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Classical Algebraic Geometry Xinwen Zhu Lecture notes by Tony Feng Fall 2011
Classical Algebraic Geometry
Xinwen ZhuLecture notes by Tony Feng
Fall 2011
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Contents
1 Algebraic Varieties 7
1.1 Affine algebraic varieties . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Regular functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Morphisms of algebraic varieties . . . . . . . . . . . . . . . . . . . . 11
2 Dimension Theory 13
2.1 Tangent Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Dimension theory of rings . . . . . . . . . . . . . . . . . . . . . . . . 16
2.3 Local structure of smooth points . . . . . . . . . . . . . . . . . . . . 17
3 Projective varieties 21
3.1 Abstract algebraic varieties . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 Projective space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 Regular functions on projective varieties . . . . . . . . . . . . . . . . 24
3.4 Products of Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.5 The Grassmannian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4 Morphisms 35
4.1 Rational maps and correspondences . . . . . . . . . . . . . . . . . . 35
4.2 Blowing up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.3 Rational maps and function fields . . . . . . . . . . . . . . . . . . . . 39
4.4 Complete varieties and projections . . . . . . . . . . . . . . . . . . . 40
4.5 Separable morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
4.6 Smooth morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
5 Divisors and Line Bundles 51
5.1 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
5.2 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
5.3 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
5.4 The canonical bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5.5 The ramification divisor . . . . . . . . . . . . . . . . . . . . . . . . . 64
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6 The Hilbert Polynomial 676.1 Construction of the Hilbert Polynomial . . . . . . . . . . . . . . . . 676.2 Numerical invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.3 Dimension theory of intersections . . . . . . . . . . . . . . . . . . . . 70
7 Algebraic Curves 757.1 Curves and function fields . . . . . . . . . . . . . . . . . . . . . . . . 757.2 Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757.3 The Picard Group of Curve . . . . . . . . . . . . . . . . . . . . . . . 797.4 Jacobians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 817.5 The Riemann-Roch Theorem . . . . . . . . . . . . . . . . . . . . . . 83
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Disclaimer
These are lecture notes from a course (Math 232a: Algebraic Geometry I) offeredat Harvard University in the fall of 2011 by Xinwen Zhu. I have (lightly) editedthem, but there are inevitably still typos and mistakes stemming from my ownmisunderstanding, for which I take full responsibility.
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Chapter 1
Algebraic Varieties
1.1 Affine algebraic varieties
Let k be an algebraically closed field.
Definition 1.1.1. We define affine n-space to be the (for now) set
Ank = (a1, . . . , an) | ai ∈ k.
Definition 1.1.2. An algebraic set X ⊂ An is the set of zeros of a collection ofpolynomials f1, . . . , fm ∈ k[x1, . . . , xn], i.e.
X = (a1, . . . , an) ∈ Akn | f1(a1, . . . , an) = . . . = fm(a1, . . . , an) = 0.
Remark 1.1.3. Note that this depends only on the ideal I = (f1, . . . , fm) ⊂ k[x1, . . . , xn].
Lemma 1.1.4. In the notation used above, the following relations hold.
1. I1 ⊂ I2 ⇐⇒ V (I1) ⊃ V (I2).
2. V (I1) ∪ V (I2) = V (I1 ∩ I2) = V (I1I2).
3.⋂α V (Iα) = V (
∑α Iα).
4. V (0) = An, V (k[x1, . . . , xn]) = ∅.
Definition 1.1.5. The Zariski topology on An is the topology where the closed subsetsare the algebraic sets.
Example 1.1.6. Consider the Zariski topology on A1k. The algebraic sets are ∅, the
finite subsets, and all of A1k (i.e. the topology is the cofinite topology). This follows
from the observation that k[x] is a principal ideal domain, and f splits into linearfactors.
Example 1.1.7. Consider the Zariski topology on A2k. The algebraic sets include
∅,A2k, and V (f), but there are also non-principal ideals. However, it can be showed
that every algebraic set is a finite union of these.
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Remark 1.1.8. The Zariski topology on A2 is not equal to the product topology onA1 × A1!
Theorem 1.1.9 (Hilbert’s Nullstellensatz). Let X = V (I). Then
f ∈ k[x1, . . . , xn] | f(p) = 0 ∀p ∈ X =√I.
Corollary 1.1.10. There is an order-reversing bijection between radical ideals andalgebraic sets: I 7→ V (I) and X 7→ I(X).
Example 1.1.11. Consider V (xy) ⊂ A2, which is the union of the two coordinateaxes: V (xy) = V (x)∪V (y). This algebraic set is somehow built out of other pieces.To make this notion precise, we make a definition.
Definition 1.1.12. Let X 6= ∅ be a topological space. We say that X is irreducibleif X cannot be written as the union of two proper closed subsets.
Note that this notion is trivial for the classical topology, since nearly every set canbe written as the union of two proper closed subsets. However, it’s a good notationfor algebraic geometry, since the Zariski topology is so coarse. For example, A1
k isirreducible since the proper closed subsets are finite.
Definition 1.1.13. The irreducible components of an algebraic set are its maximalirreducible closed subsets.
Proposition 1.1.14. There is a bijection between irreducible algebraic sets of Ank(with the induced topology) and prime ideals of k[x1, . . . , xn]. In addition, everyalgebraic set can be written as the union of its irreducible components.
Proof. We first show that if V (I) 6= ∅ is irreducible, then I is prime. We may assumethat I is radical. Suppose fg ∈ I and f /∈ I. Then V (I) ⊂ V (fg) = V (f) ∪ V (g),so V (I) ⊂ V (f) or V (I) ⊂ V (g), showing that f ∈ I or g ∈ I.
Conversely, suppose I is prime; we want to show that V (I) is irreducible. Sup-pose V (I) = V (I1) ∪ V (I2) = V (I1I2). Therefore, I = I1I2, so I = I1 or I = I2.
Remark 1.1.15. It is a fact that any radical ideal I can be written as I = p1∩. . .∩pr,with pr minimal prime ideals containing I, so we may write
V (I) = V (p1) ∪ . . . ∪ V (pr).
So the V (pi) are maximal irreducible components of V (I).
Definition 1.1.16 (Tentative). An affine (algebraic) variety is an irreducible alge-braic set in An with its induced topology. A quasi-affine variety is an open subsetof an affine variety.
Example 1.1.17. We give some examples of affine algebraic sets.
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(i) An.
(ii) Let l1, . . . , lm be independent linear forms of x1, . . . , xm; and let a1, . . . , am ∈ k.Then V (l1−a1, . . . , lm−am) ⊂ An. This is called a linear variety of dimensionn−m.
(iii) The variety V (x1 − a1, . . . , xn − an) “corresponds to” the point (a1, . . . , an) ∈Ank .
(iv) Let f ∈ k[x1, . . . , xn] be irreducible. Then (f) is prime, and V (f) is an affinevariety (usually called a hypersurface in An).
(v) Given f1(t), . . . , fn(t) ∈ k[t]. Then V (x1 − f1, . . . , xn − fn) ⊂ An+1, xi − fi ∈k[t, x1, . . . , xn]. For example, V (x− t2, y − t3) ⊂ A3.
1.2 Regular functions
Definition 1.2.1. Let X ⊂ An be a quasi-affine variety. A function f : X → k iscalled regular at p ∈ X if there exists an open neighborhood U ⊂ X containing pand polynomials f1, f2 ∈ k[x1, . . . , xn] such that f2(q) 6= 0 for all q ∈ U , and f = f1
f2on U . We say f is regular if it is regular at all p ∈ X.
Observe that the set of regular functions on X form a ring, denoted by O(X).
Lemma 1.2.2. Let f : X → k be a regular function; regard it as a map f : X → A1k.
Then f is continuous.
Proof. We claim that it is enough to show that f−1(0) is closed, since the closedsubsets of A1
k are just finite unions of points or the empty set. It suffices to show thatfor every p ∈ X, there is an open subset U ⊂ X containing p such that f−1(0) ∩ Uis closed in U .
We can choose U such that f = f1/f2 on U , with f1, f2 ∈ k[x1, . . . , xn]. Since f2
does not vanish on U , f−1(0)∩U = f−11 (0)∩U = V (f1)∩U , which is indeed closed
in U by definition.
Corollary 1.2.3. Let f be a regular function on a quasi-affine variety X. If f = 0on some nonempty open subset U ⊂ X, then f = 0 on X.
Proof. Let Z = x ∈ X | f(p) = 0. This is a closed subset since f is continuous.So X = Z∪X \U . Since X is irreducible, it is contained in X \U ∪Z, so X = Z.
Corollary 1.2.4. If X is a quasi-affine variety, then O(X) is an integral domain.
Proof. If fg = 0, then V (f) ∪ V (g) = X, so V (f) = X or V (g) = X.
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Definition 1.2.5. The field of rational functions on X is the fraction field of O(X),denoted by K = k(X).
The local ring at p ∈ X is OX,p = lim←−p∈U O(U).
Let f ∈ k(X) be a rational function, so we may write f = f1f2
, with f1, f2 ∈ O(X).Therefore, f defines a regular function on U = X − f2 = 0. Conversely, iff ∈ O(U), we claim that f can be regarded as a rational function on X. To seethis, pick p ∈ U , so there exists an open neighborhood V containing p such thatf = f1
f2on V , where f1 and f2 are polynomials and can be regarded as in O(X).
This construction is well-defined: for p′ ∈ V ′ ⊂ U , suppose f =f ′1f ′2
on V ′. Then
f1
f2=f ′1f ′2
on V ∩ V ′
so
f1f′2 = f ′1f2 on V ∩ V ′.
Therefore, f1f′2 = f ′1f2 on X, so
f1
f2=f ′1f ′2
in K.
In conclusion, for each open set U ⊂ X and each p ∈ U , we have inclusions
O(X) ⊂ O(U) ⊂ OX,p ⊂ k(X).
Suppose X is given as V (p). We want to explicitly determine O(X) or K = k(X).
Theorem 1.2.6. Let X = V (p) ⊂ An be an affine variety. Then
(i) There exists a natural isomorphism
A(X) := k[X1, . . . , Xn]/p ' O(X).
(ii) There is a one-one correspondence
points on X ←→ maximal ideals of A(X)p←→ mp
(iii) A(X)mp ' OX,p.
Proof. Part (ii) follows from recalling that Y ⊂ X as nonempty irreducible algebraicsets is equivalent to I(X) ⊂ I(Y ). Therefore, the points must correspond to maximalideals.
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There is a natural map A(X)→ O(X), which is injective by the Nullstellensatz.On the other hand, there is an inclusion O(X)→ Frac(A(X)) = k(X). So we havethe commutative diagram
A(X) //
O(X) //
Frac(A(X))
A(X)mp
// OX,p // K
Recall that
A(X) =⋂
m maximal
A(X)m →⋂p∈XOX,p = O(X).
But the map A(X)mp → OX,p is also a surjection since f ∈ OX,p can be written in
the form f = f1f2
on some open set U containing p, with f2 /∈ p. Therefore, A(X)surjects onto O(X) as well.
1.3 Morphisms of algebraic varieties
Definition 1.3.1. Let X,Y be two quasi-affine varieties. A morphism ϕ : X → Y isa continuous map such that for any open subset V ⊂ Y , if f : V → k is regular thenf ϕ : ϕ−1(V )→ k is regular.
A morphism ϕ : X → Y is an isomorphism if there is a morphism ψ : Y → Xsuch that ψ ϕ = ϕ ψ = id.
Definition 1.3.2. A quasi-affine variety is said to be affine if it is isomorphic to anaffine variety.
Example 1.3.3. A1−0 is a quasi-affine variety, but is also an affine variety becauseit is isomorphic to V (xy − 1) ⊂ A2.
Lemma 1.3.4. Let ϕ : X → Y ⊂ An be a map (of sets) of two quasi-affine varieties.Then ϕ is a morphism if and only if xi ϕ is regular.
Here xi represents the function given by projection to the ith coordinate.
Proof. If ϕ is a morphism, then xi ϕ is regular since xi is regular. Conversely,suppose xi ϕ is regular for each xi. For each f ∈ k[x1, . . . , xn], we have
f ϕ = f(x1 ϕ, . . . , xn ϕ) ∈ O(X).
It follows that ϕ is continuous, since for any closed subset Z =⋂V (fα), we have
ϕ−1(Z) =⋂α
V (fα(xi ϕ, . . . , xn ϕ)).
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If V ⊂ Y , p ∈ ϕ−1(V ), and f ∈ O(V ), we have f = f1f2
in some neighborhood ofϕ(p) contained in V , where f1, f2 ∈ k[x1, . . . , xn], and f2(p) 6= 0. Therefore, in someneighborhood of p,
f ϕ =f1 ϕf2 ϕ
is regular at p.
Example 1.3.5. Consider the map (of sets) Gm = A1 − 0 → V (xy − 1) ⊂ A2
sending t 7→ (t, t−1). Since each coordinate is a regular function, this is a morphism.The inverse morphism is given by (x, y) 7→ x.
Proposition 1.3.6. Let X be quasi-affine and Y be affine. Then there is a naturalisomorphism
Hom(X,Y ) ' Homk−alg(O(Y ),O(X)).
Proof. If ϕ : X → Y is a morphism, then we obtain ϕ∗ : O(Y ) → O(X) given byf 7→ f ϕ.
Now suppose that we are given a ψ ∈ Homk−alg(O(Y ),O(X)). Let Y ⊂ An,and define a map ϕ : X → An by p 7→ (ψ(x1)(p), . . . , ψ(xn)(p)) ∈ Y . Note thatϕ(p) ∈ Y since ϕ factors through I(Y ) by definition.
Corollary 1.3.7. There is an equivalence of categories between affine varieties andfinitely generated integral k-algebras.
Example 1.3.8. The quasi-affine variety A2 − (0, 0) has ring of regular functionsk[x, y], so the coordinate ring is not sufficient to describe quasi-affine varieties.
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Chapter 2
Dimension Theory
2.1 Tangent Space
Definition 2.1.1. Let A→ B be a homomorphism of commutative algebras, and Ma B-module. We define the derivations of B into M to be
DerA(B,M) = D | D : B →M map of abelian groups
such that:
(i) D(b1b2) = b1D(b2) + b2D(b1),
(ii) D(a) = 0 for all a ∈ A.
Observe that DerA(B,M) is a B-module.
Example 2.1.2. Let A = k, B = k[x1, . . . , xn], and M = k. Then
Derk(B, k) = D =∑
λi∂
∂xi| λ1, . . . , λn ∈ k,
where λi is the image of xi. By definition,
∂
∂xi(xj) = δij .
Now consider A = k, B = OX,p. Then we have a map B → OX,p/mp ' k.
Definition 2.1.3. The Zariski tangent space of X at p is defined to be
TpX = Derk(OX,p, k).
A priori TpX is a OX,p-module, and m acts trivially, so it is an OX,p/m ' k vectorspace.
Lemma 2.1.4. TpX ' (m/m2)∗.
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Proof. Suppose we have a derivation D : OX,p → k. Then D factors through m2.Since Ox,p ' k ⊕m, and D is zero on k, D depends only on its value on m.
Exercise 2.1.5. Complete the proof.
Exercise 2.1.6. Show that m/m2 is a finite-dimensional k-vector space.
Theorem 2.1.7. Let X be a quasi-affine variety. Let K = k(x) be its functionfield.
(i) For all p ∈ X, we have
dimK Derk(K,K) ≤ dimk TpX.
(ii) There exists a nonempty open set U ⊂ X such that
dimk TpX = dimK Derk(K,K) for all p ∈ U.
Definition 2.1.8. A point p ∈ X is called smooth (or non-singular) if dimTpX =dimX. Otherwise, p is called a singular point. X is called smooth (or non-singular)if all points p ∈ X are smooth.
Proof. After replacing X with an open subset, we can assume X is affine. We have
Derk(K,K) ' Derk(O(X),K).
If X = V (p) ⊂ An, then O(X) = k[x1, . . . , xn]/p, then
Derk(O(X),K) = D =∑
λi∂
∂xiλi ∈ K | D(f) = 0 for all f ∈ p.
Choosing a set of generators (f1, . . . , fm) = p, we identify the above with
Derk(O(X),K) = D =∑
λi∂
∂xiλi ∈ K |
∑λi∂fi∂xi
= 0..
This is the kernel of the map J : Kn → Kn given by the Jacobian J . So
dimK Derk(K,K) = n− rankKJ.
By the same reasoning,
dimk TpX = n− rankkJ(p).
So we need to show that (i) rankkJ(p) ≤ rankKJ and (ii) there exists a non-emptysubset U such that rankkJ(p) = rankKJ on U .
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The first point (i) is clear, as the rank can only go down under specialization.For the second point, assume that rankKJ = r. Then there exist n × n matricesA,B with entries in K such that
AJB =
(Ir∗
).
Let A = A0/α, B = B0/β, for α, β ∈ O(X) such that A0, B0 have entries inO(X). Let f = αβ detA0 detB0 ∈ O(X), so for all p ∈ U = X − V (f) we haverankkJ(p) = r.
Lemma 2.1.9. The function p 7→ dimk TpX is upper semi-continuous, i.e.
Xl = p ∈ X | dimTpX ≥ l
is closed.
Proof. We know
dimTpX = n− rank k
(∂fi∂xj
(p)
).
So Xl is V (p + ideal generated by (n− l + 1)× (n− l + 1) minors of J).
Example 2.1.10. Suppose chark 6= 2. Consider the curves defined by
• y2 = x3,
• y2 = x3 + x2, and
• y2 = x3 + x.
These are plane curves, so intuitively they should have dimension 1.
• For the first curve f = y2 − x3, we have fx = −3x2 and fy = 2y, so
J = (−3x2, 2y).
Then J(0, 0) = (0, 0); the Zariski tangent space at the origin is two-dimensional.Thus the curve is singular at the origin.
• For the second curve f = y2 − x3 − x2, fx = −3x2 − 2x and fy = 2y, so
J = (−x2 − 2x, 2y).
Then J(0, 0) = (0, 0); the Zariski tangent space at the origin is two-dimensional,so this curve is also singular at the origin is singular.
• For the third curve f = y2 − x3 − x, we have fx = 3x2 − 1 and fy = 2y, so
J = (3x2 − 1, 2y).
In this case the rank of J is 1 at every point, and the curve is smooth.
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2.2 Dimension theory of rings
Definition 2.2.1. Let K/k be a (finitely generated) field extension. We say that Kis separably generated over k if there exists L satisfying k ⊂ L ⊂ K, L/k is purelytranscendental, and K/L is a finite, separable extension.
Theorem 2.2.2. If K/k is finitely separably generated, then
dimK Derk(K,K) = tr . degkK.
Theorem 2.2.3. Let k be a perfect field (e.g. algebraically closed). Then everyfinitely generated K/k is separably generated.
Proposition 2.2.4. Let X be an affine variety in An. Then dimX = n− 1 if andonly if X = V (f) for some irreducible f ∈ k[x1, . . . , xn].
Remark 2.2.5. dimX = n− 2, then it is not necessarily true that I(X) is generatedby two elements. We’ll encounter an example of this in the homework.
Proof. Suppose X = V (f). We want to show that dimX = k − 1. By definition,
dimX = n− rank
(∂f
∂xi
)≥ n− 1.
So we want to show that
rank
(∂f
∂xi
)6= 0 in OX
Well, ∂f∂xi
is 0 only if it lies in (f), but f is irreducible so this implies ∂f∂xi
= 0 for alli. This is only possible when k has characteristic p > 0 and
f = g(xp1, . . . , xpn) = g(x1, . . . , xn)p.
But f is irreducible by assumption, so this is impossible.
Now assume that dimX = n − 1. Let g ∈ p = I(X). We can assume thatg is irreducible, so dimV (g) = n − 1. Therefore, the proposition follows from theproceeding lemma.
Lemma 2.2.6. Let A be an integral domain over k and p ⊂ A be a prime ideal.Then
tr .degk A/p ≤ tr . degk A
with equality holding if and only if p = 0 or both sides are infinity.
Remark 2.2.7. Note that this lemma also implies that the dimension of a hypersur-face is n− 1.
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Proof. Assume to the contrary that tr . degk A/p = tr .degk A = n. Then there existx1, . . . , xn ∈ A such that x1, . . . , xn are algebraically independent over k.
Choose some non-zero element y ∈ p. There exists P ∈ k[Y,X1, . . . , Xn] suchthat P (y, x1, . . . , xn) = 0. We may assume that P is irreducible, and not a powerof Y . Therefore,
P (0, x1, . . . , xn) = 0
is an algebraic relation in A/p.
2.3 Local structure of smooth points
Theorem 2.3.1. Let f1, . . . , fr ∈ k[x1, . . . , xn] have no constant terms and inde-pendent linear terms. Let p = (f1, . . . , fr)OAn,0 ∩ k[x1, . . . , xn]. Then:
(i) p is a prime ideal,
(ii) X = V (p) has dimension n− r, and 0 ∈ X is smooth.
(iii) V (f1, . . . , fr) = X ∪ Y , where Y is some algebraic set such that 0 /∈ Y .
The theorem tells us that locally around a smooth point, an n − r-dimensionalaffine variety in affine n-space is cut out by r equations.
Corollary 2.3.2. Let X be an affine variety of dimension n− r and suppose 0 ∈ Xis smooth. Then there exists f1, . . . , fr ∈ I(X) such that
I(X) = (f1, . . . , fr)OAn,0 ∩ k[x1, . . . , xn].
Proof. We just need to find r polynomials in I(X) with independent linear terms, atwhich point we may apply the Theorem 2.3.1. Let m = (x1, . . . , xn) ⊂ k[x1, . . . , xn]be the maximal ideal corresponding to the origin, and m0 the maximal ideal of O(X)corresponding to 0. These fit into a short exact sequence
0→ p→ m→ m0 → 0.
Looking “to second order” gives an exact sequence
0→ (p + m2)/m2 → m/m2 → m0/m20 → 0.
But(p + m2)/m2 ' p/(p ∩m2),
so we choose a basis (f0, . . . , fr) for p/(p∩m2) and lift it to (p+m2)/m2, which thenhas the property that the linear terms are independent. We then set
q = (f1, . . . , fr)OAn,0 ∩ k[x1, . . . , xn]
= ∑ gifi
ki| ki(0) 6= 0,
∑ gifiki⊂ k[x1, . . . , xm] ⊂ I(X).
So Y = V (q) ⊃ X, whiledimY = dimX = n− r, implying that X = Y .
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Recall that if (A,m, k) is a local ring, then we define the completion
A = lim←−A/mn.
Example 2.3.3. We have seen that
OAn,0 = k[x1, . . . , xn](x1,...,xn).
ThenOAn,0 = k[[x1, . . . , xn]].
Proof of Theorem 2.3.1. Let
p = (f1, . . . , fr)OAn,0 ∩ k[x1, . . . , xn],
p′ = (f1, . . . , fr)OAn,0
p′′ = (f1, . . . , fr)k[[x1, . . . , xn]].
It suffices to show:
(i) p′′ is prime in k[[x1, . . . , xn]].
(ii) p′′ ∩ OAn,0 = p′.
By definition,
p ⊂ p′′ ∩ OAn,0 =∑
gifi | gi ∈ k[[x1, . . . , xn]],∑
gifi ∈ k[x1, . . . , xn](x1,...,xn)
⊂∞⋂n=0
(p′ + mN ).
We have an exact sequence
0→ mN → OAn,0 → OAn,0/mN → 0,
where m = (x1, . . . , xn) is the maximal ideal of OAn,0. Thus
OAn,0/mN ' k[x1, . . . , xn]/(x1, . . . , xn)N
This gives a splitting of k-vector spaces
OAn,0 ' mN ⊕ polynomials of deg < N.
Now, any element f ∈ p′′ ∩ OAn,0 may be written as f =∑hifi with hi ∈
k[[x1, . . . , xn]]. We may write hi = h′i + (deg ≥ N terms). Then
f −∑
h′ifi = (f ′i −∑
h′ifi) + mNOAn,0 ∈ mNOAn,0.
Therefore,
p′′ ∩ OAn,0 ⊂⋂N≥0
(p′ + mN ) = p′
by Krull’s intersection theorem, stated below.
18
Theorem 2.3.4 (Krull intersection theorem). If (A,m, k) is Noetherian and local,then for any ideal I,
I =∞⋂N=0
(I + mN ).
We will show below that k[[x1, . . . , xn]]/(f1, . . . , fr) ' k[[xr+1, . . . , xn]]. This willimply that p′′ is prime, and
OAn,0/p′ → OAn,0/p
′′
is injective. In particular, xr+1, . . . , xn are algebraically independent in OAn,0 sincetheir images are algebraically independent. Therefore, dimV (p) ≥ n− r.
We have TpV (p) =(mp/m
2p
)∗. We have a short exact seqauence
0→ p/(p + (x1, . . . , xn))→ m/m2 → mp/m2p → 0.
So TpV (p) is cut out by the linear terms of f1, . . . , fr in TpAn. Since the dimensionof the tangent space is always at least the dimension of the variety, we conclude thatdimV (p) = n− r and p is smooth.
Finally, we wish to show that V (f1, . . . , fr) = V (p) ∪ Y where 0 /∈ Y . Letg1, . . . , gs be a set of generators of p. Then
gi ∈ (f1, . . . , fr)OAn,0 ∩ k[x1, . . . , xn]
so there exists hi /∈ (x1, . . . , xn), i.e. not vanishing at 0, such that higi ∈ (f1, . . . , fr).Let h = h1h2 . . . hs, so hp ⊂ (f1, . . . , fr). Then
V (f1, . . . , fr) ⊂ V (hp) = V (h) ∪ V (p).
So V (f1, . . . , fr) = V (f1, . . . , fr, h)∪ V (p). We may take the first term in the unionto be Y , since h(0) 6= 0 implies 0 /∈ Y .
It only remains to complete the proof of the claim above, which is accomplishedby the following two results.
Proposition 2.3.5 (Formal implicit function theorem). Let f =∑aixi+ higher
terms ∈ k[[x1, . . . , xn]] such that ai 6= 0. Then every g ∈ k[[x1, . . . , xn]] can beuniquely written as
g = uf + h(x2, . . . , xn).
Proof. We aren’t going to prove this, but it’s actually pretty trivial: just attemptto solve for the power series term-by-term.
Corollary 2.3.6. k[[x1, . . . , xn]]/(f) ' k[[x2, . . . , xn]].
By induction, we conclude the following.
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Corollary 2.3.7. Let f1, . . . , fr be r power series with fi =∑aijxj+ higher terms.
If det(aij)1≤i,j≤r 6= 0 then
k[[x1, . . . , xn]]/(f1, . . . , fr) ' k[[xr+1, . . . , xn]].
This wraps up our proof of the theorem. Let’s recall the corollary from last time.
Corollary 2.3.8. Let p = (0, . . . , 0) ∈ X = V (p) be a smooth point of X, anddimX = n − r. Choose f1, . . . , fr ∈ p with independent linear terms (i.e. cuttingout the tangent space at 0). [Note that this is possible because we have the shortexact sequence
0→ p/(p + (x1, . . . , xn))→ m/m2 → mp/m2p → 0.
The third term has dimension n − r, the second n, so the first has dimension r.This amounts to choosing r polynomials that cut out the tangent space at p.] Thenp = (f1, . . . , fr)OAn,0 ∩ k[[x1, . . . , xn]].
In other words, in the neighborhood of smooth point on an affine variety ofdimension n− r, the variety is cut out by r equations. If we trace through our proofof the theorem, we obtain the following corollary.
Corollary 2.3.9. Let p ∈ X be a smooth point on a quasi-affine variety of dimensionr. Then
OAn,0 ' k[[x1, . . . , xr]].
This is analogous to differential geometry, where a manifold is locally of dimensionr. Note, however, that we must work with completions of the local rings.
20
Chapter 3
Projective varieties
3.1 Abstract algebraic varieties
So far, we’ve established the notion of affine varieties. Just like Euclidean space isused to build up general manifolds, affine varieties are used to build up abstractvarieties. We now discuss these more general notions. Fix an algebraically closedfield k.
Definition 3.1.1. A (pre)variety over k is a connected irreducible topological spacetogether with a covering U = Uα, for each alpha a homeomorphism ϕα : Uα → Xα
with Xα a quasi-affine variety, and for each α, β a morphism of quasi-affinevarieties
ϕβ ϕ−1α : ϕα(Uα ∩ Uβ)→ ϕβ(Uα ∩ Uβ).
In addition, we require U to be maximal, i.e. if V ⊂ X is open, ψ : V∼→ Y is a
map to a quasi-affine variety such that ϕα ψ−1 : ψ(Uα ∩ V ) → ϕα(Uα ∩ V ) is amorphism for all α implies V ∈ U . This is just a technical point to make this choiceof atlas canonical.
Definition 3.1.2. A function f : X → k is called regular if for all α the compositionf ϕ−1
α : Xα → k is regular.
Definition 3.1.3. We denote by O(X) the ring of regular functions on X.
OX,p = lim−→U
O(U),
where the direct limit ranges over open sets U containing p. Finally, we define k(X)to be k(Xα) for any α. In particular, k(X) is not necessarily the fraction field ofO(X).
We also define dimension, tangent space, and smoothness, etc. affine-locally,since these should be local notions.
Definition 3.1.4. Let X be a pre-variety. An irreducible closed subset Y ⊂ X withthe canonical variety structure is called a closed subvariety.
21
In this case, we need to check that the restrictions of the transition maps arestill morphisms, but this is clear by using the criterion that the maps are morphismsif and only if the coordinate functions are morphisms.
Definition 3.1.5. Let X,Y be two varieties. A continuous map ϕ : X → Y is calleda morphism if for any p ∈ X and U containing ϕ(p)
f ∈ O(U) =⇒ f ϕ ∈ O(ϕ−1(U)).
This stuff is too abstract. Let’s move on to projective varieties - the most im-portant class of varieties in geometry.
3.2 Projective space
Definition 3.2.1. The projective n-space Pn is the set
(kn+1 − 0)/ ∼
where (a0, . . . , an) ∼ (b0, . . . , bn) if and only if bj = λaj .
Remark 3.2.2. Alternatively, we may view Pn as the set of 1-dimensional subspacesin kn+1. This is an important perspective!
An element p ∈ Pn is called a point, although it is really an equivalence class.Any (a0, . . . , an) ∈ p is called a set of homogeneous coordinates of p.
Definition 3.2.3. An algebraic set Z in Pn is the set of zeros of a set of homogeneouspolynomials f1, . . . , fm.
Z = V (f1, . . . , fm) = p ∈ Pn | f1(p) = . . . = fm(p) = 0
where we evaluate the fj on a set of homogeneous coordinates of p.
Note that in general the value of a polynomial at a point p ∈ Pn is not well-defined,since p has many different sets of homogeneous coordinates. However, the notion ofzeros of a homogeneous polynomial is well-defined.
Let I = (f1, . . . , fr) where the fi are homogeneous and V (I) = V (f1, . . . , fr).Then I is a homogeneous ideal of k[x0, . . . , xn], i.e. if f ∈ I, f =
∑d fd with fd the
homogeneous piece of degree d, then fd ∈ I.
Lemma 3.2.4. (i) V (k[x1, . . . , xn]) = V (x1, . . . , xn) = ∅ and V (0) = Pn.(ii) V (
⋃α I) =
⋂α V (Iα).
(iii) V (I1 ∩ I2) = V (I1) ∪ V (I2).
Proof. Left as an easy exericse.
Definition 3.2.5. The Zariski topology on Pn has as its closed subsets the algebraicsets.
22
Now we mention the extension of some results on affine varieties to projectivespace.
Theorem 3.2.6 (Hilbert’s Nullstellensatz). For all homogeneous ideals I and allhomogeneous polynomials f satisfying deg f ≥ 1, f ∈
√I if and only if f vanishes
on V (I).
Corollary 3.2.7. There is a 1− 1 correspondence between algebraic sets in Pn andhomogeneous radical ideals contained in S+ := (x0, . . . , xn).
Proposition 3.2.8. Let I be a homogeneous ideal in S = k[x1, . . . , xn], regardingS as a graded ring. If
√I = p1 ∩ . . . ∩ pr with pi the minimal primes containing I,
then each pi is a homogeneou ideal.
Corollary 3.2.9. V (p) is irreducible if and only if p is a homogeneous prime inS+. Furthermore, every algebraic set in Pn can be uniquely written as the union ofits irreducible components.
Definition 3.2.10. A projective variety is an irreducible algebraic set in Pn. A quasi-projective variety is an open subset of a projective variety.
Proposition 3.2.11. Pn together with the Zariski topology has a natural varietystructure over k.
Corollary 3.2.12. Every quasi-projective variety has a natural variety structureover k.
Proof of Theorem 3.2.11. Let Hi = (a0, . . . , an) | ai = 0/ ∼= V (xi). This aclosed subset by definition, so Ui := Pn \Hi is open. Note also the Ui cover Pn sinceevery point in Pn has at least one non-zero coordinate by definition. Define
ϕi : Ui → An
(a0 : . . . : an) 7→(a0
ai, . . . ,
aiai, . . . ,
anai
).
[The hat means the element is omitted.] Now we need to show:
(i) ϕi is a homeomorphism.
(ii) ϕi ϕ−1j is a morphism.
Without loss of generality suppose i = 0. For (i), since this is a bijection, we justneed to show that it is closed. Let I ⊂ S be a homogeneous ideal. Denote by Iih theideal of polynomials of the form f(1, x1, . . . , xn) where f ∈ I. We want to considerϕi : U0 ∩ V (I)→ An. But
U0 ∩ V (I) = (a0, . . . , an) | f(a0, . . . , an) = 0 ∀ f ∈ I,
23
which is mapped to(a1
a0, . . . ,
ana0
)| g(a1
a0, . . . ,
ana0
)= 0 ∀ g ∈ I ih
.
On the other hand, let I ⊂ k[x1, . . . , xn] be any ideal. We define the homogeneousideal
Ih =
xdeg f
0 g
(x1
x0, . . . ,
xnx0
)| g ∈ I
.
Then ϕ0(U0∩V (Ih)) = V (I). This completes the proof that ϕ0 is a homeomorphism.For (ii), assume that i = n, j = 0. We wish to show that the map
ϕn ϕ−10 : An \ xn 6= 0 → An \ xn 6= 0
(a1, . . . , an) 7→(
1
an,a1
an, . . . ,
an−1
an
)is an algebraic morphism. For this, it suffices to check that the pullback of thecoordinate functions is regular. This is obvious, since the pullback of xi is xi+1
xn,
which is regular on Un.
In particular, every projective variety V (p) ⊂ Pn is a variety.
Example 3.2.13. Consider V (xy − z2) ⊂ P2. Then Ux = (x, y, z) | x 6= 0 ' A2.On this chart,
V (xy − z2) ∩ Ux = V (y − z2) ⊂ A2.
On the chart Uz = (x, y, z) | z 6= 0 ' A2,
V (xy − z2) ∩ Uz = V (xy − 1) ⊂ A2.
3.3 Regular functions on projective varieties
Let Y = V (p) ⊂ Pn be a projective variety. We denote by S(Y ) = S/p the homoge-neous coordinate ring of Y . This is a graded ring
S(Y ) =⊕d
S(Y )d.
In the affine setting, we saw that there was an equivalence of categories betweenaffine varieties and affine k-algebras via the coordinate ring. However, this is notthe case with projective varieties: there are non-isomorphic projective varieties withisomorphic coordinate rings.
Let p′ be a homogeneous prime ideal of S(Y ). We denote by S(Y )p′ the local-ization of S(Y ) at p′, i.e.
S(Y )p′ =
f
g
∣∣ f, g homogeneous, deg f = deg g, g /∈ p′.
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Proposition 3.3.1. Let Y = V (p) be a projective variety, p ∈ Y , and mp thehomogeneous maximal ideal corresponding to p. Then
(i) OY,p ' S(Y )mp.
(ii) K = k(Y ) = S(Y )(0).
(iii) O(Y ) = k.
Proof. Let Yi = Y ∩ Ui. This is an affine variety since it is a closed, irreduciblesubset of an affine variety. There is a natural isomorphism
ϕ∗i : A(Yi)→ S(Y )xi
The map sends f(x1, . . . , xn) 7→ f(x1x0, . . . , xnx0
). This induces an inverse map of
spectra pSx0 7→ pih, and Y0 = V (pih). (Here the “ih” stanfors for “inhomogeneous.”This means to replace f(x0, x1, . . . , xn) with f(1, x1, . . . , xn).)
Now (i) and (ii) follow from the corresponding facts for affine varieties: withoutloss of generality, we may assume that p ∈ U0 ∩ Y = Y0. Then by definition
OY,p = OY0,p = A(Yi)mihp' (S(Y )x0)ϕ∗0(mih
p ) ' S(Y )mp .
For part (iii), let f ∈ O(Y ) ⊂ k(Y ) = S(Y )(0) ⊂ L, where L is the fraction fieldof S(Y ). Since Y =
⋃Yi, we get a map
O(Y )→∏
O(Yi) = A(Yi) ' S(Y )xi .
So for each i, there exists ni such that xnii f ∈ S(Y ). Now let N ≥
∑Ni. Then
S(Y )Nf ⊂ S(Y )N . So for any m, S(Y )Nfm ⊂ S(Y )N .
Regarding f as a map between finite-dimension k-vector spaces S(Y )N → S(Y )N ,there exists m such that f satisfies the polynomial relation
fm + a1fm−1 + . . .+ am = 0 : S(Y )N → S(Y )N ,
implies that fm+a1fm−1+. . .+am = 0 in L. In particular, this is a finite polynomial
relation over k, but k is algebraically closed, so f ∈ k.
3.4 Products of Varieties
Definition 3.4.1. Let X,Y be two varieties. The product
X ×k Y
is a variety together with two morphisms
X ×k Yprx→ X and X ×k Y
pry→ Y,
25
such that the natural map
Hom(Z,X ×k Y )→ Hom(Z,X)×Hom(Z, Y )
is a bijection.
Theorem 3.4.2. The product of X and Y exists, and is unique up to unique iso-morphism.
Proof. The second property follows by general abstract nonsense. We will show thatthere is a natural variety structure on X × Y making it the product. We do this intwo steps.
(i) If X and Y are affine, then X ×k Y exists.
(ii) Let Ui be a cover of X and each Ui ×k Y exists, then X ×k Y exists.
These two claims establish the theorem, since for any two varieties X and Y , wehave open covers by affines X =
⋃Ui and Y =
⋃Vj . Then Ui ×k Vj exists, so
Ui ×k Y exists, so X ×k Y exists.For (i), recall that we have an equivalence of categories between affine varieties
over k and finitely generated integral k-algebras given by
X 7→ O(X) = A(X) and A 7→ Spec A.
Lemma 3.4.3. Let Z be any variety, and X be affine. Then
Hom(Z,X) ' Homk−alg(O(X),O(Z)).
Proof. Cover Z by quasi-affines: Z =⋃Ui. Then
Hom(Z,X)→∏I
Hom(Ui, X)⇒∏
Hom(Ui ∩ Uj , X)
(In categorical language, Hom(Z,X) is the equalizer of the second map.) Concretely,this means that giving a morphism ϕ : Z → X is equivalent to giving morphismsϕi : Ui → X such that ϕi|Ui∩Uj = ϕj |Ui∩Uj . On the other hand, we have
O(Z)→∏I
O(Ui)⇒∏I×I
O(Ui ∩ Uj)
by definition. Since the Ui are quasi-affine, the first sequence is equivalent to∏I
Homk−alg(O(X),O(Ui))⇒∏I×I
Homk−alg(O(X),O(Ui ∩ Uj)).
and using the above, we deduce that Hom(O(X),O(Z)) is the equalizer, so we havean isomorphism.
26
Now we return to proving (i). Let X,Y be affine. Let A = O(X) and B = O(Y ).Then A⊗kB is a finitely generated k-algebra, since if A = k[x1, . . . , xn]/(f1, . . . , fm)and B = k[y1, . . . , ys]/(g1, . . . , gr), then
A⊗k B = k[x1, . . . , xn, y1, . . . , ys]/(f1, . . . , fm, g1, . . . , gt).
Consider Spec A⊗kB = V (f1, . . . , fm, g1, . . . , gt) ⊂ An+s. We claim that Spec (A⊗kB) is X ×k Y . Well,
Hom(Z, Spec (A⊗k B)) ' Homk−alg(A⊗k B,O(Z))
= Homk−alg(A,O(Z))×Homk−alg(B,O(Z))
= Hom(Z,X)×Hom(Z, Y ).
The rest of the proof is delegated to homework.
Example 3.4.4. This tells us that An ×k Am = An+m. (But recall that the topologyis not the product topology).
As an extended example, we study the product variety Pn ×k Pm. We have anatural open covering Pn =
⋃Ui and Pm =
⋃j Vj .
Lemma 3.4.5. If X ×k Y exists and U ⊂ X is an open subset, then U ×k Y existsand is equal to U × Y ⊂ X ×k Y with the induced topology.
Pn × Pm =⋃
(Ui ×k Pm) =⋃i,j
(Ui ×k Uj)
has an open cover by affines of the isomorphic to An+m.
Lemma 3.4.6. The topology on Pn×kPm can be described as follows: closed subsetsare V (f1, . . . , fN ) where fi ∈ k[x0, . . . , xn, y0, . . . , ym] that are bihomogeneous, i.e.
fi =∑
cIxa00 . . . xann y
b00 . . . ybmm
where a0 + . . . + an = d and b0 + . . . + bm = e. (In this case we say the bidegree is(d, e)).
Proof. We have maps ϕi × ϕj : Ui × Vj∼→ An × Am ' An+m. We claim that
(ϕi × ϕj) ((Ui × Vj) ∩ V (f1, . . . , fN ))
are exactly the algebraic sets in An+m. Given this, the proof is the same as thatshowing ϕi : Ui → An is a homeomorphism.
This establishes that Pn ×k Pm is a variety, but we want to show that it is aprojective variety, i.e. can be realized as a subset of some PN .
27
Definition 3.4.7. Let ϕ : X → Y be a morphism of (pre)-varieties. We say say ϕ isa closed embedding if ϕ is one-to-one onto an irreducible closed subset ϕ(X) ⊂ Y(giving ϕ(Y ) and induced variety structure) and ϕ : X → ϕ(X) is an isomorphism.
Theorem 3.4.8. Let N = mn+m+ n. Then there is a natural closed embedding
S : Pn ×k Pm → PN .
This is called the Segre map.
Corollary 3.4.9. Let X,Y be projective. Then X ×k Y is projective.
Proof. Immediate from the theorem and the following lemma.
Lemma 3.4.10. Let ϕ′ : X ′ → X and ψ : Y ′ → Y be closed embeddings. Thenϕ× ψ : X ′ ×k Y ′ → X × Y is a closed embedding.
Proof. Exercise.
Proof of Theorem 3.4.8. The map s is given by
([a0 : . . . : an], [b0 : . . . : bm])→ [. . . : aibj : . . .].
First we need to check that the map is a morphism. It suffices to check this onan open cover of affines. Put coordinates [x0 : . . . : xn] on Pn and coordinates[y0 : . . . : ym], and denote the coordinates in PN by [. . . : zij : . . .]. Set
Uij = Pn −Hij = zij 6= 0.
Then we have a diagram
Ui × Uj s //
ϕi×ϕj
Uij
ϕij
An × Am // An+m
For convenience, we assume that i = j = 0. Then the map on the bottom is givenby
((x1, . . . , xn), (y1 : . . . : ym)) 7→ (x1, . . . , xn, y1, . . . , ym, xiyj).
This shows that s is a morphism. Next, we show that s is injective. Supposes(a, b) = s(a′, b′), i.e. aibj = a′ib
′j for each i, j. This means that
aibj = λa′ib′j∀ i, j.
There exists i0, j0 such that ai0 6= 0 and bj0 6= 0, so we may assume that (a, b) ∈Ui0 × Vj0 . Then
ai0bj0 = λa′i0b′j0
28
implies that (a′, b′) ∈ Ui0 × Vj0 . Let µ =ai0a′i0
and ν =bj0b′j0
, so λ = µν. Then
aibj0 = λa′ib′j0 =⇒ ai = µa′i =⇒ a = a′.
Similarly, b = b′. This verifies that s is injective.Let p = ker(k[zij ] → k[x0, . . . , xn, y0, . . . , ym]) sending zij 7→ xiyj . Then p is
prime, and p ⊃ p′ = (zijzkl−zilzkj). In fact, one can show that p = p′. Obviously,
s(Pn × Pm) ⊂ V (p) ⊂ B′.
We now prove that S(Pn × Pm) = V (p). Let
(cij) ∈ V (p′) =⇒ cijckl = cilckj .
There exists ci0j0 6= 0, so (cij) ∈ Ui0j0 . Let ai =(cij0ci0j0
)∈ Pn and bj =
(ci0jci0j0
)∈ Pm.
Then
S(a, b) = (aibj) =
(cij0ci0jc2i0j0
)=
(cijci0cj0
)= (cij) ∈ PN .
This also lets us conclude that s−1(Uij) = Ui ×k Vj .Finally, we show that s is an isomorphism. We want a map
s−1 : V (p) ∩ Ui0j0 → Ui0 × Vj0 .
Let’s consider the diagram
V (p) ∩ Ui0j0 // _
Ui0 × Vj0
Ui0j0
77
The bottom map is given by projection to the first n+m coordinates:
(a1, . . . , an, b1, . . . , bm, bij) 7→ (a1, . . . , an, b1, . . . , bm).
3.5 The Grassmannian
Example 3.5.1. (Linear subvarieties)
• A hyperplane in Pn is defined by a linear polynomial V (a0x0 + . . .+ anxn).
• A linear variety in Pn is the intersection of hyperplanes.
• A linear variety of domension r is also called an r-plane.
29
• A 1-plane is called a line.
Lemma 3.5.2. (i) Let X ⊂ Pn be an r-plane. Then X ' Pr and I(X) can begenerated by n− r linear polynomials.
(ii) There is a natural bijection between (r + 1)-dimensional sub-vector spaces inkn+1 and r-planes in Pn:
L 7→ PL.
Proof. Left as exercise.
Definition 3.5.3. The Grassmanian G(r, n) is the set of all r-dimensional subspacesin kn, (equivalently, the set of all (r − 1)-planes Pn−1).
A couple of remarks on notation:
(i) We also use the notation G(r − 1, n− 1) := G(r, n).
(ii) Let V be a finite-dimensional k-vector space. Then G(r, V ) is the set of r-dimensional subspaces in V . In particular, G(1, V ) is denoted P(V ).
Theorem 3.5.4. G(r, V ) is naturally a projective variety of dimension r(n − r),where n = dimV .
Proof. Let e1, . . . , en be a basis for V . For each I ⊂ 1, . . . , n, let kI = Spanei, i ∈I ⊂ V . Now, fix I ⊂ 1, . . . , n with |I| = n− r. Let
VI = L ⊂ V | dimL = r, L ∩ kI = 0 ⊂ G(r, V ).
We claim that
G(r, V ) =⋃
I⊂1,...,n,|I|=n−r
VI .
This is clear because for each r-dimensional space in G(r, V ) we can just choose Ito represent a complementary subspace. There is a natural bijection
ϕI : VI∼→ Ar(n−r).
In fact,
VI ' Hom(V/kI , kI) ('M(n−r)r(k) ' A(n−r)m).
In one direction, we can take any ϕ ∈ Hom(V/kI , kI) 'M(n−r)r(k) and consider its
graph Γϕ ⊂ V . In the other direction, given L ⊂ V , we may project to V/kI . Thisis an isomorphism since L ∩ kI = 0.
30
Explicitly, this means each L ⊂ VI uniquely determines
(v1, . . . , vr) = (e1, . . . , en)
1. . .
1ar+1,1 . . . ar+1,r
......
an,1 . . . an,r
It is not hard to show that ϕJ ϕ−1
I is a morphism, giving a variety structure onG(r, n). Details are left to the reader.
We now construct a closed embedding ψ : G(r, n)→ PN . Let W =∧r V . Then
dimW =(nr
). For each L ∈ G(r, n), we choose a basis v1, . . . , vr ∈ L and associate
to it v1 ∧ . . . ∧ vr ∈ w. This is unique up to scalars since∧r L is one-dimensional.
Now define
ψ : G(r, n)→ P(W )
L 7→ v1 ∧ . . . ∧ vr = [L].
This is called the Plucker embedding.Let I = r + 1, . . . , n. We have ψ : VI → P(W ). The coordinates on P(W ) are
(aJ), where J ⊂ 1, . . . , n and |J | = r. Using the basis
(v1, . . . , vr) = (e1, . . . , en)
1. . .
1ar+1,1 . . . ar+1,r
......
an,1 . . . an,r
,
we see that ψ maps to the r × r minors of the matrix
1. . .
1ar+1,1 . . . ar+1,r
......
an,1 . . . an,r
.
This is transparently a polynomial map, therefore a morphism.Next we show that ψ is injective. But this follows from the fact that we can
recover L as
L = v ∈ V | v ∧ [L] = 0 ∈r+1∧
V .
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Let w ∈ W . When can we write w = v1 ∧ . . . ∧ vr? (Such a w is calleddecomposable.) An obvious condition is that the map
− ∧ w : V →r+1∧
V
has kernel with dimension ≥ r. The following result in linear algebra shows thatthis condition is also sufficient.
Lemma 3.5.5. Let w ∈W be non-zero. Then
dim ker(∧w) ≤ r
with equality holding if and only if w = v1 ∧ . . . ∧ vr.
Proof. Left as exercise.
So we have a map δ : W → Hom(V,∧r+1 V ) ⊃ ϕ | rankϕ ≤ n− r sending
w 7→ (v 7→r∧v ∧ w).
This induces a map on projectivizations δ : P(W )→ PHom(V,∧r+1 V ). The latter
set contains
Zn−r(V,
r+1∧V ) := ϕ : V →
r+1∧V | rankϕ ≤ n− r.
In general, let V1 and V2 be two vector spaces and let
Zd(V1, V2) = ϕ : V1 → V2 | rankϕ ≤ d ⊂ P(Hom(V1, V2)).
This is an algebraic set in PHom(V1, V2) determined by the vanishing of the (d +1)× (d+ 1) minors; such a variety is called a “determinantal variety.” Anyway, wehave ψ(G(r, n)) = δ−1Zn−r(V,
∧r+1 V ).
Remark 3.5.6. In this language, the Segre embedding of Pn × Pm is just the deter-minental variety Z1(kn+1, km+1).
Now we want to show that ψ is a homeomorphism, and in fact an isomorphismof varieties. This is something we can check locally, so consider the restriction
ψ : VI → UI := P(W )− aIc = 0.
For example, supposet I = r + 1, . . . , n. The map sends
1. . .
1ar+1,1 . . . ar+1,r
......
an,1 . . . an,r
→ aJ = r × r minor for J.
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Suppose J = 1, 2, i, . . . , r ∪ j for some r + 1 ≤ j ≤ n. Then aJ = aji, so theprojection to this coordinate on Ui is an inverse morphism back to Vi.
We remark that the ideal of ψ(G(r, n)) can be generated by quadratic polynomials,called the “Plucker” relations.
33
34
Chapter 4
Morphisms
4.1 Rational maps and correspondences
Definition 4.1.1. A correspondence from a variety X to a variety Y is a relationgiven by a closed subset Z ⊂ X × Y .
• Z is said to be a rational map if Z is irreducible and there is an open subsetX0 ⊂ X such that prX(Z ∩ pr−1
X (X0))→ X0 is an isomorphism.
• Z is said to be birational if Z is rational and Z−1 = (y, x) ∈ Y ×X ⊂ Y ×Xis also rational.
We will also use 99K to denote a rational map.
Example 4.1.2. Let X be projective, and let f ∈ k(X) be a rational function. Thenf gives rise to a rational map f : X 99K P1. Recall that k(P1) = k(t), a rational mapf : X → P1 is a map k(t) → k(X). Now we see that f ∈ k(X) uniquely specifiesthe map k(t)→ K(X) sending t 7→ f , so
k(X) \ 0 = Homk−alg(k(t),K(X)).
Since X is projective, we may write any f ∈ K(X) as
f =g
hg, h ∈ k[x0, . . . , xn] homogeneous of the same degree.
Now we define X0 = X − (g = 0 ∩ h = 0). This is a non-empty open subset ofX. Then we define
F : X0 → P1
by sending p 7→ [g(p) : h(p)]. It is obvious after restricting to local affines thatthis is a morphism. Let y0, y1 be the homogeneous coordinates on P1. Consider
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V (y0h−y1g); this is a bihomogeneous polynomial of bidegree (d, 1) defining a closedsubset of Pn × P1.
Z ⊂ X × P1
yy %%X P1
Let Z = V (y0h− y1g) ∩X × P1. Over X0, pr−1(X0) ∩ Z is just the graph of F , i.e.
pr−1(X0) ∩ Z = (x, F (x)) | x ∈ X.
In fact, pr : Z∩pr−1(X0)→ X0 is an isomorphism since there is an inverse morphismby the universal property.
Now let Z∗ be the closure of Z ∩ pr−1(X0) in X × P1. Then Z∗ is irreduciblebecause it is isomorphic to X0, and Z∗∩pr−1(X0) = Z∩pr−1(X0) since Z∩pr−1(X0)is already closed in X × P1. Therefore, Z∗ is a rational map from X to P1.
Later, we will see that for any two varieties X and Y , any inclusion k(Y ) ⊂ k(X)gives a rational map X 99K Y .
4.2 Blowing up
Let O = [0 : . . . : 1] ∈ Pn and let p : Pn − O → Pn−1 be defined by
[a0 : . . . : an] 7→ [a0 : . . . : an−1].
Then p is a morphism called projection from O. Geometrically, it works as follows:for a point q ∈ Pn−O, take the line through q and O; this intersects Pn−1 in aunique point. Note that p−1(p(x)) is the line joining O and x, less O.
Let Z = V (xiyj − yixj0≤i,j≤n−1) ⊂ Pn × Pn−1. We have a correspondence
Z = V (xiyj − yixj)
ww ((Pn Pn−1
Then Z = Γp ∪ pr−1(O) = Γp ∪Pn−1. This Pn−1 is called the exceptional divisor E.We claim that Z is irreducible. This is because
Z ∩ pr−1(Pn − O) = Γp = (x, p(x)) ∈ (Pn − O)× Pn−1.
But the graph is isomorphic to Pn − O, so Γp is irreducible. To prove the claim,it suffices to show that Z = Γp.
For q ∈ Pn−1, let `q = p−1(q) ∪ O, which is the line in Pn through O and q.Then
lq × q ⊂ Pn × q ⊂ Pn × Pn−1
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and(lq − O)× q ⊂ Γp ⊂ Z.
But (O, q) ∈ `q × q ⊂ Γp ⊂ Z, so O × Pn−1 = E ⊂ Γp, implying that Z ⊂ Γp.So Z is irreducible; since it is a closed subspace of projective space, it is a projectivevariety. We call it the blow-up of Pn at O. Note that projection to Pn is one-to-onefrom Z − E to Pn (since it is the graph of the projection map at these points) andpr−1(O) = E ' Pn−1.
Definition 4.2.1. Let X ⊂ Pn be quasi-projective and O ∈ X. The blow-up of X atO is
BlOX := pr−1(X − O) ⊂ Z.Example 4.2.2. Let X : y2 = x3 + x2 ⊂ A2 ⊂ P2. Let O = [0 : 0 : 1] ∈ X.Let X be the blowup of X at O. Let (u, v) be the coordinates on P1. ThenZ = V (xu − yv) ⊂ P2 × P1. So pr−1(X) ∩ Z ⊃ X is given by the solutions of thesystem
xu− yv = 0
y2 − x3 − x2 = 0
Restricting to the open subset A2 × A1 ⊂ A2 × P1 where v 6= 0, this system isxu− y = 0 =⇒ y = xu
y2 − x3 − x2 = 0 =⇒ x2(u2 − x− 1) = 0.
This is reducible, given by V (y−xu, x2(u2−x−1)) = V (y−xu, u2−x−1)∪Y (y−xu, x). The second term is the exceptional locus E ∩ (A2 × A1). The first thing isX.
Example 4.2.3. Last time we defined a rational map from X to Y to be a correspon-dence Z ⊂ X × Y which is bijective over some open subset of X. This is right incharacteristic 0, but not quite enough in characteristic p. Let char k > 0. Considerthe Frobenius morphism F : A1 → A1. Then ΓF ⊂ A1 × A1 is a correspondence,which is bijective over an open subset of each A1, but we want to consider it as notbirational (it does not induce an isomorphism on function fields).
Now we summarize the geometric picture in our discussion of projections andblowups. Let O = (0, . . . , 0, 1) ∈ Pn. There is a projection p : Pn − 0 → Pn−1,which gives a correspondence
Z = V (xiyj − yixj)
ww ((Pn Pn−1
The variety Z is called BlO(Pn). If X is a subvariety of Pn, then the closure ofpr−1(X − O) is called the blowup of X, BlO(X). The picture is that the blowupseparates the lines through a singular point.
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Resolution of singularities. The basic problem is: given X an algebraic variety,can we find Y → X a birational, surjective morphism with Y smooth? Hironakaproved that this can be done over a field of characteristic 0 (and he got a Fieldsmedal for this work), but the answer is unknown over fields of positive characteristic.
Example 4.2.4 (Incidence correspondence). Let C ⊂ G(r, n)× Pn be defined as
C = (L, x) | x ∈ L .
We claim that C is a correspondence; furthemore, C is a smooth projective varietyof dimension (r + 1)(n− r) + r.
Proof. Let Pn = P(V ), where dimV = n+1. Let e0, . . . , en be a basis of V . Considerthe map
G(r, n)× Pn ψ×id→ P(W )× Pn,
where W =∧r+1 V sending (L, x) 7→ ([L], x). The coordinates on W are
∑aIei ∈
w, I ⊂ 0, . . . , n, where |I| = r + 1. The x ∈ L ⇐⇒ [L] ∧ x = 0, which we canexpress in coordinates as ∑
aIeI ∧∑
xiei = 0.
This is a bihomogeneous polynomial F (aI , x) in the aI and xi of bi-degree (1, 1). So
C = V (F ) ∩ (ψ(G(r, n))× Pn) .
We claim that for each I ⊂ 0, 1, . . . , n of cardinality n− r,
pr−1(VI) ∩ C ' VI × Pr,
where the VI are the open sets in the cover of ψ(G(r, n)). Note that this gives thedesired dimension formula, since dimVI × Pr = dimVI + dimPr.
I corresponds to a vector space kI ⊂ V of dimension n−r. So kI/ ∼ correspondsto M ⊂ Pn, an n− r − 1 plane in Pn. After changing coordinates, we may assumethat M = (0, . . . , 0, ar, . . . , an−r−1). We define a map by projecting away fromM :
PM : Pn −M → Pr
(a0, . . . , an) 7→ (a0, . . . , ar)
P−1M (y) is an (n− r)-plane.
Now we define a map
C ∩ pr−1(VI)→ VI × Pr
(L, x) 7→ (L,PM (x))
This is well-defined because L ∩ kI = 0, so x ∈ P(L) =⇒ x /∈M .
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4.3 Rational maps and function fields
Definition 4.3.1. A morphism f : X → Y is called dominant if f(X) contains adense subset of Y (in fact, we can assume that it contains an open dense subset).A rational map Z ⊂ X × Y is called dominant if prY : Z → Y is dominant.
In the homework, we defined another notion of rational maps
Maps(X,Y )rat′ = (U, f) | U ⊂ X open subset, f : U → Y a morphism.
On the homework you show that there is an injection Maps(X,Y )rat′ → Maps(X,Y )rat.In the other direction, if Z ⊂ X × Y is a rational map, there is an isomorphismpr−1(X0) ∩ Z ' X0. Composing with projection to Y gives a morphism X0 → Y .Therefore, these two notions are equivalent.
Under this corresondence, we have that Z is dominant ⇐⇒ f : U → Y isdominant. Now if f : X 99K Y , then we obtain a map f∗ : k(Y )→ k(X) as follows.If g ∈ k(Y ), then g ∈ O(V ) for V an open subset of Y , so gf ∈ O(f−1(V )) ⊂ k(X).
Proposition 4.3.2. Consider the functor F defined by F (X) = k(X), and F (f :X 99K Y ) = (f∗ : k(Y ) → k(X)). Then F induces an equivalence of categoriesbetween the category of algebraic varities with morphisms the dominant rationalmaps, and the category of finitely generated field extensions of k, with morphismsk − alg homomorphism.
Proof. First, we have to show that
Maps(X,Y )ratdom → Homk−alg(k(Y ), k(X))
is a bijection. To do this, we construct an inverse. Let θ : k(Y )→ k(X). Let Y0 ⊂ Ybe affine open and Y0 ⊂ An, with coordinate functions (y1, . . . , yn). Similarly, letX0 ⊂ X be affine. So θ(yi) is a rational function on X0, and can be written as therational of two regular functions:
θ(yi) =fif0
fi ∈ O(X0), f0 6= 0.
Let U = X0 − f0 = 0. This is a non-empty open subset of X since f0 6= 0, soθ(yi) ∈ O(U). So we may regard θ as a map O(Y0) → O(U), where Y0 is affine, soby the equivalence of categories between affine varieties and affine k-algebras, thisinduces a map ϕ : U → Y0. Furthermore, ϕ is dominant because if the closure in Y0
of the image is not all of Y0, then there is a regular function f on Y0 which is sentto 0 by θ. But this is impossible under a non-zero field homomorphism.
Let K = k(x1, . . . , xn) be a finitely generated field extension of k. We want tostudy how K arises as the function field of some variety. Let A = k[x1, . . . , xn] ⊂ K,and let X = Spec A. Then k(X) = K. This seemingly trivial observation has someinteresting corollaries.
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Corollary 4.3.3. Every algebraic variety X is birational to a hypersurface in Pn .
Proof. Let K = k(X), finitely generated over k. Then K/k is separably generated,i.e. there exists x1, . . . , xr ∈ K transcendental over k such that K/k(x1, . . . , xr) isfinite separable. By the primitive element theorem, there exists y ∈ K such thatK = k(x1, . . . , xr, y), where y satisfies its minimal polynomial equation
f(x1, . . . , xr, y) = yn + a1(x1, . . . , xr)yn−1 + . . .+ an(x1, . . . , xr) = 0.
This f is irreducible by minimality. From the construction, V (f) ⊂ Ar+1 is an affinevariety of dimension r and k(V (f)) = K. Therefore, X is birational to V (f).
4.4 Complete varieties and projections
Definition 4.4.1. An algebraic variety is called complete if for any algebraic varietyY , the projection map X × Y → Y is closed.
Theorem 4.4.2. The projection pr2 : Pn × Pm → Pm is a closed map, i.e. everyZ ⊂ Pn × Pm is closed implies pr2(Z) is closed.
Remark 4.4.3. In fact, one can show that Pn is complete.
Example 4.4.4. The map A1×A1 → A1 given by projection to the second coordinateis not closed, since V (xy − 1) maps to A1 − 0.
Corollary 4.4.5. Let X,Y be projective, and let Z ⊂ X × Y be a correspondence.Then prY (Z) is closed.
Corollary 4.4.6. Let A ⊂ X × Y and B ⊂ Y × Z be correspondences. Then
C = B A := (x, z) ∈ X × Z | ∃y ∈ Y, (x, y) ∈ A, (y, z) ∈ B
is a correspondence.
Proof. We have projection maps
X × Y × Z
pr12wwpr13
pr23
&&X × Y X × Z Y × Z
So C = pr13(pr−112 (A) ∩ pr−1
23 (B)) is closed.
Corollary 4.4.7. Let Z ⊂ G(r, n) be a closed subvariety. Then⋃L∈Z
L ⊂ Pn
is a projective variety.
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Proof. We have the incidence correspondence
C
yy Z ⊂ G(r, n) Pn
Then⋃L∈Z L ⊂ Pn = pr2(pr−1
1 (Z)).
Corollary 4.4.8. Let X ⊂ Pn − O. Then PO(X) is a projective variety in Pn−1
and PO : X → PO(X) = X ′ has finite fibers. In addition, dimX = dimPO(X) andS(X) = Sn/I(X) is a finite module over S(X ′) = Sn−1/I(X ′).
Proof. If y ∈ Pn−1, then P−1O (y) ' A1, so P−1
O (y) ∩ X is closed in A1. Therefore,P−1O (y)∩X 6= P−1
O (y) =⇒ P−1O (y)∩X is finite (since the only closed subsets of A1
are finite points).
Because O /∈ X, there exists f ∈ I(X) depending on xn, which we can assumeis of the form
f = xdn + . . .+ a1(x0, . . . , xn−1)xd−1n + . . .+ ad(x0, . . . , xn−1).
So S(X) is generated over S(X ′) by 1, xn, . . . , xd−1n , i.e. S(X) is finite over S(X ′).
This also shows that
tr .degk S(X) = dimC(X) = dimC(X ′) + 1 = tr degk S(X ′).
pr1(X) ⊂ Z = BlO(Pn)
uu ((O /∈ X ⊂ Pn Pn−1
Corollary 4.4.9 (Noether’s normalization lemma). Let Xr ⊂ Pn be a projectivevariety of dimension r. Then there exists an n− r− 1 plane L such that X ∩L = ∅and prL : Xr → Pr has finite fibers and S(Xr) is finite over Sr.
Proof. We induct on n− r. If n− r = 1, project from a point to get X → PO(X) ⊂Pn−1. Since PO(X) is a closed subset of Pn−1 of dimension n−1, it is equal to Pn−1.
In general, choose O /∈ X. Project PO : X → PO(X) = X ′ ⊂ Pn−1. ChooseL′ ⊂ Pn−1 an (n− 1)− (r1) plane such that PL′ : X ′ → Pr with PL′ PO = PL′O =PL.
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Proof of Theorem 4.4.2. It’s enough to show that pr2 : Pn×Am → Am is closed. LetZ = V (f1, . . . , fr) be an algebraic set of Pn×Am. Let fi = fi(x0, . . . , xn, y1, . . . , yn),so fi is homogeneous of degree di in the x’s. If pr2(z) does not contain b, thenfi(x, b) do not have common zeros in Pn. By Hilbert’s Nullstellensatz, thereexists N such that (x0, . . . , xn)N ⊂ (f1(x, b), . . . , fr(x, b)). Let
UN = b ∈ Am | (x0, . . . , xn)N ⊂ (f1(x, b), . . . , fr(x, b)).
In other words, Am − pr2(Z) =⋃N≥1 UN , so it’s enough to show that UN is open.
Let S = k[x0, . . . , xn] =⊕Sd be the natural grading on S. For b ∈ Am, define
T(N)b : SN−d1 ⊕ . . .⊕ SN−dr → SN
(g1, . . . , gr) 7→∑
gifi(x, b).
So T(N)b is a k-linear map between two k-vector spaces. In terms of bases, the map
is given as a matrix whose entries are polynomials in b. We can view this as a map
T (N) : Am → Homk(SN−d1 ⊕ . . .⊕ SN−dr , SN )
and b ∈ UN ⇐⇒ T(N)b is surjective. We observe:
(i) T (N) is a morphism of affine varieties.
(ii) Let D = dimk SN − 1. Then ZD is a determinantal variety, so the affine coneC(ZD) ⊂ Homk(SN−d1 ⊕ . . .⊕ SN−dr , SN ) is closed.
(iii) Thus Am − UN = (T (N))−1(C(ZD)) is closed.
Proposition 4.4.10. Let ϕ : Xr → Y s be a morphism of quasiprojective varieties(r ≥ s). Let y ∈ ϕ(Xr) and W ⊂ ϕ−1(y) be an irreducible component of the fiber.Then dimW ≥ r − s.
Proof. We have a finite map Y s → Ps by Noether normalization. This fits into acomposition
Xr → Y s → Ps,
so we may assume that Y s = Ps since the second map has finite fibers. By restrictingto an affine open, we can assume Y = As and y = 0. Let fi = yi ϕ, where yi arethe coordinate functions on As, so ϕ−1(0) = V (f1, . . . , fs). Now it suffices to show:if Xr ⊂ An is affine, then every irreducible component of Xr ∩ V (f) has dimension≥ r − 1.
We can assume that Xr 6⊂ V (f); we will show that each irreducible componentof X ∩V (f) has dimension r− 1. Replace X and V (f) by their closures in Pn. Well
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V (f) = V (F ), where F is the homogenization of f . Now, after taking the d-upleembedding we can assume degF = 1.
Now we show by induction on n− r that if X ⊂ Pn is a projective variety, thenevery irreducible component of X ∩H has dimension r−1. It is clear if n−r = 0. Ifn− r = 1, then X = V (f) is a hypersurface in Pn. We can assume that H = V (xn),so X ∩H is a hypersurface in Pn−1.
If n− r ≥ 2, let O ∈ H, O /∈ X ∩H. Now PO : Pn → Pn−1 sends H ' Pn−1 →PO(H) ' Pn−2. Observe that
PO(X ∩H) = PO(X) ∩ PO(H − O).
The inclusion ⊂ is obvious; for the other direction, let y ∈ PO(X) ∩ PO(H − O).There exists x ∈ X such that PO(x) = y, and P−1
O (PO(H)) = H −O, implying thatx ∈ H.
Let W ⊂ X ∩H be an irreducible component. Of course, PO(W ) is irreduciblein PO(X) ∩ PO(H). If PO(W ) is a component of PO(X) ∩ PO(H − O), then weare done by induction. We claim that we can find O ∈ H − X ∩ H so that thiswill be thecase. Suppose X ∩ H = W ∪W ∗, where W ∗ is the union of the othercomponents. It is enough to show that there exists O such that PO(W ) 6⊂ PO(W ∗).
To this end, pick x ∈ W \ W ∗. Consider P−1x (Px(W ∗)) = C(Px(W ∗)). But
dimWi ≤ dimX − 1 = r − 1. Therefore, dimC(Px(Wi)) ≤ r < n − 1 = dimH. Sothere exists O ∈ H−C(Px(W ∗))−W . Now consider the projection from this point:we claim that PO(x) /∈ PO(W ∗), since otherwise PO(x) ∈ PO(W ∗) =⇒ O is on theline joining x and a point in W ∗, i.e. O is in the cone over the image of projectionfrom x.
4.5 Separable morphisms
Definition 4.5.1. Let ϕ : X → Y be a dominant morphism. We say ϕ is separable ifk(X) is separably generated over ϕ∗(k(Y )).
Example 4.5.2. Let Fp : X → X be the Frobenius morphism. Then F is dominantbut not separable, since the map in the other direction is k(X) → k(X) sendingx 7→ xp.
Proposition 4.5.3. Let ϕ : Xr → Y r be a separable morphism. Then there existsan open subset Y0 ⊂ Y such that ϕ−1(y) is finite and #ϕ−1(y) = [k(X) : k(Y )] forall y ∈ Y0.
Before we give the proof, let us recall the primitive element theorem.
Theorem 4.5.4 (Primitive Element theorem). Let L/K be a finite separable exten-sion. Then L is generated over K by one element. In addition, if β1, . . . , βr form aset of generators, then some linear combination α =
∑λiβi ∈ K.
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Proof of Proposition 4.5.3. We begin with the special case where k(Y ) = k(X).We can assume X,Y are affine, so Y ⊂ Am and X ⊂ An. Let x1, . . . , xm bethe coordinate functions on Am. Write (ϕ∗)−1(xi) = fi
f0, fi ∈ O(Y ). Consider
Y0 = Y − V (f0). Then
ϕ−1(y) =
(f1(y)
f0(y), . . . ,
fn(y)
f0(y)
), y ∈ Y0.
In general, we can assume that k(X) = k(Y )[f ], where f ∈ O(X). Consider themap
Xψ→ Y × A1 pr1→ Y
x 7→ (ϕ(x), f(x)) 7→ ϕ(x)
and let Z = ψ(X). Then we have a map of function fields in the other direction,k(Y ) → k(Z) → k(X). But since k(Z) contains f , it also surjects to k(X), so it isisomorphic to k(X).
k(Y ) ⊂ k(Z) ' k(X).
There exists Z0 ⊂ Z open such that #ψ−1(Z) = 1 for z ∈ Z0. Therefore, we canreplace X by Z and assume X ⊂ Y × A1. Let P (t) be the minimal polynomial off , so
P (t) = tn + a1tn−1 + . . .+ an, ai ∈ k(Y ).
By replacing Y by an open subset, we may assume that ai ∈ O(Y ). we have byassumption that k(X) = k(Y )[f ]/P (f), so by replacing X by an open subset wemay also assume htat O(X) = O(Y )[f ]/P (f). In other words,
X = (y, t) ∈ Y × A1 |∑
ai(y)tn−i = 0.
For fixed y, this is a degree n polynomial, and therefore has n solutions with multi-plicity, corresponding to the points of the fiber #ϕ−1(y). We must prove that thereis an open set over which there are exactly n solutions (without multiplicity).
Recall that by separability, p(t) and p′(t) have distinct roots in k(y)[t]. So p(t)and p′(t) do not have common zeros in k(y)[t], so there exists a(t), b(t) ∈ k(y)[t]such that
a(t)p(t) + b(t)p′(t) = 1.
Clearing denominators, we have
a(t)p(t) + b(t)p′(t) = c(y).
Therefore, over the open subset Y0 = y ∈ Y | c(y) 6= 0, p(t) and p′(t) do not havecommon zeros. Over this open subset, #ϕ−1(y) = n = [k(X) : k(Y )].
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4.6 Smooth morphisms
Let ϕ : X → Y be a morphism, ϕ(x) = y. Then there is a map ϕ∗ : OY,y → OX,x
pulling back mY,y 7→ mX,x. In particular, ϕ induces dϕx : TxX → TyY , sinceTxX = mx/m
2x.
Definition 4.6.1. The morphism ϕ is smooth at x if (i) x (resp. ϕ(x)) is smooth inX (resp. in Y) and (ii) dϕx : TxX → TyY is surjective.
Proposition 4.6.2. Let ϕ : X → Y be a dominant morphism. Then the followingare equivalent.
(i) ϕ is separable.
(ii) There exists X0 ⊂ X open such that ϕ is smooth at x ∈ X0.
(iii) There exists a point x ∈ X such that ϕ is smooth at x.
Proof. First we show that (iii) =⇒ (ii). We may assume that X and Y are affine.By replacing X by the graph of ϕ in An × Am, we assume that X ⊂ An × Am andϕ is the projection to the second coordinate.
X //
An × Am
Y // Am.
Let g1, . . . , gp be generators of I(Y ). Then ϕ∗(g1), . . . , ϕ∗(gp), f1, . . . , fq generateI(X) for some f1, . . . , fq. Let x ∈ X and y ∈ Y . Then
TxX =
(ξ1, . . . , ξn+m) ∈ kn+m |
(0 ∂g
∂y∂f∂x
∂f∂y
)(ξη
)=
(00
).
The projection to TyY maps (ξ1, . . . , ξn+m) 7→ (0, . . . , 0, ξn+1, . . . , ξn+m). The kernelof dϕx consists of
(ξ1, . . . , ξn) |n∑j=1
∂fi∂xj
ξj = 0 = ker
(∂fi∂xj
).
Now, ϕ is smooth at x ⇐⇒ dim ker dϕx = r − s. However, since the kernel alwayshas dimension at least ≥ r − s, this is the same as saying ker dϕx = r − s ⇐⇒rank
∂fj∂xi≥ n−(r−s). This defines an open subset (determined by the non-vanishing
of minors). Therefore, (iii) =⇒ (ii).Next we show (i) =⇒ (ii). We have
Derk(Y )(k(X), k(X))∼→ Derk(Y )(k(Y )⊗O(Y ) O(X), k(X)).
45
This is
D ∈ Derk(Y )(k(Y )[x1, . . . , xn], k(X)) |Dfi = 0 ∼= (ξ1, . . . , ξn) ∈ k(X)n |∑
ξi∂fi∂xj
= 0
since k(Y )⊗O(Y ) O(X) ' k(Y )[x1, . . . , xn]/(f1, . . . , fq). Therefore,
dimk(X) Derk(Y )(k(X), k(X)) = n− rank k(X)
(∂fi∂xj
).
We know that there exists X0 ⊂ X open such that
rank k(X)
(∂fi∂xj
)= rank k
(∂fi∂xj
)∀x ∈ X0.
According to (i), k(X)/k(Y ) is separable. By a fact from commutative algebra,k(X)/k(Y ) is separable generated if and only if
dimk(X) Derk(Y )(k(X), k(X)) = r − s,
where r = tr .deg k(X) and s = tr .deg k(Y ). Therefore, there exists an openX0 ⊂ X such that
rank k(X)
(∂fi∂xj
)= n− r + s.
This shows that (i) =⇒ (ii). Conversely, if ϕ is smooth on an open subset then
rank k(X)
(∂fi∂xj
)= n − r + s on an open subset, and we can run the argument in
reverse.
Remark 4.6.3. If ϕ : X → Y is smooth over X0 ⊂ X, then ϕ is dominant, so thisassumption was not necessary. Indeed, suppose that ϕ : X → ϕ(x) ⊂ Y is a properclosed subset, so we have a factorizaton
Xψ// Z = ϕ(X)
ι // Y
Then dϕx = dιψ(x) dψx. Choose z ∈ Z, Z smooth at z, and z ∈ ϕ(x), so we have afactorization of the derivative map
dϕx : TxX → TzZ → TzY
so dimTzZ < TzY , so the map on tangent spaces is not surjective, so dϕx is notsmooth.
Corollary 4.6.4 (Generic smoothness). Let char k = 0 and ϕ : X → Y be dom-inant. Then there exists some (non-empty) Zariski-open subset Y0 ⊂ Y such thatϕϕ−1(Y0) : ϕ−1(Y0) \ Sing(X)→ Y0 is smooth.
46
Proof. We can assume that X and Y are smooth. In characteristic 0, any dominantmorphism is separable. Let X0 ⊂ X be the open subset such that ϕ is smooth at allx ∈ X0. Let Z = X \X0. We need to show that ψ = ϕ|Z : Z → Y is not dominant.Suppose for the sake of contradiction that ψ : Z → Y is dominant, so there existsZ0 ⊂ Z open such that ψ|Z0 : Z0 → Y is smooth. So we have a factorization
Zι→ X
ϕ→ Y
inducing a factorization of the derivative maps
dψz : TzZ
%%
// Tψ(z)Y
TzX
dϕz::
But by the definition of z, dϕz is not surjective, contradiction.
Proposition 4.6.5. Let ϕ : Xr → Y s be smooth at x ∈ X. Then there exists aunique component Z of ϕ−1(ϕ(x)) passing through x. In addition, Z is smooth at xand dimZ = dimX − dimY .
Example 4.6.6. Consider X = V (xy−t) ⊂ A3 projecting down to A1 by (x, y, t) 7→ t.For all t 6= 0, the fiber is a hyperbola, and there is a unique component passingthrough a given point in the fiber. This breaks down at t = 0, where the fiber is theunion of two axes.
Proof. We may assume that 0 ∈ X ⊂ Am × An mapping down to 0 ∈ Y ⊂ Am.Choose g1, . . . , gm−s ∈ I(Y ) with independent linear terms. Let f1, . . . , fn−r+s ∈I(X) such that (f1, . . . , fn−r+s, g1, . . . , gm−s) have independent linear terms.
Note that dimk ker dϕ = r − s implies that rank(∂fi∂xj
)(0, 0) = n − r + s, so
f1(x, 0), . . . , fn−r+s(x, 0) also have independent linear terms.As we showed in Theorem 2.3.1, V (f1, . . . , fn−r+1, g1, . . . , gm−s) = X∪X ′ where
0 /∈ X ′. Consider
V (f1, . . . , fn−r+s, g1, . . . , gm−s, y1, . . . , yn) = pr−12 (0)∩(X∪X ′) = ϕ−1(0)∪(pr−1
2 (0)∩X ′),
the latter component not containing 0. This algebraic set may also be written as
V (f1(x, 0), . . . , fn−r+s(x, 0)) ⊂ An × 0 = An,
but by Theorem 2.3.1 again,
V (f1(x, 0), . . . , fn−r+s(x, 0)) = Z ∪ Z ′ 0 /∈ Z ′.
So Z ⊂ ϕ−1(0) is the unique component passing through 0. Z is smooth at 0, anddimZ = n− (n− r + s) = r − s.
47
Theorem 4.6.7 (Zariski’s Main Theorem, smooth case). Let ϕ : X → Y be abirational morphism of quasi-projective varieties. Assume Y is smooth. Let
Y0 = y ∈ Y | #ϕ−1(y) = 1.
Then
(i) ϕ|ϕ−1(Y0) : ϕ−1(Y0)→ Y0 is an isomorphism.
(ii) For any y ∈ Y \ Y0 and x ∈ ϕ−1(y), there exists E ⊂ X a subvariety ofdimension r − 1 passing through x such that ϕ(E) has dimension ≤ r − 2. Inparticular, for any y ∈ Y \ Y0 there exists a component in ϕ−1(y) of positivedimension.
To summarize, given a birational morphism with smooth target, the morphismrestricted to the locus where it is bijective is an isomorphism. Furthermore, if thepre-image of y ∈ Y is not a single point, it has a component of positive dimension,called the exceptional divisor.
Proof. Let x ∈ X and y ∈ Y . There is an induced map ϕ∗ : OY,y → OX,x (in fact,this is injective, since the map on function fields is an isomorphism). Let
Y1 = y ∈ Y | ∃x ∈ ϕ−1(y) such that ϕ∗OY,y ' OX,x.
Y2 = y ∈ Y | ϕ∗OY,y 6' OX,x ∀x ∈ ϕ−1(y).
We claim that Y1 ⊂ Y0 and Y2 ⊂ Y \ Y0 (this immediately implies that we haveequality in both inclusions). Let y ∈ Y1, so OY,y ' OX,x. If x′ ∈ ϕ−1(y) is anotherpoint in the fiber, then OX,x = ϕ∗OY,y → OX,x′ . Then the first part of the claimfollows from this lemma:
Lemma 4.6.8. Let X be quasi-projective, x, x′ two points such that
OX,x ⊂ OX,x′ ⊂ k(X).
Then x = x′.
We already showed on the homework that an isomorphism of local rings impliesan isomorphism of open neighborhoods, so the proof of this Lemma will finish offpart (i).
Proof. We can assume that X ⊂ Pn is projective. Choose a hyperplane not passingthrough x and x′, we can assume that X is affine. So points of X correspond tomaximal ideals of O(X); and inclusion of the form OX,x ⊂ OX,x′ =⇒ mx ⊃ mx′ .
Now we must establish the second part of the claim. We can choose f ∈ OX,x
such that f = ϕ∗(ab ) for a, b ∈ OY,y, and b(y) = 0 since ϕ∗(OY,y) 6= OX,x.
48
Theorem 4.6.9. If y ∈ Y is smooth, then OY,y is a UFD.
The proof of this theorem is omitted; it follows from generaliies on commutativealgebra
So we may assume that a and b are coprime in OY,y. Let b = βb′, where β is aprime element. Then
p = O(Y ) ∩ βOY,y is a prime ideal.
Let E be a component of V (ϕ∗β) ∩X containing x. Then dimE = r − 1. So
ϕ∗(a) = fϕ∗(b) = fϕ∗(β)ϕ∗(b′) =⇒ ϕ∗(a) vanishes on E.
So a vanishes on ϕ(E). But a, b coprime implies that a /∈ p. Therefore, ϕ(E) ⊂ V (p)is a strict inclusion, so dimϕ(E) ≤ r − 2.
Example 4.6.10. Consider the blow-up of A2, Z = V (xu− yv) ⊂ A2 × P1.
Z → A2
(x, y)× (u, v) 7→ (x, y).
We know already that this is an isomorphism outside the origin O = (0, 0), P =(0, 0, 0, 1) ∈ E, etc. but let’s trace through the proof of Zariski’s main theorem inthis case.
We have an inclusion ϕ∗ : OA2,O → OZ,p. We can find a function in OZ,p notin the image of ϕ. Indeed, consider the open subset Z − u = 0 = V (xu − y) ⊂A2 × A1. Then u, which is morally ϕ∗( yx), is not in the image of OA2,O. ThenE = V (ϕ∗x) ⊂ Z − v = 0 is defined by x = 0 and xu− y = 0, i.e. x = y = 0. Wesee that E is a projective line, and the image of E lies in the intersection of the xand y axes, i.e. O.
Remark 4.6.11. We saw on the homework that the image of ϕ : Z − v = 0 =V (xu− y)→ A2 is A2 with the y axis removed and the origin put back in.
49
50
Chapter 5
Divisors and Line Bundles
5.1 Divisors
We now turn to the theory of divisors on a smooth algebraic variety X.
Lemma 5.1.1. Let Xr be an affine variety of dimension r, x ∈ X. Let f ∈ OX,x
be an irreducible element, so fOX,x ∩ O(X) = p is a prime ideal. Then X ′ = V (p)is a subvariety of dimension r − 1 containing x.
Conversely, suppose X ′ ⊂ X is a subvariety of dimension r−1 and x ∈ X ′. Thenthere exists f ∈ OX,x irreducible such that X ′ = V (p), where p = fOX,x ∩O(X). Inother words, I(X ′)OX,x = fOX,x.
Proof. By taking an open subset of X if necessary, we may assume that f ∈ O(X).Then there exists an open subset U ⊂ X such that V (f)∩U = V (p)∩U . But eachirreducible component of V (f) has dimension r − 1, so dimX ′ = r − 1.
Conversely, choose f ∈ I(X ′). Then f is not a unit in OX,x, so we may factor finto a product of primes. So, restricting to an open subset, we may assume that fis prime in OX,x. Then X ′ ⊂ V (f), where dimX ′ = r − 1, so X ′ is an irreduciblecomponent of V (f), implying that X ′ = V (p).
Definition 5.1.2. Let X ′ ⊂ Xr be a closed subvariety of dimension r − 1, x ∈ X ′.An element of OX,x is called a local equation of X ′ at x if fOX,x = I(X ′)OX,x.
We have just seen that every co-dimension 1 subvariety has a local equation.
Definition 5.1.3. Let X be a smooth algebraic variety. The divisor group Div(X) isthe free abelian group generated by the co-dimension 1 subvarieties.
In other words, elements of the divisor group are finite formal linear combinationsof co-dimension 1 subvarieties.
51
Lemma 5.1.4. There exists a natural group homomorphism
k(X)∗ → Div(X)
f 7→ (f).
Proof. We write (f) =∑nZZ and we need to define nZ ∈ Z. Choose any x ∈ Z;
in OX,x choose a local equation fZ . Then we may write
f =g
hf rZ g, h ∈ OX,x, gcd(gh, fZ) = 1.
Then we define nZ = ordZ,x(f) := r. To see that this is well-defined, note thatdifferent choices of the local equation differ by units, and the function x 7→ ordZ,x(f)is locally constant.
In a neighborhood U of x, we have U ∩Z = V (fZ)∩U . Therefore, fZ is a localequation for every y ∈ Z ∩ U . By shrinking U if necessary, we can assume that gand h are regular on U so that the above equation still holds in the local ring of anyother y ∈ U . Therefore, ordZ,y(f) = r = ordZ,x(f) for all y ∈ U ∩ Z.
Next, we have to show that for all but finitely many Z, ordZ(f) = 0. Bytaking an open subset, we may assume that X is affine (since the complement hasonly finitely many irreducible components). Write f = g
h where g, h ∈ O(X), soordZ(f) 6= 0 =⇒ Z ⊂ V (gh). But V (gh) has only finitely many irreduciblecomponents.
Intuitively, ordZ(f) measures the order of zero or pole that f has on Z. Fromthe proof, we see that the following identities hold, as we expect.
(i) ordZ(fg) = ordZ(f) + ordZ(g).
(ii) ordZ(f + g) ≥ minordZ(f), ordZ(g) if f + g 6= 0.
Definition 5.1.5. Div0(X) = Im(k(X)∗ → Div(X)) is called the group of principaldivisors. Pic(X) = Div(X)/Div0(X) is called the Picard group (or the divisor classgroup) of X.
This is an extremely important group attached to an algebraic variety, like co-homology for topological spaces.
Example 5.1.6. Let X = An. Then Pic(X) = 0, since every subvariety of codimen-sion 1 is a hypersurface. So∑
niZi =∑
niV (fi) =∑
ni(fi) = (∏
fnii ).
More generally, we have the following proposition.
Proposition 5.1.7. Let X be a smooth affine variety. Then O(X) is a UFD if andonly if Pic(X) = 0.
52
Proof. Assume O(X) is a UFD. Let Z ⊂ X be codimension 1, choose f ∈ I(Z) (sof is not a unit). By replacing f by one of its prime factors, we may assume that fis irreducible. Therefore, (f) is prime, hence Z ⊂ V (f) is a closed subvariety of anirreducible variety, and they have the same dimension, implying that Z = V (f).
Conversely, suppose that Pic(X) = 0. Then Z = (f) for f ∈ k(X). We claimthat in fact, f ∈ O(X). To this end, it suffices to show that f ∈
⋂x∈X OX,x. By
definition, in each such local ring
f =g
hf rZ ,
but h must be a unit, since otherwise f has non-zero order at some other closedsubvariety of co-dimension 1.
If g ∈ O(X) is irreducible, then V (g) is irreducible because if Z ⊂ V (g), thenZ = V (f), so f | g. This means that if g | fh, then
V (g) ⊂ V (fh) = V (f) ∩ V (h) =⇒ V (g) ⊂ V (f) or V (g) ⊂ V (h) =⇒ g|f or g | h.
Example 5.1.8. Let X = Pn. We have a map deg : Div(X)→ Z given by
Z = V (f) 7→ deg f = deg .
This is clearly a surjective map, and Div0(X) ⊂ ker(deg) since elements of Div0(X)can be written as f
g , where f and g are homogeneous polynomials of the same degree.
Exercise 5.1.9. Check that the divisor of∏fnii∏gmjj
is∑niV (fi)−
∑mjV (gj).
Also note that the divisor of a constant function is 0. So we have a short exactsequence
0→ k∗ → k(X)∗ → Div(X)deg→ Z→ 0.
Now let D =∑niZi such that degD = 0. Since all codimension 1 subvarieties of
Pn are hypersurfaces, Zi = V (fi) for each i. So∏fnii =
∏nk>0 f
nkk∏
nj<0 f−nj
j
=f
g=⇒ D = (
f
g)
Definition 5.1.10. A prime divisor is D = Z, where Z is a subvariety of codimension1.
Definition 5.1.11. An effective divisor is D =∑niZi, ni ≥ 0.
Lemma 5.1.12. Let f ∈ k(X)∗. Then (f) is effective if and only if f ∈ O(X)− 0.
Proof. We essentially proved this already. We can assume that X is affine, where thecoordinate ring is the intersection of the local rings. But ordZ(f) ≥ 0 =⇒ f ∈ OX,x
for all x ∈ X.
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Corollary 5.1.13. Let X be smooth, F ⊂ X a closed codimension ≥ 2 subvariety.Then O(X) ' O(X − F ).
Proof. This follows from the above lemma and the observation that the divisor groupis unchanged by removing a codimension 2 subvariety.
Let U ⊂ X be open. Then we get a restriction map Div(X) → Div(U). IfX = U ∪ V , then we have an exact sequence
0→ Div(X)→ Div(U)⊕Div(V )→ Div(U ∩ V ).
5.2 Linear Systems
Definition 5.2.1. Let X be a projective variety. Let D ∈ Div(X). Define
L (D) = f ∈ k(X) | f = 0 or (f) +D ≥ 0.
Note that L (D) has a natural k-vector space structure, since ordZ(f + g) ≥minordZ(f), ordz(g). We also define |D| to be the set of all effective divsors of theform (f)+D for some f ∈ k(X). This corresponds to the one-dimensional subspacesof L (D), giving a natural identification
|D| ' P(L (D)).
This is because
(f) +D = (g) +D =⇒ (f) = (g) =⇒ (f
g) = 0 =⇒ f ∈ O(X)∗ = k∗
since X is projective. Later, we will see that dim L (D) =: `(D) < ∞. Therefore,|D| has a natural structure as a projective space.
Definition 5.2.2. A linear subvariety of |D| is called a linear system.
Definition 5.2.3. A linear system D is called complete if it is of the form |D| forsome D.
Definition 5.2.4. Let L be a linear system on X. The base points of L are definedto be ⋂
D∈LSupp(D) where if D =
∑niZi then Supp(D) =
⋃ni 6=0
Zi.
One of the most important questions in additive function theory is to calculatethe quantity `(D) = dim L (D).
Let X ⊂ Pn be a projective variety, S(X) its homogeneous coordinate ring, withthe natural grading by degree S(X) =
⊕S(X)d. Let f ∈ S(X)d, Hi = V (Xi). A
54
priori f is not a well-defined function on X, but fXd
i
defines a regular function on
X ∩ (Pn −Hi). Therefore,
(f
Xdi
) is an effective divisor on X ∩ (Pn −Hi).
Moreover, note that
(f
Xdi
)|X∩(Pn−Hi−Hj) = (f
Xdj
)X∩(Pn−Hi−Hj)
since Xi and Xj are both nonvanishing on the intersection. Therefore, there exists
an effective divisor (f) := X ·V (f) on X whose restriction to X ∩ (Pn−Hi) is ( fXd
i
).
Another way to reach this definition is to consider X ∩ V (f). This is a union ofcodimension 1 components
X ∩ V (f) =⋃Zi.
Then we define X · V (f) =∑niZi, where ni is determined as follows. Pick Xk(i)
such that Zi 6⊂ Hk(i). Then set
ni := ordZi
f
Xdk(i)
.
In this way, we define a map
S(X)d → Div(X).
We claim that the image is a linear system. This is because (exercise)
(f) = (g) + (f
g),
f
g∈ k(X).
Fix g. Every other (f) is
(f1) = (g) + (f1
g)
(f2) = (g) + (f2
g)
(f1 + f2) = (g) + (f1 + f2
g).
To summarize, this linear system, denoted by LX(d), is isomorphic to P(S(X)d).
Theorem 5.2.5. For d 0, LX(d) is complete.
Now we return to the theorem we stated before.
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Theorem 5.2.6. For d 0, LX(d) is complete.
Proof. Let D ∈ |LX(d)|. We know that LX(d) contains (xdi ) = dHi, so
D − dHi = (fi) fi ∈ k(X).
Then fifj
= (D − dHi) − (D − dHj) = d(Hj − Hi). Therefore, in the fraction field
L of S(X), we have xdi fi = xdjfj up to some constant which we normalize to be 1.
Thus F := xdi fi is well-defined.Note that (fi)|Pn−Hi is effective, so fi ∈ O(X − Hi ∩ X), i.e. xni fi ∈ S(X).
Therefore, xN−di F = xn−di xdi fi ∈ S(X)N .Now define S′d ⊂ L by
S′d = s ∈ L | ∃N such that xNi s ∈ S(X)N+d ∀i
and set S′ =⊕
d≥0 S′d. So S′ is a graded ring containing S(X). Also, F ∈ S′,
and to say D ∈ LX(d) is equivalent to saying that F ∈ S(X), since F = xdi fi andD = (fi) + dHi. So to show that the linear system is complete for all sufficientlylarge d, it suffices to show that S′d = S(X)d for d 0.
Fix s ∈ S′d, d ≥ 0. We have xNi s ∈ S(X)N+d for some N , and taking `0 =N(n+ 1) we get
S(X)`s ⊂ S(X)`+d for ` ≥ `0.Therefore,
S(X)`sq ⊂ S(X) ∀q ≥ 0.
So
x`00 sq ∈ S(X) ∀q ≥ 0 =⇒ sq ∈ 1
Xd00
S(X)∀q ≥ 0.
Consider M = 1
xl00
S(X) ⊂ L. This is a f.g. S(X)-module, so s satisfies a polynomial
relationsq + a1s
q+1 + . . .+ aq = 0 ai ∈ S(X),
i.e. s is integral over S(X). Therefore, S′ is integral over S(X).
Theorem 5.2.7. Let A be a finitely generated integral k-algebra, K = Frac(A). LetL/K be a finite field extension. Then the integral closure A′ of A in L is finite overA and A′ is also a finitely generated k-algebra.
Using this fact from commutative algebra, we know that S′ is finite over S(X).Let s1, . . . , sm be a set of homogeneous generators of degree di. There exists `0 suchthat x`isj ∈ S(X)`+d for any ` ≥ `0. Therefore, S′d = S(X)d if d ≥ maxdi+`0.
Note that if D1 ∼ D2, then L (D1) ' L (D2), since
D1 −D2 = (f) =⇒ L (D1)·f→ L (D2)
is an isomorphism.
56
Corollary 5.2.8. We have `(D) <∞.
Proof. If |D| 6= ∅, we can assume that D =∑niZi is effective. Choose Fi homo-
geneous of degree di such that Zi ⊂ X ∩ V (Fi) but X 6⊂ V (Fi). Let F =∏Fnii , a
homogeneous polynomial of degree e =∑nidi of degree e =
∑nidi. Then
(F ) = G = D +D′, where D′ is effective.
Therefore, L (D) ⊂ L (G), so `(D) ≤ `(G). By taking some multiple of G, we mayassume that L (G) is complete by the theorem. Therefore, `(G) = dimLX(e) <∞.
Let ϕ : X 99K Pn be a rational map such that ϕ(X) is not contained in anyhyperplane. We will construct a linear system Lϕ on X as follows. By definition,we have a diagram
Z = Γϕ
pr1
pr2 ##X Pn
where pr1 : Z → X is birational. Let F ⊂ X be the subvariety such that pr1 :pr−1
1 (F )→ F is not an isomorphism over F . By Zariski’s Main Theorem, codim F ≥2 (since dim pr−1
1 (F ) ≤ r − 1).Now let H = V (`) ⊂ Pn be a hyperplane. We define a divisor ϕ∗(H) on X as
follows: since F has codimension at least two, divisors on X are the same as divisorson X − F . Define ϕ∗H on X − F − ϕ−1(Hi) as (ϕ∗( `xi )). One can check the this
agrees on the intersections X − F − ϕ−1(Hi ∪Hj), so this extends to a divisor onX − F .
It is easy to see that if H = V (`) and H ′ = V (`′), then ϕ∗H = ϕ∗H ′ + (ϕ∗( ``′ )).Therefore, ϕ∗H form a linear system, denoted by Lϕ ' G(n− 1, n) = (Pn)∗.
Proposition 5.2.9. The base point locus of Lϕ is F .
Proof. If x /∈ F , then ϕ(x) ∈ Pn and there exists H such that ϕ(x) /∈ H. Therefore,x /∈ ϕ−1(H) = Supp(ϕ∗H|X−F ).
Conversely, let H ⊂ Pn be a hyperplane. We need to show that F ⊂ Supp(ϕ∗H).By changing coordinates if necessary, we may assume that H = H0. Then ϕ∗ xix0 isa rational function on X with divisor ϕ∗Hi − ϕ∗H0. Therefore, ϕ∗ xix0 is a regularfunction on X − Supp(ϕ∗H0), so ϕ|X−Supp(ϕ∗H0) is a morphism given by
p 7→ (ϕ∗(x1
x0(p)), . . . , ϕ∗(
x1
x0(p))).
But we know that this rational map is not regular on F , so F ⊂ Supp(ϕ∗(H0)).
Theorem 5.2.10. There is a 1− 1 correspondence between
57
(a) A linear system L on X, with basepoints of codimension ≥ 2, and an isomor-phism ψ : L ' (Pn)∗,
(b) A rational map ϕ : X 99K Pn such that ϕ(X) is not contained in any hyper-plane.
In addition, this correspondence is PGLn+1 equivariant.
In other words, given a linear system L on X with basepoints of codimension ≥ 2,there exists a rational map ϕ : X 99K L∗ = G(n − 1, L) not contained in anyhyperplane.
Remark 5.2.11. There is a natural map V → V ∗∗ when V is finite dimension. Well,P(V ∗) is the projectivization of the dual space, so its elements correspond naturallyto hyperplanes in P(V ), i.e. P(V )∗.
Suppose V,W are two vector spaces. We can take their projectivizations: PVand PW . If there is an isomorphism ψ : PV ' PW , then there exists a unique liftψ : V 'W which is unique up to scaling by a constant.
Proof. We have already shown how (b) gives rise to (a). Conversely, let X be smoothand projective, L a linear system on X with base points of codimension ≥ 2. ThenL = PV for some V ⊂ L (D), D ∈ L. Fix an isomorphism ψ : L ' (Pn)∗, whichgives a lift ψ : V → kx0, . . . , xn, where the xi are coordinates on Pn, which isunique up to scaling by a constant. Let Di = ψ−1(Hi), Hi = V (xi). The point isthat we can regard Di as “coordinates” on X, since they correspond to functions“up to constants.” So we write Di − D0 = (fi), where fi is a rational function,regular X − Supp(D0). Define a map
x 7→ (f0(x), f1(x), . . . , fm(x)).
It is evident that this construction is equivariant with respecto to PGLn+1(C).
Example 5.2.12. Let X = Pn, L ⊂ LX(1), where LX(1) is the linear system ofhyperplanes on Pn, i.e. (Pn)∗. Then the rational map ϕL : X 99K L∗ is the projectionPM away from M =
⋂H∈LH.
Conversely, given given M ⊂ Pn of dimension n− r − 1, we may define
L = H ∈ (Pn)∗ | H ⊃M.
Then ϕL = PM . More generally, given L1 ⊂ L2 ⊂ (Pn)∗, we have the diagram
L∗2
PM
X = Pn
ϕL2
::
ϕL1 $$L∗1
58
where M =⋂H∈L1
H ⊂ L∗2.
Example 5.2.13. Let X = Pn, L = LX(d). Then ϕL : X 99K L∗ is the d-upleembedding.
5.3 Differentials
Now we turn to the construction of a divisor class called the canonical class.
Definition 5.3.1. Let A → B be a ring homomorphism. The Kahler differentialsis a pair (ΩB/A, d), where ΩB/A is a B-module and d ∈ DerA(B,ΩB/A) are char-acterized by the universal property that for any B-module M , the natural mapHomB(ΩB/A,M)→ DerA(B,M) given by ϕ 7→ ϕ d is an isomorphism.
In categorical language, the Kahler differential represents the functor of B-modules sending M 7→ DerA(B,M).
Theorem 5.3.2. (ΩB/A, d) exists and is unique up to unique isomorphism.
Proof. Uniqueness follows from the usual abstract nonsense. For existence, we con-struct ΩB/A as the free B-module on the symbols db for b ∈ B, modded out bythe relations da = 0, d(b1 + b2) = db1 + db2 and d(b1b2) = b1db2 + b2db1. Thend : B → ΩB/A is defined by d(b) = db.
For any D : B →M , we defined ϕ : ΩB/A →M by sending ϕ(db) = D(b).
Example 5.3.3. If A = k, B = k[x1, . . . , xn] then ΩB/A ' Bdx1 ⊕ . . .⊕Bdxn.
Corollary 5.3.4. If B/A is finitely generated as an A-algebra, with generatorsb1, . . . , br, then ΩB/A is finitely generated over B with generators db1, . . . , dbr.
Lemma 5.3.5. Let S ⊂ B be a multiplicative set. Then S−1ΩB/A ' ΩS−1B/A.
Proof. DerA(S−1B,M) = DerA(B,M) for any S−1B-module M .
Corollary 5.3.6. Let X/k be an algebraic variety. Then ΩOX,x/k is a finitely gen-erated OX,x-module.
Notation. Let X/k be affine. We denote ΩX := ΩO(X)/k and ΩX,x := ΩOX,x/k.
Remark 5.3.7. If X is not affine, we denote ΩX to be the functor from the categoryof affine opens on X to abelian groups, U 7→ ΩU . In other words, ΩX is a sheaf onX.
Proposition 5.3.8. Let X/k be an algebraic variety. Then X is smooth at x if andonly if ΩX,x is a free module over OX,x of rank dimX = r.
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Proof. We claim that ΩX,x ⊗OX,xk ' T ∗xX = (TxX)∗. In other words,
Homk(ΩX,x ⊗OX,xk, k) ' HomOX,x
(ΩX,x, k) ' Derk(OX,x, k) = TxX.
If ΩX,x is free of rank r, then dimk ΩX,x ⊗OX,xk = r = dimTxX, so X is smooth at
x.Conversely, suppose X is smooth at x, so dimk ΩX,x⊗OX,x
k = r. Let K = k(X);we claim that dimK ΩX,x ⊗K = r. This is because
HomK(ΩX,x ⊗OX,xK,K) = HomOX,x
(ΩK/k,K) = Derk(K,K) = r.
We will be done after showing the following lemma.
Lemma 5.3.9. Let (A,m, k) be a Noetherian local ring and M be a finite A-module,K = Frac(A). If
dimkM ⊗A k = r = dimKM ⊗A Kthen M is a free A-module of rank r.
Our proof of this lemma uses Nakayama’s lemma, which we will state but notprove.
Lemma 5.3.10. Let (A,m, k) be a Noetherian local ring, M a finite A-module. IfM ⊗A k = 0, then M = 0.
Proof of Lemma 5.3.9. Let m1, . . . ,mr ∈M be elements whose images in k form abasis of M ⊗A k. Consider the map of A-modules
Ar →M
ei 7→ mi
We claim that this is an isomorphism. Consider the exact sequence of the cokernel
Ar →M → N → 0
then upon tensoring with k, we get
Ar ⊗ k →M ⊗ k → N ⊗ k → 0
but N ⊗k = 0 =⇒ N = 0 by Nakayama’s lemma. Therefore, the map is surjective.Now consider the exact sequence of the kernel:
0→ N → Ar →M → 0.
Tensoring with K, we get the exact sequence
0→ N ⊗K → Ar ⊗K →M ⊗K → 0
implying that N⊗K = 0 since the map on the right is a surjection of vector space ofthe same dimension. But since N is a submodule of a free module, it is torsion-free,so this is only possible if N = 0.
60
In the proof, we showed ΩX,x ⊗ k ' T ∗xX = mx/m2x. Let’s give a more explicit
construction of this isomorphism. Recall that we have a derivation d : OX,x → ΩX,x.One can check that m2
x goes to 0 in the map to ΩX,x⊗k, so the map factors throughthe quotient.
mx ⊂ OX,xd //
ΩX,x
mx/m
2x
// ΩX,x ⊗ k
In particular, if x1, . . . , xr ⊂ mx are such that x1, . . . , xr (mod m2x) are a basis of
mx/m2x then dx1, . . . , dxr form a basis of ΩX,x over OX,x. Such x1, . . . , xr are called
local parameters of X at x. So every f ∈ OX,x may be written as
df =∑ ∂f
∂xidxi
where ∂∂xi∈ HomOX,x
(ΩX,x,OX,x) = DerOX,x(OX,x,OX,x) are the dual basis.
Let X be affine. Observe that every element in ΩX can be regarded as a mapfrom X to
⊔x∈X T
∗xX. For x ∈ X, we have
Ωx → ΩX,x → ΩX,x ⊗ k ' T ∗xX
ω 7→ ωx 7→ ω(x)
Now we want to give another definition of Ωx without all of this commutativealgebra. Let f ∈ O(X); we know df should be in ΩX . The map given by df in thiscase is
x 7→ f − f(x) (mod m2x).
So we define ΩX to be the maps ω : X →⊔x∈X T
∗xX such that for every p ∈ X,
there exists an open neighborhood V containing p such that there exist fi, gi ∈ O(V )with ω =
∑fidgi. This is another way of understanding the Kahler differentials: as
a map to the disjoint union of cotangent spaces given locally by regular functions.Moreover, this makes sense for arbitrary varieties. Our old definition clearly has anatural inclusion into this set; it is left as an exercise to check that they are thesame.
5.4 The canonical bundle
Definition 5.4.1. Let Xr be a smooth variety of dimension r. The canonical form is
ωX,x := ∧rOX,xΩX,x
is a free OX,x-module of rank 1, with a basis given by dx1 ∧ . . . ∧ dxr.
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Definition 5.4.2. We define ωK/k := ∧rΩK/k. This is a 1-dimensional K-vectorspace; elements in ωK/k are called rational r-forms.
Our goal is to associate to each element in ωK/k a divisor. Let w ∈ ωK/k benon-zero. Recall that
ωK/k ' ωX,x ⊗OX,xK,
so ω = fdx1 ∧ . . . ∧ dxr, where f ∈ K and x1, . . . , xr are local parameters for Zi.Define
(ω) =∑
niZi, ni = ordZif.
We have made many non-canonical choices here: the point x and the local parame-ters x1, . . . , xr. However, one can check everything is well-defined; this is left as anexercise.
It is clear from this construction that for f ∈ K, (fω) = (f) + (ω). Therefore(ω) | ω ∈ ωK/k − 0 is a well-defined divisor class, called the canonical divisorclass (usually denoted KX). Likewise, we define the n-fold canonical class as follows:let ω ∈ ω⊗nK/k. This is a one-dimensional space over K, whose elements look like
ω = f(dx1 ∧ . . . ∧ dxr)n.
We then define
(ω) =∑
niZi ni = ordZi(f) for x ∈ Zi.
Notice that if ω1 ∈ ωn1
K/k and ω2 ∈ ωn2
K/k, then (ω1⊗ω2) = (ω1)+(ω2). In particular,
ω ∈ ωnK/k − 0 implies that (ω) ∼ nKX .Now we turn to a discussion of linear systems. Assume X is projective. For
ω ∈ ω⊗nK/k, we may define
O((ω)) = f ∈ K | (f) + (ω) = (fω) ≥ 0 = ω ∈ ω⊗nK/k | (ω) ≥ 0.
This does not depend on the choice of ω, so we denote it by L (nKX).
Definition 5.4.3. Let Pn(X) = `(nKX) is the nth plurigenera of X. In particular,P1(X) =: Pg is called the geometric genus of X.
Note that⊕
n≥0 L (nKX) has a natural k-algebra structure. This is called thecanonical ring of X. The following theorem was just proved.
Theorem 5.4.4 (Siu, Hacon, McKernan, ...). The canonical ring is a finitely gen-erated k-algebra.
Example 5.4.5. Let X = Pn. We claim that KX = −(n + 1)H. In particular, thisimplies that Pg(Pn) = 0.
Let x0, . . . , xn be homogeneous coordinates on on Pn. Let yi = xix0
. Let
ω = dy1 ∧ . . . ∧ dyn ∈ ωK/k.
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Therefore, (ω)|Pn−H0 = 0, so (ω) = mH0. Now let zi = xixn
, so yn = 1z0
and yi = ziy0
for 1 ≤ i ≤ n− 1. In these coordinates,
ω = d
(z1
z0
)∧ . . . ∧ d
(zn−1
z0
)∧ d(
1
z0
)= ± 1
zn+10
dz0 ∧ . . . ∧ dzn−1.
So we see that (ω) = −(n+ 1)H0.
Example 5.4.6. Let X = V (y2z−x3−xz2) ⊂ P2, chark 6= 2. We claim that KX = 0,so Pg(X) = 1.
On (P2 −Hz) ∩X, the curve is defined by V (y2 − x3 − x). Here
ΩK/k = 〈dx, dy | 2ydy − (3x2 + 1)dx〉.
So ΩX,(a,b) ⊗ k as a vector space over k is 〈dx, dy | 2bdy − (3a2 + 1)dx〉. Therefore,dx 6= 0 if b 6= 0, so x is a local parameter at (a, b) if b 6= 0. Let
ω =dx
y=
2dy
3x2 + 1.
From the first expression, it is clear that the divisor is supported possibly on y = 0(in this chart) and the line at infinity. If at (a, b), 3a2 + 1 6= 0, then dy 6= 0 so y is alocal parameter and ω is not supported at such points. But by the non-singularityof the curve, these two cover all of this affine chart.
Now we look at X ∩ Hz = p = (0, 1, 0). On the affine y 6= 0, the curve isV (z − x3 − xz2). So
ΩX,p ⊗ k = 〈dx, dz | dz = 3x2dx+ z2dx+ 2xzdz〉 = 〈dx, dz | dz〉
So x is a local parameter at p. The coordinate change is x 7→ x/z, and y 7→ y/z, soin the new local coordinates
ω =dx
y=d(x/z)
y/z=zdx− xdz
z.
From the relation, we know that dz = 3x2dx+ z2dx+ 2xzdz, so
dz =(3x2 + z2)dx
1− 2xz.
From the relation we know x3
z = 1− xz, so
ω = dx− x
z· 3x2 + z2
1− 2xzdx = dx− 3x3/z + xz
1− 2xzdx
= dx− 3− 3xz + xz
1− 2xzdx = dx− 3− 2xz
1− 2xzdx
=−2
1− 2xzdx.
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Proposition 5.4.7. If X,Y are birational then Pn(X) = Pn(Y ).
Proof. If X and Y are birational, then by Zariski’s Main Theorem we have a mor-phism U → V defined away from codimension 2. Since codim (X \ U) ≥ 2 andcodim (Y \V ) ≥ 2, U and V have the same divisor groups as X and Y , respectively.Note that k(X) ' k(Y ), so ωk(X)/k ' ωk(Y )/k. These two facts together imply thatthere is a canonical bijection L (nKX) ' L (nKY ).
5.5 The ramification divisor
Let ϕ : X → Y be a morphism of algebraic varieties. We wish to define ϕ∗ : ΩY →ΩX . One way is purely through commutative algebra, and another way is geometric.We will discuss the first way and sketch the second.
More generally, if we have a map of commutative rings
A→ Bϕ→ C
then there is a natural map
ΩB/A ⊗B C → ΩC/Aϕ∗→ ΩC/B → 0.
For any C-moduleM (which is also aB-module via ϕ), HomB(ΩB/A,M) ' DerA(B,M).Furthermore, there is a natural map DerA(B,M)← DerA(C,M) ' HomC(ΩC/A,M)obtain by composing with ϕ. Taking M = ΩC/A, define ϕ∗ := res(Id) : ΩB/A⊗BC →ΩC/A.
Remark 5.5.1. Some remarks.
(i) From the construction, we see that ϕ∗(df) = d(ϕ∗f).
(ii) The kernel of ϕ∗ : DerA(C,M) → DerA(B,M) is precisely DerB(C,M).Therefore, we have a right exact sequence
ΩB/A ⊗B C → ΩC/A → ΩC/B → 0.
Here is the geometric way of seeing this map. Let ω ∈ Ωy, ω : Y →⊔y∈Y T
∗y Y .
We want a map ϕ∗ω : X →⊔x∈X T
∗xX. We let ϕ∗ω(x) = dϕ∗ω(ϕ(x)) (note that
ω(ϕ(x)) ∈ T ∗ϕ(x)Y , and dϕ∗ : T ∗ϕ(x)Y → T ∗xX). If ω = df , where f ∈ O(Y ), then
ϕ∗(df) = d(ϕ∗f) = d(f ϕ). Therefore, if ω =∑fidgi, then locally on X
ϕ∗ω =∑
fi ϕd(gi ϕ).
Now specialize to the case A = k,B = k(Y ), C = k(X), and ϕ∗ : k(Y )→ k(X)is induced by a dominant morphism ϕ : X → Y . By the discussion above, we havea right exact sequence
Ωk(Y )/k ⊗ k(X)→ Ωk(X)/k → Ωk(X)/k(Y ) → 0.
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If ϕ : X → Y is separable, then
dimk(X) Ωk(X)/k(Y ) = dimX − dimY.
This is because Derk(K,K) has dimension the transcendental degree of K over k ifthe extension is separable, and HomK(ΩK/k, k). But then we can see by countingthe dimensions that the sequence is also left-exact. In particular, if dimX = dimYthen this tells us that Ωk(Y )/k ⊗k(Y ) k(X) ' Ωk(X)/k. Then we get an induced map
ϕ∗ : ωk(Y )/k ⊗k(Y ) k(X) ' ωk(X)/k.
Let ω ∈ ωk(Y )/k. Then ϕ∗(ω ⊗ 1) ∈ ωk(X)/k.In the homework we also introduced the divisor ϕ−1(ω). This was a map on
divisor groups ϕ−1 : Div(Y )→ Div(X) defined as follows. If D =∑niYi ∈ Div(Y ),
then ϕ−1D =∑niϕ
−1(Yi). Write ϕ−1(Yi) =∑mjXj . Choose x ∈ Xj and a local
equation fYi at ϕ(x), and define mj = ϕXjϕ∗(f). It is left as an exercise on the
homework to show that this process gives a well-defined divisor class and is a grouphomomorphism.
Lemma 5.5.2. B = ϕ∗(ω ⊗ 1) − ϕ−1(ω) is an effective divisor, called the branchdivisor.
Proof. Write ω = fdy1 ∧ . . . ∧ dyx, where (y1, . . . , yr) are local parameters at y =ϕ(x), so (x1, . . . , xr) are local parameters at x. Then
ϕ∗ω = (f ϕ)d(y1 ϕ) ∧ . . . ∧ d(yr ϕ) = (f ϕ) det
(∂(yi ϕ)
∂xj
)dx1 ∧ . . . ∧ dxr.
Around x,
(ϕ∗ω)− ϕ−1(ω) = (f ϕdet
(∂(yi ϕ)
∂xj
))− (f ϕ) = (det
(∂(yi ϕ)
∂xj
))
In other words,
Supp(B) = x ∈ X | det∂(yi ϕ)
∂xj∈ mx
= x ∈ X | (dϕ)∗ : T ∗ϕ(x)Y → T ∗xX not an isomorphism.
= x ∈ X | (dϕ)∗ not smooth at x.
Example 5.5.3. Let X = (y2z = x3 + xz2) ⊂ P2. The projection [x, y, z] 7→ [x, z]defines a map ϕ : X → P1.
Let [x, y, z] be homogeneous coordinates on P2, [u, v] homogeneous coordinateson P1. Then ϕ sends X ′ := z 6= 0 → A1 := v 6= 0. So ϕ∗ : k[u] →
65
k[x, y]/(y2 − x3 − x) sending u 7→ x. Let du ∈ Ωk(P3)/k. Then ϕ−1(du) = 0on X ′. On the other hand, (ϕ∗du) = (dx). Now, x is a local parameter onx− y = 0 = (1, 0), (0, 0), (−1, 0), at which y is a local parameter, and
dx =2y dy
3x2 + 1.
So (dx) = [1] + [0] + [−1] on X ′. Finally, we need to check the point at infinity,p = (0, 1, 0). There is a difficulty here: the map isn’t even defined at p. to rectifythis, consider the blowup of X at p.
X ⊂ P2 × P1
yy %%X P1
The equations y2z = x3 + xz2
xv = zu
cut out X along with an exceptional component. In A2 × A1 defined by y 6= 0 andv 6= 0, the system is
z = x3 + xz2
xv = z
So X ∩ (A2 ×A1) has coordinate ring k[x, z, v]/(xv − z, v − x2 − x2z2). This showshow to extend the map to p.
p ∈ X ∩ (A2 × A1)
uu ((p ∈ X ∩ A2(y 6= 0) A1(u 6= 0)
The form is ω = du = −dvv2
. The point p has x = 0, z = 0, v = 0. Recall from thecomputation last time that x is a local parameter in first factor. So B around p isgiven by d(vϕ)
dx = dvdx .
v = x2 + x2v2 =⇒ dv = 2xdx+ 2xv2dx+ 2x2vdv =⇒ dv
dx=
2x+ 2xv2
1− 2x2v.
This vanishes to order 1.We conclude that B = [1] + [0] + [−1] + [∞].
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Chapter 6
The Hilbert Polynomial
6.1 Construction of the Hilbert Polynomial
Theorem 6.1.1. Let M be a finitely generated graded module over S = k[x0, . . . , xn].Then there exists a polynomial Pm(t) ∈ Q[t] of degree at most n such that PM (d) =dimkMd for d 0.
Proof. We induct on n. If n = −1, then the result is trivial since M is then afinite dimensional vector space over k. Now suppose the theorem holds for < n.Let N ′ = m ∈ M | xnm = 0 and N ′′ = M/xnM . Then N ′, N ′′ are finitelygenerated graded k[x0, . . . , xn−1] modules. By induction, there exists PN ′(t) ∈ Q[t]and PN ′′(t) ∈ Q[t] with PN ′(d) = dimkN
′d and PN ′′(d) = dimkN
′′d for d 0. We
have an exact sequence
0→ N ′d →Mdxn−→Md+1 → N ′′d+1 → 0.
So dimMd+1 − dimMd = dimN ′′d+1 − dimN ′d = PN ′′(d+ 1)− PN ′(d). But R(t) :=PN ′′(t+ 1)− PN ′(t) ∈ Q[t] with degree at most n− 1.
Lemma 6.1.2. Let f(t) ∈ Q[t] of degree m. Then there exists g(t) ∈ Q[t] of degreem+ 1 such that g(t+ 1)− g(t) = f(t).
Exercise 6.1.3. Easy exercise.
So there exists P (t) ∈ Q[t] such that P (t + 1) − P (t) = PN ′′(t + 1) − PN ′(t) =dimMd+1 − dimMd for d 0. Then dimMd+1 − P (d + 1) = dimMd − P (d) ford 0, implying that dimMd−P (d) is constant for all sufficiently large d. We thenset PM (d) to be the constant plus P (d).
In particular, let M = S/I(X) where X is an algebraic set in Pn. Then PX :=PS/I(X) is called the Hilbert polynomial of X.
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Example 6.1.4. (i) Let X = Pn. Then M = S, and Md which has dimension(n+dn
).
PPn(t) =
(t+ n
n
)=
1
n!tn + . . .
(ii) Now suppose X = V (f) ⊂ Pn. Then S(X) = S/f . Suppose deg f = d.
0→ Sm×f→ Sm+d → S(X)m+d → 0
So dimk S(X)m+d = dimk Sm+d − dimk Sm =(m+d+n
n
)−(m+nn
)for m ≥ d. So
PX(t) =
(t+ n
n
)−(t− d+ n
n
)=
d
(n− 1)!tn−1 + . . .
(iii) Let X = p1, . . . , pm ⊂ Pn. We may assume that pi /∈ H0 = V (X0). We havea left exact sequence
0→ I(X)d → Sd → km
f 7→(f
xd0(p1), . . . ,
f
xd0(pm)
).
For d 0 (e.g. d > m), this is also surjective. To see this, we can just considerproducts of hyperplanes passing through one of the points and no others. So ford 0, S(X)d ' km, implying that PX(t) = m.
Remarks. Let X be a smooth, projective variety.
(i) Let LX(d) = P(S(X)d), so dimLX(d) = PX(d)− 1 for d 0. We proved thatthis linear system is complete for d 0, so dim |dH| = dimLX(d) for d 0.In other words,
PX(d) = `(dH) d 0.
(ii) We saw that the canonical ring⊕
d≥0 L (dKX) was finitely generated. Thereexists a polynomial P (t) such that
P (d) = `(dKX) d 0.
The degree of P (d) is called the Kodaira dimension of X, which can be anyinteger in [−1, d]. (Degree −1 means the polynomial is the zero polynomial)
6.2 Numerical invariants
Definition 6.2.1. Let Xr ⊂ Pn be projective. We define the arithmetic genus of Xto be
pa(X) = (−1)r(PX(0)− 1).
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Theorem 6.2.2. (i) If X ' Y as varieties, then pa(X) = pa(Y ). (ii) If X and Yare smooth and birational, then pa(X) = pa(Y ).
Example 6.2.3. (i) Since we calculated PPn(t) =(t+nn
), we have pa(Pn).
(ii) For X = V (f), deg f = d, we saw that
PX(t) =
(t+ n
n
)−(t+ n− d
n
)=⇒ PX(0) = 1−
(n− dn
).
Therefore,
pa(X) = (−1)n(n− dn
)=
0 d ≤ n(d−1n
)d > n
So pa(y2 − x3) = pa(y2z = x3 + zx2) = pa(y
2z = x3 + z2x) = 1. Note that we reallyneed smoothness in the second part of the theorem, since singular cubic curves arebirational to P1.
Theorem 6.2.4 (Serre duality). If X is a smooth projective curve, then pa(X) =pg(X).
Proposition 6.2.5. Let Xr ⊂ Pn be projective. Then
PX(t) =d
r!tr + . . . ,
where d ∈ Z is positive. This d is called the degree of X.
Proof. We saw that this is true if X is a hypersurface. Now let M ⊂ Pn be a(n−r−2)-plane such that PM : X → X ′ ⊂ Pr+1 is birational. We know that (i) S(X)is a finite S(X ′)-module, and (ii) k(X) ' k(X ′). So Frac(S(X)) ' Frac(S(X ′)) 'k(X)(x0) where x0 6= 0 in S(X ′).
Let f1, . . . , fs be the generators of S(X) over S(X ′). We can write fi := gihi
, where
gi, hi ∈ S(X ′). Setting h =∏hi, we have S(X) ⊂ 1
hS(X ′), i.e. hS(X) ⊂ S(X ′).Write h =
∑d hd, where each hd is homogeneous of degree d. Since S(X) and S(X)′
are homogeneous, hdS(X) ⊂ S(X ′) for each d. So we can assume that there existsh ∈ S(X ′) homogeneous of degree d0 such that hS(X) ⊂ S(X ′). Then
dimS(X ′)d ≤ dimS(X)d ≤ dimS(X)d+d0 .
So
PX′(d) ≤ PX(d) ≤ PX′(d+ d0).
Then PX − P ′X is a polynomial of degree less than degPX′(t). Therefore PX′(t) =Dr! t
r + . . ., so PX has the same degree with the same leading coefficient.
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6.3 Dimension theory of intersections
Proposition 6.3.1. Let Zn be an affine variety and Xr, Y s closed subvarieties. Letx ∈ Xr ∩ Y s. Assume that x is smooth in Zn, and write
Xr ∩ Y s = W1 ∪ . . . ∪Wr ∪W ∗,
where the Wi are irreducible components and x /∈W ∗. Then dimWi ≥ r + s− n.
Corollary 6.3.2. Let Xr, Y s ⊂ Pn be projective with r+ s ≥ n. Then Xr ∩Y s 6= ∅.
Proof. Take the affine cone C(Xr), C(Y s) ⊂ An+1. Then 0 ∈ C(Xr)∩C(ys) contains0, and dimC(Xr) = r + 1, dimC(Y s) = s + 1, we have dimC(Xr) ∩ C(Y s) ≥(r + 1) + (s+ 1)− (n+ 1) ≥ 1.
Proof of Proposition 6.3.1. We can consider X ∩ Y = (X × Y ) ∩∆, where ∆ is theimage of the diagonal embedding Z → Z × Z. The image has dimension r + s, andwe know that the subvariety cut out by one equation has dimension at least 1 less,so it is enough to show that around (x, x), ∆ is cut out by n equations f1, . . . , fn.
Let x1, . . . , xn be local parameters at x ∈ Z. By shrinking Z if necessary,we may assume that the xi are regular functions on Z, i.e. xi ∈ O(Z). Considerfi = xi⊗1−1⊗xi ∈ O(Z×Z). In other words, fi(p, p
′) = xi(p)−xi(p′). Then clearlyfi |∆= 0. But df1, . . . , dfn are linearly independent in T ∗(x,x)(Z × Z) = T ∗xZ ⊕ T ∗xZbecause m(x,x) = mx ⊗ O(Z) + O(Z)⊗mx, so x1 ⊗ 1, . . . , xn ⊗ 1, 1⊗ x1, . . . , 1⊗ xnare local parameters at (x, x).
Then f = (f1, . . . , fn) : Z × Z → An is smooth at (x, x), so f−1(0) = Z∗ ∪ Z∗∗where (x, x) ∈ Z∗ and (x, x) /∈ Z∗∗, dimZ∗ = n and Z∗ is smooth at (x, x). But also∆ ⊂ f−1(0) and contains (x, x) and also has the same dimension, so Z∗ = ∆.
Theorem 6.3.3. Let Xr ⊂ Pn have degree d. Then there exists an open subsetU ⊂ G(n− r,Pn) such that for any L ∈ U ,
(i) #(L ∩X) = p1, . . . , pk, and
(ii) For any pi ∈ Xr∩L, where Xr and L are smooth at pi, we have TpiXr+TpiL =
TpiPn (⇐⇒ TpiXr ∩ TpiL = 0). In this case, k = d.
Definition 6.3.4. Let Xr, Y s ⊂ Zn, where Zn is smooth. Let Xr ∩ Y s =⋃Wi,
where Wi are irreducible components.
(i) We say that X and Y intersect properly at Wi if dimWi = r + s− n.
(ii) We say that X and Y intersect transversely at Wi if there exists x ∈Wi suchthat (i) X,Y are smooth at x and (ii) TxX + TxY = TxZ.
Lemma 6.3.5. If X and Y intersect transversely at W , and x ∈ W as in thedefinition, then W is smooth at x.
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Proof. This is similar to Proposition 6.3.1. We have X ∩Y = (X×Y )∩∆. Locally,the diagonal ∆ is cut out by n equations f1, . . . , fn, and X ∩ Y = ϕ−1(0) whereϕ : X×Y → An is the map defined by f1, . . . , fn. The condition TxX+TxY = TxZimplies that ϕ is smooth at (x, x), and then we can apply our results on smoothmorphisms: there exists a unique component of the pre-image passing through x,and this component is smooth at x.
Theorem 6.3.6. Let Xr, Y s ⊂ Pn be projective varieties intersecting transversely.Then degX deg Y =
∑Wi∈X∩Y degWi.
Proof. For Wi, let W ′i ⊂ Wi be the open subset consisting of x such that X,Y aresmooth at x and TxX + TxY = TxZ. We have shown that such a point exists oneach component, but since it is an open condition there is automatically an opendense subset of such points.
Step 1. First, suppose Y is a plane Ls, so Xr∩Y s =⋃Wi with dimWi = r+s−n.
Choose M2n−r−s such that M ∩Wi transversely for any i (this is possible since thereis a Zariski open set of choices for each Wi, so the intersection is nonempty).
#(X ∩ (L ∩M)) = #((X ∩ L) ∩M) =∑
#(Wi ∩M) =∑
degWi
and L ∩M is an n− r plane.
Step 2. Suppose r+s = n. ConsiderX ⊂ Pn ⊂ P2n+1 embedded by (a0, . . . , an) 7→(a0, . . . , an, 0, . . .) and Y ⊂ Pn ⊂ P2n+1 embedding in the last coordinates. Definethe join of X and Y :
J(X,Y ) =⋃
x∈X,y∈Yxy =
(a0, . . . , a2n+1) | (a0,...,an,0,...)∈X
(...,0,an+1,...,a2n+1)∈Y
.
Note that S(J(X,Y )) = S(X)⊗ S(Y ). Therefore,
dimS(J(X,Y ))` =∑i=0
dimS(X)i dimS(Y )`−i.
By definition, we have
dimS(X)i = dx
(i+ r
r
)+ o(ir) + . . .
dimS(Y )`−i = dy
(`− i+ s
s
)+ o((`− i)s).
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So then
dimS(J(X,Y )) =∑i=0
(dx
(i+ r
r
)+ o(ir)
)(dy
(`− i+ s
s
)+ o((`− i)s)
)
= dxdy∑i=0
(i+ r
r
)(`− i+ s
s
)+ o(`r+s+1)
= dxdy
(`+ r + s+ 1
r + s+ 1
)+ o(`r+s+1).
This tells us that
PJ(X,Y )(t) =dxdy
(r + s+ 1)!tr+s+1 + . . . ,
i.e. J(X,Y ) has dimension r + s+ 1 and degree dxdy.Now comes the trick. Take L = (x0 − xn+1, x1 − xn+2, . . . , xn − x2n+1). Then
L ∩ J(X,Y ) = X ∩ Y and the intersections being transverse in both cases is equiv-alent. But L is an n-plane, so #X ∩ Y = #L ∩ J(X,Y ) = dxdy.
Step 3. We now turn to the general case. Choose a M2n−r−s plane intersectingeach Wi transversely. Again,
#(M ∩ (X ∩ Y )) = #(M ∩ (⋃Wi)) =
∑degWi.
On the other hand, M ∩ (X ∩ Y ) = X ∩ (Y ∩M). We claim that Y intersects Mtransversely: let x ∈ Y ∩M . By assumption, TxWi+TxM = TxPn so TxY +TxM =TxPn, so Y ∩M transversely. Let Y ∩M =
⋃Yi, and note that dimYi = n − r.
Then
#(X ∩ (Y ∩M)) = #(X ∩ (⋃Yi))
=∑
#(X ∩ Yi)
=∑
degX deg Yi
= degX∑
deg Yi,
by Step 2, and∑
deg Yi = deg Y by Step 1.
Suppose Xr, Y s ⊂ Pn intersect properly and transversely with irreducible com-ponents Wi, then Bezout’s theorem tells us that degX · deg Y =
∑degWi. In
particular, if X and Y are plane curves in P2 of degree d and e, then #X ∩ Y = de.What if X and Y intersect properly but not transversely? If you count the
multiplicity of each point properly, then we still get #(X ∩Y ) = de. More precisely,assume Xr, Y s ⊂ Zn where Zn is smooth and X and Y intersect properly. LetX ∩ Y =
⋃Wi. Then one defines the intersection number i(X,Y ;Wi). We won’t
define this in general, but f X = V (f), Y = V (g) and p ∈ X ∩ Y , then i(X,Y ; p) tobe the length of OA2,p/(f, g) as an OA2,p module.
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Theorem 6.3.7 (Bezout). If X,Y ⊂ Pn intersect properly, then
degX deg Y =∑
Wi∈X∩Yi(X,Y ;Wi) degWi.
73
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Chapter 7
Algebraic Curves
7.1 Curves and function fields
Definition 7.1.1. An algebraic curve is an algebraic variety of dimension 1.
Lemma 7.1.2. Let X be a smooth curve and Y be a projective variety. Then everyrational map ϕ : X → Y is a morphism.
Proof. This follows from our version of Zariski’s main theorem, that a birational mapof smooth quasiprojective varieties is defined away from codimension 2. (Technically,we only stated the theorem for X quasiprojective. One can reduce to this case bylooking at an affine cover of the graph Γϕ.)
Corollary 7.1.3. Let X and Y be two smooth projective curves. If X and Y arebirational, then they are isomorphic.
So as far as smooth curves go, they are completed determined by their functionfields k(X), which has transcendence degree 1.
Definition 7.1.4. A function field K/k is a finitely generated field over k of tran-scendence degree 1.
Question: Given a function field K, can we find a smooth, projective curve X withk(X) ' K? Equivalently, given any curve Y , can we find a smooth projective curvebirational to it?
The following theorem assures us that the answer is yes.
Theorem 7.1.5. There is an equivalence of categories between smooth projectivecurves over k and function field over k.
7.2 Normality
Before embarking on the proof, we recall the notion of normality.
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Definition 7.2.1. A variety is called normal if for all x ∈ X, OX,x is integrally closedin k(X).
Lemma 7.2.2. If X is affine, then X is normal if and only if O(X) is integrallyclosed in k(X).
Proof. Suppose O(X) is integrally closed. More generally, if R is integrally closedthen S−1R is integrally closed. To see this, suppose x ∈ Frac(R) satisfies
xn + an−1xn−1 + . . .+ a0 = 0
where ai ∈ S−1R. For some a ∈ S, aai ∈ R for all i. Then we multiply the aboveequation by an:
(ax)n + aan−1(ax)n−1 + . . .+ ana0 = 0.
Since R is integrally closed, ax ∈ R, so x ∈ S−1R.Applying this in the case where R = O(X) and S = R \ mp, we see that OX
integral =⇒ OX,p integral. Conversely, it is a general fact that
R =⋂Rm,
where the intersection ranges over the maximal ideals m of R. To see this, if x ∈ R,then the set a ∈ R : ax ∈ R is an ideal in R, not contained in any maximalideal since x ∈ Rm implies that there exists s ∈ R \m such that sx ∈ R. Therefore,this set is the whole ring; in particular, 1x = x ∈ R. Therefore, OX,p integral for allp =⇒ O(X) integral by the above remarks and the Nullstellensatz.
Definition 7.2.3. Let X be an algebraic variety. Then the normalization of X is anormal algebraic variety X together with a morphism π : X → X such that if Yis normal and f : Y → X is dominant, then there exists a unique f such that thediagram commutes
Yf //
f
X
πX
Proposition 7.2.4. The normalization exists and is unique.
Proof. If X is affine, let A be the integral closure of O(X) in k(X) and X = Spec A.We have a natural embedding O(X)→ A, hence a morphism X → X. A dominantmap Y → X corresponds to an embedding O(X) → O(Y ). Write Y =
⋃Yi where
Yi is affine, so the map factors through A.In the general case, X =
⋃Xi, where the Xi are affine. By the above construc-
tion, we get an Xi for each Xi, and we check the agreement on the intersections, sowe may construct product, etc.
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Theorem 7.2.5. If X is smooth, then X is normal.
Proof. Recall a commutative algebra fact that we stated (but did not prove) earlier:if X is smooth at x, then OX,x is a UFD. Also, UFDs are integrally closed; this isessentially the “rational root theorem.” Suppose we have an integral equation
un + a1un−1 + . . .+ an = 0, ai ∈ OX,x.
Write u = fg , where gcd(f, g) = 1. Then
fn + a1gfn−1 + . . .+ ang
n = 0.
Let p be a prime divisor of fg. Then we must have p | fn, so p | f . Contradiction.
Theorem 7.2.6. If X is normal, then Xsing has codimension ≥ 2.
Example 7.2.7. The cone xy − z2 ⊂ P3 is not smooth, but it is normal.
Proposition 7.2.8. Let X be normal and Y ⊂ X be a closed subvariety of codi-mension 1. Then there exists U ⊂ X open such that U ∩ Y = ∅ and Y = V (f) onU .
Proof. We will prove the proposition for dimX = 1, so that Y = y ∈ X. Letf ∈ O(X), where X is affine and f(y) = 0. By shrinking X, we can assume thatthe zeros of f are just y. By Hilbert’s Nullstellensatz,
√(f) = my. Therefore, there
exists r such that mry ⊂ (f) ⊂ my. We may assume that this r is minimal, so there
exist a1, . . . , ar−1 ∈ my such that
g = a1a2 . . . ar−1 /∈ (f), but gmy ⊂ (f).
Let u = gf , so that u /∈ OX,y but umy ⊂ OX,y. We claim that umy 6⊂ my. Suppose
otherwise; then multiplication by u is a OX,y-module homomorphism my → my, sothere exists F ∈ OX,y[t] monic such that F (u) = 0, implying that u is integral overOX,y. But OX,y is integrally closed, and u /∈ OX,y by assumption.
Therefore, my is strictly contained in umy ⊂ OX,y, so umy = OX,y. Thenmy = (u−1), which is what we wanted.
We have shown more generally that if X is a normal curve, OX,y is a PID. Forthe general case, recall that if x ∈ X then
OX,x = lim−→x∈U
O(U).
If Y ⊂ X is a closed subvariety, then
OX,Y = lim−→U∩Y 6=∅
O(U).
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It is easy to see that OX,Y is a local ring with residue field k(Y ) and maximal ideal
mX,Y = lim−→U∩Y 6=∅
I(U ∩ Y ).
The proof now follows along similar lines, replcaing OX,y by OX,Y . Namely, pickan open U ⊂ X such that U ∩ Y 6= ∅ and U ∩ Y is set-theoretically defined by oneequation and OX,Y is integrally closed.
Proof of Theorem 7.2.6. Suppose codim Xsing = 1, i.e. there exists Y ⊂ Xsing asubvariety of codimension 1. By shrinking X, we may assume that Y = V (f). Lety ∈ Y be a smooth point of Y . Let y1, . . . , yn−1 be local parameters of Y at y, i.e.y1, . . . , yn−1 generate mY,y.
0→ (f)→ O(X)→ O(Y )→ 0.
So the lifts y1, . . . , yn−1, f generate mX,y, i.e. dy1, . . . , dyn−1, df generate mX,y/m2X,y;
but y being singular means that the tangent space has dimension greater than n.
Proof of Theorem 7.1.5. Let K/k be a function field. We choose any projectivecurve X such that k(X) = K (take any B ⊂ K a finitely generated k-algebrahaving K as its fraction field - e.g. take a finite set of generators for K and thealgebra generated by them - U = Spec B and then take its closure in Pn). Now letX be the normalization of X, so X is smooth. We need to show that it is projective.
Cover X by affine opens Ui such that any two points of X are contained insome Ui. By construction, X =
⋃Ui, where Ui is the normalization of Ui. So Ui is
affine and smooth; we let Ui be the closure of Ui in some projective space. So wehave a rational map X → U i, since Ui is a dense open subset of X; this extends toa morphism ϕi : X → U i. Consider X →
∏U i and let ϕ : X → X be the closure of
the image of ϕ.
X // X //∏Ui
Ui?
OO
// U i?
OO
We claim that ϕ is an isomorphism, which would finish off the proof. To see that ϕ isinjective, suppose x, y ∈ X. By our choice of open cover, there exists Ui containingx and y. Therefore, ϕ(x) = ϕ(y) =⇒ x = y under the projection to Ui, so x = y.
Lemma 7.2.9. Let X be a variety and X → X its normalization. Then π issurjective.
Proof. By construction, it is enough to show this if X is affine. So assume thatA = O(X) ⊂ B = O(X) ⊂ k(X). Let x ∈ X correspond to the maximal idealmx ⊂ A. So it suffices to show that mxB is a proper ideal of B, since in this case
78
it is contained in a maximal ideal which maps to x. Recall a commutative algebraresult from earlier: if B is integral over A and finitely generated, then it is finiteover A. Then by Nakayama’s lemma, mxB 6= B.
Let X be the normalization of X. We next claim that ϕ is an open embedding,
so we may regard X ⊂ X as an open subset. So we have a rational map X → X,
which extends to a morphism since X is normal. By the universal property, this
factors through X → X, giving a map back X → X.
For x ∈ X, choose x′ ∈ X a lift of x; its image in X will map to x.
X // X ⊂ Ui
Ui?
OO
// U i
At the level of local rings, we have the diagram
OX,x
OX,ϕ(x)oo
OUi,xOU i,x
oo
OO
The map on the left is an isomorphism, as is the map on the bottom. The map onthe right is an injection, hence the map on top is an isomorphim. By a homeworkproblem, this is an isomorphism in some neighborhood of any point.
7.3 The Picard Group of Curve
Definition 7.3.1. Let ϕ : X → Y be a morphism of algebraic varieties. Then ϕ iscalled finite if for any affine open V ⊂ Y , U := ϕ−1(V ) is affine and O(U) is a finiteO(V ) module.
Proposition 7.3.2. Let ϕ : X → Y be a non-constant morphism of smooth projec-tive curves. Then ϕ is finite.
Proof. We first observe that ϕ is surjective since it is closed and nonconstant(projective varieties are complete). Let V ⊂ Y be an affine open. ConsiderA = O(V ) ⊂ k(Y ), and let B be the integral closure of A in k(X). Let U = Spec B;U is smooth since X is smooth. So we have a rational map U 99K X, which extendsto a morphsim.
We claim that ϕ−1(V ) = U . Once we show this, we are done since we know thatB is a finite A-algebra.
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To establish the claim, suppose for the sake of contradiction that there existsy0 ∈ V and x0 /∈ U such that ϕ(x0) = y0. Then there exists f ∈ k(X) with poles atx0, but which is a regular function on U . This is because ϕ−1(y0) = x0, x1, . . . , xnand we may choose f to be any function regular on x1, . . . , xn ∩ U but with poleat x0 (working in an affine open).
This means that f ∈ B, and B is integral over A, so
fn + a1fn−1 + . . .+ an = 0 ai ∈ O(V ) = A.
So f = −a1 − a2f − . . . −
anfn−1 . Since 1
f ∈ OX,x0 , we have f ∈ OX,x0 , which is acontradiction.
Definition 7.3.3. Let X be a curve. The degree map deg : DivX → Z is defined by∑nipi 7→
∑ni.
Theorem 7.3.4. Let X be a smooth projective curve. If D = (f) is principal, thendegD = 0.
Corollary 7.3.5. The degree map factors through the Picard group.
Definition 7.3.6. We define Pic0(X) = ker(deg : Pic(X)→ Z).
Proposition 7.3.7. Let ϕ : X → Y be a dominant morphism of smooth projectivecurves. Let D ∈ Div(Y ), ϕ∗D ∈ Div(X). Then degϕ∗D = degD degϕ, wheredegϕ = [k(X) : k(Y )].
Proof. It suffices to show this for D = p. Let ϕ−1(p) = p1, . . . , p` ⊂ X. Let y bea local parameter at p. In OX,pi , y = uix
rii , where ui ∈ O∗X,xi , and ϕ∗D =
∑ripi.
We wish to show that∑ri = degϕ.
Let V be an affine open containing p, A = O(V ), U = ϕ−1(V ), and B = O(U).Since ϕ is a finite map, U is also affine and B is a finite A-module. Let mp ⊂ A bethe maximal ideal corresponding to p. So OY,p ' Amp → Bmp , and Bmp is a finiteAmp-module. In fact, it is also a free Amp-module, say of rank n, since Amp is aPID and B is an integral domain which is a finite module over it (we are invokingthe structure theorem for finitely generated modules over a PID). Furthermore,n = degϕ, as is apparent from the fact that Bmp ⊗ k(Y ) = k(X). So Bmp/mpBmp isan n-dimensional k-vector space.
We claim that Bmp =⋂
OX,pi .
Bmp = fg| f ∈ B, g ∈ O(V ), g(p) 6= 0.
Now the inclusion ⊂ is obvious, since the lift of a regular function at p will be aregular function at all of the pi. On the other hand, suppose h ∈
⋂OX,pi . So
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h is regular away from q1, . . . , qm, and p ∈ W := V − ϕ(qj) ⊂ V . So h ∈O(U)⊗O(V ) O(W ) ⊂ Bmp .
Therefore,
Bmp/(y) '∏
OX,pi/(xrii ).
This implies that
n = dimk Bmp/(y) =∑
dimk OX,pi/(xrii ) =
∑ri.
Proof of 7.3.4. Let f be any nonconstant rational function on k(X), which is equiv-alent to a dominant rational map X 99K P1. So (f) = ϕ∗([0]− [∞]), pretty much bydefinition. So the proposition implies that deg(f) = degϕdeg([0]− [∞]) = 0.
Corollary 7.3.8. Let X be a smooth projective curve. If there exist x, y ∈ X suchthat x− y ∈ Div0(X), then X ' P1.
Proof. If x − y = (f), then f defines a rational map which extends to a morphismϕ : X → P1. So x = ϕ∗([0]), so degϕ = 1, implying that ϕ is an isomorphism.
Example 7.3.9. (i) Pic0(P1) = 0; we already established that the degree induces anisomorphism with Pic(P1) and Z.
7.4 Jacobians
Example 7.4.1. (ii) Let X be the elliptic curve y2z = x3 − xz2. Let p0 = (0, 1, 0)and LX(1) the linear system of hyperplanes. The hyperplane z = 0 is a divisor inLX(1). We can see that Hz ∩X = p0. Around p0, the curve is given by z = x3−xz2
and x is a local parameter. Therefore, the order of z at p0 is at least 3. Thenordp0(xz2) ≥ 7, implying that ord(x3 − xz2) = 3. The conclusion is that 3p0 is inthe hyperplane class.
Now define a map AJ : X → Pic0(X) sending p 7→ p− p0. We claim that this isinjective:
p− p0 ∼ q − p0 =⇒ p ∼ q =⇒ p = q
since X 6' P1.We next claim that AJ is surjective. Let D ∈ Div(X), degD = 0. Write
D =∑nipi with
∑ni = 0; then we may also write D =
∑ni(p− p0).
Now we claim that any such divisor is linearly equivalent to a divisor of thisform where ni ≥ 0. Given x ∈ X, xp0 ∩ X = x, p0, y. So x + p0 + y ∼ 3p0, sox− p0 ∼ p0 − y. While
∑ni > 1, D is of the form
D = (p1 − p0) + (p2 − p0) + . . .
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Writing p1p2∩X = p1, p2, q, so we can replace (p1−p0)+(p2−p0) ∼ p0−q ∼ q′−p0.The conclusion is that X is bijective with Pic0(X); this endows X with a group
structure where p0 is the identity element and the group law is defined by thehyperplane class is 0.
Theorem 7.4.2. Let X be an elliptic curve. The multiplication map m : X ×X →X and the inverse s : X → X are morphisms of algebraic varieties.
Proof. Geometrically, the inverse is obtained by intersecting with the vertical linethrough that point. The group law can also be expressed geometrically by inter-secting with lines. This can all be defined by rational functions in terms of thecoordinates.
Example 7.4.3. Let’s compute the inverse map. Let (a, b) ∈ X. The line through xand p0 is given by az− x. The third intersection point is given by intersecting withy2z = x3 + xz2, which has the solutions (0, 1, 0), (a, b, 1), (a,−b, 1). So the inversemap sends (a, b) 7→ (a,−b).Definition 7.4.4. A group variety G is an algebraic variety together an identity e ∈ Gand morphisms m : G×G→ G and s : G→ G, such that (G, e,m, s) form a groupin the usual sense. A projective group variety is called an abelian variety.
Corollary 7.4.5. The variety X defined by y2z = x3 − xz2 is an abelian variety.
Remark 7.4.6. Assume char k 6= 2, 3. Then in fact, every genus one smooth projec-tive curve is isomorphic to a cubic curve in P2 of the form y2z = x3 + axz2 + bz3,−4a2 − 27b3 6= 0. By the same argument, X is an abelian variety.
Suppose k = C. The regular differential on C is ω = dxy = dx√
x3+ax+b. People
wanted to understand this integral; they realized that one needed to pass to aRiemann surface, which was the elliptic curve. Integrating this form on an ellipticcurve X from some initial point gives a map from X to C, which is not well-definedbecause X is not simply-connected. So we need to mod out by the integral overgenerators for the first homology of the elliptic curve.
Example 7.4.7. The following are affine group varieties, called linear algebraic groups.
• GLn = A ∈Mn×n | detA 6= 0.
• PGLn = GLn/k×In.
• Gm = GL1 = (A1 − 0,×).
• Ga ' (A1, 0).
Mathematicians may not always choose to best terminology, but the following the-orem shows that they are at least somewhat reasonable.
Theorem 7.4.8. Every abelian variety is abelian.
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Theorem 7.4.9. If X is a smooth projective curve of genus g, then Pic0(X) isnaturally an abelian variety of dimension g.
Remark 7.4.10. In fact, every algebraic group is built from an abelian variety andan algebraic group. That is, for any algebraic group G there is an exact sequence
1→ N → G→ A→ 1,
where N is a linear algebraic group and A is an abelian variety.
7.5 The Riemann-Roch Theorem
Theorem 7.5.1 (Riemann-Roch). Let X be a smooth projective curve of genus gover k, D a divisor on X. Then
`(D)− `(KX −D) = degD + 1− g.
Corollary 7.5.2. degKX = 2g − 2.
Proof. Let D = KX in the theorem, and use `(0) = 1, `(KX) = g.
Corollary 7.5.3. If degD ≥ 2g − 1, then
`(D) = 1− g + degD.
Proof. deg(KX −D) < 0. If `(KX −D) > 0, then KX −D is linearly equivalent toan effective divisor, but it has negative degree, which is impossible.
Corollary 7.5.4. If degD ≥ 2g, then `(D − p) = `(D)− 1.
Corollary 7.5.5. If degD ≥ 2g, then |D| has no base points, hence the rationalmap X 99K |D|∗ extends to a morphism.
Theorem 7.5.6. Let D be a divisor on X. If for all p, q ∈ X, `(D−p−q) = `(D)−2,then the rational map Z|D| : X → |D|∗ is a closed embedding.
Proof. Note that this condition certainly implies that `(D − p) = `(D) − 1, since`(D) − `(D − p) ≤ 1. So |D| has no base points: the rational map extends to amorphism.
Next, note that ϕ is injective, since
L (D − p− q) ⊂ L (D − p) ⊂ L (D)
are all strict inclusions. So there exists f ∈ L (D − p) \ `(D − p − q), i.e. p ∈Supp((f)+D) but q /∈ Supp((f)+D). This is a hyperplane section passing throughthe image of p but not q, so their images are distinct.
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Let X ′ = ϕ(X) ⊂ |D|∗. We need to show that ϕ : X → X ′ is an isomorphism.X → X ′ is a finite morphism, and X ′ → |D|∗ is a closed embedding, hence a finitemorphism. So for an affine open V ⊂ |D|∗, the pullback U is an affine open, andB = O(U) is finite over A = O(V ).
V ∩X ′ = x ∈ V | f(x) = 0 if ϕ∗f = 0 in B.
So O(V ∩X ′) = Im(A → B) and I(V ∩X ′) = ker(A → B). Let p ∈ X; then OX,p
is finite over OX,ϕ(p).Key claim: mX′,ϕ(p) → mX,p/m
2X,p is surjective. To establish the claim, it suffices
to show thatT ∗ϕ(p)|D|
∗ → T ∗ϕ(p)X′ → T ∗pX
is surjective. Recall how this map is defined: we choose (f0, . . . , fn) a basis ofL (D). Then ϕ(p) = (f0(p), . . . , fn(p)). Let x0, . . . , xn be coordinates on |D|∗,ϕ(p) = (1, 0, . . . , 0). Then
ϕ∗(xix0
)=fif0∈ mX,p.
Surjectivity means that there exists i such that fif0/∈ m2
X,p. Because L (D − 2p) 6=L (D − p), there exists f ∈ L (D − p) \L (D − 2p).
Let n = mX′,ϕ(p)OX,p, so n ⊂ mX,p. But from the claim, n contains a localparameter of X at p, hence n = mX,p. Now consider the map of OX′,ϕ(p)-modules
OX′,ϕ(p) → OX,p → N → 0.
Tensoring by OX′,ϕ(p)/mX′,ϕ(p),
OX′,ϕ(p)/mX′,ϕ(p)//
≈
OX,p/mX′,ϕ(p)OX,p//
≈
N/mX′,ϕ(p)//
≈
0
k // k // 0 // 0
So by Nakayama’s lemma N = 0, i.e. OX′,ϕ(p) → OX,p is a surjection of OX′,ϕ(p)-modules, hence an isomorphism.
Corollary 7.5.7. If degD ≥ 2g + 1, then Z|D| : X → |D|∗ ' PdegD−g is a closedembedding.
Corollary 7.5.8. Let g ≥ 2. Then Z|2KX | is a closed embedding (to P3g−4).
What about the linear system Z|KX | : X 99K Pg−1? (Assume char k = 0.)
Example 7.5.9. g = 2. This gives a map ϕ : X → P1 which is separable. Weknow that KX − ϕ∗KP1 is the branch divisor, and ϕ∗([0]) = KX . Hence degϕ =degKx = 2g − 2, and degB = 2− 2(−2) = 6. So [k(X) : k(P1)] = 2, implying thatk(X) = k(x)[y]/(y2 − f(x)).
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Exercise 7.5.10. Show that the support of the branch divisor B are the points overthe the zeros of f(x) of odd order, and the point over ∞ if deg f is odd.
Definition 7.5.11. A curve X is called hyperelliptic if there exists ϕ : X → P1 ofdegree 2.
Corollary 7.5.12. Every genus 2 curve is hyperelliptic.
Example 7.5.13. Let g ≥ 3. If X is not hyperelliptic, then Z|Kx| : X 99K Pg−1 is aclosed embedding.
Example 7.5.14. Let ϕ : X → Y be separable of degree n. The branch divisor isB := KX − ϕ∗KY , and taking degrees we have
2gX − 2 = n(2gY − 2) + degB.
Let x ∈ X, y = ϕ(x) ∈ Y . Let tx, ty be the local parameters, ϕ∗ty = texu, forsome u ∈ O∗X,x.
dϕ∗tydtx
= ete−1x u+ tex
du
dtx.
Definition 7.5.15. ϕ is tame at x if char k = p - e,
So if ϕ is tame at x, then ordx
(dϕ∗tydtx
)= e − 1. If ϕ is tame everywhere, (e.g. in
characteristic 0 or if deg n < p), then the formula above tells us that
B =∑x∈X
(ex − 1)x =⇒ degB =∑x∈X
(ex − 1).
This is called the Riemann-Hurwitz formula.
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