An introduction to Algebraic Geometry · 10 CHAPTER 2. ALGEBRAIC VARIETIES AND REGULAR MAPS Example 2.2.2. Projective space Pn with the euclidean (classical) topology is reducible

An introduction to Algebraic Geometry

Kieran G. O’Grady

12th April 2019

2

Contents

1 Introduction 51.1 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Algebraic varieties and regular maps 72.1 The Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Decomposition into irreducibles . . . . . . . . . . . . . . . . . . . 92.3 Regular maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Regular functions on affine varieties . . . . . . . . . . . . . . . . 162.5 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6 Elimination theory . . . . . . . . . . . . . . . . . . . . . . . . . . 232.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 Rational maps and dimension 293.1 Rational maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 The field of rational functions . . . . . . . . . . . . . . . . . . . . 313.3 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4 Maps of finite degree . . . . . . . . . . . . . . . . . . . . . . . . . 363.5 Grassmannians . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.6 Degree of a closed subset of Pn . . . . . . . . . . . . . . . . . . . 453.7 Intersecting closed subsets of a projective space . . . . . . . . . . 483.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 First order approximation 574.1 Tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.2 Cotangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.3 Smooth points of quasi projective varieties . . . . . . . . . . . . . 644.4 Transverse Bezout . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5 Regular maps and their fibers 695.1 Sard’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.2 Dimensions of fibers . . . . . . . . . . . . . . . . . . . . . . . . . 745.3 Cubic surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.4 Local invertibility of regular maps . . . . . . . . . . . . . . . . . 765.5 Local factoriality . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3

4 CONTENTS

6 Holomorphic maps: local theory 796.1 Holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . 796.2 Holomorphic maps . . . . . . . . . . . . . . . . . . . . . . . . . . 826.3 Complex valued differential forms . . . . . . . . . . . . . . . . . . 836.4 Cauchy’s integral formula . . . . . . . . . . . . . . . . . . . . . . 846.5 Holomorphic functions are analytic . . . . . . . . . . . . . . . . . 876.6 A version of Hartogs’ Theorem . . . . . . . . . . . . . . . . . . . 89

7 Complex manifolds 917.1 Definition of complex manifold . . . . . . . . . . . . . . . . . . . 917.2 The category of complex manifolds . . . . . . . . . . . . . . . . . 937.3 Other examples of complex manifolds . . . . . . . . . . . . . . . 957.4 The holomorphic tangent space . . . . . . . . . . . . . . . . . . . 957.5 Differential forms on complex manifolds . . . . . . . . . . . . . . 977.6 Holomorphic differential forms . . . . . . . . . . . . . . . . . . . 997.7 The Dolbeault-Grothendieck Lemma . . . . . . . . . . . . . . . . 102

8 Vector bundles and sheaves 1098.1 Holomorphic vector bundles . . . . . . . . . . . . . . . . . . . . . 109

9 Hermitian geometry 1119.1 Hermitian metrics on complex manifolds . . . . . . . . . . . . . . 1119.2 Kahler manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

10 The Hodge decomposition 11310.1 Statement of the Hodge decomposition Theorem . . . . . . . . . 11310.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

11 Harmonic forms 115

12 Kahler identities 117

A Commutative algebra a la carte 119A.1 Noetherian rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 119A.2 The Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . 121A.3 Unique factorization . . . . . . . . . . . . . . . . . . . . . . . . . 123A.4 Extensions of fields . . . . . . . . . . . . . . . . . . . . . . . . . . 123A.5 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124A.6 Order of vanishing . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Chapter 1

Introduction

1.1 Conventions

K is an algebraically closed field.

5

6 CHAPTER 1. INTRODUCTION

Chapter 2

Algebraic varieties andregular maps

2.1 The Zariski topology

Let KrZ0, . . . , Znsd Ă KrZ0, . . . , Zns be the degree-d subspace of the algebra ofpolynomials. If F P KrZ0, . . . , Znsd, and Z P Kn`1, then F pZq “ 0 if and onlyif F pλZq “ 0 for every λ P K˚, because F pλZq “ λdF pZq. Hence, althoughF pxq is not defined, it makes to state F pxq “ 0 or F pxq “ 0 for a point x P Pn.Let Fi P KrZ0, . . . , Znsdi for i P t1, . . . , ru; we let

V pF1, . . . , Frq :“ tx P Pn | F1pxq “ . . . “ Frpxq “ 0u.

Example 2.1.1. A subset X Ă Pn is a hypersurface if it is equal to V pF q, whereF is a non constant homogeneous polynomial.

Let I Ă KrZ0, . . . , Zns be a homogeneous ideal, i.e. such that

I “8à

d“0

I XKrZ0, . . . , Znsd. (2.1.1)

We let

V pIq :“ tx P Pn | F pxq “ 0 @ homogeneous F P Iu.

If I is generated by homogeneous polynomials F1, . . . , Fr, then V pIq “ V pF1, . . . , Frq,and hence V pIq is a projective variety. Conversely, by Hilbert’s basis Theorem,i.e. Theorem A.1.6, a homogeneous ideal I is generated by homogeneous poly-nomials F1, . . . , Fr, and hence V pIq is a projective variety.

Corollary 2.1.2. The collection of subsets V pIq Ă Pn, where I runs throughthe collection of homogeneous ideals of KrZ0, . . . , Zns, satisfies the axioms forthe closed subsets of a topological space.

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8 CHAPTER 2. ALGEBRAIC VARIETIES AND REGULAR MAPS

Proof. We have H “ V pp1qq, Pn “ V pp0qq. If I, J are homogeneous ideal, thenI X J is a homogeneous ideal of KrZ0, . . . , Zns, and V pIq Y V pJq “ V pI X Jq.If tItutPT is a family of homogeneous ideals of KrZ0, . . . , Zns, then

č

tPT

V pItq “ V pxtItutPT yq,

where xtItutPT y is the ideal generated by the collection of the It’s (it is a homo-geneous ideal).

Definition 2.1.3. The Zariski topology of Pn is the topology whose closed setsare the sets V pIq Ă Pn, where I runs through the collection of homogeneousideals of KrZ0, . . . , Zns. The Zariski topology of a subset A Ă Pn is the topologyinduced by the Zariski topology of Pn.

Example 2.1.4. We will always identify An with the open subset pPnzV pZ0qq Ă

Pn. Thus An has a Zariski topology, that we describe below. Let J Ă Krz1, . . . , znsbe an an ideal, in general not homogeneous. We let

V pJq :“ tz P An | fpzq “ 0 @f P Ju. (2.1.2)

The notation is potentially confusing, but it will always be clear form the con-text whether V pJq is a subset of a projective space or of an affine space. Asubet X Ă An is closed if and only if there exist an ideal J Ă Krz1, . . . , zns suchthat X “ V pJq. In fact, if X is closed, say X “ pPnzV pZ0qq X V pF1, . . . , Frq,where Fj Ă KrZ0, Z1, . . . , Zns are homogeneous, then X “ V pf1, . . . , frq, wherefjp, z1, . . . , znq :“ F p1, z1, . . . , znq. Conversely, letX Ă An be equal to V pf1, . . . , frq(notice that by Hilbert’s basis Theorem, every ideal J is finitely generated, andif J “ pf1, . . . , frq, then V pJq “ V pf1, . . . , frq). For j P t1, . . . , ru, let dj be thedegree of fj . Then

FjpZ0, . . . , Znq :“ Zdj0 f

ˆ

Z1

Z0, . . . ,

ZnZ0

˙

is a homogegenous polynomial of degree dj . Thus V pF1, . . . , Frq Ă Pn is closed,and we have X “ pPnzV pZ0qq X V pF1, . . . , Frq.

Remark 2.1.5. It is natural to study subsets of An defined by polynomial equa-tions, i.e. Zariski closed subsets, see Example 2.1.4. One of the advantages ofconsidering closed subsets of projective space is that they are compact. Com-pactness of a closed X Ă Pn may be understood for the moment being to becompactness in the classical topology.

Notice that the Zariski topology is weaker than the classical topology of Pn.In fact, unless n “ 0, the Zariski is much weaker than the classical topology, inparticular it is not Hausdorff.

Given a subset A Ă Pn, let

IpAq :“ xF P KrZ0, . . . , Zns | F is homogeneous and F ppq “ 0 for all p P Ay,(2.1.3)

where x, y means “the ideal generated by”. Clearly IpAq is a homogeneous idealof KrZ0, . . . , Zns, and V pIpAqq is the closure of A in the Zariski topology.

2.2. DECOMPOSITION INTO IRREDUCIBLES 9

Definition 2.1.6. A quasi-projective variety is a Zariski locally closed subsetof a projective space, i.e. X Ă Pn such that X “ U X Y , where U, Y Ă Pn areZariski open and Zariski closed respectively.

Example 2.1.7. By Example 2.1.4, every subset V pJq Ă An, where J Ă

Krz1, . . . , zns is an ideal, is a quasi projective variety.

Definition 2.1.8. Let X Ă Pn be a closed subset. A principal open subset ofX is an open U Ă X which is equal to

XF :“ XzV pF q,

where F P KrZ0, . . . , Zns is a homogeneous polynomial of strictly positive degree.In general, if X Ă Pn is locally closed, a principal open subset of X is an openU Ă X which is equal to XF , for a homogeneous polynomial F P KrZ0, . . . , Znsof strictly positive degree.

Claim 2.1.9. Let X Ă Pn be locally closed. The collection of principal opensubsets of X is a basis of the Zariski topology of X.

Proof. Let U Ă X be open. Then U is open in X. Hence it suffices to provethe claim for X closed. We have U “ XzW , where W is closed. Let W “ V pIq,where I Ă KrZ0, . . . , Zns is a homogeneous ideal. Let J Ă KrZ0, . . . , Zns bethe homogeneous ideal generated by all products F ¨ Zi, where F P I, andi P t0, . . . , nu. Then V pJq “ V pIq “ W , and J is generated by a non emptyfinite set of homogeneous polynomials F1, . . . , Fr. Then

U “ XzV pF1, . . . , Frq “ XF1YXF2

Y . . .YXFr .

Remark 2.1.10. If V is a finite dimensional complex vector space, the Zariskitopology on PpV q is defined by imitating what was done for Pn: one associatesto a homogeneous ideal I Ă SymV _ the set of zeroes V pIq, etc. Similarlyone defines the Zariski topology on a finite dimensional complex affine space.Everything that we do in the present chapter applies to this situation, but forthe sake of concreteness we formulate it for Pn and An.

2.2 Decomposition into irreducibles

A proper closed subset X Ă P1 (or X Ă A1) is a finite set of points. In general,a quasi projective variety is a finite union of closed subsets which are irreducible,i.e. are not the union of proper closed subsets. In order to formulate the relevantresult, we give a few definitions.

Definition 2.2.1. Let X be a topological space. We say that X is reducibleif either X “ H or there exist proper closed subsets Y,W Ă X such thatX “ Y YW . We say that X is irreducible if it is not reducible.


Example 2.2.2. Projective space Pn with the euclidean (classical) topology isreducible except if n “ 0. On the other hand, Pn with the Zariski topologyis irreducible for any n. In fact suppose that Pn “ Y Y W with Y and Wproper closed subsets. Then there exist F P IpY q such that F ppq “ 0 for one(at least) p PW and g P IpW q such that gpqq ‰ 0 for one (at least) q P Y . Thenfg “ 0 because Pn “ Y YW ; that is a contradiction because KrZ0, . . . , Zns isan integral domain.

Definition 2.2.3. Let X be a topological space. An irreducible decompositionof X consists of a decomposition (possibly empty)

X “ X1 Y ¨ ¨ ¨ YXr (2.2.1)

where each Xi is a closed irreducible subset of X (irreducible with respect tothe induced topology) and moreover Xi Ć Xj for all i ‰ j.

We will prove the following result.

Theorem 2.2.4. Let A Ă Pn with the (induced) Zariski topology. Then Aadmits an irreducible decomposition, and such a decomposition is unique up toreordering of components.

The key step in the proof of Theorem 2.2.4 is the following remarkableconsequence of Hilbert’s basis Theorem (i.e. Theorem A.1.6).

Proposition 2.2.5. Let A Ă Pn, and let A Ą X0 Ą X1 Ą . . . Ą Xm Ą . . . be adescending chain of Zariski closed subsets of A, i.e Xm Ą Xm`1 for all m P N.Then the chain is stationary, i.e. there exists m0 P N such that Xm “ Xm0

form ě m0.

Proof. Let Xi be the closure of Xi in Pn. Then Xi “ A X Xi, because Xi isclosed in A. Hence we may replace Xi by Xi, or equivalently we may supposethat the Xi are closed in Pn. Let Im “ IpXmq. Then I0 Ă I1 Ă . . . Ă Im Ă . . .is an ascending chain of (homogeneous) ideals of KrZ0, . . . , Zns. By Hilbert’sbasis Theorem and Lemma A.1.3 the ascending chain of ideals is stationary,i.e. there exists m0 P N such that Im0 “ Im for m ě m0. Thus Xm0 “ V pIm0q “

V pImq “ Xm for m ě m0.

Proof of Theorem 2.2.4. If A is empty, then it is the empty union (of irre-ducibles). . Next, suppose that A is not empty and that it does not admit anirreducible decomposition; we will arrive at a contradiction. First A in redu-cible, i.e. A “ X0YW0 with X0,W0 Ă A proper closed subsets. If both X0 andW0 have an irreducible decomposition, then A is the union of the irreduciblecomponents of X0 and W0, contradicting the assumption that A does not admitan irreducible decomposition. Hence one of X0, W0, say X0, does not have anirreducible decomposition. In particular X0 is reducible. Thus X0 “ X1 YW1

with X1,W1 Ă X0 proper closed subsets, and arguing as above, one of X1,W1,say X1, does not admit a decomposition into irredicbles. Iterating, we get astrictly descending chain of closed subsets

A Ľ X0 Ľ X1 Ľ ¨ ¨ ¨ Ľ Xm Ľ Xm`1 Ľ ¨ ¨ ¨

2.2. DECOMPOSITION INTO IRREDUCIBLES 11

This contradicts Proposition 2.2.5. This proves that X has a decompositioninto irreducibles X “ X1 Y . . .YXr.

By discarding Xi’s which are contained in Xj with i “ j, we may assumethat if i “ j, then Xi is not contained in Xj .

Lastly, let us prove that such a decomposition is unique up to reordering,by induction on r. The case r “ 1 is trivially true. Let r ě 2. Suppose thatX “ Y1 Y . . . Y Ys, where each Yj is Zariski closed irreducible, and Yj Ć Ykif j “ k. Since Ys is irreducible, there exists i such that Ys Ă Xi. We mayassume that i “ r. By the same argument, there exists j such that Xr Ă Yj .Thus Ys Ă Xr Ă Yj . It follows that j “ s, and hence Ys “ Xr. It follows thatX1 Y . . . YXr´1 “ Y1 Y . . . Y Ys´1, and hence the decomposition is unique upto reordering by the inductive hypothesis.

Definition 2.2.6. Let X be a quasi projective variety, and let

X “ X1 Y . . .YXr

be an irreducible decomposition of X. The Xi’s are the irreducible componentsof X (this makes sense because, by Theorem 2.2.4, the collection of the Xi’sis uniquely determined by X).

We notice the following consequence of Proposition 2.2.5.

Corollary 2.2.7. A quasi projective variety X (with the Zariski topology) isquasi compact, i.e. every open covering of X has a finite subcover.

The the following result makes a connection between irreducibility and al-gebra.

Proposition 2.2.8. A subset X Ă Pn is irreducible if and only if IpXq is aprime ideal.

Proof. The proof has essentially been given in Example 2.2.2. Suppose thatX is irreducible. In particular X “ H (by definition), and hence IpXq is aproper ideal of KrZ0, . . . , Zns. We must prove that KrZ0, . . . , ZnsIpXq is anintegral domain. Suppose the contrary. Then there exist

F,G P pKrZ0, . . . , ZnszIpXqq (2.2.2)

such thatF ¨G P IpXq. (2.2.3)

By (2.2.3), we have X “ pX X V pF qq Y pX X V pGqq, and both X X V pF q,X X V pGq are proper closed subsets of X by (2.2.2). This proves that if X isirreducible, then IpXq is a prime ideal.

Next, assume that X is reducible; we must prove that IpXq is not prime.If X “ H, then IpXq “ KrZ0, . . . , Zns and hence IpXq is not prime. Thus wemay assume that X “ H, and hence there exist proper closed subset Y,W Ă Xsuch that X “ Y YW . Since Y ĆW and W Ć Y , there exist F P pIpY qzIpW qqand G P pIpW qzIpY qq. It follows that both (2.2.2) and (2.2.3) hold, and henceIpXq is not prime.


Remark 2.2.9. Let I :“ pZ20 q Ă KrZ0, Z1s. Then V pIq “ tr0, 1su is irreducible

although I is not prime. Of course IpV pIqq is prime, it equals pZ0q.

Remark 2.2.10. Let X Ă An. Let IpXq Ă Krz1, . . . , zns be the ideal of polyno-mials vanishing on X. Then X is irreducible if and only if IpXq is a prime ideal.The proof is analogous to the proof of Proposition 2.2.8. One may also dir-ectly relate IpXq with the ideal J Ă KrZ0, . . . , Zns generated by homogeneouspolynomials vanishing on X (as subset of Pn), and argue that IpXq is prime ifand only if J is.

Example 2.2.11. Let V pF q Ă Pn be a hypersurface, and let F1, . . . , Fr be thedistinct prime factors of the decomposition of F into a products of primes (recallthat KrZ0, . . . , Zns is a UFD, by Corollary A.3.2). The irreducible decom-position of V pF q is

V pF q “ V pF1q Y . . .Y V pFrq.

In fact, each V pFiq is irreducible by Proposition 2.2.8. What is not obviousis that V pFiqno Ă V pFjq if Fi, Fj are non associated primes. This followsfrom Hilbert’s Nullstellensatz, i.e. Theorem A.2.1 (or by a simpler argumentinvolving only unique factorization in the ring of polynomials).

2.3 Regular maps

Definition 2.3.1. Let X Ă Pn and Y Ă Pm be quasi projective varieties. Amap f : X Ñ Y is regular at a P X if there exist an open U Ă X containing aand F0, . . . , Fm P KrZ0, . . . , Znsd such that for all rZs P U pF0pZq, . . . , FmpZqq ‰p0, . . . , 0q, and

fprZsq “ rF0pZq, . . . , FmpZqs. (2.3.1)

The map f is regular if it is regular at each point of X.

The identity map of a quasi projective variety is regular (choose FjpZq “ Zj).If f : X Ñ Y and g : Y ÑW are regular maps of quasi projective varieties, thecomposition g ˝ f : X Ñ W is regular, because the composition of polynomialfunctions is a polynomial function. Thus we have the category of quasi project-ive varieties. In particular we have the notion of isomorphism between quasiprojective varieties.

Example 2.3.2. Let X Ă An be a locally closed subset (recall that An “ PnZ0).

Then f : X Ñ Pm is a regular map if and only if, given any a P X, there existf0, . . . , fm P Krz1, . . . , zns (in general not homogeneous) such that on an opensubset U Ă X containing a we have

fpzq “ rf0pzq, . . . , fmpzqs. (2.3.2)

(This includes the statement that V pf1, . . . , fmq X U “ H.) In fact, if f is reg-ular there exist homogeneous F0, . . . , Fm P KrZ0, . . . , Znsd such that fpr1, zsq “

2.3. REGULAR MAPS 13

rF0p1, zq, . . . , Fmp1, zqs, and it suffices to let fjpzq :“ Fjp1, zq. Conversley,if (2.3.4) holds, then

fprZ0, Z1, . . . , Znsq “ rZd0 , Z

d0f1

ˆ

Z1

Z0, . . . ,

ZnZ0

˙

, . . . , Zd0fm

ˆ

Z1

Z0, . . . ,

ZnZ0

˙

s,

(2.3.3)and for d is large enough, each of the rational functions appearing in (2.3.3) isactually a homogeneous polynomial of degree d.

Example 2.3.3. Let X Ă An be a locally closed subset and let f : X Ñ Pm be amap such that fpXq Ă PmT0

(we let rT0, . . . , Tms be homogeneous coordinates onPm). Then f is regular if and only if locally there exist f0, . . . , fm P Krz1, . . . , zns(in general not homogeneous) such that, in affine coordinates pT1

T0, . . . , TmT0

q, wehave

fpzq “

ˆ

f1pzq

f0pzq, . . . ,

fmpzq

f0pzq

˙

. (2.3.4)

Example 2.3.4. Let f P Krz1, . . . , zns. Let Y :“ V pfpz1, . . . , znq ¨ zn`1 ´ 1q ĂAn`1. The map

AnzV pfq ÝÑ Ypz1, . . . , znq ÞÑ pz1, . . . , zn,

1fpz1,...,znq

q

is an isomorphism.

Example 2.3.5. Let

Cn “"

rξ0, . . . , ξns P Pn | rk

ˆ

ξ0 ξ1 ¨ ¨ ¨ ξn´1

ξ1 ξ2 ¨ ¨ ¨ ξn

˙

ď 1

*

. (2.3.5)

Since a matrix has rank at most 1 if and only if all the determinants of its 2ˆ 2minors vanish it follows that Kn is closed. We have a regular map

P1 ϕnÝÑ Kn

rs, ts ÞÑ rsn, sn´1t, . . . , tns(2.3.6)

Let us prove that ϕn is an isomorphism. Let ψn : Cn Ñ P1 be defined as follows:

ψn prξ0, . . . , ξnsq “

#

rξ0, ξ1s if rξ0, . . . , ξns P Cn X Pnξ0rξn´1, ξns if rξ0, . . . , ξns P Cn X Pnξn

Of course one has to check that the two expressions coincide for points in KnXPnξ0 X Pnξn : from (2.3.5) we get that ξ0 ¨ ξn ´ ξ1ξn´1 vanishes on Kn and thisshows the required compatibility. One checks easily that ψd ˝ ϕn “ IdP1 andϕn ˝ ψn “ IdKn ; thus ϕn defines an isomorphism P1 „

ÝÑ Kn.Unless we are in the trivial case n “ 1, it is not possible to define ψn globally

as

ψn prξ0, . . . , ξnsq “ rP pξ0, . . . , ξnq, Qpξ0, . . . , ξnqs, (2.3.7)


with P,Q P Krξ0, . . . , ξnse not vanishing simultaneously on Kn. In fact supposethat (2.3.7) holds, and let

pps, tq :“ P psn, . . . , tnq, qps, tq :“ Qpsn, . . . , tnq.

The polynomials pps, tq, qps, tq are homogeneous of degree de, they do not vanishsimultaneously on a non zero ps0, t0q, and forall rs, ts P P1 we have rpps, tq, qps, tqs “rs, ts. It follows that pps, tq “ s¨rps, tq and qps, tq “ t¨rps, tq, where rps, tq has nonon trivial zeroes, i.e. rps, tq is constant. In particular de “ deg p “ deg q “ 1,and hence d “ 1.

Example 2.3.6. We recall the formula

dimKrZ0, . . . , Znsd “

ˆ

d` n

n

˙

. (2.3.8)

(See Exercise 2.7.1 for a proof.) Let

Pn νndÝÑ Pp

d`nn q´1

rZs ÞÑ rZd0 , Zd´10 Z1, . . . , Z

dns

(2.3.9)

be defined by all homogeneous monomials of degree d - this is a Veronese map.Clearly νnd is regular. Let V n

d :“ Im νnd . Let us show that V nd is Zariski closed,

and that the map Pn Ñ V nd defined by νnd is an isomorphism. (Notice that

if n “ 1, this has been proved in Example 2.3.5.) The variety V nd is the

corresponding Veronese variety.

First1, let

PpKrT0, . . . , Tns1qµndÝÑ PpKrT0, . . . , Tnsdq

rLs ÞÑ rLds(2.3.10)

and let W nd “ Impµnd q. The above map can be identified with the Veronese

map νnd . In fact, writing L P KrT0, . . . , Tns1 as L “řni“0 αiTi, we see that

rα0, . . . , αns are coordinates on PpKrT0, . . . , Tns1q, and they give an identifica-tion Pn „

ÝÑ PpKrT0, . . . , Tns1q. Moreover, let

Ppd`nn q´1 „

ÝÑ PpKrT0, . . . , Tnsdq,r. . . , ξI , . . .s ÞÑ

ř

I“pi0,...,inqi0`...ìn“d

d!i0!¨...ïn!ξIT

I

where T I “ T i00 ¨. . .¨Tinn . By the Newton formula p

řni“0 αiTiq

d “ř

I

d!i0!¨...ïn!α

IT I ,

we see that, modulo the above isomorphisms, the Veronese map νnd is identifiedwith µnd , and hence V n

d is identified with W nd .

1This is ok if charK “ 0, or charK ą d, not in general (we have denominators which areproducts of integers smaller or equal than d).

2.3. REGULAR MAPS 15

Now let us show that W nd is closed. The key observation is that rF s P W n

d ifand only if BF

BZ0, . . . , BF

BZnspan a 1-dimensional subspace of KrZ0, . . . , Zns. This

may be proved by induction on degF and Euler’s identity

nÿ

j“0

ZjBF

BZj“ pdegF q ¨ F, (2.3.11)

valid for F homogeneous. Now, the condition that BFBZ0

, . . . , BFBZn

span a 1-dimensional subspace of KrZ0, . . . , Zns is equivalent to the vanishing of determ-inants of all 2 ˆ 2 minors of the matrix whose entries are the coordinates ofBFBZ0

, . . . , BFBZn

; thus W nd is closed.

In order to show that µnd is an isomorphism, we notice that if F “ Ld,where L P PpKrT0, . . . , Tns1 is non zero, then for each i P t0, . . . , nu the partial

derivative Bn´1FBZn´1

i

is a multiple of L (eventually equal to 0 if BLBZi

“ 0), and that

one at least of such pn´ 1q-th partial derivative is non zero. Thus, the inverseof µnd is the regular map θnd : W n

d ÝÑ PpKrT0, . . . , Tns1q defined by

θnd prF sq :“

$

’

’

&

’

’

%

r Bn´1FBZn´1

0

s if Bn´1FBZn´1

0

“ 0,

¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨

r Bn´1FBZn´1

ns if B

n´1FBZn´1

n“ 0.

(2.3.12)

Proposition 2.3.7. A regular map of quasi projective varieties is Zariski con-tinuous.

Proof. Let X Ă Pn and Y Ă Pm be Zariski locally closed, and let f : X Ñ Ybe a regular map. We must prove that if C Ă Y is Zariski closed, then f´1C isZariski closed in X. Let U ĂW be an open subset such that (2.3.1) holds. Letus show that φ´1C X U is closed in U . Since C is closed C “ V pIq X Y whereI Ă KrT0, . . . , Tms is a homogeneous ideal. Thus

φ´1C X U “ trZs P U | P pF0pZq, . . . , FmpZqq “ 0 @P P Iu.

Since each P pF0pZq, . . . , FmpZqq is a homogeneous polynomial, we get thatφ´1C X U is closed in U .

By definition of regular map X can be covered by Zariski open sets Uα suchthat (2.3.1) holds with U replaced by Uα. We have proved that Cα :“ φ´1CXUαis closed in Uα for all α. It follows that φ´1C is closed. In fact let Cα Ă X bethe closure of Cα and Dα :“ XzUα. Since Cα is closed in Uα we have

Cα X Uα “ Cα “ φ´1C X Uα. (2.3.13)

Moreover Dα is closed in X because Uα is open. By (2.3.13) we have

φ´1C “č

α

`

Cα YDα

˘

.

Thus φ´1C is an intersection of closed sets and hence is closed.


The following lemma will be useful later on. The easy proof is left to thereader.

Lemma 2.3.8. Let f : X Ñ Y be a map between quasi projective varieties.Suppose that Y “

Ť

iPI Ui is an open cover, that f´1Ui is open in X for eachi P I and that the restriction

f´1Ui ÝÑ Uix ÞÑ fpxq

is regular for each i P I. Then f is regular.

Definition 2.3.9. A quasi projective variety is

‚ an affine variety if it is isomorphic to a closed subset of an affine space(as usual we view An as the open subset PnZ0

Ă Pn),

‚ a projective variety if it is isomorphic to a closed subset of a projectivespace.

Example 2.3.10. Let F P KrZ0, . . . , Zns be a homogeneous polynomial of strictlypositive degree. The principal open subset PnF (see Definition 2.1.8) is an affine

variety. In fact, let νnd : Pn ÝÑ Ppd`nn q´1 be the Veronese map, see (2.3.9), and

let V nd :“ Im νnd be the corresponding Veronese variety. As shown in Example

2.3.6 the map Pn Ñ V nd defined by νnd is an isomorphism. It follows that the

restriction of νnd to PnF defines an isomorphism between PnF and V nd zH, where

H Ă Ppd`nn q´1 is a suitable hyperplane section. Equivalently, PnF is isomorphic

to the intersection of the affine space Ppd`nn q´1

zH and the closed set V nd , and

hence is an affine variety.If Y Ă Pn is closed, and F P KrZ0, . . . , Zns is homogeneous of strictly

positive degree d, it follows that the principal open set YF “ Y zV pF q is anaffine variety. In fact, since νnd is an isomorphism νnd pYF q is closed in the affinevariety V n

d zH, and hence is itself affine. Moreover, the restriction of νnd to YFdefines an isomorphism YF and the affine variety νnd pYF q.

Claim 2.1.9 and Example 2.3.10 give the following result.

Proposition 2.3.11. The open affine subsets of a quasi projective variety forma basis of the Zariski topology.

In a certain sense, open affine subsets of a quasi projective variety are similarto the open subsets of a complex manifold given by charts of a holomorphic atlas.

2.4 Regular functions on affine varieties

Definition 2.4.1. A regular function on a quasi projective varietyX is a regularmap X Ñ K.

2.4. REGULAR FUNCTIONS ON AFFINE VARIETIES 17

Let X be a non empty quasi projective variety. The set of regular functionson X with pointwise addition and multiplication is a K-algebra, named the ringof regular functions of X. We denote it by KrXs.

If X is a projective variety, then it has few regular functions. In fact wewill prove (see Corollary 2.6.6) that every regular function on X is locallyconstant. On the other hand, affine varieties have plenty of functions. In fact ifX Ă An is closed we have an inclusion

Krz1, . . . , znsIpXq ãÑ KrXs. (2.4.1)

Theorem 2.4.2. Let X Ă An be closed. Then (2.4.1) is an equality, i.e. everyregular function on X is the restriction of a polynomial function on An.

Before proving Theorem 2.4.2, we notice that, if X Ă An is closed, theNullstellensatz for Krz1, . . . , zns implies a Nullstellensatz for Krz1, . . . , znsIpXq.First a definition: given an ideal J Ă pKrz1, . . . , znsIpXqq we let

V pJq :“ ta P X | fpaq “ 0 @f P Ju .

The following result follows at once from the Nullstellensatz.

Proposition 2.4.3 (Nullstellensatz for a closed subset of An). Let X Ă An beclosed, and let J Ă pKrz1, . . . , znsIpXqq be an ideal. Then

f P pKrz1, . . . , znsIpXqq | f|V pJq “ 0(

“?J.

(The radical?J is taken inside Krz1, . . . , znsIpXq.) In particular V pJq “ H

if and only if J “ p1q.

The following example makes it clear that Proposition 2.4.3 must play arole in the proof of Theorem 2.4.2. Let X Ă An be closed. Suppose thatg P Krz1, . . . , zns and that gpaq ‰ 0 for all a P Z. Then 1g P KrXs andhence Theorem 2.4.2 predicts the existence of f P Krz1, . . . , zns such thatg´1 “ f|X . By Proposition 2.4.3, pgq “ p1q in Krz1, . . . , znsIpXq, becauseV pgq “ H, where g :“ g|X . hence there exists f P Krz1, . . . , zns such that

f ¨ g “ 1, where f :“ f|X , i.e. g´1 “ f|X

Proof of Theorem 2.4.2. Let ϕ P KrXs. We claim that there exist fi, gi PKrz1, . . . , zns for 1 ď i ď d such that

1. X “Ť

1ďiďdXgi , i.e. V pg1, . . . , gdq XX “ H,

2. for all a P Xgi we have ϕpaq “ fipaqgipaq

,

3. for 1 ď i ď j we have pgjfi ´ gifjq|X “ 0.

(Notice: the last item implies that on Xgi X Xgj we have figi “ fjgj .) Fori “ 1, . . . , d let gi :“ gi|X and fi :“ fi|X . Then

giϕ “ f i. (2.4.2)


In fact by Item (1) it suffices to check that (2.4.2) holds on Xfj for j “ 1, . . . , d.For j “ i it holds by Item (2), for j “ i it holds by Item (3). (Notice: if wedo not assume that Item (3) holds we only know that (2.4.2) holds on Uj XUi.) By Proposition 2.4.3 we have that pg1, . . . , gdq “ p1q, i.e. there existh1, . . . , hd P Krz1, . . . , zns such that

1 “ h1g1 ` ¨ ¨ ¨ ` hdgd.

where hi :“ hi|X . Multiplying by ϕ both sides of the above equality and re-membering (2.4.2) we get that

ϕ “ h1g1ϕ` ¨ ¨ ¨ ` hdgdϕ “ h1f1 ` . . .` h1fd “ ph1f1 ` ¨ ¨ ¨ ` hdfdq|X . (2.4.3)

It remains to prove that there exist fi, gi P Krz1, . . . , zns with the propertiesstated above. By definition of regular function there exist an open covering ofX, and for each set U of the open cover a couple α, β P Krz1, . . . , zns such thatϕpxq “ αpxqβpxq for all x P U (it is understood that βpxq “ 0 for all x P U).By Remark 2.4.4 we may cover U by open affine sets Xγ1 , . . . , Xγr . Since

V pβq ĂrŞ

i“1

V pγiq the Nullstellensatz gives that, for each i, there exist Ni ą 0

and µi P Krz1, . . . , zns such that γNii “ µiβ and hence ϕpxq “ µipxqαpxqγipxqN

for all x P Xγi . Since Xγi “ XγNiwe get that we have covered X by principal

open sets Xg1 such that ϕ “ f 1g1 for all x P Xg1 , where f 1 P Krz1, . . . , zns (ofcourse f 1 depends on g1). By Corollary 2.2.7, the open covering has a finitesubcovering, corresponding to f 11, g

11, . . . , f

1d, g

1d. Now let

fi :“ f 1ig1i, gi :“ pg1iq

2.

Clearly Items (1) and (2) hold. In order to check Item (3) we write

pgjfi ´ gifjq|X “ ppg1jq

2f 1ig1i ´ pg

1iq

2f 1jg1jq|X “ ppg

1ig1jqpf

1ig1j ´ f

1jg1iqq|X .

Since ϕpzq “ f 1ipzqg1ipzq “ f 1jpzqg

1jpzq for all z P Xg1i

X Xg1jthe last term

vanishes on Xg1iXXg1j

, on the other hand it vanishes also on pXzXg1iXXg1j

q “

X X V pg1ig1jq because of the factor pg1ig

1jq.

We end the present section with a couple of consequences of Theorem2.4.2.

First we give a more explicit version of Proposition 2.3.11 in the case thatthe quasi projective variety itself is affine. Given a quasi projective variety X,and f P KrXs, let

Xf :“ XzV pfq, (2.4.4)

where V pfq :“ tx P X | fpxq “ 0u. The following remark is easily verified.

Remark 2.4.4. Let X Ă An be closed (and hence an affine variety). Let f P

KrXs, and hence by Theorem 2.4.2 there exists rf P Krz1, . . . , zns such thatrf|X “ f . Let Y Ă An`1 be the subset of solutions of gpz1, . . . , znq “ 0 for allg P IpXq, and the extra equation fpz1, . . . , znq ¨ zn`1 ´ 1 “ 0. Then the map

Xf ÝÑ Ypz1, . . . , znq ÞÑ pz1, . . . , zn,

1fpz1,...,znq

q

2.4. REGULAR FUNCTIONS ON AFFINE VARIETIES 19

is an isomorphism. In particular Xf is an open affine subset of X. Moreover,the open affine subset Xf , for f P KrXs form a basis for the Zariski topology ofX.

Notice that, by Theorem 2.4.2 and the above isomorphism, every regularfunction on Xf is given by the restriction to Xf of g

fm , where g P KrXs andm P N.

Next, we give a few remarkable consequences of Theorem 2.4.2.

Proposition 2.4.5. Let R be a finitely generated K algebra without nilpotents.There exists an affine variety X such that KrXs – R (as K algebras).

Proof. Let α1, . . . , αn be generators (over K) of R, and let ϕ : Krz1, . . . , zns Ñ Rbe the surjection of algebras mapping zi to αi. The kernel of ϕ is an idealI Ă Krz1, . . . , zns, which is radical because R has no nilpotents. Let X :“V pIq Ă An. Then KrXs – R by Theorem 2.4.2.

In order to introduce the next result, consider a regular map f : X Ñ Y of(non empty) quasi projective varieties. The pull-back f˚ : KrY s Ñ KrXs is thehomomorphism of K-algebras defined by f˚pϕq :“ ϕ ˝ f .

Proposition 2.4.6. Let Y be an affine variety, and let X be a quasi projectivevariety. The map

tf : X Ñ Y | f regularu ÝÑ tϕ : KrY s Ñ KrXs | ϕ homomorphism of K-algebrasuf ÞÑ f˚

(2.4.5)is a bijection.

Proof. We may assume that Y Ă An is closed; let ι : Y ãÑ An be the inclusionmap. Suppose that f, g : X Ñ Y are regular maps, and that f˚ “ g˚. Thenf˚pι˚pziqq “ g˚pι˚pziqq for i P t1, . . . , nu, and hence f “ g. This proves injectiv-ity of the map in (2.4.5). In order to prove surjectivity, let ϕ : KrY s Ñ KrXs bea homomorphism of K algebras. Let fi :“ ϕpι˚pziqq, and let f : X Ñ An be theregular map defined by fpxq :“ pf1pxq, . . . , fnpxqq for x P X. Then fpxq P Y forall x P X. In fact, since Y is closed, it suffices to show that gpfpxqq “ 0 for allg P IpXq. Now

gpf1pxq, . . . , fnpxqq “ gpϕpι˚pz1qq, . . . , ϕpι˚pznqq “ ϕpgpι˚pz1qq, . . . , ι

˚pznqq “ ϕp0q “ 0.

(The second and last equality hold because ϕ is a homomorphism of K-algebras.)Thus f is a regular map f : X Ñ Y such that f˚pι˚pziqq “ ϕpι˚pziqq for i Pt1, . . . , nu. By Theorem 2.4.2 the K-algebra KrY s is generated by ι˚pz1q, . . . , ι

˚pznq;it follows that f˚ “ ϕ.

Corollary 2.4.7. In Proposition 2.4.5, the affine variety X such that KrXs –R is unique up to isomorphism.


2.5 Products

We will prove that the category of quasi projective varieties has (finite) products.First let X, Y be affine varieties. Thus, we may assume that X Ă Am

and Y Ă An are closed subsets. Then X ˆ Y Ă Am ˆ An – Am`n is aclosed subset, and the maps X ˆ Y Ñ X and X ˆ Y Ñ Y given by the twoprojections are regular. One checks easily that X ˆ Y with the two projectionmaps is the product of X and Y in the category of quasi projective varieties(use Proposition 2.4.6). The ring of regular functions of XˆY is constructedfrom KrXs and KrY s as follows. Let πX : X ˆ Y Ñ X and πY : X ˆ Y Ñ Y bethe projections. The K-bilinear map

KrXs ˆKrY s ÝÑ KrX ˆ Y spf, gq ÞÑ π˚Xpfq ¨ π

˚Y pgq

(2.5.1)

induces a linear map

KrXs bK KrY s ÝÑ KrX ˆ Y s. (2.5.2)

Proposition 2.5.1. The map in (2.5.2) is an isomorphism.

Proof. We may assume that X Ă Am and Y Ă An are closed subsets. ThenX ˆ Y Ă Am`n is closed subset, and hence the map in (2.5.2) is surjectiveby Theorem 2.4.2. It remains to prove injectivity, i.e. the following: if A ĂKrXs and B Ă KrY s are finite-dimensional complex vector subspaces, then themap A b B Ñ KrX ˆ Y s obtained by restriction of (2.5.2) is injective. Lettf1, . . . , fau, tg1, . . . , gbu be bases of A and B. By considering the maps

X ÝÑ Kaz ÞÑ pf1pzq, . . . , fapzqq

Y ÝÑ Kbz ÞÑ pg1pzq, . . . , gbpzqq

(2.5.3)

we get that there exist p1, . . . , pa P X and q1, . . . , qb P Y such that the squarematrices pfippjqq and pgipqjqq are non-singular. By change of bases, we mayassume that fippjq “ δij and gkpqhq “ δkh. Computing the values of π˚Xpfiq ¨π˚Y pgjq on pps, qtq for 1 ď i, s ď a and 1 ď j, t ď b we get that the functions. . . , π˚Xpfiq ¨ π

˚Y pgjq, . . . are linearly independent. Thus A b B Ñ KrW ˆ Zs is

injective.

Since every quasi projective variety has an open cover by affine varieties,one could try to define the product of quasi projective varieties X and Y bygluing together the products of the affine varieties in open coverings of X and Y .This is done in scheme theory, where schemes are algebriac varieties defined byatlases with charts given by affine schemes. However, one wants to show more,for example that the product of projective varieties is a projective variety. Thisis why we need the more elaborate construction presented below.

Let Mm`1,n`1 be the vector space of complex pm ` 1q ˆ pn ` 1q matrices.Let

Σm,n :“ trAs P PpMm`1,n`1q | rkA “ 1u.

2.5. PRODUCTS 21

Then Σm,n is a projective variety in PpMm`1,n`1q “ Pmn`m`n. In fact theentries of a non zero matrix A P Mm`1,n`1 define homogegeous coordinates onPpMm`1,n`1q, and Σm,n is the set of zeroes of determinants of all 2ˆ 2 minorsof A. Let rW s P Pm and rZs P Pn; then W t ¨ Z is a complex pm` 1q ˆ pn` 1qmatrix of rank 1, determined up to recsaling. Thus we have the Segre map

Pm ˆ Pn σm,nÝÑ Σm,n

prW s, rZsq ÞÑ rW t ¨ Zs(2.5.4)

Proposition 2.5.2. The map in (2.5.4) is a bijection.

From now on, we identify Pm ˆ Pn with the projective variety Σm,n. Inparticular Pm ˆ Pn has a Zariski topology.

Claim 2.5.3. A subset X Ă Pm ˆ Pn is closed if and only if there exist biho-mogeneous polynomials 2

F1, . . . , Fr P KrW0, . . . ,Wm, Z0, . . . , Zns

such that

X “ V pF1, . . . , Frq :“ tprW s, rZsq P Pn ˆ Pm | 0 “ F1pW ;Zq “ ¨ ¨ ¨ “ FrpW ;Zqu .(2.5.5)

Remark 2.5.4. If m “ 0 and n “ 0, then the Zariski topology on the productPmˆPn is not the product topology. In fact it is finer than the product topology

Example 2.5.5. The diagonal ∆Pn Ă Pn ˆ Pn is closed. In fact, ∆ is the set ofcouples prW s, rZsq such that the matrix with rows W and Z has rank less than2, and hence it is the zero locus of the bihomogeneous polynomials WiZj´WjZifor pi, jq P t0, . . . , nu. Notice that this is not in contrast with the fact that, ifn “ 0, the Zariski topology on Pn is not Hausdorff, because of Remark 2.5.4.

Claim 2.5.6. The projections of Pm ˆ Pn on its two factors are regular maps.

Proof. Let aij , where pi, jq P t0, . . . ,mu ˆ t0, . . . , nu, be the homogeneous co-ordinates on PpMm`1,n`1q given by the entries of a matrix A P Mm`1,n`1.Then

Pm ˆ Pn “ď

0ďiďm0ďjďn

pPm ˆ Pnqaij . (2.5.6)

On the open subset pPmˆPnqaij , the projections PmˆPn Ñ Pm, PmˆPn Ñ Pnare given by

Pm ˆ Pn ÝÑ PmrAs ÞÑ ra0j , . . . , amjs

Pm ˆ Pn ÝÑ PnrAs ÞÑ rai0, . . . , ains

respectively.

2A polynomial F P KrW ;Zs is bihomogeneous of degree pd, eq if F “ř

deg I“ddeg J“e

aI,JWIZJ .


Proposition 2.5.7. Let X be a quasi projective variety, and let f : X Ñ Pmand g : X Ñ Pn be regular maps. Then

X ÝÑ Pm ˆ Pnx ÞÑ pfpxq, gpxqq

(2.5.7)

is a regular map.

Proof. We have the open cover of Pm ˆ Pn given by (2.5.6), with open setsindicized by t0, . . . ,mu ˆ t0, . . . , nu. By Lemma 2.3.8, it suffices to provethat, for each pi, jq P t0, . . . ,mu ˆ t0, . . . , nu, the following hold:

1. pf ˆ gq´1pPm ˆ Pnqaij q is open in X.

2. The restriction

pf ˆ gq´1pPm ˆ Pnqaij q ÝÑ pPm ˆ Pnqaijx ÞÑ pfpxq, gpxqq

(2.5.8)

is regular.

We have

pf ˆ gq´1ppPm ˆ Pnqaij q “ Xzpf´1V pWiq Y g´1V pZjqq.

Both f and g are continuous, because they are regular, and hence f´1V pXiq

and g´1V pYjq are closed. It follows that Item (1) holds. The map

AmÂn ÝÑ pPmˆPnqaijppw0,..., pwi,...,wmq,pz0,...,pzj ,...,znqq ÞÑ prw0,...,wi´1,1,wi`1...,wms,rz0,...,zj´1,1,zj`1,...,znsq

is an isomorphism commuting with the projections. Item (2) follows.

It follows that Pm ˆ Pn with the two projections is the product of Pm andPn in the category of quasi projective varieties.

Now suppose that X Ă Pm and Y Ă Pn are locally closed sets. It followsfrom Claim 2.5.3 that Y ˆY Ă PmˆPn is locally closed, i.e. we have identifiedW ˆZ with a quasi-projective set. Moreover, the projections of XˆY to X andY are regular, because they are the restrictions of the projections of Pm ˆ Pnto X ˆ Y .

The proof of the following result is easy; we leave details to the reader.

Proposition 2.5.8. Keep notation as above. The quasi projective variety X ˆY , with the projections to the two factors, is the product of X and Y in thecategory of quasi projective sets.

Notice that if X Ă Pm and Y Ă Pn are closed then X ˆ Y is closed inPm ˆ Pn. Hence the product of projective varieties is a projective variety. Onthe othar hand, we have already observed that the product of affine varieties isan affine varietry.

2.6. ELIMINATION THEORY 23

Remark 2.5.9. Let X Ă Pm and Y Ă Pn be locally closed sets. Let ϕ : X„ÝÑ X 1,

ψ : Y„ÝÑ Y 1 be isomorphisms, where X 1 Ă Pa and Y 1 Ă Pb are locally closed

sets. ThenX ˆ Y ÝÑ X 1 ˆ Y 1

pp, qq ÞÑ pϕppq, ψpqqq(2.5.9)

is an isomorphism. This follows from the formal property of a categoricalproduct. Thus the isomorphism class of XˆY is independent of the embeddingsX Ă Pm and Y Ă Pn. This is why we say that X ˆ Y is the product of X andY .

Since the product of two quasi projective varieties exists, also the productX1ˆ . . .ˆXr of a finite collection X1, . . . , Xr of quasi-projective varieties exists;it is given by pX1 ˆ pX2 ˆ pX3 . . .ˆXrq . . .q (we may rearrange the parenthesisarbitrarily, and we will get an isomorphic variety).

Let X be a quasi projective variety, and let ∆X Ă X ˆX be the diagonal.It follows from Example 2.5.5 that ∆X is closed in X ˆ X (this is not incontradiction with the fact that, if X is not finite, then it is not Hausdorff,see Remark 2.5.4). This property of quasi projective varieties goes under thename of properness. The following is a consequence of properness.

Proposition 2.5.10. Let X, Y be quasi projective varieties, and let f, g beregular maps X Ñ Y . If fpxq “ gpxq for x in a dense subset of X, then f “ g.

Proof. Let ϕ : X Ñ Y ˆ Y be the map defined by ϕpxq :“ pfpxq, gpxqq. Thenϕ is regular, because Y ˆ Y is the categorical square of Y . Since ∆Y is closed,ϕ´1p∆Y q is closed. By hypothesis ϕ´1p∆Y q contains a dense subset of X, henceit is equal to X, i.e. fpxq “ gpxq for all x P X.

2.6 Elimination theory

Let M be a topological space. Then M is quasi compact, i.e. every open coveringhas a finite subcovering, if and only if M is universally closed, i.e. for anytopological space T , the projection map T ˆ M Ñ T is closed, i.e. it mapsclosed sets to closed sets. (See tag/005M in [dJc].)

A quasi projective variety X is quasi compact, but it is not generally truethat, for a variety T , the projection T ˆX Ñ T is closed. In fact, let X Ă Pnbe locally closed; then ∆X , the diagonal of X, is closed in X ˆPn, because it isthe intersection of X ˆX Ă Pn ˆ Pn with the diagonal ∆Pn Ă Pn ˆ Pn, whichis closed. The projection X ˆ Pn Ñ Pn maps X to X, hence if X is not closedin Pn, then X is not universally closed. This does not contradict the result intopology quoted above, because the Zariski topology of the product of quasiprojective varieties is not the product topology.

The following key result states that projective varieties are the equivalent ofcompact topological spaces in the category of quasi projective varieties.


Theorem 2.6.1 (Main Theorem of elimination theory). Let T be a quasi-projective variety and let X be a closed subset of a projective space. Then theprojection

π : T ˆX Ñ T

is closed.

Proof. By hypothesis we may assume that X Ă Pn is closed. It follows thatT ˆ X Ă T ˆ Pn is closed. Thus it suffices to prove the result for X “ Pn.Since T is covered by open affine subsets, we may assume that T is affine, i.e. Tis (isomorphic to) a closed subset of Am for some m. It follows that it sufficesto prove the proposition for T “ Am. To sum up: it suffices to prove that ifX Ă AmˆPn is closed, then πpXq is closed in Am, where π : AmˆPn Ñ Am isthe projection. We will show that pAmzπpXqq is open. By Claim 2.5.3 thereexist Fi P Krt1, . . . , tm, Z0, . . . , Zns for i “ 1, . . . , r, homogeneous as polynomialin X0, . . . , Xn such that

X “ tpt, rZsq | 0 “ F1pt, Zq “ . . . “ Frpt, Zqu.

Suppose that Fi P Krt1, . . . , tmsrZ0, . . . , Znsdi i.e. Fi is homogeneous of degreedi in Z0, . . . , Zn. Let t P pT zπpXqq. By Hilbert’s Nullstellensatz, there existsN ě 0 such that

pF1pt, Zq, . . . , Frpt, Zqq Ą KrZ0, . . . , ZnsN . (2.6.10)

We may assume that N ě di for 1 ď i ď r. For t P Am let

KrZ0, . . . , ZnsN´d1 ˆ . . .ˆ rZ0, . . . , ZnsN´drΦptqÝÑ KrZ0, . . . , ZnsN

pG1, . . . , Grq ÞÑřri“1Gi ¨ Fi

Thus Φptq is a linear map: choose bases of domain and codomain and let Mptqbe the matrix associated to Φptq. Clearly the entries of Mptq are elementsof Krt1, . . . , tms. By hypothesis Φptq is surjective and hence there exists amaximal minor of Mptq, say MI,Jptq, such that detMI,Jptq “ 0. The openpAmzV pdetMI,Jqq is contained in pT zπpXqq. This finishes the proof of The-orem 2.6.1.

We will give a few corollaries of Theorem 2.6.1. First, we prove an ele-mentary auxiliary result.

Lemma 2.6.2. Let f : X Ñ Y be a regular map between quasi-projective vari-eties. The graph of f

Γf :“ tpx, fpxqq | p P Xu

is closed in X ˆ Y .

Proof. The mapf ˆ IdY : X ˆ Y Ñ Y ˆ Y

is regular, and Γf “ pf ˆ IdXq´1p∆Y q. Hence Γf is closed because ∆Y is closed

in Y ˆ Y .

2.7. EXERCISES 25

Proposition 2.6.3. Let X Ă Pn be closed, and let Y be a quasi-projective set.A regular map f : X Ñ Y is closed.

Proof. Since closed subsets of X are projective it suffices to prove that fpXq isclosed in Y . Let π : X ˆ Y Ñ Y be the projection map. Then fpXq “ πpΓf q.By Lemma 2.6.2 and the Main Theorem of elimination theory we get thatfpXq is closed.

Corollary 2.6.4. A locally-closed subset of Pn is projective if and only if it isclosed.

Proof. Let X Ă Pn be a locally closed subset. If it is closed, then it is projectiveby definition. Conversely, suppose that X is projective. Hence there exist aclosed subset Y Ă Pm and an isomorphism f : Y

„ÝÑ X. Composing f with the

inclusion X ãÑ Pn, we get a regular map g : Y Ñ Pn. Then X “ gpY q is closedby Proposition 2.6.3.

Remark 2.6.5. By way of contrast, notice that it is not true that a locally-closedsubset of An is affine if and only if it is closed. In fact the complement of ahypersurface V pfq Ă An is affine but not closed.

Corollary 2.6.6. Let X be a projective variety. A regular map f : X Ñ K islocally constant.

Proof. Composing f with the inclusion j : K ãÑ P1 we get a regular map f : X Ñ

P1. By Proposition 2.6.3 fpXq is closed. Since fpXq S r0, 1s it follows thatfpXq “ fpXq is a finite set.

2.7 Exercises

Exercise 2.7.1. Let k be a field. Given a finite-dimensional k-vector space V definethe formal power series pV P krrtss as

PV :“8ÿ

d“0

pdimk Symd V qtd

where Symd V is the symmetric product of V . Thus if V “ krx1, . . . , xns1 thenSdpkrx1, . . . , xns1q “ krx1, . . . , xnsd.

1. Prove that if V “ U ‘W then PV “ PU ¨ PW .

2. Prove that if dimk V “ n then PV “ p1´ tqń and hence (2.3.8) holds.

Exercise 2.7.2. We recall that if φ : B Ñ A is a homomorphism of rings, and I Ă A,J Ă B are ideals, the contraction Ic Ă B and the extension Je Ă A are the idealsdefined as follows:

Ic :“ φ´1I, Je :“

#

rÿ

i“1

λiφ pbiq | λi P A, bi P J @i “ 1, . . . , r

+

(2.7.11)

(In other words, Je is the ideal of A generated by φpJq.)Let f : X Ñ Y be a regular map between affine varieties and suppose that f˚ : KrY s ÝÑ

KrXs is injective.


1. Let p P X. Prove that the contraction mcp “ pf˚q´1pmpq is maximal, and

mcp “ mfppq.

2. Let q P Y . Prove that

f´1pqq “

p P X | mp Ą meq(

.

In particular f´1pqq is not empty if and only if meq ‰ KrXs.

3. Give an example of f and q such that meq “ KrW s.

Exercise 2.7.3. The left action of GLnpKq on An defines a left action of GLnpKq onKrz1, . . . , zns as follows. Let φ P Krz1, . . . , zns and g P GLnpKq. Let z be the columnvector with entries z1, . . . , zn: we define gφ P Krz1, . . . , zns by letting

gφpXq :“ φpg´1¨ zq.

Now let G ă GLn pKq be a subgroup. The algebra of G-invariant polynomials is

Krz1, . . . , znsG :“ tφKrz1, . . . , zns P| gφ “ φ @g P Gu .

(it is clearly a K-algebra). Now suppose that G is finite. One identifies AnG withan affine variety proceeding as follows.

1. Define the Reynolds operator as

Krz1, . . . , zns ÝÑ Krz1, . . . , znsGφ ÞÑ 1

|G|

ř

gPG gφ.

Prove the Reynolds identity

R pφψq “ φR pψq @φ P Krz1, . . . , znsG.

2. Let I Ă Krz1, . . . , zns be the ideal generated by homogeneous φ P Krz1, . . . , znsGof strictly positive degree (i.e. non-constant). By Hilbert’s basis theorem thereexists a finite basis tφ1, . . . , φdu of I; we may assume that each φi is homogen-eous and G-invariant. Prove that Krz1, . . . , znsG is generated as K-algebra byφ1, . . . , φd. Since Krz1, . . . , znsG is an integral domain with no nilpotents it fol-lows that there exist an affine variety X (well-defined up to isomorphism) suchthat KrXs „

ÝÑ Krz1, . . . , znsG. One sets AnG “: X.

3. Let ι : Krz1, . . . , znsG ãÑ Krz1, . . . , zns be the inclusion map. By Proposition2.4.6, there exist a unique regular map

An πÝÑ X “ AnG. (2.7.12)

such that ι “ π˚. Prove that

π ppq “ π pqq if and only if q “ gp for some g P G,

and that π is surjective. [Hint. Let J Ă Krz1, . . . , znsG be an ideal. Show thatJe XKrz1, . . . , znsG “ J where Je is the extension relative to the inclusion ι.]

Exercise 2.7.4. Keep notation and hypotheses as in Exercise 2.7.3. Describeexplicitly AnG and the quotient map π : An Ñ AnG for the following groups G ă

GLn pKq:

2.7. EXERCISES 27

1. n “ 2, G “ t˘12u.

2. n “ 2, G “

Bˆ

ωk 00 ω´1

k

˙F

where ωk is a primitive k-th rooth of 1.

3. G “ Sn, the group of permutation of n elements viewed in the obvious way asa subgroup of GLn pKq (group of permutations of coordinates).

Exercise 2.7.5. We recall that, if f P Krts and a P K, the multiplicity of f at a isdefined to be the maximum integer m such that pt´ aqm divides f . We denote it bymultapfq.

(a) Let F P KrZ0, Z1sd. Let p P P1. Let L P KrZ0, Z1s1 be such that Lppq “ 0,and let t be an affine coordinate on P1

L. Let a be the affine coordinate of p. Wehave F Ld “ fptq P Krts. Show that multapfq does not depend on the choicesmade above (of course it does depend on the choice of F and p); we denote itby multppF q.

(b) Prove that V pF q consists of d points if and only if multppF q ď 1 for all p P P1.

(c) Let ∆d Ă PpKrZ0, Z1sdq be the subset of rF s such that there exists p P P1

for which multppF q ě 2. Prove that ∆d is a closed irreducible subset ofPpKrT0, T1sdq. (Hint: let r∆d Ă PpKrT0, T1sdq ˆ P1 be the subset of couples

prF s, rZsq such that F has a multiple root at Z. Show that r∆d is closed inPpKrT0, T1sdq ˆ P1, and then project to the first factor.)

(d) Assume that charK “ 0, or more generally that charK does not divide d “degF . Let p “ ra0, a1s P P1. Prove that multppF q ě 2 if and only if

BF pa0, a1q

BZ0“BF pa0, a1q

BZ1“ 0. (2.7.13)

(Hint: use Euler’s relation d ¨ F “ Z0BFBZ0

` Z1BFBZ1

.)


Chapter 3

Rational maps anddimension

3.1 Rational maps

Let X and Y be quasi projective varieties. We define a relation on the set ofcouples pU,ϕq where U Ă X is open dense and ϕ : U Ñ Y is a regular map, asfollows: pU,ϕq „ pV, ψq if the restrictions of ϕ and ψ to U X V are equal. Onechecks easily that „ is an equivalence relation.

Definition 3.1.1. A rational map f : X 99K T is a „-equivalence class ofcouples pU,ϕq where U Ă X is open dense and ϕ : U Ñ Y is a regular map. Letf : X 99K Y be a rational map.

1. The map f is regular at x P X (equivalently x is a regular point of f), ifthere exists pU,ϕq in the equivalence class of f such that x P U . We letRegpfq Ă X be the set of regular points of f . By definition Regpfq is anopen subset of X.

2. The indeterminancy set of f is Indpfq :“ XzRegpfq (notice that Indpfqis closed). A point x P X is a point of indeterminancy if it belongs toIndpfq.

From now on we will consider only rational maps between irreducible quasiprojective varieties. Let f : X 99K Y and g : Y 99KW be rational maps between(irreducible) quasi projective varieties. It might happen that for all x P Regpfqthe image fpxq does not belong to Regpgq, and then the composition g ˝ fmakes no sense. In order to deal with compositions of reational maps, we givethe following definition.

Definition 3.1.2. A rational map f : X 99K Y between irreducible quasi pro-jective varieties is dominant if it is represented by a couple pU,ϕq such thatϕpUq is dense in Y .

29

30 CHAPTER 3. RATIONAL MAPS AND DIMENSION

Notice that if f : X 99K Y is dominant and pV, ψq is an arbitrary represent-ative of f then ψpV q is dense in Y .

Definition 3.1.3. Let f : X 99K Y be a dominant rational map, and letg : Y 99K W be a rational map (X,Y,W are irreducible). Let pU,ϕq and pV, ψqbe representatives of f and g respectively. Then ϕ´1V is open dense in X. Welet g ˝ f : X 99K W be the rational map represented by pϕ´1V, ψ ˝ ϕq. (Theequivalence class of pϕ´1V, ψ ˝ ϕq is independent of the representatives pU,ϕqand pV, ψq.)

Definition 3.1.4. A dominant rational map f : X 99K Y between irreduciblequasi projective varieties is birational if there exists a dominant rational mapg : Y 99K X such that g˝f “ IdX and f˝g “ IdY . An irreducible quasi projectivevariety X is rational if it is birational to Pn for some n, it is unirational if thereexists a dominant rational map f : Pn 99K X.

Example 3.1.5. 1. Of course isomorphic irreducible quasi projective varietiesare birational. On the other a quasi projective (irreducible) variety isbirational to any of its dense open subsets. In particular Pn is birationalto An, although they are not isomorphic if n ą 0 (if they were isomorphic,they would be diffeomorphic as C8 manifolds, but Pn is compact, An isnot).

2. Let 0 “ F P KrZ0, . . . , Zns2, and let Qn´1 :“ V pF q Ă Pn. Suppose thatF is prime, i.e that rkF ě 3, and hence Qn´1 is irreducible. We claimthat Qn´1 is rational. In fact, after a suitable change of coordinates, wemay assume that F “ Z0Zn ´ G, where 0 “ G P KrZ1, . . . , Zn´1s2. Therational maps

Qn´1 f99K Pn´1

rZ0, . . . , Zns ÞÑ rZ0, . . . , Zn´1s

and

Pn´1 g99K Qn´1

rT0, . . . , Tn´1s ÞÑ rT 20 , T0T1, . . . , T0Tn´1, GpT1, . . . , Tn´1qs

are dominant, and they are inverses of each other. Notice that if n “ 2,then f and g are regular (see Example 2.3.5), while for n ě 3, the quadricQn´1 is not isomorphic to Pn´1, because the underlying C8 manifolds arenot homeomorphic.

Proposition 3.1.6. Irreducible quasi varieties X, Y are birational if and onlyif there exist open dense subsets U Ă X and V Ă Y that are isomorphic.

Proof. An isomorphism ϕ : U„ÝÑ V clearly defines a birational map f : X 99K

Y . Conversely, suppose that f : X 99K Y is birational with inverse g : Y 99K X.Let pU,ϕq represent f and pV, ψq represent g. Then ϕ´1V Ă U and ψ´1U Ă Vare open dense. By hypothesis the composition

ψ ˝`

ϕ|ϕ´1V

˘

: ϕ´1V Ñ U

3.2. THE FIELD OF RATIONAL FUNCTIONS 31

is equal to the identity on an open non-empty subset of ϕ´1V . By Proposition2.5.10, we get that ψ ˝

`

ϕ|ϕ´1V

˘

“ Idϕ´1V . In particular ψ ˝ ϕ`

ϕ´1V˘

Ă

U i.e. ϕ`

ϕ´1V˘

Ă ψ´1U , and similarly

ϕ ˝`

ψ|ψ´1U

˘

“ Idψ´1U , ψ`

ψ´1U˘

Ă ϕ´1V.

Thus we have isomorphisms ϕ´1V„ÝÑ ψ´1U and ψ´1U

„ÝÑ ϕ´1V .

Many natural invariants of projective varieties do not separate between (pro-jective) birational varieties. This fact gives practical criteria that allow to es-tablish that certain projective varieties are not birational. On the other hand,it leads us to approach the classification of isomorphism classes of projectivevarieties in two steps: first we classify equivalence classes for birational equival-ence, then we distinguish isomorphim classes within each birational equivalenceclass.

3.2 The field of rational functions

If we consider the category whose objects are irreducible quasi projective vari-eties, and morphisms are dominant rational maps, we get a familiar algebraiccategory. In order to explain this, we introduce a key definition. Let X be anirreducible quasi projective variety. The field of rational functions on X is

KpXq :“ tf : X 99K K | f is a rational mapu . (3.2.1)

Addition and multiplication are defined on representatives. Let f, g P KpXq berepresented by pU,ϕq and pV, ψq respectively. Then

f ` g :“ r`

U X V, ϕ|UXV ` ψ|UXV˘

s,

f ¨ g :“ r`

U X V, ϕ|UXV ¨ ψ|UXV˘

s.

Example 3.2.1. • KpPnq – Kpz1, . . . , znq is the purely transcendental exten-sion of K of transcendence degree n.

• Let p P Krzs be free of square factors (and deg p ě 1). Then t2 ´ ppzq isprime and hence X :“ V

`

t2 ´ ppzq˘

Ă A2 is irreducible. Then Kpzq ĂKpXq is an extension of degree 2. We may ask whether KpXq is a purelytrascendental extension of K. The answer is yes if deg p “ 1, 2 (see Ex-ample 2.3.5), no if deg p ě 3 (this requires new ideas).

Let f : X 99K Y be a dominant rational map of irreducible quasi projectivevarieties. We have a well-defined pull-back

KpY q ϕ˚

ÝÑ KpXqϕ ÞÑ ϕ ˝ f

(The composition is well defined because by hypothesis f is dominant.) Themap f˚ is an inclusion of extensions of K. Suppose that f : X 99K Y and


g : Y 99KW are dominant rational maps of irreducible quasi projective varieties.Then g ˝ f : X 99KW is dominant and

f˚ ˝ g˚ “ pg ˝ fq˚. (3.2.2)

Of course Id˚X : KpXq Ñ KpXq is the identity map. We will prove the followingresult.

Theorem 3.2.2. By associating to each quasi projective variety its field offractions, and to each dominant rational map f : X 99K Y of irreducible quasiprojective varieties the pull back, we get an equivalence between the categoryof irreducible quasi projective varieties with homomorphisms dominant rationalmaps, and the category of finitely generated field extensions of K.

What must be proved are the following two statements:

1. An extension of fields K Ă E is isomorphic to the filed of rational functionsKpXq of a quasi projective variety X if and only it it is finitely generatedover K.

2. Let E, F be finitely generated field extensions of K, and let α : E Ñ Fbe a homomorphism of K extensions (i.e. an inclusion E ãÑ F which isthe identity on K). Let Y,X be irreducible quasi projective varieties suchthat KpY q,KpXq are isomorphic to E and F respectively as extensions ofK (they exist by Item (1)). Then there exists a unique dominant rationalmap f : X 99K Y such that f˚ “ α.

Item (1) is proved in Proposition 3.2.4. Item (2) is proved in Proposition3.2.5.

We start by observing that we may restrict our attention to affine (irredu-cible) varieties. In fact, let X be an irreducible quasi projective variety, and letY Ă X be an open dense affine subset (e.g. a prinipal open subset). We have awell-defined restriction map

KpXq 99K KpY q. (3.2.3)

In fact, let f P KpXq, and let pU,ϕq be a couple representing an element. ThenU X Y is an open dense subset of Y , and the couple pU X Y, ϕ|UXY q represents

an element f P KpY q, which is independnet of the representative of f . Therestriction map in (3.2.3) is an isomorphism of K extensions. Hence, whendealing with the field of fractions of a quasi projective variety, we may assumethat the variety is affine.

Let X be an irreducible quasi projective variety. We have an inclusion of Kextensions:

(field of fractions of KrXs) ãÑ KpXqαβ ÞÑ rpXzV pβq, αβ qs

(3.2.4)

Claim 3.2.3. Let X be an affine irreducible variety. Then (3.2.4) is an iso-morphism.

3.2. THE FIELD OF RATIONAL FUNCTIONS 33

Proof. We must prove that the map in (3.2.4) is surjective. Let f P KpXq, andlet pU,ϕq represent f . By Remark 2.4.4, there exists 0 “ γ P KrXs such thatthe dense principal open subset Xγ is contained in U . Moreover, by Remark2.4.4 and Theorem 2.4.2, KrXf s is generated as K-algebra by KrXs andγ´1, hence φ is represented by pXγ ,

αγm q where α P KrXs. Let β :“ γ. Since

Xγ “ Xβ , we have proved that f belongs to the image of (3.2.4).

Proposition 3.2.4. A field extension of K is isomorphic to the field of fractionsof an irreducible quasi projective variety if and only if it is finitely generated overK.

Proof. Let X be a quasi projective variety. The field KpXq is isomorphic tothe field of fractions of an open dense affine subset of X. Thus we may assumethat X Ă An is closed. By Claim 3.2.3 KpXq is the field of quotients ofKrXs, and by Theorem 2.4.2 KrXs is generated (over K) by the restrictionsof the coordinate functions z1, . . . , zn. Hence the restrictions of the coordinatefunctions z1, . . . , zn to X generate KpXq over K.

Now assume that E is a finitely generated field extension of K. In particularthe transcendenece degree of E over K is finite, say d. Let f1, . . . , fm P KpXqbe a transcendence basis of KpXq over K. Then KpXq is a finitely generatedalgebraic extension of K pf1, . . . , fmq. By the Theorem on the primitive ele-ment, i.e. Theorem A.4.1, there exists g P KpXq algebraic of degree d overK pf1, . . . , fmq and such that KpXq is generated over K by f1, . . . , fm, g. LetP P K pf1, . . . , fnq rys be the minimal polynomial of g over K pf1, . . . , fnq. Thus

P pyq “ yd ` c1yd´1 ` ¨ ¨ ¨ ` cd, ci P K pf1, . . . , fmq .

Write ci “aibi

where ai, bi P Krf1, . . . , fms. Let rQ P Krf1, . . . , fmsrys be ob-

tained from P by clearing denominators, i.e. rQ “ pb1 ¨ . . . ¨ bdqP . Let Q P

Krf1, . . . , fmsrys be obtained from rQ by factoring out the maximum commondivisor of the coefficients (recall that Krf1, . . . , fms is a UFD). Notice that Q isirreducible and hence prime. Write

Q “ e0yd ` e1y

d´1 ` ¨ ¨ ¨ ` ed, ei P Krf1, . . . , fms, e0 “ 0.

Let θ : Krf1, . . . , fm, ys„ÝÑ Krz1, . . . , zm, ys be the isomorphism of K-algebrae

mapping fi to zi and y to itself. Let Φ :“ θpQq. Then X :“ V pΦq Ă An`1 is anirreducible hypersurface because Φ is prime. Let zi :“ zi|X . We claim that therational functions on X represented by tz1, . . . , zmu are algebraically independ-ent over K. In fact suppose thatR P Krt1, . . . , tms andRpz1, . . . , znq “ 0. By thefundamental Theorem of Algebra, for any pξ1, . . . , ξmq P pAmzV pe0qq there existsξm`1 P K such that pξ1, . . . , ξm, ξm`1q P X. It follows that Rpξ1, . . . , ξmq “ 0for all pξ1, . . . , ξmq P pAnzV pe0qq, and hence R ¨ e0 vanishes identically on Am.Thus R ¨ e0 “ 0, and since e0 “ 0 it follows that R “ 0. This proves thattz1, . . . , zmu are algebraically independent over K. On the other hand y :“ y|Xis algebraic over Kpz1, . . . , zmq and its minimal polynomial equals Φ. Since the


field of fractions of X is the field of quotients of KrXs “ Krz1, . . . , zm`1spΦq,we get that

E – K pf1, . . . , fmq ryspQpyqq – K pz1, . . . , zmq ryspΦq – KpXq.

Proposition 3.2.5. Let X and Y be irreducible quasi projective varieties. Sup-pose that α : KpY q ãÑ KpXq is an inclusion of extensions of K. There exists aunique dominant rational map f : X 99K Y such that f˚ “ α.

Proof. We may assume that X Ă An and Y Ă Am are closed. By Claim3.2.3 KpXq, KpY q are the fields of fractions of KrXs and KrY s respectively, andby Theorem 2.4.2, KrXs “ Krz1, . . . , znsIpXq and KrY s “ Krw1, . . . , wmsIpY q.Given p P Krz1, . . . , zns and q P Krw1, . . . , wms we let p :“ p|X and q :“ q|Y .We have

α pwiq “f igi, fi, gi P Krz1, . . . , zns, gi ‰ 0.

Let U :“ XzpV pg1q Y . . .Y V pgmqq. Then U is open and dense in X. Let

UrφÝÑ Am

a ÞÑ

´

f1paqg1paq

, . . . , fmpaqgmpaq

¯

We claim that rφpUq Ă Y . In fact let h P IpY q. Since α is an inclusion ofextensions of K,

hpf1g1, . . . , fmgmq “ hpαpw1q, . . . , αpwmqq “ αphpw1, . . . , wmq “ αp0q “ 0.

This proves that if h P IpY q then h vanishes on rφpUq , i.e. rφpUq Ă Y . Thus rφinduces a regular map φ : U Ñ Y . Let f : X 99K Y be the equivalence class ofpU, φq. Then f˚ “ α.

It is clear by the above construction that f is the unique rational (dominant)map such that f˚ “ α.

The result below follows at once from what has been proved above.

Corollary 3.2.6. Irreducible quasi projective varieties are birational if and onlyif their fields of rational functions are isomorphic as extensions of K.

The result below follows from the above corollary and the proof of Propos-ition 3.2.4.

Proposition 3.2.7. Let X be an irreducible quasi projective variety and letm :“ Tr.degK KpXq. Then X is birational to an irreducible hypersurface inAm`1.

3.3. DIMENSION 35

3.3 Dimension

Let X be an irreducible quasi projective variety. The dimension of X is definedto be the transcendence degree of KpXq over K. Next, let X be an arbitraryquasi projective variety, and let X “ X1 Y ¨ ¨ ¨ Y Xr be its irreducible decom-position.

1. The dimension of X is the maximum of the dimensions of its irreduciblecomponents. We say that X has pure dimension n if every irreduciblecomponent of X has dimension n.

2. Let p P X. The dimension of X at p is the maximum of the dimensionsof the irreducible components of X containing p.

Notice that the dimension of X is equal to the dimension of any open densesubset U Ă X.

Example 3.3.1. The dimension of An is equal to n because tz1, . . . , znu is atranscendence basis of Kpz1, . . . , znq over K.

Remark 3.3.2. Let f : X 99K Y be a dominant map of irreducible quasi project-ive varieties. Then dimY ď dimX, because we have the inclusion f˚ : KpY q ãÑ

KpXq of field extensions of K.

Proposition 3.3.3. Let X be an irreducible quasi projective variety and Y Ă Xbe a proper closed subset. Then dimY ă dimX.

Proof. We may assume that Y is irreducible. Since X is covered by open affinevarieties, we may assume that X is affine. Thus X Ă An is a closed (irreducible)subset, and so is Y . We may choose a transcendence basis tf1, . . . , fdu of KpY q,where each fi is a regular function on Y (for example a coordinate function).

Let f1, . . . , fd P KrXs such that fi|W “ fi. Since Y is a proper closedsubset of X, there exists a non zero g P KrXs such that g|Y “ 0. It suffices

to prove that f1, . . . , fd, g are algebraically independent over. We argue bycontradiction. Suppose that there exists 0 ‰ P P KrS1, . . . , Sd, T s such thatP pf1, . . . , fd, gq “ 0. Since X is irreducible we may assume that P is irreducible.Restricting to Y the equality P pf1, . . . , fd, gq “ 0, we get that P pf1, . . . , fd, 0q “0. Thus P pS1, . . . , Sd, 0q “ 0, because f1, . . . , fd are algebraically independent.This means that T divides P . Since P is irreducible P “ cT , c P K˚. ThusP pf1, . . . , fd, gq “ 0 reads g “ 0, and that is a contradiction.

Corollary 3.3.4. A (non empty) closed subset X Ă An`1 has pure dimensionn if and only if it is an irreducible hypersurface. Similarly, a closed subsetX Ă Pn`1 has pure dimension n if and only if it is an irreducible hypersurface.

Proof. Let X Ă An`1 be an irreducible hypersurface. Let IpXq “ pfq. Reorder-ing the coordinates pz1, . . . , zn, zn`1q we may assume that

f “ c0zdn`1 ` c1z

d´1n`1 ` ¨ ¨ ¨ ` cd, ci P Krz1, . . . , zns, c0 ‰ 0, d ą 0.


In proving Proposition 3.2.7 we showed that the restrictions to X of the zi’s,for i “ 1, . . . , d give a transcendence basis of KpXq. Thus dimX “ n. Since theirreducible components of a hypersurface are hypersurfaces (if f “

ś

fmii is thedecomposition of f into prime factors, the irreducible components of V pfq arethe hypersurfaces V pfiq), it follows that a hypersurface X Ă An`1 is of puredimension n.

In order to prove the converse, let X Ă An`1 be a closed subset of puredimension n. Thus every irreducible component of X is a closed subset of An`1

of dimension n. Since the union of hypersurfaces in An`1 is a hypersurface inAn`1, it suffices to prove that each irreducible component of Xis a hypersurface,i.e we may assume that X is irreducible. Since dimX “ n ă dimAn`1, thereexists a non zero f P IpXq Ă Krz1, . . . , zn`1s. Since X is irreducible, the idealIpXq is prime, and hence there exists a prime factor g of f which vanishes onX. Thus X Ă V pgq, dimX “ n “ dimV pgq (by the result that we just proved),V pgq is irreducible, and X is closed in V pgq. By Proposition 3.3.3 we get thatX “ V pgq. This finishes the proof for closed subsets of An`1.

The result for closed subsets of Pn`1 follows by a smilar proof, or by inter-secting with standard open affine subsets PnZi .

Proposition 3.3.5. Let X, Y be quasi projective varieties. Then dimpXˆY q “dimX ` dimY .

Proof. We may assume that X and Y are irreducible affine varieties. There existtranscendence bases tf1, . . . , fdu, tg1, . . . , geu of KpXq and KpY q respectivelygiven by regular functions. Let πX : X ˆ Y Ñ X and πY : X ˆ Y Ñ Y bethe projections. We claim that tπ˚Xpf1q, . . . , π

˚Xpfdq, π

˚Y pg1q, . . . , π

˚Y pgequ is a

transcendence basis of KpX ˆ Y q.First, by Proposition 2.5.1 KrX ˆ Y s is algebraic over the subring gener-

ated (over K) by π˚Xpf1q, . . . , π˚Y pgeq.

Secondly, let us show that π˚Xpf1q, . . . , π˚Y pgeq are algebraically independent.

Suppose that there is a polynomial relation

ÿ

0ďm1,...,meďN

Pm1,...,mepπ˚Xpf1q, . . . , π

˚Xpfdqq ¨ π

˚Y pg1q

m1 ¨ . . . ¨ π˚Y pgeqme “ 0,

where each Pm1,...,me is a polynomial. Since g1, . . . , ge are algebraically inde-pendent we get that Pm1,...,mepf1paq, . . . , fdpaqq “ 0 for every a P X. Sincef1, . . . , fd are algebraically independent, it follows that Pm1,...,me “ 0 for every0 ď m1, . . . ,me ď N , and hence P “ 0. This proves that π˚Xpf1q, . . . , π

˚Y pgeq

are algebraically independent.

3.4 Maps of finite degree

Let f : X 99K Y be a rational map of irreducible quasi-projective varieties.

3.4. MAPS OF FINITE DEGREE 37

Definition 3.4.1. The degree of f is given by

deg f :“

#

0 if f is not dominant,

rKpXq : f˚pKpY qqs if f is dominant.

We recall that

rKpXq : f˚KpY qs “ dimKpY qKpXq.

Thus 0 ă deg f ă 8 if and only if f is dominant and dimX “ dimY .

Example 3.4.2. Let pz1, . . . , zn, wq be affine coordinates on An`1. Let X Ă An`1

be an irreducible hypersurface and IpXq “ P . Write

P “ a0wd ` a1w

d´1 ` ¨ ¨ ¨ ` ad, ai P Krz1, . . . , zns, a0 ‰ 0

and let f : X Ñ An be the forgetful (projection) map fpz, wq “ z. Thendeg f “ d. In fact suppose that d “ 0. Then Im f “ V pa0q and hence fis not dominant. If d ą 0, then KpXq “ Kpz1, . . . , znqrwspP q, and hencerKpXq : Kpz1, . . . , znqs “ d.

Below is the main result of the present section.

Proposition 3.4.3. Let f : X Ñ Y be a regular map between irreducible quasi-projective varieties. Suppose that deg f ă 8. There exists an open dense Y 0 Ă

Y such that|f´1tqu| “ deg f @q P Y 0.

Let us check the statement of Proposition 3.4.3 for the map f : X Ñ Anof Example 3.4.2. Let p P Krx1, . . . , xn, ys be as in that example. ThenV pp, BpByq is a proper closed subset of X and hence it has dimension strictlysmaller than dimX “ n. Thus ∆ :“ fpV pp, BpByqq is a proper closed sub-set of AnK and hence pAnKz∆zV pa0qq is an open dense subset of AnK. Let x PpAnKz∆zV pa0qq: then ppx, yq P Krys is a degree-d polynomial with simple rootsand hence |f´1pxq| “ d.

Before giving the proof of Proposition 3.4.3 we consider a more generalversion of Example 3.4.2. Let Y be an affine variety. Let P P KpY qrts be anirreducible polynomial:

P “ td ` a1td´1 ` ¨ ¨ ¨ ` ad, ai P f

˚pKpY qq.

Since Y is affine KpY q is the field of fractions of KrY s. Thus there exists0 ‰ b P KrY s such that b ¨ ai P f˚pKrY sq for all 1 ď i ď d. Let c0 :“ b,ci :“ b ¨ ai, 1 ď i ď d and

Q :“ c0td ` c1t

d´1 ` ¨ ¨ ¨ ` cd P KrY srts. (3.4.1)

Let π : Y ˆ A1 Ñ Y be the projection.

Lemma 3.4.4. Keep hypotheses and notation as above.


(a) There is one and only one irreducible component V pQqi of V pQq whichdominates Y , i.e. such that πpV pQqiq “ Y ; call it V pQq0.

(b) There is an open dense U Ă Y such that |π´1pqq| “ d and π´1pqq Ă V pQq0for every q P U .

Proof. (a): We have πpV pQqq Ą Y zV pc0q, and Y zV pc0q is dense in Y becausec0 ‰ 0. It follows that there exists at least one irreducible component V pQq0 ofV such that πpV pQq0q “ Y . Let g P IpV pQq0q. We claim that

Q|g in KpY qrts. (3.4.2)

(Notice: we do not claim that Q|g in KrY srts.) In fact suppose that (??) doesnot hold. Then Q and g are coprime (in KpY qrts) because Q is prime, andhence there exist α, β P KpY qrts such that α ¨ Q ` β ¨ g “ 1. Multiplying by0 ‰ γ P KrY srts such that α ¨ γ, β ¨ γ P KrY srts we get that

pα ¨ γqQ` pβ ¨ γqg “ γ.

It follows that if q P V pQq0 then γpqq “ 0. Since γ ‰ 0 we get that πpV pQq0q ‰Y : that is a contradiction. This proves (3.4.2). Let IpV pQq0q “ pg1, . . . , grq.From (3.4.2) we get that there exist h1, . . . , hr P KrY srts and m1, . . . ,mr P KrY ssuch that

mi ¨ gi “ Q ¨ hi, mi ‰ 0, i “ 1, . . . , r. (3.4.3)

Set m “ m1 ¨ ¨ ¨ ¨ ¨mr. Then V pQq0zV pmq “ V pQqzV pmq by (3.4.3), and henceV pQq0 is the unique irreducible component of V pQq dominating Y .

(b): Let

Q1 :“ dc0td´1 ` pd´ 1qc1t

d´2 ` ¨ ¨ ¨ ` cd´1 P KrY srts. (3.4.4)

be the derivative of Q with respect to t. Then Q1 “ 0 and degtQ1 “ d ´ 1 ă

d “ degtQ. Thus Q and Q1 are coprime in KpY qrts and hence there existµ, ν P KpY qrts such that

µ ¨Q` ν ¨Q1 “ 1.

Arguing as above we get that there exists a proper closed C Ă Y such that

π´1pY zCq X V pQq X V pQ1q “ H. (3.4.5)

Now let U :“ pY zCzV pc0qzV pmqq: then |π´1pqq| “ d and π´1pqq Ă V pQq0 forevery q P U .

Remark 3.4.5. If KrY s is a UFD we may factor out the gcd tc0, . . . , cdu andhence by renaming the ci’s we may assume that gcd tc0, . . . , cdu “ 1. It followsthat V pQq is irreducible (the proof is the same as the one for hypersurfaces inAn). The problem is that in general KrY s will not be a UFD (an example:Y “ V px1x2´x3x4q Ă A4 and Q “ x1y´x3), and hence there might be no wayof “reducing” the polynomial of (3.4.4) in order to get that V pQq is irreducible.

3.4. MAPS OF FINITE DEGREE 39

Proof of Proposition 3.4.3. Suppose that deg f “ 0. Then fpXq ‰ Y andY 0 :“ Y z ¯fpXq will do. Now suppose that d :“ deg f ą 0. Since Y is coveredby open affine sets we may assume that Y itself is affine. By definition we havean inclusion f˚ : KpY q ãÑ KpXq (of extension fields of K) and KpXq as vectorspace over KpY q has dimension d. Since we are in characteristic zero there existsξ P KpXq primitive over f˚pKpY qq. Let

P “ td ` a1td´1 ` ¨ ¨ ¨ ` ad, ai P f

˚pKpY qq

be the minimal polynomial of ξ. Let Q P KrY srts be the polynomial in (3.4.4),obtained by clearing denominators of a1, . . . , ad. By Lemma 3.4.4 there existsa unique irreducible component V pQq0 of V pQqdominating Y . Let π0 : V pQq0 ÑY be the restriction of the projection π : Y Â1 Ñ Y . We have a commutativediagram

Xφ

//

f

V pQq0

π0

||Y

with φ birational. By Proposition 3.1.6 there exist open dense subsetsX 1 Ă Xand V pQq10 Ă V pQq0 fitting into a commutative diagram

X 1ψ

//

f 1:“f|X1

V pQq10

π10:“π0|V pQq10||Y

(3.4.6)

with ψ an isomorphism. Since XzX 1 ‰ X and dimX “ dimY we havefpXzX 1q ‰ Y . On the other hand f´1pqq “ pf 1q´1pqq if q P Y zfpXzX 1q.Since (3.4.6) is commutative and ψ is an isomorphism, we get that

|pf 1q´1 tqu | “ |pπ10q´1 tqu |, q P Y zfpXzX 1q.

Hence the proposition follows from Lemma 3.4.4.

Definition 3.4.6. We introduce some terminology. Let Y be a quasi-projectiveset and P a property that might or might not hold for a given y P Y (formallyP is a subset of Y ). We say that property P holds for the generic point of Yif there exists an open dense Y 0 Ă Y such that property P holds for all y P Y 0.

Example 3.4.7. 1. The generic point of Y is smooth.

2. If f : X Ñ Y is a map of quasi-projective varieties and deg f ă 8 then|f´1 tqu | “ deg f for the generic q P Y .

Proposition 3.4.3 gives that if f : X Ñ Y is a dominant map of quasi-projective varieties and deg f ă 8 then fpXq contains an open dense subset ofY “ ¯fpXq. A similar result holds in general.


Proposition 3.4.8. Let f : X Ñ Y be a map of quasi-projective varieties. ThenfpXq contains an open dense subset of fpXq.

Proof. It suffices to prove the proposition for X and Y irreducible. ReplacingY by fpXq we may assume that f is dominant. Thus we have a well-definedinclusion of K-extensions f˚ : KpY q ãÑ KpXq. Suppose that dimX “ dimY :then deg f ă 8 and the proposition follows from Proposition 3.4.3. Nowsuppose that dimX ą dimY : we will prove that there exists an irreduciblelocally closed W Ă X such that

fpW q “ Y and dimW “ dimY. (3.4.7)

We may assume that Y is affine. Let m :“ dimY and φ1, . . . , φm P KpY q be atranscendence basis of KpY q over K. Replacing Y by an open dense susbset wemay assume that φ1, . . . , φm P KrY s. Since f is dominant f˚pφ1q, . . . , f

˚pφmqare algebraically independent over K; extend them to a transcendence basistf˚pφ1q, . . . , f

˚pφmq, ψ1, . . . , ψku of KpXq. Thus m ` k “ dimX. By the the-orem on the primitive element there exist an irreducible polynomial

Q :“ a0td ` a1t

d´1 ` ¨ ¨ ¨ ` ad, ai P Krz1, . . . , zm, u1, . . . , uks, a0 ‰ 0

and a birational map α : X 99K V pQq Ă Am`k`1 (here pz1, . . . , zm, u1, . . . , uk, tqare affine coordinates on Am`k`1) such that

α˚pziq “ f˚pφiq, α˚pujq “ ψj , 1 ď i ď m, i ď j ď k.

There exist open dense subsets X0 Ă X and V pQq0 Ă V pQq such that α isregular on X0, αpX0q Ă V pP q0 and α|X0 defines an isomorphism α|X0 : X0 „

ÝÑ

V pQq0. Let β : V pQq0 Ñ X0 be the inverse α|X0 . Let

V pQqπÝÑ Am`k

pz, u, tq ÞÑ pz, uq

be the projection map. Let B :“ πpV pQqzV pQq0q. Since V pQqzV pQq0 is aproper closed subset of V pQq we have dimpV pQqzV pQq0q ă dimV pQq “ m` k:it follows that dimB ă m` k and hence B ‰ Am`k. Now let

Am`k ρÝÑ Am

pz, uq ÞÑ z

be the projection map. There exists an affine subspace L Ă Am`k such that thefollowing hold:

(I) dimL “ m and ρpLq “ Am.

(II) L Ć B Y V pa0q (because B ‰ Am and a0 ‰ 0).

By (I)-(II) π´1pLq is isomorphic to a hypersurface in Am`1, moreover π´1pLqXV pQq0 contains a (unique) irreducible component H0 such that dimH0 “ m

3.5. GRASSMANNIANS 41

and πpH0q “ L, by Lemma 3.4.4. Let W :“ βpH0q Ă X0. Then dimW “

dimH0 “ m because β is an isomorphism. Moreover

f˚pφiq “ pα|X0q˚pπ˚pziqq.

It follows that f˚pφ1q|W , . . . , f˚pφmq|W are algebraically independent elements

of KpW q. Hence the restriction of f to W is a map of finite non-zero degreeW Ñ Y , i.e. (3.4.7) holds.

By Proposition 3.4.3 we get that fpW q contains an open dense subset ofY , and this proves the proposition.

Example 3.4.9. Let f : A2 Ñ A2 be given by fpx, yq “ px, xyq;

A2 fÝÑ A2

px, yq ÞÑ px, xyq

Then Im f “ tp0, 0qu Y pA2zV pxqq is neither closed nor open.

Definition 3.4.10. A subset of a quasi-projective set Y is constructible if it isa finite union of locally closed subsets of Y .

The result below follows from Proposition 3.4.8 (we leave the proof to thereader).

Proposition 3.4.11. Let f : X Ñ Y be a map of quasi-projective varieties.Then fpXq is a constructible subset of Y .

Remark 3.4.12. If f : M Ñ N is a smooth map of C8 manifold it might verywell be that fpMq does not contain any non-empty open subset of fpMq. Forexample, let

R fÝÑ T2 :“ R2Z2

t ÞÑ rpt,?

2tqs

We have fpRq “ T2 but fpRq does not contain any non-empty open subset ofT2 because it is a subset of measure 0. Notice also that the analogue of Pro-position 3.4.8 does not hold if we consider real quasi-projective sets (with theZariski topology) and real regular maps: consider the projection

A2R Ą V px2 ` y2 ´ 1q ÝÑ A1

Rpx, yq ÞÑ x

3.5 Grassmannians

Let V be a complex vector space of finite dimension, and let 0 ď h ď dimV .The Grassmannian of h-dimensional vector subpaces of V is the set of (complex)subvector spaces of V of dimension h:

Gr ph, V q :“ tW Ă V | dimW “ hu .


Notice that if h P t0,dimV u, then Gr ph, V q is a singleton, that Gr p1, V q “PpV q, and that we have a bijection

PpV _q ÝÑ Gr pdimV ´ 1, V qrf s ÞÑ kerpfq

We will identify the elements of Gr ph, V q with the points of a projective variety.Consider the Plucker map

Gr ph, V qPÝÑ P

´

ŹhV¯

W ÞÑŹh

W.

(this makes sense:Źh

W is a 1-dimensional subspace ofŹh

V because dimW “

h).

Proposition 3.5.1. Keep notation as above. Then P is injective, and Im P

is a closed subset of P´

ŹhV¯

.

Before proving Proposition 3.5.1, we prove the result below.

Lemma 3.5.2. Let v1, . . . , va P V be linearly independent, and let α PŹh

V .Then

vi ^ α “ 0 @i P t1, . . . , au (3.5.1)

if and only if there exists β PŹhá

V such that

α “ v1 ^ . . .^ va ^ β. (3.5.2)

Proof. The non trivial statement is that if (3.5.1) holds, then (3.5.2) holds.Extend v1, . . . , vs to a basis v1, . . . , vm of V . Given a subset I Ă t1, . . . ,mu ofcardinality s, we let vI “ vi1 ^ . . . ^ vis , where I “ ti1, . . . , isu and 1 ď i1 ă. . . ă is ď m. The collection of the vI ’s is a basis of the exterior algebra

Ź‚V .

Henceα “

ÿ

|I|“h

cIvI ,

where cI are complex numbers. Since

0 “ vi ^ α “ÿ

|I|“hiRI

cIvi ^ vI ,

it follows that t1, . . . , au Ă I for all I such that cI “ 0. Now, if t1, . . . , au Ă I,then vI “ v1 ^ . . .^ va ^ γ. It follows that (3.5.2) holds.

Proof of Proposition 3.5.1. For α PŹh

V , let mα be the linear map

VmαÝÑ

Źh`1V

v ÞÑ v ^ α

3.5. GRASSMANNIANS 43

It follows from Lemma 3.5.2 that if α “ 0, then the kernel of mα has dimensionat most h, and that dim kerpmαq “ h if and only if α is decomposable, i.e. α “w1 ^ . . .^ wh, where w1 ^ . . .^ wh P V are linearly independent. Thus

ImpPq “

!

rαs P P´

ľhV¯

| dimpkermαq ě h)

, (3.5.3)

and if rαs P ImpPq, then rαs “Źh

kerpmαq. The latter equality shows thatP is injective. Morover, the equality in (3.5.3) shows that ImpPq is closed

in PpŹh

V q. In fact, choose a basis v1, . . . , vm of V , and let x1, . . . , xn be theassociated dual basis. Notice that the basis of V determines the basis . . . , vI , . . .(where |II “ h) of

ŹhV , and hence projective coordinates r. . . , ZI , . . .s (where

|II “ h) on PpŹh

V q. Then mα is described (with respect to the chosen bases)by a matrix of order

`

nh`1

˘

ˆh with entries linear functions in x1, . . . , xn. Hencethe right hand side of (3.5.3) is the set of points where all determinants ofminors of order pn ´ h ` 1q ˆ pn ´ h ` 1q of mα vanish. Thus ImpPq is equalto the common zeroes of homogeneous polynomials (of degree n´ h` 1) in thehomogeneous coordinates r. . . , ZI , . . .s, it follows that is closed.

Remark 3.5.3. In the proof of Proposition 3.5.1 we exhibited polynomialsdefining Gr ph, V q which are of high degree. In fact, the ideal of Gr ph, V q Ă

PpŹh`1

V q is generated by quadrics. In the first non-trivial case, i.e. h Rt0, 1,dimV ´ 1,dimV u, i.e. Gr p2, V q with dimV ě 4, we can easily describethe Plucker quadrics generating the (homoheneous) ideal of the Grassmannian;

in fact α PŹ2

V is decomposable if and only if α^ α “ 0.

Remark 3.5.4. We have a bijection between Gr pk ` 1, V q and the set of linearsubspaces of PpV q of dimension k:

Gr pk ` 1, V q ÝÑ Grpk,PpV qq :“ tL Ă PpV q | L linear subspace, dimL “ kuW ÞÑ PpW q.

Thus by Proposition 3.5.1 we may identify Gr pk,PpV qq with a projective set.

In order to do computations, we will need to write explicitly homogeneousof the Plucker image of elements W P Gr ph, V q. This is done as follows. Let

v1, . . . , vm be a basis of V , and let p. . . , vI , . . .q be the associated basis ofŹh

V ,where I runs through subsets of t1, . . . ,mu of cardinality h (notation as in theproof of Lemma 3.5.2). Thus we also have associated homogeneous coordin-

ates r. . . , TI , . . .s on PpŹh

V q. By associating to linearly independent vectorsw1, . . . , wh P V the matrix with rows the coordinates of the wi’s in the chosenbasis, we get a matrix

¨

˚

˝

a11 ¨ ¨ ¨ a1m

.... . .

...ah1 ¨ ¨ ¨ ahm

˛

‹

‚

of rank h. Viceversa, every such matrix determines the coordinates of lin-early independent vectors w1, . . . , wh P V . Now, the homogeneous coordinates


r. . . , TI , . . .s of Ppxw1, . . . , whyq are given by

TI “ det

¨

˚

˝

a1,i1 ¨ ¨ ¨ a1,ih...

. . ....

ah,i1 ¨ ¨ ¨ ah,ih

˛

‹

‚

,

where, as usual, I “ ti1, . . . , ihu with 1 ď i1 ă . . . ă ih ď dimV .

Proposition 3.5.5. The Grassmannian Grph, V q has an open covering by pair-wise intersecting open subsets isomorphic to an affine space of dimension h ¨pdimV ´ hq.

Proof. We identify Grph, V q with its image by the Plucker map PpGrph, V qq Ă

PpŹh

V q. Let m :“ dimV , and let v1, . . . , vm be a basis of V . Keep the notationintroduced above. In particular r. . . , TI , . . .s are homogeneous coordinates on

PpŹh

V q, where I runs through subsets of t1, . . . ,mu of cardinality h. Thus wehave the open covering

Grph, V q “ď

|I|“h

Grph, V qI , (3.5.4)

where, as usual Grph, V qI Ă Grph, V q is the open subset of points such thatTI “ 0. Let I “ t1, . . . , hu. The map

Mh,m´hpKq ÝÑ Grph, V qI¨

˚

˝

a1,1 ¨ ¨ ¨ a1,m´h

.... . .

...ah,1 ¨ ¨ ¨ ah,m´h

˛

‹

‚

ÞÑ x. . . , vi `řm´hj“1 ai,jvh`j , . . .y1ďiďh

(3.5.5)

is an isomorphism. We have similar isomorphisms

Ahpm´kq – Mh,m´hpKq„ÝÑ Gr ph, V qJ

for any other multiindex J . One easily checks that for all subsets I, J Ă

t1, . . . ,mu of cardinality h the interesection Gr ph, V qIXGr ph, V qJ is non empty.

Corollary 3.5.6. The Grassmannian Grph, V q is irreducible, of dimension h ¨pdimV ´ hq.

Remark 3.5.7. Let E Ă Grph, V q ˆ V be the subset of couples pv,W q suchthat v P W , and let π : E Ñ Grph, V q be the defined by pv,W q ÞÑ W . Oneeasily checks that E is closed, and that π is a regular map. The inverse imageπ´1pGrph, V qIq is described as follows. For A P Mh,m´hpKq, let wipAq P Vfor i P t1, . . . , hu be the vector appearing in (3.5.5). Then (3.5.5) gives anisomorphism

Mh,m´hpKq ˆKh ÝÑ π´1pGrph, V qIq

pA, tq ÞÑ pxw1pAq, . . . , whpAqy,řhi“1 tiwipAqq

(3.5.6)

where t “ pt1, . . . , thq P Kh.

3.6. DEGREE OF A CLOSED SUBSET OF PN 45

3.6 Degree of a closed subset of Pn

Let X be an irreducible quasi-projective variety. The codimension of a closedsubset Y Ă X is equal to dimX ´ dimY , and is denoted by codpY,Xq. Belowis the main result of the present section.

Theorem 3.6.1. Let X Ă Pn be closed, and let c :“ codpX,Pnq.

1. If 0 ď k ă c and Λ P Grpk,Pnq is generic, then Λ does not intersect X.

2. If c ď k ď n and Λ P Grpk,Pnq, then Λ does intersect X.

3. There exists a strictly positive integer degX such that for a generic Λ PGrpc,Pnq the intersection ΛXX has cardinality degX.

The proof of the Items in Theorem 3.6.1 will follow from some preliminaryresults.

Example 3.6.2. Let X Ă Pn be a hypersurface. Thus c “ codpX,Pnq “ 1.Item (1) of Theorem 3.6.1 is trivially verified, because PnzX is an open densesubset of Pn. It is also straightforward to check that Item (2) holds. In fact letΛ “ PpW q, where W is a vector subspace of dimension at least 2. If X “ V pF q,thenΛ X X “ V pF|W q, and since dimW ě 2, the non constant homogeneouspolynomial F|W has non trvial zeroes, i.e. Λ X X is not empty. RegardingItem (3): let F be a generator of the homogeneous ideal IpXq; thus F is de-termined up to multiplication by a non zero factor. Then degX “ degF - seeExercise 3.8.1.

Given 0 ď k ď n let ΓXpkq Ă X ˆGrpk,Pnq be defined by

ΓXpkq “ tpp,Λq P X ˆGrpk,Pnq | p P Λu .

Restricting to ΓXpkq the projections of X ˆGrpk,Pnq, we get regular maps

ΓXpkq

π

||

ρ

%%

X Grpk,Pnq

(3.6.1)

If Λ P Grpk,Pnq, then ρ´1pΛq is identified with ΛXX. Hence Theorem 3.6.1is a statement about the fibers of the map ρ. Hence we must start by studyingΓXpkq. The result below is essentially obtained by considering the fibers of themap π, which are all alike.

Proposition 3.6.3. Let X Ă Pn be closed and irreducible. Then ΓXpkq isclosed irreducible of dimension

dim ΓXpkq “ dimX ` kpn´ kq. (3.6.2)


Proof. A straightforward computation shows that ΓXpkq is closed. Let 0 ď i ďn. We identify PnZi with Kn, as usual. We have an isomorphism

XZi ˆGrpk,Knq αiÝÑ ΓXpkq X pPnZi ˆGrpk,Pnqq

pp,W q ÞÑ pp, p`W q

Notice that W is a k-dimensional vector subspace of Kn. Moreover p`Wdenotes the closure in Pn of the affine subspace p `W Ă PnZi » Kn. Omittingthose indices i such that X Ă V pZiq, we get that ΓXpkq is covered by openirreducible subsets of dimension

dimpXZi ˆGrpk,Knqq “ dimX ` dim Grpk,Knq “ dimX ` kpn´ kq.

Since X is irreducible XZiXXZj ‰ H for every couple pi, jq of indices such thatXZi and XZj are non empty. It follows that ΓXpkq is irreducible, of dimensiongiven by (3.6.2).

Corollary 3.6.4. Let X Ă Pn be closed. Then ΓXpkq is closed of dimension

dim ΓXpkq “ dimX ` kpn´ kq. (3.6.3)

If k ď codpX,Pnq then

dim ΓXpkq ď dim Grpk,Pnq (3.6.4)

with equality if and only if k “ codpX,Pnq.

Proof. Let X “ X1 Y ¨ ¨ ¨ YXr be the irreducible decomposition of X. Then

ΓXpkq “ ΓX1pkq Y ¨ ¨ ¨ Y ΓXr pkq.

Thus (3.6.3) follows from Proposition 3.6.3. Let’s prove (3.6.4). Let c :“codpX,Pnq and Xi such that c “ n´ dimXi. Then

dim ΓXipcq “ n´ c` cpn´ cq “ pc` 1qpn´ cq “ dim Grpc,Pnq.

This gives (3.6.4).

Proof of Item (a) of Theorem 3.6.1. By Corollary 3.6.4, the image of themap ρ in (3.6.1) is a proper closed subset of Grpk,Pnq. Hence for generic Λ PGrpk,Pnq, the fiber ρ´1pΛq “ ΛXX is empty.

The result below will be useful in proving Items (b), (c) of Theorem 3.6.1,and also in other circumstances.

Proposition 3.6.5. Let X Ă Pn be closed. Suppose that p P PnzX and thatH Ă Pnz tpu is a hyperplane. Let

pPnztpuq πÝÑ H

x ÞÑ xp, xy XH

be the projection. Then πpXq is a closed subset of H and dimπpXq “ dimX.

3.6. DEGREE OF A CLOSED SUBSET OF PN 47

Proof. We may assume that X is irreducible. Since π|X is regular and X isprojective πpXq is closed by Proposition 2.6.3. It remains to prove thatdimπpXq “ dimX. We may assume that p “ r0, . . . , 0, 1s, H “ V pZnq, and Xis not contained in V pZ0q. We have

πprZ0, . . . , Znsq “ rZ0, . . . , Zn´1s.

Let Y :“ πpXq. We have an injection of fields π˚ : KpY q ãÑ KpXq, and we mustprove that rKpXq : π˚pKpY qqs ă 8. The field KpY q is generated (over K) by

pZ1Z0q|Y , . . . , pZn´1Z0q|Y .

On the other hand KpXq is generated by

pZ1Z0q|X , . . . , pZn´1Z0q|X , pZnZ0q|X .

Since π˚ppZiZ0q|Y q “ pZiZ0q|X , it suffices to prove that pZnZ0q|X is algebraicover pZ1Z0q|X , . . . , pZn´1Z0q|X . There exists F P IpXq such that F ppq ‰ 0because p R X. Since p “ r0, . . . , 0, 1s we get that

F “ a0Zdn ` a1Z

d´1n ` ¨ ¨ ¨ ` ad, ai P KrZ0, . . . , Zn´1si, a0 ‰ 0.

Dividing by Zd0 and restricting to X we get that

a0 ¨ ppZnZ0q|Xqd ` a1 ¨ ppZnZ0q|Xq

d´1 ` ¨ ¨ ¨ ` ad “ 0 (3.6.5)

where aj :“ pajZj0q|X for 1 ď j ď d. Since a0 ‰ 0, Equation (3.6.5) shows that

pZnZ0q|X is algebraic over pZ1Z0q|X , . . . , pZn´1Z0q|X .

Proof of Item (b) of Theorem 3.6.1. The proof is by induction on codpX,Pnq.If codpX,Pnq “ 0 the result is trivial (if you don’t like to start from codpX,Pnq “0 you may begin from codpX,Pnq “ 1, i.e. X a hypersurface). Let’s prove theinductive step. Let p P Λ. If p P X there is nothing to prove; thus we mayassume that p R X. Choose a hyperplane H Ă Pn not containing p, and let πbe projection from p, as in (3.6.6). Then Y :“ πpXq Ă H » Pn´1 is closed anddimY “ dimX by Proposition 3.6.5. Thus codpY,Pn´1q “ pcodpX,Pnq´ 1q.Let Λ1 :“ πpΛz tpuq. Then Λ1 Ă H is a linear subspace with dim Λ1 “ pdim Λ´1q,and hence dim Λ1 ě codpY,Pn´1q. By the inductive hypothesis it follows thatΛ1 X Y is not empty. Let y P Λ X Y . Since y P πpXq there exists x P X suchthat πpxq “ y. By definition of π, we have x P xp, yy. Since p P Λ and y P Λ(because Λ1 “ ΛXH), it follows that x P Λ. Thus x P ΛXX.

Proof of Item (c) of Theorem 3.6.1. We start by defining the degree of aclosed X Ă Pn. First assume that X is irreducible. Let c :“ codpX,Pnq.Let

ΓXpcqπÝÑ Grpc,Pnq

pp,Λq ÞÑ Λ(3.6.6)


Since ΓXpcq and Grpc,Pnq are varieties we have a well-defined deg π. By Co-rollary 3.6.4 we have dim ΓXpcq “ dim Grpc,Pnq: thus deg π ă 8. The degreeof X is defined to be

degX :“ degpΓXpcqπÝÑ Grpc,Pnqq. (3.6.7)

In general let X “ X1 Y ¨ ¨ ¨ YXr be the irreducible decomposition of X. Thedegree of X is defined to be

degX :“ÿ

dimXi“dimX

degXi. (3.6.8)

If X is irreducible, Item (c) of Theorem 3.6.1 follows from Proposition3.4.3 applied to the map π of (3.6.6). In general let X “ X1 Y ¨ ¨ ¨ YXr be theirreducible decomposition of X. By Item (a) of Theorem 3.6.1, for genericΛ P Grpc,Pnq

ΛXXi “ H if dimXi ă dimX, ΛX pXi XXjq “ H if i ‰ j.

It follows that for Λ generic

ΛXX “ğ

dimXi“dimX

ΛXXi

and hence Item (c) follows from the case X irreducible.

Remark 3.6.6. Theorem 3.6.1 gives a characterization of the dimension of aclosed X Ă Pn via its intersections with linear subspaces.

3.7 Intersecting closed subsets of a projectivespace

Theorem 3.6.1 characterizes the dimension of a closed subset of a projectivespace via its intersection with linear subsapces. In the persent section we deriveinteresting consequences of that result.

First we introduce the join of two closed subsets X,Y Ă PN such that

xXy X xY y “ H, (3.7.1)

where xXy and xY y are the linear subspaces generated by X and Y respectively.

Definition 3.7.1. The join of X and Y is the subset of PN swept out by thelines joining a point of X to a point of Y :

JpX,Y q :“ď

xPX,yPY

xx, yy. (3.7.2)

Lemma 3.7.2. Let X,Y Ă PN be closed subsets such that (3.7.1) holds. Then

3.7. INTERSECTING CLOSED SUBSETS OF A PROJECTIVE SPACE 49

1. JpY,W q is closed,

2. if X and Y are irreducible JpX,Y q is irreducible,

3. dim JpX,Y q “ dimX ` dimY ` 1.

Proof. Let m :“ dimxXy and n :“ dimxY y. There exist homogeneous coordin-ates

rS0, . . . , Sm, T0, . . . , Tn, U0, . . . , Ups

on PN such that

xXy “ trS0, . . . , Sm, 0, . . . , 0su, xY y “ tr0, . . . , 0, T0, . . . , Tn, 0, . . . , 0su.

Then

JpX,Y q “ trS0, . . . , Sm, T0, . . . , Tn, 0, . . . , 0s | rSs P X, rT s P Y u. (3.7.3)

Item (1) follows at once. Let p P pJpY,W qzXzY q. By (3.7.1) there is uniquecouple pϕ1ppq, ϕ2ppqq P X ˆ Y such that p P xϕ1ppq, ϕ2ppqy, and the map

pJpX,Y qzXzY qϕÝÑ X ˆ Y

p ÞÑ pϕ1ppq, ϕ2ppqq(3.7.4)

is regular, with fibers isomorphic to K˚. Moreover for any 0 ď i ď m and0 ď j ď n the inverse image ϕ´1pXSi ˆ YTj q is isomorphic to XSi ˆ YTj ˆK˚.Items (2) and (3) follow.

Proposition 3.7.3. Let X Ă Pn be closed, irreducible of strictly positive di-mension. Let H Ă Pn a hyperplane not containing X. Then XXH is not emptyand every irreducible component of X XH has dimension equal to pdimX ´ 1q.

Proof. The intersection is non empty by Theorem 3.6.1. First we will provea weaker result, namely that

dimX XH “ dimX ´ 1. (3.7.5)

Let c :“ codpX,Pnq. Then (3.7.5) is equivalent to codpX X H,Hq “ c. SinceX XH Ĺ X we have dimX XH ă dimX and hence codpX XH,Hq ě c. ByTheorem 3.6.1 applied to the closed pX XHq Ă H it suffices to prove that ifL Ă H is an arbitrary linear subspace with dimL “ c then LX pX XHq ‰ H.By Theorem 3.6.1 applied to X we have L XX ‰ H: since L Ă H we haveL XX Ă L X pX XHq. This proves (3.7.5). The proposition states a strongerresult namely that every irreducible component of X XH has dimension equalto pdimX ´ 1q. The proof is by induction on codpX,Pnq, the initial case beingcodpX,Pnq “ 1 (Notice that if codpX,Pnq “ 0 the statement of the prosition istrivially true). If codpX,Pnq “ 1 then X is a hypersurface by Corollary 3.3.4and hence XXH is a hypersurface in H: by Corollary 3.3.4 every irreduciblecomponent of X X H has coddimension one in H. Let’s prove the inductivestep. We assume that codpX,Pnq “ c ě 2. Suppose that W1 is an irreducible


component of XXH. Pick a point p P HzX and a hyperplane H 1 not containingp and different from H. Let

Pnztpu πpÝÑ H 1

q ÞÑ xp, qy XH 1

be the projection. We will consider πppXqXπppHq. Let XXH “W1Y¨ ¨ ŸWr

be the irreducible decomposition of X X H. Let us prove that there exists psuch that

πppW1q Ć πppWiq @i P t2, . . . , ru. (3.7.6)

In fact, let q P W1zŤri“2Wi, and let i P t2, . . . , ru. Then Jpq,Wiq is defined,

and by Lemma 3.7.2, it is closed irreducible. Moreover, by Lemma 3.7.2

dim Jpq,Wiq “ dimWi ` 1 (3.7.7)

Since H Č X, dimWi ď dimX ´ 1 and since codpX,Pnq ě 2 we have dimWi ď

dimH ´ 2. Thus (3.7.7) gives that Jpq,Wiq ‰ H. Hence there exists p PHz

Ťri“2 Jpq,Wiq. Then πppqq R πppWiq for i P t2, . . . , ru, and hence (3.7.6)

holds.Each of πppW1q, . . . , πppWrq is closed, and

πppXq X πppHq “ πppW1q Y . . .Y πppWrq.

Moreover, by (3.7.6) it follows that πppW1q is an irreducible component ofπppXq X πppHq. By Proposition 3.6.5 we have dimπpX “ dimX and hencecodpπppXq, H

1q “ pcodpX,Pnq ´ 1q. By the inductive hypothesis we get thatcodpπppW1q, πppXqq “ 1. Since dimπppW1q “ dimW1 and dimπppXq “ dimX(by Proposition 3.6.5) we get that codpW1, Xq “ 1.

Corollary 3.7.4. Let X Ă Pn be closed of codimension c. Let Λ P Grpc,Pnq.Then XXΛ is not empty and every irreducible component of XXΛ has dimensionat least pdimX ´ cq.

The proposition below is a remarkable generalization of the well-known linearalgebra result: “a system of n homogeneous linear equations in pn`1q unknownshas at a non-trivial solution”.

Proposition 3.7.5. Let Y,W Ă Pn be closed and suppose that pdimY `

dimW q ě n. Then Y X W “ H and moreover every irreducible componentof Y XW has dimension at least pdimY ` dimW ´ nq.

Proof of Proposition 3.7.5. Let rs0, . . . , sn, t0, . . . , tns be homogeneous co-ordinates on P2n`1. We have two embeddings

Pn iÝÑ P2n`1

rX0, . . . , Xns ÞÑ rs0, . . . , sn, 0, . . . , 0sPn j

ÝÑ P2n`1

rX0, . . . , Xns ÞÑ r0, . . . , 0, X0, . . . , Xns

(3.7.8)

3.8. EXERCISES 51

Since the images of i and j are disjoint linear subspaces of P2n`1 the joinJpipY q, jpW qq is defined. We will intersect JpipY q, jpW qq with the linear sub-space of P2n`1 defined by

Λ :“ V ps0 ´ t0, . . . , sn ´ tnq.

We have an isomorphism

Y XW„ÝÑ ΛX JpipY q, jpW qq

rX0, . . . , Xns ÞÑ rX0, . . . , Xn, X0, . . . , Xns(3.7.9)

By Lemma 3.7.2 the closed JpipY q, jpW qq Ă P2n`1 has dimension pdimY `dimW ` 1q. On the other hand Λ is a codimension-pn ` 1q linear subspace ofP2n`1; by Corollary 3.7.4 ΛX JpipY q, jppW qq is not empty and every irredu-cible component of ΛXJpipY q, jppW qq has dimension at least pdimY `dimW ´

nq. Isomorphism (3.7.9) gives that Y XW is not empty and every irreduciblecomponent of Y XW has dimension at least pdimY ` dimW ´ nq.

Example 3.7.6. Let n ě 2 and X Ă Pn be a smooth hypersurface. Then X isirreducible. In fact suppose that X “ Y YW where Y,W are proper closedsubsets of X. Then Y and W are of pure dimension pn´ 1q and hence Y XWis not empty by Proposition 3.7.5. Let p P Y XW : as is easily checked X issingular at p, that is a contradiction.

3.8 Exercises

Exercise 3.8.1. Let X Ă Pn be a hypersurface, and let IpXq “ pF q.

(a) Let ∆pF q Ă Grp1,Pnq be the subset of lines PpW q such that there exist p P PpW qfor which multppF q ě 2. - see Exercise 2.7.5. Prove that ∆pF q is a properclosed subset of Grp1,Pnq. (Hint: for the proof that ∆pF q is closed, see Exercise2.7.5, for the proof that it is a proper subset, see the proof of Lemma 3.4.4.)

(b) Prove that degX “ degF . (Hint: recall Item (b) of Exercise 2.7.5.)

Exercise 3.8.2. Let X Ă Pn be a hypersurface. Prove that

degX “ maxt|ΛXX| | Λ P Grp1,Pnq such that ΛXX is finiteu. (3.8.10)

(An analogous result holds for a closed pure dimensional X Ă Pn of any codimension,

see Proposition ??.)

Exercise 3.8.3. Let ∆d Ă PpKrT0, T1sdq be the subset of rF s for which there existp P P1 such that multppF q ě 2 - see Exercise 2.7.5.

(a) Prove that ∆d is an irreducible hypersurface in PpKrT0, T1sdq.

(b) Prove that ∆d has degree 2d´ 2.

Exercise 3.8.4. Let R be an integral domain. Let F P RrT0, T1sm and G P

RrT0, T1sn; we assume throughout that m,n are not both 0. The resultant


Rm,npF,Gq is the element of R defined as follows. Consider the map of freeR-modules

RrT0, T1sn´1 ‘RrT0, T1sm´1Lm,npF,GqÝÑ RrT0, T1sm`n´1

pΦ,Ψq ÞÑ Φ ¨ F `Ψ ¨G(3.8.11)

and let Sm,npF,Gq be the matrix of Lm,npF,Gq relative to the basis

pTn´10 , 0q, pTn´2

0 T1, 0q, . . . , p0, Tm´10 q, p0, Tm´2

0 T1q, . . . , p0, Tm´11 q

of the domain and the basis

Tm`n´10 , Tm`n´2

0 T1, . . . , T0Tm`n´21 , Tm`n´1

1

of the codomain. Then

Rm,npF,Gq :“ detSm,npF,Gq. (3.8.12)

Explicitly: if

F “mÿ

i“0

aiTmí0 T i1, G “

nÿ

j“0

bjTn´j0 T j1 (3.8.13)

then

Rm,npF,Gq “ det

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

a0 0 ¨ ¨ ¨ 0 b0 0 ¨ ¨ ¨ 0a1 a0 ¨ ¨ ¨ 0 b1 b0 ¨ ¨ ¨ 0...

......

......

......

......

... ¨ ¨ ¨ a0

...... ¨ ¨ ¨ b0

am am´1 ¨ ¨ ¨... bn bn´1 ¨ ¨ ¨

...

0 am ¨ ¨ ¨... 0 bn ¨ ¨ ¨

...

0 0 ¨ ¨ ¨... 0 0 ¨ ¨ ¨

......

... ¨ ¨ ¨...

...... ¨ ¨ ¨

...0 0 ¨ ¨ ¨ am 0 0 ¨ ¨ ¨ bn

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

. (3.8.14)

Now let k be a field and let k Ă K be an algebraic closure of k. Let F P

krT0, T1sm and G P krT0, T1sn.

(a) Prove that Rm,npF,Gq “ 0 if and only if there exists H P krT0, T1sd withd ą 0 which divides F and G in krT0, T1s.

(b) Prove that Rm,npF,Gq “ 0 if and only if there exists a common non-trivial root of F and G in P1

K, i.e. a non zero pT0, T1q P K2 such thatF pT0, T1q “ GpT0, T1q “ 0.

3.8. EXERCISES 53

(c) Suppose that charK does not divide d. Give an explicit homogeneouspolynomial of degree p2d´ 2q in the coefficients ci of

F “dÿ

i“0

ciTi0 ¨ T

dí1

which vanishes if and only if there exists p P P1K such that multppF q ě 2.

(Hint: recall Item (d) ofExercise 2.7.5.)

Exercise 3.8.5. Let Cd Ă Pd be the the image of the Veronese map νd : P1Ñ Pd

given by νdprs, tsq “ rsd, sd´1t, . . . , tds. Prove that deg Cd “ d.

Let X Ă Pn be a closed subset. For k P t0, . . . , nu, we let

FkpXq :“ tΛ P Grpk,Pnq | Λ Ă Xu

be the set of k dimensional linear spaces contained in X. Thus F0pXq “ X. Thefirst interesting case is F1pXq, i.e. the set of lines contained in X. By solvingthe following exercises, one proves interesting results about FkpXq.

Exercise 3.8.6. Let X Ă Pn be a closed subset. Prove that FkpXq is a closed subsetof Grpk,Pnq, arguing as follows:

1. If X “ Pn, then FkpPnq “ Grpk,Pnq. If X is not Pn, then X “ V pP1q X . . . XV pPrq and FkpXq “ FkpV pP1qqX . . .XFkpV pPrqq. Hence it suffices to prove theresult for X “ V pP q Ă Pn a hypersurface.

2. Since we have the open covering of Grpk,Pnq given by (3.5.4), it suffices to showthat the intersection FkpV pP qq X Grpk,PnqI is closed for every multiindex I Ăt0, . . . , nu of cardinality k ` 1. Prove by explicit computation that FkpV pP qq XGrpk,PnqI is closed.

Exercise 3.8.7. Let LkpKrZ0, . . . , Znsdq Ă Grpk,Pnq ˆ PpKrZ0, . . . , Znsdq be

LkpKrZ0, . . . , Znsdq :“ tpΛ, rP sq | Λ Ă V pP qu.

Prove that LkpKrZ0, . . . , Znsdq is closed, arguing as follows:

1. Since we have the open covering

Grpk,Pnq ˆ PpKrZ0, . . . , Znsdq “ď

|I|“k`1

Grpk,PnqI ˆ PpKrZ0, . . . , Znsdq,

in order to prove that LkpKrZ0, . . . , Znsdq is closed it suffices to show that theintersection of LkpKrZ0, . . . , Znsdq with the open subset indicized by I, call itLkpKrZ0, . . . , ZnsdqI , is closed.

2. Let I “ t0, . . . , ku. Identify Grpk,PnqI with Mk`1,n´kpKq via the isomorphismin (3.5.5). Going back to the proof of Proposition ??, we see that

LkpKrZ0, . . . , ZnsdqI “ tpA, rP sq | CJpA,P q “ 0 @Ju. (3.8.15)

Since each CJpA,P q is a polynomial in the entries of A and (the coefficients)of P , homogeneous in P , we get that LkpKrZ0, . . . , ZnsdqI is closed. The prooffor the other subsets I Ă t0, . . . , nu of cardinality k ` 1 is the same. ThusLkpKrZ0, . . . , Znsdq is closed in Grpk,Pnq ˆ PpKrZ0, . . . , Znsdq.


Exercise 3.8.8. Prove that if k ă n then LkpKrZ0, . . . , Znsdq is irreducible and

dimLkpKrZ0, . . . , Znsdq “ dimPpKrZ0, . . . , Znsdq`pk`1q¨pn´kq´

˜

d` k

k

¸

, (3.8.16)

arguing as follows:

1. Show that

LkpKrZ0, . . . , ZnsdqI X LkpKrZ0, . . . , ZnsdqI1 “ H

for any two subsets I, I 1 Ă t0, . . . , nu.

2. Since each LkpKrZ0, . . . , ZnsdqI is open, and any two have non empty intersec-tion by the previous item, it will suffice to show that each LkpKrZ0, . . . , ZnsdqIis irreducible of dimension given by (3.8.16). For P P KrZ0, . . . , Znsd, let

P “ÿ

degK“d

PKZK ,

whereK runs through multiindicesK “ pk0, . . . , knq of degree d. Rewrite (3.8.15)as

LkpKrZ0, . . . , ZnsdqI “ tpA, rP sq |ÿ

degK“d

DJ,KpAqPK “ 0 @Ju,

where DJ,KpAq is a polynomial in the entries of the matrix A.

3. By the previous item, LkpKrZ0, . . . , ZnsdqI is the set of couples pA, rP sq, whereP is any non trivial solution of

`

d`kk

˘

homogeneous linear equations. Show thesystem of linear equations has maximum rank for each A by observing that therestriction map

KrZ0, . . . , Znsd ÝÑ Krλ0, . . . , λksdP ÞÑ P pλ0w0pAq ` . . .` λkwkpAqq

is surjective, where wipAq :“ vi `řm´hj“1 ai,jvh`j for i P t0, . . . , ku, so that ΛA

(the linear subspace corresponding to A) is the span of rw0pAqs, . . . , rwkpAqs.

4. Given A, by the previous item there exists a`

d`kk

˘

ˆ`

d`kk

˘

minor of the matrixpDJ,kpAqqJ,K , call it mpAq with non zero determinant. Let Mk`1,n´kpKqm Ă

Mk`1,n´kpKq be the open subset of points such the minor mpAq has non zerodeterminant. Show that the open subset

LkpKrZ0, . . . , ZnsdqI X tA PMk`1,n´kpKq | detmpAq “ 0u ˆ PpKrZ0, . . . , Znsdq(3.8.17)

is isomorphic to tA PMk`1,n´kpKq | detmpAq “ 0u ˆ Pr, where

r “ PpKrZ0, . . . , Znsdq ´

˜

d` k

k

¸

.

Conclude from this that LkpKrZ0, . . . , Znsdq is irreducible, of dimension givenby (3.8.16).

Exercise 3.8.9. Let k ă n. Prove that the subset of PpKrZ0, . . . , Znsdq defined by

trP s P PpKrZ0, . . . , Znsdq | FkpXq “ Hu (3.8.18)

3.8. EXERCISES 55

is closed, irreducible, of dimension at most equal to

dimPpKrZ0, . . . , Znsdq ` pk ` 1q ¨ pn´ kq ´

˜

d` k

k

¸

.

In particular, show that for all d ě 2n´2 there exist hypersurfaces V pP q Ă Pn defined

by a degree d homogeneous P which do not contain a line.


Chapter 4

First order approximation

4.1 Tangent space

One definition of tangent space of a C8 manifold M at a point x PM is as thereal vector space of derivations of the space EM,x of germs of C8 functions at x.We will give an analogous definition of the tangent space of a quasi projectivevariety. A fundamental difference between quasi projective varieties and smoothmanifolds is that the dimension of the tangent space at a point might dependon the point. Intuitively, the reason is that a quasi projective variety can havenon smooth points, meaning that in a neighborhood of such a point the varietyis not a complex submanifold of the ambient projective space.

Let X be a quasi projective variety. We start by defining the ring of germsof regular functions at x P X.

Definition 4.1.1. Let X be a quasi projective variety, and let x P X. LetpU, φq and pV, ψq be couples where U, V are open subsets of X containing x,and φ P KrU s, ψ P KrV s. Then pU, φq „ pV, ψq if there exists an open subsetW Ă X containing x such that W Ă U X V and φ|W “ ψ|W .

One checks easily that „ is an equivalence relation: an equivalence class forthe realtion „ is a germ of regular function of X at x. We may define a sumand a product on the set of germs of regular functions of X at x by setting

rpU, φqs ` rpV, ψqs :“ r`

U X V, φ|UXV ` ψ|UXV˘

s, (4.1.1)

and

rpU, φqs ¨ rpV, ψqs :“ r`

U X V, φ|UXV ¨ ψ|UXV˘

s. (4.1.2)

Of course one has to check that the equivalence class of the sum and productis independent of the choice of representatives: this is easy, we leave details tothe reader. With these operations, the set of germs of regular functions of X atx is a ring.

57

58 CHAPTER 4. FIRST ORDER APPROXIMATION

Definition 4.1.2. Let X be a quasi projective variety, and let x P X. Thelocal ring of X at x is the ring of germs of regular functions of X at x, and isdenoted OX,x.

We have a natural homomorphism of rings

KrXs ρÝÑ OX,x

f ÞÑ rpX, fqs(4.1.3)

Lemma 4.1.3. Suppose that X is an affine variety, and let x P X. If ϕ P OX,xthen there exist f, g P KrXs, with gpxq “ 0, such that ϕ “ ρpfq

ρpgq .

Proof. Let ϕ be represented by pU, hq, where U Ă X is open, and x P U . Sincethe principal open affine subsets of X form a basis of the Zariski topology, thereexists α P KrXs such that Xα Ă U and x P Xα (see Remark 2.4.4). Thusϕ “ rpXα, h|Xαqs. By Remark 2.4.4, there exist f P KrXs and m P N such

that h is the restriction to Xα of fαm . Then ϕ “ ρpfq

ρpαmq .

There is a well-defined surjective homomorphism

OX,x ÝÑ KrpU, φqs ÞÑ φpaq

(4.1.4)

The kernel

mx :“ trpU, φqs | φpxq “ 0u

of (4.1.4) is a maximal ideal, because (4.1.4) is a surjection to a field.

Proposition 4.1.4. With notationas above, mx is the unique maximal ideal ofOX,x, and hence OX,x is a local ring. Moreover, OX,x is Noetherian.

Proof. Let f “ rpU, φqs P pOX,xzmxq. Then W :“ pUzV pφqq is an open subsetof X containing x and hence g :“ rpW, pφ|W q

´1s belongs to OX,x. Since gf “ 1we get that f is invertible. It follows that mx contains any proper ideal of OX,xand hence is the unique maximal ideal of OX,x.

In order to prove that OX,x is Noetherian, we notice that if U Ă X isZariski open and contains x, then the natural homomorphism OU,x Ñ OX,x is anisomorphism. Since X is covered by open affine subsets, it follows that we mayassume that X is affine. Let I Ă OX,x be an ideal. Let ρ be the homomorphismin (4.1.3). Then ρ´1pIq is a finitely generated ideal, because KrXs is Noetherian.Let f1, . . . , fr be generators of ρ´1pIq. Then ρpf1q, . . . , ρpfrq generate I. In factlet ϕ P I. By Lemma 4.1.3, there exist f, g P KrXs, with gpxq “ 0, such that

ϕ “ ρpfqρpgq . We have f “

řri“1 aifi, and hence ϕ “

řri“1

ρpaiqρpgq ρpfiq.

The homomorphism (4.1.4) equips K with a structure of OX,x-module. MoreoverOX,x is a K-algebra. Thus it makes sense to speak of K-derivations of OX,x toK.

4.1. TANGENT SPACE 59

Definition 4.1.5. Let X be a quasi projective variety, and let x P X. TheZariski tangent space to X at x is DerKpOX,x,Kq, and will be denoted by ΘxX.Thus ΘxX is an OX,x-module (see Section A.5), and since mx annihilatesevery derivation OX,x Ñ K, it is a complex vector space.

Lemma 4.1.6. Let a P An. The complex linear map

ΘaAn ÝÑ KnD ÞÑ pDpz1q, . . . , Dpznqq

(4.1.5)

is an isomorphism.

Proof. The formal partial derivative BBzm

defined by (A.5.1) defines an elementof ΘaAn by the familiar formula

B

Bzm

ˆ

f

g

˙

paq :“

BfBzm

paq ¨ gpaq ´ fpaq ¨ BgBzm

paq

gpaq2.

(See Example A.5.3.) Since BBzm

pzjq “ δmj , the map in (4.1.5) is surjective.Let’s prove that the map in (4.1.5) is injective. Assume that D P ΘX,x

is mapped to 0 by the map in (4.1.5), i.e. Dpxjq “ 0 for j P t1, . . . , nu. Letf, g P Krz1, . . . , zns, with gpaq “ 0. Then

D

ˆ

f

g

˙

“Dpfq ¨ gpaq ´ fpaq ¨Dpgq

gpaq2.

(See Example A.5.3.) Hence it suffices to show that Dpfq “ 0 for everyf P Krz1, . . . , zns. Consider the first-order expansion of f around a i.e. write

f “ fpaq `nÿ

i“1

cipzi ´ aq `R, R P m2a. (4.1.6)

Since D is zero on constants (because D is a K-derivation) and Dpzjq “ 0 forall j it follows that Dpfq “ DpRq, and the latter vanishes by Leibniz’ rule andthe hypothesis Dpzjq “ 0 for all j.

The differential of a regular map at a point of the domain is defined bythe usual procedure. Explicitly, let f : X Ñ Y be a regular map of quasiprojective varieties, let x P X and y :“ fpxq. There is a well-defined pull-backhomomorphism

OY,yf˚

ÝÑ OX,xrpU, φqs ÞÑ rpf´1U, φ ˝ pf|f´1U qqs

(4.1.7)

The differential of f at x is the linear map of complex vector spaces

TxXdfpxqÝÑ TyY

D ÞÑ pφ ÞÑ D pf˚φqq(4.1.8)


The differential has the customary functorial properties. Explicitly, supposethat we have

X1f1 // X2

f2 // X3 , x1 P X1, x2 “ f1 px1q .

Since pf2 ˝ f1q˚“ f˚1 ˝ f

˚2 we have

d pf2 ˝ f1q px1q “ df2 px2q ˝ df1 px1q . (4.1.9)

Moreover d IdXpxq “ IdTxX for x P X.

Remark 4.1.7. It follows from the above that if f is an isomorphism, thendfpxq : TxX Ñ TfpxqY is an isomorphism, in particular dimTxX “ dimTyY .

The next result shows how to compute the Zariski tangent space of a closedsubset of An. Since every point x of a quasi projective variety X is containedin an open affine subset U , and ΘxX “ ΘxU (because restriction defines anidentification OX,x “ OU,x), the result will allow to compute the Zariski tangentspace in general.

Proposition 4.1.8. Let ι : X ãÑ An be the inclusion of a closed subset anda P X. The differential dιpaq : ΘaX Ñ ΘaAn is injective and, identifying ΘaAnwith Kn via (4.1.5), we have

Im djpaq “

#

v “ pv1, . . . , vnq P Kn |nÿ

i“1

Bf

Bzipaq ¨ vi “ 0 @f P IpXq

+

. (4.1.10)

Proof. The differential dιpaq is injective because the pull-back ι˚ : OAn,a Ñ OX,ais surjective. Let D P DerKpOX,a,Kq. If f P IpXq Ă Krz1, . . . , zns, thendιpDqpfq “ Dpι˚fq “ Dp0q “ 0. Hence Im dιpaq is contained in the right-hand side of (4.1.10). Let’s prove that Im dιpaq contains the right-hand side

of (4.1.10). Let rD P DerKpOAn,a,Kq belong to the right hand side of (4.1.10),

i.e. rDpfq “ 0 for all f P IpXq. By Item (3) of Example A.5.3 it follows thatrDp fg q “ 0 whenever f, g P Krz1, . . . , zns and f P IpXq (of course we assume

that gpaq “ 0). Thus rD descends to a K-derivation D P DerKpOX,a,Kq, andrD “ dι˚paqpDq.

Remark 4.1.9. With the hypotheses of Proposition 4.1.8, suppose that IpXqis generated by f1, . . . , fr. Then

Im djpaq “

#

v “ pv1, . . . , vnq P Kn |nÿ

i“1

BfkBzipaq ¨ vi “ 0 k P t1, . . . , ru

+

.

In fact, the right hand side of the above equation is equal to the right hand side

of (4.1.10), because if f “řrj“1 gjfj , then Bf

Bzipaq “

řrj“1 gjpaq

BfjpaqBzi

.

4.1. TANGENT SPACE 61

Example 4.1.10. Let f P Krz1, . . . , zns be a polynomial without multiple factors,i.e. such that

a

pfq “ pfq, and let X “ V pfq. Let a P X; by Remark 4.1.9Zariski’s tangent space to X is the subspace of Kn defined by

nÿ

i“1

Bf

Bzipaq ¨ vi “ 0.

Hence

dim ΘaX “

#

n´ 1 if p BfBz1paq, . . . , Bf

Bznpaqq “ 0,

n if p BfBz1paq, . . . , Bf

Bznpaqq “ 0.

Let us show that

XzV

ˆ

Bf

Bz1, . . . ,

Bf

Bzn

˙

(4.1.11)

is an open dense subset of X (it is obviously open, the point is that it is dense),i.e. dim ΘaX “ n´ 1 for a in an open dense subset of X.

First assume that f is irreducible. Reordering the coordinates if necessary,we may assume that

f “ a0zdn ` a1z

d´1n ` ¨ ¨ ¨ ` ad, ai P Krz1, . . . , zn´1s, a0 ‰ 0, d ą 0.

ThusBf

zn“ da0z

d´1n ` pd´ 1qa1z

d´2n ` ¨ ¨ ¨ ` ad´1 “ 0.

The degree in zn of f is d (i.e. f has degree d as element of Krz1, . . . , zn´1srzns)while the degree in zn of Bfzn is pd ´ 1q and hence f - Bfzn . This shows that theset in (4.1.11) is dense in X if f is irreducible.

In general, let f “ f1 ¨ ¨ ¨ ¨ ¨fr be the decomposition of f as product of primefactors. Let Xi “ V pfiq. Then

X “ X1 Y ¨ ¨ ¨ YXr

is the irreducible decomposition of X. As shown above, for each i P t1, . . . , ru

XjzV

ˆ

Bfjz1, . . . ,

Bfjzn

˙

“ H.

Hence there exists a P Xj such thatBfjzhpaq ‰ 0 for a certain 1 ď h ď n. We may

assume in addition that a does not belong to any other irreducible componentof X. It follows that

Bf

zhpaq “

Bfjzhpaq ¨

ź

k‰j

fkpaq ‰ 0.

This proves that the open set in(4.1.11) has non empty intersection with everyirreducible component of X, and hence is dense in X.

Notice also that if a belongs to more than one irreducible component of X,then all partial derivatives of f vanish at a. In other words, any point in theopen dense subset of points a such that dim Θa “ n ´ 1 belongs to a singleirreducible component of X.


The result below shows that the behaviour of the tangent space examinedin the above example is typical of what happens in general.

Proposition 4.1.11. Let X be a quasi projective variety. The function

X ÝÑ Nx ÞÑ dim ΘxX

(4.1.12)

is Zariski upper-semicontinuous, i.e. for every k P N

Xk :“ tx P X | dim ΘxX ě ku

is closed in X.

Proof. Since X has an open affine covering, we may suppose that X Ă An isclosed. Let IpXq “ pf1, . . . , frq. For x P An let

Jpf1, . . . , fsqpxq :“

¨

˚

˝

Bf1z1pxq ¨ ¨ ¨

Bf1znpxq

.... . .

...Bfrz1pxq ¨ ¨ ¨

Bfrznpxq

˛

‹

‚

be the Jacobian matrix of pf1, . . . , fsq at x. By Proposition 4.1.8 we havethat

Xk “ tx P X | rk Jpf1, . . . , frqpxq ď n´ ku . (4.1.13)

Given multi-indices I “ t1 ď i1 ă . . . ă im ď su and J “ t1 ď j1 ă . . . ăjm ď nu let Jpf1, . . . , fsqpxqI,J be the m ˆ m minor of Jpf1, . . . , frqpxq withrows corresponding to I and columns corresponding to J (if m ą mintr, nu weset Jpf1, . . . , fsqpxqI,J “ 0). We may rewrite (4.1.13) as

Xk “ X X V p. . . ,det Jpf1, . . . , frqpxqI,J , . . .q|I|“|J|“n´k`1 .

It follows that Xk is closed.

4.2 Cotangent space

Let X be a quasi projective variety, and let x P X. The cotangent space to Xat x is the dual complex vector space of the tangent space ΘxX, and is denotedΩXpxq:

ΩXpxq :“ pΘxXq_. (4.2.1)

We define a map

OX,xdÝÑ ΩXpxq (4.2.2)

as follows. Let f P OX,x be represented by pU, φq. The codomain of the dif-ferential dφpxq : ΘxU Ñ ΘφpxqK is identified with with K, because of the iso-morphism in (4.1.5), and hence dφpxq P pΘxUq

_. Since U Ă Z is an open subsetcontaining x, the differential at x of the inclusion map defines an identificationΘxU

„ÝÑ ΘxX. Thus dφpxq P pΘxXq

_ “ ΩXpxq. One checks immediately thatif pV, ψq is another representative of f then dψpxq “ dφpxq. We let

dfpxq :“ dφpxq, pU, φq any representative of f .

4.2. COTANGENT SPACE 63

Remark 4.2.1. We equip ΩXpxq with a structure of OX,x-module by composingthe evaluation map OX,x Ñ K given by (4.1.4) and scalar multiplication of thecomplex vector-space ΩZpaq. With this structure (4.2.2) is a derivation over K.

Remark 4.2.2. Let f P Krz1, . . . , zns and a P An. Then the familiar formula

dfpaq “nÿ

i“1

Bf

Bzipaqdzipaq

holds. In fact this follows from the first-order Taylor expansion of f at a:

f “ fpaq`nÿ

i“1

Bf

Bzipaqpziáiq`

ÿ

1ďi,jďn

mijpziáiqpzjájq, mij P Krz1, . . . , zns.

(4.2.3)

Remark 4.2.3. Let X Ă An be closed, and let a P X. Identify ΘaAn with Knvia Lemma 4.1.6. By Remark 4.2.2 we have the identification

TaX “ Anntdfpaq | f P IpXqu.

Let X be a quasi projective variety, and let x P X. Let mx Ă OX,x be themaximal ideal. By Leibiniz’ rule dφpxq “ 0 if φ P m2

x (recall that d : OX,x ÑΩXpxq is a derivation over K). Thus we have an induced K-linear map

mxm2x

δpxqÝÑ ΩXpxq

rφs ÞÑ dφpaq(4.2.4)

Proposition 4.2.4. Keep notation as above. Then δpxq is an isomorphism ofcomplex vector spaces.

Proof. First we prove that δpxq is surjective. If X “ An, surjectivity follows atonce from Lemma 4.1.6. In general, we may assume that X is a closed subsetof An, and surjectivity follows from Proposition 4.1.8.

In order to prove injectivity of δpxq, we must show that if φ P mx is suchthat dφpxqpDq “ 0 for all D P ΘxX, then φ P m2

x. We may suppose that X isa closed subset of An. In order to avoid confusion, we let x “ a “ pa1, . . . , anq.Let pU, fgq be a representative of φ, where f, g P KrXs, and fpaq “ 0, gpaq “ 0.It will suffice to prove that f P m2

a. Since 0 “ dφpaq “ gpaq´1dfpaq we havedfpaq “ 0. By Theorem 2.4.2 there exists f P Krz1, . . . , zns such that f|X “ f .By Proposition 4.1.8 we may identify ΘaX with the subspace of TaKn “ Kngiven by (4.1.10). By hypothesis dfpaqpDq “ 0 for all D P ΘaX, i.e.

dfpaq P Ann pΘaXq Ă ΩAnpxq.

By (4.1.10) there exists h P IpXq such that dfpaq “ dhpaq. Then pf ´ hq|X “ f

and dpf ´ hqpaq “ 0. Thus pf ´ hq P Krz1, . . . , zns has vanishing value anddifferential at a. It follows (first-order Taylor expansion of f ´ h at a) that

pf ´ hq P pz1 ´ a1, . . . , zn ´ anq2.

Since h P IpXq we get that f P m2a.


4.3 Smooth points of quasi projective varieties

Definition 4.3.1. Let X be a quasi projective variety, and let x P X. ThenX is smooth at x if dim ΘxX “ dimxX, it is singular at x otherwise. The setof smooth points of X is denoted by Xsm. The set of singular points of X isdenoted by singX.

Example 4.3.2. Let X Ă An be a hypersurface. By Corollary 3.3.4, thedimension of X is equal to n´ 1, and hence the set of smooth points of X is anopen dense subset of X by Example 4.1.10. By the last sentence in Example4.1.10, X is locally irreducible at any of its smooth points.

The main result of the present section extends the picture for hypersurfacesto the general case.

Theorem 4.3.3. Let X be a quasi projective variety. Then the following hold:

1. The set Xsm of smooth points of X is an open dense subset of X.

2. For x P X we have dim ΘxX ě dimxX.

3. X is locally irreducible at any of its smooth points, i.e. if X is smooth ata, there is a single irreducible component of X containing a.

We will prove Theorem 4.3.3 at the end of the section. First we go throughsome preliminary results.

Our first result proves a weaker version of Item (1) of Theorem 4.3.3, andproves Item (2) of the same theorem.

Proposition 4.3.4. Let X be a quasi projective variety. Then the followinghold:

1. The set of smooth points of X contains an open dense subset of X.

2. For x P X we have dim ΘxX ě dimxX.

Proof. Suppose that X is irreducible of dimension d. By Proposition 3.2.7there is a birational map g : X 99K Y , where Y Ă Ad`1 is a hypersurface.By Proposition 3.1.6 there exist open dense subsets U Ă X and V Ă Y suchthat g is regular on U , and it defines an isomorphism f : U

„ÝÑ V . By Example

4.3.2, the set of smooth points Y sm of Y is open and dense in Y . Since V isopen and dense in Y the intersection Y sm X V is open and dense dense in Yand hence f´1pY sm X V q is an open dense subset of X. Since f´1pY sm X V q iscontained in U sm, we have proved that the set of smooth points of X contains anopen dense subset of X. We have proved that Item (1) holds if X is irreducible.In general, let X “ X1 Y ¨ ¨ ¨ YXr be the irreducible decomposition of X. Let

X0j :“ pXz

ď

i‰j

Xiq “ pXjzď

i‰j

Xiq

By the result that was just proved, pX0j q

sm contains an open dense subset of

smooth points. Every smooth point of X0j is a smooth point of X, because X0

j

4.3. SMOOTH POINTS OF QUASI PROJECTIVE VARIETIES 65

is open in X. ThusŤ

ipX0i q

sm is an open dense subset of X, containing an opendense subset of X. This proves Item (1).

Let us prove Item (2). Let x0 P X, and let X0 be an irreducible componentof X containing x0 such that dimX0 “ dimx0

X. By Item (1) Xsm0 contains

an open dense subset of points x such that dim ΘxX0 “ dimxX0, and henceby Proposition 4.1.11 we have dim ΘxX0 ě dimxX0 for all x P X. Inparticular dim Θx0X0 ě dimx0 X0 “ dimx0 X. Since Θx0X0 Ă Θx0X, it followsthat dim Θx0

X ě dimx0X.

The next result involves more machinery. We will give an algebraic versionof the (analytic) Implicit Function Theorem. The algebraic replacement forthe ring of analytic functions defined in a neighborhood of 0 P An is the ringKrrz1, . . . , znss of formal power series in z1, . . . , zn with complex coefficients. Wehave inclusions

Krz1, . . . , zns Ă OAn,0 Ă Krrz1, . . . , znss. (4.3.1)

(The second inclusion is obtained by developing fg as convergent power series

centered at 0, where f, g P Krz1, . . . , zns and gp0q “ 0.) We will need thefollowing elementary results.

Lemma 4.3.5. Let m Ă Krz1, . . . , zns, m1 Ă OAn,0 and m2 Ă Krrz1, . . . , znssbe the ideals generated by z1, . . . , zn in the corresponding ring. Then for everyi ě 0 we have pm2qi X OAn,0 “ pm

1qi, and pm1qi XKrz1, . . . , zns “ mi.

Proof. By induction on i. For i “ 0 the statement is trivially true. The proofof the inductive step is the same in both cases. For definiteness let us showthat pm2qi`1 XOAn,0 “ pm

1qi`1, assuming that pm2qi XOAn,0 “ pm1qi. The non

trivial inclusion is pm2qi`1XOAn,0 Ă pm1qi`1. Assume that f P pm2qi`1XOAn,0.

Then f P pm2qi XOAn,0, and hence f P pm1qi by the inductive hypothesis. Thuswe may write

f “ÿ

|I|

αJzJ ,

where the sum is over all multiindices J “ pj1, . . . , jnq of weight |J | “řns“1 js “

i, and αJ P OAn,0 for all J . Since f P pm2qi`1, we have αJp0q “ 0 for all J . Itfollows that αJ P m

1 for all J , and hence f P pm1qi`1.

Proposition 4.3.6 (Formal Implicit Function Theorem). Let ϕ P Krrz1, . . . , znss,and suppose that

ϕ “ z1 ` ϕ2 ` . . .` ϕd ` . . . , ϕd P Krz1, . . . , znsd. (4.3.2)

Given α P Krrz1, . . . , znss, there exists a unique p P Krrz1, . . . , znss such that

pα´ p ¨ ϕq P Krrz2, . . . , znss. (4.3.3)

Proof. Write p “ p0`p1` . . .`pd` . . ., where pd P Krz1, . . . , znsd, and the pd’sare the indeterminates. Expand the product p ¨ϕ, and solve for p0 by requiringthat p¨ϕ have have the same linear term modulo z2, . . . , zn as α, then solve for p1

by requiring that p ¨ ϕ have have the same quadratic term modulo pz2, . . . , znq2

as α , etc. By (4.3.2) there is one and only one solution at each stage.


Corollary 4.3.7. With hypotheses as in Proposition 4.3.6, the natural mapKrrz2, . . . , znss Ñ Krrz1, . . . , znsspϕq is an isomorphism.

Proposition 4.3.8. Let f1, . . . , fk P Krz1, . . . , zns and a P An. Suppose that

(i) each fi vanishes at a, and

(ii) the differentials df1paq, . . . , dfkpaq are linearly independent.

Then V pf1, . . . , fkq “ X Y Y , where

1. X,Y are closed in An, a P X, while Y does not contain a;

2. X is irreducible of dimension n ´ k, it is smooth at a, and TapXq “Annpxdf1paq, . . . , dfkpaqyq (as subspace of TaAn).

Moreover, there exists a principal open affine set Ang containing a such thatf1|Ang , . . . , fk|Ang generate the ideal of X X Ang .

Proof. By changing affine coordinates, if necessary, we may assume that a “ 0,and that dfip0q “ zi for i P t1, . . . , ku. Let J 1 Ă OAn,0 be the ideal gener-ated by f1, . . . , fk (to be consistent with our notation, we should write J 1 “pϕpf1q, . . . , ϕpfkqq), let J :“ J 1 X Krz1, . . . , zns, and let J2 Ă Krrz1, . . . , znssbe the ideal generated by f1, . . . , fk. Lastly, let I Ă Krz1, . . . , zns be the idealgenerated by f1, . . . , fk. We claim that

J ¨ g Ă I Ă J. (4.3.4)

for a suitable g P Krz1, . . . , zns with gp0q “ 0. In fact, the second inclusion istrivially true. In order to prove the first inclusion, let h1, . . . , hr be generators ofthe ideal J Ă Krz1, . . . , zns. By definition of J , there exist ai, gi P Krz1, . . . , zns,for i P t1, . . . , ru, such that ai P I, gip0q “ 0, and hi “

aigi

. Hence the second

inclusion in (4.3.4) holds with g “ g1 ¨ . . . ¨ gr. This proves (4.3.4), and hence wehave V pJq Ă V pIq Ă pV pJq Y V pgqq. It follows that, letting X :“ V pJq, thereexists a closed Y Ă V pgq such that

V pf1, . . . , fkq “ X Y Y, 0 R Y. (4.3.5)

Let us prove that J is a prime ideal, so that in particular X is irreducible. First,we claim that

J2 X OAn,0 “ J 1. (4.3.6)

The non trivial inclusion to be proved is J2 X OAn,0 Ă J 1. Let f P J2 X OAn,0.

Then there exist α1, . . . , αk P Krrz1, . . . , znss such that f “řkj“1 αjfj . Given

s P N, let αsj be the MacLaurin polynomial of αj of degree s, i.e. such that

pαj ´ αsjq P pm

2qs`1, where m2 is as in Lemma 4.3.5. Then

f “kÿ

j“1

αpsqj fj `

kÿ

j“1

pαj ´ αsjqfj .

4.3. SMOOTH POINTS OF QUASI PROJECTIVE VARIETIES 67

Both addends are in OAn,0. In addition, the first addend belongs to J 1, and thesecond one belongs to pm2qs`1. By Lemma 4.3.5, it follows that the secondone belongs to pm1qs`1. Hence f P

Ş8

s“0pI1 ` pm1qs`1q. By Corollary A.6.2,

it follows that f P I 1. This proves (4.3.6). By (4.3.6) and the definition of J ,we have an inclusion

Krz1, . . . , znsJ Ă Krrz1, . . . , znssJ2.

Hence, in order to prove that J is prime, it suffices to show that Krrz1, . . . , znssJ2

is an integral domain. In fact we will see that the natural map

Krzk`1, . . . , zns ÝÑ Krrz1, . . . , znssJ2 (4.3.7)

is an isomorphism of rings. This follows from the algebraic version of the Impli-cit Function Theorem, i.e. Proposition 4.3.6. In fact, by Proposition 4.3.6,the natural map Krrz2, . . . , znss Ñ Krrz1, . . . , znsspf1q is an isomorphism. Leti P t2, . . . , ku. Given the identification Krrz1, . . . , znsspf1q “ Krrz2, . . . , znss,the image of fi under the quotient map Krrz1, . . . , znss Ñ Krrz1, . . . , znsspf1q

is an element zi ` f 1i , where f 1i P pm2q2 (notation as in Lemma 4.3.5). It-

erating, we get that the map in (4.3.7) is an isomorphism of rings. As ex-plained above, this proves that J is a prime ideal. In particular X is irredu-cible. Moreover, since zk`1, . . . , zn P KrXs, the isomorphism in (4.3.7) showsthat KpXq has transcendence degree n ´ k, i.e. X has dimension n ´ k. Sincef1, . . . , fk vanish on X, and their differentials are linearly independent, it followsthat dim Θ0pXq ď pn´kq “ dim0X. Hence dim Θ0pXq “ pn´kq “ dim0X, byItem (2) of Proposition 4.3.4, i.e. X is smooth at 0, and Θ0pXq Ă Θ0An is theannihilator of df1p0q, . . . , dfkp0q. This proves Items (1) and (2). The last state-ment in the proposition holds with the polynomial g appearing in (4.3.4).

Corollary 4.3.9. Let X Ă An be a Zariski closed subset. Let a be a smoothpoint of X, and let k “ n´ dimaX. Then following hold:

1. there exist f1, . . . , fk P Krz1, . . . , zns with linerly independent differentialsdf1paq, . . . , dfkpaq, and a Zariski open affine subset U Ă An containing a,such that IpX X Uq “ pf1|U , . . . , fk|U q;

2. there is a unique irreducible component of X containing a.

Proof. Since X is smooth at a, and dimaX “ n ´ k, there exist f1, . . . , fk PIpXq such that df1paq, . . . , dfkpaq are linearly independent. Of course X Ă

V pf1, . . . , fkq. By Proposition 4.3.8 there is a unique irreducible componentof V pf1, . . . , fkq containing a, call it Y , and dimY “ n ´ k. Every irreduciblecomponent of X containing a is contained in Y . Since dimaX “ n ´ k, thereexists (at least) one irreducible component of X containing a of dimension n´k.Let X 1 be such an irreducible component; by Proposition 3.3.3, X 1 “ Y . Itfollows that there is a single component of X containing a, and it is equal to theunique irreducible component of V pf1, . . . , fkq containing a. Hence the corollaryfollows from Proposition 4.3.8.


Proof of Theorem 4.3.3. Item (2) has been proved in Proposition 4.3.4.Item (3) follows at once from Corollary 4.3.9, because X is covered by openaffine subset.

In order to prove Item (1), let X “Ť

iPI Xi be the irreducible decompositionof X. Since X is covered by open affine subset, Corollary 4.3.9 gives that

Xsm Ă Xzď

i,jPIi “j

pXi XXjq. (4.3.8)

The right hand side of (4.3.8) is an open dense subset of X. Let X0i be an

irreducible component of the right hand side of (4.3.8). Thus X0i Ă Xi is the

complement of the intersection of Xi with the other irreducible componets of X.The set of smooth points of X0

i is non empty by Proposition 4.3.4, and it isopen by upper semicontinuity of the dimension of ΘxX (Proposition 4.1.11),because dimxX is independent of x P X0

i . Hence Xsm is an open dense subsetof the open dense subset of X given by the right hand side of (4.3.8), and henceis open and dense in X.

4.4 Transverse Bezout

Chapter 5

Regular maps and theirfibers

5.1 Sard’s theorem

Let f : X Ñ Y be a regular map of quasi projective varieties.

Definition 5.1.1. Let p P X. The map f is smooth at p if

1. p is a smooth point of X, q :“ fppq is a smooth point of Y ,

2. and the differential dfppq : ΘpX Ñ ΘqY is surjective.

Claim 5.1.2. Let f : X Ñ Y be a regular map of quasi-projective sets. The setof smooth points of f is open in X.

Proof. The set of points p P X such that (1) of Definition 5.1.1 holds is equalto Xsm X f´1pY smq and hence is open by Theorem 4.3.3. Thus it remains toprove that the set of p P Xsm X f´1pY smq such that dfppq is not surjective isclosed in Xsm X f´1pY smq. It suffices to prove it for X and Y affine, smooth.By Corollary 4.3.9 we may assume that Y is irreducible of dimension d. Thuswe must check that the set

tp P X | rk dfppq ď pd´ 1qu (5.1.1)

is closed in X. By hypothesis X Ă Am and Y Ă An are closed. Via the iden-tification provided by Proposition 4.1.8 the differential dfppq gets identifiedwith the Jacobian matrix Jfppq. It follows that (5.1.1) is the set of zeroes ofdeterminants of dˆ d minors of Jfppq, and hence is closed.

The following result has a well-known analogue in the category of smooth oranalytic manifolds.

69

70 CHAPTER 5. REGULAR MAPS AND THEIR FIBERS

Proposition 5.1.3. Let f : X Ñ Y be a regular map of quasi-projective variet-ies. Suppose that p P X and that f is smooth at p. Then f´1 tfppqu is smoothat p and

dimp f´1pfppqq “ dimp x´ dimfppq Y.

Proof. We may assume that X and Y are affine. Let n :“ dimY , and let q :“fppq. There exists r such that Y Ă An`r is closed. By Corollary 4.3.9 thereexist a Zariski open U Ă An`r containing q and φ1, . . . , φr P Krz1, . . . , zn`rssuch that

1. dφ1pqq, . . . , dφrpqq are linearly independent, and

2. V pφ1, . . . , φrq X U “ Y X U .

Let ψ1, . . . , ψn P Krz1, . . . , zn`rs be such that 0 “ ψ1pqq “ . . . “ ψnpqq and

tdφ1pqq, . . . , dφrpqq, dψ1pqq, . . . , dψnpqqu

is a basis of the cotangent space of An`r at q (we may choose the ψi’s to be co-ordinate functions if we wish). By Proposition 4.3.8 V pφ1, . . . , φr, ψ1, . . . , ψnqhas dimension zero at q. Thus shrinking the open set U above, if necessary, wemay assume that

V pψ1, . . . , ψn, φ1, . . . , φrq X U “ tqu. (5.1.2)

Let ψi :“ ψi|Y . By (5.1.2) we have that

f´1pqq “ V pf˚pψ1q, . . . , f˚pψnqq. (5.1.3)

We have dpf˚pψiqqppq “ f˚dψipqq. By hypothesis dfppq is surjective, i.e. thetranspose

ΩY pqqdfppqt

ÝÑ ΩXppq

is injective. Since dψ1pqq, . . . , dψnpqq are linearly independent, it follows thatdpf˚ψ1qppq, . . . , dpf

˚ψnqppq are linearly independent. Let m :“ dimpX. SinceX is affine, there exists s such that that X Ă Am`s is closed. By hypothesisX is smooth at p, and hence by Corollary 4.3.9 there exist a Zariski openU Ă Am`s containing p and ψ1, . . . , ψs P Krz1, . . . , zm`ss such that

1. V pψ1, . . . , ψsq XU “ X XU , and

2. dψ1ppq, . . . , dψsppq are linearly independent.

Since X is closed in Am`s there exist ϕ1, . . . , ϕn P Krz1, . . . , zm`ss such thatϕi|X “ f˚φi. By (5.1.3) we have that

f´1tqu XU “ V pψ1, . . . , ψs, ϕ1, . . . , ϕnq XU .

Applying Proposition 4.3.8 we get that V pψ1, . . . , ψs, ϕ1, . . . , ϕnq is smoothat p of dimension m´ n “ dimpX ´ dimp Y .

5.1. SARD’S THEOREM 71

Definition 5.1.4. Let f : W Ñ Z be a regular map between quasi-projectivesets. A point q P Z is a regular value of f if f is smooth at each p P f´1tqu.

Remark 5.1.5. If q R fpW q then q is a regular value of f .

Below are the main results of the present section.

Proposition 5.1.6. Let f : X Ñ Y be a regular dominant map of irreduciblequasi-projective varieties. The set of smooth points of f is an open dense subsetof X.

Theorem 5.1.7 (Sard’s theorem for quasi-projective varieties). Let f : X Ñ Ybe a regular map of quasi-projective varieties. Suppose that X is smooth. Thenthe set of regular values of f contains an open dense subset of Y .

Before proving the above results we will recall results on derivations of anextension of fields k Ă K. We will assume for simplicity that

char k “ 0 and K is finitely generated over k. (5.1.4)

We will study DerkpK,Kq where the structure of K-module of K is given bymultiplication in the field K. The case of interest for us is k “ K and K “ KpXqwhere X is an irreducible quasi-projective variety. Let us associate a geometricobject to a derivation D P DerKpKpXq,KpXqq. Let p P X. We recall thatOX,p Ă KpXq. Suppose that

DpOX,pq Ă OX,p. (5.1.5)

Then we may define a tangent vector Dppq P ΘpX by setting

OX,pDppqÝÑ K

φ ÞÑ Dpφqppq.(5.1.6)

The result below shows that an element of DerKpKpXq,KpXqq may be thoughtof as a vector field on an open dense subset U Ă X (the open U depends on D).

Claim 5.1.8. Let X be an irreducible quasi-projective variety and D P DerKpKpXq,KpXqq.There exists an open dense U Ă X such that for all p P U Equation (5.1.5) holdsand hence the tangent vector Dppq P ΘpX is defined.

Proof. We may assume that X is affine. Thus KpXq is the fraction field of KrXs.Let f1, . . . , fr be generators of the K-algebra KrXs. There exists 0 “ g P KrXssuch that g ¨ Dpfiq P KrXs for i “ 1, . . . , r. Let U :“ Xg “ pXzV pgqq. ThenXg is an affine open dense subset of X, and its ring of regular functions is thesubring of KpXq given by

KrU s “ thgk | h P KrXs, k ě 0u.

Thus (A.5.4) gives that DpKrU sq Ă KrU s. Applying (A.5.4) again it followsthat (5.1.5) holds for p P U .


Notice that DerkpK,Kq is a K-vector space: in fact addition in DerkpK,Kqis the usual addition of derivations, and given α P K and D P DerkpK,Kqone defines αD by setting pαDqpβq :“ αDpβq (as is easily checked pαDq PDerkpK,Kq).

Proposition 5.1.9. Keep hypotheses and notation as above. Let x1, . . . , xn P Kbe algebraically independent over k. The map of K-vector spaces

DerkpK,Kq ÝÑ Kn

D ÞÑ pDpx1q, . . . , Dpxnqq(5.1.7)

is surjective. If in addition we assume that tx1, . . . , xnu is a transcendence basisof K over k then it is an isomorphism.

Proof. It suffices to prove that (5.1.7) is an isomorphism when tx1, . . . , xnu isa transcendence basis of K over k. Let F :“ kpx1, . . . , xnq Ă K. Since F is apurely transcendental extension of k we have an isomorphism of K-vector spaces

DerkpF,Kq„ÝÑ Kn

D ÞÑ pDpx1q, . . . , Dpxnqq.

Equivalently every D P DerkpF,Kq is given by

Dpφq “nÿ

i“1

αiBφ

Bxi, αi P K,

and the αi may be chosen arbitrarily. Thus it remains to prove that restriction

DerkpK,Kq ÝÑ DerkpF,KqD ÞÑ D|F

(5.1.8)

defines an isomorphism of K-vector spaces. Since K is finitely generated overk and char k “ 0 there exists a primitive element y over F , i.e. K is generatedby y over F . Let

P “ zm ` a1zm´1 ` ¨ ¨ ¨ ` am P F rzs

be the minimal polynomial of y over F . In particular

ym ` a1ym´1 ` ¨ ¨ ¨ ` am “ 0, ai P F. (5.1.9)

Suppose that D P DerkpK,Kq; by (5.1.9) we get that

mym´1Dpyq ` ym´1Dpa1q ` pm´ 1qa1ym´2Dpyq ` ¨ ¨ ¨ `Dpamq “ 0

i.e.P 1pyq ¨Dpyq `Dpa1qy

m´1 ` ¨ ¨ ¨ `Dpamq “ 0.

Since P 1 ‰ 0 (because char k “ 0) and P is the minimal polynomial of y over Fwe have P 1pyq ‰ 0 and hence

Dpyq “ ´

˜

mÿ

i“1

Dpaiqymí

¸

¨ P 1pyq´1. (5.1.10)

5.1. SARD’S THEOREM 73

This proves that (5.1.8) is injective. We also get surjectivity: in fact givenD P DerkpF,Kq we extend it to a derivation in DerkpK,Kq by defining its valueon y via (5.1.10).

Corollary 5.1.10. Keep hypotheses and notation as above. Then Tr degkK “

dimK DerkpK,Kq.

Proof of Proposition 5.1.6. We know that the set of smooth points of f isopen, we must prove that it is non-empty. We are free to replace X andY by open dense subsets X0 and Y 0 respectively (of course we require thatfpX0q Ă Y 0): in the course of the proof we will rename X0 and Y 0 by X andY respectively. In particular we may assume that X and Y are smooth. Letφ1, . . . , φm be a transcendence basis of KpY q over K. Replacing Y by the opendense subset Y 0 of points where each of φ1, . . . , φm is regular we may assumethat φ1, . . . , φm are regular (of course we replace X by f´1Y 0). Since f is dom-inant f˚φ1, . . . , f

˚φm are algebraically independent in KpXq. By Proposition5.1.9 there exist Dj P DerKpKpXq,KpXqq for j “ 1, . . . ,m such that

Djpf˚φiq “ δij “

#

1 if i “ j,

0 if i ‰ j.

By Claim 5.1.8 there exists an open dense X0 Ă X such that DjpOX,pq Ă OX,pfor every p P X0: thus we may assume that DjpOX,pq Ă OX,p for every p P X.Then Dj defines a tangent vector Djppq P ΘpX for each p P X. Let p P X: weclaim that dfppq is surjective. In fact let q :“ fppq. Then df˚φ1pqq, . . . , df

˚φmpqqare linearly independent because

xDjppq, f˚φiy “ δij . (5.1.11)

In particular dφ1pqq, . . . , dφmpqq are linearly independent. Since m “ dimYand Y is smooth it follows that tdφ1pqq, . . . , dφmpqqu is a basis of ΩY pqq. Thisproves that the transpose of dfppq is injective and hence dfppq is surjective.

Proof of Sard’s theorem. One checks easily that it suffices to prove the theoremfor X and Y quasi-projective varieties. If f is not dominant then every point ofthe open dense set pY zfpXqq is a regular value of f - see Remark 5.1.5. Nowsuppose that f : X Ñ Y is dominant. By Proposition 5.1.6 the open set

X0 :“ tp P X | dfppq is surjectiveu

is dense in X. Let C be an irreducible component of XzX0; we claim that¯fpCq ‰ Y . In fact suppose the contrary. Applying Proposition 5.1.6 to f|C

we get that there exists an open dense C0 Ă C sm such that

dfppq|TpC : TpC Ñ TfppqY

is surjective. That contradicts the definition of X0. This proves that ¯fpCq ‰ Y .It follows that

¯fpXzX0q ‰ Y.

Thus Y z ¯fpXzX0q is an open dense subset of regular values of Y .


5.2 Dimensions of fibers

The following key result is a particular case of Krull’s HauptidealSatz (valid forarbitrary Noetherian rings).

Theorem 5.2.1. Let X be an irreducible quasi projective variety and 0 “ f PKrXs. Every irreducible component of V pfq has dimension pdimX ´ 1q.

Proof. We may assume that X is affine. Thus there exists n such that X Ă

An is closed. By Theorem 2.4.2 there exists f P Krz1, . . . , zns such thatf “ f|X . We must prove that, if Y is an irreducible component of V pfq, thendimY “ dimX ´ 1. We view An as the open affine set PnX0

Ă Pn, and we let

X,¯

V p ˜ qf, Y Ă Pn be the closures of X, V pfq and Y respectively. Let d be the

degree of the hypersurface¯

V p ˜ qf Ă Pn. Let N :“ p`

d`nn

˘

´ 1q and

Pn νndÝÑ PN

rX0, . . . , Xns ÞÑ rXd0 , X

d´10 X1, . . . , X

dns

be the Veronese map. Then νnd defines an isomophism X„ÝÑ νpXq. Since

V pfq is a hypersurface of degree d there exists a hyperplane H Ă PN such that

ν´1pHq “¯

V p ˜ qf . Thus νnd defines an isomorphism X X¯

V p ˜ qf„ÝÑ νpXq XH. It

follows that νpW q is an irreducible component of νpXq XH. By Proposition3.7.3 we have

dimW “ dim W “ dim νpW q “ dim X ´ 1 “ dimX ´ 1.

Corollary 5.2.2. Let f : X Ñ Y be a regular dominant map of quasi-projectivevarieties and let p P X. Every irreducible component of f´19fppqq has dimen-sion at least equal to dimX ´ dimY .

Proof. We may assume that X and Y are affine. Let dimY “ m. As iseasily checked there exist φ1, . . . , φm P KrY s such that fppq is an irreduciblecomponent of V pφ1, . . . , φmq (in fact choose 0 ‰ φ1 P Iptfppquq, then chooseφ2 P Iptfppquq not vanishing on any irreducible component of V pφ1q etc.). Then

V pf˚φ1, . . . , f˚φmq “ f´1pfppqq \ C

where C is closed in X. Thus f´1pfppqq is a union of irreducible components ofV pf˚φ1, . . . , f

˚φmq. By repeated application of Theorem 5.2.1 every irredu-cible component of V pf˚φ1, . . . , f

˚φmq has dimension at least pdimX ´mq “pdimX ´ dimY q.

Proposition 5.2.3. Let f : W Ñ Z be a regular map of quasi-projective sets.The function

WαÝÑ N

p ÞÑ dimp f´1tfppqu

5.2. DIMENSIONS OF FIBERS 75

is upper-semicontinuous, i.e. given k P N the set

Wk :“ tp PW | dimp f´1tfppqu ě ku

is closed.

Proof. As is easily checked it suffices to prove the proposition for W and Zvarieties and f dominant. Let W 0 ĂW be the set of smooth points of f . ThenW 0 is open dense in W and

dimp f´1 tfppqu “ dimW ´ dimZ @p PW 0. (5.2.1)

Let k P N. If k ď dimW ´dimZ then Wk “W by Corollary 5.2.2 and henceWk is closed. Let k ą dimW ´ dimZ. By (5.2.1) we get that

Wk “

p PW zW 0 : dimp f´1

fppq(

ě k(

where f :“ f|W zW 0 . Since W zW 0 is closed in W the proposition follows byinduction on the dimension of the domain.

Example 5.2.4. The function α of Proposition 5.2.3 is not constant in general.A typical example is provided by the blow-up of Pn at p0 P Pn, i.e. the set

Blp0Pn :“ tpp, `q P Pn ˆGrp1,Pnq : ` Ą tp0, puu .

As is easily checked Blp0Pn is closed in Pn ˆGrp1,Pnq. Let

f : Blp0Pn Ñ Pn, pp, `q ÞÑ p

be projection; then

f´1 tpu “

#

xp0, py if p ‰ p0,

t` P Grp1,Pnq : p0 P ù if p “ p0.

Thus

dimp f´1 tpu “

#

0 if p ‰ p0,

n´ 1 if p “ p0.

Proposition 5.2.5. Let f : W Ñ Z be a dominant morphism (i.e. regular map)of quasi-projective varieties. There exists an open dense Z0 Ă Z such that

f´1 tqu has pure dimension dimW ´ dimZ for all q P Z0,

i.e. the generic fiber of f has pure dimension equal to dimW ´ dimZ.

Proof. By induction on dimW . By Proposition 3.4.8 there exists an opendense Z 1 Ă Z contained in fpW q. Applying Sard’s theorem to the dominantmap

f|W smXf´1Z1 : W smXf´1Z 1 Ñ Z 1

we get that there exists an open dense Z2 Ă Z 1 Ă Z such that f´1 tqu XW smhas pure dimension pdimW ´ dimZq for q P Z2. Now we apply the induct-ive hypothesis and Corollary 5.2.2 to the restriction of f to each irreduciblecomponent of W zpW smXf´1Z2q.


5.3 Cubic surfaces

5.4 Local invertibility of regular maps

5.5 Local factoriality

The result below is of fundamental importance.

Theorem 5.5.1. Let X be a smooth quasi projective variety. Let D Ă X be aclosed subset of pure codimension 1, and let a P D. There exists an open affinesubset U Ă X containing a such that the ideal IpD X Uq Ă KrU s is principal.

Remark 5.5.2. If we assume that D is smooth at a, then Theorem 5.5.1follows from Proposition 4.3.8 and Corollary 4.3.9. In fact, replacing Xby a suitable open affine subset containing a, we may assume that X is affine.Hence there exists an embedding X Ă An as closed subset. Thus D Ă An is alsoclosed. Applying Proposition 4.3.8 and Corollary 4.3.9 to X and D, weget that there exist an open affine subset U Ă An containing a, and functionsf1, . . . , fk`1 P KrU s, such that

IpX X Uq “ pf1, . . . , fkq, IpD X Uq “ pf1, . . . , fk`1q.

Since principal open affine sets form a basis for thge Zariski topology, we mayassume that U is a principal open set, say U “ AnzV pϕq. Hence also U XX isan open principal set, in particular it is affine. Moreover the image of fk`1 inKrX X U s is a generator of the ideal of D XX X U .

Proof of Theorem 5.5.1.

The statement of Theorem 5.5.1 is summarized by stating that a smoothquasi projective variety is locally factorial.

EXPLAINThe result below follows from Theorem 5.5.1, actually the weak version

in Remark 5.5.2 suffices.

Proposition 5.5.3. Let X be a smooth quasi projective variety, and let f : X 99KPn be a rational map. The indeterminacy set Indpfq has codimension at least 2in X.

First we prove the following.

Proof of Proposition 5.5.3. The indeterminacy set Indpfq is a proper closedsubset of X. We argue by contradiction. Suppose that D is a codimension 1irreducible closed subset of X contained in Indpfq. Let a be a smooth point ofD. By Lemma ?? there exist an open affine subset U Ă X containing a andϕ P KrU s such that IpDXUq Ă KrU s is generated by ϕ. Since X is smooth at a,there is a unique irreducible component of X containing a, hence we may assume

5.5. LOCAL FACTORIALITY 77

that U is irreducible. There exist f0, . . . , fn P KrU s such that V pf0, . . . , fnq isa proper subset of U , and

fpxq “ rf0pxq, . . . , fnpxqs @x P pUzV pf0, . . . , fnqq.

Exercises


Chapter 6

Holomorphic maps: localtheory

6.1 Holomorphic functions

Let U Ă Kn be an open subset, and f : U Ñ K be a function. Identifying Knwith R2n, and K with R2, we may view f as a function from the open U Ă R2n

to R2, and hence it makes sense to state that f is, or is not, differentiable ata P U .

Definition 6.1.1. Let U Ă Kn be an open subset. A function f : U Ñ Kis holomorphic if, for each a P U , it is differentiable at a, and the differentialdfpaq : Kn Ñ K is complex linear, i.e. there exists a complex linear functionL : Kn Ñ K such that

limhÑ0

||fpa` hq ´ fpaq ´ Lphq||

||h||“ 0.

Remark 6.1.2. With notation as in Definition 6.1.1, the linear function L isidentified with the differential dfpaq via the standard identifications of Kn andK with R2n and R2 respectively.

Let HomRpKn,Kq be the real vector space of R-linear maps Kn Ñ K. ThenHomRpKn,Kq contains the subspace HomKpKn,Kq of K-linear maps Kn Ñ K,and the subspace HomKpKn,Kq of K-conjugate linear maps Kn Ñ K, i.e. ho-momorphism f : Kn Ñ K of additive groups such that fpλvq “ λfpvq for λ P Kand v P Kn (equivalently, such that v ÞÑ fpvq is K-linear). We have a directsum of real vector spaces

HomRpKn,Kq “ HomKpKn,Kq ‘HomKpKn,Kq. (6.1.1)

Thus, a function f : U Ñ K is holomorphic if and only if its differential ateach point of U belongs to the direct summand HomKpKn,Kq of the above

79

80 CHAPTER 6. HOLOMORPHIC MAPS: LOCAL THEORY

decomposition. A differentiable function f : U Ñ K is antiholomorphic if dfpaq PHomKpKn,Kq for all a P U .

We rewrite the decomposition in (6.1.1) as follows. First notive that we havea natural isomorphism

HomRpKn,Rq bR K „ÝÑ HomRpKn,Kq

f b λ ÞÑ pv ÞÑ λfpvqq(6.1.2)

Remark 6.1.3. Let f : U Ñ K be differentiable, and write f “ u ` iv, whereu, v are real functions. For a P U , the decomposition dfpaq “ dupaq ` idvpaqillustrates (6.1.2), by rewriting it as dfpaq “ dupaq ` dvpaq b i.

Thus, letting

Ω1,0a pUq :“ HomKpKn,Kq, Ω0,1

a pUq :“ HomKpKn,Kq, (6.1.3)

(as above, U Ă Kn is open) we may rewrite (6.1.1) as

TapUq˚ bR K “ Ω1,0

a pUq ‘ Ω0,1a pUq. (6.1.4)

Complex bases of Ω1,0a pUq and Ω0,1

a pUq are respectively

tdz1paq, . . . , dznpaqu, tdz1paq, . . . , dznpaqu (6.1.5)

Next, let

BBz1paq, . . . , BBznpaq, BBz1paq, . . . , BBznpaq P TapKnq bK (6.1.6)

be defined by the conditions

B

B

Bzjpaq, dzkpaq

F

“ δj,k, (6.1.7)

B

B

Bzjpaq, dzkpaq

F

“

B

B

Bzjpaq, dzkpaq

F

“ 0, (6.1.8)

B

B

Bzjpaq, dzkpaq

F

“ δj,k. (6.1.9)

With the above notation, a differentiable function f : U Ñ K is holomorphic ifand only if

Bfpaq

Bzk“ 0 @k P t1, . . . , nu, @a P U. (6.1.10)

Informally: f is holomorphic if it depends on the zj ’s, but not on the zk’s.

Example 6.1.4. Let f : Kn Ñ K be a polynomial function of the zj ’s and zk’s,i.e.

fpzq “ÿ

|J|`|K|ďd

cJ,KzJzK , (6.1.11)

6.1. HOLOMORPHIC FUNCTIONS 81

where J “ pj1, . . . , jnq and K “ pk1, . . . , knq are multindices. Then

Bf

Bzspzq “

ÿ

J,Kksě1

cJ,KkszJzKés , (6.1.12)

where te1, . . . , enu is the standard basis of Rn. It follows that f is holomorphicif and only cJ,K “ 0 for all K “ p0, . . . , 0q.

Remark 6.1.5. Let f : U Ñ K be a holomorphic function of one variable, i.e. U

is an open subset of K. For a P U we let f 1paq :“ BfpaqBz .

Example 6.1.6. Let R ą 0. Let f : Bp0, Rq Ñ K be defined by an absolutelyconvergent series

fpzq “8ÿ

m“0

cmzm, (6.1.13)

i.e. the right hand side is absolutely convergent for every z P Bp0, Rq. We claimthat f is holomorphic, and that

f 1pzq “ÿ

m“0

pm` 1qcm`1zm. (6.1.14)

In fact, given 0 ă ρ ă R, there exists Mpρq ą 0 such that

|cm|ρm ďMpρq @m, (6.1.15)

because the right hand side of (6.1.13) is absolute convergent for every z suchthat |z| “ ρ. It follows that the right hand side of (6.1.14) is absolutely conver-gent for |z| ă R. Moreover, for |z| ă R we have

fpzq´fpz0q “

8ÿ

m“0

cmppz0`pz´z0qqm´zm0 q “

8ÿ

m“1

cm

˜

mÿ

j“1

ˆ

m

j

˙

pz ´ z0qjzm´j0

¸

“

“ pz ´ z0q

˜

8ÿ

m“0

pm` 1qcm`1zm0

¸

` pz ´ z0q2ϕpzq, (6.1.16)

where ϕpzq is uniformly bounded on Bpz0, εq for ε ă pR ´ |z0|q, (use (6.1.15)).Hence

Bf

Bzpz0q “

ÿ

m“0

pm` 1qcm`1zm0 ,

Bf

Bzpz0q “ 0.

The claim follows because z0 is an arbitrary point of Bp0, Rq.

Writing out (6.1.10) in real coordinates, one gets the Cauchy-Riemann equa-tions. More precisely, let zj “ xj ` iyj , where xj , yj are the real coordinatefunctions. Then (6.1.7), (6.1.8), and (6.1.9) are equivalent to

B

Bzjpaq “

1

2

ˆ

B

Bxjpaq ´ i

B

Byjpaq

˙

,B

Bzjpaq “

1

2

ˆ

B

Bxjpaq ` i

B

Byjpaq

˙

.

(6.1.17)


In particular, we may rewrite (6.1.10) as

Bf

Bxj“Bf

iByj@j P t1, . . . , nu.

Letting fpzq “ upzqìvpzq, where upzq, vpzq are the real and the imaginary partof fpzq respectively, we get that f : U Ñ K is holomorphic if and only if it isdifferentiable and for all j P t1, . . . , nu the following Cauchy-Riemann equationshold on U :

Bu

Bxj“

Bv

Byj, (6.1.18)

Bu

Byj“ ´

Bv

Bxj. (6.1.19)

6.2 Holomorphic maps

Definition 6.2.1. Let U Ă Kn be an open subset. A map f : U Ñ Km isholomorphic if, for each a P U , it is differentiable at a, and the differentialdfpaq : Kn Ñ Km is complex linear.

Write f “ pf1, . . . , fmq; then f is holomorphic if and only if each of itscomponent functions fj : U Ñ K is holomorphic. This holds because an R-linearmap V ÑW1‘W2, where V,W1,W2 are complex vector spaces is K-linear if andonly if each of the maps V Ñ Wj obtained by composing with the projectionspW1 ‘W2q ÑWj is K-linear.

Theorem 6.2.2. Let U Ă Kn be a non empty open subset.

1. The set of holomorphic functions f : U Ñ K with pointwise addition andmultiplication is a ring, with unit the constant function 1. If f : U Ñ Kis holomorphic and nowhere zero, then 1fpzq is holomorphic.

2. Let U Ă Kn and W Ă Km be open subsets. Let f : U Ñ W and g : W Ñ

Kk be holomorphic (f holomorphic means that it is holomorphic whenviewed as a map U Ñ Kn). Then the composition g ˝ f : U Ñ Kk isholomorphic.

3. Holomorphic Inverse Function Theorem: Let U Ă Kn be open, and letf : U Ñ Kn be holomorphic. Let a P U , and assume that dfpaq : Kn Ñ Knis invertible. Then f is a local diffeomorophism at a, with holomorphiclocal inverse.

4. Holomorphic Implicit Function Theorem: Let U Ă Kn be open, andlet f : U Ñ Kk be holomorphic, with components f1, . . . , fk. Write ele-ments of Kn as pz, wq, where z P Kn´k w P Kk. Let pa, bq P U . Sup-

pose that fpa, bq “ 0, and that the k ˆ k matrix´

Bflpa,bqBwh

¯

1ďl,hďkis non

degenerate. Then there exist open (non empty) balls Bpa,Rq Ă Kn´k,

6.3. COMPLEX VALUED DIFFERENTIAL FORMS 83

Bpb, rq Ă Kk and a holomorhic function ϕ : Bpa,Rq Ñ Bpb, rq such thatBpa,Rq ˆBpb, rq Ă U and

tpz, wq P Bpa,Rq ˆBpb, rq | fpz, wq “ 0u “ tpz, ϕpzqq | z P Bpa,Rqu.

Proof. All the statements above follow from corresponding results on differenti-able maps. As an example, assume that f : U Ñ K is holomorphic and nowherezero. Let u, v : U Ñ K be the real and imaginary parts of f . Then

1

fpzq“

u

u2 ` v2´ i

v

u2 ` v2.

Thus 1f is differentiable. Since

B

Bzj

ˆ

1

f

˙

“ ´

BfBzj

f2“ 0,

1f is holomorphic.

6.3 Complex valued differential forms

Definition 6.3.1. Let U Ă Kn be open. A complex valued m form ω on Uis a section of p

ŹmT pUq˚q bR K (the complexified m-th exterior power of the

cotangent bundle of U), i.e. ω “ α ` iβ, where α, β are real m forms on U .We say that ω is continuous, differentiable or Cl if each of α, β is respectivelycontinuous, differentiable or Cl.

We recall that the complexified cotangent space of an open U Ă Kn ata point a has a direct sum decomposition with addends the complex vectorsubspaces Ω1,0

a pUq and Ω0,1a pUq. There is a similar decomposition of the fiber of

pŹm

T pUq˚q bR K at a.

Definition 6.3.2. Let Ωp,qa pUq be the complex subspace of pŹm

T pUq˚q bR Kspanned by all elements of the form df1paq^ . . .^dfppaq^dg1paq^ . . .^dgqpaq,where f1, . . . , fp are holomorphic defined in an open neighborhood of a, andg1, . . . , gq are antiholomorphic defined in an open neighborhood of a.

For multindices J “ pj1, . . . , jpq and K “ pk1, . . . , kqq, let

dzJpaq :“ dzj1paq ^ . . .^ dzjppaq, dzKpaq :“ dzk1paq ^ . . .^ dzkq paq. (6.3.1)

A complex basis of Ωp,qa pUq is provided by all dzJpaq ^ dzJpaq, where the mul-tiindices J , K have p and q entries respectively.

We have a direct sum decomposition

p

mľ

TapUq˚q bR K “

à

p`q“m

Ωp,qa pUq. (6.3.2)


Hence a complex valued m form on U can be written uniquely as

ω “ÿ

|I|`|J|“m

ωI,JdzI ^ dzJ , ωI,J : U Ñ K, (6.3.3)

and ω is continuous, differentiable or Ck if and only if each of ωI,J is respectivelycontinuous, differentiable or Ck.

Definition 6.3.3. A differential form ω on an open U Ă Kn is of type pp, qq ifωpaq P Ωp,qa pKnq for all a P U .

Let U Ă Kn be open. For a differentiable complex valued m form ω “

u` iv, where u, v are the real and imaginary parts of ω, we let dω “ du` idv.It is convenient to split dω according to the decomposition in (6.3.2). For adifferentiable function f : U Ñ K, we let

Bf :“nÿ

j“1

Bf

Bzjdzj , Bf :“

nÿ

k“1

Bf

Bzkdzk. (6.3.4)

We extend B and B to linear operators on differential forms by imposing Leibinizrule. Thus

Bpÿ

|I|`|J|“m

ωI,JdzI ^ dzJq :“ÿ

|I|`|J|“mkPt1,...,nu

BωI,JBzk

dzk ^ dzI ^ dzJ (6.3.5)

Bpÿ

|I|`|J|“m

ωI,JdzI ^ dzJq :“ÿ

|I|`|J|“mkPt1,...,nu

BωI,JBzk

dzk ^ dzI ^ dzJ (6.3.6)

A straighforward computation shows that

d “ B ` B. (6.3.7)

Since d ˝ d “ 0, it follows that

B ˝ B “ 0, B ˝ B “ 0, B ˝ B ` B ˝ B “ 0. (6.3.8)

In fact, it suffices to prove that the above operators are zero on a pp, qq form ω.We have

0 “ d ˝ dpωq “ B ˝ Bpωq ` pB ˝ B ` B ˝ Bqpωq ` B ˝ Bpωq. (6.3.9)

Since B ˝ Bpωq is of type pp` 2, qq, pB ˝ B ` B ˝ Bqpωq is of type pp` 1, q` 1q, andB ˝ Bpωq is of type pp, q ` 2q, it follows that each vanishes.

6.4 Cauchy’s integral formula

Definition 6.4.1. Let U Ă Kn be open, and let ω be a continuous complexvalued 1 form on U . Write ω “ α ` iβ, where α, β are (real) 1 forms on U .

6.4. CAUCHY’S INTEGRAL FORMULA 85

Given a piecewise C1 parametrized path γ : ra, bs Ñ U , we let

ż

γ

ω :“

ż

γ

α` i

ż

γ

β “

ż b

a

γ˚pαq ` i

ż b

a

γ˚pβq.

(Since γ is piecewise C1, γ˚pαq and γ˚pβq make sense over each closed intervalover which γ is differentiable, and they are continuous 1 forms, hence they havefinite integrals.)

Remark 6.4.2. With notation as in Definition 6.4.1, the integral of ω doesnot change if we reparametrize γ by a non decreasing differentiable functionrc, ds Ñ ra, bs (by the change of variables formula). Thus we may speak of theintegral of ω over an oriented path in U .

Remark 6.4.3. Complex valued differential forms make sense on any open ofRd (but of course dzi and dzi make sense only on Kn), and one may definedifferentiation and pull-back as above, by reducing to the real and imaginary

parts. In Definition 6.4.1 we could have definedş

γω to be

şb

aγ˚pωq.

Definition 6.4.4. Given a P K and R ą 0 we let ΓapRq be the path

r0, 2πsΓapRqÝÑ K

θ ÞÑ a`R exppiθq

Example 6.4.5. We haveż

ΓapRq

dz

z ´ a“ 2πi.

In fact

ΓapRq˚

ˆ

dz

z ´ a

˙

“Rieiθdθ

Reiθ“ idθ.

(see Remark 6.4.3) and the result follows.

The following integral representation is the beginning of complex analyis inone variable.

Theorem 6.4.6 (Cauchy’s integral formula). Let f : U Ñ K be a holomorphicfunction, where U Ă K is open. Suppose that the closed disk Bpa,Rq is containedin U . Then, for all z P Bpa,Rq we have

fpzq “1

2πi

ż

ΓapRq

fptqdt

t´ z(6.4.1)

We prove Cauchy’s integral formula after a few preliminaries.

Key Observation 6.4.7. Let U Ă K be open, and f : U Ñ K differentiable.Then

dpfpzqdzq “Bf

Bzdz ^ dz `

Bf

Bzdz ^ dz “

Bf

Bzdz ^ dz.

In particular f is holomorphic if and only if the (differentiable) 1 form fpzqdzis closed.


Theorem 6.4.8 (Cauchy-Goursat). Let U Ă K be open, and f : U Ñ K beholomorphic. Let R Ă U be compact, with piecewise C1 boundary BR, withorientation induced by the standard orientation1 of Kn “ R2n. Then

ż

BR

fdz “ 0. (6.4.2)

Proof. If one assumes that f is C1, then Stokes’ Theorem applies to fpzqdz andthe Theorem follows from the Key Observation 6.4.7. For the beautiful proof(by Goursat) valid without the assumption that f is C1, see Ahlfors [Ahl78].

Proof of Cauchy’s integral formula. By Example 6.4.5, it suffices to provethat

ż

ΓapRq

fptq ´ fpzq

t´ zdt “ 0. (6.4.3)

Let δ be a very small (strictly) positive number. By applying Proposition6.4.8 to the region between the circles described by Γzpδq and ΓapRq, we getthat

ż

ΓapRq

fptq ´ fpzq

t´ zdt “

ż

Γzpδq

fptq ´ fpzq

t´ zdt.

On the other hand,ş

Γzpδq

f 1paqdt “ 0 because f 1paqdt “ dpf 1paqtq is an exact

differential, and henceż

ΓapRq

fptq ´ fpzq

t´ zdt “

ż

Γzpδq

fptq ´ fpzq ´ f 1pzq ¨ pt´ zq

t´ zdt. (6.4.4)

Since f is differentiable at a, with derivative f 1paq, the integrand in the righthand side of (6.4.4) has absolute bounded above, say by M ą 0. It follows thatthe integral in the right hand side of (6.4.4) has absolute bounded above by2πδM . Since δ is arbitrarily small, it follows that integral in the left hand sideof (6.4.4) is zero.

Corollary 6.4.9. Let f : U Ñ K be a holomorphic function, where U Ă K isopen. Then f is C8, and the derivatives of any order are holomorphic. Supposethat the closed disk Bpa,Rq is contained in U . Then for z P Bpa,Rq we have

f pnqpzq “n!

2πi

ż

ΓapRq

fptqdt

pt´ zqn`1. (6.4.5)

Proof. If the closed disk Bpa,Rq is contained in U , then (6.4.5) holds for n “ 0by Cauchy’s integral formula. By differentiation under the integral sign (thishas to be justified, we leave details to the reader) and induction on n, we getthe corollary.

1If zj “ xj ` iyj , the orientation is given by dx1 ^ dy1 ^ . . .^ dxn ^ dyn.

6.5. HOLOMORPHIC FUNCTIONS ARE ANALYTIC 87

The following result, which follows at once from (6.4.5), is again in starkcontrast with the picture for C8 functions.

Corollary 6.4.10. Let tfnunPN be a sequence of holomorphic functions fn : U ÑK, and suppose that tfnu converges uniformly on compact sets to a functionf : U Ñ K. Then f is holomorphic, and for every k ě 0, the sequence of

derivatives tfpkqn unPN converges uniformly on compact sets to f pkq.

6.5 Holomorphic functions are analytic

Definition 6.5.1. Let U Ă Kn be an open subset. A function f : U Ñ Kis analytic if, for each a P U there exist an open ball Bpa,Rq Ă U and anabsolutely convergent power series in Bpa,Rq

ÿ

mPNncmpz ´ aq

m, (6.5.1)

where cm is a complex number and pzáqm “ pz1á1qm1 ¨ ¨ ¨ pznánq

m1 , whosesum is equal to fpzq for all z P Bpa, rq.

Example 6.1.6 shows that analytic functions of one variable are holo-morphic. The same is true of analytic functions of several complex variables.What is surprising is that the converse holds, i.e. holomorphic functions areanalytic that is the main result of the present subsection.

Let a P Kn and let pR1, . . . , Rnq P Rn`. Let

B pa1, R1q ˆ ¨ ¨ ¨ ˆB pan, RnqfÝÑ K

be a continuous function. Let 0 ă ri ă Ri for i P t1, . . . , nu. We let

ż

Γa1 pr1qˆ¨¨¨ˆΓan prnq

fptqdt1 ^ ¨ ¨ ¨ ^ dtnpt1 ´ z1q ¨ ¨ ¨ ¨ ¨ ptn ´ znq

:“

ż

r0,2πsn

ϕ˚ˆ

fptqdt1 ^ ¨ ¨ ¨ ^ dtnpt1 ´ z1q ¨ ¨ ¨ ¨ ¨ ptn ´ znq

˙

,

(6.5.2)where ϕ :“ Γa1pr1q ˆ . . .ˆ Γanprnq, i.e.

r0, 2πsnϕÝÑ Kn

pθ1, . . . , θnq ÞÑ pa1 ` r1 exppiθ1q, . . . , an ` rn exppiθ1qq

(Continuity of f guarantees that the integral in the right hand side of (6.5.2) isdefined.)

Proposition 6.5.2. Let a P Kn and let pR1, . . . , Rnq P Rn`. Suppose that

B pa1, R1q ˆ ¨ ¨ ¨ ˆB pan, RnqfÝÑ K


is a continuous function which is holomorphic in each variable separately. Let0 ă ri ă Ri for i P t1, . . . , nu. Then

fpzq “1

p2πiqn

ż


f pt1, . . . , tnq dt1 ^ ¨ ¨ ¨ ^ dtnpt1 ´ z1q ¨ ¨ ¨ ¨ ¨ ptn ´ znq

(6.5.3)

for all z P B pa1, R1q ˆ ¨ ¨ ¨ ˆB pan, Rnq

Proof. By induction on n. For n “ 1 (6.5.3) is Cauchy’s formula, i.e. Theorem6.4.6. Let’s prove the inductive step. Let n ě 2. By Fubini’s theorem theright-hand side of (6.5.3) is equal to

1

2πi

ż

Γan prnq

¨

˚

˝

1

p2πiqn´1

ż

Γa1 pr1qˆ¨¨¨ˆΓan´1prn´1q

f pt1, . . . , tnq dt1 ^ ¨ ¨ ¨ dtn´1

pt1 ´ z1q ¨ ¨ ¨ ¨ ¨ ptn´1 ´ zn´1q

˛

‹

‚

dtntn ´ zn

.

By the inductive hypothesis the above integral is equal to

1

2πi

ż

Γan prnq

f pz1, . . . , zn´1, tnq dtntn ´ zn

,

and the proposition follows from Cauchy’s integral formula

Arguing as in the proof of Corollary 6.4.9, we get the following result.

Corollary 6.5.3. Let f : U Ñ K be a holomorphic function, where U Ă Kn isopen. Then f is C8, and the partial derivatives of any order are holomorphic.Suppose that the closure of B pa1, R1qˆ¨ ¨ ¨ˆB pan, Rnq is contained in U . Thenfor z P B pa1, R1q ˆ ¨ ¨ ¨ ˆB pan, Rnq we have

Bk1`...`knfpzq

Bzk11 . . . Bzknn“k1! . . . kn!

2πi

ż


f pt1, . . . , tnq dt1 ¨ ¨ ¨ dtn

pt1 ´ z1qk1`1

¨ ¨ ¨ ¨ ¨ ptn ´ znqkn`1

.

(6.5.4)

Theorem 6.5.4. Let U Ă Kn be open and f : U Ñ K. The following conditionson f are equivalent:

1. f is holomorphic.

2. f is a continuous function, and is holomorphic in each variable separately.

3. f is analytic.

Proof. (1) ùñ (2): immediate from the definitions. (2) ùñ (3): by Proposi-tion 6.5.2 and the geometric series expansion

1

tk ´ zk“

1

tk ´ ak¨

1

1´ zkáktkák

“1

tk ´ ak`

zk ´ akptk ´ akq2

`pzk ´ akq

2

ptk ´ akq3` . . . ,

6.6. A VERSION OF HARTOGS’ THEOREM 89

we get that for z P Ba1pr1q ˆ . . .ˆBanprnq

fpzq “ÿ

kPNnck1,...,knpz1 ´ a1q

k1 . . . pzn ´ anqkn , (6.5.5)

where

ck1,...,kn “1

2πi

ż


fptqdt1 ^ ¨ ¨ ¨ ^ dtn

pt1 ´ z1qk1`1

¨ ¨ ¨ ¨ ¨ ptn ´ znqkn`1

(6.5.6)

(3) ùñ (1): Since f is analytic, it is a continuous function, and it is analytic ineach variable separately. By Example 6.1.6, it follows that f is holomorphic ineach variable separately. By Proposition 6.5.2 it follows that f is holomorphic(differentiation under the inegral sign).

Corollary 6.5.5. Let U Ă Kn be open and f : U Ñ K be holomorphic. Leta P U , and let (according to Theorem 6.5.4) be an expansion in power seriesof f around a

fpzq “ÿ

mPNncmpz ´ aq

m, z P Bpa,Rq Ă U.

ThenBk1`...`knfpaq

Bzk11 . . . Bzknn“ pk1q! . . . pknq!ck1,...,kn . (6.5.7)

Proof. Follows from Corollary 6.5.3 and (6.5.6).

The following result is in stark contrast with what happens for C8 functions.

Proposition 6.5.6 (Principle of analytic continuation). Let U Ă Kn be openand connected. If f, g : U Ñ K are holomorphic, and are equal on a non emptyopen V Ă U , then they are equal on all of U .

Proof. Since the difference of two holomorphic functions is holomorphic, it suf-fices to prove that if a holomorphic function is zero on a non empty open V Ă U ,then it is zero on all of U . Let D Ă U be the subset of z such that all partialderivetives of f in z vanish. Then D is closed because it is the intersectionof the closed subsets of points where a specific partial derivative vanishes. Inaddition D is non empty because it contains Y . Since U is connected, it sufficesto show that U is also open. If a P D, then f vanishes in a neighborhood of aby Corollary 6.5.5, and hence D contains an open neighborhood of a. ThusD is open.

6.6 A version of Hartogs’ Theorem


Chapter 7

Complex manifolds

7.1 Definition of complex manifold

LetX be a topological manifold. A holomorphic atlas onX is a family tpUk, ϕkqukPK ,where

1. tUkukPK is an open covering of X,

2. ϕk : Uk„ÝÑ Vk is a homeomorphism between Uk and an open Vk Ă Kn,

3. and for each k, h P K, the transition function ϕhpUhXUkq ÝÑ ϕkpUhXUkqis holomorphic (this makes sense because domain and codomain are opensubsets of Kn).

Remark 7.1.1. If tpUk, ϕkqukPK is a holomorphic atlas onX, we say that pUk, ϕkqare the charts of the atlas. Each chart determines holomorphic coordinatespz1 ˝ϕk, . . . , zn ˝ϕkq on Uk. It is often convenient to identify Uk with its imageϕkpUkq Ă Kn, and to denote the associated coordinates by pz1, . . . , znq.

Two holomorphic atlases on X are compatible if the union is a holomorphicatlas. The relation of compatibility is an equivalence relation.

Definition 7.1.2. A complex manifold is an equivalence class of holomorphicatlases for the relation of compatibility. If the charts take values in Kn, thedimension of X is n.

Let U Ă Kn be open. The atlas on U defined by the identity map U Ñ Udetermines an equivalence class of holomorphic atlases on U , and hence givesU the structure of a complex manifold. Below are fundamental examples ofcompact complex manifolds.

Example 7.1.3. Let Pn be complex projective space, with atlas tpPnZi , fiqu0ďiďn,where PnZi is the open subset of points whose Zi homogeneous coordinate is nonzero, and

PnZifiÝÑ Kn

rZs ÞÑ pZ0

Zi, . . . , Zi´1

Zi, Zi`1

Zi, . . . , ZnZi q

91

92 CHAPTER 7. COMPLEX MANIFOLDS

The above atlas is holomorphic, hence it provides Pn a structure of complexmanifold. From now on Pn denotes the above complex manifold. More generallythe complex Grassmannian Grpd, nq of complex vector subspaces V Ă Kn ofdimension d has the following holomorphic atlas. First, given a multiindex J “pj1, . . . , jdq, where 1 ď j1 ă . . . ă jd ď n, let V pZj1 , . . . , Zjdq be the kernel of thelinear map Kn Ñ Kd defined by Z Ñ pZj1 , . . . , Zjdq. Let Grpd, nqJ Ă Grpd, nqbe the open subset defined by

Grpd, nqJ :“ tW P Grpd, nq |W X V pZj1 , . . . , Zjdq “ t0u.

A d dimensional subspace W Ă Kn belongs to Grpd, nqJ if and only if it has abasis tv1, . . . , vdu given by the rows of a matrix

»

—

—

—

—

—

—

–

z1,1 . . . z1,j1´1 1 z1,j1`1 ¨ ¨ ¨ z1,j2´1 0 z1,j2`1 ¨ ¨ ¨ 0 z1,jd`1 ¨ ¨ ¨ z1,n

z2,1 . . . z2,j1´1 0 z2,j1`1 ¨ ¨ ¨ z2,j2´1 1 z2,j2`1 ¨ ¨ ¨ 0 z2,jd`1 ¨ ¨ ¨ z2,n

......

......

......

......

......

......

......

zd,1 . . . zd,j1´1 0 zd,j1`1 ¨ ¨ ¨ zd,j2´1 0 zd,j2`1 ¨ ¨ ¨ 1 zd,jd`1 ¨ ¨ ¨ zd,n

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

We let fJ : Grpd, nqJ Ñ Kdpn´dq be the map associating to W the entries Zk,jabove. The atlas tpGrpd, nqJ , fJqu is homolorphic, and it gives Grpd, nq a struc-ture of complex manifold of dimension dpn´ dq.

Definition 7.1.4. Let X be a complex manifold. A subset Y Ă X is a complexsubmanifold of X if the following holds. There exist a covering tUkukPK of Xby the open sets of a (holomorphic) atlas of X and, for each k P K, holomorhicfunctions f1

k , . . . , frk : Uk Ñ K (we identify Uk with an open subset of Kn via

the local chart, see Remark 7.1.1) such that

1. Y X Uk is the set of zeroes of f1k , . . . , f

rk :

Y X Uk “ tz P Uk | f1k pzq “ . . . “ frk pzq “ 0u.

2. The differentials df1k pzq, . . . , df

rk pzq are linearly independent for each z P

Y X Uk.

Given a complex submanifold Y Ă X, we can define an equivalence class ofholomorphic atlases on Y , by imitating the C8 definition - we simply replacethe C8 Implicit Function Theorem by its holomorphic analogue, i.e. Item (4)of Theorem 6.2.2. Thus Y is a complex manifold, and the inclusion mapY ãÑ X is a holomorphic map.

Proposition 7.1.5. Let X Ă An be a Zariski closed subset. Let a be a smoothpoint of X, and let k “ n ´ dimaX. Then there exists an open (classicaltopology) U Ă An containing a such that X X U is a complex submanifold ofU , of dimension n´ k.

Proof. Follows from Corollary 4.3.9, because polynomial functions are holo-morphic, and the algebraic differential of a polynomial agrees with the usualdifferential.

7.2. THE CATEGORY OF COMPLEX MANIFOLDS 93

Corollary 7.1.6. Let X Ă Pn be closed. Then Xsm is a complex submanifoldof Pnz singX.

Example 7.1.7. Let X Ă Pn be a hypersurface, i.e. X “ V pF q, where F is anon zero homogeneous polynomial of strictly positive degree. Assume that

V

ˆ

BF

BZ0, . . . ,

BF

BZn

˙

“ H. (7.1.1)

Then X is a complex submanifold of Pn, of dimension n ´ 1. In fact letpz0, . . . , pzj , . . . , znq be the (customary) holomorphic coordinates on PnZj given

by zk :“ ZkZj

. Then

X X PnZj “ tz P Kn | F pz0, . . . , zj´1, 1, zj`1, . . . , znq “ 0,

and hence it suffices to show that for each z P X X PnZj , at least one of the

partial derivatives BFBzkpz0, . . . , zj´1, 1, zj`1, . . . , znq does not vanish. Suppose

the contrary. From Euler’s relation

ÿ

0ďkďn

ZkBF

BZk“ pdegF qF, (7.1.2)

we get that

0 “ pdegF qF pz0, . . . , zj´1, 1, zj`1, . . . , znq “

“ÿ

0ďkďnk “j

zkBF

Bzkpz0, . . . , zj´1, 1, zj`1, . . . , znq`

BF

Bzjpz0, . . . , zj´1, 1, zj`1 . . . , znq “

“BF

Bzjpz0, . . . , zj´1, 1, zj`1 . . . , znq, (7.1.3)

and hence also BFBzjpz0, . . . , zj´1, 1, zj`1, . . . , znq vanishes. This contradicts (7.1.1).

Notice that F :“řnj“0 Z

dj provides an example satisfying (7.1.1) in an arbitrary

number of variables and arbitrary degree.

7.2 The category of complex manifolds

Definition 7.2.1. Let X and Y be complex manifolds. A continuous mapf : X Ñ Y is holomorphic if, for any atlases tpUk, ϕkqukPK ofX and tpWh, ψhquhPHof Y , the following holds. Let pk, hq P K ˆH; then the map

ϕkpUk X f´1pWhqq ÝÑ Kn

z ÞÑ ψhpfpϕ´1k pzqqq

(7.2.1)

is holomorphic (this makes sense because the domain is an open subset of a Kn).


If the maps in (7.2.1) are holomorphic for one choice of atlas for X, thenthey are holomorphic for any other choice of compatible atlas. Similarly, ifthe maps in (7.2.1) are holomorphic for one choice of atlas for Y , they remainholomorphic for a compatible atlas of Y . Thus, in order to check whether agiven continuous function is holomorphic, it suffices to check that the mapsin (7.2.1) are holomorphic for one choice of atlas for X and one choice of atlasfor Y .

In particular, if U Ă Kn is open, the two definitions of a holomorphic mapf : U Ñ Km, i.e. Definition 6.2.1 and Definition 7.2.2, coincide. We noticethat the identity map IdX : X Ñ X is holomorphic, and that the compositionof holomorphic maps f : X Ñ Y and g : Y Ñ Z is holomorphic.

Definition 7.2.2. Let X and Y be complex manifolds. A holomorphic mapf : X Ñ Y is an isomorphism if it has a holomorphic inverse, i.e. a holomorphicmap g : Y Ñ X such that g ˝ f “ IdX and f ˝ g “ IdY . An automorphism of Xis an isomorphism between X and itself.

The set of automorphisms of X with operation given by composition is agroup that we denote by AutpXq.

Definition 7.2.3. Let X, Y be two complex manifolds, with holomorphic at-lases tpUj , fjqujPJ and tpVk, gkqukPK respectively. Then

tpUj ˆ Vk, fj ˆ gkqupj,kqPJˆK

is a holomorphic atlas of the topological manifold XˆY . Replacing tpUj , fjqujPJand tpVk, gkqukPK by compatible holomorphic atlases, we get a holomorphicatlas compatible with tpUjˆVk, fjˆgkqupj,kqPJˆK . Hence XˆY has a complexstructure induced by those of X and Y .

From now on X ˆ Y denotes the complex manifold defined in Definition7.2.3. The projections XˆY Ñ X and XˆY Ñ Y are holomorphic, because inlocal coordinates they are given by the projection the first (or last) coordinates.Moreover XˆY is the product of X and Y in the category of complex manifolds,i.e. given a complex manifold W and holomorphic maps f : W Ñ X and g : W Ñ

Y , there is a unique holomorphic map W Ñ X ˆ Y which composed with thetwo projections X ˆ Y Ñ X and X ˆ Y Ñ Y gives back f and g.

A complex manifold determines an underlying C8 manifold, because a holo-morphic atlas is also a C8 atlas, and a holomorphic map between complexmanifolds is a C8 map of the underlying C8 manifolds. One distinctive featureof the C8 manifolds underlying holomorphic manifolds (beyond having evendimension) is that they are oreintable.

Proposition 7.2.4. The C8 manifold underlying a complex manifold is ori-entable.

Proof. Let X be a complex manifold. Let z “ pz1, . . . , znq be holomorphiccoordinates on a holomorphic chart pU, fq of X. Then

ωz :“ indz1 ^ dz1 ^ . . .^ dzn ^ dzn “ 2ndx1 ^ dy1 ^ . . .^ dxn ^ dyn

7.3. OTHER EXAMPLES OF COMPLEX MANIFOLDS 95

is a volume form on U . Let u “ pu1, . . . , unq be holomorphic coordinates onanother holomorphic chart pV, gq of X, and let ωu be the corresponding volumeform on V . Let z “ ϕpuq be the transition function, and let

Jpϕq :“

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

Bϕ1

Bu1. . . Bϕ1

Bun...

......

BϕnBu1

. . . BϕnBun

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

be the holomorphic Jacobian determinant. Then ωz “ |Jpϕq|2ωw. Thus theholomorphic atlas of X is oriented, and hence X is orientable.

Remark 7.2.5. The proof of Proposition 7.2.4 shows that a complex manifoldhas a well-defined complex orientation. An isomorphism f : X Ñ Y of complexmanifolds maps the complex oientation of Y to the complex orientation of X.Moreover the complex orientation of the product of complex manifolds X andY is the product of the complex orientation of X and the complex orientationof Y .

One goal that we would like to reach when studying complex manifolds isto determine the isomorphism classes of complex manifolds. A necessary con-dition for two complex manifolds to be isomorphic is that the underlying C8

manifolds be diffeomorphic. The latter condition is far from being sufficient.The simplest example is provided by K and the unit disc ∆ :“ tz P K | |z| “ 1u.A holomorphic map K Ñ ∆ is constant by Louville’s Theorem (see Exercise7.7.1), and hence K and ∆ are not isomorphic, although they are clearly dif-feomorphic. A richer family of such examples is provided by annuli in K, seeAhlfors [Ahl78]. We will give plenty of compact examples later on.

7.3 Other examples of complex manifolds

7.4 The holomorphic tangent space

Let M be a C8 manifold, and a P M . The ring of germs of smooth functionsat a, denoted EM,a, is the set of equivalence classes of couples pU, fq, whereU Ă M is an open subset containing a, f P K8pUq, and couples pU, fq, pV, gqare equivalent if there exists a couple pW,hq such that W Ă U X V and h “f|W “ g|W . Given ϕ “ rpU, fqs P EM,a, the evaluation ϕpaq :“ fpaq is welldefined. Hence we may give the abelian group R a structure of module overEM,a by setting ϕ ¨ x “ ϕpaqx, for ϕ P EM,a and x P R.

One may define the tangent space of M at a as the real vector space ofR derivations D : EM,a Ñ R, where R has the EM,a module structure definedabove. Explicitly, we require that for ϕ,ψ P EM,a

Dpϕ` ψq “ Dpϕq `Dpψq, Dpϕ ¨ ψq “ ϕpaq ¨Dpψq ` ψpaq ¨Dpϕq. (7.4.1)


In local coordinates px1, . . . , xrq centered at a (meaning that a has coordinatesp0, . . . , 0q), a basis of tangent space of M at a is given by

B

Bx1p0q, . . . ,

B

Bxrp0q, (7.4.2)

where BBxjp0q is the derivation mapping ϕpxq to Bϕ

Bxjp0q. We denote the tangent

space to M at a by TRa pMq. Let f : M Ñ N be a C8 map between C8

manifolds. Let a PM . Then we have the pull back homomorphism of rings

EN,fpaqf˚

ÝÑ EM,a

rpV, ξqs ÞÑ rpf´1V, ξ ˝ pf|f´1pV qqs(7.4.3)

If D P TRa pMq, then we get a derivation dfpaqpDq : EN,fpaq Ñ R by setting

EN,fpaqdfpaqpDqÝÑ R

ϕ ÞÑ Dpf˚pϕqq(7.4.4)

The map dfpaq : TRa pMq Ñ TR

fpaqpNq is linear. In local coordinates px1, . . . , xrq

on M centered at a, and py1, . . . , ysq on N centered at fpaq, the map f is given byx ÞÑ pf1pxq, . . . , fspxqq, where each fi is C8. The matrix of dfpaq with respectto the bases of TR

a pMq and TRfpaqpNq associated to the given local coordinates

(see (7.5.2)) is the Jacobian matrix

Jfpaq :“

ˆ

BfiBxj

p0q

˙

1 ď i ď s1 ď j ď r

Now we let TKa pMq :“ TR

a pMq bR K be the complexified tangent space to M ata. Moltiplication on the right hand side by complex numbers, gives TK

a pMq astructure of complex vector space (of dimension dimM). Let E K

M,a :“ EM,abRKbe the ring of germs of complex valued smooth functions at a. Concretely, anelement of E K

M,a is represented by pU, f ` igq, where f, g P K8pUq, and a basis

over K of TKa pMq is provided by the basis of TR

a pMq given above. One has acanonical identification of TK

a pMq with the complex vector space of derivationsDerKpE K

M,a,Kq (these are maps D : E KM,a Ñ K such that (7.4.1) holds for all

ϕ,ψ P E KM,a Ñ K). All of the considerations above valid for real derivations

go through unchanged for complex valued derivations. In particular we have acomplexified pull back f˚ : E K

N,fpaq Ñ E KM,a, and correspondingly the complexi-

fied differential dfpaq : TKa pMq Ñ TK

fpaqpNq. (The notation does not distinguish

between the real differential and the complexified differential.)Going from the tangent space to the complexified tangent space does not

give anything new, but if M is the C8 manifold underlying a complex manifold,then the complexified tangent space has a canonical splitting into a direct sumof complex vector spaces of equal dimensions. In order to explain this, we givea couple of definitions.

7.5. DIFFERENTIAL FORMS ON COMPLEX MANIFOLDS 97

Definition 7.4.1. Let X be a complex manifold, and x P X. The ring of germsof holomorphic functions at x is the set of ϕ P E K

M,a which are represented bycouples pU, fq such that f is holomorphic, and is denoted Oan

X,x. (It is clear that

OanX,x is a subring of E K

M,a.)

Definition 7.4.2. Let X be a complex manifold, and x P X. A (complex)tangent vector v P TK

x pXq is holomorphic if vpϕq for every ϕ P OanX,x, it is anti

holomorphic if vpϕq for every ϕ P OanX,x.

In local holomorphic coordinates pz1, . . . , znq centered at x, a basis of thecomplexified tangent space of X at x is given by

B

Bz1p0q, . . . ,

B

Bznp0q,

B

Bz1p0q, . . . ,

B

Bznp0q.

The first n tangent vectors are holomorphic, the last n are anti holomorphic.Let TxpXq Ă TK

x pXq be the subspace of holomorphic tangent vectors (noticethe potential confusion!). Then we have a direct sum decomposition

TKx pXq “ TxpXq ‘ T xpXq. (7.4.5)

Now let X,Y be complex manifolds, and let f : X Ñ Y be a holomorphic map.Let x P X. The pull back homomorphism f˚ : E K

Y,fpxq Ñ E KX,x maps Oan

Y,fpxq to

OanX,x. It follows that

dfpxqpTxpXqq Ă TfpxqpY q, dfpxqpT xpXqq Ă T fpxqpY q. (7.4.6)

In other words, holomorphic maps respect the direct sum decomposition of the(complexified) tangent space in (7.4.5).

7.5 Differential forms on complex manifolds

Let M be a C8 manifold. Let a PM . The real cotangent space to M at a is thedual TR

a pMq_ of the real vector space TR

a pMq. Similarly the complex cotangentspace to M at a is the dual TK

a pMq_ of the complex vector space TK

a pMq - itis canonically isomorphic to TR

a pMq_ bR K. In local coordinates px1, . . . , xrq

centered at a, a basis of the real or complex cotangent space to M at a is givenby

dx1p0q, . . . , dxrp0q. (7.5.1)

Now assume that X is a complex manifold, and let x P X. A basis of TKx pXq

_

at x in local holomorphic coordinates pz1, . . . , znq centered at x is given by

dz1p0q, . . . , dznp0q, dz1p0q, . . . , dznp0q. (7.5.2)

Let Ω1,0x pXq,Ω

0,1x pXq Ă TK

x pXq_ be the (complex) subspaces defined by

Ω1,0x pXq :“ AnnpT xpXqq, Ω0,1

x pXq :“ AnnpTxpXqq. (7.5.3)


By the direct sum decomposition in (7.4.5), we have

TKx pXq

_ “ Ω1,0x pXq ‘ Ω0,1

x pXq. (7.5.4)

A basis of the first summand is given by dz1p0q, . . . , dznp0q, and a basis of thesecond summand is given by dz1p0q, . . . , dznp0q.

The decomposition in (7.5.6) induces a direct sum decomposition of theexternal algebra of TK

x pXq_ as follows. The complex bilinear wedge product

mapŹp

Ω1,0x pXq ˆ

ŹqΩ0,1x pXq Ñ

Źp`qTKx pXq

_ descends to a linear mapŹp

Ω1,0x pXq bK

ŹqΩ0,1x pXq Ñ

Źp`qTKx pXq

_, which is injective. Thus we maydenote the image by

ŹpΩ1,0x pXq b

ŹqΩ0,1x pXq. We have

pľ

Ω1,0x pXq b

qľ

Ω0,1x pXq “

$

’

’

&

’

’

%

ÿ

|J| “ p|K| “ q

aJ,KdzJ ^ dzK | aJ,K P K

,

/

/

.

/

/

-

Clearly we have a direct sum decomposition

mľ

TKx pXq

_ “à

p`q“m

pľ

Ω1,0x pXq b

qľ

Ω0,1x pXq. (7.5.5)

The union ofŹp

Ω1,0x pXq b

ŹqΩ0,1x pXq for x P X is a C8 (complex) sub

vector bundle of the m-the exterior product of the complex cotangent bundleŹm

TKpXq_, isomorphic to the tensor product of the C8 vector bundlesŹp

Ω1,0pXqand

ŹqΩ0,1pXq. Hence we have an injection of spaces of C8 global sections

ΓpX,pľ

Ω1,0x pXq b

qľ

Ω0,1x pXqq ãÑ ΓpX,

mľ

TKx pXq

_q.

The image of the above injection is denoted A p,qpXq. The elements of A p,qpXqare called pp, qq forms. Thus we have a direct sum decomposition

ΓpX,mľ

TKx pXq

_q “à

p`q“m

A p,qpXq. (7.5.6)

If X is an open subset U Ă Kn, the above direct sum reduces to that in (6.3.2).Notice that if f : X Ñ Y is a holomorphic map of complex manifolds, then thepull back on differential forms takes A p,qpY q to A p,qpXq:

f˚pA p,qpY qq Ă A p,qpXq. (7.5.7)

Let ϕ P A p,qpXq. By (6.3.7), we have dϕ P A p`1,qpXq ‘ A p,q`1pXq. Wedefine

A p,qpXqBÝÑ A p`1,qpXq, A p,qpXq

BÝÑ A p,q`1pXq

by setting dϕ “ Bϕ`Bϕ, where Bϕ P A p`1,qpXq and Bϕ P A p,q`1pXq. In localcoordinates, B and B are given by (6.3.5) and (6.3.6). Since d ˝ d “ 0, we haveB ˝ B “ 0 (see (6.3.8)). Thus we have the Dolbeault complex

0 Ñ A p,0pXqBÝÑ . . .

BÝÑ A p,qpXq

BÝÑ A p,q`1pXq

BÝÑ . . . . . . (7.5.8)

7.6. HOLOMORPHIC DIFFERENTIAL FORMS 99

The corresponding cohomology is the Dolbeault cohomology

Hp,q

BpXq :“ tϕ P A p,qpXq | Bpϕq “ 0utBpψq | ψ P A p,q´1pXqu. (7.5.9)

Let f : X Ñ Y be a holomorphic map of complex manifolds. By (7.5.10), thepull back on differential forms defines a homomorphism of the correspondingDolbeault complexes, and hence it descends to a homomorphism of vector spaces

Hp,q

BpY q

Hp,qBpfq

ÝÑ Hp,q

BpXq

rϕs ÞÑ rf˚pϕqs(7.5.10)

7.6 Holomorphic differential forms

Dolbeault’s cohomology group Hp,0

BpXq consists of the subspace of A p,0pXq

whose elements are the ϕ such that Bϕ “ 0. In local coordinates z1, . . . , zn wehave ϕ “

ř

|J|“p appzqdzJ , and

Bϕ “ÿ

|J|“p1ďkďn

Bappzq

zkdzkdzJ

Hence Bϕ “ 0 if and only if all the ap’s are holomorphic functions of z. Theseare known as holomorphic p forms.

Example 7.6.1. Let us show that if n ą 0 then Hn,0

BpPnq “ 0. In fact, let

α P Hn,0

BpPnq. Then, for j P t0, . . . , nu we have

α|PnZj“ fj

˜

Z0

Zj, . . . ,

xZjZj, . . . ,

ZnZj

¸

d

ˆ

Z0

Zj

˙

^ . . .^

d

ˆ

ZjZj

˙

^ . . . d

ˆ

ZnZj

˙

,

where fj : Kn Ñ K is an entire function. It suffices to prove that fj “ 0 for all j.Let j “ k P t0, . . . , nu. On PnZj X PnZk we have two sets of holomorphic coordin-

ates, namely pZ0Zj , . . . , ZnZjq and pZ0Zk, . . . , ZnZkq, which are related bythe equations

ZsZj“ZsZk¨

ˆ

ZjZk

˙´1

, s P t0, . . . ,pj, . . . , nu.

Hence

d

ˆ

ZsZj

˙

“

ˆ

ZjZk

˙´1

d

ˆ

ZsZk

˙

´ZsZk¨

ˆ

ZjZk

˙´1

d

ˆ

ZjZk

˙

, s P t0, . . . ,pj, . . . , nu.

Thus we get the following equality of differential forms on PnZj X PnZk :

fj

ˆ

Z0Zk¨

´

ZjZk

¯´1,...,ZnZk

¨

´

ZjZk

¯´1˙

´

ZjZk

¯´pn`1qd´

Z0Zk

¯

^...^d´

ZnZk

¯

“

“p´1qk´jfk

´

Z0Zk,...,ZnZk

¯

d´

Z0Zk

¯

^...d´

ZnZk

¯

.


Hence, letting tj :“ ZjZk for j P t0, . . . ,pk, . . . , nu, we have

fj`

t0 ¨ t´1j , . . . , tn ¨ t

´1j

˘

¨ t´pn`1qj “ p´1qk´jfkpt0, . . . , tnq. (7.6.1)

Notice that in the left hand side of (7.6.1) there is no variable with index j, andthat the variable with index k is t´1

j . Equation (7.6.1) holds for tj “ 0, but theright hand side is continuous (even holomorphic) on all of Kn. It follows that forfixed t0, . . . , ptj , . . . , ptk, . . . , tn, the left hand side of (7.6.1) has a well defined limitfor tj Ñ 0. This contradicts the maximum principle for holomorphic functionsunless fj “ 0.

Definition 7.6.2. Let Y be a complex manifold of dimension n, and X Ă Y acomplex submanifold of codimension 1, i.e. dimX “ n ´ 1. Let U :“ Y zX. Ameromorphic n form on Y with (at most) a simple pole alongX, is a holomorphicn form ϕ on U such that for every a P X there exist

1. an open V Ă Y containing a,

2. a holomorphic function f : V Ñ K such that X X V “ tx P V | fpxq “ 0u,and dfpxq “ 0 for every x P V ,

3. a holomorphic n form ψ on V such that ϕ|pV zXq “ψf .

If ϕ is a meromorphic n form on Y with (at most) a simple pole along X,then there is an associated holomorphic pn´1q form on X defined as follows. Leta P X, and let V Ă Y be an open set containing a such that (1)-(3) above hold.By the hypothesis on the differential of f , we can extend f to a holomorphiccoordinate system pz1, . . . , zn´1, fq on V , after shrinking V around a if necessary.Thus, letting ψ be as in Item (3), we have ψ “ ωdz1 ^ . . . dzn´1 ^ df , i.e.

ϕ|pV zXq “ ωdz1 ^ . . .^ dzn´1 ^df

f“ ωdz1 ^ . . .^ dzn´1 ^ d log f. (7.6.2)

Claim 7.6.3. Keeping notationa as above, the holomorphic pn ´ 1q form onV XX defined by

pωdz1 ^ . . .^ dzn´1q|VXX (7.6.3)

depends only on the meromorphic n form ϕ, not on the other choices that havebeen made.

Proof. First notice that we have chosen a local equation f “ 0 of V XX withnon vanishing differential df and the extension of f to a holomorphic coordin-ate system pz1, . . . , zn´1, fq. Suppose that we replace f “ 0 by another localequation g “ 0 with non vanishing differential dg. It follows that f “ ug, whereu is holomorphic and nowhere zero. Thus

ϕ|pV zXq “ ωdz1^. . .^dzn´1^dpugq

ug“ ωdz1^. . .^dzn´1^

ˆ

dg

g` u´1du

˙

“

“ p1` u´1gqωdz1 ^ . . .^ dzn´1 ^dg

g.

7.6. HOLOMORPHIC DIFFERENTIAL FORMS 101

Since g vanishes on X X V we get that

pωdz1 ^ . . .^ dzn´1q|VXX “ pp1` u´1gqωdz1 ^ . . .^ dzn´1q|VXX .

This proves that the formula in (7.6.3) does not depend on the choice of a localequation f “ 0 of V XX (with non vanishing differential df). The proof thatthe formula in (7.6.3) does not depend on the choice of the extension of f to aholomorphic coordinate system pz1, . . . , zn´1, fq is left to the reader.

By Claim 7.6.3, given a complex manifold Y of dimension n and a complexsubmanifold X Ă Y of dimension n´ 1, there is a well defined homomorphismof vector spaces

tϕ meromorphic n form on Y with (at most) a simple pole on XuResÝÑHn´1,0

BpXq, (7.6.4)

named the Poincare residue map. The following observation follows at oncefrom the definition of the Poj care residue map.

Remark 7.6.4. Let ϕ be a meromorphic n form on Y with (at most) a simplepole on X, given locally near a P X by the formula in (7.6.2). Then Respϕqvanishes at a if and only if ωpaq “ 0.

Example 7.6.5. Let X Ă Pn be a hypersurface, i.e. X “ V pF q, where F is a nonzero homogeneous polynomial of strictly positive degree d. Assume that (7.1.1)holds, and hence X is a complex submanifold of Pn, of dimension pn ´ 1q.Suppose that d ě pn`1q. Then there exists non zero holomorphic pn´1q formson X, i.e. non zero elements of Hn´1,0

BpXq. In fact, let Ω be the holomorpbic n

form on Kn`1zt0u defined by

Ω :“nÿ

j“0

p´1qjZjdZ0 ^ . . .^ydZj ^ . . .^ dZn. (7.6.5)

Let a P pKn`1zt0uq. The tangent space to the K˚ equivalence class K˚a of a isgenerated by the tangent vector θa :“

ř

k “ 0np´1qkakBBZk

. Now notice that if

v1, . . . , vn´2 P TapKn`1q, then

xΩ, θa ^ v1 ^ . . .^ vn2y “ 0. (7.6.6)

Hence Ω descends to an element ofŹnpTapKn`1xθayq

˚. Notice that the latteris canonically identified with

ŹnpTrasPnq˚. The n form does not descend to an

n form on Pn because it is not invariant under the action of K˚, or in otherwords, the element of

ŹnpTrasPnq˚ defined above depends on the lift of ras to

a point of pKn`1zt0uq. We can remedy to this by considering

rϕ :“G

FΩ, (7.6.7)

where G P KrZ0, . . . , Znsdń´1 (this explains why we need d ě pn` 1q). The nform rϕ on pKn`1zV pF qq is invariant under the action of K˚, hence it descends


to a holomorphic form ϕ on PnzV pF q. Actually ϕ is a meromorphic n form onPn with (at most) a simple pole along X. Thus Respϕq is a holomorphic n formon X. If ras P X and Gpaq “ 0, then Respϕq is non zero at ras by Remark7.6.4.

Notice that if d “ 1, i.e. X is a hyperplane (and n ě 2) then there are nonnonzero holomorphic n forms on X by Example 7.6.1. The same statementholds if d ď n, see ??

7.7 The Dolbeault-Grothendieck Lemma

7.7.1 Statement of the result

Given a P Kn and r P p0,`8sn, we let ∆pa, rq Ă Kn be the polydisk

∆pa, rq :“ Bpa1, r1q ˆ . . .ˆBpan, rnq. (7.7.1)

In the present section we will prove the following result.

Lemma 7.7.1 (Dolbeault-Grothendieck). Let a P Kn and r P p0,`8sn. Ifq ě 1, then Hp,q

Bp∆pa, rqq “ 0.

7.7.2 A general Cauchy integral formula in one variable

Throughout the present subsection U Ă K is an open set, f : U Ñ K is acontinuous map, and the closed ball Bpa, rq is contained in U .

Claim 7.7.2. Let z P Bpa, rq. The improper integral

ĳ

Bpa,rqztzu

fpwq

w ´ zdw ^ dw (7.7.2)

is absolutely convergent.

Proof. All we must do is examine the behaviour of the integrand near z. Let r0 ą

0 be such that Bpz, r0q Ă Bpa, rq. On Bpz, r0q we switch to polar coordinatespρ, θq P r0, r0q ˆ r0, 2πq, where w “ z ` ρeiθ. Then the integrand becomes

´2ifpz ` ρeiθqeíθdρ^ dθ, (7.7.3)

which is continuous on r0, r0q ˆ r0, 2πq. The result follows.

Theorem 7.7.3 (Cauchy’s integral formula, 2nd version). Keep notation andassumptions as above. Then

fpzq “1

2πi

ż

Γpa,rq

fpwq

w ´ zdw `

1

2πi

ĳ

Bpa,rq

Bfpwq

Bw

dw ^ dw

w ´ z(7.7.4)

(The last integral is the improper integral in (7.7.9).)

7.7. THE DOLBEAULT-GROTHENDIECK LEMMA 103

Proof. Let ε ą 0 be such that Bpz, εq Ă Bpa, rq. Applying Stokes’ Theorem tothe differential form fpwqdw on the region Bpa, rqzBpz, εq, we get that

ż

Γpa,rq

fpwq

w ´ zdw ´

ż

Γpz,εq

fpwq

w ´ zdw “

ĳ

Bpa,rqzBpz,εq

Bfpwq

Bw

dw ^ dw

w ´ z. (7.7.5)

Now let ε Ñ 0. The second integral in the left hand side of (7.7.5) tends to2πifpzq, while the double integral in the right hand side of (7.7.5) tends to thevalue of the absolutely convergent improper integral in Claim 7.7.2.

7.7.3 The B equation in one variable

Let U Ă K be an open set, and let f : U Ñ K be a Cs map, where s ě 1,possibly s “ 8.

Proposition 7.7.4. Keep notation and assumptions as above, and suppose thatBpa, rq is contained in U . Let g : Bpa, rq Ñ K be defined by

gpzq “1

2πi

ĳ

Bpa,rq

fpwq

w ´ zdw ^ dw. (7.7.6)

Then g P CspBpa, rqq, and BgBz “ fpzq on Bpa, rq.

Proof. Notice that the double integral is absolutely convergent by Claim 7.7.2.We will prove that g is of class Cs, and that Bg

Bz “ fpzq.

Let z0 P Bpa, rq, and let ε ą 0 be such that Bpz0, 2εq Ă Bpa,Rq. By using asuitable partition of unity, we may write f “ f1 ` f2, where f1, f2 are of classCs, and

supp f1 Ă Bpz0, 2εq, supp f2 Ă pUzBpz0, εqq. (7.7.7)

For j P t1, 2u and z P Bpz0, εq, let

gjpzq “1

2πi

ĳ

Bpa,rq

fjpwq

w ´ zdw ^ dw. (7.7.8)

Then g “ g1 ` g2. The integrand defining g2 is of class Cs, hence

Bg2

Bz“

1

2πi

ĳ

Bpa,rq

B

Bz

ˆ

fjpwq

w ´ z

˙

dw ^ dw “ 0, z P Bpz0, εq. (7.7.9)

Thus BgBz “

Bg1Bz for z P Bpz0, εq. In order to compute Bg1

Bz , we notice that wemay view g1 as a function on all of K (vanishing outside Bpz0, 2εq). Making thesame change of variables as in the proof of Claim 7.7.2 (see (7.7.3)), we get

g1pzq “1

2πi

ĳ

K

f1pwq

w ´ zdw ^ dw “ ´

1

π

`8ż

0

2πż

0

f1pz ` ρeiθqeíθdρ^ dθ.


Taking the derivative with respect to z, and going backwards, we get that

Bg1

Bz“ ´

1

π

`8ż

0

2πż

0

Bf1

Bzpz ` ρeiθqeíθdρ^ dθ “

1

2πi

ĳ

K

Bf1

Bw

dw ^ dw

w ´ z.

Now we apply the second version of Cauchy’s integral formula, i.e. Theorem7.7.3 to the function f1 and the disk Bpz0, 2εq. Since f1 vanishes on Γpz0, 2εq,we get that for z P Bpz0, εq

Bg

Bz“Bg1

Bz“

1

2πi

ĳ

Bpz0,2εq

Bf1

Bw

dw ^ dw

w ´ z“ f1pzq “ fpzq.

This proves that BgBz “ fpzq for z P Bpz0, εq. Since z0 is an arbitrary point of

Bpa, rq, this proves that BgBz “ fpzq on Bpa, rq. The arguments given above also

show that g1 and g2 are of class Cs, and hence g is of class Cs.

7.7.4 Proof of the Dolbeault-Grothendieck lemma

First one proves that if q ě 1, a B-closed β P A p,qp∆pa,Rqq is B exact on anypolydisk ∆pa, rq Ă ∆pa,Rq with radii strictly smaller than the correspondingradii of R.

Proposition 7.7.5. Let a P Kn and r,R P p0,`8sn with ri ă Ri for i Pt1, . . . , nu. Let q ą 0, and let β P A p,qp∆pa,Rqq be B-closed. Then there existsα P A p,q´1p∆pa, rqq such that

Bα “ β|∆pa,rq. (7.7.10)

Proof. First assume that p “ 0. The proof is by induction on the maximuminteger m such that dzm appears in the expression

β “ÿ

|K|“q

βKdzK , βK P K8p∆pa, rqq. (7.7.11)

If m “ 1, then β “ β1dz1. Let

αpzq “1

2πi

ĳ

Bpa1,r1q

β1pw1, z2, . . . , znq

w1 ´ z1dw1 ^ dw1. (7.7.12)

Since Bβ “ 0 the function β1 is holomorphic in each of the variables zj withj ą 1, and hence so is α. Moreover Bα

Bz1“ β1 by Proposition 7.7.4. Hence

Bα “ β on ∆pa, rq.Next, we prove the induction step. We assume that (7.7.10) has a solution

if m is the maximum integer such that dzm appears in (7.7.11), and we showthat (7.7.10) has a solution if the maximum such integer is m` 1. We write

β “ÿ

|K|“qkiăm`1

βKdzK `ÿ

|K|“q´1kiăm`1

γKdzK ^ dzm`1. (7.7.13)


For a multiindex K appearing in the second summation in the right hand sideabove, let

gK :“1

2πi

ĳ

Bpam`1,rm`1q

γKpz1, . . . , zm, wm`1, zm`2, . . . , znq

wm`1 ´ zm`1dwm`1 ^ dwm`1.

(7.7.14)Since Bβ “ 0 the functions γK are holomorphic in each of the variables zj with

j ą pm` 1q, and hence so is gK . Moreover BγKBzm`1

“ γK by Proposition 7.7.4.

It follows that m is the maximum integer such that dzm appears in

β ´ B

¨

˚

˚

˝

ÿ

|K|“q´1kiăm`1

gKdzK

˛

‹

‹

‚

. (7.7.15)

By the inductive hypothesis, the above differential is equal to Bpξq for a suitableξ, and we are done. This prove the proposition if p “ 0.

In order to prove the result in general, suppose that

β “ÿ

|J|“p|K|“q

βJKdzJ ^ dzK (7.7.16)

is B-closed. For each multiindex J with |J | “ p, the p0, qq form

βJ :“ÿ

|K|“q

βJKdzK (7.7.17)

is B-closed, hence there exists αJ P A 0,q´1p∆pa, rqq such that BpαJq “ βJ (thecase p “ 0 of the proposition has been proved). Then

Bpÿ

|J|“p|K|“q

dzI ^ αJq “ β. (7.7.18)

Proof of the Dolbeault-Grothendieck lemma. Arguing as in the proof of Pro-position 7.7.5, we see that it suffices to prove the result for p “ 0. Thus,let q ą 0, and let β P A 0,qp∆pa,Rqq be B-closed. Choose a sequence trp`q Pp0,`8snu`PN converging to R, and such that rp`qj ă rp` ` 1qj ă R for allj P t1, . . . , nu. Let ` P N. By Proposition 7.7.5, there exists α``1 P

A 0,q´1p∆pa, rp` ` 1qqq such that Bpα``1q “ β|∆pa,rp``1qq. Multiplying α``1

by a suitable C8 function on ∆pa,Rq equal to 1 on ∆pa, rp`qq and vanishingoutside ∆pa, prp`q ` rp`` 1qq2q, we get γ` such that

Bpγ`q “ β|∆pa,rp`qq, γ` P A 0,q´1p∆pa,Rqq. (7.7.19)


We inductively modify γ` so that the modified sequence trγù`PN consists ofsolutions of (7.7.19) converging to a C8 solution α of the equation Bpαq “ β.

Let rγ` “ γ` for ` “ 0, 1. Now let ` ě 2. Assuming that rγ0, . . . , rγ`´1 havebeen defined such that (7.7.19) holds with γ` replaced by rγ`, we define rγ` asfollows.

We distinguish between the case q “ 1, and the case q ą 2. First, supposethat q “ 1. On ∆pa, rp`´ 1qq the function γ`´ rγ`´1 is holomorphic by (7.7.19),and hence it is equal to an absolutely convergent power series on any strictlysmaller poldisk, in particular on ∆pa, rp`´2qq. Hence there exists a polynomialfunction p on ∆pa,Rq such that we have

|γ`pzq ´ rγ`´1pzq ´ ppzq| ď1

2`, @z P ∆pa, rp`´ 2qq. (7.7.20)

Let rγ`pzq :“ γ`pzq´ppzq. The Cauchy sequence trγù`PN is absolutely convergenton compact subsets of ∆pa,Rq, hence it converges to a continuous functionα on ∆pa,Rq. In fact α is C8, because on ∆pa, rp`0qq it is the limit of asequence tγ`0 ` fù`ě`0`2, where tfù is a sequence of holomorphic functionsconverging uniformly on compact sets (recall Corollary 6.4.10). We also seethat Bpαq “ β.

Now suppose that q ą 1. On ∆pa, rp` ´ 1qq the p0, q ´ 1q form γ` ´ rγ`´1 isB-closed, hence by Proposition 7.7.5 and an easy argument involving a bumpfunction, there exists ω P A 0,q´2p∆pa,Rqq such that

Bpωq|∆pa,rp`´2qq “ pγ` ´ rγ`´1q|∆pa,rp`´2qq. (7.7.21)

Let rγ`pzq :“ γ` ´ Bpωq. The sequence trγù`PN is eventually stationary on every∆pa, rp`q, hence it converges to a C8 p0, q ´ 1q form α, and clearly Bpαq “ β.

Exercises

Exercise 7.7.1. Let f be an entire function, i.e. a holomorphic function f : KÑK. Suppose that there exists an integer d such that

lim|z|Ñ`8

|fpzq|

|z|d`1“ 0. (7.7.22)

Prove that f is a polynomial of degree at most d, i.e. there exist a0, . . . ad P Ksuch that fpzq “ a0z

d ` . . . ` ad. (Hint: prove that f pnqp0q “ 0 for n ą d.) Inparticular one gets Liouville’s Theorem: a bounded entire function is constant.

Exercise 7.7.2. Let U Ă K be open, and a P U . Suppose that f : pUztauq Ñ Kis holomorphic, and that there exists r ą 0 such that f is bounded on Bpa, rqXU .Riemann’s extension Theorem states that f extends to a holomorphic functionrf : U Ñ K. Prove it as follows. Let r ą 0 be such that Bpa, rq Ă U . Show that


the usual Cauchy integral formula holds for all z P pBpa, rqztauq:

fpzq “1

2πi

ż

Γaprq

fptq

t´ zdt,

and then notice that the right hand side of the above euqation extends to aholomorphic function over a as well.

Exercise 7.7.3. Let U Ă K be open and connected and let f : U Ñ K beholomorphic non constant. Prove that f is open, i.e. it maps open sets to opensets, proceeding as follows. Let a P U , and let

fpzq “8ÿ

m“0

cmpz ´ aqm

be a power series expansion of f in a neighborhhod of a, say Bpa, rq. Let m0

be the minimum strictly positive natural number such that cm0 “ 0 (since f isnot constant on U , such an m0 exists by the Principle of analytic prolungation).Then, on Bpa, rq we have

fpzq “ c0 ` cm0pz ´ aqm0gpzq,

where g is holomorphic and gpaq “ 0.

1. Prove that for a sufficiently small positive δ, there exists a homolorphicfunction h : Bpa, δq such that g|Bpa,δq “ hm0 (use the Inverse functionTheorem, i.e. Item (3) of Theorem 6.2.2).

2. Let ϕ : Bpa, δq Ñ K be the holomorphic function ϕpzq “ c1m0m0 pzáq¨hpzq.

By Item (1), on Bpa, δq we have fpzq “ c0`ϕpzqm0 . Check that ϕ1paq “ 0,

and hence ϕpBpa, δqq Ą Bp0, δ1q, for some δ1 ą 0 by the Inverse functionTheorem.

3. Conclude that fpBpa, δqq Ą Bpc0, δm01 q.

Notice that the analogous statement for differentiable (or even analytic) realfunctions of a real variable is false.

Exercise 7.7.4. Prove the Maximum modulus priciple: Let U Ă Kn be openand connected, and let f : U Ñ K be holomorphic non constant. If K Ă U iscompact, any z0 P K achieving the maximum of the absolute value function|fpzq| is not an interior point of K, i.e z0 P BK. (Hint: if n “ 1 the resultfollows at once from Exercise 7.7.3. If n ą 1 reduce to the case n “ 1 byrestricting f to lines in Kn.)

Exercise 7.7.5. Let

ˆ

a bc d

˙

be an invertible 2ˆ 2 matrix. Then

P1 fÝÑ P1

rZ0, Z1s ÞÑ rcZ1 ` dZ0, aZ1 ` bZ0s(7.7.23)


is an automorphism of P1. (The weird choice of formula in (7.7.23) is explainedby the formula fpzq “ az`b

cz`d valid when using the affine coordinate z “ z1z0.)

Prove that every automorphism of P1 (as complex manifold!) is of the aboveform, and hence

AutpP1q – PGL2pKq,

by arguing as follows.

1. Let ϕ P AutpP1q. Composing with a suitable automorphism in (7.7.23), wemay replace ϕ by an automorphism ψ0 of P1 such that ψpr0, 1sq “ r0, 1s.The restriction of ψ to the affine line P1ztr0, 1su defines a (holomorphic)automorphism ψ0 P AutpKq. It suffices to prove that there exists pα, βq PK˚ ˆK such that ψ0pzq “ αz ` β.

2. Prove that (7.7.22) holds for f “ ψ0 and d “ 1. Conclude that ψ0 is apolynomial function of degree 1 by Exercise 7.7.1.

Exercise 7.7.6. Let f : K Ñ K be an automorphism. Prove that the maprf : P1 Ñ P1 defined by setting

rfprZ0, Z1sq :“

#

r1, fpZ1

Z0qs if Z0 “ 0,

r0, 1s if Z0 “ 0,

is an automorphism of P1. Conclude that there exists pα, βq P K˚ˆK such thatfpzq “ αz ` β.

Exercise 7.7.7. Prove that the upper half plane

H :“ tz P K | Impzq ą 0u

is isomorphic (as complex manifold) to the unit disc ∆ Ă K. (Hint: find anautomorphism f of P1 which takes the closure of the real line to the boundaryof the unit disc. Either f or 1

f defines an isomorphism between H and ∆.)

Chapter 8

Vector bundles and sheaves

8.1 Holomorphic vector bundles

109

110 CHAPTER 8. VECTOR BUNDLES AND SHEAVES

Chapter 9

Hermitian geometry

9.1 Hermitian metrics on complex manifolds

9.2 Kahler manifolds

111

112 CHAPTER 9. HERMITIAN GEOMETRY

Chapter 10

The Hodge decomposition

10.1 Statement of the Hodge decomposition The-orem

Usually the Hodge decomposition Theorem is formulated in terms of harmonicforms. Here we state it with no mention of harmonicity. Let X be a complexmanifold. We let

Hp,qd pXq :“ tϕ P A p,qpXq | dϕ “ 0utdψ | ψ P A p`q´1pXqu (10.1.1)

be the subspace of Hp`qDR pX;Kq of classes represented by closed pp, qq-forms.

Theorem 10.1.1 (Hodge decomposition). Let X be a compact Kahler manifold.The complex De Rham cohomology of X decomposes as a direct sum

HmDRpX;Kq “

à

p`q“m

Hp,qd pXq. (10.1.2)

Remark 10.1.2. The proof of Theorem 10.1.1 gives (via harmonic forms) thatthe group Hp,q

d pXq is isomorphic to the corresponding Dolbeault cohomology -this is part of the full statement of the Hodge Decomposition Theorem. If X isprojective, it implies (via GAGA) that Hp,q

d pXq is isomorphic to the cohomologygroup of the sheaf of algebraic p-forms in the Zariski topology. We will go overthis important point when we will explain the proof of the Hodge decompositionTheorem.

The cohomology of smooth projective varietiese enjoys many special proper-ties. One such property follows at once from the Hodge decomposition Theoremand the observation that complex conjugation on Hm

DRpX;Kq maps isomorph-ically Hp,qpXq to Hq,ppXq.

Corollary 10.1.3. The odd Betti numbers of a smooth projective variety areeven.

113

114 CHAPTER 10. THE HODGE DECOMPOSITION

10.2 Examples

Holomorphic differentials on compact Kahler manifolds. Curves. Abelian vari-eties. Poincare residue for hypersurfaces.

Chapter 11

Harmonic forms

115

116 CHAPTER 11. HARMONIC FORMS

Chapter 12

Kahler identities

117

118 CHAPTER 12. KAHLER IDENTITIES

Appendix A

Commutative algebra a lacarte

A.1 Noetherian rings

In what follows, rings are always commutative with 1. The proofs of the resultsbelow are contained in most Algebra textbooks (e.g. Lang [?]).

Definition A.1.1. A (commutative unitary) ring R is Noetherian if every idealof R is finitely generated.

Example A.1.2. A field K is Noetherian, because the only ideals are t0u “ p0qand K “ p1q. The ring Z is Noetherian, because every ideal has a singlegenerator.

Lemma A.1.3. A (commutative unitary) ring R is Noetherian if and only ifevery ascending chain

I0 Ă I1 Ă . . . Ă Im Ă . . .

of ideals of R (here Im is defiend for all m P N, and Im Ă Im`1 for all m P N)is stationary, i.e. there exists m0 P N such hat Im “ Im0

for m ě m0.

Proof. Suppose that R is Noetherian. The union I :“Ť

mPN Im is an idealbecause the tImu form a chain. By Noetherianity I is finitely generated, sayI “ pa1, . . . , arq. There exists m0 such that aj P Im0

for j P t1, . . . , ru, andhence I “ Im0

. Let m ě m0; then Im Ă I and I Ă Im, hence I “ Im. ThusIm0

“ Im for m ě m0.Now suppose that every ascending chain of ideals of R is stationary. Let

I Ă R be an ideal. Suppose that I is not finitely generated. Let a1 P I. Thenpa1q Ĺ I because I is not finitely generated; let a2 P pIzpa1qq. Then pa1, a2q Ĺ Ibecause I is not finitely generated. Iterating, we get a non stationary chain ofideals (contained in I)

pa1q Ĺ pa1, a2q Ĺ . . . Ĺ pa1, . . . , amq Ĺ

119

120 APPENDIX A. COMMUTATIVE ALGEBRA A LA CARTE

This is a contradiction.

Example A.1.4. The ring HolpKq of entire functions of one variable is not No-etherian. In fact let fm P HolpKq be defined by

fmpzq :“8ź

n“m

ˆ

1´z2

n2

˙

, m ě 1.

Then pfmq Ĺ pfm`1q. Thus pf1q Ă pf2q Ă . . . Ă pfmq Ă . . . is a non-stationaryascending chain of ideals, and hence HolpKq is not Noetherian by LemmaA.1.3.

Theorem A.1.5 (Hilbert’s Basis Theorem). Let R be a Noetherian commutat-ive ring. Then Rrts is Noetherian.

Proof. For a non zero f P Rrts, we let `pfq be the leading coefficient of f , i.e. iff “

řmi“0 cit

i with cm “ 0, then `pfq “ cm.Let I Ă Rrts. We must prove that I is finitely generated. If I “ p0q there is

nothing to prove and hence we may assume I ‰ p0q. Thus the set

`pIq :“ t`pfq | 0 ‰ f P Iu

is non-empty and it makes sense to define

J :“ x`pIqy Ă R

as the ideal of R generated by `pIq. By hypothesis J is finitely generated:J “ pc1, . . . , csq. Since J is generated by `pIq we may assume that each generatoris the leading coefficient of a polynomial in I, i.e. for each 1 ď i ď s there existsfi P I such that `pfiq “ ci. Let

d :“ max1ďiďs

tdeg fiu .

LetH :“ IXtf P Rrts | deg f ď du. ThenH is a submodule of tf P Rrts | deg f ď du »Rd`1 (as R-modules). Since R is Noetherian every submodule of Rd`1 is finitelygenerated (argue by induction on d; if d “ 0 it holds by definition of Noetherianring, if d ą 0 consider the projection Rd`1 Ñ R) and hence

H “ pg1, . . . , gtq.

Let us prove thatI “ pf1, . . . , fs, g1, . . . , gtq.

In fact let f P I. If deg f ď d then f P H and hence f P pg1, . . . , gtq Ăpf1, . . . , fs, g1, . . . , gtq. Now suppose that deg f ą d. Then `pfq “

řsi“1 aici.

Let

h :“ f ´sÿ

i“1

aitdeg f´deg fifi.

Then deg h ă deg f . Sinceřsi“1 ait

deg f´deg fifi P pf1, . . . , fs, g1, . . . , gtq it suf-fices to prove that h P I. If deg h ď d we are done, otherwise we iterate untilwe get down to a polynomial of degree less or equal to d.

A.2. THE NULLSTELLENSATZ 121

Theorem A.1.6 (Hilbert’s basis Theorem). Every ideal of Krx1, . . . , xns isfinitely generated.

Proof. By induction on n. If n “ 0, the ring is a field, and hence is Noetherian.The inductive step follows from Theorem A.1.5, because Krx1, . . . , xns –Krx1, . . . , xn´1srts.

A.2 The Nullstellensatz

If Y Ă An is a subset, we let IpY q :“ tf P Krz1, . . . , zns | f |Y “ 0u. We recallthat the radical of an ideal I ina ring R, is the set of elements a P R such thatam P I for some m P N. As is easily checked, the radical is an ideal; it is denotedby?I,

Theorem A.2.1 (Hilbert’s Nullstellensatz). Let I Ă Krz1, . . . , zns be an ideal.Then IpV pIqq “

?I.

Before giving the proof of the Nullstellensatz, we introduce some notation,and prove auxiliary results. For pa1, . . . , anq P An, let

ma :“ pz1 ´ a1, . . . , zn ´ anq “ tf P Krz1, . . . , zns | fpa1, . . . , anq “ 0u . (A.2.1)

Notice that ma is the kernel of the surjective homomorphism

Krz1, . . . , znsφÝÑ K

f ÞÑ fpa1, . . . , anq,

and hence is a maximal ideal.

Proposition A.2.2. An ideal m Ă Krz1, . . . , zns is maximal if and only if thereexists pa1, . . . , anq P An such that m “ ma.

Proof. We know that ma is maximal. Now suppose that m Ă Krz1, . . . , zns is amaximal ideal. Let

Krz1, . . . , znsφÝÑ Krz1, . . . , znsm “: K

be the quotient map. Notice that m X K “ t0u because m “ p1q. Thus φpKqis a copy of K and hence K is a field extension of K. Let 1 ď i ď n andzi :“ φpziq.We claim that

there exists ai P K (meaning ai P φpKq) such that zi “ ai. (A.2.2)

In fact suppose that zi R K. Let c P K; since zi “ c and K is a field pzi ćq´1 exists. The field K is a quotient of Krz1, . . . , zns - a K-vector space ofcountable dimension - thus K as vector space over K has a countable basis.Since K is uncountable we get that tpxićq

´1ucPK is a set of linearly dependent


elements. Thus there exist pairwise distinct complex numbers c1, . . . , cs P Kand λ1, . . . , λs P K˚ (each λh is non-zero) such that

sÿ

h“1

λhpzi ´ chq´1 “ 0. (A.2.3)

Multiplying both sides byśsj“1pzi ´ cjq we get that

sÿ

h“1

λh

sź

j “h

pzi ´ cjq “ 0. (A.2.4)

The polynomial ϕ P Krts defined by

ϕ :“sÿ

h“1

λh

sź

j “h

pt´ cjq

is non-zero. In fact ϕpc1q “ λ1

śs1ăjďspc1ćjq “ 0. Since ϕpziq “ 0 (see (A.2.4))

and ϕ “ 0 we get that zi is algebraic over K, and since K is algebraically closedit follows that zi P K, that is a contradiction. We have proved that (A.2.2)holds. Thus

pzi ´ aiq P kerφ “ m, i “ 1, . . . , n.

Since ma is generated by pz1 ´ a1q, . . . , pzn ´ anq it follows that ma Ă m. Theideal ma is maximal and so is m: this implies that m “ ma.

Corollary A.2.3 (Weak Nullstellensatz). Let I Ă Krz1, . . . , zns be an ideal.Then V pIq “ H if and only if I “ p1q.

Proof. If I “ p1q, then V pIq “ H. Assume that V pIq “ H. Suppose thatI ‰ p1q. Then there exists a maximal ideal m Ă Krz1, . . . , zns containing I.Since I Ă m, V pIq Ą V pmq. By Proposition A.2.2 there exists a P Kn suchthat m “ ma and hence V pmq “ V pmaq “ tpa1, . . . , anqu. Thus a P V pIq andhence V pIq ‰ H. This is a contradiction, and hence I “ p1q.

Proof of Hilbert’s Nullsetellensatz (Rabinowitz’s trick). Let f P IpV pIqq. ByHilbert’s basis theorem I “ pg1, . . . , gsq for g1, . . . , gs P Krz1, . . . , zns. LetJ Ă Krz1, . . . , zn, ws be the ideal

J :“ pg1, . . . , gs, f ¨ w ´ 1q.

Since f P IpV pIqq we have V pJq “ H and hence by the Weak NullstellensatzJ “ p1q. Thus there exist h1, . . . , hs, h P Krx1, . . . , xn, ys such that

sÿ

i“1

higi ` h pf ¨ w ´ 1q “ 1.

Replacing w by 1fpzq in the above equality we get

sÿ

i“1

hi

ˆ

z,1

fpzq

˙

gipzq “ 1. (A.2.5)

A.3. UNIQUE FACTORIZATION 123

Let d ąą 0: multiplying both sides of (A.2.5) by fd we get that

sÿ

i“1

hi pzq gipzq “ fdpzq, hi P Krz1, . . . , zns.

Thus f P?I.

What if instead of Krz1, . . . , zns we consider an ideal of krx1, . . . , xns wherek is an arbitrary field ? The analogue of Theorem A.2.1 is not true. Infact take k “ R and consider the ideal px2 ` 1q Ă Rrxs: then V pIq “ H butI “ p1q. If the field is algebraically closed then the analogue of Theorem A.2.1holds (the proof that we gave above goes through with no change if the field isuncountable.) There is a modified version of Proposition A.2.2 which holdsfor an arbitrary field: it states that if m Ă krx1, . . . , xns is a maximal ideal thenkrx1, . . . , xnsm is an algebraic extension of k, see Cor. 5.24 or Prop. 7.9 of [?].

A.3 Unique factorization

Theorem A.3.1. Let R be a UFD. Then Rrts is a UFD. Moreover a polynomialp “ a0t

d ` a1td´1 ` . . .` ad is prime if and only if

1. p is prime when viewed as element of Krts, where K is the field of fractionsof R,

2. and the greatest common divisor of a0, a1, . . . , ad is 1.

Corollary A.3.2. The ring Krx1, . . . , xns is a unique factorization domain.

Proof. By induction on n. If n “ 0, the ring is a field, and hence it istrivially a UFD. The inductive step follows from Theorem A.3.1, becauseKrx1, . . . , xns – Krx1, . . . , xn´1srts.

A.4 Extensions of fields

Let F Ă E be an extension of fields. Elements α1, . . . , αn P E are algebraic-ally dependent over F is there exists a non zero polynomial f P F rz1, . . . , znssuch that fpα1, . . . , αnq “ 0 (strictly speaking, we should say that the settα1, . . . , αnu is algebraically dependent over F ). A collection tαiuiPI of ele-ments of E is algebraically independent over F if there does not exist a nonempty finite ti1, . . . , inu Ă I such that αi1 , . . . , αin are algebraically dependent(with the usual abuse of language, we also say that the αi’s are algebaricallyindependent). A transcendence basis of E over F is a maximal set of algebra-ically independent elements of E over F . There always exists a transcendencebasis, by Zorn’s Lemma. One proves that any two transcendence bases have thesame cardinality, which is by definition the transcendence degree of E over F ;we denote it by Tr.degF pEq. An extension F Ă E is algebraic if the transcend-ence degree is 0. Every finitely generated extension F Ă E can be obtained


as a composition of extensions F Ă K and K Ă E, where F Ă K is a purelytranscendental extension, i.e. there exists a transcendence basis tα1, . . . , αnu ofK over F such that K “ F pα1, . . . , αnq (thus F pα1, . . . , αnq is isomorphic tothe filed of rational functions in n indeterminates with coefficients in F ), andF Ă K is a finitely generated algebraic extension.

Theorem A.4.1. Let F Ă E be a finite extension of fields, i.e. the dimensionof E as F -vector space is finite. Suppose that F is of characteristic 0. Thenthere exists a primitive element of E over F , i.e. α P E such that E “ F pαq.

A.5 Derivations

Let R be a ring (commutative with unit), and let M be an R-module.

Definition A.5.1. A derivation from R to M is a map D : R Ñ M such thatadditivitity and Leibinitz’ rule hold, i.e. for all a, b P R,

Dpa` bq “ Dpaq `Dpbq, Dpabq “ bDpaq ` aDpbq.

If k is a field and R is a k-algebra a k-derivation (or derivation over k) D : RÑM is a derivation such that Dpcq “ 0 for all c P k. We let DerpR,Mq be the setof derivations from R to M . If R is a k-algebra we let DerkpR,Mq Ă DerpR,Mqbe the subset of k-derivations.

Example A.5.2. Let k be a field, and let f “ř

I aIzI be a polynomial in

krz1, . . . , zns, where the summation is over multiindices I, aI P K for every I,and aI is almost always zero. The formal derivative of f with respect to zm isdefined by the familar formula

Bf

Bzm“

ÿ

I s.t. im ą 0

ihaIzi11 ¨ . . . ¨ z

im´1

m´1 ¨ zim´1m ¨ z

im`1

m`1 ¨ . . . zinn . (A.5.1)

The map

krz1, . . . , znsBBzmÝÑ krz1, . . . , zns

f ÞÑBfBzm

(A.5.2)

is a k-derivation of the k algebra to istelf. We claim that Derkpkrz1, . . . , zns, krz1, . . . , znsqis freely generated (as krz1, . . . , zns module) by B

Bz1, . . . , B

Bzn. In fact there is no

relation between BBz1, . . . , B

Bznbecause

BzjBzm

“ δjm, and moreover, given a k de-rivation

D : krz1, . . . , zns Ñ krz1, . . . , zns

we have D “řnm“1 αm

BBzm

, where αm :“ Dpzmq.

Example A.5.3. Let D : RÑM be a derivation.

1. By Leibniz we have Dp1q “ Dp1 ¨ 1q “ Dp1q `Dp1q and hence Dp1q “ 0.

A.6. ORDER OF VANISHING 125

2. Suppose that g P R is invertible. Then

0 “ Dp1q “ Dpg ¨ g´1q “ g´1Dg ` fDpg´1q (A.5.3)

and hence Dpg´1q “ ´g´2Dpfq.

3. Suppose that f, g P R and that g is invertible. By Item (2) we get thatthe following familiar formula holds:

Dpf ¨ g´1q “ g´2pDpfq ¨ g ´ f ¨Dpgqq. (A.5.4)

Let D,D1 P DerpR,Mq and z P R we let

RD`D1ÝÑ M

a ÞÑ Dpaq `D1paq(A.5.5)

andR

zDÝÑ M

a ÞÑ zDpaq(A.5.6)

Both D ` D1 and zD are derivations and with these operations DerpR,Mqis an R-module. If R is a k-algebra then DerkpR,Mq is an R-submodule ofDerpR,Mq.

A.6 Order of vanishing

The prototype of a Noetherian local ring pR,mq is the ring OX,x of germs ofregular functions of a quasi projective variety X at a point x P X, with maximalideal mx, see Proposition 4.1.4. The following result of Krull can be inter-preted as stating that a non zero element of OX,x can not vanish to arbitraryhigh order at x. In other words, elements of OX,x behave like analytic functions(as opposed to C8 functions).

Theorem A.6.1 (Krull). Let pR,mq be a Noetherian local ring. Thenč

iě0

mi “ t0u.

Proof. SinceR is Noetherian the ideal m is finitely generated; say m “ pa1, . . . , anq.Let b P

Ş

iě0 mi. Let i ě 0; since b P mi there exists Pi P RrX1, . . . , Xnsi

such that Pipa1, . . . , anq “ b. Let J Ă RrX1, . . . , Xns be the ideal gener-ated by the Pi’s. Since R is Noetherian so is RrX1, . . . , Xns. Thus J is fi-nitely generated and hence there exists N ą 0 such that J “ pP0, . . . , PN q.Thus there exists QN`1í P RrX1, . . . , XnsN`1í for i “ 0, . . . , N such that

PN`1 “řNi“0QN`1íPi. It follows that

b “ PN`1pa1, . . . , anq “Nÿ

i“0

QN`1ípa1, . . . , anqPipa1, . . . , anq “ bNÿ

i“0

QN`1ípa1, . . . , anq.

(A.6.7)


NowQN`1ípa1, . . . , anq P m for i “ 0, . . . , N and hence ε :“řNi“0QN`1ípa1, . . . , anq P

m. Equality (A.6.7) gives that p1 ´ εqb “ 0: since ε P m the element p1 ´ εq isinvertible and hence b “ 0.

Corollary A.6.2. Let pR,mq be a Noetherian local ring, and let I Ă R be anideal. Then

č

iě0

pI`miq “ t0u.

Proof. Let S :“ RI. Then S is a Noetherian local ring, with maximal idealmS :“ I`m. The corollary follows by applying Theorem A.6.1 to pS,mSq.

Bibliography

[Ahl78] Lars V. Ahlfors, Complex analysis, third ed., McGraw-Hill Book Co.,New York, 1978, An introduction to the theory of analytic functions ofone complex variable, International Series in Pure and Applied Math-ematics. MR 510197

[dJc] A. J. de Jong & c., The stacks project.

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