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# Algebraic Surfaces

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CHAPTER 1

### Prelimanary

1. An Example

This section is devoted to illustrate the connection between compact Riemann surface and complex algebraic curve. We will basically work out the example of elliptic curves and leave the general discussion for interested reader.

Let Λ ⊂ C be a lattice, that is, Λ = Zλ1+Zλ2with C = Rλ1+Rλ2. Then an elliptic curve E := C/Λ is an abelian group.

On the other hand, it’s a compact Riemann surface. One would like to ask whether there are holomorphic function or meromorphic functions on E or not.

To this end, let π : C → E be the natural map, which is a group ho- momorphism (algebra), holomorphic function (complex analysis), and covering (topology). A function ¯f on E gives a function f on C which is double periodic, i.e.,

f (z + λ1) = f (z), f (z + λ2) = f (z), ∀z ∈ C.

Recall that we have the following well-known result:

Theorem 1.1 (Liouville). A bounded entire function, i.e. a holo- morphic function on C with bounded image, is constant.

Corollary 1.2. There is no non-constant holomorphic function on E.

Proof. First note that if ¯f is a holomorphic function on E, then it induces a doubly periodic holomorphic function f on C such that f ◦π = f . It then suffices to claim that a a doubly periodic holomorphic¯ function on C is bounded.

To do this, consider D := {z = α1λ1 + α2λ2|0 ≤ αi ≤ 1}. The image of f is f (D). However, D is compact, so is f (D) and hence f (D) is bounded. By Liouville theorem, f is constant and hence so is ¯f . ¤ The next hope is to ask if there is a meromorphic function on E with only a simple pole or not. The answer is NO, which can be proved by residue theorem.

Exercise 1.3. Prove that there is no non-zero meromorphic func- tion on E with only a simple pole

1

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2 1. PRELIMANARY

Hint. If the pole is not on the boundary of D, then consider the path integral along the boundary of D, otherwise consider the path integral along a translation of D. By using Residue theorem, one can see it.

Then the next hope is to look for functions with pole of order 2.

Luckily, we have one, which is the Weierstrass P-function, P(z) := z−2+ X

ω∈Λ−{0}

((z − ω)−2− ω−2).

And P0 is a functions with pole of order 3. By direct computation, one sees that

{wp_equation}

Lemma 1.4.

P0(z)2 = 4P(z)3− g2P(z) − g3. Where

g2 = 60 X

ω∈Λ−{0}

ω−4

and

g3 = 140 X

ω∈Λ−{0}

ω−6.

Exercise 1.5. Verify that P is doubly peiodic, i.e. P(z+λ) = P(z) for all z ∈ C and λ ∈ Λ.

Work out the computation in Lemma 1.4.

Theorem 1.6. An elliptic curve can be embedded into P2C as a non- singular cubic.

Sketch. We considering the map ϕ : E − {0} → C2 given by

¯

z 7→ (P(z), P0(z)).

This map can be extended to ϕ : E → P2 as

½ ϕ(¯z) = [P(z), P0(z), 1] if ¯z 6= 0, ϕ(¯z) = [0, 1, 0] if ¯z = 0.

The affine defining equation in C2 is y2 = 4x3− g2x − g3. And the projective defining equation is

y2z = 4x3− g2xz2− g3z3.

One can verify that this cubic is non-singular and the map ϕ is an

embedding. ¤

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2. DIVISORS 3

2. Divisors

Similar phenomena occurs for any compact Riemann surface. As the above section suggested, the essential point is to find enough func- tions and then determine the algebraic relation between those func- tions. All these can be done for any compact Riemann surface. Thus the purpose of this section is to show the following

Theorem 2.1. Any compact Riemann surface can be embedded into a projective space as an algebraic curve.

To study functions more systematically, it’s natural to consider di- visors.

Definition 2.2. Let X be a compact Riemann surface. A divisor, denoted D = P

niPi, is a finite formal sum of finite points (codim=1).

Given divisors D1 =P

niPi and D2 =P

miPi, one can define D1+ D2 :=X

(ni+ mi)Pi.

Let Div(X) be the set of all divisors on X. It’s clear that Div(X) is a free abelian group under the addition defined above. In fact, one can think Div(X) as the free abelian group on the set X.

Given a meromorphic function f on X, one can count its zeros and poles with multiplicity. This give rise to the following group homomor- phism

div : M(X) − {0} → Div(X),

where M(X) denotes the field of meromorphic functions on X.

Exercise 2.3. A meromorphic function on a compact Riemann surface X has at most finitely many zeros and poles.

Hence div is well-defined.

The idea for divisors is to collect information on poles and zeros.

We denote the functions with prescribed poles and zeros, collected in D =P

niPi, as

L(D) := {f ∈ M(X) − {0}|div(f ) + D ≥ 0}.

It’s clearly a vector space over the ground field and its dimension is denoted l(D). Another important notion for divisor is the degree, which is

deg(D) :=X ni.

Example 2.4. Let X = C ∪ {∞} be the Riemann sphere. Then M(X) ∼= C(z). We use the notation of [x] to denote the divisor of the point with coordinate z = x.

Let f1(z) = z, f2(z) = 1/(z − 1), f3(z) = z/(z − 1)2. By easy computation, one finds that div(f1) = 1[0] − 1[∞] since it has a zero

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4 1. PRELIMANARY

at 0 and a pole at ∞. Similarly, div(f2) = 1[∞] − 1[1], div(f3) = 1[0] + 1[∞] − 2[1].

Now fix a divisor D = 2[1]. What is L(D)? What does it mean? It is nothing but the set of meromorphic functions with at most a pole of order 2 at [1] and no other poles. More precisely,

L(D) = {g(z)/(z − 1)2|deg(g(z)) ≤ 2}.

Because if deg(g(z)) ≥ 3 then it gives a pole at ∞.

We recall some definitions and properties of divisors.

Definition 2.5. Let D = P

niPi be a divisor. We said that D is effective if ni ≥ 0 for all i, denoted D ≥ 0.

WE write D1 ≥ D2 if D1− D2 ≥ 0.

Definition 2.6. Given divisors D, D0, they are said to be linearly equivalent, denoted D ∼ D0, if D − D0 = div(f ) for some f ∈ M(X).

The linear series |D| is thus defined to be the set

|D| := {D0 ∈ Div(X)|D0 ≥ 0, D0 ∼ D}.

Notice that we have a induced natural map π : L(D) − {0} → |D|,

f 7→ div(f ) + D.

If one identify L(D) ∼= Cn, then |D| can be identified as Pn−1. In particular, one has that if L(D) 6= 0, then

dim |D| = dim L(D) − 1.

It is interesting an important to determine the dimension of L(D), denoted l(D). We recall the following fact.

{p1} Proposition 2.7. Let D = P

niPi be a divisor. We said that D is effective if ni ≥ 0 for all i, denoted D ≥ 0. Suppose now that D is an effective non-zero divisor then L(−D) = {0}.

We leave the proof as an exercise. (Hint: prove that a holomorphic function on a compact Riemann surface must be constant).

Among all divisors, there is a most important one, called canonical divisor, denoted KX. It is a divisor associate to meromorphic 1-form.

Example 2.8. Let X = C ∪ {∞} be the Riemann sphere. X can be covered by coordinate charts (U0, z), (U1, w), where U0 = X − {∞} ∼= C and U1 = X − {0} ∼= C. Note that w = 1z.

We consider a holomorphic 1-form dz on U0. This can be extended into a meromorphic 1-form on U1, because near ∞, z = w1 and hence dz = −1w2dw. Therefore, the extended meromorphic 1-form gives rise to a divisor −2[∞]. This is a canonical divisor of X.

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2. DIVISORS 5

Warning. One can pick other meromorphic 1-form and then pro- duce a different canonical divisor. However, all these are linearly equiv- alent. (Exercise: check it). Therefore, to be more precise, one should consider the equivalent classes of divisors of 1-forms. Nevertheless, in most real application, the quantity that we are going to compute is the same inside the equivalent class. Therefore, we usually abuse the notation by picking any one inside the class and call it the canonical divisor KX.

In general, l(D) can be computed or estimated by the most impor- tant theorem for curves, the Riemann-Roch theorem:

Theorem 2.9 (Riemann-Roch).

l(D) − l(KX − D) = deg(D) + 1 − g(X),

where KX denotes the canonical divisor, l(D) denotes the dimension of L(D), and g(X) is the genus of the curve X.

We will leave the sketch of the proof to next section. In this section, we concentrate on its applications.

We now redo the example of elliptic curve with the help of Riemann- Roch theorem.

Let E = C/Λ be an elliptic curve. Then the genus is 1. The 1-from dz on C is doubly periodic, hence induced a 1-form dz on E. Thus KX = 0.

Let D 0 be an effective divisor. By Prop. 2.7, K = 0, and Riemann-Roch, we have:

l(D) = deg(D).

Let ¯0 be the image of π(0) in E.The vector space L(k[¯0]) has a nat- ural basis {1}, {1, P}, {1, P, P0} respectively when k = 1, 2, 3. How- ever, when k = 6, one finds that l(6[¯0]) = 6 and thus {1, P, P0, P2, PP0, P3, P02} must be linearly dependent. Hence there must be a relation between them involving P3, P02.

By using L(3[¯0]), one produces a meromorphic map ϕ : E 99K P(L(3[¯0])) = P2.

The linear dependency shows that the image satisfies a cubic polyno- mial. This recover the embedding given in the previous section.

Turning to a compact Riemann surface X in general. Following the above construction, we would like to ask if there is a divisor D such that L(D) has enough sections in the sense that

ϕD : X 99K P(L(D)), is an embedding.

By Riemnann-Roch Theorem, one can prove that:

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6 1. PRELIMANARY

Theorem 2.10. Let X be a compact Riemann surface of genus g.

Suppose that D is a divisor with deg(D) ≥ 2g + 1, then ϕD is en embedding.

Sketch. ¤

3. Riemann-Roch Theorem 4. Algebraic Varieties

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CHAPTER 2

### Algebraic Surfaces

1. Riemann-Roch Theorem on Surfaces

Let D be a divisor on X. If D0 ∼ D, then O(D) ∼= O(D0). Thus sometime it’s useful to pick a better element in |D| instead of looking at D itself.

Theorem 1.1 (Bertini). Let Bs|D| be the base locus of |D|. If dim|D| ≥ 1, then the general member of |D| is non-singular away from the Bs|D|.

In particular, if |D| is base point free, then general member in |D|

is non-singular.

Proof. Fix D0, D1 ∈ |D| with local equation f, f + g on an affine open set U ⊂ X. We have a subseries (a pencil) Dλ ∈ |D| locally defined by f + λg. Suppose that Pλ ∈ U is a singular point of Dλ not in the base locus B := Bs|D|. We may assume that g(Pλ) 6= 0. We have

f + λg(Pλ) = 0,

and

∂zi(f + λg)(Pλ) = 0, for all zi. Where z1, ..., zn are the local coordinates.

These equations defines a subvariety in Z ⊂ U × P1. And let V = pr1(Z) ⊂ U.

One note that f /g is locally constant (−λ) on V − B. Hence for λ different from value of f /g, Dλ is non-singular away from B (in U).

Next one notice that one can cover X by finitely many affine open

sets. ¤

Remark 1.2. If |D| is very ample, then it is base point free.

Definition 1.3. A divisor D is said to be ample if mD is very ample for some m > 0.

Our first aim is the following:

{decomp}

Theorem 1.4. Let D be a divisor on a projective variety X. There exist a very ample divisor A such that A + D is very ample.

Corollary 1.5. Let D be a divisor on a projective variety X.

Then there are non-singular very ample divisor Y1, Y2 such that D ∼ Y1− Y2

7

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8 2. ALGEBRAIC SURFACES

Lemma 1.6. The following are equivalent:

(1) D is ample.

(2) For every coherent sheaf F on X, we have Hi(X, F⊗O(nD)) = 0 for all i > 0 and n À 0.

(3) For every coherent sheaf F on X, F⊗O(nD) is globally gen- erated for all n À 0.

{crit1}

Remark 1.7. A sheaf is globally generated if the natural map H0(X, F)⊗OX F is surjective. If F = O(D) for some divisor D, then O(D) is globally

generated if and only if D is base point free.

Exercise 1.8. Show that if D1 is very ample and D2 is base point free, then D1+ D2 is very ample.

(Hint: consider the subspace L(D1)⊗L(D2) ⊂ L(D1+ D2). Show that the map defined by this subspace is everywhere defined and an embedding. Thus the map defined by D1+ D2 is an embedding.

proof of theorem 1.4. X is projective, then X ,→ Pn for some n. Take H a hyperplane in Pn , then H ∩ X is a very ample divisor on X. By abuse the notation, we still called it a hyperplane section and denote it as H.

Note that very ample is clearly ample. Hence by the Lemma 1.7 (3), there is an n0such that D+n0H is base point free. By the exercise,

D + (n0 + 1)H is very ample. ¤

We are now able to define intersection of subvarieties. We start by considering intersection on surface.

Theorem 1.9. Let X be a non-singular projective surface. There is a unique pairing Div(X) × Div(X) → Z, denoted by C.D for any two divisor C, D, such that

(1) if C and D are non-singular curves meeting transversally, then C.D = #(C ∩ D),

(2) it is symmetric. i.e. C.D = D.C,

(3) it is additive. i.e. (C1+ C2).D = C1.D + C2.D,

(4) it depends only on the linear equivalence classes. i.e. if C1 C2 then C1.D = C2.D.

Proof. See [Ha, V 1.1]. ¤

Remark 1.10. Let X be a projective variety. An 1-cycle is a formal linear combination of irreducible curves. The group of all 1-cycles is denoted Z1(X) (=free abelian group on irreducible curves). One can similarly define a pairing Z1(X) × Div(X) → Z.

Two curves C1, C2 are said to be numerically equivalent if C1.D = C2.D for all D, denoted C1 ≡ C2. We define

N1(X) := Z1(X)⊗R/ ≡ .

It’s a famous theorem asserts that N1(X) is finite dimensional. Its dimensional is called Picard number, denoted ρ(X).

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1. RIEMANN-ROCH THEOREM ON SURFACES 9

Remark 1.11. Let V ⊂ X be a subvariety of codimension i, and D is a divisor. Then it make sense to consider V.Di by decomposing D ∼ H1− H2 and then compute (V ∩ Hi).Di−1 in V ∩ Hi inductively on dimension. One can simply set

V.Di := (V ∩ H1).Di−1− (V ∩ H2).Di−1.

Remark 1.12. Let X be a variety over C. A divisor D gives a class c1(D) ∈ H2(X, Z) via Div(X) → H1(X, O) → H2(X, Z). And a curve C give rise to a class [C] ∈ H2(X, Z). The pairing H2(X, Z) × H2(X, Z) → Z gives an intersection theory.

An important feature of ampleness is that it’s indeed a ”numerical property”.

Theorem 1.13 (Nakai’s criterion). Let X be a projective variety.

A divisor D is ample if and only V.Di > 0 for all subvariety of codi- mension i.

In particular, if dimX = 2, then D is ample if and only if D.D > 0 and D.C > 0 for all irreducible curve C.

Another important criterion is due to Kleiman. Let NE(X) ⊂ N1(X) be the cone generated by effective curves. And let NE(X) be its closure.

For any divisor D, it defines a linear functional on N1(X) and we set D>0 = {x ∈ N1(X)|(x.D) > 0}.

Theorem 1.14 (Kleiman’s criterion). D is ample if and only if D>0 ⊃ NE(X) − {0}.

Before we revisit the Riemann-Roch theorem on surface, we need the useful adjunction formula:

Proposition 1.15 (Adjunction formula). Let S ⊂ X be a non- singular subvariety of codimension 1 in a non-singular variety X. Then KS := KX + S|S. In particular, if dimX = 2 then 2g(S) − 2 = (KX + S).S

Given a codimension 1 subvariety Y ⊂ X and a divisor D ∈ Div(X) . One can consider the restriction D|Y, which is supposedly to be a divisor. However, this is not totally trivial. For D = P

niDi, one might want to consider naively that D|Y :=P

ni(Di ∩ Y ). But what if Di = Y for some i? That is, how to define Y |Y?

One way to think of this is that we deform Y such that limt→0Yt = Y , then we take Y |Y := limt→0Yt|Y. (This needs some extra care).

proof of adjunction formula. Recall that a canonical divisor is a divisor defined by n-forms if dimX = n. Thus one has ΩnX = OX(KX). Where ΩnX denote the sheaf of n-forms on X. Also one can

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10 2. ALGEBRAIC SURFACES

consider sheaf of n-forms on X with pole along S, denoted ΩnX(S). It’s clear that ΩnX(S) ∼= OX(KX + S).

One has the following exact sequence

0 → OX(KX) → OX(KX + S) → OS(KX + S|S) → 0. (2.2.1) On the other hand, one has the Poincar´e residue map

nX(S) ³ Ωn−1S

with kernel ΩnX. Hence we have an exact sequence

0 → ΩnX → ΩnX(S) ³ Ωn−1S → 0 (2.2.2) Comparing these two sequences, one sees that Ωn−1S ∼= OS(KX + S|S).

Hence the canonical divisor KS = KX + S|S.

We now describe the Poincar´e residue map. (cf. [G-H, p147]). The problem is local in nature, it suffices to describe it locally. We may assume that on a small open set U, S is defined by f . And let z1, .., zn be the local coordinates of U.

The sheaf ΩnX(S) on U can be written as ω = g(z)dzf (z)1∧...∧dn. Since S is non-singular, then at least one of ∂z∂f

i 6= 0. The residue map send ω to

ω0 := (−1)i−1g(z)dz1∧ ... ∧ cdzi∧ ... ∧ dn

∂f /∂zi |f =0. This is independent of choice of i since on S

df = ∂f

∂z1dz1+ ... + ∂f

∂zndzn = 0.

Another way to put it is that the residue map sends ω to ω0 such that ω = dff ∧ ω0.

It’s clear that the ω0 = 0 if and only if f (z)|g(z), which means that ω is indeed in ΩnX.

¤ Theorem 1.16 (Riemann-Roch theorem for divisors on surfaces).

Let X be a non-singular projective surface and D ∈ Div(X) a divisor on X, then one has

χ(X, D) = χ(X, OX) + 1

2D.(D − KX).

Proof. Write D ∼ H1− H2 with Hi are non-singular very ample divisor. We consider the sequences:

0 → O(D) ∼= O(H1− H2) → O(H1) → OH2(H1) → 0, 0 → O → O(H1) → OH1(H1) → 0.

It’s clear that

χ(X, D) = χ(X, H1) − χ(H2, OH2(H1))

= χ(X, OX) + χ(H1, OH1(H1)) − χ(H2, OH2(H1)).

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2. BLOWING-UP AND BLOWING-DOWN 11

By Riemann-Roch on curves and adjunction formula, χ(H1, OH1(H1)) = H1.H1+ 1 − g(H1) = H1.H1+ 1 −1

2(KX+ H1).H1, χ(H2, OH2(H1)) = H1.H2+ 1 − 1

2(KX + H2).H2. Collecting terms, one has

χ(X, D) = χ(X, OX)+1

2(H1−H2).(H1−H2−KX) = χ(X, OX)+1

2D.(D−KX).

¤ 2. Blowing-up and Blowing-down

Remark 2.1. The construction of blowing up can be found almost in any book. (Some called it σ-process however). We refer [Beauville, complex algebraic surfaces, chap. II]. However, Beauville only proved that the map h is a bijective morphism. It would be a good exercise to prove that h indeed an isomorphism.

In this section, we introduce the important notion of blowing-up.

This process is essential in studying singularities and hence birational geometry in general.

We first introduce the local version. Let An be the affine space with coordinates z0, ..., zn−1 and 0 ∈ An be the ”origin”. We construct a variety Y ⊂ An× Pn−1 by {ziXj = zjXi}i6=j, where X0, ..., Xn are the homogeneous coordinates of Pn−1. There is a natural morphism π : Y → An by projection. One sees that π−1(0) ∼= Pn−1 and π : Y − π−1(0) ∼= An− {0}. We say Y is the blowing-up of An at 0 and denoted Bl0(Y ).

In general, let x ∈ X be a point in a variety X. Pick an open affine neighborhood U of x. We identify (U, x) with an open set (U0, 0) ⊂ An. Then one has eU := π−1(U0) → U0 which is the blowing-up of U0 at 0.

Glue X − U and eU together, we get πX : eX → X. Which is called the blowing-up of X at x. Note that one has similarly that π−1X (x) ∼= Pn−1 and πX : eX − πX−1(x) ∼= X − {x}. The divisor πX−1(x) is called the exceptional divisor, and usually denoted E.

Exercise 2.2. Let π : X = Blx(P2) → P2 be the blowing-up of P2 at a point x ∈ P2. Prove that

KX = πKP2 + E by local coordinate computation.

In fact, if dimX = 2, π : eX → X is a blowing-up at a point x ∈ X, then

KX˜ = πKX + E.

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12 2. ALGEBRAIC SURFACES

More generally, if dimX = n, and π : eX = Blx(X) → X is the blowing- up at x, then

KX˜ = πKX + (n − 1)E.

Let’s play a little bit around the blowing-ups. Let’s restrict our- selves to surfaces. One might expect that there are similar higher- dimensional formulation. Let X be a surface, and C ⊂ X be a curve.

Let f be the local equation of C around x. By fixing local coordinates z1, z2, we can write

f = f (z1, z2) = fm+ fm+1+ ...

with fm 6= 0. We define the multiplicity of C at x to be mx(C) := m.

One can have an equivalent definition by vanishing order of partial differentials. Hence one can check the mx(C) is well-defined.

We consider π : eX = Blx(X) → X. And let C be a curve passing through x ∈ X. Then π−1(C) consists of irreducible components, E and the other part maps onto C. The part maps onto C can be defined

as C := πe −1(C − {x}),

which is called the proper transform of C. Thus we have π−1(C) = C ∪ E. More precisely, by computing the equations, one hase

πC = eC + mx(C)E, this is called the total transform of C.

We here collect some properties regarding the blowing-up on sur- face.

Proposition 2.3. Let π : eX = Blx(X) → X be the blowing-up at x ∈ X. Then one has:

(1) There is a natural isomorphism Div(X) ⊕ ZE → Div( eX) by (D, nE) 7→ πD + nE. And the isomorphism induces an iso- morphism PicX ⊕ ZE → Pic eX.

(2) Let D, D0 ∈ Div(X), then (πD).(πD0) = D.D0. (3) Let D ∈ Div(X), then (πD).E = 0.

(4) E.E = −1.

Proof. It’s easy to check the isomorphism given in (1).

For (2) and (3), it follows by choosing ∆, ∆0 which are linear equiv- alent to D, D0 respectively but not passing through x.

For (4), by adjunction formula and the fact the E ∼= P1,

−2 = deg(KE) = (KX˜ + E).E = (πKX + 2E).E = 2E.E.

¤ The blowing-up gives the first example of binational morphism.

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2. BLOWING-UP AND BLOWING-DOWN 13

Definition 2.4. By a rational map f : X 99K Y from X to Y , we mean a regular function on a dense Zariski open (or simply non-empty Zariski open) set U ⊂ X.

More precisely, a rational map can be written as (U, f ) where U ⊂ X is a dense Zariski-open set and f : U → Y is regular.

We say (U, f ) ∼ (V, g) if f = g on U ∩ V . In fact, a precise definition of rational map should be the equivalent class of the pairs (U, f ). However, we usually abuse the notation if no confusion is likely.

Definition 2.5. A rational map φ : X 99K Y is said to be bira- tional if it admits an inverse. That is, there is an ψ : Y 99K X such that ψ ◦ φ = idX, φ ◦ ψ = idY

Example 2.6. Let π : Y = Bl0(A2) → A2, take ψ : A2−{0} → Y ⊂ A2×P1 such that ψ(x, y) = ((x, y), [x, y]). Then ψ◦π = idY, π◦ψ = idX. Hence π is a birational morphism.

Exercise 2.7. The following are equivalent:

(1) X and Y are birationally equivalent.

(2) there are non-empty open subset U ⊂ X and V ⊂ Y such that U, V are isomorphic.

(3) K(X) ∼= K(Y ) as k-algebra.

Given a variety X, one can obtain various birational equivalent varieties

... → Xn → Xn−1 → ... → X1 → X

by successive blowing-ups. It’s also a natural question to ask if X is obtained by blowing-ups? Another way to put it is if X minimal or not? The precise formulation of minimal model in any dimension is quite subtle.

We start by working on contraction on surfaces. In order to produce a minimal object, we need to tell whether a surface X is obtained from blowing-ups.

Definition 2.8. Let C ⊂ X be a curve on X, we say that C is a (−1)-curve if C ∼= P1 and C2 = −1

We seen that we can have a (−1)-curve by blowing-up. In fact we will prove that any (−1)-curve comes from blowing-ups.

Theorem 2.9 (Caltelnuovo). Let X be a surface (non-singular complex projective surface) with E ⊂ X a (−1)-curve. Then there is a morphism π : X → X0 with X0 non-singular such that π is the blowing-up of X0 with exceptional divisor E.

Proof. The idea is to construct a morphism which is identical at E but isomorphic outside E.

First pick H any very ample divisor on X, Let k := H.E > 0. We consider H0 = H + kE, then H0.E = 0. Notice that the restriction map

H0(X, O(H0)) → H0(E, O(H0|E) = OE) ∼= H0(P1, O) ∼= C.

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14 2. ALGEBRAIC SURFACES

Hence the map ϕH0 produce by |H0| is constant on E. We need to refine H so that ϕH0 is isomorphic outside E.

To this end, we first pick any very ample H0. It’s clear that nH0 is very ample for all n > 0. On the other hand, H0 is ample, one can arrange that H := nH0 is very ample with H1(X, O(H)) = 0. 1

Consider the exact sequence

0 → OX(H+(i−1)E) → OX(H+iE) → OE(H+iE|E) = OE(k−i) → 0.

Claim. H1(X, O(H + iE)) = 0 for all 1 ≤ i ≤ k.

Grant this for the time being, then one has an exact sequence

0 → H0(X, OX(H+(i−1)E)) → H0(X, OX(H+iE)) → H0(E, OE(k−i)) → 0.

Note that H0(E, OE(k−i)) is of dimension k−i+1, let ai,0, ..., ai,k−i H0(X, O(H + iE)) be the lifting of a basis in H0(E, O(k − i)).

Remark. Before we move on, we would like to remark the dif- ference between H0(X, O(D)) and L(D). It actually comes from two possible definition of O(D). If we define the sheaf O(D) as O(D)(U) = {f ∈ K(X)|div(f ) + D|U ≥ 0 on U}. Then H0(X, O(D)) = L(D).

However, another way to look at the sheaf O(D) is to consider it as the sheaf of sections line bundle associate to D. Then under this con- sideration, for s ∈ H0(X, O(D)), div(s) gives an effective divisor Ds linearly equivalent to D. To view it as L(D) is the classical treatment.

The modern viewpoint tends to think it as section of line bundles. We take the convention that H0(X, O(D)) represents the global section of line bundle of D from now on.

Let me describe the correspondence in more detail. Given a divisor D, one has a system of local equations (Uifi). The basic idea behind the notion of line bundle is instead of looking at functions, we look at local functions satisfying given patching conditions. The correspondence is given as

L(D) → H0(X, O(D)), f 7→ (Ui, f fi) = s.

And the correspondence between their divisor is given by div(s) = div(f ) + D,

which is an effective divisor Ds ∈ |D|.

Turning back to the proof, let s ∈ H0(X, O(E)) be a section such that div(s) = E. Then the map H0(X, OX(H + (i − 1)E)) → H0(X, OX(H + iE)) is given by multiplying s. Therefore, by working on the sequence inductively, one can have a basis of H0(X, O(H +kE)), given as

{s0sk, ..., snsk, a1,0sk−1, ..., a1,k−1sk−1, ..., ak−1,0s, ak−1,1s, ak}.

1This follows by picking n À 0 such that nH0− KX is ample. By Kodaira vanishing theorem, we have H1(Xω⊗O(nJ0− KX)) = 0.

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2. BLOWING-UP AND BLOWING-DOWN 15

We consider the map ϕH0 : X → PN given by the above basis.

Note that ak∈ H0(X, O(H0)) whose restriction to E is a non-zero con- stant. Hence one has ϕ is well-defined along E and ϕ(E) = [0, ..., ak] = [0, ..., 1]. Moreover, for x 6∈ E, s(x) 6= 0, hence

[s0sk(x), ..., snsk(x)] = [s0(x), ..., sn(x)] = ϕH(x).

Since H is very ample, ϕH defines an embedding on X and hence on X − E. One sees that the first n + 1 coordinate of ϕH0 gives an embedding on X −E already, so it follows that ϕH0 gives an embedding on X − E.

It remains to show that X0 := ϕH0(X) is non-singular. Let U ⊂ X be the open subset defined by ak 6= 0. It’s clear that E ⊂ U. We want to identify U with an open set V ⊂ fA2 ⊂ A2×P1. This can be achieved by considering

h : U → A2× P1, x 7→¡

(ak−1,0s

ak (x),ak−1,1s

ak (x)), [ak−1,0(x), ak−1,1(x)]¢ .

We might need to shrink U so that ak−1,0(x) and ak−1,1(x) are not simultaneously vanishing. It’s obvious that h factor through fA2. Let V = h(U) ⊂ fA2. Moreover, one has the commutative diagram

U −−−→ fh A2

ϕH0



y π

 y ϕH0(U) −−−→ A¯h 2, where ¯h = (ak−1,0a s

k ,ak−1,1a s

k ) is a rational map on PN defined on ϕH0(U).

Another remark is that h clearly maps E ⊂ U onto E ⊂ fA2. It suffices to show that h : U → V is an isomorphism. Because, the induced map

¯h is an isomorphism . Therefore, ϕH0(U) is non-singular at ϕH0(E), which is the only possible singularity.

However, to show that h is an isomorphism is not trivial. One can first prove that it’s a hemeomorphism, hence in particular, bijective.

Then one prove the h induces isomorphism on all local rings. (cf. [Ha.

Ex I.3.2, I.3.3]) ¤

{neg_int}

Lemma 2.10. [KM 3.40] Let Y → X be a resolution of a normal surface with Y projective. Let Ei be the exceptional divisors. Then binary form (Ei· Ej) is negative definite.

Proof. Let D = P

aiEi. We would like to prove that D2 < 0.

Suppose on the contrary that D2 ≥ 0. Assume that D is effective.

Pick H an ample divisor on Y such that H − KY is ample and H2(Y, O(H + nD)) = 0 for all n À 0.

Then by Riemann-Roch, we have

h0(Y, O(H + nD)) − h1(Y, O(H + nD)) = χ(Y, O(H + nD)),

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16 2. ALGEBRAIC SURFACES

goes to infinity as n goes to infinity.

However, let f(nD + H) be the Weil divisor by push-forward. It’s clear that

H0(Y, O(nD + H)) ⊂ H0(X, OX(f(nD + H))) = H0(X, OX(f(H))), which is bounded by a finite dimensional space. This is the required contradiction.

If D = D++ D, then D2 ≤ (D2++ D2) < 0. We are done. ¤ 3. Minimal Model Program for Surfaces

We will need the following terminology and statements from mini- mal model program.

Definition 3.1. Let X be a smooth projective variety. It is said to be minimal if KX is nef.

Suppose that −KX is not nef then the Cone Theorem and the Contraction Theorem in minimal model program asserts the following:

Theorem 3.2 (Cone Theorem). Let X be a nonsingular projective variety. Suppose that −KX is not nef. Then there is a rational curve l in X such that −(dim X + 1) ≤ −KX · l < 0.

Theorem 3.3 (Contraction Theorem). Let X be a nonsingular pro- jective variety. Suppose that −KX is not nef. Then there is a contrac- tion map ϕ : X → Y which is an algebraic fiber space such that ϕ(l) is a point. We also have that the relative Picard number ρ(X/Y ) = 1.

Moreover, a curve C ⊂ X is contracted if and only if C ≡num λl for some λ ∈ Q.

By applying these two theorem to surfaces with −KX is not nef, we end up with the following three situations.

I. KX · l = −1.

Then −2 = (KX + l) · l gives that l2 = −1. In other words, l is a (−1) curve. It’s easy to verify that ϕ contracts only l. Moreover, one can check that ϕ is the blowing down.

II. KX · l = −2.

Then similarly, we have l2 = 0. We claim that dim Y = 1. Suppose that dim Y = 0, then every curve in X is contracted. In particular, an ample curve h is contracted and h ≡num λl. However,

0 < h · l = λl2 = 0.

This is a contradiction. Suppose that dim Y = 2. We take a Stein factorization X → Z → Y . We may assume Z is normal and hence X → Z is a resolution. By Lemma 2.10, l is contracted hence l2 < 0, a contradiction.

We have seen that dim Y = 1 with connected fibers. Let f be a general fiber. One sees that −KX· f < 0 and f2 = 0. It follows that f must be a rational curve. Thus ϕ : X → Y is a ruled surface.

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4. CANONICAL SURFACE SINGULARITIES 17

III. KX · l = −3.

Then we have l2 = 1. We claim that dim Y = 0. Suppose that dim Y = 2, then we argue as above. Suppose that dim Y = 1, then one sees that l2 = 0, a contradiction. Thus we have dim Y = 0.

Now ρ(X) = 1, hence every divisor is proportional to l numerically.

In particular, KX ≡ λl. Adjunction formula gives KX ≡ −3l. We verify that l is ample. To see this, note that l2 = 1 > 0. Also, for any curve l 6= C ⊂ X, we have C ≡ αl for some α 6= 0. Since l · C = αl2 = α ≥ 0, thus α > 0 and l · C > 0.

Next we claim that |l| = |KX + 4l| gives an isomorphism to P2. To see this, we verify that h2(OX) = h0(Ω2X) = 0 for KX ≡ −3l can not be effective. Also h1(OX) = h0(Ω1X) = 0, otherwise there is a non- trivial Albanese map. If the Albanese image has dimension ≥ 2, then one has a non-zero two-form. If the Albanese image has dimension 1, then one has a fibration map with fiber f2 = 0. Both cases lead to a contradiction. We conclude that χ(O) = 1. Thus by vanishing theorem and Riemann-Roch theorem we have h0(l) = χ(l) = 3.

Now by Reider’s theorem, one has |KX+ 4l| is base point. In total,

|l| produce a morphism

ψ : X → P2.

Again, one can claim that image of ψ has dimension 2. That is, ψ is surjective and generically finite. Let h be the hyperplane section in P2, we have

1 = l2 = ψ(h)2 = deg(ψ)h2 = deg(ψ).

Thus ψ is birational. It’s easy to see that there is no exceptional curve since Picard number is one. Hence ψ is an isomorphism. This concludes the proof.

4. Canonical Surface Singularities

Given a non-singular surface of general type, by contracting (−1) curves, we might assume that it is non-singular minimal surface. Its canonical divisor KX is nef and big. That is KX · C ≥ 0 for all C and KX2 > 0.

In this section, we would like to discuss the canonical model for a surface of general type. In general, it might be singular. We shall discuss he possible singularities as well.

Theorem 4.1 (Base point freeness theorem). Let X be a non- singular projective variety. D is a nef divisor such that D − KX is nef and big. Then |mD| is base point free for m À 0.

Proof. This is a special case of the well-known Base-Point-Freeness Theorem in Minimal Model Program. We refer the reader to [?] for a

general statement and proof. ¤

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18 2. ALGEBRAIC SURFACES

An immediate application to surface of general type is that we have a morphism ϕm : X → P for m À 0. Indeed, one sees that ϕm has only connected fiber hence birational by increasing m if necessary (cf.

[?]). In total, we end up with a birational map ϕ : X → Y := ϕm(X) ⊂ PN.

Y is called the canonical model of X, sometimes denoted Xcan. One notices that ϕ(C) = pt if and only if C · KX = 0.

We would like to discuss this in a more general setting.

Definition 4.2. A point y ∈ Y is said to be a Du Val singularity if there is a resolution f : X → Y such that KX·Ei = 0 for all exceptional curve Ei.

We need to require that the above resolution to be a minimal res- olution in the sense there there is no (-1) curve over y.

We have seen that a canonical model for a surface of general type has at worst DuVal singularities.

{DuVal}

Proposition 4.3. Let f : X → Y 3 y be a resolution of a Du Val singularity y. Let f−1(y) = ∪Ei be the exceptional set. Then Ei2 = −2 for all i and P

Ei is a normal crossing divisor.

Proof. For an exceptional curve Ei ⊂ X that f (Ei) = y, by Lemma 2.10, Ei2 < 0. Thus 2pa(Ei) − 2 = (KX+ Ei) · Ei < 0. It follows that Ei is a rational curve and Ei2 = −2.

Moreover, if Ei∩ Ej 6= 0 for some i 6= j. We consider C = Ei+ Ej. By Lemma 2.10, we have

0 > C2 = Ei2 + Ej2+ 2Ei· Ej = −4 + 2Ei· Ej. Hence Ei· Ej = 1.

Furthermore, suppose P = Ei∩ Ej then P 6∈ Ek for any k 6= i, j.

To see this, suppose on the contrary that P ∈ Ek. Then we consider C = Ei+ Ej+ Ek. By Lemma 2.10, we have

0 > C2 = −6+2Ei·Ej+2Ei·Ek+2Ej·Ek = −4+2Ei·Ek+2Ej·Ek ≥ 0,

We will need some easy combinatorics to explore some geometry of surfaces.

Include the file

Proposition 4.4. Let Γ be a graph and Q be its associate qua- dratic form. Then q is negative definite if and only if Γ is of the type An, Dn, E6, E7, E8.

A rational curve C with C2 = −2 is called (−2)-curve. Given an exceptional set f−1(y) = ∪Ei, we associate a Dykin diagram Γ as following: vertex are Ei, edge connecting Ei, Ej if Ei∩ Ej 6= ∅.

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4. CANONICAL SURFACE SINGULARITIES 19

The associated quadratic form Q(Γ) is the same as the intersection form on {Ei}. A singularity which associated Dykin diagram of its minimal resolution is of A − D − E type is called an A-D-E singularity.

Therefore, Du Val singularities are A-D-E singularities as well.

Next, we would like to give an explicit description of A-D-E singu- larities. We recall some result of Artin’s.

Theorem 4.5. Let f : X → Y 3 y be a resolution of a surface singularity with exceptional set f−1(y) = ∪Ei.

(1) There is a cycleP

aiEi such that ai > 0 for all i and Z ·Ei ≤ 0 for all i. The minimal one with this property is called the

”fundamental cycle” Z.

(2) y is rational, i.e. R1fOX = 0, if and only if pa(Z) = 0.

(3) mult(Q) = −Z2.

(4) the embedding dimension dim m/m2 = −Z2+ 1.

Definition 4.6. A surface singularity y ∈ Y is called a rational double point if it is a rational singularities with multiplicity 2.

It’s easy to write down a fundamental cycle for each A-D-E surface singularities and show that they are rational double points by Artin’s result. Now we would like to show that a rational double point is ana- lytically equivalent to one of the following hypersurface singularities.











An x2+ y2+ zn+1 = 0 Dn x2+ y2z + zn−1 = 0 E6 x2+ y3+ z4 = 0 E7 x2+ y3+ y3z = 0 E8 x2+ y3+ z5 = 0

(∗)

We will need the following four methods repeatedly:

(1) The Weierstrass preparation theorem

(2) The elimination of the term yn−1 from the polynomial yn+ ....

(3) Hensel’s lemma: let f be a formal power series with leading term fd. Assume that fd = gh, then f = GH where G, H having leading terms g, h respectively.

(4) Let M be a monomial, then uM is equivalent to M by a suit- able coordinate change.

Step 1. By (1) and (2), we have F = (unit) · (x2+ f (y, z)).

Step 2. If mult0f (y, z) ≤ 2, then by (1) and (2), we have f = (unit) · (y2 + (unit) · zm). This gives equation of An type by (3) and (4).

Step 3. If mult0f (y, z) ≥ 4. Set X := (p2+r−4f (qr, r) = 0). Then π : (p, q, r) 7→ (x = pr2, y = qr, z = r)

maps X → Y is a resolution. Computation shows that it’s not Du Val.

We thus assume that mult0f (y, z) = 3, we write f = f3 + ....

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20 2. ALGEBRAIC SURFACES

Step 4. Assume that f3 is not a cube. We may assume that f = z(y2+ (unit) · zm) by (1),(2). This gives Dn type by (3),(4).

Step 5. We thus assume that f3 = y3. We write f = y3+ yzaua+ zbub where a ≥ 3, b ≥ 4 and ua, ub are either units or zero.

Step 6. Either a ≤ 3 or b ≤ 5.

To see this, suppose on the contrary that a ≥ 4 and b ≥ 6, then set X := (p2+ q3+ qra−4ua(pr3, qr2, r) + rb−6(pr3, qr2, r) = 0. Then

π : (p, q, r) 7→ (x = pr3, y = qr2, z = r)

gives a birational morphism. Computation shows that Y is not canon- ical.

Step 7. a ≥ b−1, b = 4, 5. We have f = y3+yzaua+zbub. If a ≥ b, then we make it into f = y3+ zbub, then we are done. If a = b − 1, then we make it into f = y3+ y2zb−2u + zbu0 by (2). We can make it into f = y3 + zbu because 2(b − 2) ≥ b. Thus we get E6, E8 types.

Step 8. Finally, we consider f = y3 + yz3ua+ zbub. We consider a blowup by y = y1z1, z = z1, one sees that f is reducible for it is reducible after blow-up. Hence we assume that f = y(y2 + yz2ua +

z3u0) ¤

Direct computation by these equations shows that there are Du Val. Take Y = (x2 + y2 + zn+1 = 0) ⊂ A3 =: A for example. Let π : B → A be the blowing-up with exceptional divisor E and X be the proper transform of Y . One has KB = πKA+2E, and X = πY −2E.

By adjunction, one sees that KX = πKY. Note that on X, there is a singularity defined by (x2+ y2+ zn−1 = 0) if n ≥ 2. Note also that E ∩ X = D1+ D2 defined by (x2 + y2 = 0). Keep blowing-up, we end up with a resolution ˜π : ˜X → Y such that KX˜ = ˜πKY. Hence Y has only Du Val singularities. One also verify that its Dykin diagram is An.

Therefore we summarize the following:

Theorem 4.7. The following are equivalent:

(1) q ∈ Y is a Du Val singularity.

(2) q ∈ Y is an A-D-E singularities.

(3) q ∈ Y is a rational double point.

(4) q ∈ Y is analytically isomorphic to a hypersurface singularity with equations given above in (∗).

We have also seen that the singularities appeared in the canonical model of a surface of general type must be Du Val.

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