Rational curves on algebraic varieties

(1)

Rational curves on algebraic varieties

Wibrich Drost

August 14, 2015

Master thesis

Supervisor: dr. M. Shen

(2)

Abstract

The Graber-Harris-Starr theorem is a useful theorem: it shows that a map from a variety to a curve has a section if a general fibre is separably rationally connected. In this thesis we state and prove this theorem. To reach this point, we first define the Hilbert scheme, secondly we determine when a scheme contains a rational curve, and finally we give a method to deform such a curve. This deformation theory is the main part in the proof of the Graber-Harris-Starr theorem.

Title: Rational curves on algebraic varieties Author: Wibrich Drost

Supervisor: dr. M. Shen

Second grader: prof. dr. L.D.J. Taelman Date: August 14, 2015

Korteweg-De Vries Institute for Mathematics University of Amsterdam

Science Park 105-107, 1098 XG Amsterdam http://www.science.uva.nl/math

(3)

Introduction

In the beginning of the 21th century the mathematicians Tom Graber, Joe Harris and Jason Starr proved a very useful theorem, now this theorem is known as the Graber-Harris-Starr theorem. This theorem shows that a morphism from a scheme to a smooth curve has a section if a general fibre is separably rationally connected. The goal of this thesis is to give the formal statement and a proof of this theorem.

To reach this statement we show the existence of a rational curve in a variety and give a way to deform an arbitrary curve using rational curves. To be able to implement this, some knowledge of the Hilbert scheme is required. Therefore, the first chapter introduces the Hilbert scheme of a scheme X. If X is a scheme over a field k, then we have that the Hilbert scheme of X is a scheme in which the underlying set contains all closed subschemes. In this chapter we also introduce the scheme of morphisms, which is a subscheme of the Hilbert scheme. Subsequently we calculate the tangent space of both schemes and then extend this construction to give estimates of the local dimension of these schemes. These estimates appear to be very useful in the remainder of this thesis. In the second chapter we give and prove the bend and break theorem. This theorem shows that, under certain conditions, we can deform a curve inside a scheme X in several components such that one component is a rational curve. The degree of the rational curve can be huge. We show that, if this degree is above some bound, there is a way to split this rational curve into several rational curves, which gives that we can assume that the degree of the rational curve is below the given bound. Actually the conditions of the bend and break theorem are too strong, we give this stronger version of the bend and break theorem. In the proof of this theorem, we use the upper bound of the constructed rational curve in the bend and break theorem.

Let X be a scheme and let D ⊂ X be a curve in X. The main condition the schemes in chapter 3 have to satisfy is the following condition.

(∗) Through a general point x ∈ D there is a rational curve C on X in a general direction v ∈ ND/X,x such that H1(C, NC/X) = 0.

Since this implies that the scheme X must contain a lot of rational curves, the discussion in the second chapter gives an idea whether our scheme satisfies this condition. Next this condition is used to give a method for deforming a curve into a smooth curve. First we assume that C is a smooth curve in a scheme X. Most of the time we cannot deform C directly, since the obstruction space is non-vanishing. We show that we can add rational curves C1, . . . , Cn to our curve C such that the resulting curve is a comb and the

obstruction space is zero. The following figure visualizes the idea of this comb.

C

(6)

Since the obstruction space of this comb vanishes, we can deform it into a smooth curve. This gives a way to deform a smooth curve in a scheme X. As a next step we extend this deformation to nodal curves; curves with finitely many singular points, such that the local model at each singular point is of the form xy = 0.

With these constructions in mind we are able to formulate the more advanced Graber-Harris-Starr theorem in the last chapter. In the first part of chapter 4 we introduce the notion of a (separably) rationally connected variety. These two notions are the same if the variety is a variety over a field with characteristic 0. Finally, we conclude with the proof of the Graber-Harris-Starr theorem. In the proof we use the theory on deforming a curve to construct a section.

Prerequisites

The reader should be familiar with the notion of a scheme and some other basic algebraic geometry, a good reference for this is [Har77]. In this thesis I refer back to this book for some elementary results.

The terminology in this thesis follows [Har77] as well. For example, a variety is an integral separated scheme of finite type over a field k (not necessary algebraically closed). Acknowledgment

I would like to thank my supervisor Mingmin Shen for his support, useful comments, good explanation of the problems and the pleasant way of cooperation during my thesis.

(7)

1. The Hilbert scheme

The Hilbert scheme of a scheme X contains important information about the scheme X. In this chapter we derive the Hilbert scheme from the Hilbert functor. We also define for two schemes X and Y the scheme of homomorphisms X → Y , which is a subscheme of the Hilbert scheme of X × Y . In the second section we calculate the tangent space of both schemes. From this we can construct some important estimates about the local dimension of both schemes. The Hilbert scheme, the scheme of homomorphism and the estimates of the local dimension will play an essential role in the later chapters.

1.1. Construction of the Hilbert scheme

In this section we define the Hilbert scheme. If X is a scheme over a field k, then we have that the Hilbert scheme of X, Hilb(X), is a scheme in which the underlying set contains all closed subschemes of X. Furthermore we define for schemes X and Y a subscheme Hom(X, Y ) of Hilb(X × Y ) that consists of all morphisms from X to Y .

Definition 1.1. Let g : X → S a projective morphism of schemes and assume that OX(1) is an relative ample line bundle. Let Z ⊂ X be a closed subscheme of X such

that it is flat over S. Then the Hilbert polynomial of Z is a polynomial P such that for n 0 we have that the rank of g∗OZ(n) is equal to P (n).

Let Z be defined as in the above definition. Then we have that the Hilbert polynomial exist, for a proof see [Har77, I Theorem 7.5, p. 51]. With the above definition in mind, we can define the Hilbert functor and a subfunctor that gives subschemes with a certain Hilbert polynomial.

Definition 1.2. Let X be a projective scheme over a scheme S. (i) The Hilbert functor of X

Hilb (X) : SchS → Set

is defined for a scheme Z over S by

Hilb (X)(Z) = subschemes V ⊂ X ×S Z which are proper and flat over Z.

(ii) Let P be a polynomial. The functor

HilbP(X) : SchS → Set

is defined for a scheme Z over S as follows, HilbP(X)(Z) =

subschemes V ⊂ X ×_SZ which are proper and flat over Z and have Hilbert polynomial P.

(8)

If Z is a connected scheme we have that Hilb (X)(Z) =[

P

HilbP(X)(Z), (1.1)

where the union is disjoint. Since we want to define the Hilbert scheme, we want to prove that the Hilbert functor is representable. The following theorem gives the precise statement.

Theorem 1.3. Let X be a projective scheme over a scheme S. The Hilbert functor of X is representable by a scheme Hilb (X), called the Hilbert scheme of X. Let P be a polynomial. Then also the functor HilbP(X) is representable by a scheme, HilbP(X),

which is projective over S.

Proof. For the proof of the second claim see [Kol96, I Theorem 1.4, p. 10, proof p. 12-16]. Then the first claim in the theorem follows from equation (1.1).

The representability of the Hilbert functor of a scheme X gives that there is a natural isomorphism Hilb (X)(−) ∼= HomS(−, Hilb (X)). Let [U ] ∈ Hilb (X)(Hilb (X)) be the

element that corresponds with the identity morphism in HomS(Hilb(X), Hilb(X)). Let

T be an arbitrary scheme over S and let [Z] ∈ Hilb (X)(T ) be an element corresponding with a morphism ϕ : T → Hilb(X). In this setting we have that the following diagram is a pushout diagram. Z T U Hilb(X) ϕ

By this reason we call [U ] the universal family of the functor Hilb (X). To give an idea what the Hilbert scheme of a scheme X is, we give an explicit example.

Example 1.4. Let X = Pn

k for a field k and let g : X → Speck be a projective morphism.

We are going to give some examples of the Hilbert scheme of X for some fixed Hilbert polynomials.

Let p ∈ X be a point, then we have that rank (g∗Op(m)) = 1 for all m ∈ Z. Now let

Z ⊂ X be a closed subscheme, such that rank (g∗OZ(m)) = 1 for every m ∈ Z. Then we

have that

k ∼= g∗OZ(m)(Spec k) = OZ(m)(g−1(Spec k)) = OZ(m)(Z).

This shows that Z must be a point from which it follows that Hilb1(X) = {points in X} ∼= X.

Now let l = P1 ⊂ X be a line in X. We have for m ∈ Z that rank (g∗Ol(m)) = dimkH0(l, Ol(m))

= dimkH0(P1, OP1(m))

(9)

Since all possible subschemes Z ⊂ Pn _{with the property that rank (g}

∗OZ(m)) = m + 1

are lines we have that

Hilbt+1(X) = {lines in X} ∼= Grass(2, kn+1).

In a similar way as for lines on Pn _{we could prove that for a hyperplane H = P}n−1 _{⊂ X}

the rank of g∗OH(m) is equal to m+n−1_n−1 . From this it follows that

Hilb(t+n−1

n−1 )(X) = {hyperplanes in X}

∼

= P kn+1∗ .

Now we define an important subscheme of the Hilbert scheme, the Hom scheme. This scheme is also defined by the representability of a functor, namely the following functor. Definition 1.5. Let X, Y be schemes over a base scheme S. The functor

HomS(X, Y ) : SchS → Set

is defined for Z ∈ SchS by

HomS(X, Y )(Z) = HomZ(X ×SZ, Y ×SZ).

We can identify a morphism f ∈ HomS(X, Y ) with its graph Γf ⊂ X ×S Y . With

this identification we have that HomS(X, Y ) is a subfunctor of Hilb (X ×S Y ). For this

subfunctor we have the following theorem.

Theorem 1.6. Let X and Y be projective schemes over a scheme S and suppose that X is flat over S. Then we have that the functor HomS(X, Y ) is represented by an open

subscheme HomS(X, Y ) ⊂ Hilb(X ×SY ).

Proof of theorem 1.6. For the proof see [Kol96, I Theorem 1.10, p. 16].

The idea behind this theorem is that all points of Hilb(X ×SY ) that correspond with

a graph of a morphism f : X → Y over S form a subscheme of Hilb(X ×SY ).

Now we give an example of the scheme of morphism in order to given an idea of the elements of this scheme.

Example 1.7. Let X = Y = P1

k, we want to calculate Hom (X, Y ). We have that a

morphism ϕ : P1 → P1 _{is given by [x}

0 : x1] 7→ [ϕ0 : ϕ1] for ϕ0, ϕ1 ∈ k[x0, x1] homogeneous

polynomials of the same degree. First assume that ϕ has degree 0. Then we have that ϕ is a constant map, which gives that the scheme of morphisms ϕ ∈ Hom(P1_{, P}1_{) of degree}

0 is isomorphic to P1k.

Now assume that the degree of ϕ is equal to 1. Then we have that ϕ0 = a00x0+ a01x1

and ϕ1 = a10x0+ a11x1, for a00, a01, a10, a11 ∈ k such that a00a11− a10a016= 0. This shows

that

{ϕ ∈ Hom(P1

, P1) : deg(ϕ) = 1} ∼_{= P}3\ {a00a11− a10a01 = 0}

∼ = PGL2

(10)

Finally, assume that deg ϕ = n for n ∈ N. Then we have that ϕ0 and ϕ1 are given by ϕ0 = a0nxn₀ + a₀n−1₁xn−1₀ x₁+ · · · + a₀₁n−1x₀xn−1₁ + a₁nxn₁ ϕ1 = b0nxn₀ + b₀n−1₁xn−1₀ x₁+ · · · + b₀₁n−1x₀xn−1₁ + b₁nxn₁ such that det              a0n a₀n−1₁ . . . a₁n 0 0 . . . 0 0 a0n a₀n−1₁ . . . a₁n 0 . . . 0 .. . . .. . .. . .. 0 . . . a0n a₀n−1₁ . . . a₁n b0n b₀n−1₁ . . . b₁n 0 0 . . . 0 0 b0n b₀n−1₁ . . . b₁n 0 . . . 0 .. . . .. . .. . .. 0 . . . b0n b₀n−1₁ . . . b₁n              6= 0

This shows that all morphisms of degree n are isomorphic with the subscheme of P2n+1

defined by the above equation. So we have that Hom (P1_{, P}1_{) is the union of all these}

affine projective spaces.

The Hom scheme allows all morphisms between two schemes. Now we construct a closed subscheme of the Hom scheme that consist of all morphisms that are equal on a closed subscheme.

Definition 1.8. Let X and Y be schemes over S. Let Z be a closed subscheme of X that is a scheme over S and let p : Z → Y be an S-morphism. The functor

Hom (X, Y ; p) : SchS → Set

is defined for a scheme T ∈ SchS by

Hom (X, Y ; p)(T ) = T -morphisms f : X ×ST → Y ×ST such that f |Z×ST = p ×SidT.

Also this functor is representable by a scheme, see the following theorem.

Theorem 1.9. Let X be an projective scheme that is flat over a scheme S and let Y be a quasi-projective scheme over S. Let Z be a closed subscheme of X that is projective and flat over S. Let p : Z → Y be an S-morphism. Then we have that the functor Hom (X, Y ; p) is representable by a closed subscheme Hom(X, Y ; p) ⊂ Hom(X, Y ). Proof. For the proof see [Mor79, Prop. 1, p. 595].

To give an idea why this scheme is a closed subscheme of the scheme of homomorphism we calculate the scheme Hom(P1_{, P}1_{, 0 7→ 0). We construct this scheme in a similar way}

(11)

Example 1.10. Let again X = Y = P1

k. We are going to calculate Hom(X, Y, 0 7→ 0).

Let ϕ ∈ Hom(P1_{, P}1_{, 0 7→ 0) be an element. First assume that deg ϕ = 0, then we have}

that ϕ is a constant map. Since we must have that 0 7→ 0, this implies that ϕ = 0. Now assume that deg ϕ = 1. Example 1.7 shows that ϕ is given by ϕ0 = a00x0+ a01x1

and ϕ1 = a10x0 + a11x1. In projective coordinates we have that 0 = (0 : 1). Since ϕ

maps 0 to 0, we conclude that a01= 0 and a11= 1. This shows that the subscheme of all

morphisms of Hom(P1_{, P}1_{, 0 7→ 0), that has degree 1, form the closed subscheme of PGL} 2

defined by a01 = 0 and a11 = 1.

Finally, assume that deg ϕ = n for n ∈ N. The condition 0 7→ 0 gives that a1n = 0

and b1n = 1 in the notation of example 1.7. So we have again that the scheme of all

morphisms of Hom(P1_{, P}1_{, 0 7→ 0) with degree n form the closed subspace of the scheme}

of all morphisms of Hom(P1, P1) of degree n defined by a1n = 0 and b₁n = 1.

So in this example the scheme Hom (X, Y, 0 7→ 0), which the union is of the above constructed subschemes, is indeed a closed subscheme of Hom(X, Y ).

1.2. Local structure of the Hilbert scheme

Now we are ready to give some important estimates of the local dimension of the Hilbert and Hom-scheme. First we calculate the tangent space of both schemes. Then we gen-eralize the setup to show what the obstruction is of extending a subscheme of X over a ring A to a subscheme of X × Spec B, where B is an extension of A. Finally, we use these results to give some estimates about the local dimension of the Hilbert and Hom-scheme. We will use these estimates in the following chapters about the existence of rational curves and deforming curves.

1.2.1. The tangent space

In this section we are going to calculate the tangent space of the Hilbert scheme and the scheme of morphisms. First we show that the tangent space at a point is isomorphic to a set of morphisms. For this we look at the scheme Spec k[ε], where k[ε] = k[t]/t2 for a field k.

We have the projection k[ε] → k induces a closed embedding Spec k → Spec k[ε]. Since there is exactly one k-morphism k[ε] → k, we have that Spec k → Spec k[ε] is the unique closed embedding. This shows that Spec k[ε] contains Spec k as its unique closed point. With this in mind we can prove the following lemma.

Lemma 1.11. Let Y be a scheme and y ∈ Y (k) be a point of Y .Then we have that TY,y ∼= {f ∈ Homk(Spec k[ε], Y ) : f (Spec k) = y}

Proof. First we prove that there is a map

TY,y → {f ∈ Homk(Spec k[ε], Y ) : f (Spec k) = y}.

Let ψ ∈ TY,y = my/m2y

∗

. Then we get a homomorphism ψ : my → k such that m2y is a

subspace of the kernel. Now define ϕy : OY,y → k[ε] by ϕy(a + z) = a + ψ(z)ε for a ∈ k

and z ∈ my. This induces a homomorphism ϕ : OY → k[ε] by OY(U ) → OY,y ϕy

−→ k[ε], which gives an element f ∈ Hom (Spec k[ε], Y ).

(12)

Now we construct a map

{f ∈ Homk(Spec k[ε], Y ) : f (Spec k) = y} → TY,y,

that is the inverse of the map constructed above. Let f ∈ Homk(Spec k[ε], Y ) be such

that f (Spec k) = y. We have the induced homomorphism ϕ : OY → k[ε], which gives

a homomorphism ϕy : OY,y → k[ε]. Since ϕy is a local homomorphism we have that

ϕy(my) ⊂ (ε). Let z ∈ my and assume that ϕy(z) = aε, then define ψ : my → k by

ψ(z) = a. Since ϕ(m2

y) ⊂ (ε2) = 0 we have that ψ is an element of TY,y. It is easy

to see that these maps are the inverse map of each other, which proves the required statement.

Let X be a scheme and let [Z] ∈ Hilb(X). We want to know what the tangent space of Hilb(X) at [Z] is. From the above lemma it follows that

THilb(X),[Z]∼= {f ∈ Hom(Spec k[ε], Hilb(X)) : f (Spec k) = [Z]}

Combining this with the representability of the Hilbert functor, we get that the tangent space of Hilb(X) at a point [Z] consist of all extensions Zε flat over Spec k[ε] such that

the following diagram commutes.

Spec k Spec k[ε]

Z Zε X

First we calculate the tangent space of Hilb (X) at a special point [Z]. Before we consider this case we give the following definition.

Definition 1.12. Let Y be a closed subscheme of a variety X over k. We say that Y is of local complete intersection in X if the ideal sheaf IY of Y in X is locally generated by

r = codim(Y ) elements at every point of Y .

Assume that Z is of local complete intersection in X. Then we have locally that IZ = (t1, . . . , tr), where r = codim(Z). Let Zε be an extension of Z in X, so Zε is an

element of THilb(X),[Z]. The scheme Zε corresponds to the ideal sheaf IZε. Since we have

that Zε is an extension of Z in X × Spec k[ε] we have that IZε is locally of the form

(t1 + s1ε, . . . , tr+ srε) for si ∈ OX. Let Zε0 be an other extension, such that the ideal

sheaf IZ0

ε is locally generated by (t1 + s

0

1ε, . . . , tr + s0rε). If for all i ∈ {1, . . . , r} we

have that si − s0i = fi ∈ IZ, then we have that there exist fi1, . . . , fir ∈ IZ such that

fi = fi1t1+ · · · + firtr. From this it follows that IZε is determined by si mod IZ for all

i ∈ {1, . . . , r}. Now define a morphism ϕ : IZ → OZ by ti 7→ si. The above construction

shows that we get an inclusion THilb(X),[Z],→ HomZ(IZ, OZ).

Now let ϕ ∈ HomX(IZ, OZ) be a morphism. Define si = ϕ(ti) and let Zε be the scheme

that corresponds to the ideal IZε = (t1+ s1ε, . . . , tr+ srε). We want to show that Zε is

an element of the tangent space of Hilb(X) at Z. So we must have that Zε is flat over

(13)

Lemma 1.13 (Local criteria of flatness). Let (R, mR) → (S, mS) be a morphism of local

noetherian rings and let M be a finitely generated module over S. Then M is flat over R if and only if

(a) M ⊗ R/mR is flat over R/mR and

(b) mR⊗RM → M is injective.

Proof. First assume that M is flat over R. Then [Har77, III Prop. 9.1A, p. 253] gives that both (a) and (b) are true.

Now assume that (a) and (b) are satisfied. By [Har77, III Prop. 9.1A, p. 253] it is enough to prove that for every ideal a ⊂ R the map a ⊗ M → M is injective.

Let a ⊂ R be an ideal. Then we have the following exact sequence 0 → mR∩ a → a

q

−

→ R/mR.

Let B be the image of q and apply ⊗RM to the given sequence.

(mR∩ a) ⊗RM a⊗RM B ⊗RM

M R/mR⊗RM

0

α β

γ

Since B ⊗RM = (B ⊗RR/mR) ⊗R/mR M and M ⊗ R/mR is R/mR-flat, we have that γ

is injective. From this it follows that if α is injective, then also β is injective. So to prove that M is flat over R, we show that α is injective.

We have that (mR∩ a) ⊗RM → mR⊗RM is injective, since mRis the maximal ideal of

R. By assumption we have that mR⊗RM → M is injective, thus α : (mR∩a)⊗RM → M

is injective. This shows that β is injective.

Back to the situation of the constructed scheme Zε by ϕ, we have in the notation of

the above lemma that R = k[ε] and S = k. This shows that IZε is flat over Spec k[ε],

which proves that Zε ∈ THilb(X),[Z]. Now we can conclude that in the case that Z is of

local complete intersection in X we have that

THilb(X),[Z]∼= HomX(IZ, OZ).

In general we have the following theorem.

Theorem 1.14. Let X be a projective scheme over a field k and Z ⊂ X be a closed subscheme. Then we have that the Zariski tangent space of Hilb(X) at [Z] is naturally isomorphic to

THilb(X),[Z]∼= HomX(IZ, OZ) ∼= HomZ(IZ/IZ2, OZ). (1.2)

Proof. For the proof see [Kol96, I Theorem 2.8, p. 31].

For the subscheme Hom(X, Y ) of Hilb(X × Y ) we have the following isomorphism of the tangent space.

(14)

Theorem 1.15. Let X be a projective variety and let Y be a quasi-projective variety over a field k. Let f : X → Y be a morphism and assume that Y is smooth along f (X). Then we have that

THom(X,Y ),[f ] ∼= HomX(f∗ΩY, OX).

Proof. Since Hom(X, Y ) is an open subscheme of Hilb(X ×S Y ) we have that at f the

schemes Hom(X, Y ) and Hilb(X ×SY ) are locally isomorphic. So from theorem 1.14 it

follows that

THom(X,Y ),[f ] ∼= HomX×SY(IΓf, OΓf) ∼= HomΓf(IΓf/I

2

Γf, OΓf),

where Γf is the graph of f . So we only need to show that

HomΓf(IΓf/I

2

Γf, OΓf) ∼= HomX(f

∗

ΩY, OX).

By [Mat70, 26.I, p. 187] we have the following short exact sequence 0 → IΓf/I

2

Γf → ΩX×Y ⊗ OΓf → ΩΓf → 0.

By construction also the following sequence is a short exact sequence. 0 → f∗ΩY → ΩX ⊕ f∗ΩY → ΩX → 0

Since ΩX⊕ f∗ΩY ∼= ΩX×Y ⊗ OΓf and ΩY ∼= ΩΓf the 5-lemma shows that these sequences

are isomorphic, which proves that HomΓf(IΓf/I

2

Γf, OΓf) ∼= HomX(f

∗_Ω

Y, OX).

1.2.2. Local structure of Hilb(X) and Hom(X, Y )

In the previous section we have looked at the extension 0 → (ε) → k[ε] → k → 0, which gives a way to calculate the tangent space. In this section we look at arbitrary small extensions 0 → M → B → A → 0, and the problem of extending a closed subscheme Z of a scheme X over A to a closed subscheme of X × Spec B that is flat over Spec B. Definition 1.16. A small extension of a local ring A is a short exact sequence

0 → M → B → A → 0

such that B is a local ring with maximal ideal mB and M · mB = 0.

Let X be a scheme over a field k and Y a closed subscheme of X. Let A be a local ring over the field k and let

0 → M → B → A → 0

be a small extension of the local ring A. Suppose that there exist a subscheme YA of

X ×Spec kSpec A that is a scheme over A such that YA is an extension of Y that is flat

over Spec A. Then we have the following commutative diagram.

Spec k Spec A Spec B

X X × Spec A X × Spec B

Y YA ?

f

(15)

The question is, when does there exist an extension YB of YA such that YB is a

sub-scheme of X ×Spec kSpec B flat over Spec B filling in the question mark in diagram (1.3).

For this we have the following lemma. Lemma 1.17. Let the notation be as above.

(i) The obstruction of extending YA to YB such that YB is flat over Spec B is a subspace

of Ext1_O_X(IY, OY ⊗ M ).

(ii) If there exist an extension YB, then the set of extensions forms a homogeneous space

over HomX(IY, OY ⊗ M ).

Before we give the proof of this lemma, we first give some other results. The following proposition gives a criterion for the flatness of YB.

Proposition 1.18. Let the notation be as above and let YB be a closed subscheme of

X × Spec B such that diagram (1.3) commute. Let ϕ : OX×Spec B → OX×Spec A be the

induced morphism from f . The following are equivalent. (i) YB is flat over Spec B.

(ii) IY ⊗kM ⊂ IYB ⊂ ϕ

−1_(I

YA) and the map ¯ϕ : IYB/(IY ⊗kM ) → IYA induced by ϕ

is an isomorphism.

Proof. We have the following commutative diagram. 0 0 IY ⊗kM OX ⊗kM OY ⊗kM 0 0 IYB OX×Spec B OYB 0 0 IYA OX×Spec A OYA 0 0 0 ϕ (1.4)

All rows in this diagram are exact and the second column is exact. By diagram chasing one can prove that the first column is exact if and only if the third column is exact. For more details see [FG05, Prop. 4.1, p. 9].

If we now assume that YB is flat over B, we have that the third column of diagram 1.4

is exact. So the first column is exact, which proves (ii). We have that (ii) implies that the first column of the diagram is exact, so the third column is exact. Now lemma 1.13 gives that YB is flat over Spec B.

(16)

Lemma 1.19. Let 0 → E → F −→ G → 0 be a short exact sequence of coherentf sheaves. The morphism f has a section if and only if the extension class of the se-quence in Ext1_{(G, E ) vanishes. In this case the sections of f form an affine space over}

Hom(G, E ).

Proof. The short exact sequence gives the following long exact sequence

0 Hom(G, E ) Hom(G, F ) Hom(G, G)

Ext1_{(G, E )} _Ext1_{(G, F )} _{. . .} α

Let η ∈ Ext1(G, E ) be the element such that id ∈ Hom (G, G) maps to η. If f has a section, then we have that η = 0. Now if η = 0, then we have that id ∈ Hom (G, G) is in the image of α. This shows that there exist a morphism g ∈ Hom (G, F ) such that α(g) = f ◦ g = idG, which proves that f has a section.

Finally, we want to prove that the sections of f form an affine space over Hom(G, E ). Let g, g0 be two sections of f . Then we have that g − g0 ∈ ker(Hom(G, F ) → Hom(G, G)) and thus that g − g0 ∈ Hom(G, E). From this it follows that the sections of f form an affine space over Hom(G, E ).

With these two lemma’s in mind, we now could give the proof of lemma 1.17. Proof of lemma 1.17. The exact sequence

0 → OX ⊗kM → OX×Spec B ϕ

−

→ OX×Spec A → 0

induces the following exact sequence

0 → OX ⊗kM → ϕ−1(IYA) → IYA → 0

From diagram (1.4) we get the following short exact sequence.

0 → OY ⊗kM = (OX ⊗kM )/(IY ⊗kM ) → ϕ−1(IYA)/(IY ⊗kM )

¯ ϕ

−

→ IYA → 0

Claim 1.20. Giving IYB such that YB is flat over Spec B is equivalent to finding a section

of ¯ϕ : ϕ−1(IYA)/(IY ⊗kM ) → IYA.

Proof. The existence of a section is equivalent to the existence of a subsheaf G of ϕ−1(IYA)/(IY ⊗kM ) such that ¯ϕ|G is an isomorphism. If YBexist such that YBis flat over

Spec B, then proposition 1.18 shows that G := IYB/(IY ⊗kM ) is the required sheaf. If ϕ

has a section, then define IYB as the inverse image of ϕ

−1_(I

YA) → ϕ

−1_(I

YA)/(IY ⊗kM ).

Then proposition 1.18 shows that YB is flat over Spec B.

Lemma 1.19 now shows that the obstruction of extending YA to YB flat over Spec B is

a subspace of Ext1

OX×Spec A(IYA, OY ⊗kM ). These extensions form a homogeneous space

(17)

To finish the proof of this lemma, we only have to show that Exti

OX×Spec A(IYA, OY⊗kM )

is isomorphic to Exti_O_X(IY, OY ⊗kM ) for i = 0, 1. Since YA is flat over Spec A we have

that Exti_O X×Spec A(IYA, OY ⊗kM ) ∼= Ext i OX⊗kOSpec A(IY ⊗kOSpec A, OY ⊗kM ) ∼ = ExtiOX(IY, OY ⊗kM ).

Now we apply lemma 1.17 to give an estimate for the local dimension of the Hilbert scheme.

Theorem 1.21. Let X be a projective scheme over a field k and let Y ⊂ X be a closed subscheme. Then

dim[Y ]Hilb(X) ≥ dim HomX(IY, OY) − dim Ext1OX(IY, OY).

Proof. Let R be the completion of OHilb(X),[Y ] and let 0 → M → B → A → 0 be a small

extension. Lemma 1.17 shows that we have the following exact sequence

0 → T ⊗kM → hR(B) → hR(A) → Ext1OX(IY, OY ⊗kM ), (1.5)

where T = mR/m2R ∼= THilb(X),[Y ] and hR: Ring → Set is defined by hR(A) = Hom(R, A).

For the proof of the exactness of this sequence see [FG05, Theorem 1.7, p. 2].

Since the behaviour of a small extension gives an idea of the local dimension of Hilb (X) at [Y ], the estimate follows from sequence (1.5). By theorem 1.14 we have that THilb(X),[Y ] ∼= HomX(IY, OY). So sequence (1.5) shows that

dim[Y ]Hilb(X) ≥ dim HomX(IY, OY) − dim Ext1OX(IY, OY),

because Ext1

OX(IY, OY ⊗kM ) ∼= Ext

1

OX(IY, OY) ⊗ M .

For the bend and break theorem we want to know the local structure of Hom(X, Y ) and Hom(X, Y ; f ) for schemes X and Y and f : B → Y a morphism from a closed subscheme B ⊂ X to Y . The following two theorems give estimates for the local dimension of both schemes.

Theorem 1.22. Let X be a projective variety and let Y be a quasi-projective variety. Let f : X → Y be a morphism such that Y is smooth along f (X). Then we have that any irreducible component of Hom(X, Y ) at [f ] has at least dimension

dim[f ]Hom(X, Y ) ≥ dim H0(X, f∗TY) − dim H1(X, f∗TY).

Proof. First we show that Exti_(f∗_Ω

Y, OX) ∼= Hi(X, f∗TY) for i = 0, 1. Since f∗ΩY is

a locally free sheaf we have that f∗ΩY → f∗ΩY → 0 is a free resolution of f∗ΩY. Now

from [Har77, III Prop. 6.5, p. 234] it follows for i = 0, 1 that Exti(f∗ΩY, OX) ∼= Hi(X, Hom (f∗ΩY, OX))

∼

(18)

Now since Hom (X, Y ) is a open subscheme of Hilb (X × Y ) we have that also the obstruction of extending along a small extension 0 → M → B → A → 0 is a subspace of Ext1_O

X×Y(IΓf, OX×Y). We have by the proof of theorem 1.15 that

Exti_O

X×Y(IΓf, OX×Y) ∼= Ext

i_(f∗

ΩY, OX)

∼

= Hi(X, f∗TY).

Now in the same way as in theorem 1.21 we have that

dim[f ]Hom(X, Y ) ≥ dim H0(X, f∗TY) − dim H1(X, f∗TY).

Theorem 1.23. Let X be a projective variety and let Y be a quasi-projective variety. Let B ⊂ X be a closed subscheme and let g : B → Y be a morphism. Let f : X → Y be a morphism such that f |B = g and Y is smooth along f (X). Then we have that any

irreducible component of Hom(X, Y ; g) at [f ] has at least dimension

dim[f ]Hom(X, Y ; g) ≥ dim H0(X, f∗TY ⊗ IB) − dim H1(X, f∗TY ⊗ IB).

Proof. In a similar way as for theorem 1.22 we could prove the given estimate of the dimension of Hom(X, Y ; g). For a detailed proof see [Kol96, II Theorem 1.7, p. 95].

Let C be a curve, Y a scheme and f : C → Y a morphism. Then we want to give an estimate of the dimension of Hom(C, Y ) and Hom(C, Y ; f ) in terms of the canonical divisor of Y and the Euler characteristic of C. Before giving the statement we first give an extension of the Riemann-Roch theorem.

Theorem 1.24 (Riemann-Roch). Let C be a curve and let F be a locally free coherent sheaf on C, then we have that

χ(C, F ) = deg(det F ) + rank (F )χ(C, OC).

Proof. If L is an invertible sheaf we have the Riemann-Roch theorem for divisors [Har77, IV Theorem 1.3, p. 295]. Let D be the associated divisor of L, then we have that

l(D) − l(K − D) = deg D + χ(C, OC),

where l(D) = dimkH0(C, L), l(K − D) = dimkH0(C, KC ⊗ L∗) and K is the canonical

divisor. By Serre duality [Har77, III Theorem 7.6, p. 243] we have that χ(C, L) = l(D) − l(K − D) = deg D + χ(C, OC).

Now let L1 and L2 be two invertible sheaves on C. Then by the above equation we have

that

χ(C, L1⊕ L2) = χ(C, L1) + χ(C, L2)

= deg L1+ χ(C, OC) + deg L2+ χ(C, OC)

= deg(L1⊗ L2) + 2 · χ(C, OC)

= deg(det(L1⊕ L2)) + rank (L1 ⊕ L2)χ(C, OC).

Since F is a locally free coherent sheaf we have that F is locally the sum of invertible sheaves. So the above calculation shows that

(19)

From this theorem we can deduce the following estimates for the local dimensions of Hom(C, Y ) and Hom(C, Y ; f ).

Theorem 1.25. Let C be a smooth curve and let f : C → Y be a morphism to a smooth variety Y of pure dimension n. Then we have that

dim[f ]Hom(C, Y ) ≥ χ(C, f∗TY)

≥ − deg_Cf∗KY + nχ(C, OC),

where KY the canonical sheaf of Y . Let B ⊂ C be a finite closed subscheme of C with

morphism g : B → Y such that f |B= g. Let l(B) be the length of B, then we have that

dim[f ]Hom(C, Y ; f |B) ≥ χ(C, f∗TY ⊗ IB) (1.6)

= − deg(f∗KY) + n(−l(B) + χ(C, OC)).

Proof. By theorem 1.22 and the Riemann-Roch theorem (theorem 1.24) we have that dim[f ]Hom(C, Y ) ≥ dim H0(C, f∗TY) − dim H1(C, f∗TY)

= χ(C, f∗TY)

= deg(det f∗TY) + rank (f∗TY)χ(C, OC)

= deg(−f∗KY) + nχ(C, OC).

By theorem 1.23 we have that

dim[f ]Hom(C, Y ; g) ≥ dim H0(C, f∗TY ⊗ IB) − dim H1(C, f∗TY ⊗ IB)

=χ(C, f∗TY ⊗ IB)

= deg(det(f∗TY ⊗ IB)) + rank (f∗TY ⊗ IB)χ(C, OC)

= deg((det f∗TY) ⊗ I_B⊗n) + rank (f∗TY)χ(C, OC)

= deg(f∗TY) + n · −l(B) + nχ(C, OC)

(20)

(21)

2. Existence of rational curves in

varieties

We give a way to construct a rational curve from arbitrary curves in a scheme, a rational curve is a curve that is rationally equivalent with P1_{. The most important tool in this}

construction is the bend and break theorem (theorem 2.2). The idea behind this theorem is to break the original curve in several irreducible components, such that one irreducible component is a rational curve. As assumption for the bend and break theorem we must have that the degree of the canonical bundle is small enough. Actually, this estimate is too small. We make this more precise in section 2.2, where we prove a more general version of the bend and break theorem (theorem 2.7). These two theorems give a good idea when a scheme contains a rational curve. The main ideas of the proofs in this chapter are coming from [Deb07] and [Mor79].

2.1. Bend and break theorem

In the first chapter about the Hilbert scheme we have constructed important tools for stating the bend and break theorem. For the proof of this theorem we need the following lemma, which is a very important lemma in algebraic geometry.

Lemma 2.1 (Rigidity lemma). Let f : X → Y be a proper morphism that satisfies f∗OX = OY. Let g : X → Z be a morphism. Assume that for some y ∈ Y there is a

factorization y f−1(y) Z f g hy

Then there is an open neighbourhood y ∈ U ⊂ Y and a factorization

U

f−1(U ) Z

f

g hU

Proof. Let Z0 be the image of (f, g) : X → Y × Z and let p1 : Z0 → Y , p2 : Z0 → Z

be the projections. By assumption we have that f−1(y) is contracted by g. Since f−1(y) = (f, g)−1(p−1₁ (y)) we have that (f, g)−1(p−1₁ (y)) is contracted by g and thus by (f, g). This shows that (f, g)((f, g)−1(p−1₁ (y))) = p−1₁ (y) is a point. So there exist an open neighbourhood y ∈ U ⊂ Y such that p−1(U ) is finite.

(22)

If g factors through f on U we must have that p1 is an isomorphism on U . Since

f∗OX = OY we have that f is surjective. This shows that p1 : Z0 → Y is surjective.

We have already proved that p is finite, so if (p1)∗O_p−1

1 (U ) = OU we have that p1 is an

isomorphism. We have that OU ⊆ (p1)∗O_p−1

1 (U ) ⊆ (p1)∗(f, g)∗Of

−1_{(U )} = f_∗O_f−1_{(U )}= O_U.

This shows that (p1)∗O_p−1

1 (U )= OU and thus that p1 is an isomorphism.

The rigidity lemma plays an essential role in the proof of the bend and break theorem. So with this lemma in mind we can now state and prove the theorem.

Theorem 2.2 (Bend and break). Let C be a smooth curve and let X be a projective variety. Let f : C → X be a morphism and let c ∈ C be a point. If we have that dim[f ]Hom(C, X; f |{c}) ≥ 1, there exist a rational curve on X through f (c).

Proof. If C is already a rational curve, then there is nothing prove. Hence we assume that g(C) ≥ 1. First we give an idea of the proof. For this let B0 be a 1-dimensional curve of Hom(C, X; f |{c}) through [f ] and let B be the compactification of B0. We have the

natural morphism F : C × B0 _{→ X, which extends to a rational map F : C × B 99K X.} The rigidity lemma shows that this map is not a morphism. We blow up C × B several times to get a morphism G : D → X, where D is the blowing up. Then we choose a fibre that consist of a rational curve. This proves that X contains a rational curve. Figure 2.1 gives the above described idea of the construction in the proof.

C × {b} C × B F C × {b} D E G f (c) f (C) G(E) X

Figure 2.1.: Idea of the proof of the bend and break theorem.

Suppose that F : C × B 99K X is a morphism. Since F ({c} × B) is a point, the rigidity lemma gives that there exist an open neighbourhood c ∈ V ⊂ C such that F factors through V , which is visualized in the following diagram.

V

V × B X

p1

F hV

(23)

Since f |V must be equal to hV we have that F and f ◦ p1 coincide on V × B. But then

f ◦ p1 and F are equal on C × B0, which means that the image of B0 in Hom(C, X; f |{c})

is 0-dimensional. This is in contradiction with our assumption, so F is a rational map on C × B. Hence there exist a point b ∈ B such that F is not defined on (c, b).

Now blow up C × B several times to get a morphism G : D → X, where D is the blowing up, such that the following diagram commutes.

C × B D

X

G

F

Since D is the blowing up, we have that the fibre of b under the projection D → B is the union of C × {b} and finitely many rational curves. Since F is not defined at (c, b), we have that at least one of the rational curves is not contracted by G. Let E be this rational curve. We have that c × B is contracted by G to the point f (c), which proves that the rational curve G∗E passes through f (c).

The rational curve constructed in the bend and break theorem can have a huge degree. We want to have an upper bound of the degree of this rational curve. Suppose that deg f∗KX ≤ − dim X − 2 and let n = dim X. Then by theorem 1.25 we have that

dim[f ]Hom(P1, X; f |{0,∞}) ≥ − deg f∗KX + n −l({0, ∞}) + χ(P1, OP1)

= − deg f∗KX + n(−2 + 1) ≥ 2. (2.1)

With this estimate in mind the next proposition gives an upper bound of the degree of the rational curve.

Proposition 2.3. Let X be a projective variety and let f : P1 _{→ X be a rational curve.}

If dim[f ]Hom(P1, X; f |{0,∞}) ≥ 2, the curve f∗P1 can be deformed into a sum of ν rational curves with ν ≥ 2.

Proof. In this proof we first construct a space such that the image in X has dimension 2. Then we prove that the fibres are rational curves. Now if there exist a fibre that is not irreducible, we are done. This idea is visualized in figure 2.2.

Let Aut P1

0,∞ be the group of automorphisms on P1 fixing 0 and ∞. Then we have

that Aut P1

0,∞ is the multiplicative group Gm. Denote by H the group of morphisms

Hom (P1, X; f |{0,∞}). Since dim[f ]H ≥ 2 we can find a non-singular curve D ⊂ H and

a finite morphism α : D → H such that [f ] ∈ α(D) and α(D) is not contained in the Gm-orbit of [f ]. Let F : P1× D → X × D be the induced morphism.

Claim 2.4. Let p1 : X × D → X be the first projection. Then we have that

dim(p1◦ F )(P1× D) = 2.

Proof. By construction we have that dim(p1 ◦ F )(P1 × D) ≤ 2. Now assume that the

dimension of (p1◦ F )(P1× D) is less than or equal to 1. Let C = f∗P1. Since [f ] ∈ α(D), we have that C = (p1 ◦ F )(P1 × D). From these statements it follows that p1 ◦ F is a

composition of a morphism F0 _{: P}1 _{× D → P}1 _{and the normalization f : P}1 _{→ C ⊂ X.}

(24)

P1 0 ∞ s0 s∞ P1 0 ∞ f (0) f (∞) X

Figure 2.2.: Idea in the proof of bounding the degree of the rational curve.

F (∞ × D) = f (∞). This shows that F0 _{is induced from D → Aut P}1

0,∞, which

con-tradicts the fact that α(D) is not contained in the Gm-orbit of [f ]. So we have that

dim(p1◦ F )(P1× D) = 2.

Let ¯_{D be the smooth compactification of D, and let Y be the closure of F (P}1 _{× D)}

in X × ¯D. Let ˜Y be the normalization of Y and let π be the following morphism π : ˜Y → Y ⊂ X × ¯D−p→ ¯2 D, where p2 is the second projection.

Claim 2.5. P1_{× D ∼}_{= π}−1_(D).

Proof. Since Y is the closure of F (P1_{× D) in X × ¯}_{D we have that π}−1_{(D) is isomorphic}

to P1× D if F is an immersion. By construction we have that F is a finite morphism, so it is enough to prove that F is an immersion on an open subset U ⊂ P1_{× D. Let t ∈ D}

be such that α(t) = [f ]. We have that F is unramified at 0 × t. So there exist an open neighbourhood W ⊂ X × D of F (0 × t) such that F is unramified on V = F−1(W ). Since the first projection p1 : V ×X×DV → V is proper, the diagonal map ∆ : V → V ×X×DV

is an open immersion. The fact that F−1F (0 × t) = {0 × t} gives that p−1₁ (0 × t) is one point. This shows that p1(V ×X×DV \ ∆(V )) is a closed set that does not contain {0×t}.

Let U = V \ p1(V ×X×DV \ ∆(V )), then we have that U is an open neighbourhood of

{0 × t} and F is unramified at U . Since U = U ×X×DU we have also that F is radical at

U . This proves that F is an open immersion at U and thus that P1× D ∼= π−1(D). This claim shows that the general fibres of π are isomorphic to P1, from which it follows that any irreducible component of a fibre of π is a rational curve. So we have that any fibre of Y over ¯D is a union of rational curves. Now if there exist a t ∈ ¯D such that the fibre of t over π is not irreducible, then we are done.

Suppose that all fibres of π are irreducible. The singular locus of Y does not contain a fibre of π, since all fibres are unions of rational curves. This shows that for any t ∈ ¯D the morphism ˜Yt→ Yt induces an isomorphism of an open dense subset of ˜Yt with its image.

Since the fibre of t over π is irreducible by assumption we have that ˜Yt is an irreducible

reduced subscheme of ˜Y . From claim 2.5 and the flatness of π it follows by [Har77, III Prop. 9.3, p. 255] that

dim H1( ˜Yt, OYt) = dim H

1

(25)

which shows that ˜Yt∼= P1, which gives that ˜Y is a P1-bundle.

We have that F (0 × D) and F (∞ × D) give two sections s0 and s∞ of π. From [Har77,

V Prop. 2.3, p. 370] it follows that there exist a divisor S on ¯D such that s0 − s∞ is

linearly equivalent to π∗S. This shows that the self-intersection number (s0− s∞)2 = 0.

Now let H be an ample divisor on (p1◦ F )(P1× X). Since s0 and s∞are both sections

we have that s0 6≡ 0, s∞ 6≡ 0 and s0.H = s∞.H = 0. Now the Hodge Index theorem

(theorem A.12) shows that s2

0 < 0 and s2∞< 0. Since s0 and s∞ are two different sections

we have that s0.s∞= 0. So we have that

0 = (s0− s∞)2 = s02− 2s0.s∞+ s2∞< 0,

which is impossible. This shows that there exist a t ∈ ¯D such that the fibre of t over π is not irreducible, which finishes the proof.

2.2. Stronger version of the bend and break theorem

The bend and break theorem (theorem 2.2) shows that X contains a rational curve if dim[f ]Hom(C, X; f{c}) ≥ 1 for f : C → X a morphism from a curve C to a scheme X.

So by equation (1.6) we have that X contains a rational curve if deg f∗KX < n(−1 + χ(C, OC)).

Actually, it is enough if deg f∗KX < 0. Before we give the precise statement we first

define the following.

Definition 2.6. A divisor D of a complete variety X is called numerically effective if for every irreducible curve Z ⊂ X we have that deg(i∗D) ≥ 0, where i : Z → X is the inclusion.

This gives a way to reformulate that deg f∗KX < 0, because now we could say that KX

is not numerically effective. With this in mind we give the stronger version of the bend and break theorem, which is a theorem of S. Mori.

Theorem 2.7 (Mori). Let X be a non-singular projective variety over an algebraic closed field k. Then X contains a rational curve if the canonical divisor is not numerically effective.

We prove this theorem in two steps. First we assume that the characteristic of k is greater than 0, the proof of this statement is given in the lemma below. Then we extend this to the case that the characteristic of k is equal to zero.

Lemma 2.8. Let X be a non-singular projective variety over an algebraically closed field k with characteristic p > 0. Then X contains a rational curve if the canonical divisor is not numerically effective.

Proof. Since KX is not numerically effective there exist an irreducible curve C on X with

inclusion i : C → X such that deg i∗KX < 0. Let ˜C be the normalization of C and let

a ∈ N be such that

(26)

We have that a ∈ N exist since deg i∗KX < 0. Let q = pa and let π : Spec k → Spec k

be the (1/q)-the power endomorphism. We have that π is a flat morphism, because k is a field. Let D be the base change of ˜C over π and denote by f : D → ˜C the induced morphism. Then we have the following pushout diagram.

Spec k D Spec k ˜ C π f

Since π is flat we have by [Har77, III Prop 9.3, p. 255] that H1_{(D, O}

D) ∼= π∗H1( ˜C, O_C˜).

Let P be a rational point of D and j : P → (i ◦ f )(P ) ∈ X be the inclusion. Since the degree of f is q we have by theorem 1.25 that

dim[i◦f ]Homk(D, X; j) ≥ − deg((i ◦ f )∗KX) + dim X(−l(P ) + χ(D, OD))

= − q deg(i∗KX) + dim(X)(−1 + 1 − dimkH1( ˜C, O_C˜))

= − dim(X) dimkH1( ˜C, O_C˜) − q deg(i∗KX) ≥ 1.

Now from the bend and break theorem (theorem 2.2) it follows that X contains a rational curve.

So if the characteristic is positive we have proved the existence of a rational curve. Now we prove the case where the characteristic is zero. For this we construct a field with positive characteristic and prove that there exist a rational curve by the above lemma. Then we show that this curve gives actually a rational curve in X.

Proof of theorem 2.7. We prove the theorem of Mori under the extra assumption that K−1_X is ample. For the proof without this extra assumption see [Kol96, II Theorem 5.8, p. 143]. By lemma 2.8 we have that X contains a rational curve if the characteristic of k is positive. Now assume that the characteristic of k is equal to zero.

Since X is a projective variety we have that there exist an N ∈ N such that we have

a morphism X → PN

k. From this it follows that there exist homogeneous polynomials

Fi ∈ k[x0, . . . , xN] for i ∈ {1, . . . , r} where r ∈ N such that X is the zero locus of

{F1, . . . , Fr}.

Let A0 _{be the ring generated over Z by all coefficients of {F}1, . . . , Fr}. Let X0 be the

zero locus of F1, . . . , Fr over A0. Now it could be that some fibres of X0 → Spec A0 are

singular or has dimension greater than the dimension of the fibre of X → Spec k. To remove these points we add the inverse of these elements to A0, then we get a new ring A. Let X be the zero locus of F1, . . . , Fr over A. This construction gives the following

pushout diagram. Spec k X Spec A X f (2.2)

(27)

Claim 2.9. The canonical divisor of X is not numerically effective.

Proof. By assumption we have that the canonical divisor of X is not numerically effective. So there is a curve C on X such that deg i∗KX < 0 where i : C → X is the inclusion.

Since Spec k → Spec A is a closed immersion and the diagram (2.2) is a pushout diagram we have that f is also a closed immersion. This shows that C is also a curve on X. Let j = f ◦ i : C → X be the inclusion, then we have that

deg j∗KX= deg(f ◦ i)∗KX= deg(i∗(f∗KX))

= deg(i∗KX) < 0.

So the canonical divisor of X is not numerically effective.

Define S = Spec A and let s ∈ S be a closed point. By construction of A we have that the function field k(s) has characteristic > 0. Lemma 2.8 shows that the fibre Xs of s

contains a rational curve C. By equation (2.1) and proposition 2.3 we could assume that the deg i∗KXs ≥ − dim X − 1 = − dim X − 1, where i : C → Xs is the inclusion.

Claim 2.10. K−1_X is relative ample.

Proof. Since X is constructed from the polynomials that define X we have that there exists a morphism j : X → PNA such that we have the following commutative diagram.

PNk PNA Spec A Spec k X X g f i j

Since KX is ample we have that there exists a d > 0 such that KdX ∼= i ∗_O

PNk(1). The both

squares in the above diagram are both pushout diagrams. From this it follows that we also have that Kd

X ∼= j ∗_O

PNA(1), which proves that KX is ample.

Let n = dim X and define Hom(P1, X)≤n+1 as the subspace of Hom(P1, X) consisting

of all morphisms f such that deg f∗K−1_X _{≤ n + 1. Let Hom(P}1_{, X)}

≤n+1 be defined on the

same way. Since K−1_X _{is relative ample we have that Hom(P}1_{, X)}

≤n+1 has finitely many

irreducible components. Look at the following diagram. Hom(P1, X)≤n+1

Spec k

Hom(P1, X)≤n+1

S = Spec A We want to prove that Hom(P1_{, X)}

≤n+1is not empty. Suppose that Hom(P1, X)≤n+1= ∅.

Then we have that Hom(P1, X)≤n+1 is supported over a proper closed subscheme Y ⊂ S,

because Hom(P1_{, X)}

≤n+1has finitely many components. Since A is finitely generated over

(28)

We have already proved that at every closed point s of S, the variety Xs contains a

rational curve of degree less than n + 1. This is in contradiction with the existence of closed points in S not in Y . So we have that Hom(P1, X)≤n+1 6= ∅, which proves that X

(29)

3. Deforming curves

Let X be a smooth projective variety over an algebraic closed field k. Let D ⊂ X be a curve in X. The goal of this chapter is to give deformations of the curve D. Most of the time it is not possible to deform D directly. Therefore we add rational curves to D such that we can deform the resulting curve C to a smooth curve C0. This is visualized in figure 3.1. D C1 C2 . . . Cn C = D C1 C2 . . . Cn D C0

Figure 3.1.: Deforming a curve D.

To add these rational curves, we must have that the variety X contains a lot of rational curves. The previous chapter gives an idea when this is satisfied.

That we could not deform D directly is almost always a consequence of the fact that the obstruction space is non-vanishing. So we first show what the obstruction space is, and in section 3.3 we construct a curve that has as an irreducible component the curve D and has no obstruction.

We explain in several steps how we smooth a curve as visualized in figure 3.1. For this we first consider the case of a node in X, which we want to smooth. Then we construct a comb with handle D such that the obstruction space is zero and the normal bundle of this comb is globally generated. Finally, we smooth this comb, which gives a deformation of the curve D in X.

(30)

3.1. Obstruction of deforming a curve

In this section we give the obstruction space for deforming a curve. By a deformation of a curve in this chapter we always mean a first-order deformation of that curve.

Definition 3.1. Let C be a curve on a variety X over an algebraically closed field k. Then we have that a first order deformation of C is an extension of C in Spec k[ε].

So the goal of this chapter is to give the obstruction of extending C to a curve C0 for the small extension

0 → (ε) → k[ε] → k → 0. (3.1)

From lemma 1.17 it follows that the obstruction space of deforming C over the small extension (3.1) is equal to Ext1

OX(IC, OC ⊗ (ε)) ∼= Ext

1

OX(IC, OC). If the curve C is of

local complete intersection, we have that IC is a locally free sheaf of finite rank. In this

case we can use [Har77, III Prop. 6.5, p. 234], which shows that Ext1OX(IC, OC) ∼= H

1

(C, HomOX(IC, OC)) (3.2)

= H1(C, NC/X).

For a smooth curve we automatically have that it is of local complete intersection. Since we are going to add curves to our curve C such that we get a nodal curve, we need to prove that a nodal curve is of local complete intersection. Before we prove this, we give the definition of a nodal curve.

Definition 3.2. A nodal curve is a complex algebraic curve C with finitely many singular points, such that at each singular point the local model is xy = 0, where x and y are the local generators of the two curves intersecting in that point.

If we look at figure 3.1 we have that the curve C is a nodal curve, the construction of this curve is given in detail in section 3.3. Now we prove that the curve C is of local complete intersection. At the non-singular points of C, it is clear that C is locally generated by codim(C) elements. Now let P ∈ C be a singular point. Then we have that it locally looks like xy = 0, where x and y are the local generators of the two curves that intersect in P . This implies that also locally around P the curve C is generated by codim(C) elements, which shows that C is of local complete intersection.

From this it follows that the calculation (3.2) holds, which shows that the obstruction space for a deformation of a curve C is isomorphic to H1_{(C, N}

C/X).

3.2. Smoothing a node

Let C1, C2 be two curves on a scheme X such that C1 and C2 intersect in a single point

P and the union C = C1∩ C2 is a nodal curve. We call such a curve a node. This node

is visualized in figure 3.2.

Under the assumption that the obstruction space vanishes, we give the idea how to smooth this curve C in a few steps. First we assume that C is the node in A2 that has as irreducible components the x and y-axis. Then we extend this to an arbitrary surface. Finally, we give a way to smooth the curve C on an arbitrary scheme X.

(31)

C1

C2 X

P

Figure 3.2.: Node in a scheme X.

3.2.1. Smoothing a node in a surface

First we consider the case where X = A2 and where C1 is the y-axis and C2 is the x-axis.

Then we have that IC1 = (x) and IC2 = (y). This gives that the curve C is generated by

IC = (xy). Denote by O the origin, which is the singular point of C. This situation is

visualized in figure 3.3.

A2

x-axis y-axis

O

Figure 3.3.: Standard node in A2_.

Now we want to deform C to a smooth curve. Let C0 be a first order deformation of C. Then we have that C0 is generated by xy + ε ˜ϕ(xy) = 0, where ˜ϕ is the pullback of ϕ ∈ Hom(IC, OC) in the following exact sequence.

0 → IC → OA2 → OC → 0

˜

ϕ(xy) 7→ ϕ(xy)

We have that C0 is still a nodal curve if we have that there exist ˜ϕ1 and ˜ϕ2, coming from

ϕ1, ϕ2 ∈ Hom(IC, OC) in the same way as above for ϕ, such that

xy + ε ˜ϕ(xy) = (x + ε ˜ϕ1(x))(y + ε ˜ϕ2(y)).

This shows that if ˜ϕ(xy) = x ˜ϕ2(y) + y ˜ϕ1(x), the constant term of ˜ϕ is zero, we have that

C0 still is a nodal curve. Since Hom (IC, OC) = NC/A2 we have that ˜ϕ comes from an

element of the normal bundle. So we want to state the property, that the constant term of ˜ϕ is zero, in a condition on the element ϕ as an element of the normal bundle. Before we give this condition on ϕ, we first deduce some properties of the normal bundle. We have that

(32)

This shows that we get the following diagram with exact rows.

k(O) k(O) 0 NC2/A2 NC/A2 NC1/A2 ⊗ OC1(O) 0 0 NC2/A2 K NC1/A2 0 0 0 id id

The sheaf K is the subsheaf of N_C/A2 such that restriction to C₁ gives N_C

1/A2. The rows

in this diagram gives the following commutative diagram in cohomology where the rows are long exact sequences.

0 H0(C2, NC2/A2) H 0_{(C, N} C/A2) H0(C₁, N_C 1/A2 ⊗ OC1(O)) . . . 0 H0_(C 2, NC2/A2) H 0_{(C, K)} _H0_(C 1, NC1/A2) . . . id

Now we are going back to the element ϕ ∈ Hom(IC, OC) = NC/A2 that has constant

term zero. The above diagram shows that ϕ has a vanishing constant term, if and only if restricting to C1 gives an element in NC1/A2, so it has no pole at O. Since

H0(C, NC/X) ∼= THilb(X),[C]we have that the subset H0(C, K) ⊂ H0(C, NC/A2) corresponds

to the first order deformation of C that keeps the singular point. From this it follows that an element v ∈ H0_{(C, N}

C/X) corresponds to a smoothing of

the node (a first-order deformation that smooths the node) if and only if v 6∈ H0(C, K). We have that v 6∈ H0_{(C, K) if and only if v|}

C1 ∈ H

0_(C

1, NC1/A2⊗OC1(O))\H

0_{(C, N} C1/A2).

This shows that we can smooth C if and only if such an element exist.

Now assume that X is an arbitrary surface, and let C = C1∪ C2 be a node on X with

singular point P = C1∩ C2. Then there exist an open affine subspace U ⊂ X of P , such

that IC1|U = (x) and IC2|U = (y). From this it follows in a similar way as for X = A

2

that we can smooth the curve C if there exists an element v ∈ H0(C, NC/X) such that

v|C1 ∈ H

0_(C

1, NC1/X ⊗ OC1(P )) \ H

0_{(C, N} C1/X).

(33)

3.2.2. Smoothing a node in a general scheme

Let X be a scheme and let C1, C2 be curves on X such that C = C1∪ C2 is a node as

vizualised in figure 3.2. Let P = C1∩ C2 be the singular point of this node. Our goal is

to show when we can smooth such a curve C. Since in the previous subsection we have showed how we can smooth C if dim X = 2, we assume in this subsection that dim X > 2. Let Q ∈ C be a point, then by construction of the normal bundle we have that NC/X,Q =

TX,Q

TC,Q, where TC

∼

= TC1 ⊕ TC2. This shows that

NC/X,Q|C1 = TX,Q TC,Q C1 = TX,Q|C1 (TC1,Q⊕ TC2,Q)|C1 = TX,Q|C1 TC1,Q|C1 ⊕ TC2,Q|C1 = NC1/X,Q if Q ∈ C1\ {P } NC1/X,PhTC2,Pi if Q = P

where NC1/X,PhTC2,Pi is the normal bundle of C1 at P where we allow a pole at P in

the direction of C2. So the above equation gives that NC/X|C1 = NC1/XhTC2,Pi, it is the

normal bundle of C1 where we allow a pole at P in the direction of C2.

Now we give two important short exact sequences, which give some properties of NC1/XhTC2,Pi.

0 → NC2/X → NC/X → NC/X|C1 ∼= NC1/XhTC2,Pi → 0 (3.3)

0 → NC1/X → NC1/XhTC2,Pi → k(P ) → 0 (3.4)

The first short exact sequence (sequence (3.3)) gives the following long exact sequence.

0 H0(C2, NC2/X) H 0_{(C, N} C/X) H0(C1, NC1/XhTC2,Pi) H1(C2, NC2/X) H 1_{(C, N} C/X) . . .

Now if there exists an element [ϕ] ∈ H0(C, NC/X) such that restricting to C1 gives an

element [ϕ|C1] ∈ H

0_(C

1, NC1/XhTC2,Pi) \ H

0_(C

1, NC1/X), we have that we can smooth our

curve C. This follows in the same way as described above in subsection 3.2.1, where we smooth a node in a surface.

3.3. Construction of the comb with handle D

Let D ⊂ X be a curve in a scheme X. Now we want to construct a comb C with handle D such that the obstruction space is zero. For this we first give the definition of a comb. Definition 3.3. A comb with n teeth is a connected and reduced nodal curve C with irreducible components D, C1, . . . , Cn such that

(i) D is smooth and every Ci is isomorphic to P1,

(ii) the only singularities of C are the intersection points,

(34)

We call D the handle and the Ci the teeth. A subcomb C0 of C is a subcurve of C that

contains the handle D, so C0 = D ∪ Ci1 ∪ · · · ∪ Cip with i1, . . . , ip ∈ {1, . . . , n}.

Let D ⊂ X be a curve in a smooth projective variety X. Then we assume the following important condition on X.

(∗) Through a general point x ∈ D there is a rational curve C on X in a general direction v ∈ ND/X,x such that H1(C, NC/X) = 0.

Under this condition we prove that we can add teeth to the curve D such that the obstruction space vanishes. This is given in the following proposition.

Proposition 3.4. Let X be a smooth projective variety over an algebraically closed field k. Let D ⊂ X be a smooth irreducible curve and assume that condition (∗) is satisfied. Then there are curves C1, . . . , Cn on X such that C0 = D ∪ C1∪ · · · ∪ Cn is a comb with

H1(C0, NC0_/X) = 0.

Proof. We prove this theorem in two steps. First we construct a comb C0 such that H1_{(D, N}

C0_/X|_D) = 0. Then we show that this comb actually satisfies the condition. The

main idea, of adding teeth to the curve D, is from [AK03, Theorem 27, p. 33]. Step 1. First we construct a comb C0 ⊂ X such that H1_{(D, N}

C0_/X|_D) = 0.

Let D0 = D and assume that there exist a comb Di = D ∪ C1 ∪ · · · ∪ Ci such that

di = h1(D, NDi/X|D) 6= 0 and H

1_(C

j, NCj/X) = 0 for j = 1, . . . , i. Then we show

that there is a curve Ci+1 distinct from C1, . . . Ci with H1(Ci+1, NCi+1/X) = 0 such that

di+1< di, where Di+1 = Di∪ Ci+1. We continue this until dn= 0.

By Serre duality we have that H1(D, NDi/X|D) ν ∼ = Hom(NDi/X|D, KD) = Hom((IDi/I 2 Di) ν_| D, KD) = Hom(K_D−1, IDi/I 2 Di|D)

Now take a rational curve Ci+1 distinct from C1, . . . , Ci such that Ci+1 intersects D

transversally in exactly one point Pi+1 ∈ X. Let v ∈ ND/X be the tangent direction of

Ci+1 at Pi+1, then we have the following exact sequence.

0 7→ IDi+1/I 2 Di+1|D → IDi/I 2 Di|D → (k · v) ν _{→ 0} _(3.5)

So we have an inclusion Hom(K−1_D , IDi+1|D/I

2 Di+1) → Hom(K −1 D , IDi/I 2 Di|D), which proves

that di+1≤ di. Now we want to find an element ϕ ∈ Hom(K−1_D , IDi/I

2

Di|D) such that it

is not in the image of the inclusion, because then we have that di+1< di.

Let ϕ ∈ Hom (K_D−1, IDi/I

2

Di|D) be some non-zero element, this element exists since

di 6= 0. We are done if ϕ does not come from an element ˜ϕ ∈ Hom(K−1_X , IDi+1/I

2

Di+1|D).

We are going to choose a general point Pi+1∈ D such that a rational curve Ci+1through

Pi+1 satisfies this condition. For each general P ∈ D let ϕP be the composition of ϕ with

the restriction map IDi/I

2

Di|D → (ND,P)

ν_{. Since ϕ 6= 0, there is an open subset V ⊂ D}

such that ϕP has rank 1 for every P ∈ V . Denote by ξP ∈ ND,P∗ the element that spans

the image of ϕP. Choose a general point P ∈ V not contained in C1 ∪ · · · ∪ Ci and a

(35)

(i) H1_(C

i+1, NCi+1/X) = 0,

(ii) Ci+1 intersects Di only at P and the intersection is transversal,

(iii) The projection of the tangent direction of Ci+1 at P to ND,P, denoted by v, is not

contained in the kernel of ξP : ND,P → k.

We can choose this general point and the rational curve since condition (∗) is satisfied. From sequence (3.5) it follows that ϕ comes from an element ˜ϕ ∈ Hom(K−1_D , IDi+1/I

2 Di+1|D)

if and only if ξPi+1|v = 0. By construction of Ci+1 this is not the case, so di+1 < di.

Via the described construction we can construct a comb C0 ⊂ X with handle D that satisfies H1_{(D, N}

C0_/X|_D) = 0.

Step 2. Finally, we show that the comb C0 actually satisfy H1(C0, NC0_/X) = 0.

We prove this by using a long exact sequence in cohomology. In order to do this, first look at the following short exact sequence.

0 → ⊕n_i=1NCi/X → NC0/X → NC0/X|D → 0

This gives the following long exact sequence. 0 ⊕ni=1H0(Ci, NCi/X) H 0_(C0_{, N} C0_/X) H0(D, N_C0_/X|_D) ⊕n i=1H1(Ci, NCi/X) H 1_(C0_{, N} C0_/X) H1(D, N_C0_/X|_D) 0 (3.6)

Since for all i we have that H1_(C

i, NCi/X) = 0 and H

1_{(D, N}

C0_/X|_D) = 0 it follows that

H1_(C0_{, N}

C0_/X) = 0.

Remark 3.5. If we add extra teeth to C0, which forms a new comb C, this comb C also satisfies H1_{(D, N}

C/X|D) = 0. By a similar argument as in step 2 we also have that

H1_{(C, N}

C/X) = 0.

In a similar way as in section 3.2.2 we can prove that NC0_/X|_D ∼= N_D/XhT_C i,Pii n i=1, where ND/XhTCi,Pii n

i=1 is the normal sheaf of D alowing single poles at the points Pi in

the tangent direction TCi,Pi.

For smoothing the curve C0, we need that there exists a section in H0_(C0_{, N}

C0_/X) that

has poles at every singular point Pi of C0 = D ∪ C1∪ · · · ∪ Cn. If we have that NC0_/X is

generated by global sections, then we show that such a section exists for a subcomb of C0. In order to do this we first give the definition of a globally generated sheaf.

Definition 3.6. Let X be a scheme of finite type over a field k. A coherent sheaf F is generated by its global sections at a point x ∈ X, or globally generated at x, if the images of the global sections of F in the stalk Fx generate that stalk as a OX,x-module.

We say that F is generated by its global sections, or globally generated, if it is globally generated at every point of X.

So we have that if F is a globally generated sheaf of a scheme X, then H0(X, F )−→ F ⊗ k(x) → 0f

(36)

is exact, i.e. f is surjective, for all x ∈ X.

Now we extend the constructed comb C0 in proposition 3.4 to a comb C such that NC/X is globally generated. By remark 3.5 we have that C then also satisfies that the

obstruction space vanishes.

Lemma 3.7. Let the assumptions be as in proposition 3.4 and let C0 be the constructed comb with handle D. Then we can extend C0 to a comb C such that NC/X is globally

generated.

Proof. Let M be a line bundle on D. In a similar way as in proposition 3.4 we can add teeth to C0 to get a comb C = D ∪ C1 ∪ · · · ∪ Cn such that H1(D, NC/X|D ⊗ M ) = 0.

Since M is a line bundle on a curve we have that M embeds in OD(−P ) for every P ∈ D.

From this it follows that H1_{(D, N}

C/X|D⊗ OD(−P )) = 0 for all P ∈ D. Now look at the

following exact sequence, where NC/X|D(−P ) = NC/X|D⊗ OD(−P ).

0 → NC/X|D(−P ) → NC/X|D → NC/X|D⊗ k(P ) → 0

This sequence is exact since NC/X|D(−P ) consist of all elements of NC/X that has at

least a zero at P . Now we have the following long exact sequence.

0 H0_{(D, N}

C/X|D(−P )) H0(D, NC/X|D) H0(D, NC/X|D⊗ k(P ))

H1_{(D, N}

C/X|D(−P )) H1(D, NC/X|D) . . . f

If f is surjective, we have that NC/X|D is globally generated at P , because

H0(D, NC/X|D ⊗ k(P )) ∼= NC/X|D⊗ k(P ).

By construction we have that H1(D, NC/X|D ⊗ OD(−P )) = 0, which implies that f is

surjective. Since this holds for every P the sheaf NC/X|D is globally generated.

Finally, we want to show that actually NC/X is globally generated. Since NC/X|D is

globally generated it is enough to prove that NC/X|Ci is generated by global sections for

each i ∈ {1, . . . , n}. Since all the rational curves Ci are globally generated it is enough to

check the surjectivity of H0_{(C, N}

C/X) → H0(C, NC/X⊗ k(P )) for P ∈ {Pi : i = 1, . . . , n},

the set of intersection points. Let i ∈ {1, . . . , n}, then we have the following short exact sequence.

0 → NC/X|Ci(−Pi) → NC/X|Ci → NC/X|Ci ⊗ k(Pi) → 0

By a similar argument as above we have that NC/X|Ci is globally generated if

H1(Ci, NC/X|Ci(−Pi)) = 0.

Look at the following short exact sequence.

Rational curves on algebraic varieties