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CAMBRIDGE STUDIES IN ADVANCED MATHEMATICS 92
Editorial Board
B. Bollob´as, W. Fulton, A. Katok, F. Kirwan, P. Sarnak, B. Simon
RATIONAL AND NEARLY
RATIONAL VARIETIES
The most basic algebraic varieties are the projective spaces, and rational vari-
eties are their closest relatives. In many applications where algebraic varieties
appear in mathematics and the sciences, we see rational ones emerging as the
most interesting examples. The authors have given an elementary treatment
of rationality questions using a mix of classical and modern methods. Arising
from a summer school course taught by J´anos Koll´ar, this book develops the
modern theory of rational and nearly rational varieties at a level that will par-
ticularly suit graduate students. There are numerous examples and exercises,
all of which are accompanied by fully worked out solutions, that will make this
book ideal as the basis of a graduate course. It will act as a valuable reference
for researchers whilst helping graduate students to reach the point where they
can begin to tackle contemporary research problems.

CAMBRIDGE STUDIES IN ADVANCED MATHEMATICS
All the titles listed below can be obtained from good booksellers or from Cambridge
University Press. For a complete series listing visit
http://publishing.cambridge.org/stm/mathematics/csam
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24 H. KunitaStochastic ﬂows and stochastic differential equations
25 P. WojtaszczykBanach spaces for analysts
26 J.E. Gilbert & M.A.M. MurrayClifford algebras and Dirac operators in harmonic analysis
27 A. Fr¨ohlich & M.J. TaylorAlgebraic number theory
28 K. Goebel & W.A. KirkTopics in metric ﬁxed point theory
29 J.F. HumphreysReﬂection groups and Coxeter groups
30 D.J. BensonRepresentations and cohomology I
31 D.J. BensonRepresentations and cohomology II
32 C. Allday & V. PuppeCohomological methods in transformation groups
33 C. Soul´eetal.Lectures on Arakelov geometry
34 A. Ambrosetti & G. ProdiA primer of nonlinear analysis
35 J. Palis & F. TakensHyperbolicity, stability and chaos at homoclinic bifurcations
37 Y. MeyerWavelets and operators I
38 C. WeibelAn introduction to homological algebra
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42 E.B. DaviesSpectral theory and differential operators
43 J. Diestel, H. Jarchow, & A. TongeAbsolutely summing operators
44 P. MattilaGeometry of sets and measures in Euclidean spaces
45 R. PinskyPositive harmonic functions and diffusion
46 G. TenenbaumIntroduction to analytic and probabilistic number theory
47 C. PeskineAn algebraic introduction to complex projective geometry
48 Y. Meyer & R. CoifmanWavelets
49 R. StanleyEnumerative combinatorics I
50 I. PorteousClifford algebras and the classical groups
51 M. AudinSpinning tops
52 V. JurdjevicGeometric control theory
53 H. VolkleinGroups as Galois groups
54 J. Le PotierLectures on vector bundles
55 D. BumpAutomorphic forms and representations
56 G. LaumonCohomology of Drinfeld modular varieties II
57 D.M. Clark & B.A. DaveyNatural dualities for the working algebraist
58 J. McClearyA user’s guide to spectral sequences II
59 P. TaylorPractical foundations of mathematics
60 M.P. Brodmann & R.Y. SharpLocal cohomology
61 J.D. Dixon et al.Analytic pro-p groups
62 R. StanleyEnumerative combinatorics II
63 R.M. DudleyUniform central limit theorems
64 J. Jost & X. Li-JostCalculus of variations
65 A.J. Berrick & M.E. KeatingAn introduction to rings and modules
66 S. MorosawaHolomorphic dynamics
67 A.J. Berrick & M.E. KeatingCategories and modules with K-theory in view
68 K. SatoLevy processes and inﬁnitely divisible distributions
69 H. HidaModular forms and Galois cohomology
70 R. Iorio & V. IorioFourier analysis and partial differential equations
71 R. BleiAnalysis in integer and fractional dimensions
72 F. Borceaux & G. JanelidzeGalois theories
73 B. Bollob´asRandom graphs
74 R.M. DudleyReal analysis and probability
75 T. Sheil-SmallComplex polynomials
76 C. VoisinHodge theory and complex algebraic geometry, I
77 C. VoisinHodge theory and complex algebraic geometry, II
78 V. PaulsenCompletely bounded maps and operator algebras
79 F. Gesztesy & H.Holden Soliton Equations and Their Algebro-Geometric Solutions, I
81 S. MukaiAn Introduction to Invariants and Moduli
82 G. TourlakisLectures in Logic and Set Theory, I
83 G. TourlakisLectures in Logic and Set Theory, II
84 R. A. BaileyAssociation Schemes
85 J. Carlson, S. M¨uller-Stach & C. PetersPeriod Mappings and Period Domains
89 M. Golumbic & A. TrenkTolerance Graphs
90 L. HarperGlobal Methods for Combinatorial Isoperimetric Problems
91 I. Moerdijk & J. MrcunIntroduction to Foliations and Lie Groupoids

RationalandNearlyRational
Varieties
J´ANOS KOLL ´AR, KAREN E. SMITH,
AND
ALESSIO CORTI

published by the press syndicate of the university of cambridge
The Pitt Building, Trumpington Street, Cambridge, United Kingdom
cambridge university press
The Edinburgh Building, Cambridge CB2 2RU, UK
40 West 20th Street, New York, NY 10011–4211, USA
477 Williamstown Road, Port Melbourne, VIC 3207, Australia
Ruiz de Alarc´on 13, 28014 Madrid, Spain
Dock House, The Waterfront, Cape Town 8001, South Africa
http://www.cambridge.org
CεCambridge University Press 2004
This book is in copyright. Subject to statutory exception
and to the provisions of relevant collective licensing agreements,
no reproduction of any part may take place without
the written permission of Cambridge University Press.
First published 2004
Printed in the United Kingdom at the University Press, Cambridge
TypefaceTimes 10/13 pt.SystemL
ATEX2ε[tb]
A catalog record for this book is available from the British Library
ISBN 0 521 83207 1 hardback

Contents
Introduction 1
Description of the chapters 2
Prerequisites 4
Notation and basic conventions 5
1 Examples of rational varieties 7
Rational and unirational varieties 7
Rational curves 9
Quadric hypersurfaces 13
Quadrics over ﬁnite ﬁelds 15
Cubic hypersurfaces 20
Further examples of rational varieties 29
Numerical criteria for nonrationality 31
2 Cubic surfaces 35
The Segre–Manin theorem for cubic surfaces 36
Linear systems on surfaces 37
The proofs of the theorems of Segre and Manin 42
Computing the Picard number of cubic surfaces 47
Birational self-maps of the plane 49
3 Rational surfaces 60
Castelnuovo’s rationality criterion 61
Minimal models of surfaces 63
Rational surfaces over perfect ﬁelds 68
Field of deﬁnition of a subvariety 76
Del Pezzo surfaces 81
v

vi Contents
4 Nonrationality via reduction modulop 93
Nonrational cyclic covers 94
Construction of cyclic covers 97
Differential forms in characteristicp 101
Reduction to characteristicp 108
Matsusaka’s theorem and Abhyankar’s lemma 111
Relative Canonical modules and Jacobian ideals 114
Explicit examples (by J. Rosenberg) 118
5 The Noether–Fano method for proving nonrationality 122
The Noether–Fano method 123
Numerical consequences of maximal centers 129
Birationally rigid Fano varieties 134
Quartic threefolds 144
6 Singularities of pairs 149
Discrepancies 150
Canonical and log canonical pairs 155
Computing discrepancies 157
Inversion of adjunction 160
The log canonical threshold of a plane curve singularity 164
Zero–dimensional maximal centers on threefolds 175
Appendix: proof of the connectedness theorem 177
7 Solutions to exercises 182
Exercises in Chapter 1 182
Exercises in Chapter 2 194
Exercises in Chapter 3 200
Exercises in Chapter 4 211
Exercises in Chapter 5 215
Exercises in Chapter 6 217
References 228
Index 232

Introduction
The most basic algebraic varieties are the projective spaces, and rational vari-
eties are their closest relatives. Rational varieties are those that are birationally
equivalent to projective space. In many applications where algebraic varieties
appear in mathematics, we see rational ones emerging as the most interesting
examples. This happens in such diverse ﬁelds as the study of Lie groups and their
representations, in the theory of Diophantine equations, and in computer-aided
geometric design.
This book provides an introduction to the fascinating topic of rational, and
“nearly rational,” varieties. The subject has two very different aspects, and we
treat them both. On the one hand, the internal geometry of rational and nearly
rational varieties tends to be very rich. Their study is full of intricate construc-
tions and surprising coincidences, many of which were thoroughly explored by
the classical masters of the subject. On the other hand, to show that particular
varieties arenotrational can be a difﬁcult problem: the classical literature is
riddled with serious errors and gaps that require sophisticated general methods
to repair. Indeed, only recently, with the advent of minimal model theory, have
all the difﬁculties in classical approaches to proving nonrationality based on
the study of linear systems and their singularities been ironed out.
While presenting some of the beautiful classical discoveries about the ge-
ometry of rational varieties, we pay careful attention to arithmetic issues. For
example, we consider whether a variety deﬁned over the rational numbers is
rationaloverQ, which is to say, whether there is a birational map to projective
space given locally by polynomials with coefﬁcients inQ.
The hardest parts of the book focus on how to establish nonrationality of
varieties, a difﬁcult problem with many basic questions remaining open today.
There are good general criteria, involving global differential forms, that can be
used in many cases, but the situation becomes very difﬁcult when these tests fail.
Forexample, using simple numerical invariants called the plurigenera, it is easy
1

2 Introduction
to see that a smooth hypersurface in projective space whose degree exceeds its
embedding dimension can not be rational. However, it is a very delicate problem
to determine whether or not a lower degree hypersurface is rational.
Rationality of quadric and cubicsurfaceswascompletely settled in the nine-
teenth century, but rationality for threefolds occupied the attention of algebraic
geometers for most of the twentieth century. In the 1970s, Clemens and Grifﬁth
identiﬁed a new obstruction to rationality for a threefold inside its third topo-
logical (singular intermediate Jacobians
provided the ﬁrst proof that no smooth cubic threefold is rational. Because this
approach ﬁts better in a book about Hodge theory, we do not discuss it here.
Instead, we prove that no smooth quartic threefold in projective four-space is
rational, drawing on ideas from the minimal model program. Beyond this, very
little is known: no one knows whether or not all smooth cubic fourfolds are ratio-
nal, or indeed, whether there exists any nonrational smooth cubic hypersurface
of any dimension greater than three.
On the other hand, in this book we do present a technique for proving non-
rationality of “very general” hypersurfaces of any dimension greater than two
whose degree is close to their dimension. Like other approaches to proving
nonrationality, this technique uses differential forms; the novelty here is that
the differential forms we use are deﬁned on varieties of prime characteristic.
Our biggest omission is perhaps never to deﬁne precisely what we mean by
a “nearly rational variety.” Current research in birational algebraic geometry
indicates that the most natural class of nearly rational varieties is formed by
rationally connected varieties, introduced in Koll´aret al. (1992
is easy to state the deﬁnition, it is harder to appreciate why we claim that this
is indeed the most natural class of nearly rational varieties to consider. Our aim
in this book is more modest; we hope to inspire the reader to learn more about
rationality questions. As a next step, we recommend the general introduction
to rationally connected varieties given in Koll´ar (2001
a detailed treatment for the technically advanced.
Description of the chapters
Chapter 1 describes some basic examples of rational varieties, concentrating on
quadric and cubic hypersurfaces. We give fairly complete answers for quadric
hypersurfaces, but many open questions remain about cubics. We also discuss
the simplest nonrationality criteria in terms of differential forms.
Cubic surfaces are examined in detail in Chapter 2. This is a classical topic
that began with the works of Schl¨aﬂi and Clebsch and culminated with the

Description of the chapters 3
arithmetic studies of Segre and Manin. The main results here are about smooth
cubic surfaces of Picard number one: no such cubic surface is rational. This is
essentially an arithmetic result, since cubic surfaces over an algebraically closed
ﬁeld never have Picard number one and are always rational. On the other hand,
the techniques are quite geometric, and show many of the higher dimensional
methods in simpler form.
A general study of rational surfaces is given in Chapter 3. For instance,
we prove Castelnuovo’s criterion for rationality, giving a simple numerical
characterization of rationality for smooth complex surfaces. Although this result
is classical, we develop it within the modern framework of the minimal model
program. This allows us to also treat surfaces that are not deﬁned over an
algebraically closed ﬁeld.
In Chapter 4, we construct examples of higher dimensional smooth nonra-
tional hypersurfaces of low degree. The constructed varieties are all Fano, which
means in particular that the naive numerical invariants introduced in Chapter 1
all vanish here even though the varieties are not rational. Our proof is based on
the method of reduction to prime characteristic, where we are able to exploit
some of the quirks of differential forms arising from the peculiarity that the
derivative of apth power is zero in characteristicp. These positive character-
istic varieties are then lifted to get examples overC. While this method yields
many examples of smooth nonrational varieties, it is not capable of producing
complete families such that every smooth member is nonrational.
Chapter 5 develops the Noether–Fano method, a technique for proving non-
rationality of higher dimensional varieties, analogous to the ideas presented in
Chapter 2 to treat cubic surfaces. Using this approach, we produce complete
families of Fano varieties in which no smooth member is rational. This example,
presented in Section 5.3, is by far the simplest higher dimensional application
of the Noether–Fano method. We also start the proof that no smooth quartic
threefold in projective four-space is rational. This fact about quartic threefolds
wasﬁrst claimed by Fano (1915
with the work of Iskovskih and Manin (1971
In Chapter 6, we present more advanced machinery, namely the theory of sin-
gularities of pairs, for carrying out the general method developed in Chapter 5.
Our main application is the proof of a particular numerical result which is a key
ingredient in the proof that no quartic threefold is rational. These techniques
also have numerous applications to diverse problems of higher dimensional
geometry.
Chapter 7 contains the solutions of the exercises. The reader is strongly
urged to try to work them out ﬁrst instead of going to the solutions straight
away.

4 Introduction
This book began with a series of lectures by J. Koll´ar given at the European
Mathematical Society Summer School in Algebraic Geometry in Eger, Hungary
in August 1996. The notes were written up by K. E. Smith. Later new chapters
were added and the old ones have been revised and reorganized. Section 4.7
(by J. Rosenberg) answers a problem raised in the original lectures.
Wethank C. Araujo, A.-M. Castravet, J. Keum, S. Kov´acs, R. Lazarsfeld,
J. Parsons, P. Vojta, and C. Werner, for many comments and corrections, N. Katz
and P. Swinnerton-Dyer for some nice examples.
Prerequisites
Wehave devoted considerable effort to making our exposition as elementary as
possible. Chapters 1 and 2 should be accessible to students who have completed
a year long introductory course on classical algebraic geometry, for instance
along the lines of Shafarevich (1994, vol.1). In particular, we use the language
of linear systems of curves on surfaces, including their intersection theory, but
we do not use cohomology.
In Chapter 3, we use basic facts about intersection theory on surfaces and
cohomology for line bundles on curves and surfaces, including the Riemann–
Roch theorem, Serre duality, the adjunction formula, and the Kodaira vanishing
theorem. We use the most rudimentary aspects of the theory of schemes of
ﬁnite type over a ﬁeld in our discussion of the ﬁeld of deﬁnition of a variety.
Hodge theory is also mentioned in a peripheral way. Reid’s lectures (1997
an excellent and concise summary of much of the material needed in Chapter 3
and later in the book. Students familiar with Sections IV and V of Hartshorne’s
book (1977
In Chapter 4, we work with schemes over SpecZand their K¨ahler differen-
tials, but we carefully explain all that is used beyond the most basic deﬁnitions.
Wehope that this chapter will help those familiar with classical algebraic ge-
ometry to appreciate the theory of schemes.
Chapters 5 and especially 6 are somewhat harder. We assume more sophisti-
cation in manipulatingQ-divisors, and use two major theorems that the reader
is asked to accept without proof, namely the Lefschetz theorem on the Picard
group of hypersurfaces and Hironaka’s results on the resolution of singulari-
ties. One technically more demanding proof (of Theorem 6.32) is relegated to
an Appendix. Chapter 6 may be the hardest, mainly because of the number of
new concepts involved. It serves as a good introduction to some more advanced
books on birational geometry or the minimal model program, for instance to
Koll´ar and Mori (1998

Notation and basic conventions 5
Notation and basic conventions
Letkbe a ﬁeld. Our main interest is in the casesk=Cfor studying geometric
properties andk=Qfor investigating the arithmetical questions. We occasion-
ally encounter other cases too, for instance the ﬁnite ﬁeldsF
q, the real numbers
R,p-adic ﬁeldsQ
pand algebraic number ﬁelds. The algebraic closure ofkis
denoted¯k.
The notation for the ground ﬁeld is suppressed when the ﬁeld is clear from
the context or irrelevant, but occasionally we writeX
kto emphasize that the
varietyXis deﬁned over the ground ﬁeldk.IfL⊃kis a ﬁeld extension
thenX
Ldenotes the varietyX kviewed as being overL.Technically speaking,
X
L=Xk×SpeckSpecL.
IfLis any ﬁeld containingk, then anL-point,oran L-rational point,is
one having all of its coordinates deﬁned overL. That is, thinking of a variety
overkas locally a subvariety ofA
n
given by the vanishing of polynomials with
coefﬁcients ink, then anL-rational point is given by ann-tuple of elements ofL
satisfying the deﬁning polynomials. Thinking more scheme-theoretically, anL-
point on a schemeXcan be deﬁned as a morphism SpecL→X.Inparticular,
ak-rational point on ak-scheme corresponds to a maximal ideal whose residue
ﬁeld isk. The symbolX(L) denotes the set ofL-points ofX.
Morphismsandrational mapsbetween varieties are always assumed to be
deﬁned over the ground ﬁeld, except where explicitly stated otherwise. Like-
wise, linear systems on a varietyX
kare assumed deﬁned overk.
Morphisms are denoted by solid arrows→and rational maps by dashed
arrows⊃⊃→. The “image” of a rational map is the closure of the image of the
morphism obtained by restricting the rational map to some nonempty open set
where it is deﬁned; in the same way, we deﬁne the image of a subvariety under a
rational map, provided that the map is deﬁned at its generic point. In particular,
letf:Y⊃⊃→Xbe a rational map and suppose thatfis deﬁned at the generic
point of some subvarietyZofY. Then the image ofZonX, denotedf
∗(Z), is
the closure inXof the setf|
X0
(Z∩X 0), whereX 0is some open set meeting
Zon whichfis a well-deﬁned morphism. In the case of birational maps, the
image is also called thebirational transform, especially in the case where this
image has the same dimension.
LetXbe a normal variety. An irreducible and reduced subscheme of codi-
mension one is called aprime divisor.Adivisor onXis a formal linear com-
binationD=
⊃
d
iDiof prime divisors whered i∈Z.Inusing this notation
we assume that theD
iare distinct. AQ-divisoris a formal linear combination
D=
⊃
d
iDiof prime divisors whered i∈Q. The divisorDis calledeffective
ifd
i≥0 for everyi.Adivisor (orQ-divisor)Dis calledQ-CartierifmDis

6 Introduction
Cartier for some nonzero integerm, where by Cartier we mean that it is locally
deﬁned by a single equation. On a smooth variety every divisor is Cartier. The
supportofD=
⊃
d
iDi, denoted by SuppDis the subscheme∪ di=0Di.Linear
equivalenceof two divisors is denoted byD
1∼D2.
A property of a varietyX
krefers to the variety considered overk.Weadd
the adjectivegeometricallywhen talking about a property ofX
¯k.For example,
the afﬁne plane curve deﬁned by the equationx
2
+y
2
=0isirreducible as
aQ-variety but it is geometrically reducible. For many properties (including
smoothness or projectivity), the distinction does not matter.
Varieties are assumed reduced and irreducible, except where explicitly stated
otherwise. In particular, the terms “smooth curve” and “smooth surface” always
refer toconnectedsmooth surfaces and curves. The one exception is that we
use the term “curve on a surface” to mean any effective divisor, which may or
may not be reduced and irreducible. Because we are concerned with birational
properties, there is no loss of generality in assuming all varieties to be quasi-
projective. In any case, our main interest is in smooth projective varieties.
In writing these notes, our policy was not to be sidetracked by anomalies in
positive characteristic. These usually appear when the base ﬁeld is not perfect,
that is, when it has algebraic extensions obtained by takingpth roots (in charac-
teristicp). Technical problems related to such issues are relegated to exercises
and they can be safely ignored for most of the book.
The exception is Chapter 4 where the unusual properties of such ﬁeld exten-
sions are exploited to prove several results about varieties overCorQ.

1
Examples of rational varieties
In this chapter, we introduce rational varieties through examples. After giv-
ing the fundamental deﬁnitions in the ﬁrst section and settling the rationality
question for curves in Section 2, we continue with the rich theory of quadric
hypersurfaces in Section 3. This is essentially a special case of the theory of
quadratic forms, though the questions tend to be strikingly different.
Quadrics over ﬁnite ﬁelds are discussed in Section 4. Several far-reaching
methods of algebraic geometry appear here in their simplest form.
Cubic hypersurfaces are much more subtle. In Section 5, we discuss only the
most basic rationality and unirationality facts for cubics. A further smattering
of rational varieties is presented in Section 6, together with a more detailed look
at determinantal representations for cubic surfaces.
Avery general and useful nonrationality criterion, using differential forms,
is discussed in Section 7.
1.1 Rational and unirational varieties
Roughly speaking, a variety isunirationalif a dense open subset is parametrized
by projective space, andrationalif such a parametrization is one-to-one.
Tobe precise, ﬁx a ground ﬁeldk, and letXbe a variety deﬁned overk.It
is important to bear in mind thatkneed not be algebraically closed and that all
constructions involving the varietyXare carried out over the ground ﬁeldk.
Deﬁnition1.1.Avariety isrationalif it is birationally equivalent to pro-
jective space. Explicitly, the varietyXis rational if there exists a birational map
P
n
X.
Deﬁnition1.2.The varietyXisunirationalif there exists a generically
ﬁnite dominant mapP
n
X.
7

8 1 Examples of rational varieties
Rational varieties were once called “birational,” in reference to the rational
maps between them and projective space in each direction. “Unirationality”
thus refers to the map fromP
n
to the variety, deﬁned in one direction only.
This explains the odd use of the preﬁx “uni” in referring to a map which is
ﬁnite-to-one.
Weemphasize that in both deﬁnitions above, the varieties and the maps are
deﬁned over the ﬁxed ground ﬁeldk. This means that the varietyXis deﬁned
locally by polynomials with coefﬁcients ink, and also that the map can be
described by polynomials with coefﬁcients ink.
Our guiding question throughout this book is the following:Which varieties
are rational or unirational?
The rationality or unirationality of a variety may depend subtly on the ﬁeld
of deﬁnition. For example, a variety deﬁned overQmay be considered as
avariety deﬁned overR.Itispossible that there is a birational map given by
polynomials withrealcoefﬁcients from projective space to the variety, but there
is no such birational map given by polynomials withrationalcoefﬁcients. Our
ﬁrst example nicely illustrates this point.
Example1.3.Consider the plane conicCdeﬁned by the homogeneous
equationx
2
+y
2
=pz
2
,wherepis a prime number congruent to−1 modulo
4. We claim that
1. the Diophantine equationx
2
+y
2
=pz
2
has no rational solutions (aside
from the trivial solutionx=y=z=0),
2. the curveCis not rational overQ, and
3. the curveCis rational overQ(
√
p).
Indeed, assume thatx
2
+y
2
=pz
2
has a rational solution. By clearing denom-
inators, we may assume thatx,y,andzare integers, not all divisible byp.If
neitherxnoryis divisible byp, then the congruencex
2
≡−y
2
modpleads
to a solution ofu
2
≡−1 modp. But this is impossible sincep≡−1 mod 4.
(This easy fact is sometimes called Euler’s criterion for quadratic congruences;
if you have not seen it before, check by hand the examplesp=3,7,11 before
looking it up in any elementary number theory book.) This contradiction forces
pto divide bothxandy. But thenp
2
dividespz
2
,sothatpdivideszas well,
a contradiction. This establishes (1
Now, ifCis rational (or even unirational) overQ, then images of the rational
points under the mapP
1
φφπCgive plenty of rational points onC, contradicting
(1).
Finally, (3
P
1
φφπC⊂P
2

1.2 Rational curves 9
given by
(t:1)→(t
2
−1:2t:
1
√
p
(t
2
+1)).
This is a special case of the parametrization given later in the proof of Theorem
1.11 for a general quadric hypersurface.
Wesay thatXisgeometrically rationalifXis rational over¯k. The reader
is cautioned however, that the literature is inconsistent: some authors use the
term “rational” to mean “geometrically rational.”
One must be careful about trusting intuition based on extensive study of
algebraic varieties over an algebraically closed ﬁeld. For example, even when
a mapP
n
φφπXas in Deﬁnition 1.2 is dominant, the induced map on the set
ofk-pointsP
n
(k)φφπX(k) can be very far from surjective. For instance, with
the ground ﬁeld ﬁxed to beQ, consider the mapP
1
→P
1
givenby( s:t)→
(s
2
:t
2
). The image of the rational points is a very sparse subset of the set of
all rational points of the target variety. This is typical for maps deﬁned over
algebraically non-closed ﬁelds.
1.2 Rational curves
OverC, and more generally, over any algebraically closed ﬁeld, the only smooth
projective curve remotely resembling the projective line isP
1
itself. Indeed,
as is frequently covered in a ﬁrst course in algebraic geometry, the follow-
ing are equivalent for a smooth projective curve over an algebraically closed
ﬁeld:
1. the curve is isomorphic to the projective line;
2. the curve is birationally equivalent to the projective line;
3. there is a nonconstant map from the projective line to the curve;
4. the curve has no nonzero global holomorphic (that is, K¨ahler) one-forms: in
other words, the canonical linear system is empty.
But what about curves over algebraically non-closed ﬁelds? It is still the
case that every rational map from a curve is, in fact, an everywhere-deﬁned
morphism; the usual proof of this fact does not require an algebraically closed
ground ﬁeld. So over any ground ﬁeld, a birational map from a curve is an
isomorphism, and (1
(3
not. In fact, we see that rationality questions for curves come down to the case
of plane conics, where the answers depend on the ground ﬁeld.

10 1 Examples of rational varieties
Given a smooth projective curve, how can we tell if it is rational? Of course,
if a curve is rational overk,itiscertainly rational over its algebraic closure
¯k,sowemight as well restrict our attention to geometrically rational curves.
Among all the different representations of a smooth geometrically rational curve
(for instance, as a projective line, a plane conic, a twisted cubic, and so on),
the following proposition shows that the plane conics account for all possible
birational models of the projective line over any ﬁeld.
Proposition1.4.A smooth projective geometrically rational curve is iso-
morphic to a smooth plane conic.
Again we emphasize (we will soon stop!) that the interesting part of this
statement is that all this is going on over some ﬁxed ground ﬁeldk, which need
not be algebraically closed. So any smooth curve overkthat is rational when
considered as a variety over¯kmust be isomorphic (overk)toacurve inP
2
deﬁned by a quadratic polynomial with coefﬁcients in the ground ﬁeldk. This
would be obvious ifkwere algebraically closed.
The proof uses two basic results of algebraic geometry over algebraically
nonclosed ﬁelds. Both are quite elementary but they do not always receive the
emphasis that they deserve in introductory texts.
Proposition1.5.Let X be a smooth quasi-projective variety deﬁned over
a ﬁeld k. Then it has a canonical divisor deﬁned over k. Thus we can speak of
the canonical divisor class K
Xas a linear equivalence class deﬁned over k.
Proof.Let us start with the most classical case whenkhas characteristic
zero andXis a curve.
Ifgis any function onX, the divisor ofdgis a canonical divisor. Ifgis in
k(X) then the corresponding divisor (dg)isdeﬁned overk.
Wedo something similar in higher dimensions. Chooseg
1,...,g nalge-
braically independent functions ofk(X). Then the divisor ofdg
1∧···∧dg n
is a canonical divisor deﬁned overk.
Wehave to be a little more careful in positive characteristic. The problem
is that ifgis apth power thendg=0 and its divisor (dg )isnot deﬁned. It is
not hard to show that this problem can be avoided by a careful choice of the
functionsg
i. See, for instance, van der Waerden (1991, 19.7).
Another possibility, more in keeping with modern techniques, is to construct
the sheaf of differential forms (i.e. the sheaf of K¨ahler differential one-forms)
as in Shafarevich (1994, III.5) and deﬁne the canonical class as the divisor class
corresponding to its determinant bundle.
Proposition1.6.Let D be a divisor on a smooth projective variety X
deﬁned over a ﬁeld k. Then the dimension of the complete linear system deﬁned

1.2 Rational curves 11
by D does not depend on k. That is, if K⊃kisany ﬁeld extension thendim|D|
is the same whether computed over k or K .
Thus two divisors D
1,D2, both deﬁned over a ﬁeld k, are linearly equivalent
over k if and only if they are linearly equivalent over K .
Proof.The classical argument, identifying the linear system|D|with a
projective space of functionsfsuch that (f)+D≥0isexplained in Shafare-
vich (1994, III.3.5). The last part follows by noting thatD
1andD 2are linearly
equivalent if and only if dim|D
1−D2|=dim|D 2−D1|=0.
Those who are familiar with the sheaf theoretic viewpoint should also look
at Exercise 3.34.
Proof of Proposition1.4.LetCbe a smooth geometrically rational
curve and letO
C(KC) denote its canonical line bundle, which is deﬁned overk
by Proposition 1.5.
Because the curve is isomorphic toP
1
over¯k,wehaveisomorphisms
O
C(KC)
∼
=O P
1(−2)over¯k. The global sections of the dual bundleO C(−KC)
∼
=
O
P
1(2 ¯k.ByProposition 1.6, the
space of global sections ofO
C(−KC)isalso three dimensional when consid-
ered overk. These global sections deﬁne an embedding (given by the complete
linear system|−K
C|)ofC,overk,asaconic in the projective plane.
Proposition 1.4 ensures that rationality questions for any geometrically ra-
tional curve reduce to questions about plane conics. So given a plane conic, how
can we tell whether or not it is birationally equivalent to the projective line?
Because every birational map from a smooth curve is actually an everywhere-
deﬁned morphism, we are essentially asking when a plane conic is isomorphic
toP
1
. This, in turn, is essentially equivalent to the theory of quadratic forms in
three variables. The slight difference is that in quadratic form theory one usually
does not consider the forms that differ by a scalar multiple to be equivalent,
although they determine the same conic.
Proposition1.7.The following are equivalent for any smooth projective
geometrically rational curve over a ﬁeld k:
1.the curve is isomorphic to the projective line;
2.the curve admits a k-point;
3.the curve has a point deﬁned over some odd degree ﬁeld extension of k;
4.there is an odd degree line bundle on the curve deﬁned over k.
Proof.It is clear that (1
Toshow that (3 k

-pointPon the geo-
metrically rational curveC, wherek

is an extension ofkof odd degreed.Ifk

12 1 Examples of rational varieties
is separable overk, thenPhasddistinct conjugatesP
1=P,P 2,...,P dunder
the corresponding Galois action onC. Their union{P
1,...,P d}is deﬁned over
k(as shown in Exercise 1.8 below). ThusO
C(P1+···+P d)isadegreedline
bundle deﬁned overk, proving (4
Finally, assume (4 Lbe a line bundle of degree 2r+1. AsCis
geometrically rational, the line bundleO
C(KC) has degree minus two, and so
L⊗O
C(rKC)isadegree one line bundle deﬁned overk. Its global sections
deﬁne an isomorphism fromCto the projective line.
Exercise1.8.Letk

/kbe a ﬁnite Galois extension. The Galois group
Gal(k

/k) acts onA
n
k
=Speck

[x1,...,x n] coordinatewise. LetXbe a closed
algebraic subset ofA
n
k
. Prove that the following are equivalent:
1. the setXcan be deﬁned by polynomials ink[x
1,...,x n];
2. the setXis invariant under the Gal(k

/k)-action.
(A more advanced version of this result is discussed in Section 3.4)
Exercise 1.8 indicates why we occasionally assume that our varieties are
deﬁned over a perfect ground ﬁeld in treating rationality questions. A perfect
ﬁeld is one that admits no inseparable extensions; in particular, the splitting
ﬁeld of any polynomial over a perfect ﬁeld is a Galois extension. However,
overanonperfect ground ﬁeldk(necessarily of prime characteristicp), there
are subsets ofA
n
k
that are not deﬁned overkeven though they are ﬁxed by the
group of automorphisms ofk

overk.
Exercise1.9.Letk

/kbe a purely inseparable extension of degreep
a
. Let
Cbe a smooth curve inA
n
k
and letPbe ak

-point ofC. Prove that the divisor
p
a
Pcan be deﬁned by polynomials ink[x 1,...,x n].
Finally, we point out that for curves there is no difference between rationality
and unirationality, regardless of whether or not the ground ﬁeld is algebraically
closed.
Proposition1.10 (L¨uroth theorem). A smooth projective curve is rational
if and only if there is a nonconstant mapP
1
→C.
Proof.If a curveCis rational, then the given birational equivalence can
be taken for the needed map. For the converse, it is equivalent to show that
every subﬁeld of the function ﬁeldk(t)isitself purely transcendental overk.An
elementary algebraic proof of this fact is given in van der Waerden (1991, 10.2).
Forthe more geometrically inclined reader, we show how the statement
reduces easily to the case where the curve is deﬁned over an algebraically closed

1.3 Quadric hypersurfaces 13
ground ﬁeld. Suppose that there is a nonconstant mapP
1
→C. Assuming the
statement over an algebraically closed ﬁeld, this implies thatCis geometrically
rational. On the other hand, the image of anyk-point ofP
1
produces ak-point
onC,sothe curveCis rational overkby Proposition 1.7.
1.3 Quadric hypersurfaces
In the previous section we proved a criterion, Proposition 1.7, for rationality
of plane conics. This criterion generalizes quite nicely to quadrics of arbitrary
dimension, a result due to Springer. For simplicity we assume that the charac-
teristic of our ﬁeldkis not two.
Theorem1.11. The following are equivalent for any quadric hypersurface
in projective space that is not the union of two hyperplanes:
1.the quadric is rational;
2.the quadric has a smooth k-point;
3.the quadric has a smooth k

-point for some odd degree ﬁeld extension k

of k.
Proof.The only obvious implication is that (2
That (1
is a birational mapφ:P
n−1
φφπQ, whereQdenotes the quadric hypersurface
inP
n
. Thenφis deﬁned on some Zariski open subsetU⊂P
n−1
.Ifkis inﬁnite,
thenU(k)isZariski dense and its image is a Zariski dense set ofk-points inQ.
Thus we get plenty of smoothk-points onQ. This argument breaks down when
kis ﬁnite, because then there are open sets with nok-points. Nonetheless, ifQis
smooth, we can use Nishimura’s lemma (see Exercise 1.12) below to conclude
thatQhas a smoothk-point. The singular case is reduced to the smooth case
using Exercise 1.13 below.
Conversely, letQbe a quadric inP
n
and letPbe a smoothk-point onQ.
Letπ:QφφπP
n−1
be the projection fromQto any hyperplane inP
n
deﬁned
overkbutnot containingP. Note thatπis generically one-to-one and deﬁned
overk, hence gives the desired birational equivalence. Indeed,πis one-to-one
except along the lines throughPlying onQ, andQcannot be covered by
such lines unless it is a cone with vertex atP. This proves that (1
from (2
It remains to show that (3
the singular case following from Exercise 1.13 below.
It is sufﬁcient to consider the case whenk

=k(z)isadegreed>1extension
generated by one element, because we can build upk

by successively adding

14 1 Examples of rational varieties
elements. (Note that ifk

is separable overk, then it is generated by one element
in any case.)
Pick ak

-pointPonQ⊂P
n
and write it as
(a
00+a01z+···+a 0,d−1z
d−1
:···:a n0+an1z+···+a n,d−1z
d−1
),
where thea
ijare ink. Consider the map⊂:P
1
→P
n
sending (s :t)to
(a
00s
d−1
+a01s
d−2
t+···+a 0,d−1t
d−1
:···:a n0s
d−1
+···+a n,d−1t
d−1
).
Because this map is deﬁned overk, the image is a rational curve, sayC. The
degree ofCis at mostd−1; let us denote this degree byd

.
IfCsits onQ, then anyk-point onCproduces for us the desiredk-point on
Q. Otherwise, letF(s,t)bethe pullback of the equation ofQtoP
1
under⊂.
Note thatFhas degree 2d

,sothat its dehomogenizationf(t)=F(1,t) also
has degree 2d

(unlesssdividesF,inwhich case the image of (0 : 1) under⊂
is ak-point ofQ). Becausezis a root off, its minimal polynomialgdivides
f, andf/gfactors ash
i. Since degf/g=degf−deggis odd, at least one
of the factors, sayh
1, has odd degreed

, withd

≤2d

−d≤d−2. Lett 0be
a root ofh
1. Then the image of the point (1,t 0) under⊂is a point ofQdeﬁned
overaﬁeld extensionk

/kof odd degreed

.Weare done by induction on the
degree of the extension.
Exercise1.12. Prove Nishimura’s lemma: IfYis smooth,Y

is projective,
and there is a rational mapYφφπY

, then ifYhas ak-point, so doesY

. Also,
ﬁnd a counterexample whenYis not smooth.
Exercise1.13. LetQbe a quadric hypersurface in projective space
deﬁned overk. Assume that the characteristic ofkis not two. Show that the
singular locus ofQis a linear subspace deﬁned overkand thatQis a cone over
a smooth quadricQ

(or elseQis a double plane). Prove thatQhas a smooth
k-point if and only ifQ

has a smoothk-point. Use this to ﬁnish the proof of
Theorem 1.11.
Despite Theorem 1.11, the classiﬁcation of quadrics up to birational equiv-
alence is still not complete. Indeed, it is not clear how to decide whether a
given quadric hask-points or not. The extensive theory of quadratic forms is
devoted to this question and to the classiﬁcation of quadrics up to isomor-
phism. On the other hand, there are reasonably complete answers over speciﬁc
ﬁelds. In the next section, we see that the rationality problem is solved for
quadrics over ﬁnite ﬁelds. For now, we mention some results for quadrics overR
andQ.

1.4 Quadrics over ﬁnite ﬁelds 15
Exercise1.14. Every quadric hypersurfaceQoverRis isomorphic to a
quadric deﬁned by the homogeneous form
x
2
1
+···+x
2
p
−
φ
x
2
p+1
+···+x
2
p+q
π
,
for somep≥q. The numbers (dimQ,p,q) form a complete set of invariants
for the isomorphism types of real quadric hypersurfaces. The quadric is rational
if and only ifq>0.
The following remarkable theorem says that the rationality of quadrics over
the rational numbers is essentially a real question.
Theorem1.15 (Hasse–Minkowski).The following are equivalent for a
smooth quadric hypersurface of dimension at least three, deﬁned over the ra-
tional numbersQ:
1.the quadric has aQ-point;
2.the quadric is rational overQ;
3.the quadric has anR-point;
4.the quadric is rational overR.
Theorem 1.11 here shows the equivalence of (1
(4
Serre (1973, IV.3).
1.4 Quadrics over ﬁnite ﬁelds
In this section, we completely settle the rationality problem for quadrics over a
ﬁnite ﬁeld by proving the following theorem.
Theorem1.16. A quadric hypersurface over a ﬁnite ﬁeld is rational, pro-
vided that it is not the union of two hyperplanes over the algebraic closure of
the ground ﬁeld.
According to Theorem 1.11, in order to prove this theorem, we need only
show that the quadric has a smooth point deﬁned over the given ﬁnite ground
ﬁeld. In fact, by Exercise 1.13, we might as well assume that the quadric is
smooth, so it is enough to prove the following theorem.
Theorem1.17. A positive dimensional quadric hypersurface over a ﬁnite
ﬁeld has a point over that ﬁeld.
It is interesting to note that the proof of Theorem 1.17 reduces easily to the
case of smooth conics inP
2
.Indeed, given any quadric hypersurface deﬁned

16 1 Examples of rational varieties
overkinP
n+1
,ahyperplane section is a quadric inP
n
.Ifthis section is singular,
then its singular set is a linear space deﬁned overkby Exercise 1.13, and so
has plenty ofk-rational points. If the section is smooth, then we can look for
rational points on it by induction.
Wegive two different proofs of the existence of such a rational point. The
ﬁrst, Theorem 1.18 below, is more general and elementary, and is based on
some congruence statements about the number ofk-rational points on varieties
deﬁned by low degree polynomials over ﬁnite ﬁelds. The second proof is more
special, but its ideas lead much further, eventually to the proof of the Weil
estimates for thek-rational points on any variety.
Theorem1.18 (Chevalley, 1935). Let f
1,...,f sbe homogeneous polyno-
mials in d variables over a ﬁnite ﬁeld k. If the sum of the degrees of the f
iis
less than d, then the polynomials have a nontrivial common solution in k.
Before proving Theorem 1.18, we note that it implies Theorem 1.17 simply
by considering the case of one homogeneous polynomial of degree two in at
least three variables.
Proof of Theorem1.18. Consider the afﬁne varietyXdeﬁned by the
polynomialsf
1,...,f sinA
d
. Letq=p
m
denote the cardinality ofk, and con-
sider the auxiliary polynomialP:=
⊂
i
(1−f
q−1
i
)∈k[x 1,...,x d]. Because
γ
q
=γfor allγink,wesee that
P(x)=

1ifx∈X(k), and
0ifx∈k
d
\X(k).
Thus we obtain that
γ
x∈k
d
P(x)≡number of points inX(k) modulop.
(It is tempting to say that equality holds, but the left hand side is ink,somodulo
pis the best one can claim.)
Next we compute the sum on the left. Let
⊂
x
ai
i
be any monomial appearing
in the polynomialP. Then
γ
x∈k
d
λ
i
x
ai
i
=
λ
i
σ
γ
y∈k
y
ai
≥
.
By assumption,

a
i≤degP<d(q−1), hencea i<q−1 for somei.Now,
for any 0≤a<q−1, there exists a (nonzerozinksuch thatz
a
ω=1. Because
γ
y∈k
y
a
=
γ
y∈k
(zy)
a
=z
a
γ
y∈k
y
a
,

1.4 Quadrics over ﬁnite ﬁelds 17
we conclude that

y∈k
y
a
=0 forain this range. Hence we conclude that

x∈k
d
⊂
i
x
ai
i
is zero for each monomialx
ai
i
occurring inP. Adding up over
all such monomials, we see that the number of points inX(k)isdivisible by
p. The trivial solution (0,..., 0) is always inX(k), thus we must have at least
p−1 nontrivial solutions.
Because the polynomials are homogeneous, the solutions come in lines
through the origin, so we can guarantee only one nontrivial solution up to
scalar multiple.
The next example shows that the degree bound of Theorem 1.18 is sharp. It
shows that over most ﬁelds, there are hypersurfaces inP
n
of any degreed>n
admitting no rational points.
Example1.19. Fix a ground ﬁeldkand letk

be a Galois extension ofk,
say of degreed. Letλ
1,...,λdbe ak-basis ofk

. Set
f(x
1,...,x d):=
λ
σ∈Galk

/k
φ
λ
σ
1
x1+···+λ
σ
d
xd
π
,
where the right hand side is the norm of the linear formλ
1x1+···+λ dxd, that
is, the product of all its conjugates overk. Becausefis invariant under the
Galois group action ofk

/k,itisdeﬁned over k, and so is the corresponding
degreedhypersurface inP
d−1
.Geometrically, this hypersurface is the union of
dhyperplanes deﬁned overk

,conjugate overk.
Weclaim that this hypersurface has nok-points. Indeed, ak-point would
amount to a nontrivialk-solution off, say (p
1,...,p d). This, in turn, would
mean a linear relation
λ
σ
1
p1+···+λ
σ
d
pd=0 for someσ∈Gal(k

/k).
Because the set{λ
1,...,λd}is linearly independent overk,soisthe conjugate
set. This forces eachp
ito be zero, and the solution was trivial after all. So the
constructed degreedhypersurface inP
d−1
has nok-points.
Tocreate hypersurfaces of higher degree with nok-points, simply set some
variables equal to zero to obtain examples of forms of degreedin fewer variables
which admit no nontrivial solutions. In particular, the corresponding hypersur-
faces can not be rational, or even unirational, overk.
The following exercise should convince the reader that the previous example
can be used to generate examples over many familiar ﬁelds, includingQandF
p.
Exercise1.20. LetKbe a ﬁeld ﬁnitely generated over an algebraically
closed subﬁeldk. Prove thatKhas separable extensions of any degree. Prove
that the same holds when the subﬁeldkis a prime ﬁeld, that is, whenk=Q
orF
p.

18 1 Examples of rational varieties
Of course, we should not be fooled by Exercise 1.20 into thinking that every
ﬁeld has extensions of every degree. Obviously, algebraically closed ﬁelds,
such asC,havenonontrivial algebraic extensions. Likewise, the ﬁeld of real
numbersRhas an extension of degree two, but no others.
Although Example 1.19 guarantees the existence of high degree hypersur-
faces with no rational points, it is not completely satisfying because the hy-
persurfaces we constructed were not smooth. The next exercise, suggested by
N. Katz, produces smooth examples, albeit not for every degree.
Exercise1.21. Letpbe a prime number and letXbe the hypersurface in
P
p−2
deﬁned by the homogeneous polynomialx
p−1
1
+···+x
p−1
p−1
. Show that
Xis a smooth hypersurface of degreep−1 that does not haveF
p-points.
Remark1.22. It is not the case that there existsmoothhypersurfaces with-
outF
q-points of every degreedinP
n
withd>n. Indeed, letC⊂P
2
be a
smooth curve of degree three and letkbe a ﬁnite ﬁeld ofqelements. The
Hasse–Weil estimates give that
#C(k)≥q+1−2
√
q>0.
(See, for instance, Hartshorne (1977, App. C) for a summary of the Weil con- jectures.) In general, the Weil estimates show that a smooth hypersurface of degreedhas a point inF
qwheneverqis very large relative tod.
Wenow give a different proof of Theorem 1.17. As we pointed out imme-
diately after its statement, the proof of Theorem 1.17 reduces to the case of a smooth conic inP
2
. Such a conic is isomorphic to the projective line over¯k,so
it sufﬁces to prove the following theorem in the case wheren=1.
Theorem1.23. Let X be a variety deﬁned over a ﬁnite ﬁeld k of cardinality
q.Assume that X is isomorphic toP
n
over¯k.Then the set of k-points of X has
cardinality
1+q+q
2
+···+q
n
.
Furthermore, the variety is in fact isomorphic toP
n
over k.
Akeyidea in the proof is the use of the Frobenius morphism, developed in
the next exercise.
Exercise1.24. Letkbe a ﬁnite ﬁeld of cardinalityq. Deﬁne the Frobenius
morphismF:P
n
→P
n
by
(x
0:···:x n)→
φ
x
q
0
:···:x
q
n
π
.
Prove the following properties:

1.4 Quadrics over ﬁnite ﬁelds 19
1.Fis one-to-one on closed points;
2.Fhas degreeq
n
;
3. the ﬁxed points ofFare exactly the points ofP
n
(k);
4. ifX⊂P
n
is a subvariety deﬁned overk, thenFrestricts to a morphism
F:X→X;
5. the restriction ofFtoXdoes not depend on the choice of the embedding
X⊂P
n
.
Exercise 1.24 suggests that we may be able to computeX(k)asaﬁxed
point set of the Frobenius morphism for any varietyX.Inturn, the cardinality
of the ﬁxed point set can be computed as an intersection number. Indeed, let
≥⊂X×Xdenote the diagonal and let ⊂X×Xdenote the graph of the
Frobenius morphismFonX. Then
X(k)
∼
=≥(k)=≥(¯k)∩ (¯k).
So the cardinality of the set ofk-points is equal to the intersection number
≥· , assuming the intersection is transverse. The computation of this inter-
section number hinges on our ability to give good geometric descriptions of
≥and of . This is our method for proving the next proposition, which estab-
lishes Theorem 1.23 in the curve case, and hence Theorems 1.17 and 1.16 in
general.
Proposition1.25. A smooth plane conic over a ﬁnite ﬁeld k of cardinality
q admits exactly q+1points deﬁned over k.
Proof.LetCdenote such a conic. According to the discussion above, the
cardinality ofC(k)isequal to the intersection number ·≥, assuming the
intersection is transverse. This intersection number can be computed over¯k.
SinceCis isomorphic toP
1
over¯k, the productC×Cis isomorphic toP
1
×P
1
over¯k, and≥,∼are divisors on this smooth surface. Every divisor onP
1
×P
1
is linearly equivalent to a divisor of the formaE 1+bE2, whereE 1=p1×P
1
andE 2=P
1
×p2, for any given pointsp 1,p2inP
1
.
Wecompute the linear equivalence class of both≥and over¯kby writing
each in the formaE
1+bE2and solving foraandbin each case. For≥, note
that≥∩E
1=(p 1,p1) and≥∩E 2=(p 2,p2),and that both intersections are
transverse. This implies that≥∼E
1+E2.For ,wehavethat ∩E 1and
∩E
2both consist of a single point, but transversality no longer holds. Choose
suitable afﬁne coordinatessandtfor an afﬁne chartA
2
contained inP
1
×P
1
so
thatE
1andE 2are given by the vanishing ofsandtrespectively, and is given
by the vanishing oft−s
q
. Then we compute that ·E 1=1 and ·E 2=q.
This gives that ∼qE
1+E2.

20 1 Examples of rational varieties
It is now straightforward to compute that
≥· =(E
1+E2)·(qE 1+E2)=1+q.
Furthermore, it can be checked that the intersection of≥and is transverse,
so it consists of precisely 1+qdistinctk-points.
Exercise1.26. The reader familiar with the notion of rational equivalence
and the intersection theory ofP
n
×P
n
should generalize the proof of Proposition
1.25 to higher dimension. This completes the proof of Theorem 1.23 in higher
dimensions.
Finally, there are many ways to prove the ﬁnal statement of Theorem 1.23
claiming that, in fact,Xis isomorphic toP
n
overk. The simplest is to build up
P
n
as follows.
Fix an isomorphism ofXwithP
n
deﬁned over¯k. Pick twok-pointsP 0and
P
1inX. The unique lineL 1through them inP
n∼
=Xis also deﬁned overk.Next
take ak-pointP
2not lying onL 1.Again, the unique planeL 2spanned by the
k-pointsP
0,P1,P2is deﬁned overk. Note thatL 1⊂L2and the linear system
O
L2
(L1) mapsL 2isomorphically toP
2
overk.Nowwe continue, building
up bigger and bigger linear spaces deﬁned overkinsideX
∼
=P
n
. The precise
value of the cardinality ofX(k)isneeded to prove that we can always choose
k-points not inL
ifori<n.Eventually, we see thatXis all ofP
n
,asavariety
overk.
A smooth projective variety that becomes isomorphic to projective space
overthe algebraic closure of the ground ﬁeld is called aSeveri–Brauer variety.
A concise introduction to the theory of Severi–Brauer varieties, including a
proof of the following result, can be found in Serre (1979, X.6).
Theorem1.27. ASeveri-Brauer variety is isomorphic to projective space
if and only if it admits a rational point.
1.5 Cubic hypersurfaces
After quadrics, the next natural examples are cubics. It turns out that the small
change in degree leads to a much more interesting theory with many unsolved
questions.
Cubic plane curves are usually called elliptic curves. The smooth ones are
not rational. There are many ways to see this; for example, overC, the elliptic
curves have genus one, so can not be isomorphic to the projective line, which
has genus zero. The theory of elliptic curves is very rich, and there are many

1.5 Cubic hypersurfaces 21
nice books on the subject. See, for instance, Reid (1988, I.2) or Shafarevich
(1994, III.3) for introductions.
Cubic surfaces were extensively studied in the nineteenth and early twenti-
eth centuries, but their theory still has plenty of unsolved problems. In contrast
to cubic curves, every smooth cubic surface over an algebraically closed ﬁeld
is rational. The point is that every such surface is isomorphic to the projective
plane blown-up at six points. If the six points are deﬁned over some alge-
braically non-closed subﬁeldk, then surface is rational overk,but in general,
rationality of cubic surfaces is a subtle question over algebraically non-closed
ﬁelds.
In this section, we concentrate on examples of rational and unirational cubic
surfaces over non-algebraically closed ﬁelds. Our goal is to ﬁnd simple geomet-
ric conditions that imply rationality or unirationality for cubic hypersurfaces.
One key fact we use is that every cubic surface contains exactly twenty-seven
lines (deﬁned over an algebraically closed ﬁeld). This fact, as well as other basic
facts about cubic surfaces, can be found in many introductory texts, including
such as Reid (1988,§7) or Shafarevich (1994, IV.2.5).
Weﬁrst note that rationality questions for singular cubics are relatively easy
to answer.
Example1.28. An irreducible cubic hypersurface in projective space (that
is not a cone over a cubic hypersurface of lower dimension) is rational overk
if it has asingular k-point.
Proof.LetXbe the cubic hypersurface inP
n+1
, deﬁned over the ground
ﬁeldk. Project from the singular pointP∈X(k)onto a general hyperplane
deﬁned overk. SincePhas multiplicity two onX,anyline throughPhas a
unique third point of intersection withX.Its projection onto the hyperplane
gives the one-to-one map fromXtoP
n
.Ofcourse, this makes sense only when
the line throughPdoes not lie onX. This is where we use the assumption that
Xis not a cone: in this case, the generic line throughPdoes not lie onX.
It is also easy to ﬁnd examples of cubics that are not unirational, as shown
by the next exercise.
Exercise1.29. Letf(x,y)beahomogeneous cubic equation overZthat
has no nontrivial roots modulo some primep.(Forinstance,x
3
−xy
2
+y
3
works forp=2or3.) Show that
f(x
0,x1)+pf(x 2,x3)=0
has no rational solutions. Conclude that the corresponding cubic surface is not unirational overQ.

22 1 Examples of rational varieties
It is harder to get examples of cubic surfaces which do have points but are
still not rational. The simplest examples use topological considerations over the
reals.
Example1.30. LetXbe a cubic surface in real projective three space,
deﬁned by an equation (in afﬁne coordinates)x
2
+y
2
=f3(z), wheref 3has
three distinct real roots. ThenXis not rational overR.
Proof.The set of points inRwheref
3takes positive values has two disjoint
components. The equationx
2
+y
2
=f3(z) has real solutions (in fact, a circle’s
worth) if and only iff
3≥0, so we see that, as a real manifold,X(R) has two
connected components. But ifXis birationally equivalent toP
2
overR, then
becauseP
2
(R)isconnected as a real manifold, so would beX(R) (see Exercise
1.31 below.)
Exercise1.31. LetXandYbe smooth projective varieties overR. Show
that ifXandYare birational thenX(R) andY(R)havethe same number of
connected components.
On the other hand, the cubic surfaceXof Example 1.30 is unirational overR.
Weleave the reader the pleasure of ﬁnding a mapP
2
R
φφπXthat is two-to-one
onto one of the manifold components and misses the other component entirely.
Indeed, the preimage of a point in the missed manifold component can be
interpreted as a pair of complex conjugate points in the complex manifoldP
2
(C).
The following example, due to Swinnerton-Dyer, gives a fascinating speciﬁc
example of a cubic surface which is not rational.
Example1.32 (Swinnerton-Dyer, 1962).Consider the cubic surface de-
ﬁned by
t(x
2
+y
2
)=(4z−7t)(z
2
−2t
2
).
1. The real points of this surface consist of two connected components.
2. On one manifold component,Q-points are dense.
3. On the other manifold component, there are noQ-points.
Tosee that the surface has two real components, consider the afﬁne chart
wheret=1. Settingv
2
=x
2
+y
2
,wesee that the surface is a surface of
revolution for the elliptic curve
v
2
=(4z−7)(z
2
−2).
Because the functionf(z)=(4z−7)(z
2
−2) has three distinct real roots, we
know from Example 1.30 that the curve, and hence the surface of revolution, has

1.5 Cubic hypersurfaces 23
two disjoint real components. The two real components correspond toz≥7/4
and|z|≤
√
2 respectively.
On the real component wherez≥7/4, theQ-points are dense. Indeed, this
component contains (x:y:z:t)=(1 : 1 : 2 : 1). The tangent plane to the
surface at this point intersects with the surface to produce an irreducible singular
cubic onS. This curve is rational overQ, and itsQ-points are dense among
itsR-points. In particular, theQ-values forzare dense among all real values
forz≥7/4. For each of these ﬁxedQ-valuesz
0, the planez=z 0intersects
the surfaceSin the circlex
2
+y
2
=f(z 0). This conic isQ-rational and its
Q-points are dense among itsR-points. Thus theQ-points ofSare dense on
the manifold component wherez/t≥7/4.
There are noQ-points on the component where|z/t|≤
√
2. To see this,
suppose that (x:y:z:t)issuch aQ-point, where, without loss of generality,
tandzare assumed relatively prime integers, witht>0. So
t(7t−4z)(2t
2
−z
2
)=(tx)
2
+(ty)
2
is an integer which is the sum of two rational squares. Thus any primepcon-
gruent to 3 modulo 4 that dividest(7t−4z)(2t
2
−z
2
) must divide it an even
number of times.
Because|
z
t
|≤
√
2, each of the integer factors
t,(7t−4z),(2t
2
−z
2
)
ispositive.Weclaim that none is congruent to 3 modulo 4. Indeed, no primep
congruent to 3 modulo 4 can divide any one of these factors to an odd power. Forifsome suchpdoes, then it must divide precisely two of the factors an odd
number of times. But becausetandzare relatively prime, it follows thattand
2t
2
−z
2
are relatively prime, and the only possible common prime factor oft
and (7t −4z)is2.Furthermore, ifpdivides both (7t −4z) and (2t
2
−z
2
),
thenpdivides (8t +7z)(7t−4z)−28(2t
2
−z
2
)=17tz. Since suchpdivides
neitherznort, the only possibility isp=17, which is not congruent to 3
modulo 4.
Now iftis even, thenzmust be odd, but this would force (2t
2
−z
2
)tobe
congruent to 3 modulo 4. On the other hand, iftis odd, then it must be congruent
to 1 modulo 4, but this forces (7t−4z)tobecongruent to 3 modulo 4. This
contradiction implies that there is noQ-rational point on the component of the
surface where|z/t|≤
√
2.
Wenow investigate general geometric criteria for rationality or unirationality
of smooth cubics.

24 1 Examples of rational varieties
Example1.33 (Rationality of cubics containing linear spaces).If a
smooth cubic hypersurface of even dimension contains two disjoint linear
spaces, each of half the dimension, then the cubic hypersurface is rational. In
particular, a smooth cubic surface is rational overkif it contains two skew
lines deﬁned overk(of the twenty-seven lines on the surface deﬁned over¯k).
Proof.LetX⊂P
2n+1
be the cubic hypersurface, and letL 1andL 2be the
two linear spaces onX. Consider the map
φ:L
1×L2φφπX
(P,Q)→third intersection pointX∩
PQ.
This deﬁnes a birational map fromL
1×L2toX. The map is well deﬁned
because a typical line intersectsXin exactly three points (counting multiplici-
ties). This map is birational: if the preimage ofx∈Xincludes two distinct pairs
(P
1,Q1) and (P 2,Q2)onL 1×L2, then the projections of the linear spacesL 1
andL 2fromxonto a general hyperplane would intersect each other in more
than one point, which is impossible (see Example 1.36 for a more general discussion). Because
P
2n
φφπP
n
×P
n
φφπL 1×L2φφπX
are birational equivalences, we conclude thatXis rational. Note that all maps
above are deﬁned over the ground ﬁeldk.
Exercise1.34. 1. Find examples of smooth cubic hypersurfaces inP
2n+1
containing two disjointn-planes.
2. What is the dimension of the variety of all such cubics? 3. Why have we not considered linear spaces of nonequal dimension? 4. Write down a birational map betweenP
2
and the cubic surface deﬁned by
the homogeneous polynomialx
2
y+y
2
z+z
2
v+v
2
x.
Example1.35 (Rationality of cubics containing conjugate linear spaces).
Weget an interesting variant of Example 1.33 for a cubic hypersurface of
even dimension containing a pair of disjoint linear spaces each of half the
dimension, which are deﬁned over some quadratic extensionk

=k(α) and
conjugate to each other overk. Let ¯αdenote the conjugate ofα.
As in the previous example, we obtain a birational mapL
1×L2φφπX
deﬁned overk

.Howdowe get down to k?
Here is the trick. Choose an afﬁne chartA
2n+1∼
=U⊂P 2n+1
intersectingL 1
andL 2.Wecan write thek

-points ofL 1∩Uin the form
C·w+c,

1.5 Cubic hypersurfaces 25
wherewis ak

-vector ofA
n
,Cisa(2n+1)×nmatrix overk

andcis a
k

-vector ofA
2n+1
. Writingw=u+αvwhereuandvarek-vectors ofA
n
, and
similarly forCandc,wecan rewrite this as
⊂(u,v):=[A·(u,v)+a]+α[B·(u,v)+b],
whereA,Bare (2n +1)×2nmatrices overk, anda,barek-vectors ofA
2n+1
.
This gives the representation
¯⊂(u,v):=[A·(u,v)+a]+¯α[B·(u,v)+b],
fork

-points inL 2∩U. Since⊂(u,v) and¯⊂(u,v)are conjugate points over
k, the line connecting them is deﬁned overk.Nowthe intersection of this line
with the cubic has a unique third intersection point which is necessarily deﬁned
overk. This gives a birational mapA
n
×A
n
φφπX, deﬁned overk.
Foranexplicit example consider the cubic surface inP
3
deﬁned byx
3
+
y
3
+z
3
=v
3
.
It is easy to see that this surface does not contain any disjoint pair of lines
deﬁned overQ(or even overR)but that it does contain the conjugate pair of
disjoint lines parameterized asL
i=(w,− iw, ⊗i), where ifori=1,2 are
the complex cube roots of 1 and we work in the afﬁne chartvω=0. Setting
w=t+
1swe obtain conjugate representations for the lines as
⊂
i(s,t)=(t,s,0)+ i(s,s−t,1).
The line joining them has a parametric representation with parameterλ:
(t,s,0)+λ(s,s−t,1).
Working out the third intersection point explicitly (a computation best done by
computer) gives the birational map⊂:(s,t)→(x:y:z:1)given by
x=t+sz
y=s+(s−t)z
z=
t
3
−1+s
3
−2s
3
−3st
2
+t
3
−1+3s
2
t
.
Example1.36 (Unirationality of cubics).More generally, given any two
subvarieties,UandV,of a cubic hypersurfaceX, one is tempted to form a
similar map:
φ:U×VφφπX
(u,v)→third intersection pointX∩uv.
IfUandVare disjoint, this map is a morphism except at pairs of points (u,v)
spanning a line onX;ingeneral, it is not deﬁned onU∩V.

26 1 Examples of rational varieties
The mapφcan not be dominant unless dimU+dimV≥dimXand it can
not be generically ﬁnite unless dimU+dimV=dimX. Whenφis ﬁnite, how
does one compute its degree?
Todetermine the preimage of a general pointx∈X, consider the projection
π
xfromx∈Xto a general hyperplane. The setπ x(U)∩π x(V) consists of
all pointsπ
x(u)=π x(v), withu,v,andxcollinear. In this case, assuming
thatuω=v, the points (u ,v)∈U×Vare the preimages ofxunderφ. So, if
U∩V=∅,weexpect that the degree ofφis the cardinality ofπ
x(U)∩π x(V).
More generally, we must subtract something for the intersection points ofU
andV.
In Example 1.33, we applied this idea withUandVlinear subspaces and
deduced that cubic hypersurfaces are rational if they contain two disjoint lin-
ear subvarieties of half the dimension. More generally, the idea is useful for
detectingunirationalityof some cubics, as show by the proof of the following
theorem of B. Segre.
Theorem1.37. A smooth cubic surface over an inﬁnite ﬁeld k is unira-
tional if and only if it admits a rational point over k.
Recall that a point on a smooth cubic surface is called an Eckardt point if
it is the intersection of three of the twenty-seven lines on the surface. Clearly,
Eckardt points are quite special: a cubic surface can have at most ﬁnitely many
Eckardt points and a general cubic surface contains none (Eckardt, 1876). We
prove here the following special case Segre’s theorem:A smooth cubic surface
over a perfect inﬁnite ﬁeld k is unirational if and only if it admits a k-rational
point that is not an Eckardt point.This case was proved by Segre (1943
deduce the theorem in its full strength, one must apply the following later result
of Segre (1951 kcontains ak-point,
then it contains inﬁnitely manyk-points. Since any particular cubic surface has
at most ﬁnitely many Eckardt points, this reduces the problem to our special
case.
Proof.First note that a smooth cubic surface inP
3
containing two non-
coplanar rational curves is unirational. Indeed, letC
1andC 2be rational curves
on the surfaceX, and deﬁne the mapφ:C
1×C2√√≡Xas in Example 1.36.
BecauseC
1andC 2do not lie in the same plane, their join (meaning the locus
of points lying on lines joining points onC
1to points onC 2) must be all ofP
3
.
This ensures that the mapφis dominant, and hence ﬁnite. BecauseC
1andC 2
are rational (overk), we conclude thatXis unirational (overk).
Thus we must ﬁnd two non-coplanar rational curves on our cubic surface.
Suppose ﬁrst that the surface contains twok-rational pointsp
1andp 2that are

1.5 Cubic hypersurfaces 27
not Eckardt points. IntersectingXwith the tangent planes at each of the two
k-points, we get two curvesC
1=Tp1
XandC 2=Tp2
XonXdeﬁned overk.
Typically,C
1andC 2are irreducible, hence each is rational overkby Example
1.28 (ifC
1is a cone over three points inP
1
, thenp 1is an Eckardt point).
Furthermore, in this typical case, these plane cubics are not coplanar. If they
were, thenT
p1
XandT p2
Xwould coincide, so that alsoC 1=C2would be an
irreducible plane cubic with two singular pointsp
1andp 2, which is impossible.
Even in the degenerate case whereC
1orC2is reducible, the argument often
goes through unchanged. Indeed, ifC
1breaks up as a smooth conic together
with a line (overk), then each of these components is rational overk, and by
similar reasons, can not be coplanar with the irreducible curveC
2.Agenuine
exception can occur whenC
1=C2is a union of a line and a smooth plane
conic overk,orwhen bothC
1andC 2are a union of three lines (over¯k).
However, these exceptional cases can be avoided as follows. Given one
non-Eckardtk-point on a smooth cubic surface, the plane sectionCobtained
by intersectingXwith the tangent planeT
pXcontains a rational component
deﬁned overk.Eveninthe case whereCis a union of three lines meeting
in distinct pointsp,p
1,p2over¯k,weget a line deﬁned over ak: since the
Galois action of¯k/kmust permute the three points but must ﬁxp,itmust also
stabilize the line throughp
1andp 2, whence this line is deﬁned overk.
1
The
rational component ofCgives us many rational points onX, and a general pair
of them produces a pair of non-coplanar rational curves onXas in the previous
paragraph. Thus we have shown that a smooth rational surface containing a
rational point that is not an Eckardt point is unirational.
It is only recently that Segre’s result has been extended to ﬁnite ﬁelds and
to all higher dimensional cubics.
Theorem1.38 (Koll´ar, 2002). A smooth cubic hypersurface of dimension
at least two is unirational over k if and only if it admits a k-point.
Note that Theorem 1.38 is completely independent of the ﬁeld: it holds over
any ground ﬁeld, inﬁnite or not. We do not prove this theorem here, but instead refer to Koll´ar (2002
The following two examples were explained to us by Swinnerton-Dyer.
Exercise1.39. Check that the cubic surface deﬁned by the equation
x
3
1
+x
2
1
x0+x1
φ
x
2
0
+x
2
2
+x
2
3
+x2x3
π
+x
3
2
+x
2
2
x3+x
3
3
1
This is where we use the assumption thatkis perfect; see Exercise 1.8 and the subsequent
remark.

28 1 Examples of rational varieties
overthe ﬁeldF
2of two elements has exactly oneF 2point. Show that up to a
linear change of coordinates, this is the only example of a cubic surface over
F
2which has precisely one point. (In fact, it turns out that this is the only such
example over any ﬁnite ﬁeld, see (Swinnerton-Dyer, 1981, 5.5).)
Exercise1.40. Check that, up to a linear change of coordinates, the cubic
surface deﬁned by the equation
x
2
φ
x
2
0
+x0x2+x
2
2
π
+x
3
φ
x
2
1
+x1x3+x
2
3
π
+x
2
2
x3
is the only cubic surface overF 2which contains a line deﬁned overF 2but no
otherF
2-points.
Example1.41. An interesting variation on the map discussed in Example
1.36 is obtained by allowingU=V.For example, suppose thatXis a smooth
cubic four-fold inP
5
containing a smooth surfaceS.
Consider the map
φ:
S×S
∼
φφπX
(P,Q)→third point of intersectionX∩PQ.
Here,
S×S
∼
is the symmetric product ofS, the quotient variety ofS×Sby the
action of the two-element group interchanging the factors. IfSis unirational
overk, then so isS×S, and hence so is the image
S×S
∼
under the generically
two-to-one quotient map.
Consider a general pointx∈X, say not onS. When isxin the image ofφ?
Consider the family of lines{sx}s∈S. The pointxis in the image ofφprecisely
when at least one of these lines intersectsSin a point other thans.Inparticular,
the projection fromx,π
x:S→S

⊂P
4
can not be one-to-one. Indeed,xhas
a unique preimage underφprecisely when the projectionπ
xcollapses exactly
two points ofSto a single point. On the other hand, ifπ
x:S→S

is not of
degree one, thenxhas inﬁnitely many preimages underφ.
Thusφis ﬁnite and dominant if and only if the generic projection ofSfrom
a pointx∈Xis one-to-one except on a ﬁnite set. The next exercise provides
one case where this condition can be veriﬁed.
Exercise1.42. LetP
1,P2,P3,P4be four points in the projective plane,
no three of them on a line. LetSbe the surface obtained by blowing up these
points. Show that the linear system of plane cubics through the four points gives an embeddingS→P
5
. (The image is called a degree ﬁve Del Pezzo surface.
Wewill study these in Chapter 3.) Prove that a general projection ofStoP
4
has exactly one singular point. Use this to give some more examples of rational cubic four-folds.

1.6 Further examples of rational varieties 29
Historical remark1.43. The above method has an interesting history
which illustrates the necessity of extreme care in counting dimensions.
One can see that the family of degree ﬁve Del Pezzo surfaces inP
5
has
dimension 35 and each is contained in a 24 dimensional family of cubics. Since
the space of cubics inP
5
has dimension 55, we might expect that every cubic in
P
5
contains a four-dimensional family of degree ﬁve Del Pezzo surfaces and is
therefore rational. It turns out, however, that if a cubic inP
5
contains a degree
ﬁve Del Pezzo surface, it always contains a ﬁve-dimensional family of them,
so the above argument is wrong.
A similarly incorrect “proof” of rationality of cubics inP
5
using degree four
ruled surfaces was published by Morin (1940
It is still not known if the general cubic inP
5
is rational or not. See Hassett
(2000
1.6 Further examples of rational varieties
Webegin with the rationality of determinantal varieties, which ultimately pro-
vides a different perspective on the fact that a cubic surface is geometrically
rational.
Exercise1.44. 1. Prove that the variety ofm×nmatrices of rank at most
tis rational over any ﬁeld. Find its nonsmooth locus.
2. Consider ann×narray of general linear forms onP
n
.Prove that the hyper-
surface deﬁned by the determinant of this array is a rational variety. When
is this variety smooth?
Theorem1.45. The equation of a smooth cubic surface over an alge-
braically closed ﬁeld can always be written as the determinant of a3×3
matrix whose entries are linear forms in four variables.
The earliest proof of this fact appears in an 1866 paper of Clebsch, who
credits Schr¨oter (Clebsch, 1866). We give here a classical geometric proof.
Foramore algebraic proof using the Hilbert–Burch theorem, see Geramita
(1989).
Proof.Weuse the conﬁguration of the twenty-seven lines on the cubic
surfaceS.Weclaim that there are nine lines on the surface that can be repre-
sented in two different ways as a union of three hyperplane sections. That is,
there are six different linear functionalsl
1,l2,l3,m1,m2,m3onP
3
such that the
hyperplane sections ofSdetermined by each is a union of three distinct lines,
and the nine lines obtained as hyperplane sections with thel
is are the same nine

30 1 Examples of rational varieties
lines obtained from them
is. Assuming this for a moment, the cubicsl 1l2l3and
m
1m2m3both deﬁne the same subscheme ofS, which means that up to scalar,
these cubics agree onS.Inother words, the cubic
l
1l2l3−λm 1m2m3
is in the ideal generated by the cubic equation deﬁningS, and hence it must
generate it. On the other hand, the cubicl
1l2l3−λm 1m2m3is the determinant
of the matrix


l
1m10
0l
2m2
−λm 30l 3

.
The proof will be complete upon establishing the existence of the special con-
ﬁguration of lines. First recall thatSis the blowup of six pointsP
1,P2,...,P 6
inP
2
,nothree on a line and no ﬁve on a conic. We embedSinP
3
using the
linear system of plane cubics through these six points. The twenty-seven lines
onS⊂P
3
are obtained as follows:
1. for each pair of two pointsP
iandP j, the birational transform of the line
PiPjinP
2
joining them;
2. for each pointP
i, the birational transform of the conicQ ithrough the re-
maining ﬁve points;
3. for each pointP
i, the ﬁberE ioverP i.
Foranypair of indicesi,j, the three lines
PiPj,Ei, andQ jform a (possibly
degenerate) triangle onS.Indeed, thinking of the hyperplane sections ofSas
cubics in the plane through the six points, this triangle is the hyperplane section given by the cubic obtained as the union of
PiPjandQ j.Nowit is easy to ﬁnd
such a conﬁguration. For instance, the nine lines
{E
1,Q2,
P1P2}∪{E 2,Q3,P2P3}∪{E 3,Q1,P1P3}
are the same as the nine lines
{Q
1,E2,
P1P2}∪{Q 2,E3,P2P3}∪{Q 3,E1,P1P3},
with the groupings indicating the two different conﬁgurations of triangles.
Remark1.46. Our proof of Theorem 1.45 uses the fact that a cubic surface
is a blowup ofP
2
at six points, and therefore does not give a new proof that cubic
surfaces are rational. However, the ﬁrst proof that cubic surfaces are rational did proceed by showing ﬁrst that they are determinantal. Indeed, the nineteenth century masters had such a detailed understanding of the conﬁguration of lines

1.7 Numerical criteria for nonrationality 31
on a cubic surface that they were able to see combinatorially that the special
conﬁguration of lines needed in the proof of Theorem 1.45 exists without using
blowups.
The following exercises provide more examples of rational varieties.
Exercise1.47. (1 k=C(t), and letXbe a degreedhypersurface in
P
n
deﬁned overk. Prove that ifd≤n, thenXhas at least onek-point. Find
an example with exactly onek-point. Ford>n, ﬁnd a hypersurface with no
k-points. Explain why such a hypersurface is nonrational.
(2 k=C(t,s) andd
2
≤
n.
Exercise1.48. (1X⊂P
1
×P
n
be a smooth hypersurface of bidegree
(a,2). Whennis at least two, show thatXis rational overC.
(2X⊂P
2
×P
n
be a smooth hypersurface of bidegree (a ,2). Forn≥4,
show thatXis rational overC.
Exercise1.49. LetPbe a point of multiplicitymon a hypersurface of
degreedinP
n
. Show that there is an at least (n+m−d−2)-dimensional
family of lines contained inXand passing throughP.
Exercise1.50. LetXbe a smooth hypersurface inP
n
of degreed≤n.
Assuming the ﬁeld is algebraically closed, ﬁnd a rational curve passing through
every point ofX.
(It is an open question whether or not every such smooth hypersurface con-
tains a rational surface through every point.)
Exercise1.51. LetX⊂P
3n+1
be a quartic hypersurface containing three
linear spaces, each of dimension 2n, whose common intersection is empty.
Prove thatXis rational.
1.7 Numerical criteria for nonrationality
Rationality and unirationality force strong numerical constraints on a variety.
Let
X= X/kbe the sheaf of regular differential forms (K¨ahler differentials)
on a varietyXoverk.
Theorem1.52. If a smooth projective variety X is rational, then it has
no nontrivial global K¨ahler one-forms. In fact, the space of global sections
(X,
⊗m
X
)of the sheaf
⊗m
X
is zero for all m≥1. The same holds for unira-
tional X, provided the ground ﬁeld has characteristic zero.

32 1 Examples of rational varieties
Proof.Suppose we have a generically ﬁnite, dominant mapφ:P
n
√√≡X.
LetU⊂P
n
be an open set over whichφis deﬁned; its complement may be
assumed to have codimension at least two.
Nonzero differential forms onXpull back to nonzero differential forms on
U, that is, we have an inclusionφ
∗

⊗m
X
→
⊗m
U
. This is obvious whenφis
birational, and easy to check whenφis ﬁnite (assumingkhas characteristic
zero). Because the complement ofUhas codimension at least two, the differ-
ential forms onUextend uniquely to forms onP
n
, inducing an identiﬁcation

√
U,
⊗m
U
≡
=
√
P
n
,
⊗m
P
n
≡
.
Therefore (X,
⊗m
X
)⊂ (P
n
,
⊗m
P
n),and the problem is reduced to proving
the vanishing forP
n
, left as an exercise.
Exercise1.53. Complete the proof by showing that (P
n
,
⊗m P
n)iszero
for anm≥1.
Remark1.54. In prime characteristic, unirationality ofXdoes not neces-
sarily force the vanishing of the invariants (
⊗m
X
). Indeed, the pull-back map
for differential forms can be the zero map, so the argument above fails. For exam-
ple, consider the Frobenius mapFonA
n
sending (λ 1,...,λn)→(λ
p
1
,...,λ
p
n
),
wherep>0isthe characteristic of the ground ﬁeld. The induced map of differ-
ential formsF
∗
→sends every differentialdxtod(x
p
)=px
p−1
dx=0.
In characteristicp,weare led to the more sensible notion ofseparable
unirationality.AvarietyXis separably unirational if there is a dominant generi-
cally ´etale mapP
n
√√≡X;bygenerically´etale,wemean that the map P
n
√√≡X
is generically ﬁnite and that the induced inclusion of function ﬁelds isseparable.
A generically ﬁnite morphismf:Y→Xis separable if and only if the pull-
back map on differentialsf
∗

1
X
→
1
Y
is injective. This is the only property
that we need in the proof of Theorem 1.52. With this in mind, the proof of
Theorem 1.52 shows that (X,
⊗m
X
)=0 for a smooth projective separably
unirational varietyXof arbitrary characteristic.
ForarbitraryX, the spaces (X,
⊗m
X
) are usually hard to compute because
the
⊗m
X
have quite high rank. Therefore it is important to have similar cri-
teria which involve line bundles only. The natural candidate is the canonical
bundleω
X=∧
n
Xof highest degree K¨ahler differential forms, which is al-
ways deﬁned over the ﬁxed ground ﬁeld. For smoothX, the canonical bundle
is represented by a divisorK
Xdeﬁned over the given ground ﬁeld, and it is
convenient to denote it byO
X(KX).
Deﬁnition1.55. Themth plurigenus of a smooth varietyXis the dimen-
sion of the vector space of global section of the invertible sheafO(mK
X). We
denote this integer byP
m(X).

1.7 Numerical criteria for nonrationality 33
The plurigenera are easily computable obstructions to rationality.
Corollary1.56. If X is a smooth projective variety that is separably
unirational (for example, rational), then the plurigenera P
m(X)vanish for all
positive m.
Proof.Probably the easiest argument is to notice that the line bundles
O(mK
X) also have a pull-back mapφ
∗
OX(mKX)→O U(mKU)asinthe proof
of Theorem 1.52. The rest of the proof applies verbatim.
Conceptually it is neater to use the injectionO
X(KX)→
⊗dimX
X
which
comes from the vector space map∧
dimV
V→V
⊗dimV
identifying the deter-
minant with the corresponding multilinear form.
Looking at the proof of Corollary 1.56 we see that we have not used the
full strength of our unirationality assumption. This leads to the following two
weaker notions.
Deﬁnition1.57. AvarietyXisruledif there exists a varietyYand a
birational mapφ:Y×P
1
√√≡X.AvarietyXisuniruledif there exists a
varietyYand a generically ﬁnite dominant mapφ:Y×P
1
√√≡X.
Loosely speaking, a variety isuniruledif it is covered by rational curves. Of
course, every rational varietyXis ruled, sinceP
n
is birationally equivalent to
P
n−1
×P
1
.
As before, in positive characteristic there is another variant. The varietyXis
separably uniruledif there is a separable, generically ﬁnite mapY×P
1
√√≡X.
Every ruled variety is separably uniruled, and every unirational variety
Xis uniruled. In characteristic zero, separably uniruled is equivalent to
uniruled.
Theorem1.58. The plurigenera vanish for any smooth projective separa-
bly uniruled variety.
Proof.Letφ:Y×P
1
√√≡Xbe a separable uniruling of a smooth variety
X.Weneed to show that (X,
⊗m
X
)=0 for all positivem.Asbefore, we are
reduced to proving that
(Y×P
1
,O(mK Y×P
1))=0 for allm≥1.
Notice thatK
Y×P
1=π
∗
1
KY+π
∗
2
KP
1where theπ idenote the coordinate pro-
jections. Thus
(Y×P
1
,O(mK Y×P
1))
∼
= (Y,O(mK Y))⊗ (P
1
,O(mK P
1)).
Finally, (P
1
,O(mK P
1))
∼
= (P
1
,O(−2m))=0 and we are done.

34 1 Examples of rational varieties
Exercise1.59. Show that the plurigenera of a smooth hypersurface of de-
greedinP
n
do not vanish whend>n.Conclude that no smooth hypersurface
whose degree exceeds its embedding dimension is separably uniruled. In par-
ticular, no such hypersurface is rational.
In positive characteristic, unirational but not separably unirational varieties
exist, and in fact, there are unirational hypersurfaces of arbitrary degree.
Exercise1.60. Show that a purely inseparable cover of a unirational vari-
ety over a perfect ﬁeld is unirational.
Remark1.61. In characteristicp,the Fermat-type hypersurface deﬁned
byx
n
0
+x
n
1
+···+x
n
m
inP
m
is unirational ifmis odd and some power ofp
is congruent to−1 modn. This is due to Shioda; the proof is elementary but
tricky (Shioda, 1974). For further examples, see Shioda and Katsura (1979

2
Cubic surfaces
In Chapter 1, we produced many examples of rational cubic surfaces. We also
discussed the fact that a smooth cubic surface is unirational if and only if
it has a rational point. In this chapter, we treat the subtle issue of rational-
ity for smooth cubic surfaces more systematically. In particular, we show
that there are many cubic surfaces deﬁned overQwhich are not rational
overQ.
Webegin in Section one by stating our main results, which provide a com-
plete understanding of rationality issues for smooth cubic surfaces of Picard
number one. These results are Segre’s theorem, stating that such a surface is
never rational, and the related result of Manin stating that any two such bi-
rationally equivalent surfaces are actually isomorphic. In the second section,
we set up the general machinery of linear systems to study birational maps
of surfaces. This technique, called theNoether–Fano method,isquite pow-
erful and ultimately leads to a proof that smooth quartic threefolds are not
rational, in Chapter 5. In this chapter, however, we apply this method only
to the case of cubic surfaces, proving the theorems of Segre and Manin in
Section 3.
Over an algebraically closed ﬁeld, every cubic surface has Picard number
seven, and it is not obvious that there is any cubic surface with Picard number
one. Thus, in Section 4, we develop criteria for checking whether a cubic
surface has Picard number one. Using this, we show that a typical diagonal
cubic surface has Picard number one over the rational numbers. Combined
with Segre’s theorem, we obtain many smooth cubic surfaces which are not
rational overQ.
In Section 5, we move away from cubic surfaces to take a closer look at
birational self-maps of the projective plane. In particular, we prove a classical
result of Noether and Castelnuovo stating that, over an algebraically closed
35

36 2 Cubic surfaces
ﬁeld, every birational self-map ofP
2
factors as a composition of projective
changes of coordinates and quadratic transformations.
2.1 The Segre–Manin theorem for cubic surfaces
Rationality for cubic surfaces is quite subtle. Our goal is to clarify the situation
by proving the following theorem of B. Segre.
Theorem2.1 (Segre, 1942). Let S be a smooth cubic surface over a ﬁeld k.
Assume that every curve C⊂Sislinearly equivalent to a hypersurface section.
Then S is not rational (over k).
It is not at all obvious that the assumptions of this theorem are ever satisﬁed.
Toclarify the situation, we consider thePicard groupPic(S)ofS, that is, the
group of divisors modulo linear equivalence.
The Picard group of a smooth cubic surfaceSover an algebraically closed
ﬁeld is isomorphic toZ
7
: thinking ofSas the blowup of the projective plane at
six points, the Picard group is freely generated by the six exceptional lines and
the pull-back of the hyperplane class. An alternative derivation of this fact is
given in Shafarevich (1994, IV.2.5). Related results are treated in Exercise 2.17.
On the other hand, the cubic surfaceS
kmay be deﬁned over some non-
algebraically closed ﬁeldk,evenifthe individual points we blow up are not
deﬁned overk.Inthis case, the Picard group ofS
kmay be smaller. Indeed, by
Proposition 1.6, there is an injection
Pic(S
k)⊂→Pic(S ¯k)
∼
=Z
7
,
and Pic(S
k)isfrequently much smaller. A convenient measure of the size of
Pic(S
k)isthePicard number, denoted by ρ k.Bydeﬁnition, the Picard number
of a smooth cubic surface overkis the rank of its Picard group.
1
Wecan thus restate Theorem 2.1 as follows:
No smooth cubic surface of Picard number one is rational.
In Exercise 2.18, we outline how to construct examples of cubic surfaces
overQwith Picard number one.
In Section 3, we prove the theorem of Segre. Essentially the same argument,
with minor modiﬁcations to be made afterwards, proves the following stronger
theorem of Manin (1966
1
Foranarbitrary normal projective variety the Picard number is deﬁned as the rank of the
N´eron–Severi group, the group of Cartier divisors up to numerical equivalence. For varieties
withh
1
(X,O X)=0 the two deﬁnitions coincide.

2.2 Linear systems on surfaces 37
Theorem2.2.Twosmooth cubic surfaces deﬁned over a perfect ﬁeld, each
of Picard number one, are birationally equivalent if and only if they are pro-
jectively equivalent.
Caution2.3.Manin’s theorem does not assert that every birational equiv-
alence is a projective equivalence, and this is not at all true. It guarantees only
that if two surfaces are birationally equivalent, then there exists an automor-
phism ofP
3
which maps one cubic surface into the other.
Remark2.4.The hypothesis of smoothness can not be weakened. For
instance, consider a plane conic deﬁned overk, together with six points on it
conjugate, but not individually deﬁned, overk.Byblowing up the six points
and then contracting the conic, we obtain a singular cubic surface with Picard
number one. All such surfaces are birationally equivalent to each other, but two
such are projectively equivalent if and only if the corresponding six-tuples of
points are projectively equivalent inP
2
.
Similarly, the hypothesis that the Picard number is one is essential. For
example, any two smooth cubic surfaces are birational overC,but the isomor-
phism classes of smooth cubic surfaces form a four-dimensional family. Since
cubic surfaces are embedded inP
3
by the anticanonical linear system, every
isomorphism between cubic surfaces is a projective equivalence.
2.2 Linear systems on surfaces
LetSbe a smooth projective surface. A rational map
S
φ
⊂⊂→P
n
is given by ann-dimensional linear system of curves onS;infact, such a map
corresponds to a uniquemobilelinear system, where mobile means that the
system has no ﬁxed curves. Recalling that thebase locusof a linear system is
the intersection of all its members, we see that the base locus of a mobile linear
system on a surface is simply the ﬁnite sets of points where the corresponding
rational map is not deﬁned.
Consider a birational morphismf:S

→S. Thebirational transformof
the mobile linear systemonSunderfis the mobile linear system

onS

whose general member is the birational transform of the general member of
. Equivalently,

is the linear system obtained by pulling backand then
throwing away any ﬁxed curves. In particular, to the extent to whichSandS

are “the same,”and

determine “the same” map toP
n
:

38 2 Cubic surfaces
An important theme in this chapter, as well as in Chapter 5 where these
methods are generalized to higher dimension, is the relationship of the linear
systemsandK
SonSto the linear systems

andK S
onS

.For surfaces, this
relationship is expressed in terms of intersection numbers. The self-intersection
number
2
and the canonical intersection number·K Sare of particular rele-
vance.
2.5. Self-intersection numbers and birational maps.Letbe a
mobile linear system on a smooth surfaceS. The self-intersection number of
–by which we mean the intersection number of two general members of–is
an important invariant of the mapφgiven by.
Whenis base point free, the mapφis a morphism and the self-intersection
number ofis simply the degree of the image ofφtimes the degree ofφ:

2
=deg(φ)·deg(φ(S)). (2.5.1)
In particular, whenφis a morphism ontoP
2
, the self-intersection number
2
is
equal to the degree ofφ.
Whenhas base points, the self-intersection number takes into account
their multiplicities. Themultiplicityof a linear system at a point is simply the
multiplicity of a general member there. Thus the expected contribution to
2
of a base pointPof multiplicitymism
2
.However, this is valid only when two
general curves inhave distinct tangents atP. The number is even higher if
the curves share tangents atP; this is the case wherehas base pointsinﬁnitely
neartoP.
Consider the blowupπ:S

→SofSat a base pointPof. The birational
transformC

of a general memberCofintersects the exceptional ﬁberEin
points corresponding to the tangent directions to the curveCatP. The base
points of the birational transform

ofthat lie inEare calledbase points
ofinﬁnitely near to P.They represent the tangent directions atPthat are
shared by all members of. (It is possible that

has base points lying in
E. Blowing them up, the base points of the birational transform of

lying
in the new exceptional ﬁber are also base points ofinﬁnitely near toP.
These correspond to higher order shared tangent directions of the members
of.)

2.2 Linear systems on surfaces 39
Let us consider what happens to the self-intersection and canonical intersec-
tion numbers ofafter blowing up one of its base points. For a general member
Cof,itiseasy to check thatπ
∗
C=C

+mEwhereEis the exceptional
ﬁber of the blowup andmis the multiplicity ofatP. Thus

=π
∗
−mEandK S
=π
∗
KS+E,
whereK
SandK S
denote the canonical classes ofSandS

respectively. This
leads to the following numerical relationship between intersection numbers on
SandS

:
Exercise2.6.With notation as above, verify that

2
=
2
−m
2
and

·KS
=·K S+m.
2.7. Resolution of indeterminacy.Because the self-intersection num-
ber of the birationally transformed linear system drops with each blowup, the
process of blowing up base points can be iterated until we have gotten rid of
all base points. In this way, we arrive at a smooth surface¯S, and a base point
free linear system¯deﬁning the “same map” (i.e. after composition with the
blowing up maps) to projective space as.
¯S
⊂
⊂
⊂
⊂
φ
¯
→
∼
.
.
.
S
∼
φ
P n
This process is calledresolving the indeterminacyof the rational mapφ .In
this commutative diagram, each morphism in the tower on the left is a blowup
at a base point.
Now suppose thatdeﬁnes abirationalmapφ
, and letTdenote the image
ofSunder this map. Resolving the indeterminacy ofφ
,wearrive at a birational
morphism
φ
¯:¯S→T⊂P
n
onto the surfaceT. From formula (2.5.1
number of¯is equal to the degree ofT. Therefore, after iterating the com-
putation of Exercise 2.6, we arrive at the following formulas for intersection
numbers onS.
Exercise2.8.Verify that

2
−
⊂
m
2
i
=degTandK S·+
⊂
m i=H·K T

40 2 Cubic surfaces
where them
iare the multiplicities of all base points of, including the inﬁnitely
near ones, andHis the restriction toTof the hyperplane class inP
n
.
Having carried out these computations, the following theorem is easy to
prove.
Theorem2.9.Let S be a smooth projective surface over k. Then S is
rational over k if and only if S admits a mobile two-dimensional linear system
deﬁned over k satisfying

2
−
⊂
m
2
i
=1and K S·+
⊂
m i=−3,
where the m
iare the multiplicities of all base points ofover¯k,including the
inﬁnitely near ones.
Wesee from the proof that the second conditionK
S·+
→
m i=−3 can
in fact be omitted. In our applications, the two numerical conditions together
turn out to be very useful.
Proof.Assume thatSis rational overk, and letφ:S⊂⊂→P
2
be a birational
map. Letbe the mobile linear system onSobtained as the birational transform
of the hyperplane system onP
2
. The dimension ofis two and the desired
numerical conditions follow immediately from the formulas of Exercise 2.8.
Conversely, given a linear systemsatisfying the given numerical condi-
tions, it determines a rational mapφ
:S⊂⊂→P
2
deﬁned overk.Weneed only
verify that this map is actually birational. Because the map isa priorideﬁned
overk,tocheck that it is birational we can assume thatkis algebraically closed
since whether or not the map is dominant and degree one is unaffected by
replacingkby its algebraic closure.
Blow up the base points of, including the inﬁnitely near ones, to obtain
a morphism¯φ:¯S→P
2
resolving the indeterminacy ofφ. The corresponding
mobile linear system¯has dimension two, and the numerical conditions¯
2
=
1 and¯·K
¯S=−3 hold. BecauseSand¯Sare birationally equivalent, it is
sufﬁcient to show that the morphismφ
¯:¯S→P
2
is a birational equivalence.
Tocheck that the mapφ
¯:¯S→P
2
is surjective, assume, on the contrary,
that its image is a plane curve,B. Then every member of¯is a union of ﬁbers
of¯S→B. This forces the self-intersection number¯
2
to be zero, contradicting
¯
2
=1. Henceφ ¯is surjective.
Finally, since¯
2
=1, the formula (2.5.1 φ ¯=1. Thusφ ¯
is birational, and our surfaceSis rational.
Caution2.10. It is possible for a linear system on a variety to be deﬁned
overkeven though its base points are not. For example, letF(X)∈Q[X]. As

2.2 Linear systems on surfaces 41
λandµvary throughC, the linear system of divisors given by the vanishing of
the polynomialsλY+µF(X)isaone dimensional linear system onA
2
deﬁned
overQ. The zeros ofF(X) determine the base points, since (x,y)isabase point
if and only if (x,y)=(α,0) whereαis a root ofF. These base points may not
be deﬁned overQ, although the linear system is deﬁned overQ. The map to
projective space determined by this linear system is deﬁned overQeven when
its base points are not.
Caution2.11. Given a mobile linear system on a smooth surface, it is not
always possible to compute the degree of the image variety from the numerical
data of Exercise 2.8 as we did in Theorem 2.9. For instance, ifis a three-
dimensional linear system satisfying

2
−
⊂
m
2
i
=4 and K S·+
⊂
m i=0,
then there are two possibilities for. Either it is a birational map whose image
is a degree four surface inP
4
or it is a degree two map to a quadric surface in
P
3
.
Wenext record an important corollary of Theorem 2.9. Although it looks
simple, it is the starting point for our later construction of families of higher
dimensional nonrational Fano varieties in Chapter 5.
Corollary2.12. Let S be a smooth projective surface whose Picard group
is generated by the canonical class K
S. Assume also that K
2
S
=1. Then S is
not rational.
Theorem 3.36 classiﬁes surfaces whose anticanonical class is ample and
satisﬁesK
2
=1 (calleddegree one Del Pezzosurfaces). Over a algebraically
non-closed ﬁeld, many such surfaces have Picard number one, so Corollary
2.12 is not vacuous. See Chapter 5, Section 3 for a higher dimensional analog
of Corollary 2.12.
Proof of Corollary2.12. Assume the contrary. Then we obtain a linear
system⊂|−dK
S|satisfying the numerical conditions of Theorem 2.9. In
particular,
d
2
=
2
=1+
⊂
m
2
i
andd=−·K S=3+
⊂
m i.
The second equation says thatm
i<dfor everyi, hence we get a contradiction
by
d
2
=1+
⊂
m
2
i
≤1+d
⊂
m i=d
2
−3d+1<d
2
.

42 2 Cubic surfaces
2.3 The proofs of the theorems of Segre and Manin
The proof of the Segre–Manin results begins with the general observation that
the Picard group of a smooth cubic surface of Picard number one is generated
by the class of a hyperplane section. Indeed, the Picard group ofSis torsion-
free, because Pic(S)⊂Pic (S
¯k)
∼
=Z
⊕7
.Also, the hyperplane classHis not
divisible: ifH=mDfor some divisorDand some integerm, then because
H
2
=3=m
2
D
2
,itfollows thatm=1. Of course, for a cubic surface, the
canonical classK
Sis given by−H,sowecan also say that the Picard group of
a smooth cubic surface of Picard number one is generated by the canonical class.
Segre’s theorem asserts that no cubic surface with Picard number one can be
rational. If this were false, there would be a birational mapφ:S⊂⊂→P
2
k
deﬁned
overk,given by some mobile linear system. Because the Picard group is
generated by the hyperplane classH, the linear systemmust be contained in
the complete linear system|dH|for somed. Therefore, the proof of Segre’s
theorem (Theorem 2.1) will be complete upon proving the following theorem.
Theorem2.13. If S⊂P
3
k
is a smooth cubic surface, then there is no mobile
linear system on S contained in|dH|that deﬁnes a birational map to the
projective plane.
Although the statement of this theorem is less appealing than Segre’s theo-
rem, we have, in effect, reduced the proof of Segre’s theorem to the case where
the ground ﬁeld is algebraically closed: if such a linear system is deﬁned over
k, then it is also deﬁned over the algebraic closure ofk. Note that a naive re-
duction of Theorem 2.1 to the algebraically closed ﬁeld case is not possible, as
the Picard number is never one over an algebraically closed ﬁeld.
Proof of Theorem2.13. Suppose that such a linear system,,exists
and deﬁnes the birational mapφ
:S⊂⊂→P
2
k
.Without loss of generality, we
assumekis algebraically closed, as explained above.
LetP
1,... ,P rbe the base points of, including the inﬁnitely near ones,
and letm
1,... ,m rbe their multiplicities. We ﬁrst claim that some base point
must have multiplicity greater thand. Indeed, from Theorem 2.9 and the fact
thatK
S=−H,wehave
ρ
m
2
i
=
2
−1=3d
2
−1 and
ρ
m
i=−K S·−3=3d−3.
(2.13.1)
If allm
iare less than or equal tod, then
3d
2
−1=
⊂
m
2
i
≤d
⊂
m i=3d
2
−3d<3d
2
−1.

2.3 The proofs of the theorems of Segre and Manin43
This contradiction ensures that at least one base point has multiplicity greater
thand.
LetPbe a base point ofof multiplicity greater thand. There is no loss
of generality in assuming thatP∈S, that is, thatPis an actual base point, not
an inﬁnitely near one. This is because the multiplicity of a base point is greater
than or equal to the multiplicity of any base point inﬁnitely near it. (In fact, the
multiplicity ofP∈Sas a base point of the linear systemis greater than or
equal to the sum of the multiplicities of all base points of

which lie overP,
where

is the birational transform of the linear systemunder the blowing
up map ofSatP.)
Furthermore, the high multiplicity base pointPcan not lie on any line on
S. Indeed, since⊂|dH|,wemust have thatL·C≤dfor all linesLonS
and allC∈. ComputingC·Las the sum over all points (with multiplicities)
inC∩L,wesee thatCcan not have a multiple point of order more thand
onL.
The proof of Theorem 2.13 proceeds by induction ond. The inductive step
is accomplished by ﬁnding a birational self-map ofS—in fact, a birational
involution—that takesto a linear system contained in the linear system|d

H|
withd

<d.Wenowconstruct this involution.
2.14.An involution of the cubic surface.First recall the following
involution of a plane cubic curveE: ﬁxing a pointPonE, deﬁne the mapτ
which sendsQ∈Eto the third point of intersection ofEwith
PQ. The map
τextends to an involution deﬁned everywhere onEby sending the pointPto
the intersection ofEwith the tangent line throughP.
Weattempt to construct a similar involution of the cubic surfaceSinP
3
.
Deﬁne a self-mapτofSas follows: ﬁx a pointPonSand for eachQinS,
letτ(Q)bethe third point of intersection ofSwith the line
PQ. This deﬁnes
a rational mapτ:S⊂⊂→Ssuch thatτ
2
=id.Ifweassume thatScontains no
lines throughP, thenτis deﬁned everywhere onS,except atP.However,
unlike the situation of the plane cubic, there is a whole plane of tangent lines to
SatP,sothere is no way to extendτto a morphism atP. Indeed,τcontracts
the entire curveD=T
PS∩Sto the pointPonS.
As usual, the best way to sort out different tangent directions at a point is to
blow up. Letπ:S

→Sbe the blowup ofSatP.Obviously, the involutionτ
extends to an involution ˜τofS

: each point in the exceptional ﬁber corresponds
to a tangent direction through the pointPonS, and so there is a unique third
point of intersection of the corresponding line withS.
Tounderstand ˜τbetter, consider the following construction. By deﬁni-
tion, the blowup ofP
3
atPconsists of those points (x, )inP
3
×P
2
, where

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Treysa and informed the chief of police, who showed him a
photograph of the prisoner—a copy of one circulated by
Scotland Yard.
“And do you see in Court the man who stole the pendant?”
asked the magistrate.
“Yes; he is there,” the Jew replied in German—“the younger
of the two.”
“You have not recovered your property?”
“No, sir.”
The court was not crowded. The London public take little or
no interest in the Extradition Court. The magistrate glanced
across at the well-dressed lady in dark grey who sat alone
upon one of the benches, and wondered who she might be.
Afterwards one of the detectives informed him privately that
she had been with the men at Worthing when they were
arrested.
“I do not know, your Worship, if you require any further
evidence,” exclaimed Mr Gore-Palmer, again rising. “Perhaps
you will glance at the evidence taken on commission before
the British Consul-General at Treysa, the British Consul in
Hamburg, and the British Vice-Consul at Cologne. I venture
to think that in face of the evidence of identification you
have just heard, you will be convinced that the German
Government have a just right to apply for the extradition of
these two persons.”
He then resumed his seat, while the white-headed old
gentleman on the bench carefully went through folio after
folio of the signed and stamped documents, each with its
certified English translation and green Consular stamps.

Presently, when about half-way through the documents, he
removed his gold pince-nez, and looking across at counsel,
asked,—
“Mr Gore-Palmer, I am not quite clear upon one point. For
whom do you appear to prosecute—for the Imperial German
Government, or for the Ministry of Foreign Affairs of the
Kingdom of Marburg?”
“I appear for both, your Worship, but I am instructed by the
latter.”
“By the Minister Stuhlmann himself, on behalf of the
Government—not by Herr Hirsch?”
“Yes, your Worship, by the Minister himself, who is
determined to crush out the continually increasing crimes
committed by foreign criminals who enter the Kingdom in
the guise of tourists, as in the case of the present
prisoners.”
Claire, when counsel’s explanation fell upon her ears, sat
upright, pale and rigid.
She recollected Steinbach’s warning, and in an instant the
vile, dastardly plot of Hinckeldeym and his creatures
became revealed to her.
They would condemn this man to whom she owed her life
as a low-bred thief, and at the same time declare that he
was her latest lover!
For her it was the end of all things—the very end!

Chapter Twenty Six.
Place and Power.
The grey-faced London magistrate had remanded the
prisoners in custody for seven days, and the papers that
evening gave a brief account of the proceedings under the
heading: “Smart Capture of Alleged Jewel-Thieves.”
During the return journey to Worthing Claire remained
almost silent at Leucha’s side. The girl, whose gallant lover
had thus been snatched from her so cruelly, was beside
herself in utter dejection and brokenness of heart. Surely
they were a downcast pair, seated in the corners of an
empty first-class carriage on the way back to the seaside
town which possessed no further charm for them.
To Claire the plot was now revealed as clear as day. She
had, however, never dreamed that Hinckeldeym and
Stuhlmann would descend to such depths of villainy as this.
Their spies had been at work, without a doubt. She had
been watched, and the watchers, whoever they were, had
evidently established the identity of the two men to whom
she owed so very much. And then Hinckeldeym, with that
brutal unscrupulousness that distinguished him, had
conceived the hellish plot to create a fresh scandal
regarding the jewel thief Guy Bourne and herself.
The man who had risked his life for hers had now lost his
liberty solely on her account. It was cruel, unjust, inhuman!
Night and day she had prayed to her Maker for peace and
for protection from the thousand pitfalls that beset her path
in that great complex world of which she was almost as
ignorant as little Ignatia herself. Yet it seemed as though,

on the contrary, she was slowly drifting on and on to a ruin
that was irreparable and complete.
She felt herself doubting, but instantly her strong faith
reasserted itself. Yes, God would hear her; she was sure He
would. She was a miserable sinner, like all other women,
even though she were queen of an earthly kingdom. He
would forgive her; He would also forgive those two men
who stood charged with the crime of theft. God was just,
and in Him she still placed her implicit trust. In silence, as
the train rushed southward, she again appealed to Him for
His comfort and His guidance.
Her bounden duty was to try and save the men who had
been her friends, even at risk to herself. Their friendliness
with her had been their own betrayal. Had they disappeared
from Paris with her jewels they would still have been at
liberty.
Yet what could she do? how could she act?
Twenty years’ penal servitude was the sentence which
Leucha declared would be given her father if tried in
England, while upon Bourne the sentence would not be less
than fifteen years, having in view his list of previous
convictions. In Germany, with the present-day prejudice
against the English, they would probably be given even
heavier sentences, for, according to Mr Gore-Palmer, an
attempt was to be made to make an example of them.
Ah! if the world only knew how kind, how generous those
two criminals had been to her, a friendless, unhappy
woman, who knew no more of the world than a child in her
teens, would it really judge them harshly, she wondered. Or
would they receive from the public that deep-felt
compassion which she herself had shown them?

Many good qualities are, alas! nowadays dead in the human
heart; but happily chivalry towards a lonely woman is still,
even in this twentieth century, one of the traits of the
Englishman’s character, be he gentleman or costermonger.
Alone in her room that night, she knelt beside the bed
where little Ignatia was sleeping so peacefully, and
besought the Almighty to protect her and her child from this
last and foulest plot of her enemies, and to comfort those
who had been her friends. Long and earnestly she remained
in prayer, her hands clasped, her face uplifted, her white
lips moving in humble, fervent appeal to God.
Then when she rose up she pushed back the mass of fair
hair from her brow, and paced the room for a long time,
pondering deeply, but discerning no way out of the
difficulties and perils that now beset her. The two accused
men would be condemned, while upon her would be heaped
the greatest shame that could be cast upon a woman.
Suddenly she halted at the window, and leaning forward,
looked out upon the flashing light far away across the dark,
lonely sea. Beyond that far-off horizon, mysterious in the
obscurity of night, lay the Continent, with her own Kingdom
within. Though freedom was so delightful, without Court
etiquette and without Court shams, yet her duty to her
people was, she recollected, to be beside her husband; her
duty to her child was to live that life to which she, as an
Imperial Archduchess, had been born, no matter how
irksome it might be to her.
Should she risk all and return to Treysa? The very
suggestion caused her to hold her breath. Her face was pale
and pensive in her silent, lofty, uncomplaining despair.

Would her husband receive her? Or would he, at the
instigation of old Hinckeldeym and his creatures, hound her
out of the Kingdom as what the liars at Court had falsely
declared her to be?
Again she implored the direction of the Almighty, sinking
humbly upon her knees before the crucifix she had placed
at the head of her bed, remaining there for fully a quarter
of an hour.
Then when she rose again there was a calm, determined
look on her pale, hard-set face.
Yes; her patience and womanhood could endure no longer.
She would take Leucha and go fearlessly to Treysa, to face
her false friends and ruthless enemies. They would start to-
morrow. Not a moment was to be lost. And instead of
retiring to bed, she spent the greater part of the night in
packing her trunks in readiness for the journey which was
to decide her fate.
The summer’s evening was breathless and stifling in Treysa.
Attired in Henriette’s coat and skirt, and wearing her thick
lace veil, Claire alighted from the dusty wagon-lit that had
brought her from Cologne, and stood upon the great, well-
remembered platform unrecognised.
The douaniers at the frontier had overhauled her baggage;
the railway officials had clipped her tickets; the wagon-lit
conductor had treated her with the same quiet courtesy that
he had shown to her fellow-passengers, and she had passed
right into the splendid capital without a single person
recognising that the Queen—“their Claire”—had returned
among them.

Leucha descended with Ignatia, who at once became
excited at hearing her native tongue again; and as they
stood awaiting their hand-baggage an agent of police
passed them, but even he did not recognise in the neat-
waisted figure the brilliant and beautiful soft-eyed woman
who was his sovereign.
At first she held her breath, trembling lest she might be
recognised, and premature information of her return be
conveyed to Hinckeldeym or to the Prefect of Police, who,
no doubt, had his orders to refuse her admittance. Yet
finding her disguise so absolutely complete, she took
courage, and passed out of the station to hail a closed cab.
They were all three utterly tired out after thirty-six hours of
rail, crossing by way of Dover and Ostend.
When Leucha and Ignatia had entered the cab she said to
the man sharply, in German,—
“Drive to the royal palace.”
The man, who took her for one of the servants, settled
himself upon his box and drove up the straight tree-lined
avenue to the great entrance gates of the royal park, which
were, as usual, closed.
As they approached them, however, her Majesty raised her
veil, and waited; while Leucha, with little Ignatia upon her
knee, sat wondering. She, “the Ladybird,” the accomplice of
the cleverest gang of thieves in Europe, was actually
entering a royal palace as intimate friend of its Queen!
The cab halted, the sentries drew up at attention, and the
gorgeous porter came forward and put in his head
inquisitively.

Next instant he recognised who it was, and started back;
then, raising his cocked hat and bowing low, gave orders to
the cabman to drive on. Afterwards, utterly amazed, he
went to the telephone to apprise the porter up at the palace
that her Majesty the Queen had actually returned.
When they drew up at the great marble steps before the
palace entrance, the gaudily-dressed porter stood bare-
headed with three other men-servants and the two agents
of police who were always on duty there.
All bowed low, saluting their Queen in respectful silence as
she descended, and Leucha followed her with the little
Princess toddling at her side. It was a ceremonious arrival,
but not a single word was uttered until Claire passed into
the hall, and was about to ascend the grand staircase on
her way to the royal private apartments; for she supposed,
and quite rightly, that her husband had, on his accession,
moved across to the fine suite occupied by his late father.
Bowing slightly to acknowledge the obeisance of the
servants, she was about to ascend the broad stairs, when
the porter came forward, and said apologetically,—
“Will your Majesty pardon me? I have orders from the
Minister Hinckeldeym to say that he is waiting in the blue
anteroom, and wishes to see you instantly upon your
arrival.”
“Then he knows of my return?” she exclaimed surprised.
“Your Majesty was expected by him since yesterday.” She
saw that his spies had telegraphed news of her departure
from London.
“And the King is in the palace?”

“Yes, your Majesty; he is in his private cabinet,” responded
the man, bowing.
“Then I will go to him. I will see Hinckeldeym afterwards.”
“But, your Majesty, I have strict orders not to allow your
Majesty to pass until you have seen his Excellency. See,
here he comes!”
And as she turned she saw approaching up the long marble
hall a fat man, her arch-enemy, attired in funereal, black.
“Your Majesty!” he said, bowing, while an evil smile played
upon his lips. “So you have returned to us at Treysa! Before
seeing the King I wish to speak to you in private.”
Deadly and inexorable malice was in his countenance. She
turned upon him with a quick fire in her eyes, answering
with that hauteur that is inherent in the Hapsbourg blood,—
“Whatever you have to say can surely be said here. You can
have nothing concerning me to conceal!” she added
meaningly.
“I have something to say that cannot be said before the
palace servants,” he exclaimed quickly. “I forbid you to go
to the King before I have had an opportunity of explaining
certain matters.”
“Oh! you forbid—you?” she cried, turning upon him in
resentment at his laconic insolence. “And pray, who are
you?—a mere paid puppet of the State, a political
adventurer who discerns further advancement by being my
enemy! And you forbid?”
“Your Majesty—I—”

“Yes; when addressing me do not forget that I am your
Queen,” she said firmly, “and that I know very well how to
deal with those who have endeavoured to encompass my
ruin. Now go to your fellow-adventurers, Stuhlmann,
Hoepfner, and the rest, and give them my message.” Every
word of hers seemed to blister where it fell. Then turning to
Leucha, she said in English,—
“Remain here with Ignatia. I will return to you presently.”
And while the fat-faced officer of State who had so
ingeniously plotted her downfall stood abashed in silence,
and confused at her defiance, she swept past him, mounted
the stairs haughtily, and turning into the corridor, made her
way to the royal apartments.
Outside the door of the King’s private cabinet—that room
wherein Hinckeldeym had introduced his spies—she held her
breath. She was helpless at once, and desperate. Her hand
trembled upon the door knob, and the sentry, recognising
her, started, and stood at attention.
With sudden resolve she turned the handle, and next
second stood erect in the presence of her husband.

Chapter Twenty Seven.
A Woman’s Words.
The King sprang up from his writing-table as though
electrified.
“You!” he gasped, turning pale and glaring at her—“you,
Claire! Why are you here?” he demanded angrily.
“To speak with you, Ferdinand. That scheming reptile
Hinckeldeym forbade me to see you; but I have defied him
—and have come to you.”
“Forbade you! why?” he asked, in a deep voice, facing her,
and at once noticing that she was disguised as Henriette.
“Because he fears that I may expose his ingenious intrigue
to you. I have discovered everything, and I have come to
you, my husband, to face you, and to answer any charges
that this man may bring against me. I only ask for justice,”
she added, in a low, earnest voice. “I appeal to you for that,
for the sake of our little Ignatia; for the sake of my own
good name, not as Queen, but as a woman! ”
“Then Hinckeldeym was aware that you were returning?”
“His spies, no doubt, telegraphed information that I had left
London. He was awaiting me in the blue anteroom when I
arrived, ten minutes ago.”
“He told me nothing,” her husband remarked gruffly,
knitting his brows in marked displeasure.

“Because he fears the revelation of his dastardly plot to
separate us, and to hurl me down to the lowest depths of
infamy and shame.”
Her husband was silent; his eyes were fixed upon hers.
Only yesterday he had called Meyer, the Minister of Justice,
and given orders for an application to the Court for a
divorce. Hinckeldeym, by continually pointing out the
Imperial displeasure in Vienna, had forced him to take this
step. He had refrained as long as he could, but at last had
been forced to yield.
As far as government was concerned, Hinckeldeym was, he
considered, an excellent Minister; yet since that night when
the man had introduced his spies, he had had his shrewd
suspicions aroused that all he had told him concerning
Claire was not the exact truth. Perhaps, after all, he had
harshly misjudged her. Such, indeed, was the serious
thought that had a thousand times of late been uppermost
in his mind—ever since, indeed, he had given audience to
the Minister Meyer on the previous morning.
Claire went on, shining forth all her sweet, womanly self.
Her intellectual powers, her elevated sense of religion, her
high honourable principles, her best feelings as a woman,
all were displayed. She maintained at first a calm self-
command, as one sure of carrying her point in the end; and
yet there was, nevertheless, a painful, heart-thrilling
uncertainty. In her appeal, however, was an irresistible and
solemn pathos, which, falling upon her husband’s heart,
caused him to wonder, and to stand open-mouthed before
her.
“You allege, then, that all this outrageous scandal that has
been the talk of Europe has been merely invented by

Hinckeldeym and his friends?” asked the King, folding his
arms firmly and fixing his eyes upon his wife very seriously.
“I only ask you, Ferdinand, to hear the truth, and as
Sovereign to render justice where justice is due,” was her
calm response, her pale face turned to his. “I was too proud
in my own honesty as your wife to appeal to you: indeed, I
saw that it was hopeless, so utterly had you fallen beneath
the influence of my enemies. So I preferred to leave the
Court, and to live incognito as an ordinary person.”
“But you left Treysa with Leitolf, the man who was your
lover! You can’t deny that, eh?” he snapped.
“I deny it, totally and emphatically,” was her response,
facing him unflinchingly. “Carl Leitolf loved me when I was a
child, but years before my marriage with you I had ceased
to entertain any affection for him. He, however, remained
my friend—and he is still my friend.”
“Then you don’t deny that to-day he is really your friend?”
he said, with veiled sarcasm.
“Why should I? Surely there is nothing disgraceful that a
man should show friendliness and sympathy towards a
woman who yearns for her husband’s love, and is lonely and
unhappy, as I have been? Again, I did not leave Treysa with
him. He joined my train quite by accident, and we travelled
to Vienna together. He left me at the station, and I have not
seen him since.”
“When you were in Vienna, a few days before, you actually
visited him at his hotel?”
“Certainly; I went to see him just as I should call upon any
other friend. I recognised the plot against us, and arranged
with him that he should leave the Court and go to Rome.”

“I don’t approve of such friends,” he snapped again quickly.
“A husband should always choose his wife’s male friends. I
am entirely in your hands, Ferdinand.”
“But surely you know that a thousand and one scandalous
stories have been whispered about you—not only in the
palace, but actually among the people. The papers, even,
have hinted at your disgraceful and outrageous behaviour.”
“And I have nothing whatever to be ashamed of. You, my
husband, I face boldly to-night, and declare to you that I
have never, for one single moment, forgotten my duty
either to you or to our child,” she said, in a very low, firm
voice, hot tears at that moment welling in her beautiful
eyes. “I am here to declare my innocence—to demand of
you justice, Ferdinand!”
His lips were pressed together. He was watching her
intently, noticing how very earnestly and how very boldly
she refuted those statements which, in his entire ignorance
of the conspiracy, he had believed to be scandalous truths.
Was it really possible that she, his wife, whom all Europe
had admired for her grace, her sweetness, and her
extraordinary beauty, was actually a victim of a deeply-laid
plot of Hinckeldeym’s? To him it seemed utterly impossible.
She was endeavouring, perhaps, to shield herself by making
these counter allegations. A man, he reflected, seldom gets
even with a woman’s ingenuity.
“Hinckeldeym has recently revealed to me something else,
Claire,” he said, speaking very slowly, his eyes still fixed
upon hers—“the existence of another lover, an interesting
person who, it appears, is a criminal!”
“Listen, Ferdinand, and I will tell you the truth—the whole
truth,” she said very earnestly. “You will remember the

narrow escape I had that day when my cob shied at a motor
car and ran away, and a stranger—an Englishman—stopped
the animal, and was so terribly injured that he had to be
conveyed to the hospital, and remained there some weeks
in a very precarious state. And he afterwards disappeared,
without waiting for me to thank him personally?”
“Yes; I remember hearing something about him.”
“It is that man—the criminal,” she declared; and then, in
quick, breathless sentences, she explained how her jewels
had been stolen in Paris, and how, when the thieves knew
of her identity, the bag had been restored to her intact. He
listened to every word in silence, wondering. The series of
romantic incidents held him surprised. They were really
gallant and gentlemanly thieves, if—if nothing else, he
declared.
“To this Mr Bourne I owe my life,” she said; “and to him I
also owe the return of my jewels. Is it, therefore, any
wonder when these two men, Bourne and Redmayne, have
showed me such consideration, that, lonely as I am, I
should regard them as friends? I have Redmayne’s daughter
with me here, as maid. She is below, with Ignatia. It is this
Mr Bourne, who is engaged to be married to Leucha
Redmayne, that Hinckeldeym seeks to denounce as my
lover!”
“He says that both men are guilty of theft within the
Empire; indeed, Bourne is, it is said, guilty of jewel robbery
in Eugendorf.”
“They have both been arrested at Hinckeldeym’s instigation,
and are now in London, remanded before being extradited
here.”
“Oh! he has not lost very much time, it seems.”

“No. His intention is that Mr Bourne shall stand his trial
here, in Treysa, and at the same time the prisoner is to be
denounced by inspired articles in the press as my lover—
that I, Queen of Marburg, have allied myself with a common
criminal! Cannot you see his dastardly intention? He means
that this, his last blow, his master stroke, shall crush me,
and break my power for ever,” she cried desperately. “You,
Ferdinand, will give me justice—I know you will! I am still
your wife!” she implored. “You will not allow their foul lies
and insinuations to influence you further; will you?” she
asked. “In order to debase me in your eyes and in the eyes
of all Europe, Hinckeldeym has caused the arrest of this
man to whom I owe my life—the man who saved me, not
because I was Crown Princess, but because I was merely a
woman in peril. Think what betrayal and arrest means to
these men. It means long terms of imprisonment to both.
And why? Merely in order to attack me—because I am their
friend. They may be guilty of theft—indeed they admit they
are; nevertheless I ask you to give them your clemency,
and to save them. You can have them brought here for trial;
and there are ways, technicalities of the law, or something,
by which their release can be secured. A King may act as he
chooses in his own Kingdom.”
Every word she spoke was so worthy of herself, so full of
sentiment and beauty, poetry and passion. Too naturally
frank for disguise, too modest to confess her depth of love
while the issue remained in suspense, it was a conflict
between love and fear and dignity.
“I think you ask me rather too much, Claire,” he said, in a
somewhat quieter tone. “You ask me to believe all that you
tell me, without giving me any proof whatsoever.”
“And how can I give you proof when Mr Bourne and his
friend are in custody in London? Let them be extradited to

Treysa, and then you may have them brought before you
privately and questioned.”
For some moments he did not speak. What she had just
alleged had placed upon the matter an entirely different
aspect. Indeed, within himself he was compelled to admit
that the suspicions he had lately entertained regarding
Hinckeldeym had now been considerably increased by her
surprising statements. Was she speaking the truth?
Whenever he allowed his mind to wander back he
recollected that it had been the crafty old President who had
first aroused those fierce jealous thoughts within his heart.
It was he who had made those allegations against Leitolf;
he who, from the very first weeks of his marriage, had
treated Claire with marked antipathy, although to her face
he had shown such cordiality and deep obeisance that she
had actually believed him to be her friend. Yes, he now
recognised that this old man, in whom his father had
reposed such perfect confidence, had been the fount of all
those reports that had scandalised Europe. If his calm,
sweet-faced wife had, after all, been a really good and
faithful woman, then he had acted as an outrageous brute
to her. His own cruelty pricked his conscience. It was for her
to forgive, not for her to seek forgiveness.
She saw his hesitation, and believed it due to a reluctance
to accept her allegations as the real truth.
“If you doubt me, Ferdinand, call Hinckeldeym at this
instant. Let me face this man before you, and let me
categorically deny all the false charges which he and his
sycophants have from time to time laid against me. Here, at
Court, I am feared, because they know that I am aware of
all my secret enemies. Make a clearance of them all and
commence afresh,” she urged, a sweet light in her

wonderful eyes. “You have clever men about you who would
make honest and excellent Ministers; but while you are
surrounded by such conspirators as these, neither you nor
the throne itself is safe. I know,” she went on breathlessly,
“that you have been seized by a terrible jealousy—a cruel,
consuming jealousy, purposely aroused against me in order
to bring about the result which was but the natural outcome
—my exile from Treysa and our estrangement. It is true
that you did not treat me kindly—that you struck me—that
you insulted me—that you have disfigured me by your
blows; but recollect, I beg, that I have never once
complained. I never once revealed the secret of my dire
unhappiness; only to one man, the man who has been my
friend ever since my childhood—Carl Leitolf. And if you had
been in my place, Ferdinand, I ask whether you would not
have sought comfort in relating your unhappiness to a
friend. I ask you that question,” she added, in a low,
intense, trembling voice. “For all your unkindness and
neglect I have long ago really forgiven you. I have prayed
earnestly to God that He would open your eyes and show
me in my true light—a faithful wife. I leave it to Him to be
my judge, and to deal out to my enemies the justice they
deserve.”
“Claire!” he cried, suddenly taking her slim white hand in his
and looking fiercely into her beautiful eyes, “is this the real
truth that you have just told me?”
“It is!” she answered firmly; “before God, I swear that it is!
I am a poor sinner in His sight, but as your wife I have
nothing with which to reproach myself—nothing. If you
doubt me, then call Leitolf from Rome; call Bourne. Both
men, instead of being my lovers, are your friends—and
mine. I can look both you and them in the face without
flinching, and am ready to do so whenever it is your will.”

All was consummated in that one final touch of truth and
nature. The consciousness of her own worth and integrity
which had sustained her through all her trials of heart, and
that pride of station for which she had contended through
long years—which had become more dear by opposition and
by the perseverance with which she had asserted it—
remained the last strong feeling upon her mind even at that
moment, the most fateful crisis of her existence.
Her earnest, fearless frankness impressed him. Was it really
possible that his wife—this calm-faced woman who had
been condemned by him everywhere, and against whom he
had already commenced proceedings for a divorce—was
really, after all, quite innocent?
He remembered Hinckeldeym’s foul allegations, the
damning evidence of his spies, the copies of certain letters.
Was all this a tissue of fraud, falsehood, and forgery?
In a few rapid words she went on to relate how, in that
moment of resentment at such scandalous gossip being
propagated concerning her, she had threatened that when
she became Queen she would change the whole entourage,
and in a brief, pointed argument she showed him the strong
motive with which the evil-eyed President of the Council
had formed the dastardly conspiracy against her.
“Claire,” he asked, still holding her soft hand with the
wedding ring upon it, “after all that has passed—after all my
harsh, inhuman cruelty to you—can you really love me still?
Do you really entertain one single spark of love for me?”
“Love you!” she cried, throwing herself into his arms in a
passion of tears; “love you, Ferdinand!” she sobbed. “Why,
you are my husband; whom else have I to love, besides our
child?”

“Then I will break up this damnable conspiracy against you,”
he said determinedly. “I—the King—will seek out and punish
all who have plotted against my happiness and yours. They
shall be shown no mercy; they shall all be swept into
obscurity and ruin. They thought,” he added, in a hard,
hoarse voice, “to retain their positions at Court by keeping
us apart, because they knew that you had discovered their
despicable duplicity. Leave them to me; Ferdinand of
Marburg knows well how to redress a wrong, especially one
which concerns his wife’s honour,” and he ran his hand over
his wife’s soft hair as he bent and kissed her lips.
So overcome with emotion was she that at the moment she
could not speak. God had at last answered those fervent
appeals that she had made ever since the first year of their
marriage.
“I have wronged you, Claire—deeply, very deeply wronged
you,” he went on, in a husky, apologetic voice, his arm
tenderly about her waist, as he again pressed his lips to
hers in reconciliation. “But it was the fault of others. They
lied to me; they exaggerated facts and manufactured
evidence, and I foolishly believed them. Yet now that you
have lifted the scales from my eyes, the whole of their
devilishly clever intrigue stands plainly revealed. It utterly
staggers me. I can only ask you to forgive. Let us from to-
night commence a new life—that sweet, calm life of trust
and love which when we married we both believed was to
be ours for ever, but which, alas! by the interference and
malignity of our enemies, was turned from affection into
hatred and unhappiness.”
“I am ready, Ferdinand,” she answered, a sweet smile
lighting up her beautiful features. “We will bury the past; for
you are King and I am Queen, and surely none shall now
come between us. My happiness to-night, knowing that you

are, after all, good and generous, and that you really love
me truly, no mere words of mine can reveal. Yet even now I
have still a serious thought, a sharp pang of conscience for
those who are doomed to suffer because they acted as my
friends when I was outcast and friendless.”
“You mean the men Bourne and Redmayne,” the King said.
“Yes, they are in a very perilous position. We must press for
their extradition here, and then their release will be easy.
To-morrow you must find some means by which to reassure
them.”
“And Hinckeldeym?”
“Hinckeldeym shall this very night answer to his Sovereign
for the foul lies he has spoken,” replied the King, in a hard,
meaning tone. “But, dearest, think no more of that liar. He
will never cross your path again; I shall take good care of
that. And now,” he said, imprinting a long, lingering caress
upon her white, open brow—“and now let us call up our
little Ignatia and see how the child has grown. An hour ago
I was the saddest man in all the kingdom, Claire; now,” he
laughed, as he kissed her again, “I admit to you I am the
very happiest!”
Their lips met again in a passionate, fervent caress.
On her part she gazed up into his kind, loving eyes with a
rapturous look which was more expressive than words—a
look which told him plainly how deeply she still loved him,
notwithstanding all the bitterness and injustice of the black,
broken past.

Chapter Twenty Eight.
Conclusion.
The greatest flutter of excitement was caused throughout
Germany—and throughout the whole of Europe, for the
matter of that—when it became known through the press
that the Queen of Marburg had returned.
Reuter’s correspondent at Treysa was the first to give the
astounding news to the world, and the world at first
shrugged its shoulders and grinned.
When, however, a few days later, it became known that the
Minister Heinrich Hinckeldeym had been summarily
dismissed from office, his decorations withdrawn, and he
was under arrest for serious peculation from the Royal
Treasury, people began to wonder. Their doubts were,
however, quickly set at rest when the Ministers Stuhlmann
and Hoepfner were also dismissed and disgraced, and a
semi-official statement was published in the Government
Gazette to the effect that the King had discovered that the
charges against his wife were, from beginning to end, a
tissue of false calumnies “invented by certain persons who
sought to profit by her Majesty’s absence from Court.”
And so, by degrees, the reconciliation between the King and
Queen gradually leaked out to the English public through
the columns of their newspapers.
But little did they guess that the extradition case pressed so
very hard at Bow Street last August against the two jewel-
thieves, Redmayne, alias Ward, and Guy Bourne, had any
connection with the great scandal at the Court of Marburg.

The men were extradited, Redmayne to be tried in Berlin
and Bourne at Treysa; but of their sentences history, as
recorded in the daily newspapers, is silent. The truth is that
neither of them was sentenced, but by the private request
of his Majesty, a legal technicality was discovered, which
placed them at liberty.
Both men afterwards had private audience of the King, and
personally received the royal thanks for the kindness they
had shown towards the Queen and to little Ignatia. In order
to mark his appreciation, his Majesty caused a lucrative
appointment in the Ministry of Foreign Affairs, where a
knowledge of English was necessary, to be given to Roddy
Redmayne, while Guy Bourne, through the King’s
recommendation, was appointed to the staff of an important
German bank in New York; and it has been arranged that
next month Leucha—who leaves her Majesty and Ignatia
with much regret—goes to America to marry him. To her
place, as Ignatia’s nurse, the faithful Allen has now
returned, while the false de Trauttenberg, who, instantly
upon Hinckeldeym’s downfall, went to live in Paris, has been
succeeded by the Countess de Langendorf, one of Claire’s
intimate friends of her days at the Vienna Court, prior to
her marriage.
What actually transpired between Hinckeldeym and his
Sovereign on that fateful night will probably never be
known. The people of Treysa are aware, however, that a few
hours after “their Claire’s” return the President of the
Council was commanded to the royal presence, and left it
ruined and disgraced. On the following day he was arrested
in his own mansion by three gendarmes and taken to the
common police-office, where he afterwards attempted
suicide, but was prevented.

The serious charges of peculation against him were, in due
course, proved up to the hilt, and at the present moment he
is undergoing a well-merited sentence of five years’
imprisonment in the common gaol at Eugendorf .
Count Carl Leitolf was recalled from Rome to Treysa a few
days later, and had audience in the King’s private cabinet.
The outcome was, however, entirely different, for the King,
upon the diplomat’s return to Rome, signed a decree
bestowing upon him di moto proprio the Order of Saint
Stephen, one of the highest of the Marburg Orders, as a
signal mark of esteem.
Thus was the public opinion of Europe turned in favour of
the poor, misjudged woman who, although a reigning
sovereign, had, by force of adverse circumstances, actually
resigned her crown, and, accepting favours of the criminal
class as her friends, had found them faithful and devoted.
Of the Ministers of the Kingdom of Marburg only Meyer
retains his portfolio at the present moment, while Steinbach
has been promoted to a very responsible and lucrative
appointment. The others are all in obscurity. Ministers,
chamberlains, dames du palais and dames de la cour, all
have been swept away by a single stroke of the pen, and
others, less prone to intrigue, appointed.
Henriette—the faithful Henriette—part of whose wardrobe
Claire had appropriated on escaping from Treysa, is back
again as her Majesty’s head maid; and though the popular
idea is that little real, genuine love exists between royalties,
yet the King and Queen are probably the very happiest pair
among the millions over whom they rule to-day.
Her Majesty, the womanly woman whose sweet, even
temperament and constant solicitude for the poor and

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Rational And Nearly Rational Varieties Jnos Kollr Karen E Smith

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