Commutative Algebra and

Algebraic Geometry II

Dmitry Gourevitch. Monday 10:00-13:00, room 104. Website.

(please send corrections to: clara.shikhelman@gmail.com )

Curves

Denition A curve is an irreducible separated 1 dimensional algebraic variety.

Denition For a variety X and x ∈ X we dene OX,x =rational functionsdened at x = (U,f)|f∈O(U),x∈U/∼ where (U, f) ∼ (V, g) ⇐⇒ f |(U∩V ) =g|(U∩V ), (U, f) + (V, g) = (U ∩ V, f + g) and (U, f) · (V, g) = (U ∩ V, f · g).

Lemma 1 Let x1, x2 ∈ X, X irreducible separated. If OX,x1⊂ OX,x2

⊂ K(x)(where K(x) are all the rational functions), and mx1

⊂ mx2then x1 = x2.

Lemma 2 mX,x ≡ (U, f) ∈ OX,x | f(x) = 0 is the only maximal ideal inOX,x.

Proof 2 (U, f) ∈ OX,x is invertible ⇐⇒ f(x) 6= 0

Exercise

• Show that OX,x is Noetherian.

• X = sepcA, mx ⊂ A, show OX,x = Amx≡ S−1A where S = A\mx.

• Is it possible that OX,x1 ⊂ OX,x2 ⊂ K(x) but mx1 6⊂ mx2?

Proof (lemma 1) x1 ∈ V1 ⊂ X, V1 open ane. Let f1, .., fd ∈ O(V1) be thegenerators. Since fi ∈ OX,x1

⊂ OX,x2, ∃ open ane V2 3 x2 s.t. fi ∈ O(V2),

thus O(V1) ⊂ O(V2) and V1∩V2 = specO(V2) = V2, since X is separated ⇐⇒ ∀ane open U, V : U∩V = specO(U)⊗O(V ). So V1 ⊃ V2, thus x1, x2 ∈ V1. Nowif f ∈ O(V1), f(x1) = 0⇒ f ∈ mx1 ⇒ f ∈ mx2 so f(x2) = 0⇒ x2 ∈ x1 = x1.

[8.4.13]

We will go through some algebraic lemmas that will help us:

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Denition Let A ⊂ B be rings, x ∈ B is called integral over A if it satisesa monic polynomial, i.e. ∃n, ai s.t. xn + an−1x

n−1 + ...+ a0 = 0.

A is called integrally closed in B if x is integral (over A)⇐⇒ x ∈ A.

A is called integrally closed if it's integrally closed in it's fraction eld.

Example

1. Z is integrally closed: ( rs )n+an−1(

rs )n−1+ ...+a0 = 0, rn = s(...)⇒ s = 1

2. Any UFD is integrally closed (the same proof).

Lemma If M is a f.g. A−module, ϕ : M → M then ∃ monic p ∈ A[t] s.t.p(ϕ) = 0.

Proof

1. We can assume M = An:An

ϕ′

→ An

↓ ↓M

ϕ→ M

.

2. For An: let p be the characteristic polynomial of ϕ, p(t) = det(R− t · Id).Hamilton Cayley shows us that p(ϕ) = 0.

Proof (lazy proof for Cayley-Hamilton)

1. R is diagonal - it's obvious.

2. A = C, R is diagonalizable.

3. A = C, any R - since diagonalizable matrices are dense.

4. A = Z[x1, ..., xd] by sending xi to transcendental number in C (with re-spect to each other to).

5. Any A: Consider Z < Rij >⊂ A and use step 4.

Denition An A−module M is called faithful if ∀a 6= 0 ∈ A ∃m ∈ M s.t.a ·m 6= 0

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Lemma A ⊂ B 3 x. B is an integral domain. TFAE:

1. x is integral over A.

2. A < x > is nite over A (as a module).

3. ∃ C s.t. A ⊂ C ⊂ B, x ∈ C and C is nite over A.

4. ∃ a faithful A < x >- module M which is f.g. over A.

Proof Obvious directions: 1 ⇒ 2 ⇒ 3 ⇒ 4. So we just need to prove 4 ⇒ 1:let's take ϕ(a) = x · a, ϕ : M → M . By previous lemma ∃ monic p s.t.p(x) = 0 ∈ End(M)⇒ p(x) = 0 ∈ A.

Corollary The set of all x ∈ B that are integral over A is a subring.

Proof Let x, y be integral, then A < x > is nite over A, then A < x, y >is nite over A < x >. Thus A < x, y > is nite over A, and we can use theequivalence 1 ⇐⇒ 3 from the lemma.

Denition The ring of x ∈ B that are integral over A is called the integralclosure of A in B, Aint,B

The integral closure of A is its integral closure in its eld of fractions.

Denition An irreducible variety X is called normal if all local rings OX,xare integrally closed (in K[x]).

Exercise Integral closure commutes with localization.

Exercise ∀ irreducible variety Y and ∀ nite ext. L ⊃ K(Y ) ∃! normalirreducible variety X and a nite surjective map ν : X → Y s.t. ν∗ : K(Y ) →K(X) is the imbedding K(Y ) ⊂ L.

Exercise

1. For x2 = y3 we look at K[x,y]/<x2−y3>. Show that neither it nor its local-ization at (0, 0) are integrally closed.

2. Show that its localization at any other point is integrally closed.

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Denition A discrete valuation ring (DVR) is a UDF with one class ofirreducible elements. (There is only one irreducible element up to multiplicationby R∗)

Exercise Let R be DVR and π ∈ R be irreducible, then any r ∈ R has aunique representation as u · πn,u ∈ R∗ and n ≥ 0. n is called the valuation ofr. Show that it satises:

1. val(ab) = val(a) + val(b)

2. val(a+ b) ≥ min(val(a), val(b))

Lemma Let C be a curve, c ∈ C. TFAE

1. C is smooth at c

2. OC,c is int. closed.

3. OC,c is a DVR.

Proof 3⇒ 2: UDF is int. closed

1 ⇒ 3: dimmc/m2c = 1 take π ∈ mc −m2

c . Then mc(mc/(π)) = mc/(π): mc/m2c =

spanπ ⇒ m2c + (π) = mc ⇒ mc/(π) = m2

c/m2c∩π ⇒ mc(mc/π) = mc/π. By

Nakayama, this implies(π) = mc. Suppose f is ∞ divisible by π: fn = f/πn

(f0) ⊆ (f1) ⊆ .. ⊆ (fN ) = (fN+1), thenf

πN+1 = a fπN ⇒ 1 = aπ in contradiction.

Thus any for any non-invertible f there exists some maximal (positive) powerof π by which f is divisible. Thus OC,c is a DVR.

2 ⇒ 1: We can assume that C is ane and c = Z(f). From NSS we knowmNc ⊂ fOC,c.

Claim: either mc is principal or mN−1c ⊂ fOC,c. Indeed, Let g ∈ mN−1

c , thenymc ⊂ (f), so y

fmc ⊂ OC,c. Thus either yfmc = OC,c or y

fmc ⊂ mc. In the rst

case fy ∈ mc and mc = ( fy ) is principal. In the second case y

f is integral andthus belongs to OC,c.Using the claim by induction we get that mc is principal. mc = (π), π generatesmc/m2

c over OC,c/mc = k.

Discussion If we look at a polynomial and sum the poles and zeroes, we geta pole of degree n in ∞, and the sum of zeroes is n, so all together n− n = 0.Now we look Z, and specZ. it's ane and not closed, we can imagine itscompletion,specZ. The points of specZ are the prime numbers p, each denesa p-adic valuation. By raising p to the power inverse to the valuation we get anon-archimedean absolute value. We can view the point at innity as the usual,archimedean, absolute value. Then we will get again that the product of allabsolute values is one.

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[14.4.13]

Lemma Let ν : C → D be a morphism of curves, which is birational. SupposeD is smooth. Then ν(C) is open and ν : C

∼→ ν(C) is isomorphism.

Proof

1. ν(C) contains an open set ⇒ C\ν(C) is nite ⇒ ν(C) is open.

2. Take c ∈ C, ν(c) ∈ D then ν : OD,ν(c) → OC,c but OD,ν(c) is DVR, thusthere is no ring between it and its eld of fractions, thus OD,ν(c) = OC,c

Using Chow's lemma we obtain

Corollary. Any smooth complete curve is projective.

Valuation Criteria

We don't have the concept of a sequence, but we can use curves to dene closure,compactness etc.

Lemma (curve selection lemma) Let X be an algebraic variety, U anopen subset, x ∈ X. Then x ∈ U ⇐⇒ ∃ a smooth curve C and a morphismν : C → X s.t. x ∈ ν(C) and ν(C) ∩ U 6= ∅.

Corollary (Pf=Exc) Let X be algebraic variety, U constructable subset x ∈X. Then x ∈ U ⇐⇒ ∃ a smooth curve C and ν : C → X s.t. x ∈ ν(C) and#ν(C) ∩ U =∞.

Proof (Lemma)

1. For X = An . Choose a point in U and connect it to x by a straight line.

2. For general X we can assume X is ane, and use NNL: ∃µ : X Annite.

Lemma (valuation criterion for separateness) A variety X is separated⇐⇒ ∀ smooth curve C and c ∈ C and open (non-empty) subset U ⊂ C,∀µ, ν : C → X: if µ = ν on U then µ = ν.

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Proof ⇒: If X is separated then µ × ν : C → X × X. µ agrees with ν on(µ× ν)−1(∆X), which is closed and includes U , thus is all C.

⇐: Choose (x, y) ∈ ∆X, and choose ν : C → X ×X and C ⊃ Uν→ ∆X. Let

pr1, pr2 denote the two projections of X×X on X .Looking at νi = pri ν thenν1|U = ν2|U ⇒ ν1 = ν2 thus (x, y) ∈ ν(C) ⊂ ∆X.

Lemma (valuation criterion for completeness) A variety X is complete⇐⇒ ∀ smooth curve C, open (non-empty) U ⊂ C, ν : U → X ∃! ν : C → Xs.t. ν|U = ν.

Proof ⇒: We dened X to be complete if the projectionX × Y → Y isclosed for any separated Y . Let X be complete, we will take X × C and then:graph(ν) ⊂ X × C∪ ↓U ⊂ C

. Thus the projection prC maps graph(ν) onto C. On

the other hand, it is an isomorphism on U and thus is an isomorphism. Itsinverse, composed with the projection on X denes ν.

⇐:In the proof of Chow's lemma we have shown that ∃ projective variety Y andopen O ⊂ Y , τ : O → X s.t. graph(τ) is closed in Y × X and projects in toX. Let y ∈ Y and ν′ : C → Y s.t. y ∈ Imν′, V = ν′−1(U) is dense in C. Letν = τ ν′|V : V → X. By condition ∃ν : C → X. µ := (ν′, ν) : C → Y ×X.Imµ ⊃ graph(τ |ν′(V )). Thus y ∈ Pr(graph(τ))⇒ O = Y and thus X = τ(O) iscomplete.

Theorem ∀ nite eld ext. K ⊃ k(t) ∃ smooth projective curve C s.t. k(C) =K. Furthermore, any smooth curve D with K(D) = K is uniquely isomorphicto an open subset of C in a way compatible with identications.

Proof (based on Kempf) Existence: C is the normalization of P1 in k(C) =K.

Intrinsically:

1. c ∈ C ⇐⇒ valuation of K, that is, each point denes a valuation, and avaluation denes a point uniquely. (note: if A = OC,c ⊂ K, so ν : K → Z,and we haveab , then OC,c =

ab ∈ K | ν(a) ≥ ν(b) =

ab ∈ K | ν(b) = 0).

Let us show that any valuation ν denes some point. By replacing t by

t−1 we can assume ν(t) ≥ 0. Let B := k(t)K⊂ R = x ∈ K | ν(x) ≥ 0.

Then p := b ∈ B | ν(b) > 0 is a prime ideal. Note that B = O(U) whereU ⊂ C is the inverse image of A1 under the normalization map C → P1.Thus p is the maximal ideal of a point c ∈ U , and c denes the valuationν.

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2. Topology: a set is closed if it's either nite or all of C.

3. O(U) = ∩c∈UOC,c = ∩c∈Ux ∈ K | valc(x) ≥ 0.

Functoriality: K ⊂ L, D corr. to L/k(t), ∃τ : D → C. Explicitely, it maps a val-uation on L to its restriction to K. Functoriality now follows from functorialityof normalization.

Now let E be an ane smooth curve with k(t) = K. Let ti be generators ofK[E]. Then E ∼= c ∈ C | valCti ≥ 0.

Exercise If C is a smooth curve, k[C] 6= K then C is ane.

[29.4.13]

Today we are following the last chapter of Fulton's book. (http://www.math.lsa.umich.edu/~wful-ton/CurveBook.pdf)

Let C be a smooth complete curve. Then it has no regular functions, exceptconstants. Let us then allow poles at some xed points, up to some xed orders.Apparently, we will get a nite-dimensional space of functions. Its dimensionwas studied by Riemann.

Denitions

1. Div(C) is the free abelian group generated by (the points of) a curve C.

2. deg : Div(C) → Z , dened to be the sum of the (formal) coecients.Div(C)0- subgroup of divisors of degree zero.

3. Given D ∈ Div(C),L(D) are the functions in k(C) with at most thegiven poles,l(D) := dimL(D).

4. ϕ : K(C)∗ → Div(C), f 7→ (f) :=∑P∈C ordP (f)P - the divisor of poles

and zeros of f with the corresponding multiplicities (a zero is a pole withnegative multiplicity). Imϕ are called the principal divisors. The Picardgroup is Pic(C) := Div(C)/Imϕ. We will show that Imφ ⊂ Div(C)0 anddenotePic0(C) = Div0(c)/Imϕ.

5. Given ν : c → γ non-constant morphism of smooth projective curves.ν∗ : k(γ) → k(C). deg ν := dimk(γ) k(C).

We dene a morphism: ν−1 : Div(γ)→ Div(C) . It is enough to dene iton the generators, i.e. points. We look at the pre-image of a point p ∈ γ,which is for example q1, q2, q3 ∈ c, where ν(qi) = p. We have the mappingOc,q3 ← Oγ,p and q3 enters ν−1(p) with coecient valq3ν

∗t, wheret is themaximal ideal in Oγ,p.

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Exercise If f ∈ k(C), f : C → P1 then (f) = (f)−1(∞− 0)

Exercise deg(ν−1(D)) = deg ν · deg(D)

Corollary

1. deg(f) = 0 ∀f

2. deg : Pic→ Z is well dened.

Corollary ν−1 : Pic(γ)→ Pic(c)

Exercise If z, z′ ∈ K∗(C), Div(z) = Div(z′)⇒ z = λz′ for some λ ∈ K

Denition

1. D ≡ D′ if D −D′ = (f) for some f .

2. D ≥ 0 if all coecients are ≥ 0. Such D is called eective divisor.

3. D ≥ D′ ⇐⇒ D −D′ ≥ 0

Proposition 1

1. If D ⊆ D′ then L(D) → L(D′) and dimK(L(D′)/L(D)) ≤ deg(D′ −D)

2. L(0) = K, L(D) = 0 if deg(D) < 0

3. l(D) ≤ deg(D) + 1

4. If D ≡ D′ then l(D) = l(D′)

Proof

1. L(D) → L(D′) is obvious. We prove the other statement by induction:We can assume D′ = D + p where p is some point (maybe already in Dand maybe not). Let t be the local parameter as p and f ∈ L(D′). Thenf = u · t−n and where n is the coecient of p in D′.

We dene ϕ : L(D′)→ K by taking ϕ : f 7→ (f ·tn)(p). Then kerϕ = L(D)so ϕ : L(D

′)/L(D) → K.

2. A function must have the same degree of zeroes or poles.

3. Follows from 1 and 2 by adding points.

4. D ≡ D′ ⇒ D = (f) + D′ for some f . Multiplication by f gives theisomorphism.

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Exercise f ∈ L(D) ⇐⇒ D + (f) ≥ 0

What is the asymptotics of l(D)? We will need some denitions and statements.

Def For any subset S ⊂ C and any divisor D =∑nPP dene degS(D) =∑

P∈S nP and LS(D) = f ∈ k(C)|ordP (f) ≥ nP∀P ∈ S. Remark. Inthis way, we obtain a sheaf O(D)dened by O(D)(U) := LU (D). We will showthat this sheaf is locally isomorphic to the sheaf O of regular functions. Suchsheaves are called invertible sheaves. We will show that all invertible sheaves arepobtained in this way. Moreover, all invertible sheaves form an abelian groupunder tensor product, and this group is isomorphic to the Picard group Pic(C).

Exercise IfD ≤ D′, then LS(D) ⊂ LS(D′). If S is nite then dim(LS(D′)/LS(D)) =degS(D′ −D).

Def Let f ∈ k(C)\k. Dene the divisor of zeros of f by (f)0 :=∑P.s.t.ordP f>0 ordP (f)P .

Proposition 2 Let x ∈ k(C) \ k, Z := (x)0, n := [k(C) : k(x)]. ThendegZ = n and ∃τ ∈ Z∀r ∈ Z.l(rZ) ≥ rn− τ

Proof

We rst show m := degZ ≤ n. Let S be the underlying set of Z. Choosev1, ..., vm ∈ LS(0) whose classes [vi] in LS(0)/LS(−Z) form a basis for thisvector space. Let us show that viare linearly independent over k(x). If not,there would be polynomials gi = λi + xhi ∈ k[x] withλi ∈ k,

∑givi = 0,

not all λi = 0. But then∑λivi = −x

∑hivi ∈ LS(−Z), so

∑λi[vi] = 0,

a conradiction. So m ≤ n. Let w1, ..., wn be a basis of k(C) over k(x). Wemay assume that each wi satises an equation wni

i + ai1wni−1i + ... = 0, aij ∈

k[x−1]. Then ordP (aij) ≥ 0 if P /∈ S. If ordP (wi) < 0 for some P /∈ S, thenordP (w

nii ) < ordP (aijw

ni−ji ), which is impossible. It follows that for somet > 0,

div(wi)+ tZ ≥ 0, 1 ≤ i ≤ n. Then wix−j ∈ L((r+ t)Z) for 1 ≤ i ≤ n, 0 ≤ j ≤ r.Since the wi are independent over k(x), and 1, x−1, ..., x−r are independent overk, wix−j |1 ≤ i ≤ n, 0 ≤ j ≤ rare independent over k. So l((r+t)Z) ≥ n(r+1).But l((r+ t)Z) = l(rZ)+dim(L((r+ t)Z/L(rZ)) ≤ l(rZ)+ tm (by Proposition1).

Therefore l(rZ) ≥ n(r + 1)− tm = rn− τ , as desired.

Finaly, since rn− τ ≤ l(rZ) ≤ rm+ 1, if we let r grow we see that n ≤ m.

Riemann's Theorem ∃g ∈ Z s.t. l(D) ≥ deg(D) + 1 − g ∀D. The smallestsuch integer is non-negative and is called the genus of c.

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Proof Dene s(D) := deg(D)− l(D) + 1.

1. s(0) = 0, so g ≥ 0, if exists.

2. D ≡ D′ ⇒ s(D) = s(D′) - clear.

3. If D ≤ D′ then s(D) ≤ s(D′) - by Proposition 1.

4. ∀x ∈ k(C) \ k∀D ∃D′ ≡ D , and r ∈ Z s.t. D′ ≤ r · (x)0. Proof:let (x)0 =

∑nPP,D =

∑mPP , We want D = (f) + D′, so we need

mP − ordP (f) ≤ rnP for all P . Let y = x−1 , and let T = P ∈ C|mP >0and ordP (y) ≥ 0. Let f =

∏P∈T (y−y(P ))mP . ThenmP−ordP (f) ≤ 0

whenever ordP (y) ≥ 0. If ordP (y) < 0, then nP > 0, so a large r will takecare of this.

5. The theorem follows now from the previous steps, using Proposition 2.Proof: let x ∈ k(C) \ k, and choose (the smallest non-negative) τ fromProposition 2. For any D, choose D′ and r from step 4. Then s(D) =s(D′) ≤ s(r(x)0) = deg(r(x)0)− l(r(x)0) + 1 ≤ τ + 1

After Riemann's death, his student Roch gave a precise formula for l(D).

Theorem (Riemann-Roch) ∃ a divisor W , called the canonical divisor s.t.∀D, l(D) = deg(D) + 1− g + l(W −D).

Remark. The sheaf corresponding to W is the sheaf of cotangent spaces.

In order to prove this theorem we will develope some high-level techniques. Thiswill take several lectures.

[06.5.13]

Category Theory

A very good language for algebra, and very helpful. Sometimes the relationbetween objects will be as important as the objects itself, or even more.

Denition A category C is:

1. A class Ob(C) of objects.

2. ∀A,B ∈ Ob(C) a set Mor(A,B), whose elements are called morphisms

3. ∀A,B,C ∈ Ob(C) a function o : Mor(A,B)×Mor(B,C)→ Mor(A,C) ,called composition s.t.

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(a) ∀A ∃IdA ∈Mor(A,A) s.t. IdAϕ = ϕ, ψIdA = ψ ∀ϕ ∈Mor(B,A),ψ ∈Mor(A,C)

(b) (ϕ ψ) φ = ϕ (ψ φ)

To be careful as Categories are not sets and we might run in to problems.Categories which are sets will be called small.

Examples

1. Sets

2. Linear spaces

3. Groups

4. Abelian Groups

5. Algebraic Varieties

6. Topological spaces

Denition Let C,D be two categories, a functor F : C → D is:

1. A correspondence (less carefully - a function) Ob(C)→ Ob(D)

2. ∀A,B ∈ Ob(C), function Mor(A,B) → Mor(F (A), F (B)) s.t. F (ϕ) F (ψ) = F (ϕ ψ) and IdA = IdF (A)

Example

1. Forgetful functor, for example linear space → sets.

2. Embedding: abelian group → groups.

3. Abelization: group → abelian groups. (by G 7→ G/[G,G])

4. For any X ∈ Ob(C), FX : C → sets by Y 7→ Mor(X,Y ). Now given

Yϕ→ Z we need to show Mor(X,Y )

ϕ→ Mor(X,Z) and we do it bysimply composing.

Denition ∀ category C dene the dual category C∗ by Ob(C∗) = Ob(C),and MorC∗(A,B) :=MorC(B,A).

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Example

1. F ∗X : C → sets∗ by Y 7→Mor(Y,X).

2. FC : linear spaces→ linear spaces∗ (going from linear spaces to dual linearspaces).

Denition ∀ C,D dene a category Func(C,D) by: Ob(Func(C,D)) =func-tors C → D. For F,G ∈ Ob(Func(C,D)) a ϕ ∈Mor(F,G) is for any A ∈ Ob(C),ϕA ∈Mor(F (A), G(A)) s.t. ∀ψ : A→ B the following diagram is commutative:

F (A)F (ψ)→ F (B)

ϕA ↓ ↓ ϕBG(A)

G(ψ)→ G(B)

.

Denition A morphism ϕ ∈ Mor(A,B) is called isomorphism if ∃ψ ∈Mor(B,A) s.t. ϕ ψ = IdB , ψ ϕ = IdA.

A and B are called isomorphic if there exists an isomorphism from A to B.

Denition A functor F : C → D is called an equivalence of categories if ∃ afunctor G : D → C s.t. F G ∼= IdD, G F ∼= IdC .

Exercise Linf.d.(C) ∼= C where Ob(C) = N and Mor(n,m) =MatC(m× n).

Denition A functor F : C → D is called fully faithful if ∀A,B ∈ Ob(C):

1. If F (A) ∼ F (B) then A ∼ B

2. F :Mor(A,B)→Mor(F (A), F (B)) is an injection.

Denition A functor F : C → D is called essentially surjective if:

1. ∀X ∈ D ∃Y ∈ C s.t. F (Y ) ∼ X

2. ∀A,B ∈ C, F :Mor(A,B)→Mor(F (A), F (B)) is a surjection.

Theorem F is an equivalence of categories ⇐⇒ F is fully faithful andessentially surjective.

Exercise Prove⇒. We won't prove the other direction.

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Example Category of ane algebraic varieties over k is equivalent to the dualcategory of f.g. k−algebras without nilpotents.

Example Let X be a topological space, dene a category Top(X) by:

1. Ob(Top(X)) =open sets U ⊂ X

2. For U, V ∈ Ob(Top(x)), Mor(U, V ) =

pt(= 1) U ⊂ V∅ o.w.

More generally, one can build suc a category from any partially ordered set.

Example A pre-sheaf of vector spaces on X is a functor: F : Top(x)→ vect.

Lemma (Yoneda) Let X,Y ∈ C. Suppose that ∀z ∈ C we have a bijectionψZ : Mor(X,Z) ∼= Mor(Y, Z) s.t. ∀ϕ : Z1 → Z2 the following diagram is

commutative:Mor(X,Z1)

ψZ1∼= Mor(Y, Z1)↓ ϕ ↓ ϕ

Mor(X,Z2)ψZ2∼= Mor(Y, Z2)

then X ∼= Y .

Proof Taking Z1 = Z2 = X and ϕ = Id : X → X, we getMor(X,X) ∼= Mor(Y,X)

‖Mor(X,X) ∼= Mor(Y,X)

and

obtain ψ : Y → X. In a similar way we obtain ψ′ : X → Y . To show thatthey are inverse to each other we take ϕ := ψ : Y → X and now takingZ1 = Y , Z2 = X we take ψ′ ∈ Mor(X,Y ) and use the above diagram again:

ψ′

Mor(X,Y ) ∼=Id

Mor(Y, Y )↓ ψ ↓ ψ

Mor(X,X)Id

∼= Mor(Y,X)ψ

. Walking along the diagram in two ways from

Mor(X,Y ) to Mor(X,X) we get that ψ ψ′ = Id. In the same way we showψ′ ψ = Id.

Equivalent formulation (Yoneda embedding) The functor C → Funct(C, sets)given by X 7→ FX , where FX =Mor(X, ·) is fully faithful.

Denition A functor F : C → D denes a functor F ∗ : D → Func(C, sets)by: x ∈ D, F ∗(X)(Y ) =Mor(F (Y ), X). Y onC : C → Func(C, sets)

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Denition If G : D → C s.t. Y onC G ∼= F ∗ then G is called right adjointto F , and F is called left adjoint to G. In other words Mor(F (X), Y ) ∼=Mor(X,G(Y )).

Exercise

1. An adjoint functor is unique up to isomorphism of functions.

2. If F G ∼= Id, G F ∼= Id then F and G are both left and right adjoint.

Examples

left-adjoint right-adjoint

F:sets → abelian, free abelian group Forgetfull: Abelian→setsAbelization Inclusion: Abelian → Groups

Ext Res:B −mod→ A−modM ⊗A · Hom(M, ·):A−mod→ A−mod

1. Take G to be the forgetful functor. We are looking for its left adjointF : sets → Abelian s.t. for A ∈ groups and Z ∈ sets. Mor(F (Z), A) ∼=Mor(Z,A). So we take the free group.

2. We haveA ∈ Abelian andB ∈ Groups and we want F s.t. Mor(F (B), A)Abelian groups

∼=

Mor(B,A)Groups

.

3. ϕ : A→ B morphism of algebras with 1. Res : B −mod→ A−mod.Theadjoint is Ext(M) := B ⊗AM

4. Exercise.

Exercise Find left and right adjoint to Forget:Top. Spaces→ sets

Youtube lectures Channel: catsters.

Exercise

1. Partially ordered set, functors are monotone functions. Find left and rightadjoint.

Denition Groupoid is a category in which all morphisms are invertible.

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Examples:

1. Group= groupoid with one object.

2. Fundamental groupoid.

Denition X ∈ C is called:

1. Initial if ∀Y Mor(X,Y ) = pt

2. Final if ∀Y Mor(Y,X) = pt

3. X is called the zero object if it's both nal and initial.

Example

1. Commutative ring with 1: the initial ring will be the free ring Z, the nalring is the zero ring 0.

2. A−modules: the zero module is both.

Denition Let X,Y ∈ Ob(C). A direct product of X and Y is X × Y ∈Ob(C), together with two morphisms (called projections) X×Y → X, X×Y →Y s.t. Y onC(X × Y ) ∼= Y onC(X) × Y onC(Y ), we have Y onC(X) that comesfrom Y onC(PX) and same for Y .

In other words, for any other object Z with two morphisms Z → X, Z → Ythere exists a unique morphism Z → X × Y , such that the following diagram is

commutativeZ

↓ X ← X × Y → Y

.

Reverting all the arrows we get the denition of direct sum (also called co-product). Let X,Y ∈ Ob(C). A direct sum of X and Y is X ⊕ Y ∈ Ob(C),together with two morphisms X → X ⊕ Y , Y → X ⊕ Y s.t.

for any other object Z with two morphisms X → Z, Y → Z there exists a unique

morphismX⊕Y → Z, such that the following diagram is commutativeZ

↑ X → X ⊕ Y ← Y

.

15

Exercise

1. In sets X ⊕ Y is disjoint union.

2. Give a Yoneda like denition.

3. For a nite number of A−modules: direct sum is the same as direct prod-uct. For innite number, nd out.

Now we would like to dene an additive category. In short, this is a categorywhere the set of morphisms between any two objects forms an abelian group,and compositions are bilinear (over Z). Apparently, this is a property, and not astructure! Namely, thereis no choice in the denition of addition of morphisms.To show that, we will give a slightly longer denition.

Lemma Suppose a category C has ⊕, × and 0−object. Then ∀A,B ∈ C ∃natural A⊕B → A×B.

Proof A→ A⊕B, we map this by Id : A→ A and as we have 0−object, wecan get the 0−map 0 : A → 0 → B. For B we dene the same way, and getA⊕B → A×B.

Lemma Suppose a category C has ⊕, × and 0−object, and ∀A,B ∈ C thenatural morphism A ⊕ B → A × B is an isomorphism. Then ∀A,B ∈ C,Mor(A,B) has a natural structure of an abelian semigroup.

Proof ϕ,ψ : A → B, we can make a map A ⊕ A → B, but we know thatA × A ∼= A ⊕ A, and we can take the formal mapping A → A × A, so we canget a mapping A → B by: A → A × A ∼= A ⊕ A → B. The neutral element inthis semi-group is the (unique) morphism that factors through the zero object.

Denition An additive category is a category that

1. has ⊕, × and 0−object.

2. the natural map A⊕B → A×B is an isomorphism, ∀A,B ∈ C

3. The semigroup Mor(A,B) is a group, i.e. any morphism has a negative,∀A,B ∈ C.

In additive categories, the group Mor(A,B) is usualy denoted Hom(A,B)

Exercise

1. Show that the composition is Z-bilinear (i.e. the composition with a xedmorphism is a homomorphism of groups).

2. Give an example of a category that satises 1,2 but not 3.

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[13.5.13]

We will have an extra meeting next Tuesday (21.5) at 14:30-16:30

Denition

• ϕ ∈ Mor(X,Y ) is calledmonomorphism if ϕ : Mor(Z,X)→ Mor(Z, Y )is an injection.

• ϕ ∈ Mor(X,Y ) is called epimorphism if ϕ : Mor(Y, Z)→ Mor(X,Z) isan injection.

Examples

1. In sets, abelian groups etc. this will be what we expect.

2. Hausdor topological spaces: monomorphism is injective, epimorphismhas a dense image.

3. Abelian groups with division. Q → Q/Z it's both monomorphism andepimorphism.

Exercise

1. What is monomorphism and epimorphism in PreSh(X)?

2. (∗) What is monomorphism and epimorphism in Sh(X)?

Notation From now on let C be a category.

Denition ϕ : A→ B, then:

The kernel, kerϕ, is the nal object in the category of pairs (D,ψ) s.t. D ∈ C,ψ : D → A, ϕ ψ = 0.

The cokernel, cokerϕ, is the initial object in the category of pairs (D′, ψ

′) s.t.

D′ ∈ C, ψ′

: B → D′, ψ

′ ϕ = 0.

Denition A pre-abelian category is an additive category s.t. every mor-phism has a kernel and a cokernel.

An abelian category is a pre-abelian category s.t. if ϕ is both monomorphismand epimorphism then ϕ is an isomorphism.

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Example

1. Abelian groups, A−modules are abelian categories.

2. The category of Banach spaces is a pre-abelian, but not an abelian cate-gory.

Denition (equivalent) A category is abelian category if:

1. ∃0 object

2. There are ⊕,× for any X,Y .

3. Any monomorphism is a kernel of some morphism.

Any epimorphism is a cokernel of some morphism.

Exercise Proof the equivalence.

Denition A complex is a sequence of objects Ci and morphisms di : Ci →Ci+1 s.t. ∀i di+1 di = 0. This gives us a natural mapping from Ci to ker di+1

that we'll denote d′

i.

An exact sequence is a complex s.t. Imdi = ker di+1 or in other words d′

i :Ci → ker di+1 is an epimorphism.

Denition The cohomology of a complex C are Hi(C) = ker(di+1)/Im(di)∼=

coker(d′

i).

Examples 0→ Aϕ→ B is exact ⇐⇒ ϕ is monomorphism

Aϕ→ B → 0 is exact ⇐⇒ ϕ is epimorphism.

0→ Aϕ→ B → 0 is exact ⇐⇒ ϕ is isomorphism.

Left exact sequence: 0→ A→ Bϕ→ C is exact ⇐⇒ A ∼= kerϕ.

Right exact sequence: Aϕ→ B → C → 0is exact ⇐⇒ C ∼= cokerϕ.

Short exact sequence: 0→ A→ B → C → 0 ⇐⇒ C ∼= B/A, A ⊂ B

Denition

1. A functor F between two additive categories is called additive if ∀A,B ∈Ob(C) F (A ⊕ B) = F (A) ⊕ F (B) and ∀ϕ,ψ ∈ MorC(A,B) F (ϕ + ψ) =F (ϕ) + F (ψ).

2. An additive functor between two abelian categories is called exact if itmaps exact sequences to exact sequences.

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Example

1. Exact functor: Mod(A)→ Mod(S−1A)

2. M 7→M ⊗Z Z/2 is right-exact but not exact.

Exercise Show that M 7→M ⊗A N is right exact ∀N ∈ Mod(A)

Lemma (5-lemma) Given the following commutative diagram of an abelian

category:

A1 B1

↓ ↓A2

∼= B2

↓ ↓A3

ϕ3→ B3

↓ ↓A4

∼= B4

↓ ↓A5 → B5

, if the rows are exact then ϕ3 is an isomorphism.

Proof We will use Mitchell's theorem that every small abelian category canbe imbedded into the category of modules over a ring.

Denot ψi : Ai → Ai+1, φi : Bi → Bi+1 and ϕi : Ai → Bi.

We will show that ϕ3 is both a monomorphism and an epimorphism:

Assume x ∈ A3 s.t. ϕ3(x) = 0 ∈ B3 (1) then it maps to 0 in B4 (2), and to 0in A4 (3), so x came from some y ∈ A2 (4) so we have y′ ∈ B2 (5) that comesfrom some z′ ∈ B1 (6) and as we have A1 B1 there is z ∈ A1 (7). we seethat ϕ2(ψ1(z)) = y′ and ϕ2(y) = y′ and as ϕ2 is an isomorphism we get thatψ1(z) = y (8).

Thus x = ψ1(ψ2(z)) = 0. Thus ϕ3 is a monomorphism.

z ∈ A1(7)← z′ ∈ B1

↓(8) ↓(6)y ∈ A2

(5)→ y′ ∈ B2

↓(4)x ∈ A3

(1)→ 0 ∈ B3

↓(2)0 ∈ A4

(3)← 0 ∈ B4

.

Assume x′ ∈ B3, then it maps to y′ ∈ B4, which is isomorphic to y ∈ A4 andmaps to z ∈ A5 which maps to z′ ∈ B5. From commutativity z′ = 0, butthen z = 0. As it's exact there is x ∈ A3 that goes to y. Then φ3(ϕ3(x)) =ψ3(ϕ4(x)) = y′ = φ3(x

′) ⇒ φ3(x′ − ϕ3(x)) = 0 ∈ B4 so there is w′ ∈ B2 that

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maps to x′ − ϕ3(x) ∈ B3, and w′ is isomorphic to w ∈ A2. Now ϕ3(ψ2(w)) =

x′−ϕ3(x)⇒ x′ = ϕ3(ψ2(w))+ϕ3(x) = ϕ3(ψ2(w)+x) so ψ2(w)+x ∈ A3 mapsto x′ and ϕ3 is an epimorphism.

Denition Given objects Ai and morphisms between them ϕij : Ai → Aja colimit (or direct limit, or injective limit), denoted lim

→Ai, is an initial object

A s.t. all of the objects map to it with ψi : Ai → A, and the following diagram

stays commutative:

A

ψi ψj

Aiϕij→ Aj

.

Example

1.⊕Ai → B

Ai Ai+1

2. Fix an object M , Σ =subojects of M with the inclusion order, thenlimσ∈Σ→

Mσ =M .

3. cokerϕ is the direct limit over a category C of objectsDi s.t.A

ϕ→ B↓ ϕi

D → Di

is

commutative and ϕi ϕ = 0.

4. Every abelian category with arbitrary direct sums has arbitrary directlimits (it is a quotient of the direct sum of the objects).

Denition The inverse limit (or projective limit, limit), denoted lim←, has

the dual denition to direct limit: it is a nal object that maps to the givenobjects. It generalizes direct product, kernel.

Claim Every abelian category with arbitrary direct products has inverse limits(it is a sub of the direct product of the objects).

Claim

1. Hom(lim→Aα, B) = lim

←Hom(Aα, B)

2. Hom(A, lim←Bα) = lim

→Hom(A,Bα)

The (obvious) proof of the claim is left as an exercise. In fact, this can be takenas the denition of the limit and the colimit.

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Exercise IfM is a nitely generatedA−module then Hom(M,⊕Nα) = ⊕Hom(M,Nα)

Denition A functor is called strongly right-exact if it commutes withdirect limits.

A functor is called strongly left-exact if it commutes with inverse limits.

Lemma Strongly left (right) exact is left (right) exact.

Exercise Localization is exact but not strongly left exact.

Proposition Let C,D be abelian categories. F : C → D an additive functor,G : C → D right adjoint to F . Then F is strongly right exact and G is stronglyleft exact.

Proof F (lim→Aα)

?∼= lim→F (Aα)

HomD(F (lim→Aα), B) = HomC(lim→

Aα, G(B)) = lim←HomC(Aα, G(B)) = lim

←HomD(F (Aα), B) =

HomD(lim→F (Aα), B)

Corollary Hom(M, ·) is strongly left-exact, M ⊗A · is strongly right exact.

Denition

• If Hom(M, ·) is exact then M is projective.

• If M ⊗ · is exact then M is at.

[20.5.13]

We go back to the proposition from last class, and the denition that followedit.

Looking at the short sequence: 0 → K → L → N → 0, we can have: 0 →Hom(M,K)→ Hom(M,L)→ Hom(M,N) but the problem is for the last mapto be onto. If we can't send some elements of M in to L, but we can sendthem to N because of some relation, we will have a problem because not allhomomorphisms can come from the previous. This problem will not happenwhen M is free.

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Example The free module is projective: Hom(A,N) = N and Hom(AI , N) =N I

Lemma TFAE

1. M is projective

2. M is a direct summand of a free module (M ⊕N ∼= AI)

3. Any epimorphism NϕM has a section s s.t. ϕ s = IdM .

4. Any short exact sequence splits. (given: 0 → L → N → M → 0 we havethat N ∼= N

′= L⊕M)

Proof 1⇒ 3:

Ms ↓Id

0 → L → Nϕ→ M → 0

3 ⇒ 4: choose s from (3), L ⊕M ∼= N by (l,m) 7→ ψ(l) + s(m) ∈ N . Thismap is clearly onto. To see that it is injective, assume ψ(l) = s(m). Thenm = ϕ(s(m)) = 0⇒ l = 0

4⇒ 2: AI M and 0→ K → AI →M → 0 splits, so M is a direct summand.

2⇒ 1: M ⊕N ∼= AI , Hom(AI , ·) = Hom(M, ·)⊕Hom(N, ·). LHS is exact, thusso is RHS, so Hom(M, ·) is exact.

Exact Show (4)⇒ (1) thus (1) ⇐⇒ (3) ⇐⇒ (4) in any abelian category.

Corollary Projective modules are at.

Example S−1A is at for any S.

Exercise Let M be f.g. then projective ≡ locally free.

Exercise TFAE

1. M is at

2. K → L⇒ K ⊗AM → L⊗AM

3. I → A⇒ I ⊗M →M

4. ∀m ∈ Mk, r ∈ Ak s.t. rTm =∑ki=1 rimi = 0 ∃X ∈ Mat(A, k × k),

n ∈Mk s.t. Xn = m, rTA = 0

5. ∃ directed system Fα s.t. Fα is f.g. and free ∀α and lim→Fα =M .

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Corollary Flat nitely presented ⇒ projective.

Denition An object I in an abelian category is called injective if Hom(·, I)is exact.

Exercise Hom(·,M) is always right-exact

Exercise TFAE

1. I is injective

2. Any monomorphism I → N has a section

3. Any exact sequence 0→ I → N → K → 0 splits.

Denition Given x ∈ X and a sheaf F we dene a stalk lim→x∈u

F(u) =: Fx

Exercise OX,x = (OX)x

Exercise Fx ∼= (u,ξ)|x∈U⊂X,ξ∈F(u)/∼ where (u, ξ) ∼ (v, ψ) if ξ|u∩v = ψ|u∩v.

Example X = A2\0, 0. OA1\0 ← OX . This map is an epimorphism inthe category of sheaves, but not in the category of presheaves. Also, it is notan epimorphism on global sections. This shows us that the naive image andcokernel constructions are not suitable for sheaves - they give presheaves, notsheaves. We need a procedure of sheacation - a functor from presheaves tosheaves, left adjoint to the forgetful functor.

Construction of sheacation F 7→ Fsh

Fsh(U) := ϕ = (ϕp)p∈u with ϕp ∈ Fp ∀P ∈ U s.t. ∀Q ∈ U ∃V.P ∈ V ⊂U,ϕ

′ ∈ F(V ) s.t. ϕQ = ϕ′

Q ∈ FQ ∀Q ∈ V

Lemma

1. sh is left adjoint to the forgetful functor

2. The natural morphism F → sh(F) is an isomorphism on stalks.

3. If F is a sheaf, then the natural morphism F → sh(F) is an isomorphism.

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Proof

1. Have to identify Hom(sh(F),G) with Hom(F ,G). From the denition ofsh(F) we have a natural morphism F → sh(F). This gives a map fromHom(sh(F),G) to Hom(F ,G). Let us show that this map is a bijection.Suppose there are dierent φ, φ

′ ∈ Hom(sh(F),G) that result in the sameψ ∈Hom(F ,G). Then they dier on some ξ ∈ sh(F)(U) for some U ⊂ X.However, there is a cover of U by Uion which ξis given by some ξi ∈ F(Ui).Then φ(ξ)|Ui

= ψ(ξi) = φ′(ξ)|Ui∈ G(Ui). Since G is a sheaf, we get

that φ(ξ) = φ′(ξ) which is a contradiction. In a similar way we provethat our map is onto: given ψ ∈Hom(F ,G) we construct the preimageφ ∈ Hom(sh(F),G).Take an open subset U \subset X and ξ ∈ sh(F)(U).By denition of sh(F), there is a cover of U by Uion which ξis given bysome ξi ∈ F(Ui). Dene φ(ξ) ∈ G(U) to be the section locally given byψ(ξi).

2. By denition and by sheaf axioms.

3. Clear.

Denition Let ϕ : F → G be a morphism if sheaves. Let ker′ϕ and coker

′ϕ

be the presheaves dened by ker′ϕ(u) = ker(ϕu), coker

′ϕ(u) = coker(ϕu).

Dene kernal and coker to be kerϕ = ker′ϕ and cokerϕ = sh(coker

′ϕ).

Exercise Show that Sh(X) is an abelian category.

Exercise LetX be an ane varietyO′

X(U) := fg | f, g ∈ Pol(X), g doesn't vanish on U.Show that sh(O′

X) = OX .

Lemma A sheaf F is zero ⇐⇒ Fp is zero ∀P ∈ X.

Proof By sheaf axioms.

Lemma A sequence of sheaves ... → Fi → Fi+1 → ... is exact ⇐⇒ ∀p ∈ Xthe corresponding sequence of stalks is exact.

Proof By the denition of kernel and cokernel, it is easy to see that theycommute with taking stalks, and this shows that all stalks of an exact sequenceare exact sequences. Vice versa, suppose all stalks of a complex of sheavesare exact sequence are exact sequences. To show that the complex is exact,consider a cohomology. All its stalks are zero, thus the cohomology is zero, sothe sequence is exact.

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Example F(U) = C constant presheaf.

sh(F(U)) :=locally constant function U → C- constant sheaf.

Exercise Γ is left exact, but not exact.

Γ : sh(X)→ vect, Γ(F) := F(X)

Denition C is a smooth curve, D ∈ Div(C). OC(D)(U) := f ∈ K(C) s.t.∀p ∈ U ordp(f) ≤ np.

Example C ∼= P1, D = p = ∞, t = x0

x1. So OC(D), OC are not isomorphic,

but are locally isomorphic.

Denition Given a point P ∈ X and a vector space L we dene a skyscraper

sheaf LP by Lp :=

Lp(u) = L p ∈ uLp(u) = 0 p /∈ u

.

Exercise 0 → OP1(∞) → OP1 → k∞ → 0, where k∞ is the 1-dimensionalskyscraper at the point ∞ ∈ P1.

Denition Let X be a topological space, OX -sheaf of ringsM(OX) category

of sheaves of OX modules. It has ⊕,⊗ (by sh(F ⊗ G)) and dual F := sh(F ′)

where F ′(u) := Hom(F(u),O(u)).

Exercise

1. D1 ≡ D2 ⇒ O(D1) ∼= O(D2)

2. O(D1)⊗O(D2) ∼= O(D1 +D2)

3. O(D) ∼= O(−D)

Exercise* Pic(C) ∼=invertible sheaves, mult= ⊗

Denition A sheaf of Ox−modules is called invertible if it is locally isomor-phic to Ox.

Denition X = specA, M is A−module, M → M(Xf ) := Mf , M′(U) =

M ⊗O(X) O(U) and so M := sh(M′)

25

Exercise

1. M(X) =M

2. Mp =Mp

Denition Let X be an algebraic variety. A sheaf of OX−modules is calledquasi-coherent if there exists an open ane cover X = ∪iUi and O(Ui) mod-ules Mi s.t. F|Ui

= Mi

It is called coherent if the Mi can be chosen to be nitely generated.

Lemma Let X = SpecA. Then the functors Γ : Shq.coh.(OX) → A − mod,˜: A−mod→ Shq.coh.(OX) are mutually inverse, preserve tensor products anddual and dene exact equivalence of categories: Shq.coh.(OX) ∼ A −mod andShcoh.(OX) ∼ f.g. A−mod

Proof Having F , M = Γ(F). There exists a nite cover of X by basic open

subsets Xf such that F|Xf= F(Xf ). Deneφf : M(Xf ) =Mf = F(X)⊗O(X)

O(X)f → F(Xf ). Let us show that this map is an isomorphism for every f, and

this will imply that this denes a natural isomorphism M ' F . Let us show,for example, that φf is onto. Take a section ξ ∈ F(Xf ). For any other

Xg from the cover, we have Xf ∩Xg = Xfg , F(Xfg) = F(Xg)f and thus thereexists n s.t. fnξ|Xf∩Xg can be lifted to a section in F(Xfg). Continuing in thisway for other elements of the cover, we get a family of sections ξg ∈ F(Xg).Is this family compatible? Possibly not. It is possible that ξg|Xgh

6= ξh|Xgh.

However, ξg|Xfgh= ξh|Xfgh

= ξ|Xfgh. Thus, if we multiply the whole family by

high enough power of f , we get a compatible family, which by the sheaf axiomslifts to a global section ξ′ ∈ F(X) such that ξ′|Xf

= fNξ. This shows that φfis onto.

Since the functors are mutually inverse, they are adjoint on both sides and thusexact.

It is easy to see that˜preserves tensor products and dual, and thus so does Γ.

[21.5.13]

Solutions of HW

Exercise 1

Question 1

26

1. The question is interesting as OX,p = Amp is not f.g. For a chain of idealsI1 ⊂ ... ⊂ In ⊂ ...Amp , the chain of numerators J1 ⊂ ... ⊂ Jn ⊂ ...Astabilizes and thus the original chain stabilizes.

2. Obvious

3. We can nd take an ane X s.t. x1, x2 ∈ X . Then OX,x1⊂ OX,x2

⇒mx2

⊂ mx1and then by NSS x1 = x2.

Question 2 Lemma: S−1Aint = (S−1A)int.

S−1Aint ⊂ (S−1A)int: If xn+an−1xn−1+ ..+a0 = 0 then (x

n

s )+ an−1

s (xs )n−1+

..+ ansn .

(S−1A)int ⊂ S−1Aint: s := sn−1sn−1..s0 and then (sx)n+(xs)n−1an−1sn−2..s0+..+ aos1..sn−1s

n−1 = 0

Question 3 We showed in class that the local ring at the origin is not inte-grally closed, since its closure includes x

y .

Now x, y 7→ xy and t 7→ t3, t2 gives an isomorphism between the complement to

the origin in C and on the line A1. This shows that the local ring at any otherpoint is integrally closed.

Question 6 k[C] 6= k. Assume t ∈ k[C] regular function, this denes mappingfrom C to A1. Now we can look at k[t] ⊂ k[C] ⊂ k(C) and let A be itsnormalization in k(C). Let γ be the unique smooth projective curve with k(γ) =k(C). Then the point of γ are valuations on k(C) and C and U := SpecA areidentied with open subsets consisting of valuations that are non-negative onk[C]and on A respectively. Since A ⊂ k[C] (why?), we get C ⊂ U . Now U is anane curve, thus any open subset of U is basic open subset and thus is ane.

Exercise 2

Question 3 deg ν is dened as the degree of k(γ) over k(C).

Taking p ∈ γ, and an ane U s.t. p ∈ U ⊂ γ. A := O(u), mp / A,B := O(ν−1(u)) and B → A. We are looking at ν−1(p) = Q1, .., Qk.DenoteordQi

ν∗t =: ni. We want to nd a basis for k(γ) over k(C) of size n :=∑ni.

1. By Chinese remainder theorem B/mpB∼= ⊕iB/mni

Qi. This shows that the

dimension of B/mB over A/mp = k is n. Choose a basis T of B/mB overA/mp = k and lift it to a subset T of B.

2. T generates Bm over Am. This is true as M =< T >⊂ Bm, Bm/n+M = 0

so Bm/M = m(Bm/M)Nakayama⇒ Bm/M = 0.

27

3. T generates k(C) over k(γ). Take f ∈ k(C) and (ν∗t)kf ∈ Bm for somek big enough and thus (ν∗t)k =

∑aiti. Now we just need to show that

there are no relations.

4. Suppose there is a relation on T with coecients in k(γ). Multiplyingby denominators we get a relation

∑aiti = 0, ai ∈ Ai, so we can get∑

aiti = 0 and thus all ai are divisible by t. Dividing the equality by twe get that all ai are divisible by any power of t, and thus are zeros.

Question 4 ls(D) = dimLs(D), and degS(D) =∑p∈S np. Let D

′= D + p.

If D =∑niQi we take an open ane subset U = SpecA that includes all Qi.

Let ti ∈ A s.t. ti(Qi) = 1 and ti ∈ Qnj

j for j 6= i. Let si ∈ mQi be a generator.

Then ski ti, |Pi ∈ S, 0 ≤ k ≤ ni − 1is a basis for lS(D).

Question 6 We want to show that l(np) = n + 1. First l(np) ≤ n + 1 fromwhat was shown in class. Now take t = x1

x0, and looking at k[t], we take1, 1t , ..,

1tn .

Question 7 Let D0 be s.t. l(D0) = deg(D0) + 1 − g. N := deg(D0) + g.

The deg(D) ≥ NRiemann⇒ l(D − D0) ≥ deg(D − D0) + 1 − g > 0. Take

f ∈ L(D−D0), then D−D0 + (f) ≥ 0⇒ D ≥ D0 − (f). So s(D) ≤ s(D0) = 0as needed. (Where s(D) := deg(D)− l(D) + 1)

Question 8 R-R: l(D)− deg(D) = 1− g + l(W −D)

1. take D = 0 then we get 1−0 = 1−g+l(W )⇒ l(W ) = g. Now substitutingD =W we get deg(W ) = 2g − 2.

2. deg(D) ≥ 2g − 1 ⇒ deg(W − D) < 0 ⇒ l(W − D) = 0 ⇒ l(D) =deg(D) + 1− g

3. deg(D) ≥ 2g ⇒ l(D − P ) = l(D)− 1 ∀P ∈ C - in the same way.

4. Have to show l(D) > 0 , l(W − D) > 0 ⇒ l(D) ≤ deg(D)2 + 1. We will

assume that D > 0 and instead of W − D take D′> 0 s.t. (D + D

′ 'W ). We can assume that l(D − P ) 6= l(D) for some point P, and thusl(D) = l(D − P ) + 1. Let f ∈ L(D) − L(D − p). Multiplication by f

denes ϕ : L(D′)/L(0) → L(W )/L(D). Thus l(D

′)−1 ≤ l(W )− l(D). On the

other hand, by R-R, l(D)− l(D′) = deg(D) + 1− g. Thus g + 1− l(D) ≤

l(D)− deg(D) + g − 1⇒ deg(D) + 2 ≤ 2l(D).

[27.5.13]

Lemma Γ(M) =M

28

Theorem ˜and Γ are mutually inverse

Proof Have to show that Γ(F) = F , ∀F ∈MX . X = ∪Xfi , F|Xfi= F(Xfi).

Taking M := Γ(F) and M(Xfi) = Mfi . Mfi → F(Xfi) as O(Xfi) = Afi form ∈ F(X) the operation m

fki

is dened. ∃j s.t. m|Xfj6= 0. DeneMj := F(Xfj ),

F(Xfj ∩Xfi) = (Mj)fi , m′= m|Xfj

∩Xfi= m|Xfifj

= 0. Thus if Mj 7→ (Mj)fi

then m′ 7→ 0 fki m = 0 and ∃N ∀j fNi m|Xfj

= 0 ⇒ fNi = 0 ∈ F(X) = M and

thus mfki

= 0 ∈Mfi and the map is one to one.

[The proof is given in the end of 20.5]

Corollary Γ, are exact equivalence of categories, Γ :Mf.g.(A)→McoX

Denition Fiber of F at p, Mmp/mp- n. dimension vector space over A/mp∼=

K

Lemma Let F be a coherent sheaf on X. Then the function p 7→ dimF|p isupper semi continuous, i.e. f−1(m,∞) is open ∀m

Proof Akϕ→ An →M → 0, whereM = cokerϕ, ϕ is given by T ∈ Mat(k, n,A).

We have (A/m)k → (A/m)n → M/m → 0, F|p = coker(T |p) and dimF|p =n− rank(T |p).

Lemma Let F be a coherent sheaf, p ∈ X. TFAE:

1. ∃p ∈ U ⊂ X s.t. F(U) = 0

2. Fp = 0

3. F|p = 0

Proof (3) ⇒ (2) having Am,M f.g. we know by Nakayama M/mM = 0 ⇒M = 0

(2) ⇒ (1) p ∈ U ⊂ X,Fp is f.g. over OX,p by ξ1, .., ξn. F(Xf ) is gen by0 = ξ1|Xf

, ..., ξk|Xfso F(Xf ) = 0.

Exercise Both equivalences are wrong for general q. coherent sheaves.

Lemma A coherent sheaf is locally free ⇐⇒ dimF|p is locally constant.

29

Proof ⇒ ok

⇐ Suppose X = specA, take m ⊂ A. M/m, f1, .., fk ∈ M s.t. fi form a basisfor M/m they dene ϕ : Ak →M .

Claim: ∃Xf s.t. f /∈ m and ϕf : Akf∼→ Mf . This is true as cokerϕ|m = 0 ⇒

∃p ∈ U s.t. cokerϕ(u) = 0, p ∈ Xg ⊂ U s.t. cokerϕg = 0. Thus fi generateMg over Ag. M |Q ∀Q ∈ Xg since dimM |Q = dimM |p. Suppose ∃ a relation∑aifi = 0 but ai(Q) = 0 ∀Q ∈ Xf ⇒ ai ≡ 0.

Lemma Let A,B be algebras, and T :M(A)→M(B) be right exact functor.Then T ∼= T (A)⊗B ·

Proof We can have Aβϕ→ Aα → M → 0, M = cokerϕ, T (M) = cokerT (ϕ).

T (Aβ)T (ϕ)→ T (Aα)

T (A)β → T (A)α, T (A)⊗AM coker of ϕ⊗1 and we have

Aβ ⊗A T (Aβ) → Aα ⊗A T (Aα)

T (A)β → T (A)α.

Discussion In some categories we have an inner Hom Hom(A,B) ∈ C, forexample Modules, vector spaces and we have Hom(A,B) ∈ Ab. What is therelation between the two?

We would like to dene an inner Hom on sheaves.

Hom(F ,G)(u) is like Hom(F(u),G(u)) but then it will not be a sheaf, wecan deal to it, or following direct solution Hom(F|u,G|u) =: H(F ,G)(u) =(Hom(F(u),G(u))sh.

Lemma (Exercise) A coherent sheaf is projective ⇐⇒ it's locally constant.(Assume X is ane)

Denition Looking the projective spaces, we can dene for n = 1 OP1(m ·∞)and for a general n we know Pn = ∪Ani (xi 6= 0) and then OPn(m)|xi 6=0 =xmi OPn |xi 6=0

Denition For n = 1 OP1(m) = OP1(m · ∞) ∀m ∈ Z.

Denition M is a graded K[x0, .., xn] module if M = ⊕Mi where Mi arevector spaces, if p ∈ K[x0, .., xn]i,m ∈Mj then pm ∈Mi+j

Denition Let M be a graded module over K[x0, .., xn]. Dene a sheaf MPn

on Pn by MPn(u) = MAn+1(π−1(u))deg 0 where π : An+1\0→ Pn.

30

Example M = k[x0, .., xn], MPn = OPn

Exercise M =? s.t. MPn = OPn(m)

Theorem ∀ F ∈ MPn∃ graded module M over k[x0, .., xn] s.t. F = MPn .

Proof M := ⊕m≥0Γ(F ⊗OPn(m)).

O(m) → O(m+ 1), F ⊗O(m) → F ⊗O(m+ 1)

Exercise Find M s.t. MPn = 0.

Exercise Generalize the theorem to any X ⊂ Pn

Denition F(m) := F ⊗O (m)

Lemma OPn(m) are at.

Corollary if F is a coherent sheaf on a projective var. X then ∃ m, k and asurjection O(m) F .

[3.6.13]

Next week we will start at 10:30 and nish at 13:15

Recall that we dened:F → ⊕d≥0Γ(F(d)) where F(d) = F ⊗O(d). We can get

M ←M where M is graded over k[x0, ..xn]

Corollary If F is coherent then ∃n, k and an epimorphism OkX F(n)

Proof F = M , M is f.g. (since a chain of submodules of M denes a chainof subsheaves of F , which stabilizes locally and thus globally. Let m1, ..,mk behomogeneous generator of M , d = max(deg(mi)).

This denes a map (k[x0, .., xn])k M satisfying deg(a1, ..ak) < d+max(deg(a, 1)).

Lemma Given M → HomA(M,A) one has M ⊗A Hom(M,A) ∼= A ⇐⇒ Mis f.g.

31

Denition

1. Let X be irreducible variety, Pic(X) = the group of invertible sheaves upto isomorphism.

2. RatX =constant sheaf k[x]

3. A sheaf of fractional ideals is a subsheaf of RatX .

4. A sheaf of fractional ideals is called principle if it has the form (f) =f · O(x) ⊂ RatX .IFI := group of invertible sheaves of X . (invertible fractional ideals)

P (X) := subgroup of principle fractional ideals.

Lemma 1→ P (X)→ IFI(X)ψ→ Pic(X)→ 1 is a short exact sequence.

Proof

1. Let kerψ = P (X), let F ∈ IFI s.t. ψ(F)ϕ∼= OX . Then ϕ−1(1) ∈ P (F).

F = ϕ−1(1)OX , a ∈ F(u), then a = ϕ−1(1) · ϕ(a) = ϕ−1(1 · ϕ(a)) =ϕ−1(ϕ(a)) = a. f · Ox ∼= OX by 1

f .

2. ψ is on to, F ∈ Pic(X), x some open dense U ⊂ X, 0 6= σ ∈ F(U).∀W dene I(W ) := f ∈ k(X) | fσ comes from a section of F over W.F(ui) ∼= O(ui).f = p

q , f ∈ I(u) ⇐⇒ ∃ξ ∈ F(w), (qξ − pσ)|u∩w = 0.

Denition Div(X) is a free abelian group generated by irreducible sub-varietiesof codimension 1 in X.

Proposition Suppose that ∀x, OX,x is a UFD. Then ∃ canonical isomorphismDiv(X) ∼= IFI(X) and 1 ∈ Γ(OX(D)) ⇐⇒ D ≥ 0.

Proof Divisor D 7→ I−1D ⊂ OX .

Claim: ID is an invertible sheaf.

For ane U ⊂ X ID := I(U ∩D) ⊂ O(U)

(U, f) ∈ ID,x non-zero divisor, f = 0 = D ∪ C, g|C = 0, g|D 6= 0 takingh ∈ I(D) then from NSS f | (h · g)n ⇒ f | (h · g)n ∈ OD,X . Thus f | n inOD,x ⇒ k generates ID,x ⇒ on some open ane W 3 x, f generates ID,x(W ).

Claim: Let I ⊂ OX be an invertible ideal. Then ∃D s.t. I = O(−D) := ID.

32

Zeroes(I) ⊂ X. By principal ideal theorem, any component of Zeroes(I) hasdimension dimX − 1. Let E be one of these components. I ′

= I · OX(E) ) I.We may assume (by Noetherian induction) that I ′

= ID′ then I = IE+D′ .

Let I ∈IFI, T := I ∩ OX , then T is invertible and T I−1 ⊂ OX , I =T (JI−1)−1 ⇒ T = O(D)⇒ I = O(D − E)

Lemma

1. Pic(An) = OAn

2. Pic(Pn) = OPn(M) |M ∈ Z ∼= Z

Proof

1. O(An) is UFD ⇒ OAn,x is UFD. I ⊂ O(An). I is invertible ⇐⇒ I isprinciple.

2. If E is an irreducible divisor then E = f = 0, f is an irreduciblepolynomial of degree e. ⇒ IE = f · OP1(−e) ⇒ IF ∼= OPn(−e). It'sleft to show OPn(m) OPn(k) for m 6= k. (OPn(m) OPn(k) ⇐⇒OP1(n− k) OPn ⇐ dimOPn(l) = 1 ⇐⇒ l = 0.

[10.6.13]

Denition Assume we have a sheaf F inX, and a mapping ν : X → Y and wewish to dene a sheaf in Y . We will dene push forward by ν?F = F(ν−1(U))for U ⊂ Y .

We dene the pull back in the same case by ν?F(U) := lim→F(U) = (v,ξ)/∼

Example We already saw one example Xν→ pt, ν?F = Γ(F).

If we have i : pt→ X then i?F = Fp.

Lemma

1. (ν µ)? = µ?ν?, (ν µ)? = ν?µ?.

From this if we have Xν→ Y and pt, we get Γ(F) = Γ(ν?F), and ν?(G)x =

Gν(x) (where F is a sheaf in X and G in Y )

2. Hom(G, ν?F)

33

Lemma F is quasi-coherent then ν?(F) and µ?(F) are quasi-coherent. ∀ν :X → Y , µ : Y → Z

Exercise ν? is left adjoint to ν? also in alg. category

Corollary ν? is right exact, ν? is left exact

Exercise ν : X → pt⇒ ν?(k) = OX .

Exercise ν : X → Y , ν?(OY ) = OX .

Lemma ν = ν?top is exact

Proof Exactness can be checked per stock. Let 0 → F → G → H → 0 be anexact sequence of sheaves on Y . ∀p ∈ X, 0→ (ν?F)p → (ν?G)p → (ν?H)p → 0and can be mapped to the exact sequence 0→ Fν(p) → Gν(p) → Hν(p) → 0.

Lemma Let X = specA, Y = specB, ϕ : B → A denes ν : X → Y . Let Mbe a B module, N be an A module. Then

1. ν?M = M ⊗B A ∈MX

2. ν?N = ResN ∈MY

Proof

1. Bα → Bβ →M → 0, OαY → OβY → M → 0 then we can use the 5-lemma

on:

ν?OαY ∼= OαX↓ ↓

ν?OβY ∼= OβX↓ ↓

ν?M → M ⊗A↓ ↓0 → 0

2. It's enough to show ν?N(Yg) = Ng. ν?N(Yg) = N(ν−1(Yg)) = N(Xgν)

Exercise F is coherent ⇒ ν?F is coherent.

34

Proof OnY F , ν?OnY ∼= OnX ν?F .

Denition

1. An invertible sheaf is called very ample if ∃σ0, ..σn ∈ Γ(α) s.t. σ0|p, .., σn|pspan α|p ∀p ∈ X

2. α is called ample if ∃k s.t. α⊗ α⊗ ..⊗ α︸ ︷︷ ︸k

is very ample.

Lemma α is very ample ⇐⇒ ∃ν : X → Pn s.t. α = ν?O(1)

Proof ⇐: α|p = O(1)|ν(p)⇒: Let p ∈ X. Choose isomorphism ϕ : α|p ∼= K. Dene ν(p) := (ϕ(σ0|p), ..., ϕ(σn|p))

Exercise

1. X is ane ⇒ any invariant sheaf is very ample

2. αi is very ample on Xi ⇒ π1α1 ⊗ π2α2 is very ample in X1 ×X2

3. L is any inv. sheaf on a projective variety X, then L(m) is very ample form 0

Exercise

1. O(m)on Pn is very ample. Thus it denes Pn → PN , nd this map

2. Pn, O(1), Pk, O(1) 7→ Pn × Pk, O(1)⊗O(1) denes Pn × Pk → Pnk+n+k.Show that this is the Segre embedding.

Exercise Let 0 → A → B → C → 0 be an exact sequence of complexes:0 0 0↓ ↓ ↓

... → Ai → Ai+1 → Ai+2 → ...↓ ↓ ↓

... → Bi → Bi+1 → Bi+2 → ...↓ ↓ ↓

... → Ci → Ci+1 → Ci+2 → ...↓ ↓ ↓0 0 0

. Then ∃δi: Hi(C) → Hi+1(A)

that gives a long exact sequence ...→ Hi(A)→ Hi(B)→ Hi(C)→ Hi+1(A)→...

35

[17.6.13]

Homework 4

Question 1 Given in the lecture

Question 2 Will take us a bit out of scope, so we'll not solve it

Question 3 It's right exact asMs =M ⊗AAs, and left exact as 0→ Lϕ→M .

Looking at Lsϕs→ Ms, assume there is something in the kernel ϕs(

ls ) = 0 ⇒

ϕs(l) = 0⇒ ∃t : tϕ(l) = 0⇒ tl = 0⇒ l ∼ 0⇒ ls = 0.

It doesn't commute with innite direct sum as (∏∞i=1 Z)2 →

∏∞i=1(Z)2 the right

side has the element (1, 12 ,14 , ..)

Question 4 we can map ⊕Hom(M,Nα) → Hom(M,⊕Nα) easily by coor-dinates. It's injective, and we can get that it's surjective by looking at thegenerators

Question 5 ⇒: AssumeM is projective. Then there exists an N s.t. M⊕N =Ak. M and N have bers constant dimension bers, so M is locally free.

⇐: Lets show right exactness:

Lemma T : A−mod → B−mod, strongly right exact, then T ∼= T (A)⊗A ·

Proof a : A→ A and T (a) : T (A)→ T (A). Then:

Aα T (A)α ∼= T (A)α ∼= T (A)⊗Aα↓ ↓ ↓ ↓Aβ T (A)β ∼= T (A)β T (A)⊗Aβ↓ ↓ ↓M T (M) ← T (A)⊗AM↓ ↓ ↓0 0 0

where m : A→M by a 7→ a ·m, and thus we have T (A)→ T (M)

Sheaf Cohomology

1. ∀i, a functor Hi : Sh(X)→ Ab

2. ∀i, ∀0→ F → G → H → 0 a morphism. δi : Hi(H)→ Hi+1(F) s.t.

(a) Hi are additive,H0 = Γ

36

(b) δi are functorial (Hi(H) → Hi(H′

)↓ δi ↓ δi

Hi+1(F) → Hi+1(F)commutative)

(c) ∀ short exact sequences 0→ F → G → H → 0 the sequence is exact:0→ Γ(F)→ Γ(G)→ Γ(H)→ Hi(F)→ Hi(G)→ Hi(H)→ 0

(d) ∀F , ∃0→ F → I s.t. Hi(I) = 0 ∀i.

Denition I is called acyclic ⇐⇒ Hi(I) = 0 ∀i.

Denition Let X be an algebraic variety, F ∈ MX , Uii∈I open anecover of X. For p ≥ 0, I ordered set Cp(F) =

∏i0<..<ip

F(Ui0 ∩ ... ∩ Uip). dp :Cp(F)→ Cp+1(F). dP (α)i0,..,ip+1 =

∑p+1k=0(−1)kαi0,..,ik−1,ik+1,..,ip+1 |U0∩..∩Up+1 .

Denition Cp is called the Cech complex and Hi(F) := Hi(Cp(F))

Example

1. X is ane ⇒ the complex is 0→ F(x)→ 0→ 0..

2. X is projective Pn = U0 ∪ .. ∪ Un, Ui ∼= An. Thus ∀i > n : Hi = 0. Startwith x = P1, H0 = Γ, H2 = H3 = .. = 0, so we just needH1. X = U0∪U1,Γ = O. C1(O) = O(U0 ∩ U1) = f

xa0x

b1: f is homogeneous of deg a+ b =

spanxm0 x

n1

xa0x

b1: m < n ≤ a+b, 0→ C0(O) : O(U0)×O(U1)→ O(U0∩U1)→

0 so H0(P1,O) = Γ(O) = K , H1(P1,O) = 0

Exercise Show that O(d), d ≥ 0 is also acyclic

Question H1(P1,O(−2)) =?

Answer C1(O(−2)) = O(−2)(U0 ∩U1) =<xm0 x

n1

xa0x

b1: m+ n = a+ b− 2⇒ m ≥

a−1 or n ≥ b−1 the only intresting option is equality and thenH1(P1,O(−2)) =span < 1

x0x1>

Lemma 0→ F → G → H → 0 ⇒0→ Cp(F)→ Cp(G)→ Cp(H)→ 0

37

Proof We have 0 → F(Ui) → G(Ui) → H(Ui) → 0 since Ui is ane and wehad 0 → F|Ui → G(Ui) → H(Ui) → 0, and in the last lesson we saw that thenthere is a δi as needed.

Now we sawO on P1 is acyclic. SetX = P1, take 0→ O(−2)→ O → Kp⊕Kq →0 ⇒ 0→ H0(X,O(−2))︸ ︷︷ ︸

=0

→ H0(X,O)︸ ︷︷ ︸=K

→ H0(X,Kp ⊕Kq)︸ ︷︷ ︸Kp⊕Kq

→ H1(X,O(−2))︸ ︷︷ ︸gen by 1/x0x1

→

H1(X,O)︸ ︷︷ ︸0

.

24.6.13

Let F be a coherent sheaf on Pn. Then ∃ ⊕ O(di) → F → 0. Let X = Pn,S = k[x0, .., xn], X = Uo ∪ ... ∪ Un.

Proposition

1. ⊕d∈ZH0(X,O(d)) = S as graded k−algebra.

2. ⊕d∈ZHn(X,O(d)) = S′where S′ ∼= S with grading shifted by S′d

∼=S−(n+1)−d .

3. Hi(X,O(d)) = 0 ∀0 < i < n.

Note Hi commute with ⊕.

Proof

1. H0 = Γ

2. F = ⊕d∈ZO(d) Ui0,..,ik := Ui0 ∩ ... ∩ Uik = xi0 · ... · xik 6= 0 ⇒F(Ui0,..,ik) = Sxi0

..xik. Thus the Chech complex of F is:

∏i0Sxi0

→∏i0<i1

Sxi0xi1→ ... →

∏j Sx0..xj−1xj+1..xn → Sx0..xn thus Hn(X,F) =

coker(∏j Sx0..xj−1xj+1..xn → Sx0..xn) = <x

j00 ...xjn

n >/<xj00 ...xjn

n :ji≥0 for some i> =<

xj00 ...xjnn : ji < 0 ∀i >= 1

x0...xnk[x−10 , .., x−1n ] so we know d 7→ −(n+1)− d

3. By induction on n. For n = 1 it is clear.

Assume n > 1, dene H = xn = 0 ∼= Pn−1, we have 0→ OX(d− 1)·xn→

OX(d)xn=0→ OH(d) → 0. Now take ⊕d to get 0 → F(−1) ·xn→ F →

FH → 0, where F(−1) = F ⊗OxOx(−1). Thus (for n > 2) we have

0 → H0(X,F(−1)) → H0(X,F) → H0(X,FH) → H1(X,F(−1)) →H1(X,F)→ 0.

38

H1(x,OH(d) ∼= H1(Pn−1,O(d)) = 0 by induction we know 0→ Hi(X,F(−1))→Hi(X,F)→ 0 for 1 < i < n−1 and 0→ Hn−1(X,F(−1))→ Hn−1(X,F)→Hn−1(X,FH)→ Hn(X,F(−1))→ Hn(X,F)→ 0.

Now take i = 1 and: 0→ k[x0, .., xn]xn→ k[x0, .., xn]

onto→ k0[x0, .., xn−1]0→

0, so 0→ H1(X,F(−1))→ H1(X,F)→ 0⇒H1(X,F(−1)) ∼= H1(X,F).Similarly, Hi(X,F(−1)) ∼= Hi(X,F) ∀1 ≤ i ≤ n− 1. Taking localizationin xn: H

i(X,F)xn = Hi(Un,F|Un) = 0, thus: ∀α ∈ Hi(X,F), ∃k s.t.Xkn · α = 0, so α = 0.

Theorem Let X be a projective variety F coh. sheaf on X. Then:

1. Hi(X,F) is f.d.

2. Hi(X,F(d)) = 0 ∀i > 0 ∀d 0

Proof

1. X → Pn. The cover will be taken to be intersection of X with the stan-dard cover of Pn. Hi(X,F) = Hi(Pn, i∗F). Thus can assume X = Pn.We move by descending induction on i.

For i > n O.K.

Now we have: 0→ R→ ⊕ki=1O(di)→ F → 0, thus: 0→ R(d)→ ⊕O(di+

d)→ F(d)→ 0.

0 → H0(R) → H0(⊕ki=1O(di) → H0(F)→ H1(R) → H1(⊕ki=1O(di) → H1(F)→ ... ... .. .. ..→ Hi(R) → Hi(⊕ki=1O(di) → Hi(F) → Hi+1(R)

⇒

dimHi(F) <∞.

2. Take d very big, we get 0 = Hi(⊕O(di+d))→ Hi(F(d))→ Hi+1(R(d))by induction on i

=0

Remark. If we just wanted to prove that Γ(F) is nite-dimensional we wouldnot have a shorter proof.

Now we shall show that the cohomologies do not depend on the cover chosen:

Lemma X is ane, X = ∪ni=1Ui nite cover. F ∈ MX ⇒ HiUi(F) = 0.

∀i > 0. (for H0 we know it's Γ)

39

Proof Dene a Chech complex of sheaves by CpUi(F) :=∏i∗F(Ui0 ∩ ..∩Uip)

.

Its global sections form the usual Chech complex CUi(F). However, CpUi(F)

is exact, and Γis an exact functor (since X is ane), thus CUi(F) is also exact,and Hi

Ui(F) = 0.

Corollary(Exc) X = ∪ki=1Ui, U0is any other open ane subset, then theChech cohomologies dened by U0, U, .., Un and U1, .., Un coincide.

Corollary Chech cohomologies do not depend on the cover.

[1.7.2013]

Lemma ∀X,Y ∀π : X → Y ane morphism.F ∈MX ,Hi(X,F) ∼= Hi(Y, π∗F)

Proof Let U i be ane cover of Y , V i := π−1(U i). CUi(π∗F) ∼= CV i(F)

Lemma If π : X → Y and X,Y are ane then π∗ is an equivalence ofcategoriesMX

∼=MY .

Proof If we compose with the equivalence ΓY we get the equivalence ΓX

Corollary π ane ⇒ π∗ is exact.

Denition Riπ∗ :MX →MY sheacation of V 7→ Hi(π−1(V ),F|π−1(V ))

These functors are called higher direct images, we can view them as relativecohomologies.

Lemma(Exc) Long exact sequence

0 → π∗F → π∗G → π∗H→ R1π∗F → R1π∗G → R1π∗H→ ...

Proposition

1. Riπ∗, F ∈MY

2. ∀ open ane V ⊂ Y , (Riπ∗F)|V ∼= Hi(π−1(V ),F)∼ ∈ MY , where ~means localization.

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Proof We can assume that Y is ane, V = Y . We have a natural mapHi(X,F)→ Γ(Y,Riπ∗F) (since ∃F → Sh(F) for all presheaves). Hi(X,F)∼ →Riπ∗F . Hi(X,F)my

→ lim−−−→n3y

Hi(π−1(u),F). We will show that this is an iso-

morphism on any basic open subset: Hi(X,F)gi → Hi(π−1(Yg),F). Theinclusionπ−1(Yg) ⊂ X is ane, thusHi(π−1(Yg),F) ' Hi(Xg,F) ' Hi(X,Fgi) 'Hi(X,F)gi.

LemmaX × Y

X Y

, OX ⊗k Hi(Y,F) ∼= Ri(πX)∗(π∗Y F)

Proof The morphism is dened byHi(Y,F)→ Hi(X×Y, π∗Y F)→ Γ(X,Ri(πX)∗(π∗Y F)).

To prove that it is an isomorphism we may argue locally, thus assume that Xis ane and prove isomorphism of global sections. Note that in this case theprojection πY is also ane, and thusΓ(X,Ri(πX)∗(π

∗Y F)) ∼= Hi(X×Y, π∗Y F) ∼=

k[X]⊗k Hi(Y,F).Remark. Sheaves were invented by Lere in order to compute cohomologies offamilies of topological spaces. Higher direct image computes cohomology of afamily of algebraic varieties with coecients in a familiy of sheaves. This lemmatreated the (trivial) case of constant families.

Denition F is abby if ∀U ⊂ V ⊂ X Res : F(U)→ F(U) is onto.

Exercise Flabby ⇒ acyclic

Denition Rat(F) := lim−−−−→U⊂XF(U). Rat(F) := constant sheaf with sections

Rat(F). F B→Rat(F). F ′:= ImB

Denition Let F be a coherent sheaf on a curve C. A section is called torsionif supp(s) is nite.

F is called torsion if any section is torsion (⇐⇒ supp(F) is nite)F is called torsion-free if no s ∈ F(u) non-zero section is torsion.

Exercise 0→ Ftorsion → F → F′ → 0 is a s.e.s.

Lemma F is (coherent and) torsion ⇐⇒ F = ⊕ki=1Fci , i.e. F is a directsum of skyscrapers.

We dene Div(F) :=∑ki=1(dimFci)ci

From now on we assume that C is smooth.

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Lemma TFAE:

1. F is torsion-free and dimRat(F) = 1

2. F ∼=sheaf of fractional ideals

3. F is invertible

Proof (1)⇒ (2) F ∼= F′ ⊂ Rat(F) ∼= Rat(O)

(2) ⇒ (3) ∀c ∈ C, Fc is rank 1 and torsion-free over OC,c, OC,c is DVR,thusFc∼= OC,c.

(3)⇒ (1) o.k.

Let D be an eective divisor, F ⊂ Rat(O) invertible sheaf. F ⊗O O(D) =F(D) = F·OC(D) ⊂ Rat(F). F|D ∼= F/F(D)is torsion sheaf, and its divisor isD.

From now on, let F be an invertible sheaf.

Denition PrincF := Rat(F)/Fc. Prin(F)(U) := ⊕c∈UPrinc(F)

Lemma 0→ F → Rat(F)→ Prin(F)→ 0 so 0→ H0(C,F)→ Rat(F)α →Γ(Prin(F))→ H1(C,F)→ 0(= Hi(Rat(F))), i.e. H1(C,F) = coker(α)

From now on, let C be complete.

Lemma H1(C,OC) = 0 ⇐⇒ c = P1

Proof H1 = 0 ⇒ α is surjective ⇒ ∃ rational function with single pole ⇒C ∼= P1. Now assume C ∼= P1, note that for C = A1, α is onto. Let p ∈ Prin(O),∃f ∈ Rat s.t. s = p−α(f) is zero except at ∞. s =

∑1≤i≤k ait

i ⇒ s ∈ Imα⇒p ∈ Imα.

Lemma F is acyclic ⇐⇒ ∀ eective divisor D, ∀c ∈ C, dimk Γ(C,F(D +c)) = dimΓ(C,F(D)) + 1

Proof ∀ eective divisor E, Γ(F(E)/F) is a subspace in Prin(F) of dimensiondeg(E). Imα ∩ Γ(F(E)/F) ∼= Γ(F(E))/Γ(F)

H1(C,F) = 0 ⇐⇒ α is onto ⇐⇒ ∀ E deg(E) = dimΓ(F(E)/F) = dimΓ(F(E))−dimΓ(F)⇐⇒ ∀ eective divisorD, ∀c ∈ C, dimk Γ(C,F(D+c)) = dimΓ(C,F(D))+1

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Lemma (Exc)

1. 0 → V0 → V1 → ... → Vn → 0 e.s. of vector spaces of nite dimensionthen

∑(−1)i dimVi = 0

2. for a complex,∑

(−1)i dimVi =∑

(−1)i dimHi. This number is calledthe Euler characteristic of the complex.

Denition χ(F) := dimH0(C,F) − dimH1(C,F) . This number is calledthe Euler characteristic of the sheaf.

Lemma (Exc) 0→ F → G → H → 0 then χ(G) = χ(F) + χ(H)

Denition g := dimH1(C,OC) - the genus

Proposition χ(F) = deg(F) + 1− g

Proof For F = O by denition. Thus it's enough to show χ(F(c)) = χ(F)+1∀c ∈ C. This follows from the standard s.e.s. 0 → F → F(c) → F(c)|c → 0,since χ(F (c)|c) = Γ(F(c)|c) = 1.

Corollary

1.dimH0(C,F) ≥ deg(F) + 1− g

2.dimH1(C,F) ≥ − degF − (1− g)

Denition Consider the diagonal embedding C ∼= ∆C ⊂ C × C, and denethe cotangent sheaf byΩC := OC×C(−∆C)|∆C .

Lemma Γ(C,ΩC ⊗F) = 0⇒ H1(C,F) = 0

Proof Dene G := F(D) for some eective divisor D. Γ(C,Ωc ⊗ G) =

Γ(C, (Ωc ⊗F(−D)) ⊂ Γ(C,Ωc ⊗F) = 0

Let c ∈ C 0→ G → G(c)→ G(c)|C → 0. δc : H0(C,G(c)|C)→ H1(C,G)

Step 1 δc = 0 ∀c

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Proof ∆C ⊂ C×C, 0→ π∗1G → π∗1G(∆)→ π∗1(G)(∆)|∆ → 0. δ : π2∗(π∗1G(∆)|∆)→

R1π2∗π∗1G ∼= H1(C,G)⊗K OC the ber of δ at c is δc.

OC×C(−∆)|∆ ∼= ΩC ⇒ π2∗(π∗1G(∆)|∆) ∼= G ⊗Oc ΩC . δ : H1(G) ⊗K OC →

ΩC ⊗Oc G, (δ)c : H1(C,G) → Γ(ΩC ⊗ G) = 0⇒ δ = 0

Step 2 H1(C,F) = 0

Proof From step 1, 0 → Γ(G) → Γ(G(c)) → Γ(G(c)c)0→ thus dimΓ(G(c)) =

dimΓ(G) + 1⇒ from previous lemma, F is acyclic.

Corollary degF > degΩc ⇒ H1(C,F) = 0

Corollary Any proper open subset U ⊂ C is ane.

Proof take c ⊂ U , dimH0(O(n, c)) > 1 for n 0

Lemma H1(C,Ωc) ∼= k

Proof Take F of the maximal degree s.t. H1(C,F) 6= 0. Then for any point

c,Γ(F(c)|c)δc→ H1(C,F)→ H1(C,F(c)) = 0 so δc is isom. ⇒ dimH1(C,F) = 1

and δ : F ⊕ Ωc∼→ H1(C,F)⊗OC so Ωc ∼= F .

Assume Now considerα ∈ Hom(F ,ΩC). Then we have H1(α) : H1(C,F) →H1(C,Ωc) = k. Then we haveH1 : Hom(F ,Ωc)→ Homk(H

1(C,F),H1(C,Ωc)) =

H1(C,F)

Serre proved that this is an isomorphism.

Theorem (Serre) H1 : Hom(F ,Ωc)∼→ H1(C,F)

This implies

Theorem (Weil-Riemann-Roch) dimk Γ(F)−dimHom(F ,Ωc) = deg(F)+1− g

Proof LHS = dimΓ(F)− dimH1(F) = χ(F) = RHS

From this we get Riemann-Roch as Hom(F ,Ωc) = Γ(Hom(F ,Ωc)) = Γ(F ⊗Ωc)

Now let us prove Serre's theorem

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Lemma Assume that we have a commutative diagram:δ : Ωc ⊗Oc

Ωc∼= H1(C,Ωc)⊗k Oc.

↑ α⊗ 1 ↑ λ⊗ 1

δF F ⊗Oc Ωc → H1(C,F)⊗k Oc

then α determines λ and conversely

Lemma⇒Theorem Given α ∈ Hom(F ,Ωc), let λ = H1(α). From thelemma, H1 is injective.

Now take λ ∈ Homk(H1(C,F),H1(C,Ωc)), and take α that forms together

with it a diagram as in the lemma. Then the lemma implies λ = H1(α).

Proof (lemma) Step 1 λ determines α: since the upper horizontal arrow isan isomorphism.

Step 2 α determines λ: α determines λ|Σc∈Cimδ|c . It's enough to nd c1, ..., cnon C s.t. H1(C,F(

∑ci)) = 0. For that it is enough to have Γ(c,Ωc ⊗

F(−∑ci) = 0. For any section we can nd a point where it does not van-

ish. Adding this point we reduce the dimension of the vector space of globalsections, and eventualy make it vanish. 0 < .. < dimΓ(c,Ωc ⊗ F(−c1) <dimΓ(c,Ωc ⊗F∗) <∞.

Remark Both Serre's duality and Riemann-Roch theorem have far-going gen-eralizations to higher dimensions.

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