the results of the previous lecture, each hi is finite, and there are only finitely many nonzero terms, so the sum makes sense. For such F, we also define the Hilbert polynomial of F to be the function pF : Z → Z given by

pF (n) = χ(F(n))

0 Note that for suﬃciently large n, we have pF (n) = h (X, F(n)), again using the results of the previous lecture. To justify calling this function a polynomial, we prove the following

Theorem 33.1. pF is a polynomial function.

r Proof. First we reduce to the case where X = PA. Pick a closed immersion r ∗ r i: X,→ PA such that i OP ' OX (1). This can be done since OX (1) is very ample. Then as we saw last time, the projection formula implies r (i∗F)(n) = i∗(F(n)), so pi∗F = pF , and we may replace X by PA. r Now pick a hyperplane H ⊂ X = P ; this gives us an exact sequence

0 → OX (−1) → OX → OH → 0 where we are omitting the pushforward of OH from the notation. Since tensoring with F is right exact we get another exact sequence

0 → → F(−1) → F → F → 0, K H where K is deﬁned as the kernel of F(−1) → F. Let I denote the image of the map F(−1) → F. We can break this sequence into two short exact sequences

0 → K → F(−1) → I → 0 0 → → F → F → 0 I H Since χ is additive on short exact sequences (we proved this for curves a few weeks ago), we get two equations

pK + pI = pF(−1)

pI + pF|H = pF . The diﬀerence of these two equations is the equality

pF|H − pK = pF − pF(−1).

By induction, we may assume pF|H is a polynomial. Also, the sheaf K is supported on H, and a coherent sheaf supported on a closed subscheme

106 can be realized as the pushforward of a coherent sheaf on that closed subscheme (see lemma below for details). Thus we may also assume by induction that pK is a polynomial. Finally, note that pF(−1)(n) = pF (n − 1). Thus the equation above shows that pF (n) − pF (n − 1) is a polynomial function. In fact (check) this implies that pF itself is a polynomial function.

To explain our treatment of F in the above, we have the following H lemma and its corollary. Lemma 33.1. Let X be a Noetherian scheme, F a coherent sheaf on X with set-theoretic support Z ⊂ X. Give Z the reduced scheme structure, r and let I ⊂ OX be the ideal sheaf of Z. Then there exists r such that I annihilates F. Corollary 33.1. F is isomorphic to the pushforward of a coherent sheaf on V (I r).

Proof. The question of whether, for given r, I r annihilates F can be checked locally. Having produced such r locally, we may take a finite cover of X by open affines and use the maximum of all the r obtained in our local constructions. Thus we may as well let X = Spec A. For I ⊂ A an ideal and F a finitely generated A-module, the condition that Fe is supported on V (I) means that, if g1, . . . , gs are generators for I, then

Fe is 0 on D(g1) ∪ ...D(gs), i.e., Fgi = 0 for each i. Let m1, . . . , mt be ni generators for F . Since Fgi = 0, there is an ni such that gi mj = 0 for all j. Let n be the maximum of these ni as i varies from 1 to s. Thus n gi mj = 0 for all i, j. a1 as We can choose r big enough so that, for any monomial g1 ··· gs of degree r, at least one of the ai is at least n. Then for this r, any monomial of degree r annihilates F , so IrF = 0. This proves the lemma. To see the corollary, observe that if IrF = 0, then F has the structure of an A/Ir-module, so the sheaf Fe on Spec A can be viewed as the pushforward of a sheaf Fe0 on Spec A/Ir along the ring map A → A/Ir. Globalizing, we have a closd immersion i: V (Ir) ,→ X. The action of −1 −1 −1 i OX on i F factors through OV (Ir ), and i F is still coherent as −1 an OV (Ir )-module. Moreover, the map F → i∗i F is an isomorphism of OX modules, so we can view F as the pushforward of the coherent −1 OV (Ir )-module i F.

107