Topics in Galois Theory JeanPierre Serre College de France Paris, France
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Topics in Galois Theory JeanPierre Serre College de France Paris, France
Notes written by Henri Damon Mathematics Department Princeton University Princeton, New Jersey
ta �
Jones and Bartlett Publishers Boston
London
Contents Foreword
ix
Notation
xi xiii
Introduction
1
Examples in low degree
1. 1 1.2 1.3
The groups Z/2Z, Z/3Z, and S3' The group 04 •
3
•
•
•
•
•
•
•
•
•
exponent 2) 3,4) 6
6
Nilpotent and solvable groups as Galois groups over Q
2.1 2.2
A theorem of Schols�Reich&rdt. The Frattini subgroup of a. finite
.
. . . .
group
.
. . .. . . . . . .
3.2.1 Extension of scalars 3.2.2 Intersections with linear subvarieties Irreducibility theorem and thin sets . . . . Hilb ert ' s irreducibility theorem ..... . Hil bert property and weak approxima.tion Proofs of prop. 3.5.1 and 3.5.2 ......
3.3 3.4 3.5 3.6
.
.
•
.
.
.
.
Galois extensions of Q(T): first examp1es
4.1 The property GalT . . . . . . . . . 4.2 Abelian groups .. . . . . . . . . . 4.3 Example: the quaternion group q. 4.4 Symmetric groups . . . . . . . . . . 4.5 The alternating group At. 4.6 Finding good specializations of T . .
.
v
.
•
.
.. . . . . . . .
.
9
9 16 19
HUb ert's irred udbility theorem The Hilbert property .......... . Propert ies of thin sets ......... .
3.1 3.2
.
4
•
Application of tori to abelian
Galois groups of
2
•
1 1 2
.
19 21 21 22 23 25 27 30 35
35 36 38 39 43 44
Contents
vi
5
6
Galois extensions of Q(T) given by torsion on elliptic curves
5.1 5.2 5.3 5.4 5.5
An a.uxiliary construction Proof of Shihts theorem. A complement.
Further results
on PSL2(F,.)
SL2(F'l)
and
as Galois
gr oups .
Galois extensions of C(T)
6.1 6.2 6.3 6.4
The GAGA principle
.
.
.
.
.
Coverings of Riemann surfaces
From C to Q .
Appendix:
.
.
.
.
.
.
.
.
. . .
.
universal ramified coverings of Riem&nn
with signature 7
. .
.. . . . . .
. . . .
.
.
.
.
.
. .
.
.
. ..
.
.
.
.
.
.
surfaces .
Counting solutions of equa.tions in finite groups
of & family of conjugacy classes .. .. . . The symmetric group Sft The alterna.ting group As The groups PSL2(Fp) The group SL2(Fa) .. . The J anko group J1 The HallJanko group J2
Rigidity
7.4.1 7.4.2 7.4.3 7.4.4
7.4.5 7.4.6 7.4.7
.
.
•
•
The main theorem
M
First variant .
8.4.2 8.4.3 8.4.4
.
The symmetric group 8ft The alternating group The group
As
.
PSL2(Fp) ...
The GalT property for the smallest si mple groups
Local properties .
8.4.1
.
Second variant
Examples ......
8.3.1 8.3.2 8.3.3 8.3.4 8.4:
81
.
Two variants .....
8.2.1 8.2.2 8.3
.
.
Preliminaries
.
.
.
.
.
.
.
.
.
.
.
_
.
.
.
.
.
.
.
A problem on good reduction
The
real case
.
.
.
.
.
.
.
..
The padic case: a theorem of Harbater
60
78 78
8 Construction of Galois extensions of Q(T) by the rigidity method
8.1 8.2
.
77
•
The FischerGriess Monster
.
65 67 70 72 72 73 74 75
.
Examples of rigidity
.
55 55 57 57
65
Rigidity and rationality on finite groups 7.1 Rationality .. .. . . . . . . . . . . .
7.2 7.3 7.4
.
47
47 48 49 52 53
Statement of Shih's theorem .
.
81 83 83 84 85 85 86 87 88 88 88 89 90
93
vii
ContentB
9
form Tr(x2) and its applications Preliminaries ............ 9.1.1 Galois cohomology (mod 2) 9.1.2 Quadratic forma. . 9. L3 Cohomology of 8ft . 9�2 The quadratic form Tr (Z2) 9.3 Application to extensions with Galois group A,.
95
The
9.1
95 95 95 97 98
.
.
.
.
.
•
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
.
.
.
.
10 Appendix: the large sieve inequality 10.1 Statement of the theorem .
•
.
.
.
.
99 103
10.2 A lemma on finite groups. . . . . . . . 10.3 The Davenport.. Halberatam theorem 10.4 Combining the information . . .
.
103 105 105 107
Bibliography
109
Index
117
Foreword These notes are based on "Topics in Galois Theoryt It
Serre
a.
course given by
JP.
at Harvard University in the Fall semester of 1988 and written down by
H. Darmon. The coutse focused on the inverse problem of Galois theory: the construction of field extensions having & given finite group G as Galois group, typically over Q but also over fields such as Q(T). Chapter
1
discusses examples for certain groups G of small order.
The
method of Scholz and Reichardt) which works over Q when G is a. pgroup
of odd order, is given in chapter 2. Chapter
3
is devoted to the Hilbert irre
ducibility theorem and its connection with weak approximation and the large sieve inequality. Chapters 4: a.nd
5
describe methods for showing tha.t G is
the Galois group of a regular extension of Q(T) ( one then says that G has property GalT)' Elementary constructions (c.g. when G is a symmctric or alternating group) are given in chapter 4:t while the method of Shih, which works for G = PSL1(P) in some cases, is outlined in chapter
5.
Chapter
6
describes the GAGA principle and the relation bctween the topological and
algebraic fundamental groups of complex curves. Chapters 7 and 8 are devoted to the ra.tionality and rigidity criterions and their application to proving the
property GalT for certain groups (notably, many of the sporadic simple groups,
including the FischerGriess Monster). The relation between the HasseWitt invariant of the quadratic form Tr (:z:2) and certain embedding problems is the
topic of c hapter 9, and
an
a.pplication to showing that
is given. An appendix (chapter
used in cha.pter
3.
10)
An
haa property GalT
gives a proof of the large sieve inequality
The reader should be warned that most proofs only give the main ideas; details have been left out. Moreover, a number of relevant topics ha.ve
been
omitted, for lack of time (and understanding), namely: a.) Thc theory of generic extensions, d. [Sal].
b) Shafa.revich's theorem On the existence of extensions of Q with
solva.ble Galois group,
d.
a
given
[ILK).
c) The Hurwitz schemes which parametrize extensions with a given Galois group and
a.
given ramification structure,
d.
[Frl], [Fr21, [Ma.31.
d) The computation of explicit equations for extensions with Galois group
ix
Foreword
x
d. [LMJ, [Ma3]J [Ma,4J, [MIl], ... published) on extensions of Q(T) with group 6·�, 6· A." and SL2(F.,). PSL2 (F1)t SL2(F.), MUJ
• •
"'
e) Mestre's results (not yet
We
wish to thank
Larry Washingt on for his helpful comments on an earlier
version of these notes. Paris, August
Galois
1991 H. Darmon
JP. Serre
Notation If V is an algebraic variety over the field K. and L is an extension of K, we denote by V(L) the set of Lpoints of V and by l'JL the Lvariety obtained from V by base change from K to L. All the varieties are supposed reduced and quasi·projective. A'" is the affine n·space; A'" ( L) = Lft. Pn is the projective nspa.cej Pn(L) = (L(ft+l)  {O})/L·; the group of a.utomorphisms of Pta is PGLta = GLn/Gm• If X is a finite set, IXI denotes the cardinality of X.
xi
Introduction The question of whether all finite groups can occur as Galois groups of an extension of the rational s (known as the inverse problem of Galois theory) is still unsolved, in spite of substantial progress in recent years. In the 1 9 30 ' s , Emmy Noether proposed the f ollowing stra tegy to attack the inverse problem [Noel: by embedding G in the
S�h
permuta.tion group on n. Q(X). Let E be Galois extension of E with
the
one defines a G·action on the field Q,X1, fixed field under this action. Then Q(X) is a
letters,
• • •
,Xft)
=
Galois group G. In g eometri c terms, the
extension Q(X) of E corresponds to the projection of varieties: 'K : A" + A"/G, where Aft is a.ffine nspace over Q. Let P be a. Qrational point of Aft /0 for which W' is unramified, and lift it to Q E An(Q). The conjugates of Q under the action of Ga.l(Q/Q) are the sQ where s E Hq C G, and Hq is the decomposition group at Q. If Hq G, then Q genera.tes a field extension of Q with Galois group G. A variety is said to be rational over Q 'or Q rationBl) if it is bira.tiona.lly =
isomorphic over
Q
to the
affine space An
function field is isomorphic to Q(T 1,
• • •
for some
n.,
or equivalently,
[Hi]) If An IG is QTfltio'R4l, then th�Te above such tha.t Hq = O.
Theorem 1 (Hilbert,
ma.ny points P, Q
as
This follows from
Hilbert's
if its
, Tft}, where the T" are indetermina.tes.
irreducibility theorem, d.
Gre
infinitely
§3.4.
Let 0 = Sft, acting on Q(Xh .. . , X,,). The field E of Sninvariants Tn), where Ti is the ith symmetric polynomial, and Q(X1J"" Xn) has Galois group S" over E: it is the splitting field of the polynomial 2 X"'  T1X"1 + T2X,, + ... + (1)ftT". Hilbert'! irreducibility theorem says tha.t the Ti can be specialized to infinitely many values ta E Q (or even ti E Z) BUch that the equation xn t1X"1 + t,X"' + ... + (1)"� = 0
Example:
is Q(T1,
• • •
)
�
has Galois group Sft OVer Q. In fact, "most" ti with t, E Z, 1 � t, � NJ only O(Nft\ log N) {Coh], [Se9]. xiii
work: of ma.y
the N" ntuples (t,)
fail to give 8ft1 d. [Gal,
Introduction
xiv
In addition to the symmetric groups, the method works for the alternating
groups An wi th
n
5 5, cf. [Mae] ( For
n � 6,
it is not known whether the
field of Aninvariants is rational.) Somewhat surprisingly, there are groups for which the method fails (i.e. An/G is not Q.rational): •
Swan [Swl] has shown that the field of Ginvariants is no t rational when
G
is a cyclic group of order 47. The obstruction is related to the automorphism
group of G which is a cyclic group of order Q((23) does not have class number •
In
1
(since
46 = 2 X 23, and h( 23) 3) .
to the fact that
=
(Le] H. Lenstra gives a general criteri on for the field of G·invariants to be when G is an abelian group: in particular, he shows that this criterion
rational
is not satisfied when G is cyclic of order
8.
(The a.bove counterexamples are over Q. Counter�exa.mples over C (involving
a nonabelian group G) are given by the following result of Saltman [Sa2]: if
there is a nonzero subgroups
H
a
E
H'(G, Q/Z)
ResS} (a) = 0 for all abelian An /G is not C  rati onal . It is
such that
generated by two elements, then
G satisfying
not hard to construct groups
the hypothesis of Saltman's theo
rem: for example, one may take a suitable extension of abelian groups of type
(P,,,, ,p).)
It is easy to map
see
(e.g., using the normal basis theorem) that the coveri n g 'K =
An
t
An/G
is generic (or versal) in the sense that every extension of Q (or of any field of characteristic zero) with Galois group G can be obtained by taking the 'l"fibre
a rational point of An/G over which 'Jl" is unramified. Hence, if AnIG is Q rational, then the set of all Gextensions of Q can be described by a syst em of
of
n rational
parameters. Such
a
parametrization implies the following property
of extensions with Galois group G [Sa1]� 2 A.!8ume API /G is Qrational. Let {Pi} be a finite set of primes, L. eztensions of Q", with Go.lois group G. Then there i8 an eztension L of Q with Gal(L/Q) = G such that L ® QJfj = Li.
Theorem
Remark:
There is a more general statement, where the L, are allowed to
be Galois algebras, a.nd Q is replaced by independent absolute values.
Proor
a field
(sketch): Each Li is parametrized by
parameter
(X)
endowed with finitely many
c�y»
E An(Q) which is sufficiently close to
Q",topology gives
an
E A "(Q�)'
A uglobal"
ea.ch of the (X(i)) i n
the
extension of Q with group G having the desired local
behaviour (Krasner's lemma ) .
QED.
xv
Introduction
The cyclic group of order 8 does not satisfy the property of tho 2. Indeed, if 2! L Q2 is the unique unr&mified extension of Q:a of degree 8, there is no cyclic extension L of degree 8 over Q such tha.t L2 � L ® Q:a (an easy exercise on characten,
see
[WaD.
perhaps extend Hilbert's theorem to a. more general class of va. riet ie s. There is an interesting suggestion of Ekedahl and Colliot.TheIene in this direction [Ek], (CT] (lee S3.5). One could
Since An/G is not always Q r ati on al, one has to settle for less:
Queztion: If G it G finite group, CAn it be reaJueG 4f tI Gtllou group of some TegUlar ezten.non F oj Q(T) 1 (Recall that '"F is regular" means that FnQ= Q .) Remarkz:
1. H F is a function field of a variety V defined over Q, then F is regular if and only if V is a.bsolutely irreducible. The regularity assumption i. included to rule out uninteresting examples such as the extCIll ion EeT) of Q(T) where E is a Galois extension of Q. 2. If such an F exi.ts, then there are infinitely many linearly di sj oint extensions of Q with Galois group G.
The existence of regular extensions of Q(T) wi th Galois group G is known when G is: • Abelian; • One of the 26 sporadic simple grou p s (with the possible exception of the Mathieu group M:a3); ePSL2(Fp ) , where at least one of is 1 (Shihl1, [Shih2]; • An, or Sn, d. [Hi); • An, d. N. Vila [Vi1 and JF. Mestre [Me21i • G2(Fp) [Th2], where G2 is the a.utomorphism group of t he octonions, and the list is not exhaustivc.
, (�), (�), (�)
The
method for finding F proceeds as follows:
1. Construction (by analytic and topological methods) of an of C(T) with Galois group G. descent from C to Q. This il the s a.ti sfies a socalled rigidity criterion.
2. A
The outline of the
1.
Elementa.ry
course will
examples,
hardest part, and requires that G
be:
and
the ScholzReichardt theorem.
2. Hilbert's irreducibility theorem and
exten sion Fe
applications.
Introduction
3.
4.
The "rigidity method" used to obtain extensions Galois groups.
of Q(T) with given
The quadratic form z ...... Tr (z2), and its applications to embedding problems, e.g., construction of extensions with Galois group An.
Chapter 1 Examples in low degree 1.1
The groups
Z/2Z, Z/3Z,
and
S3
• G = Z/2Z: all quadratic exten sions can be ob tai ned by taking s qu are roots: the map PI + PI given by X ..... X2 is generic (in characteristic ditTerent from 2  for a charaderistic·free equation, one should use X2  TX + 1 = 0
instead). •
G
=
Z13Z:
A "generic equation" for G is:
3,
with discriminant A = (T2  3T + 9)2. (In characteristic the ArtinSchreier equation y3 � Y = lIT by putting Y = group G acts on Pl by
where
u is a
this reduces to The
l/(X + 1).)
1 X u =l�X'
generator of G .
2 T=X+uX+uX=
.K33X+l X2_X
is Ginvariant and gives a map Y = Pt + PIIG. To check: gmericity, observe that a.nyextension L/ K with cyclic Galois group of order defines a homomorphism � : GK � G � Aut Y which ca.n be viewed as a lcocycle with values in Aut Y. The extensi on L/K is liven by a rational point on
3
PI/ G if and only if the twist ot Y by this coc:yde has a rational point not
invariant by u. This i.s a general property of Galois twists . But this twist has a rational point over a cubic extension of K, and every curve of genus 0 which
has a point over an odddegree extension is a projective line, and hen ce has at least one rational point distinct from the ones Jixed by
1
Chapter 1. Examples in
2 •
G
low degree
:;:. S3: The ma.p
gives
a.
projection PI
+
PI/S3 :;:. PI
which is generic, although the reasomng for 03 cannot be applied, as the order of S3 is even. But 53 can be lifted from PGL2 to GL2, and the vanishing of Hl(GKt GL2) can be used to show that P1
+
PI! 53
is generic. Exercise: Using the above construction (o r a direct argument), show that every separable cubic extension of K is given by an equation of the form x3 + TX + T::::: 0,
1.2
with T f:. 0,21/4.
The group 04
Let K4/ K be Galois and cyclic of degree 4, and suppose that Char K 1: 2. The extensi on K" is obtained from a unique tower of quadratic extensions:
J
where K2 = K(.y'E), and K" = K2( IJ, + by'E). Conversely, let K, K( 'V"E) be a quadratic extension of K, where f E Kis not a square. If II, b E K i1.D.d K4 = K2( IJ, + by'E), then K4 may not be Galois over K (its Galois closure could have Galois group isomorphic to D", the dihedral group of order 8). =
Theorem 1.2.1
for some c E K·.
The field K"
.j
is
cyclic of degree 4 if find only if 0.2  £b"J
=
Ec?
Proof: Let G be a group, £ a nontrivial homomorphism from G to Z/2Z, and X a homomorphism from H Ker E to Z/2Z. Let Hx denote the kernel =
of )(.
Lemma 1.2.2
(a) Hx
is
The following
are
equi1)tdent:
normal in G, and GJ Hx
is
cyclic of order 4.
1.2.
The group 04
(b)
Corix
map
Hl(H)
= E,
3
where
(We ab brevi at e
+
i it the
Cor
constriction
mllp.
Hl(G� Z/2Z) = Hom (G, Z/2Z) to Hl(G). Hl(G) cm b e defined b y
The corestriction
(Corix}(g) = x(Verig); wh · ere
The proof that ( a) =>
check that the
Now, assume
(b).
is immediate: replacing
(b)
transfer 04
Select
s
a by G/HXt it suffices to by s � 82•
O2 is onto: but this map is given E G  H. The transfer is given by:
+
Veri( h) = II· shs1 m�d (H, H). Hence fo r
all II E H:
�
x(Ver h) But if
h
=
==
0
(mod
E
normal in
Hx, then X(h) = O. It follows that X(shs1) G. Now, applying the hypothesis to B shows X(S2)
so
X(h) + X(shs1) = E(h)
=
i
Cor x ( s)
s2 :F 1
=
(mod H)(). It follows that le tes the proof of lemma 1.2.2.
com p
Now, let G
=
GK
:::::
E(S)
=:
af Hx
lemma. Via the
where c E
Remark:
K,
2,
that
H)( is
.
4,
and
this
define homo�
Hl(GK) with
HI (GK) is equal to the
E�t
This completes the proof of tho
In charac terist ic
K2 a.nd K4
identification of
i
=
10
is cyclic of order
Ke / K2, the corestriction map Cor : HI (GK:a) norm, and the criterion Cor x = E becomes: N(II + ov'E)
0,
(mod 2).
1
Gal(K /K). The extensions
morphism s C! and X as in the
=
2).
1.2.1.
Art in Schreier theory gives an iso morphism
where pz = z2 + z, and the c:orestriction map corresponds to the trace. Hence the analogue of tho 1.2.1 in characteristic 2 is:
BI(GK)
=:!
K/pK,
Theorem 1.2.3 Su.ppos e Char K = 2, a.nd let K2 = K (z), K4 K2 (y), where pz = e, py :;;: II + oz . Then K4 it GGlGiI over K Gnd cyclic of degree 4: if lind =
only ifTr ( II + bz)( = b) i3 of the form � + :&2
Observe
that the
variables
E,
4, z of
+
%,
with z
E
K,
tho 1.2,3 parametrize O,,extensions of
K.
In particular, it is possible in characteristic 2 to embed any quad ratic extension in a cyc lic extension of de gree 4. This is a special case of a general result: the embedding problem for pgroups always has a solu tion in charac teri sti c I' (aa
Chapter 1. Examples in low degree
4 can be seen
from the triviality of H2(G, P) when G is the absolute Galois of characteristic p and P is an abelian pgroup with GactioD.
group of a field
See for example [Sel}.)
The situation is different in characteristic :I 2: the criterion 0.2 b2€ that E must be a sum of 2 squares in K: jf b2 + c2 # 0, then: �
implies
e=
(
o.b
b2+c2
) ( 2
+
o.c
b2+c2
)2
::;: fC2
.
v=r E K, and any element of K can be expressed as a sum of 2 Conversely, if «: is the sum of two squares, t = l2 + iJ2, t hen setting
Otherwise squares.
a
solves the equation 0.2

=
b2€
�2 + ",2, b =�, =
c2e_ Hence we have
Theorem 1.2.4 A quadratic eztension eztension oj degree
c =
4 if and only if E
K(v"f)
p.,
shown:
ca.n
be embedded in K.
'" " Bum of two "quare.! in
II
cyclic
tho 1.2.4: the quadratic extension K'J can be 4 if and only if t he homo· Z/2Z given by K2 factors through a homomorphism
Here is an alternate proof of embedded in
a
cyclic extension K4 of degree
e : GK + Z/4Z. Thi s suggests that one apply Galois cohomology
morphism
GK
+
quence:
o
+
Z/2Z
+
Z/4Z
+
Z/2Z
+
to
the se
0,
obtaining:
f E Hl(GKJ Z/2Z) to BI(GK, Z/4Z) is given H'l(GKI Z/2Z) = Br:z (K), where Br (K) denotes the 2torsion in 2 the Brauer group of K. It is wellknown that the connecting homomorphism 6 : HI + H2 also known as the Bockstein map, is given by 6z = :z: :z: (cupproduct). This can be proved by computing on the "universal exam K(Z/2Z,1) which is the classifying space for Z/2Z. Th e cup ple" P ao(R) product can be computed by the formula:
The obstruction to lifting
by 6£ E
.
I
=
.
a J3 where
=
(a,{3),
Hl(G,Z/2Z) is identified with K/K·2 and (a,.B) denotes the class of
the quaternion algebra given by
1.2. The
But
(e, f )
only if
5
group C. = 0 (in additive notation), so ( ft
(1, e) = 0, i.e., e
Similarly, one
E)
could ask when the extension
extenlion of degree
8.
=
(1, f).
il a sum of two squares in
The
obstruction is
One can prove (e,g., by uling
Hence,
K.
6f is 0 if and
K... can be embedded in a cyclic an element of Br2(K),
again given by
[Sea}):
ob.ttM£ction to embedding the cyclic utensioft K4 in eI cyclic =ten.noft 0/ degree 8 it gil1en by the clus of (2, e ) + (1, eI) in Br2(K)J if eI# 01 elM by the clas" of(2, e) if eI = O.
Theorem 1.2.5 The
Hen ce, when G # 0, the OSgembedding problem is possible if quaternion algebra (2, e ) is isomorphic to (1, G ) .
and only if
the
1.2.1. Let K2 and K.. be as before, with + hvlE). Let :z: = K(yE), K... :: K:I( + b..fi. and y = G  bv'E. If Ga.l(K... /K) = 04, then we may choose a generator u of Gal(K4/ K) taking z to 11. and hence 11 to Z. Setting There i. also a direct proof of tho
K2
Ja
=
Ja
c= we have
(lC =
y( z)/( v'i) = c, so
zy/..fi,
c
E
K·, Also
and one obtainl the lame criterion as before. equation (J,2  b2e = t?e, one verifies that. K... group 0....
Remarks:
1.
The minimal polynomial for z over
x'"
J
K
Conversely, if
at
bt c, e
satisfy the
is a cyclic: extension with Galois
is
+ AXI + B = 0,
A = 2", B = ,,2  d:l. The condition for a general polynomial of this form to have Galois group C... is that A:I  4B il not a square and that
where
2. The C,,·extensionl are parametrised by the solutions tion
(J,2
_
d2
=
( E, (J" 1.1 u) of the equa
E1&:I,
with u � 0 and e not a. square. This represents a rational variety: one can solve for e in terms of ,,' I., and u. Hence the class of C...·extensioDl of Q
6
Chapter
1.
Examples in low d e gree
satisfies the conclusion of tho 2: there are C,,extensions of Q with arbit r a rily prescribed local behaviour at finitely many placesj recall that this is not true for the cyclic group of order 8. Exercises:
1. The group C4
generator" of C.
t s faithfully on PI via the map C4  PGL2 which sends a
ac
to (!l �).
The conesponding map PI
>
4 + 6z2 + 1)/(z(z2  1». This gives rise to the equation: 4 Z  TZ3 + 6Z2 + TZ + 1 = 0
by z ...... (Z
PI/C.
i.
gi""D
with Galois group C4 over Q(T). If i E K, show that this equation is generic: in fact, it is equivalent to the Kummer equation. 1.2 If i � K I show that there does not exist any one�dimensiona1 generic family fo r 1.1
C4extensions. l.3 H a C4exte nsion is de scrib ed
it
as
before by parameters
comes from the equation above if and only
if 1, a)
(
=
0
E',
a, b a.nd o r a = O.
c,
show that
2. Assume K contains a primitive 2"th root of unity z. Let L K( 2�) be a cyclic extension of K of degree 2". Show that the obstruction to the embedding of L in a cyclic extension of degree 2"+1 is (a, z) in Br2(K). =
1.3
Application of tori to abelian Galois groups of exponent 2,3,4,6
A Kt oru 8 is an algebraic gro up over K which. becomes isomorphic to a product of multiplicative groups Gm X X Gm over the algebraic closure K of K. If this isomorphism is defined over K, then the tor us is said to be split. Let T be a K· to rus and denote by X(T) its character group, • • .
X(T)
=
Homg ( T, Gm).
It is well known that X(T) i s a free Zmodule of rank n = dimT endowed wi t h the natural a ctio n of GK. The functor T ..... X(T) defines an antiequivalence between the category of finite dimensional tori over K and t he category of free Zmodul� of finite rank with GK action. A split Kto ru s is clearly a Krational variety; the same holds for tori which split over a quadratic ext ensi o n K' of K. This fol lows from the classification of tori which split over a quadratic extension (whose proof we shall omit  see [CR}): Lemma 1.3.1 A /ree Zmodule of finite rank 'With an action of
direct sum of indecomposable modules 0/ the form:
Z/2Z
is a
1.3. Application of tori to abelian G alois
groups of exponent
7
2,3,4,6
1. Z 'With trivial Action.. B. Z with the nontrivicU ACtion. 3. Z x Z with the "regular representation" of Z/2Z which interchan.gea the
two fa.c;tOf's.
The cofTesponding
tori
Are:
1. GlIl !. A "twisted form " of Gm, wh.ich correapcm4s to element.! of norm 1 in (9) below. 9. Th.e algebf'tl.ic gf'Oup RK'/KGm obtained from Gra/K, by "restriction of scAlars" to K ( c/. §9.S.1). It
is not difficult to show that the three cases give rise to Krational varieties,
a.nd the result follows.
A
HG
=
is a finite group, the group
K[G]
g of invertible elements of the group algebra
an algebraic group
defines
over K.
g�IIG� where the product
is taken
In
ovcr
char&eteristic 0, we
K,
over all irreducible representations of
denote the dimensions of these representations.
In particular, if G i.
Z[G},
where
G
=
wmmutative, then
Homg(G, Gm).
by the values of the characters of
groups:
and the covering
G. H G
group
m
ap
1
g
+ t
G
+
giG
g splits
and the 71&
There is
an
g + gIG
exact
�
4 or 6, then
group
over the field generated sequence
of algebraic
1,
is generic for extensions of
is of exponent 2.3,
G
9 is a torus with character
Therefore,
G.
have
g
K
with Galoi s
splits over a. quadr ati c
Q� Q( �)t or Q(i). By the previous g  and hence " fortiori gIG  is Qra.tional. So the abelian groups of exponent 2.3,4 or 6 yield to Noether's method (but not those of exponent 8).
extension, since the characters values lie in
result .
Exercise:
Show tha.t
all tori decomposed by a cyclic extension of
degree 4 are
rational varieties, by making a list of indecomposable integer representations of the
cyclic
[Vo2].
group of order 4 (there are nine of thete,
of degrees 1,1,2,2,3,3,4,4,4). See
Chapter
2
Nilpotent and solvable groups as Galois gron ps over Q A theorem of ScholzReichardt
2.1
Our goal will be to Rei chard t [Re):
Theorem
Q.
2.1.1
prove
the fo ll owing theorem which is due to Scholz a.nd
Every Igroup, I :/: 21
CAn
be re41ized
(Equivalently, every finite nilpotent group of
over Q.)
odd
AS
A
order
G4iois group over is
a.
Galois group
Remarks! 1.
This is a special case of a. theorem of Sha.farevich: every solvable group can be u a Galois group over Q. [The proofs of tha.t theorem given in [Shall a.nd [Is1 are known to cont ain a. mista.ke rela.tive to the prime 2 (see [Sha.3]). In the notes append.ed to his Collected Papers, p.752, Shafa.revich sketches a. method to correct this. See also [ILKl, ch. 5.J 2. The proof yields somewhat more tha.n the statement of the theorem. For example, IN, then the extension of Q wit h Galois group G can be chosen to be ramified if IGI a.t a.t most N primes . It also follows from the proof that any separable prol.group of finite exponent is a. Galois group over Q. 3. The proof does not work for I = 2. It would be interesting to see if there is a wa.y of a.d a.pting it to this I;:ue. 4. It is not known whether there is a regular Galois extension of Q(T) with Galois group G for an arbitrary Igroup G. realized
=
An ,...group can be built up from a series of central extensions by groups of orde r I. The natura.! approach to the problem of reali zi ng an I· group G as a Galois group over Q is to construct a tower of ext e nsion s of degree I which ultimately give the desired G·e.xt ension . When carried out naively, this ap p roa.ch does not workt because the embeddi n g proble m cannot always 9
Cha.pter 2. Nilpotent
10
and
solvable groups
be solved. The idea. of Scholz and Reicha.rdt is to introduce more stringent conditions on the extensions which are made at ea.ch stage, ensuring that the embedding problem has a. positive answer. Let K/Q be an extension with Galois group G, where G is a.n l�oup. Choose N � 1 such that IN is a multiple of the exponent of G, i.e., s' 1 for all s E G. The property introduced by Scholz is the following: =
Definition 2.1.2 The eztension L/Q is said to hAve property (SN) if every prime p 'Which is ra.mified in L/Q satisfies: 1. p ;; 1 (mod IN). §. If tJ is II plAce of L di1liding P, the inertia group Iv at u is equal to the decomposition group Dv. Condition 2 is equivalent to saying that the local extension Lu/Qp is totally ramified, or tha.t its residue field is Fp' Now, let 1
+
C,
�
G
+
G
+
1
be an exa.d sequence of lgroups with Ot ce nt ral . cyclic of order I. The "em bedding problem" for G is to find a Galois extension L of K containing L, with isomorphisms Gal(L/ L) � 0, and Gal(i./ K) � G such that the dia.gra.m + 4
C,
II Gal( L/ L )
G II Gal(£/K)
+ 4
+
G
+
Gal(L/K)
II
1
is commuta.tive. The ScholzReichardt theorem is a consequence of the following (a.pplied indudi vely):
Theorem 2.1.3 Let L/Q be Galois with GAlois group GJ And assume that L h48 property (SN). Assume further that IN is a multiple of tAe e:tpDnent of G. Then the embedding problem for L lind G halJ a solution £, 'Which stJ.tisjies (SN) And is ramified at at most one mon: prime thA'II L. (Furthermore, one can require tha.t this prime be taken from any set of prime numbers of density one.) The proof of tho 2.1.3 will be divided into two parts: firat, for split extensions, then for nonsplit ones.
First part: the case t;
�
G
X
0,
Let (P1, . . . ,p"') be the prime numbers ramified in L. Select a. prime number with the following properties:
q
1.
q ::::
1 (mod IN),
2.1. A 2. q
theorem
of ScholzReichardt
splits completely in the
3. Every prime Pi,
ext ensio n
11
L/Q,
(I S i � m) is an lth power in FIf'
Taken toget her, these conditions mean that the prime q splits completely in the field L( lV'f,1P\, ... J�)' The following wellk.nown lemma guarantees the existence of such a. q: LeJIiIna 2.1.4 If E /Q is
II finite eztension of Q, then there aTe in.finitely many primes which split completely in E. 1710 fact, every set of density anI! contains such II prime.
Proof: The second statement in the lemma is a consequence of Chebotarev's density theorem; the first part C&D be proved by a direct argument, without invoking Chebotarev. Assume E is Galois, and let f be a minimal polynomial with integral coefficients of a pri mi t i ve element of E. Suppose there are only finitely many prim es Pi whi ch split completely in E or are ramified. Then fez) is of the form ±p�' .. 'Pkll, for z E Z. When:c is between 1 and X, the number of distin ct values taken by f( z) is at leut .; X. But the number of values of f(:z:) which can be written in the form ±p;"l ... pr' is bounded by a power of log X. This yields a contradiction. H aving ch ose n a momorphism
q
which satisfies the l
:
(Z/qzt
ns above, fix a surjective ho�
conditi o + C,.
(Such a � exists because q == 1 (mod I).) We view ). as a Galois character. This defines a Crextension MA of Q which is ramified only a.t q, and is linea.rly disjoin t from L. The compo si t um LM), therefore has Galois group G = G xC,. Let us check that LM>. sa.tisfies pro p erty (SN)' By our choice of q, we have q ;:; 1 (mod IN). It r em ai n s to show that Iv Dv at all ramified primes. If p is ramified in L/Qt it splits completely in MA, and hence Dv = I" for all primes vip. The only prime ramified in M>. is q, and q split s completely in L by ass umptio n . Hence, for all primes 11 which are ramified in LM>., we have Dv = Iv a.s desired. =
Second part: the case where G is a nonsplit extension The pr o of will be carried out in three stages: (i) Existence of an
(ii) Modifying L
ex
so
tensi on i, giving a soluti on to the embedding problem.
tha.t it is ramified at the same pl aces
as
L.
(iii) Modifying L further so that it has property (SN), with at most one additional ramified p rime.
Chapter 2. Nilpotent
12
and
soha.ble
groups
(i) Solvability of the embedding problem The field extension L determines a surjective homomorphism; : GQ + G. The problem is to lift tP to a homo morphism ¢ : GQ + G. (Such a. � is automatically surjective because of our assumption that G does not split.) Let e E H2(G, Cr) be the class of the extension G,and let
tP· : H2(G,O,)
+
H2(GQ,e,)
be the homomorphism defined by,p_ The existence of the lifting � is equivalent to the vanishing of tP·(e) in H2(GQ, e,). As usual in Galois cohomology, we write H2( GQ,  ) as H2(Q, ) , and similarly for other fields. The following well·known lemma reduces the statement tP·e = 0 to a purely local question:
Lemma 2.1.5 Th.e restriction mAp W(Q, e,)
+
:i II H (Q�, 0,) p
is injective.
(A similar result holds for any number fiel d.)
of Proof. Let K = Q(I'I). Since [K : Q] is prime to I, the map H2(Q, e,) + H2(K, e,) is injective. Hence, it is enough to prove the lemma with Q replaced by K. In that case, H2(K, e,) is isomorphic to Brl(K), the Itorsion of the Brauer group of K. The l emma then follows from the Brauer HasseNoether theorem: an element of Br(K} which is 0 locally is O. (Note thatJ since I 'I 2, the archimedean places can be ignored.) By the a.bove lemma, it suffices to show tha.t ,p*e = 0 locally at all primes. In other words, we must lift t he map tPP : GQ. + D� C G to ¢� : GQ. + G. There are two easel:
Sketch
1. pis unramified in L, Le., tPP is trivial on :he inertia group Ip of GQp' Then cPP factors through the quotient GQp/Ip = Z. But one can always lift a map Z + G to a map Z + G: just lift the generator of Z.
2. pis ramified in L. By cons truction , 11 == 1 (mod IN), hence p ':f I and Lv/Qp is tamely ramified ( as in 2.1.2, v denotes a place of Labove p); since its Galois group Dv is equal to its inertia group Iv, it is cyclic. The homomorphism QQp + Dv c G factors through the map GQp + Gal(E /Qp), where E is the maximal abelian tame extension of Q� with exponent dividing IN. The ex tension E can be described explicitly: it is composed of the uni que unramified extension of Q� of degree IN, (obtained by taking t he fraction field of the ring of Witt vectors over FpiN) and the totally ramified extension Qp( Iyrp) (which
2.1.
is
a.
A theorem of ScholzReichardt Kummer extension since p ==
13 1 ( mo d IN) ) .
It follows that
Gal(E/ Qp)
is
a.belian group of type (IN, IN)j it is projective in the category of a.belian groups of exponent dividing IN . The inverse image of Dv in G belongs to that category ( a central extension of a cyclic group is abelian). This shows that a.n
the local lifting is possible.
(ii) Modifying the extension L so t h at it becomes unramified outside ram(L/Q) of primes ramified in L/Q
the set
Lemma 2 . 1 .6
For every prime p, let Ep be
Q,
continuous homomorphism from
Gal( Qp/ Qp) to a finite abelian group O . Suppose that almost IIll £p are un ra.mified. Then there is 4 unique f: : Ga1( Q/Q) .. O J such thtJ.t for all PI the mllps
f:
lind £p IIgree on the inertia groups
1p .
( The decomposition and inertia groups DpI Ip are only defined up to conj ugacy inside GQ . We shall implicitly assume throughout that a fixed place tI has been chosen above each p, so that Dp and
Proof of lemma:
Ip are
welldefined subgroups of GQ . )
By local class field theory, the Ep
Q; �
ca.n
be canonically identified
O. The restrictions of Ep to Z; are trivial on a closed + ,.. Zp, where n" is the cDnductor of Ep. Since almost all 'np are zero, there is a homomorphism f: : (Z/MZ)· + 0, with M = n p"" , and f:(k) = n �(kl). If we view f: as a Galois character, class field theory shows with maps subgroup 1
that it has the required properties.
(Equivalently, one may use the direct
product decomposition of the ideIe grOup IQ of [Q
=
(I:z; R.i.) X
X
Q,
as:
Q
)
•.
Proposition 2 . 1 . 7 Let 1 t C t .. t CJ t 1 be a central �zt�nsion of II group • J ana rp he Q, continuous homomorphism from GQ to � which has G lifting .,p ; GQ .. i. Let �p : GQ .. . i be liJtingll of �p = ; I D.. , such that the �p are unra.mified for almost all p. Then there is a lifting � : GQ + i such that, for every P, � is equlll to �p on the inertia group at p. Such a lifting is unique. This proposition is
also
P GL", (Tate, see [Se7,
Proof of prop. S. 1 . 7:
useful for relating Galois representations in GL,. and
§6)).)
For every p , there is
a.
Wlique homomorphism
Chapter 2. Nilpotent and solvable groups
14 such that for all which
s
=
..p(s)
Ep(s)�p(s)
a unique E : G Q + C q, = ,pf1 has the required
E GQ,. . B y the ptevious lemma, there exists
agrees
with Ep on 11"
The homomorphism
property. This proves the existence assertion. The uniqueness is proved simi
la.rly.
Corollary 2. 1.8 As.!Jumi39 the hypotheses of prop. !. 1 . 7, chosen u3ra.mijied at every prime where ¢ is unra.mijied.
Proof:
Choose local liftings
�" of ¢ which
are
possible since there is no obstruction to lifting
a
unramifie d
tJ
lijti39 of ¢ ca.n be
where
¢
is; this is
homomorphism defined on
Then, apply prop. 2. 1 . 7.
(ii): L
The corollary completes the proof of part is ramified a.t the same places as
L.
Z.
can be modified so tha.t i t
(iii) Modifying i so that i t satisfi es property (SN) We have obtained
an
extension
i which is ramified at the same places as L and G. Let p be in ram(L / Q ) = ram(l/Q).
which solves the extensi�n �roblem for
Ip (resp D", I,,) the decomposition an d inertia. groups for L (resp Ip = Dp C G; this is a cyclic grou p of order lIZ , say. Let I; inverse image of I" in C. We have Ip C D" c I;. If I� is a nonsplit
Denote by Dp,
L)
at p. We have
be the
extension of I"
( i . e.,
is cyclic of order
the Scholz condition is satisfied
at
p.
lcz+l ) Let
we even have
5
which J� is a split extension of I,,; since ip is cyclic, we
Frobenius element Frob" E
D"I ill
c
Ip
=
be the set of p E
1;1 ip may be
D"
=
I�,
and
r� LIQ) for have I� = ip X C, . The
identified with
an
element
if and only if c" = 1 . If all c,, ' s are equal to 1, i satisfies (SN ) . If not I we need to correct � : GQ + G by a Galois character X : (ZjqZ)· + Cr which satisfies the following proper ties: Cp of e'i the Scholz condition is satisfied at
1.
2.
3.
p
q == 1 (mod IN ). For every p in S, X(p) Cp . The prime q splits completely i n L I Q. =
Conditions 1 , 2, and 3 impose conditions o n the behaviour of
'Vi), Q (�, ¥p, p E 5) ,Nand L respectively. Wri te Q( where F i s cyclic of order I l and totally ramified at I. Q(
Lemma 2.1A9 The over Q.
fields L, F, and Q (V'f, �, P
E
I�)
q in =
the fields
Q (�) . F ,
S) are linearly disjoint
2.1. A t heorem of ScholzReichardt
15
Proof: Since L and F have distinct ramification, L and F are linearly dis joint: L · F has Galois group G x C,N� . The extension Q(�,�, P E S) has Galois group V :;;::: 0, x C, . . . X 0, ( 1 51 times) over q(�i) . The action of Gal( Q(�)/Q) Fi on V by conjugation is the natural action of multiplica tion by scalars. The Galois group of Q(�, ..yp, p E S) over Q is a semidirect product of Fi with V. Since 1 # 2, this group has no quotient of order I: there is no Galois subfield of Q(�' �l P E S) of degree l over q. This implies that L . F and Q( �2 �, p E S) are lin�arly disjoint. QED. =
If S :;;::: {PI ) , " ,P.} , define integers Vi , 2 :5 i :5 k, by C?i = c: . (This is possible if Cl i: 1, which we may assume . ) In order to satisfy conditions 1, 2 and 3, the prime q must have the following behaviour in the extension L · F · Q( � , � , p E S ):
in L · F and q( in q (� , �j in q(� ,
IVt);
.:jP1 /p� ), i
=
2, . . . , k.
By the Chebotarev density theorem and lemma 2. 1.9, such a q exists. One can then define the c�aracter X so that X (Pi) �. T�is completes part (ii): the homomorphism
The proof allows us to generalize the theorem somewhat. Let us ma.ke the following definition:
Definition 2 . 1. 10 If G is a profinite groUPI the following aN: equivalent : 1 . The topology of G is metriz(lble. !. G htU (I cDunteble dense .subset. 3. G C4n be written tU (I denumeroble projective limit
G
=
l�. · ·
+
G.,.
+
G.,.l ...... . . . ) ,
wheN: the G.,. 's 4re finite (a.nd the connecting homomorphisms 4N: surjective).
4 . The set of open .subgroups of G is denumerable. A group G
'Which
sa.tisfie.s these equillalent properties is .said to be separa.ble.
Gal(L/K), these properties are equivalent to [L ; K] � No; if G is a pro Igroup, they are equivalent to dim Hl (G, � No . The proof that the four properties in the definition are equivalent is elemen tary.
If G
=
Z/IZ)
Theorem 2 . 1 . 1 1 IfG is 4 separoble p ro Igroup of finite ezponent, theft there is
(I
Galois eztension of q with Galois group G.
16
Chapter
Proof: a finite
If
of ord er
IN
is t he
Igrou p ,
I.
exponent of G, write G
2 . Nilpotent an d solvable
as proj.lim(GR) where
groups
each GR is
the connect ing homomorphi sm b e i ng surj ect ive , with kernel
By tho
2.1 .3,
one
ca.n
G alois extensions LR /Q with
construct inductively an
increasing family of
Galois group G", whi ch have the ( SN) pro pert y;
the union of t he LR's h as G aloi s group G.
Remark: The finiteness condition on the exponent cannot be dropped: for example, Z, X Z, is not a Galois group over A
more
Q.
general. result ha.s been proved by Neukirch [Ne] for prosolvable
of odd order and finite exponent.
2.2
groups
The Frat tini subgroup of a finite group
Let G be a finite group. Definition
2.2.1 The Frattini su bgrou p 4' of G
mazimal subgroups of G.
�
is the intersection of
the
G. If G1 C G sat i sfi e s G. (Otherwise, choose a maximal subgrou p M such
The Fratt ini subgroup is normal . If G :f:. 1 , then � :f:.
.
G1
=
G,
then G1
:::::
that G1 C M e G. Since . C M, it follows that .Gl C M, which is a cont radi ction . ) In ot her words, a subset of G generates G if and on l y if it gene rat es Glf!: elements of f?, are sometimes referred t o as "nongenerators" .
Examples: 1. If G is a simple group, then 4' :::; 1. 2. If G is a p group , the maximal subgroups are the kernels
of the surjective Cpo Hence f?, is generated by (G, G) and (]J', w here ( OJ 0) denotes the commutator subgroup of G; more precisely� we have
homomorphisms
G
.
•
= (0, 0) · (]P.
The group 0 / . is the maximal abelian quot ient of G of type (p, p, . . . , p ).
P roposition 2. 2. 2 ( [Hu] , p. 168) Let G he a finite group, • its Frattini subgroup, N (1 normal subgroup of G with 4' c N e G. Assume N/4' is nilpotent. Then N is nilpotent. C o rollary
2.2.3 The group 4' is nilpotent.
This follows by a.pplying prop.
2.2.2
to N
=
•.
Let us p rove prop . 2.2.2. Recall that a finite group is nil pot e nt if and only has only one Sylow psu bgrou p for every p. Choose a Sylow psu bgro up P of N, and let Q �P. The ima.ge of Q by the quotient ma.p N + N/�
if it
=
2.2. The Frattini subgroup of a finite
group
17
is a Sylow psubgroup of N/� which is unique b y assu m ptio n. Hence t hi s image is a characteristic subgroup of N/�; i n particular i t is preserved by inner conj ugation by elements of
G, Le., Q is
normal in G . Let
NG(P ) ;::: {g ig E G, gPg1
=.
P}
be ti.e normalizer of P i n G. If 9
E OT then gPgl is a Sylow psubgroup of Q � Applying the Sylow theorem II in Q . there is a q E Q such that
qgPgl q l = P. Hence qg E N G( P ). G = N G ( P ), and P is
It follows that G = QNG( P)
normal in
Sylow psubgroup of N.
Application to
solvable
O. hence in Nj this
=
f)Na( P).
Therefore
implies that P is the only
groups
Proposition 2.2.4 Let G be a finite solvllble group ':1 1 . Then G is isomor phic to A quotient oj A group H which is A sem:i.·dirt:ct product U . S, whert: U is a nilpotent normlll sub!JToup oj H, And S is solvAhle with 151 < IGI · Let � be the Frattini subgroup of
Proof:
Gj
since G/� is solvable and #: 1 , it
contains a nontrivial abelian normal subgroup, e. g. , the last nontrivial term
of the descending derived series of G/�. Denote by U its inverse image in G. Since CP c U e G, with U/� abelian, U is nilpotent by prop. 2.2.2. Choose a miLXimal subgroup S of G which does not contain U: this is possible since U #: CP . Since U . 5 ':1 5 and 5 is muimal, G :::: U · S. Hen ce, writing H = U · S (with S acting by conjugation on the normal subgroup U), map H � G.
there is a surjective
The relevance of prop. 2.2.4 to Galois theory lies in the following result
which
s rt s that the embedding problem for split extensions with
as e
kernel has always a solution.
nilpotent
2.2.5 ( [Sha2], [Is]) Let L/ K be lin ezten.non of numher fields with GAlois group 51 let U be (I ftilpotent group with Saction, and let G be the semidirt:d product U · S. Theft the embedding prohlem Jor L/ K lind for
Claim
1 . ha.s
II
U . G
..... S + 1
solution.
Theorem 2.2.6 Cl(lim !. !. S implies the ezistence oj Galois eztensions of Q with given solvoble GAlois group.
18
Chapter
2.
Nilpotent and 501va.ble groups
Proof: Let G be a solvable group. We proceed by induction on the order of G. We may asume G # 1. By prop . 2.2.4, write G as a quotient of U . S wi th U nilpotent and S solvable, 151 < 101. The induction hypothesis gives a Galois extension L/Q with Galois group S. By t he claim above, U · S can be realized as a Galois group; hence, so can i t s quotient O.
Let us give a proof of claim 2.2.5 in the elementary case where U is abelian of exponent n. Observe that: 1. If cla.im 2.2.5 is true for an extension L' of L, it is true for L: for I if SI = Gal(L/K), there is a natural quotient map US' + US. Hence we may assume I'n C L, where I'n denotes the nth roots of unity. 2. We may also assume u
�
direct swn of copies of Z/nZ [S] ,
because any abelian group of exponent n on which S acts i s a quotient o f such an Smodule. Suppose that h is the number of copies in the decomposition of U as a di rect sum of Smodules Z/nZ[S). Choose places 111 " , . , Via of K which split completely in L, WI , . , . I w" places of L which extend them; any place of L ex tending one of the Vi can be written uniquely &8 OSW�, for some oS E S. Choose elements tPi E L· such that
if oS = 1 and i otherwise.
�,
=
j.
Let M be the field generated over L by the for s E S and j = 1, . . . , h . This i s a Galois extension of K, with Gal (M/ L) � U . Its Galois group over K is a.n extension of S by Uj since U is a free Z/nZ[S}module, it is known that such an extension splits (see, e.g. , [S e2 , ch. IX] ). Hence Gal(M/ K ) is isomorphic to the semidirect product of S by U.
Chapter
3
Hilbert 's irreducibility theoreIll 3.1
The Hilb ert property
Fix a ground field K with Char K = 0, and let V be an irreducible algebraic variety over K. (In what follows, algebraic varieties will be tacitly assumed to be integral and quasiprojective.) Denote by V(K) the set of Kra.tional p oi nt s of V . A subset A of V(K) is said to be of type (01 ) if there is a closed subset W e V, W =f V, with A c W(K), i.e' t if A is not Zariskidense in V. A subset A of V(K) is said to be of type (C3) if there is an irreducible variety v', with dim V = dim V' , and a generically surjective mor phism 11" : V' + V of degree � 2, with A C 1I"(v'(K».
3.1.1
subset A of V( K} is called thin ( �ince " in French) if it is contained in a finite union of sets of type (01 ) or (03), Definition
A
Alternately, a set A is thin if there is a morphism 11" :
havi ng no
ra.tional cross section,
Example: If
in K
W . V
is thin.
Definition
V = Pl , V( K)
3.1.2 (d.
=
with
and such that A C
K U {oo}.
leTS l ] , p.
property if V(K) is not thin.
This is a birational property of Definition
3.1.3
A
dim W � dim V 'If(W( K».
The set of squares (resp. cubes, .
. .
)
189) A variety V over K satufies the Hilbert
V.
field K is Hilbertian if there emu an. i�ducible variety 1, which hu the Hilbert property.
V over K, with dim V �
19
Chapter
20
3.
Hilbert'. irreducibility theorem
(cE. exe rc. 1) that if K is Hilbertian, then the projective line Pl over K has the Hilbert property (hence, our definition is equivalent to the
It is easy to show
standard one, lee e . g.
[L]).
The fields R, Qp are not Hilbertian. A number field is Hilbertian (see
§3.4).
o n irreducible varieties: The variety V is said to be absolutely ir reducible if the algebraic closure K' of K in the field K( V) of rational. functions on V is equal to K. Equivalently, V must remain irreducible upon extension
Remark
of scalars to the algebraic closure f< of K. If V is not absolutely irreducible, then V e X ) C W(K), where W is a subvariety of V, W i: V. Indeed, if V is a nonna! variety, then V(K) == 0. For, the residue field of the local ring at
P
E V contains K' , and hence no point of V is Krational. The genera.l case
follows from this by normalization. In particular, an irreducible variety which has the Hilbert property is absolutely irreducible. Therefore, in our definition of C2type subsets, we could have asked tha.t V
i
be a.bsolutely irreducible.
� : V' � V be a finite morphism (we also say t hat � is a "covering" even though it can be ramified) . Assume that V and Vi are abso lutely irreducible, and let K(V' ) ! K(V) be the corresponding field extension.
Remark: Let I
Let K(V')pl be the Galois closure of K( V' ) over K(V), and let normalis a.tion of V' in K( V')s". The variety W wit h its projection may be called the
Galois closure of V'
gebraically closed in K(V' )Sal, i.e., example, take V = V' = Ph 11'(:)
+
W need
=
V. Note that K is not
W b e the W + V always al
not be a.bsolutely irreducible.
:3; the Galois closure of V' is
and hence is not absolutely irreducible over X if K does
For
Pl/K(loill)1
not contain 1l3 '
Exercises: 1. Let V be an a:ffi.ne irreducible variety over K, with dim V � 1 . Let WI , . . . , W" be absolutely irreducible coverings of the projective line PI  Show that there exists a morphism I : V + PI such that the pullback coverings I·Wi of V are absolutely irreducible. Use this to show that if V has the Hilbert property, then so has Pl ' 2. Let V and T be absolutely irreducible varieties. and A C V(K) a. thin subset. Show that A X T( K) is thin in V X T. M ore generally, let I : W t V be a generically surjective morphism whose generic fiber is a.bsolutely irreducible ( Le. the function field extension K(W)jK(V) is regular ) . If B is a subset of W(K) such that I(B) is thin in V(K), show that B is thin in W(K). 3. Let K be a Hilbertian field, and let A be the Bet of elements of K which are sums of two squares. Show that A is not thin. (Use exerc. 2.) 4. Let K be a number field and V an abelian variety over K with dim V � 1. Show that V(K) is thin ( i.e., V does not have the Hilbert property). Hint: llse the Mordell· Well theorem. Problem: H V and V' are irreducible varieties with the Hilbert property, is that V X V' has the same property?
it true
3.2.
Propertiel of thin sets
3.2
Properties of thin sets
3.2.1 Let
21
Extension of scalars
L/ K
be a finite extension,
Extension of scalars to
L
Proposition 3.2.1 If A thin with respect to K .
V
an ablolutely irreducible variety over
yields a. variety over
c
L,
denoted
K.
ViL'
VeL) is thin with respect to L, thm A n V(K) is : (VarL) to (VarK) from Here are two equivalent definitions
The proof uses the restriction of scalarl functor RLIK
I,.. va.ri eties of RLIK :
to Kvarieties,
1.
It il the
for
every
d. [We], [Oe] .
right adj oint to the extension of scalars (VarK) Kvariety T an d L·variety W , one has: Mor K(T, RLIKW)
=
�
(VarL),
Le. ,
Mor L(T/L• Wli
In particular, taking T to be a point which is a rational over K , the above formula yields
(RL IK W ) ( K )
2.
=
We L l.
Let E L be the set of embeddings of L in some fixed algebraic closure Kj
E EL, let W" be the variety deduced from the given Lvariety W u. Then the product X = It, WIT is a f(variety. Moreover, one has natural isomorphisms from X to X· for e'1ery s E GK. By a
for each
by extension of scalars via.
K.va.riety from which RLI K W.
Weil's descent theory, these isomorphisms give rile to a
X
comes by extension of scalars; this variety is
type (01 ), then A n V(K) is clearly of type (01), Hence we A C 'If(W(L », that W is absolutely irreducible over L with dim W == dim V, and 'If is a. covering W + V with deg 'lf > 1. By restricting suitably V, we ma.y assume that 11' is finite etale. The functor RL/K then gives an etale covering RL1KW � RL/K "iL o Using the diagonal embedding ' A : V � RL/KVL , we obtain an etate covering 'If' : V + V, and a Cartesian
If A
is of
may assume that
dia.gram:
The set
' V
An V( K) is contained in 'If' (V ' ( K», and it is easy to check that all the V at least equal to deg 'If . Hence 'If' (V (K) ) and the lame is true for A n V(K).
components of V' ha.ve degree over
is thin,
Corollary
3.2.2 If K is Hilbertillft, then so is
L.
Chapter 3.
22
Hilbert's irreducibility theorem
Suppose L were not. Then A = P 1 ( L) is thin in PI with respect to L. This implies tha.t AnP J ( K) = P l ( K) is thin in PI with respect to Ki contradiction. Remark: The converse to cor. 3.2.2 is not true; see e.g. [Ku1 . 3.2.2
Intersections with linear subvarieties
Let V be the projective space Pn of dimension n, and let A C V( K) be a. thin set. We denote by Grass� the Grassma.nn variety of dlinear subspaces of Pn J where 1 � d � n. 3.2.3 There is 4 nonempty Zari.skiopen subset U such that if W belongs to U(K) J then A n W is thin in W .
Prop o sitio n
C Grass
�
It is enough to prove this when A is either of type � CI ) or of type (C2 ). The first case is easy. In the second, there is a map 11'" : V + Pn l with V' absolutely irreduciblet deg ... � 2, and A C ... (V'(K)). By Bertini's theorem (see e.g. , (Jou, ch. I, §6], [Ha, p. 1 79], [De2] , [Z», there exists a nonempty open set U in Grass! such that ...  J (W) is absolutely irreducible for all W E U. Hence, if W E U (K) , then W n A is of type ( C2 ), a.nd hence is thin.
An interesting case occurs when Ii. = 1 . Let '7r : V' + Pn be a generically surjective map of degree > I J and � the hypersurfa.ce of ramification of 1(" . Consi der the set U of lines which intersect c) t ransversal ly a.t smooth points. Then for £ E U, the covering ,...  1 (£) + £ is irreducible: one proves this over C by deforming the line into a generic one, and the general case follows.
Consider the "double plane" V; with equat ion t2 F(:, y), where F is the equation for a smooth quartic curve � in P2• The natural projection
Exam ple:
=
of V; onto P 2 is quadra.tic and ramified along the curve C). A line in P2 which intersects � transversally in 4 points lifts to an irredu ci bl e curve of genus 1 in V;; a line which is tangent at one point and at no other lifts to an irreducible curve of genus zero; fina.lly, if the line is one of the 28 bit angents to � I then its inverse image is two curves of genus zero. In that case, one ca.n take for U the complement of 28 points in P 2 = Grass� . Corollary 3.2.4 If Pn h4S the Hilbert property over K for s om e n � I J then all proiective spaces Pm over K have the Hilbert property. P1 has the Hilbert property over K: if not, P 1 (K) == A i s thin, and A x P I (K ) x . . . x PI(K) is thin in P I x . . . X P I . Th is cannot be the case, since Pn has the Hilbert property (and hence also PI X . . . X PI which is birationally isomorphic to Pn). This impl ies the same for P tn , with m � 1. For, if A = Ptn ( K) is thin i n P tn, then by prop. 3.2.3, there is a line £ such that £, n A L(K ) is thin in £ := Pl . But this contradi cts the fa.ct t h a.t PI (K) has the Hilbert property.
Proof:
=
3.3. Irreducibility theorem and thin sets 3.3
23
Irreducibility theorem and thin sets
V be a Galois covering with Galois group a, where V, W denote varieties, V = WI G, and G acts faithfully on W. Let us say that P E V(K) has property Irr( P) if P is "inert" , i .e., the inverse image of P (in the scheme sense) is one point, i.e. , the affine ring of the fiber is a field Kp ( or, equivalently, GK acts freely and transitively on the Xpoints of W above Pl. In this case, 'If is etale above P and the field Kp is a Galois extension of
Let
11' :
W
+
Kirreducible
K w ith Galois group
a.
Proposition 3.3. 1 There is a thin set A C V( K) such that for all P j A, the ineducibility property In( P) is satisfied.
Proof: By removing the ramification locus, we may assume that W � V is etale, i.e. , G acts freely on each fiber. Let E be the set of proper subgroups H of G. We denote by W/H the quotient of W by H, and by 'lfH the natural projection onto V. Let A The set
A
=
U 1I'H(W/H)( K). HE�
is thinl since the degrees of the
'lfH are equal t o [G
:
H]
> 1.
H
P ;. A , then Irr( P) is satisfied! for, lift P to P in W( K), and let H be the subgroup of G consisting of elements 9 E G luch that gF = ,.,F for lome i E
GlC = GaI(K/K).
since the image of P in
H = a. Otherwise, H would belong to E, and W/H is rational over K, the point P would be in A.
Then
If V has the Hilbert property otu!r K, the ezistence 0/ a G ootlering W � V (IS above im.plies that there is a Galois e:ctension of K with Gdou group G.
Corollary 3.3.2
A8sume furthermore that W, and hence V. are absolutely irreducible, and that
V has the Hilbert property. Let U8 say tha.t an extension L / K is of twe W if comes from lifting a Krational point on V to W.
it
Under these (lSsumptions, for etJery finite ezteu.non L of KJ there is a Galois ezteMon E / K of type W with Galois group G which is linea.rly disjoint from L .
Pr op o sition 3.3.3
There e.ut infinitely many linearly disjoint e%ten.sion.s with. Galoif group a, of type W .
Corollary 3.3.4
Cha.pter
24
3. Hilbert ' s irreducibility
theorem
Proof of prop. 9. 9. 9: Extend scalars to L; applying prop. 3.3. 1 there is a thin set AI. C V;L (L) such tha.t for all P ft A, the property Irr( P ) is satisfied over L. Set A :::: AL n V(K). T his is a thin set by prop . 3.2.1 . Choose P E V(K) with P f; A. Then Irr( P) is true over L, and hence a fortiori over K. This P
gi yes the desired extension E.
Exercise: Show tha.t the set of rational points P satisfying property Irr ( P) and giving a fixed G aloi s extension of K is thin (if G � 1). P olynomial interpretation
Let V be as above, let K( V ) be its function field, and let
be an irreducible polynomial over K(V). Let
f
G
c
Sn be the Galois group of
viewed as a group of permutations on the roots of f. (This group can be identified with a subgroup of Sn , up to conj ugacy i n Sn1 i.e., one needs to fix
a labelling of the roots.)
lli( t) E K,
If t
E
V(K)
and
t is no t
a pole of any of the
and one can define the specialization of f at
ft(X) P ro po sitio n
= xn + a l (t)xn1 + .
3.3. 5 There e2uts a thin set A
then :
1. t is not a pole of any of the 11. f,(X ) is irreducible otler K J 9. the Galois grou, of ft is G.
. .
C
ali,
t:
then
+ an ( t).
V(K) such that, if t � A,
lliJ
V by a dense open subset, we may assume that the a; have no V is smooth and that the discriminant II of f is invertible. The
By replacing
poles, that
au bvariety \If of V X A 1 defined by
( t, 2 ) E \If � ft(z) = is an et ale covering of degree
G. The
•
G
=
53  Let
over
Its G alois closure W
proposition follows by applying prop.
Examples:
S3
n.
0
3.3. 1
=:
to
YtIN
W.
has G alois group
f(X) = X3 + alX2 + a2 X + a3 be irreducible, with Galois group
K( V) .
The specialization a.t t has Galois group 53
properties are satisfied:
if
the following
1. t E V(K) is not a pole for any of the a., . 2. 6(t) :f. 0 ( where 6 = a�a� + 1 8al a2a3  4a�a3  4a�  27a�
discriminant
3.
of f).
Xl + al(t )X2 + a 2 (t)X + alet) has no root in K.
is the
25
3.4. Hilbert's irreducibility theorem
4. 6.(t) is not a square in K. Conditions 3 and 4 guarantee tha.t G is not contained in either of the maximal subgroup. of Sa, namely 82 and A3• It is clear that the set A of t E V(K) which fail to satisfy these conditions is thin. • G == 54 . The maximal subgroups of G are Sa, A&, and Dol, the dihedral group of order 8. The case of G c S3 or A& can be disposed of by imposing the same conditi on. as in the case G = S3i to handle the case of D.. , one requires the cubic resolvent
( X  (ZlZ2 + :1:3:1:.. » ( X  (Zl:1:3 + Z2:1:4» (X  (2:IZ4 + Z2Z3)) 3 X  4,X 2 + (4143  4(4)X + (44244  4� 41  � )
=
to ha.ve no root in K J where ZlJ
• • •
, Z.. are the roots of the polynomial
• G = Ss . The ma.ximal subgroupi of G are S4t S2 X S3. and As , which give conditions similar to the above, and the Frobenius group F20 of order 20, which is a lemidirect product OsC4 and can be viewed as the group of affi ne linear transformations of the form z ...... 4Z + b on Z/5Z. If the Galois group of the specialized polynomial is contained in F2Ch then the sextic resolvent, which is the minimal polynomial over K(V) for
( :C I Z2 + Z22:3 + Z32:4 + Z4:1:S + ZSZl )  (:1:1:1:3 + Z2:t4 n,
+
Z3:1:S + Z4 2: 1 + ZSZ2)
ha.s a root in K when specialized.
3 .4
Hilb ert 's irreducibility theorem
Theorem 3 . 4 . 1 (Hilbert [Hi)) II K is
a number field, then jor every n1 the 4ffine space A· ( or equivalently the projective space Pta) has the Hilbert prop erty over K . (In other words, K is Hilbertian.)
By cor. 3.2.2 and cor. 3.2.4, it is enough to show tha.t the projective line P1 over Q has the Hilbert property. There are several ways of proving this: 1. Hilbert's original method ([Hi]): The proof uses PWleux expansions (cf.
(1)). It showl that if A C A l (Q) is thin, then the number of integers n E Z n A with n < N is D( N' ) for some 6 < 1 , when N . 00. (But it does not give a. good estimate for 6.)
2. Proof by counting points of height smaller than N on P l (Q}: write e E PI (Q) as (:I:, y) with z , y E Z and :I: , y relatively prime. This can be done in a.
Cha.pter
26
3.
Hilbert 's irreducibility theorem
unique wa.y, up to a choice of sign. The height of ( is defined to be height(!) The number of points in P l ( Q ) � N 2 . On the other hand: Prop osition 3.4.2
height
� N
is
<: NJ
=
wi t h
sup( I :t I, Iyl) . height less than
N
is asymp totically
If A C Pl (Q) is thin, then the number of points of A with when N . 00 .
Sketch 0/ proof: We may assume A i s of type (02 ), A c ,,"(X(Q» , where X is is an absolutely irreducible curve and ?I" : X 4 PI has degree � 2. Let Ratx (N) be the number of points :t in X(Q) with height(?r(:t» :5 H, and let 9 be the genus of X. Case 1: 9 � 2. Then X(Q) is finite by Faltings's theoremj hence Rat x ( N ) is
0(1). One could also invoke an earlier result of Mumford (d. [L] ,[Se9] ) which gives Ratx( N) = O(loS log N ) . Case 2 : 9 = 1: B y the MordellWeil theorem, and the Neron Tate theory of normalized heights, (see [ L] ) , one has Ratx(N) ::::: O« log N)r/ 2 ) , where 7 is the rank of X(Q). Case 3: g = 0: let Hx I H denote the heights on X and Pl respectively. It is known tha.t Hx ;::.. (H 0 1r )tn, wher e m = deg 1r . H en ce , the number of points on X with H 0 'K most O( N1/m ) .
� N
is at
3. Proof by counting integral points: A variant of the second proof shows that the number of integral points in A with height len than N is O ( N t ), which is an optimal bound. 4. Proof by counting Sunih: let S = {Pi, . . . , PA: } be a. finite nonemp ty set of primes, Es = {±p;"L . . . p:'I1 } , ?'nt E Z. Let 0 E Q, and let 0 + Es be the set of 0 + e, where e E Es . If A is a thin s et in P1 (Q) J then A n (0 + Es ) finite number of o .
Proposition 3.4.3
for all but
a
is
finit e
Recall that, if V i s an affine variety over Q , a subset B of V(Q ) i s called quasiSintegral H for every regular function f on V, the set fe B ) has bounded denominators in the ring of Sintegers, i.e., there is a nonzero integer (J (de pending on f and B) such that Bf(b) is an Sin tege r for all b E B . Let now ?r; : Xi + PI be a finite number of coverings. and choose a outside the finite sets ram(wj ) of X at wh i ch 7fi is ram ifie d . If Bj C X; (Q) is the set of elements of Xj(Q) with '1rJ(b) E 0 + Es , then Bj is a. quasiSintegral
3.5.
Hilbert property and �a.k a.pproximation
27
 ':1r;I (OO)  ""'i l (a). Over Q, 11"; 1 (00) U 11";1(£1) has at least 3 elements, since ""'il (a) f. ram(wj). A theorem of Siegel, Mahler &Dd Lang then shows that Bi is finite ([L], [Se9 ) ) . Prop . 3.4.3 follows.
set in the affine curve Xi
Remark: The bound given in proof no. 3 can be. extended to affi ne nspace More precisely, let A be a. thin set in An(Q), and let Int A(N) be the number of integral points (:l: 1 , . . . , :en) E A with 1 :1:1 1 $' N. Then:
An.
Theorem 3.4.4 ( S . D . Cohen) IntA(N)
=
O(N" \ 10g N ) .
One c an replace the log N term in this inequality b y (log N)", where .., < 1 is a const ant depending on A. The proof is based on the large sieve inequality: one combines tho 3.6.2 with cor. 10. 1 . 2 of the a.ppendix, d. [Coh} and [Se9] .
Let X C Pn be an absolutely irreducible variety of dimension r. As above, denote by Ratx ( N ) the number of points of X(Q) with height :5 N. IT X is linear, deg X = 1 t then Ratx ( N ) ;::. N�+ l . P roblem:
IT
deg X � 2, one (:an
show, using
t ho
Ratx( N)
3.4.4 =
(d.
[Se9J), tha.t
O ( N�+\ log N).
A better result follows from results of Schmidt [Sc:hm]. namely, Ratx Can
thil eltimate b e improved to: Ratx (N)
3.5
=
( N) = O ( N,,+t ).
O ( N1'+C) , for every
t > 01
Hilbert property and weak approximat ion
Let K be a number field, EK the set of places in K (including the archimedean ones). For v e EK , let K. denote the completion of K at v , and let Nv be the cardinality of the residue field of K. in case v is nonarchimedean. If V is an absolutely irreducible integral variety o"er K, V(K,,) is
naturally
en dowed
with a Kvtopology which gives it the structure of a. K.analytic: space (resp. manifold, if V is smooth).
Proposition 3.5.1 If W C V, w 1: V, then W( Kv) 1: V(KII ) for all DU,t finite number of v E EK •
a
Chapter 3. Hilbert's irreducibility theorem
28
Pro p osition 3.5 .2 Let W be absolutely irreducible, of same dimension
W
as
V,
deg 1r > 1 . Let Kf( be the algebraic closure of K in the. ezte.nsion K(W)gal / K ( V ) . If 'U E EK splits completely in K'Jf and Nv is large enough, then 1r(W(K,,» # V( K,, ) . and
1r :
The proofs
Exam p le:
an d define
�
V
be a generically S"U1iecti'fJe morphism,
of prop. 3.5. 1 and Take K
1r :
W
=
.
Q, V
3. 5 . 2 will be
gi ven in § 3.6.
= PI, W the curve defined by
y3 = ( x 2 + 3)/(,2:2 + 12) ,
V by 'K ( :Z: . y ) = :z: . The n Kf( = Q ( yC'3) . 1 (mod 3). If p ==  l (mod 3 ) , p � 5, then
A prime
p
K.". when p == one checks that 7r : W ( Qp) + V(Qp) is an isomorphism of analytic Qp.manifolds . This shows that the con d i t i on ICV splits completely in K/' cannot be omi t te d. splits in
Theorem 3 . 5 . 3 Let A be a thin subse.t of V(K ), and let So b e. a finite subset of �K" then there is a finite set S of places of K satisfying : 0. a) s n so b) The image of A in llues V(K,, ) is not dense. =
Observe first that if the theorem holds for A l and A2, it holds also for Al U A2: for, choose Sl sat i sfyi n g the conclusion of the theorem for Al i then choose 52 satisfying the conclusion So U Sl '
Then, taking S
is not dense: the
=
of the theorem Sl U S21 one has
point ( :& 1 1 :1:2 ) ,
for A2• b ut with So rep laced by
where :Z:i E nve s.
in the closure of A l U A 2 i n ilv e s V(K1J ) ' Hence i t is enough t o prove tho 3.5.3 for
V(K., )
 closure(Ai ), is not
sets of type (el ) and (02),
V, W # V, A C W ( K ) , then ch oo se S {v} with v large enough , ;;. SOl W ( Kv) # V(K,, ) (prop. 3.5. 1 ) . Since A. C W( Kv), A i s n o t dense in W ( K., ) .
1.
If
W
so that
C
v
=
Ass ume A C
'K(W(K»), where 1r : W t V i s generically surj ective. dim V I W is absolutely irreducible1 and deg 1f � 2. We ma.y also assume t hat 1f is a finite morphism (replace V by a s ui t ab le open subvariety) . 2.
dim W
=
a v ;;. So such tha.t 1r(W(Kv» =1= V(Kv). Since closed subsets into closed sub sets. T his shows th at 1r(W(K,,)) is closed in V(Kv)j hence A is no t dense in V(K,, ) . By
1f
prop. 3.5 .2, there exists
is finite, it transforms
3.5.
Hilbert property and
29
weak approximation
Corollary 3.5.4 Assume
V is
CIi
projective 11arietYt
ancl let A
denote the clo
SUre of A in the compact space
II V(K.,) .
• t!So
Th.en the interior of A is empty, i. e., A is nowhere dense in the product.
One says that V has the weak appro:rimation property for a fini.te if V( K) is dense in nVES V( Kv ) .
set S of
places
Lemma
3.5 . 5 If V, V' are smooth, birationally equil1C1ilent, then V has the
'Weak approzimation property for S if and only i.f V hu the weak appro i2:ima non property for S. (In other words, the weak approximation property is a
birational property for smooth varieties).
It is enough to prove the lemma when Vi = V W, with W a closed subvariety, W # V. Clearly, if V has the weak approximation p roperty for S, so does V  W . Conversely, if V  W has the weak approximation property for S , one uses smoothness to prove that V(KtI)  W(K,,) is dense in V(K,,). Hence, any point in W(Kv} can be approximated by points in V(K)  W(K}. A s a special. case, any smooth Krational variety has the weak approximation property for any finite set S of places. Remark: T he smoothness assumption is necessary: for example, consider the Qrational curve 'Jl
=
(:1:2  5)2 ( 2  :1:2 ) .
Its rational points are not dense in the set of its real points (the points (  \1'5, 0 ) (../5, 0) are isolated).
a.nd
Definition 3.5.6 A 11ariety V is said to have property (WA) if it satisfies the weal approzimation property with respect to S for
CIilI finite
S e EK .
It is
property (WWA) ( "wed wed approzimation propertyn) if there a finite set So of pla.ces of K such thClit V ho.s the weak approximation
said to have
e=z:iats property with respect to S e EK, for all S with S n So
Examples:
1. A K·rational
=
0.
variety has property (WA).
2 . A Ktorus has property (WWA), but not necessarily (WA). More precisely, if it
is split by a. finite Galois extension L/ K, one can take for exceptional the places of K whose decomposition group in Gal( L/ K) is not cyclic. (See, e.g. , [Vol),[CTSl) . ) set
So
Chapter
30
3.
Hilbert's irreducibility theorem
Theorem 3.5 . 7 ([Ek] ,[CT] ) A l1ariety which has the WWA property satisfies the Hubert property. V( K ) would be thi n . B y tho 3.5.3 there would exist Proof: If not , A S disjoint from So such that V(K) would not be dense in TIues V(K,,); this contradicts WWA. =
The followi ng conjecture is due to ColliotTheHme [CT] i it is closely relate d to the questions discussed in [CTS2] :
Conjecture 3.5.8 A K uninltional smooth variety has the
property.
WWA
Recall t hat a variety is Kunirational if there exists a generically surj ect ive map Pn + V defined over K , for some n  one may always take n equal to dim V.
Theorem 3.5.9 Conjecture 9. 5.8 implies that group over Q .
every
finite group
is
a.
Galoia
Proof" Make G act faithfully on W = Aft for some n, and let V = W/ G. Then V is Kunirational. By conject ure 3.5.8, V ·lnOotb has the WWA p roperty, and hence satisfies the Hilbert property. B y cor. 3 . 3.2, G can. be realized as a Galois group over Q. 3.6
Proofs of prop . 3 . 5 . 1 and 3 . 5 . 2
Let 0 denote the ring of i ntegers of K , an d choose a sch eme .l::': of finite t y pe over 0 having V as its generic fiber; any two choices for V coin ci de outside a finite set of primes, i . e ., they become isomorphic as schemes over Spe cO[ � ] for some nonzero d. If 'l1 is a. nonarchimedea.n pla.ce, p. t he corresponding prime of 0, denot e by It(v ) = O/P" the residue field a.t 'lit which is a. finite field with N'I) elements . Let V( It( v)) be the set of ,,(v ). rat ional points of � (or, equivalently, of the fiber of V at 'l1). We shall use the followi ng known result:
Theorem 3.6. 1 (La.ngWeill If V
is
absolub!ly irreducible over K then J
The ori g inal proof of LangWei I is by re du cti on to the case of curves for which one can use the bound proved by Weil. A different method, which gives a mor e precise error term, is to use DeligIle's estimates for the eigenvalues of the Frob enius endomorphism, together with Bombieri's bounds for the number of zeros and poles of the zeta function, see [Bo] .
3.6.
Prooti of prop.
3.5.1
and
3.5.2
31
Proof of p r o p . 3.5.1 We ma.y assume that V is smooth and W = 0 (by replacing V by V  W). If N'IJ is large enough, tho 3.6. 1 implies that V( It('IJ» is not empty. Choose � E V(It('IJ». Since V is smooth at v (for N'IJ large enough). :t lifts to a point :E E V(Ov), which is contained in V( Ku). Hence V(Kw ) is not empty.
Proof of prop. 3.5.2 Recall that we are given a generically surjective map 11" : W + V, with deg 11" > I, dim W = dim V. By replacing W and V by open subsets if nec essary, we may assum e that X' is finite etale and V is smooth. We may also choose a scheme of finite typ e j£ for W over some O[�1, such that 'Ir comes from a map (also denoted 'Ir ) If: ...... V. By changing d, we may further assume that 'lr : W + J[ is etale and finite and V is smooth (see, e.g. EGA IV 18). For v prime to tI, we have a diagram W(It('IJ»
f
V(It('IJ»
f
�
1£(0.) �
V(Ou)
'+
W(K. )
'+
V(K. ).
�
The fact that 'Ir : W + V is finite implies that the right square is Cartesian, i.e. a point Z E W(K. ) is an O.point if and only if 11'(%) E V(K. ) has the same property. M oreove r, the reduction map V(O.) + V« 'IJ» is surjective, since V is smooth. To show that W(Kv) :F V(K.) it is thus enough to prove that W(It('IJ» :F V( It('IJ» . This is
a
consequence of the following more precise result:
Theorem 3.6.2 K�I one has
when
with
c= 1
c' <
Let
m = deg .,.. (m �
2) . ThenJ for
'11
splitting completely in
 �I ' (This implies:
1, for N'IJ large enough, '11 splitting completely in Kw.)
Let 1I::s.J be the Galois closure of j£; its Galois group G injects i nto S", I an d hence IGI S m!. Now, divide the points in }£:(It('IJ)) into two Bets:
Chapter 3. Hilbert's irreducibility theorem
32
where A
.
I At + I B !
H
1.1
a.t v
By tho 3.6. 1 applied to W, we have: =
b solu tely irreducible,
a.
all
the connected components of t he
letting
and hence,
components, we ha.ve
IWsal(�( 1.1»1
We ha.ve this gives
=
=
I;'
e(Nv)di:mV + O« Nv)d.im V  i ). H
m
Hence :
Wral( It ( v» gives
Nv )dhnV l ).
5
..!.IAI + IB I + O« Nv )dim V  l )
�
( Nv )dim V  (1 
m
� )IAI + O« m
Nu )di m V : ) ,
and therefore
Finally, since
(1 
acts freely,
=
� IAI + O« 'r(A)1 + Ir( B ) 1
Since
(N1.1)dimV + O« NtI )d.im V  � ) .
The same argument applied to the act i on of G on
IT( A )I
�) I�\
�
fiber
e be the number of these
A = WIIII ( It(v))/ H where H = Gal(WralIW). lA '
and B is the
( Nv )dim V + O« Nv)dbn V  t ) .
s plit s completely in K'ff' 1 then are
wralc"c v )),
is the set of points which can be lifted to
set of the remaining p oints
"'�HI
'/r(W(It(u») � (1 
�
I�I � ,!!
I
we get :
\ )(NV )diD1V m.
+
O« Nu )dim V  i ) ,
and this completes the proof of tho 3. 6.2, and hence of prop. 3.5.2.
3.6. Proof. of prop_ 3.5 . 1
and
3.5.2
33
A p plication to the distribution or Frobenius elements
Let let
E be
a.n
dliptic curve over
ap, for p prime, be the
Q without complex multiplication over QJ 4p
N� is
where
the
and
"trace of Frobenius" : =
1 + p  Np
nu mber of point s of B. over F�. The following
is
wellknown:
3.8.3 1/ / 1: 0 is Any polyn.omial in t'Wo vAriables over Q, then. the set 0/ primes p such thGt J(p, ap) = 0 ha.s de1L!lity O.
Theorem
( The
proof uses the Iaclic representa.tion
p : GQ
+
GL2(ZI)
a.ttached
to
since det p(Frobp) = p, Tr p( Frob. ) = ap, one is lead to consider the set of :z: E p(GQ) such that J(Tr z , det :z:) = O. Since p(GQ) is known to be open
E:
GL2(Z,), this .et has Haar measure zero; the theorem follows by Chebotarev'. density theorem. )
in
More
generally, 3.8.4 Let A be G thin subset of Z A ha.s d.ensity O.
Theorem
(p, tip )
E
applying
X
Z. The set oj p 's such th4t
is already proved for A of type (01): so assume A is of type (02 ). Let A, be the image of A in ZIIZ x Z/lZ and let 5, be the set of � E GL (Z/IZ) such that (Tr ( . ) . det ( � » bdongs to A, . Qne checks that 2 I S. I :5 I A,I . l2 (1 + 1 11). The theorem
By tho 3.6.2, we ha.ve
J AIl 5
c12 for c <
completely in some fixed extension K of
Is, l ::; This shows tha.t the density
S
of S,
1 Qi
crt { l +
in
and
I sufficiently
large,
splitting
hence
Ill).
GL2(ZIIZ)
c( l + 1/1)(1  1/Zt1 ( 1  1 /12 r1
is =
c(l  1 /1r
2
' " , 1m are large enough distinct pri me points of E is n GL,(Z/l.;Z). H each 'i splits completely in K the Chebotarev density theorem, a.pplied to the field of 11 l", division points of E, shows tha.t the upper density of the 2 set of primes p with (p, lip) E A is S cf'4 n�l (l  1 /ls) . Since this can b e m.ade arbitrarily small by taking m large enough, t he theorem follows.
if 1 is large enough. Recall now that, if 11 , numbers, the Ga.lois group of the 11 • . •
• •
 i.division
Chapter 3. Hilbert'.
34
irreducibility
Remar ks:
theorem
.
1.
The above implies, for example, that the set of the primes p for which E(F, )I  3 is a square has density O. I 2. One can prove more than density zero in 3.6.4j in fact, one has
1{P l p < N, (p , a.)
E
A}I = 0
This implies that
E
( log �)1+' ) ' 1
P (P,op)EA
<
for
some 6
00.
The proof (unpublished) uses the Selberg UA2 " sieve. 3. There are similar rellults for the Ramanujan 'Tfunction.
>
O.
Chapter
4
Galois extensions of examples 4. 1
Q (T) :
first
The property GalT
Let E be a. finite Galois extension of Q(T) with group G which is regular, i.e., Q n E = Q. Geometrically, E can be viewed as the function field of a smooth proj ective curve 0 which is absolutely irreducible over Q; the inclusion Q(T) '+ E corresponds to a (ramified) Galois covering C + PI defined over Q with group G.
Conjecture
4. 1 . 1 Every finite group
G
covering.
occurs a.s the Go.lois group of such a
Let us say that G has property GalT if there is a regular Gcovering C t P 1 as above. In that case, there are infinitely many linearly disjoint extensions of Q J with Galois group G (d. tho 3.3.3). Remark: If a regular Gcovering exists over Pta , n � 1 , then such a covering also exists over Ph by Bertini's theorem (d. e.g. [Jou] ).
Examples:
The
property GalT is satisfied for:
1. Abelian groups. 2. � and 9ft (Hilbert);
.A. (Vila, Mestre) . 3. Some nonabelian simple groups, such as the sporadic ones (with the possible exception of M23), most PSL2(F,), p prime, and a few others. 4. If G has property GalT, then so does every quotient of G. Proposition
G1
X
4. 1 .2 If
GI , G2
h.ave property
GalT}
then so does their prodtJ.t:t
G2 •
Let Ci t O2 be regula.r coverings of PI with groups Gl t G2 , and let Ih , E2 be their ramification loci.
Proof:
35
Chapter 4. Galois extensions of Q (T)
36
1. If EI n E2 = 0, then the extensions corresponding to Ch C2 are linearly disjoint, because PI i s algebraically simply connected ( see §4.4 below). One can take for covering C the fibered product Cl X PL Ch which has function field Q( CI ) ®Q(T) Q( C2). and hence has 01 x O2 as Galois group. 2. If El n E2 f:. 0, one modifies the covering 01 t P I by composing it with an a.utomorphism of PI s o that the new ramification locus is disjoint from E2• One is thus reduced to case 1. Remark : If G 1 and G2 have property GalT, one can show (cf. e.g. [Ma3), p. 229, Zusatz 1) that the wreath product 01 Wr G'l al s o has property GalT.
This
can
be used to give
an
alternate proof of prop. 4. 2.2 below.
Exercises:
1. Show that the profinite group Zp is not the Galois group of any regular extension of Q(T) . (Hence conjecture 4. 1.1 does not extend to profinite groups , not even when they are p adic Lie groups .)
2. Let G
b e a. finit e group having property GalT. Show that there exists a. regular Galois extension L of Q(T), with Galois group G, such that: ( a) Every subextension of L distinct from Q(T) has genus � 2. (b) Every Q�rational point P of PI has property Irr(P) with respect to L. (Use a suitable base change PI  Ph combined. with Fattings's theorem.)
4.2
Abelian groups
A torus defined over Q is said to be a "permutation torus" if its character group has a Zbasi s which is st able under the action of Gal(Q /Q) 1 or equiValently, if it can be expressed as a product of tori of the form RKi/QGm , where the K. are finite extensions of Q. A permutation torus is clearly rational over Q. Now J let A be a finite abelian group. The following proposition implies that A has property GalT: Prop osition
seq),
4.2 . 1 There emu a torus
such. that the quotient S'
S' is a Qrational variety. )
=
SIA is
S
a
over
QJ
and an embedding of A in
permut4tion torus.
(In particular,
The proof uses the functor Y which to a torus associates the Zdual of its character group. An exact sequence of the form 1 + A + S + S' 4 1 gives rise to the exact sequence 1 where
t
Y (S)
to
Y (S')
to
A
to
1,
A = Extl (A, Z) = Hom( � , A) = H om( A. , Q/ Z ) ,
4.2. Abelian group s
37
and A denotes as usual the Cartier dual Hom ( A, Gm). Choose K finite Galois over Q such that the action of GQ on A factors through Gal(K/Q), e.g. Q(Pn), where n is the exponent of A. Now express A as a quotient of K a free Z[Gal(K/Q)]Mmodule F, and let S' be a torus such th a.t Y(s') = F; it follows t h at S' is a permutation torus, and there i. an AMisogeny S t � . =
Proposition 4.2 . 2 Ltd G be a finite grou.p having property G alT 1 and let M be a finite abelian group with. G action. Th en the semidirect product G = M · G tUso ha property GalT.
We may assume without loss of generality = e gA, 80 th at
M
IIEG
G = ..(A
X
• • •
that M is an induced Gmodule,
X A) · G
101 ti me8
is the urreath. product of A and G. By prop. 4.2. 1, there is an isogeny 5 + 5' defined over Q, with S' a p ermut at ion torus and with kernel A C S( Q). By hypothesis, there is a regul ar etale Gcovering C
+
U where U is a Qrational
variety ( e.g. an open subvariety of PI)' The actions of A on S, and of G on C and on S X x S give rise to a natural a.. action on X = S x . . . x S x C. T his action is free . Let Y = X/a. P rop . 4.2.2 then follows from the following . • •
lemma.:
Lemma 4 . 2 . 3 The variety
Y
=
X/G is Qrational.
' 5 x Define X' = X/(A x . . . x A) . x s' x C . We have Y = X ' / G. This show s that Y is the fiber spa.ce over U with fiber the torus s' x . . . x s't =
"
which i s asso ci ated t o the princip al G.. bundle C + U. We may thus view Y as a torrLS O1ler U. In particular t the generic fiber Yu of Y � U is a torus x s' by a G over the function field Q ( U) (this torus is obtained from s' X twisting. using t he Galois extension Q( C)/Q(U» . Since S' is a permutation torus over Q. Yu is a permutation torus over Q ( U). Hence the function field Q ( Y) of Yu is a. pure transcendental ex tension of Q( U ) , which it self is a pure transcendental extension of Q. The lemma follows. (One can also deduce lemma 4.2.3 from lemma 4.3.1 below.) • • •
generated. by an abelian normal subgroup M and subgroup G having property GalT. Show usin g prop. 4.2.2 that H has property
Exercise: Let H be a finite group
a
GalT
Ch a.pte r 4. Galois extensions of Q(T)
38
4 .. 3
Example: the quat ernion group
We need
first:
Lemma 4.3.1
Let
fiber W . ThefJ.
E is
Y
+
Q8
X be an etale Galois covering with Galois group
G J end G + GL(W) be e lineer representation of G, where W is a finite dimensional vector space. Let E be the associated fiber bundle with base X and biratiofJ.allti isomorphic to X x W .
This follows from Hilbert's theorem 90: Hl (K(X). GLft) = 0 , where n is dim W, and K(X) is the function field of X over the ground field (which we assume to b e Q ) .
[Alternate proof: Use descent theory to show that E is a vector bundle over X hence is lcxally triviaL] This lemma. implies that Q(E) = Q (X)( TI I " " Tn), hence that Q(E) and Q(X) are stably isomorphic. (Recall that two extensions kl a.nd le2 of a fiel d Ie are stably isomorphic if there exist integers nI , n:z � 0 such that the extensions k1 (TJ , ' . . , Tn1 ) and �(Tl " ' " T", ) are Ieisomorphic. ) I
Application:
Let G act linearly on vector spaces WI a.nd Wa , n = dim WI , we have:
W2 being faithful. Letting
4.3.2
Q( WI x W2 ) / G � Q ( W2 / G ) (T1 ,
• • •
(In particular. Q ( W1 x W2 )/G i s
consequence Corollary
the action on
, Tn).
stably isomorphic to Q (Wz /G). ) of the lemma ap plied to E = WI X W:z and X = W,.
4.3.3
If
G
a.cts faithfully on
Q(Wz/G) are sta.bly isomorphic.
T hi s
is
a
WI and W2 , then Q ( W1 / G) and
e Wi b e a d e compos i tion of the regular representation of G as a sum of Q·i rred ucible ones. By the corolla.ry, if one of the Wi is a faithful Gmodule, then Q(R/G) is stably isomorphic to Q (Wt / G). If Q ( Wi/G) is a rati on al fiel d , then so is Q ( R/G), by 4. 3.2. Let R
=
Qs. Let Q s be group algebra Q (Qa] de com p oses as
Application to the quat ernion group
group of order 8.
The
Q[Qa]
=
the
qua.ternion
Q x Q x Q x Q x H,
where H d enotes the standard field of quaternions (over Q). and Qs a.ct s on H by left multiplication. By t he previous remark , the Qrationality of Q [Q.]JQs is equival ent to the Q rationality of H· /Qa) where H· is the mul tipli c ati ve group of H viewe d as a 4dimensiona.l Qalgebra.ic group . The group Qa has
4.4.
Symmetric groups
a center
{±1}
of order
39
2,
and
D
=
Q s /{±l} ia a. group of type
(2, 2).
On the
other hand, by the map
(N, 4J) : H·
.
Gill X S 03,
where N is the reduced nonn, and tP : H· + SOa mapa a quaternion % to l the rotation y � � y� (on the 3dimen8ional vector apace of quaterniona of trace
0).
Therefore, H·/Q a
and it suffices to show that
Gill
SOa/ D
x
SOa/D,
is a ra.tional variety over
Q. But the
the stabilizer in S03 of a flag in A:I. Hence 503/ D is isomorphic to &n open subvariety of the flag variety of A3 t which is rational over Q . Noether's method therefore applies to Q s ; in particular Q a has property GalT .
group
D ia
=
{JY].
For explicit formulae, s ee
Exercise (L. Schneps) Show that every pgroup of order r has property GaiT (use the exer cise at the end of J4.2). 4.4
Symmetric groups
The symmetric group Sft acts on the a.ffi.ne apace A'" with quotient A'" ( "sym metric functions theorem" ) . This shows that 9.,. has property GalT. Let give some explicit constructions of polynomials with S.,.
as
UB
Galois group. For
example, consider a polynomial
and put
I(X, T) Theorem 4. 4.1
(Hilbert
(The polynomial
1
of f(X, T) over Q(T) is
a
=
X" + Ct1 X" 1 + . . . + 4,.

T.
[Hi] ) If 1
is G Morse function, then the splittif&!J field regulGr eztension with Galois group S.,. .
ia called a Morse function if:
1. The zeros {jl t . . . , (j.,.l of the 2 . !C{3i ) #: 1(11; ) for i t= j .)
derivative
t of f are simple.
We will need the following simple fa.cts about the symmetric group Sn :
Lemma 4.4.2 Sft
is g
en.era ted by tratUpositions.
Chapter 4. Galois extensions of Q(T)
40 This is wellknown: indeed
S,.
is generated by the tra.nspositions
( 1 2), (2 3 ) , . . . , (n  1, n ) .
Lemma 4.4.3 Let G be
a transitive subgroup of S,. which contains a transpo sition. Then the follo wing are equivalent : 1. G contains an ( n  1 )cyde. !. G is doubly tra.nsitil1e. 9. G = S,. .
H G contains an ( n  1 )cycle, then the stabilizer of a point is transitive on the complement of the point, hence G i s doubly transitive . IT G is doubly transitive, then G contains all the transpositions in Sn. , hence G = S,. by lemma 4.4.2. That 3 => 1 is obvious.
Lemma 4.4.4 (d. {Hul, p. 171 ) Let G be
a tra.nsitive subgroup of Sn which is genemted by I!'JIdes of prime orders. Then: 1. G is primiti11l!l. !. 1/ G contains G tra.11Sposition, then G = Sn ' 9. If G contains a 3cyde, then G = An. or Sn. .
Let {y) , . . . ) lk}, wi t h Ie > 1, be a partition of { l , . . . I n} which is stable under G. Our assumptions imply that there is a cycle s of G, of prime order p, such U sP1Yi is fixed by s, we that Yi #: sYi. Since no element of Yi U sYi U have: IYi I + I s Yi I + . . . + I sP 1 Yi I � p, . . •
and hence l Yi I = 1 . This shows that { Yi , . . . , Yi:} is the trivial partition of {I, . . . , n} . Hence G is primitive. To show 2, let 0' be the subgroup of G generated by the transpositions belonging to G. Since Gr f:. 1 , it is transitive ( a non·trivial normal subgroup of a. primitive group is transitive). For {} C { l , . n} , let us denote by Sn (respo An) the symmetric (resp , altem&ting) group on O. Let 0 C { i , . . . I n } be maximal with the property Sn c G', and suppose tha.t n f=. { I , . . . I n}. By the transitivity of G' , there exists (%y) E d with % E O J Y r;. n. H enc e Snu{lI} c d , contradicting the maximality of {}. It follows that {} = {i, . . . t n} and hence G = 0' = Sn proving 2. The proof of 3 is similar 1 t akin g G' this time to be the subgroup of G genera.ted by the 3cycles belonging to a. The hypothesis implies that a' is nontri vial, &nd hence is transitive. Choose 0 C { I , . . . , n } maximal with the property An C a'. As before, if Q :f:. {I, . . . , n } , there is a. 3 cy cle ( x yz) which does not sta.bilize O. There are two cases: ' Case 1 : {:Z:, Yt z} n O has two elements, say y and z. Then clearly Anu{a} C O ) contradicting the maximality assumption for O. 0 . ,
I
4.4. Symmetric groups
41
Case 2: {=) lI� Z} n n hu
1 element, say z.
Choose two elements ,/, z' E n
(xyz) and (:ty' z') generate the alternating {=, Y . Z, lI', ZI}. In particular, the cycle (zy' %) is in G; since thi s
distinct from Zj it is easy to see that
group As on
3cycle meets n in two elements, we are reduced t o cue I , QED. More generally, Jordan has shown that a primitive su.bgroup of Sft which contains a cycle of prime order � Th.
4.4.1
n

3 is equal to An or Sn (see [Wi] , p.39).
will be proved in the following more general form:
4.4.5 If K ill Any field of chArGcteristic 01 or of chAmcteristic p not di1liding ft, And f(X) E K[X] is Morse, the" Gal(f(X )  T) = Sft over K(T).
Theorem
Proof.
We may usume
be viewed
&8
K
to be algebraically closed. The polynomial
a ramified covering of degree
f:
PI
Z
+
The corresponding field extension is
l+
n
X=
can
PI
t ::; fez).
K(X)
:J
K(T);
it is separable because
p Jn. Let G C Sft be the Galois group of the Galois closure K(T), i.e. , the Galois group of the equation f(X)  T = O. The ramification points of the covering f are
At
f
of
K(X)
over
00, the ramification is tame, and the inertia group is generated by an
ncycle. At the fUJi), the hypothesis on I implies that the inerti a. group is tame for p :f:. 2 and wild for p = 2, and that (in both cases) it is generated by a transposition.
Hence the theorem is a CO,nsequence of the following propositionl combined with lemma 4.4.4.
P roposition 4.4.6 Let C
+ PI be A regulAr Ga.lois covering with group G, tAlJUly ramified At 00. Then G is genemted by the inertiA subgroups of points outside 00, And their conjug4tu.
Prool:
Let H be the normal subgroup generated by the inertia subgroups
outside 00.
Then C/ H is
a
GI H·covering
of PI which is tame at 001 and
unramified outside. The RiemannHurwitz formula implies that the genus of
O/H is � l(l  IG/H!)i hence G = H.
Corollary 4.4.7 A) P I is a.lgebmica.lly simply connected. b) In chAmcterUtic 0, the 4ffine line is 4lgebra.iCGlly nmply connected.
Chapter 4. Galois extensions of Q(T)
42
In characteristic p > 0 , the affine line Al is not simply connected (as shown e.g. by the Artin·Schreier equation XP  X T). H G is the Galois
Remark:
=
group of an unramified covering of A 1 , then prop. 4.4.6 implies that G is generated by its Sylow psubgroups. There is a conjecture by Abhyankar [Abl] that, conversely, every group G ha.ving this property occur• . This is known to be true when G is solvable, and in many other cases (Nori, Abhyankar, Harbater, Raynaud, see e.g. [Ab2] , {SeU]). Example
of an
8n  ext e nsi on
of ramification
type
( n, n  1, 2)
The above method for constructing polynomials over Q(T) with Galois group Sn gives us polynomials with ramification type (n, 2, . . . , 2). For a different example, consider the polynomial
I(X) so that
I(X, T)
=
=
Th en
xn  xn 1 ,
xn
 )(" 1  T. a =
and hence the ramification is given (in char. 0) by
{
Hence
G
=
(n, n  1, 2).
Remarks:
at 00 : at 0 � at a :
The polynomial I has ramification type
Xp+t  XP  T. This polynomial has Sp+l
as Galois group in characteristic different from p (the proof is similar to the one above) . In characteristic p, one c an show that i t h as Galois group PGL2(Fp ). 2 . One might ask for an explicit polynomial In over Q such that In. has GaloiB group Sn.. Here is an example: l"eX) = xn.  X  1. Indeed, Selmer [Sell has shown that In is irreducible over Q. Assuming this, let us prove that In has Galois group 8ft • We look at the primes p dividing the discriminant of In, i.e . , those modulo which I.,.. ( X) hu a multiple root. This happens if In(X) and f�(X) = nXn1  1 have a common root mod p. Substituting xn l = l/n in the equation I(X ) == 0, one gets X ,:::::: n/(l  n). Hence there can be at most one double root mod p for each ramified prime p. This shows that the inertia subgroup at p is either trivial, or is of order two, generated by a transposition. But G Gal(/) is generated by its inertia subgroups, because Q has no non trivial unramified extension. By Selmer's result, G is transitive; we have just shown that G is generated by transpositions; hence G = Sn by lemma 4.4.4. =
=
n
cycle of order nj cycle of or?� r n  1 i a transpositlon.
Sn. by lemma 4.4.3.
1. Consider leX, T)
1
1  ,
43
At.
4.5. The alternating group
Many more examples can be found in the litera.ture. For instance, the "trunca.ted exponential" Z2
Z3
l + z + 2" + "6 + " ' + z,. n�
(mod 4), and Galois group � otherwis e
haa Galois group S", when n :/: O
(1. Schur).
Exercises: 1.
Let Y,.
1.1.
of P,. l
be the subvariety
Y,. is
an
94
=
for i =
0, 9&
==
4. 96
=
=
1 +
(n  2)!
49, 9T
=
48 1,
1 .2. The quotient of Y,. by Sn. ( actin g by
to P1• 1.3.
The Galois covering
given by
1, 2, .
absolutely irreducible smooth curve, 9n.
(e.g. 93, =
e homogeneous equations
defined by th
Y,.
+
PI
• .
whose
n2  5n + 2 4 • . •
show
genus 9n.
is
given by:
.
)
permut ation of coordina.tes) is isomorphic
is the Galois closure
the polynomial I(X) = xn.  x,. l .
2. In ciaaracteristic: 1 1 ,
, n  2.
of the
degree
n
covering
that the equation
is an unramified extension of A l whoae Galois closure has for Galois group Mathieu group Ml l . (Hint: reduce mod 1 1 the equa.tion of [Ma.41 , after divi�ing Xvariable by
4.5
the
the
111/4.)
The alternating group A n
One exhibits the alterna.ting group An as a Galois group over Q(T) by using following lemma ( " do uble group t rick" ) .
the
Lem m a 4.5 .1 Let G be the Galois group oj G regulAr eztension K/k(T), ram
ified At most tit three plAces which Are ra.tional over k, and let H be tJ. subgro'Up of G of inde� !. Then the ped field Kl of H is rAtional. (In particular, if
k
=
Q, then
H
has property GalT.)
Beca.use of the conditions on the ramification, the curve corresponding to Kl has genus sero, and has a krational point. The lemma follows.
Chapter 4.
44
G aloi s
extensions of
Q(T)
For example, the polynomials with ramification typ e ( n t n  1 , 2 ) di s cus s ed can be used to cons tru ct Anextensions of Q ( T ) . More precisely, let us change variables, an d p ut in the previous se ction
he X, T)
=
(n  l)xn  nXn1 +
T.
Then the discrimina.nt of h (with respect to X) is
Up to square factors, we have
�( h) ......
{
( l)T(n  l)(T  1) ",t
( 1 )2 nT(T  1)
if n is even;
if n is odd.
Hence the equa.tion D2 == � define s a rational curve. For ex a.mple , even, by replacing T by 1 + ( 1)i {n  1 )T2 J we get the equation
(n _ 1 )xn  nXn 1 + 1
+
(l)T ( n  1)T2
=
if
n
is
0,
gives rise to a regu l ar Galois extension of Q ( T) with Galois group An . Hilbert's original construction [Hi] was somewhat different. For th e sake of simplicity, we reproduce it here only in the case where n = 2m is even. Choose a polynomial
which
g(X)
=
ml
nX II (X  1',) 2 ,
with the Pi distinct and nonftzero. Then, take f(X) so that df/dX g. A ss ume that the fUh) are a.ll di stinct , and distinct from 1(0). Then f h as ramification type (n, 2, 3, 3, . . . , 3). Hence its Galois gr oup is S by lemma n 4.4.4. But then the quadr&tjc subfield fixed by An is only ramified at the two places 00 and 0; hence it is a rational field. =
Exereise:
4. 6
Show that the
condition that the f({3,) are t
1(0) can be suppressed.
Finding goo d specializat ions of T
Let I be a polynomial over Q( T) with splitting field a. regul ar G·extenlion of Q ( T ) . Although Hilbert's irreducibility theorem guaranteel that for "most" values of t, the speciali zed polynomial I(X, t) will have Galois group G over Q, it does not give a constructive method for finding , say. an infi ni te number of su ch t's.
4.6.
Finding good
specializ&tions of T
45
For p f S, where S denotes a suitable finite set of pri mes , the equation I( X, T) = 0 can be reduced mod p. If p is large enough (see exercise be low ) , then all conj ug ac y classes in G occur as Frobenius elements at t for some t E Fp. Letting C1, , CIt be the conjuga.cy classes of G J one can thu s find. distinct primes PI , . . . , PI. and points th . . . J t�, with ti E Flti • such that FrobJPf (f(X, t.» = Ci . Specializing T to any t E Q su ch that t == ti (mod Pi) for all i gives a polynomial f( X ) = f(X, t) whose G alois group over Q int er sects each of the conjugacy classes Ci, and hence is equal to G, by the following elementary result� • • •
Lemma 4.6. 1 (Jordan, [J2] ) Let G be G finite group, and H G which meets every conju.gAcy cla,ss oj G. Then H G.
a
subgroup 0/
=
Indeed, if H has a.t most
is a subgroup of G'I then the union of the conjugates of H in G
1 + ( G : H) ( l H I  l )
=
I G I  « G : H)  1 )
elements. Exercises: 1. Use lemma 4.6.1 to show that every fini te division
algebra is commutative (d. Bourbaki, A.Vm, §111 no.l). 2. Show th&t lemma 4.6.1 can b e reformulated as: "Every transitive subgroup of Sn, n � 2, contains a permutation without fixed point" . (This is how the lemma is stated in Jordan, [J2].) 3. Let 11" : X t Y b e a. Gcovering of absolutely irreducible projective smooth curves over Fp' Let N be the Dumber of geometric points of X where " is ramified, and let 9 be the genus of X. As su me that 1 + p  2g';p > N. Show tha.t, for every conjugacy class c: in G, there is a. point t E Y(F,,) over which 'If is unramified, and whOle Frobeni us class in G is c. (Apply WeiPs bound to the curve X twisted by an
element
of c.)
I
5
Chapter
G alois extensions of Q (T ) given by torsion on ellipt ic curves 5.1
Statement of Shih's theorem
Consider an elliptic curve E over Q(T) with iinvariant equal to T, e.g. the curve defined by the equation 2
y + zy  z _
3
T
36 1728 z
_
T
_
1 1 72 8 '
(Any other choice of E differs from this one by a quadratic twist only. ) By a.djoining to Q(T ) the coordinates of the ndivision points of E, one obtains a Galois extension Kn of Q(T) with Gal(Kn/Q(T» = GL2(Z/nZ). More precisely, the Galois group of C · Kn over C(T) is SL2(Z/nZ ), and the homo� morphism GQ(T) + G L2(Z/nZ ) � (Z/nZ)· is the cyclotomic character. Hence the extension Kn is not regular when n > 2: the algebraic closure of Q in Kn il Q(Pn). So the method does not give regular extensions of Q(T) with Galois group PGL2(F.. ) nor P SL2(Fp). Nevertheless, K.y. Shih was able to obtain the fonowing result [Shihl], [Shih2] :
.
Theorem 5 . 1 . 1 There ezists " regultl.f" eztension of Q( T ) 'With Go.lois P S L2 ( Fp) if = 1. = 1, ; =  1, o r
(�)
()
(�)
Shih's theorem will b e proved in §5.3.
Remark: It is also known that P SL2(F..) has property Ga.h when this follows from [M12] combined with tho 5. 1 . 1 .
47
(�)
group
=

1;
Chapter
48
5.2
5.
Elliptic curves
An auxiliary const ruction
Let E be an elliptic curve defined over a field k of characteristic O. The p torsion of E, denoted E [P], is a two.dimensional Fp.vector spacej let us call PE[P] the associated projective line P E[P) � P l (Fp). The actions of the Galois group Gle on E[P] and P E(P] give rise to representations P
and
:
p : G"
Ole
+
+
GL(E[PD
PG L( P E [P))
�
�
GL 3 (F,) PGL 3 ( Fp ).
The determinant gives a welldefined homomorphism PGL2(Fp ) + F;/F;3 . The group F;/F;2 is of order 2 when p is odd (which we assume from now on ) . Hence the homomorphism Ep det op is a quadratic Galois character, =
fp :
GK
�
{± 1 } .
The Weil pairing gives a canonical identification of J\ 3 ( E[PD with IJp , so that det p ::::; X, where X is th e pth cyclotomic chara.cter giving the action of G, on the pth roots of unity. Therefore Ep is the quadra.tic character associated to k(#), where p. (l )�p. Let K be a qua.dratic extension of Je, and let tr be the involution in Gal( K/k) . Let N be an integer � 1 and E an elliptic curve defined over K such that E and E' are Nisogenous (i .e. , there is a homomorphism � : E + E' with cyclic kernel of order N). Assume for simplicity that E has no complex multiplication, so that there are only two Nisogenies, ¢ and  � , from E to =
E',
If pIN, the maps �, ¢ induce isomorphisms (also denoted ¢, t/J) from E[P] to E'[P] . By passing to the projective lines , one gets an isomorphism ¢ : P E[P] + P E' [P] which is independent of the choice of the Nisogeny E + E', and commutes with the action of GK. We define a map PE,N : GIr. + PG L (P E[PD as follows: 1. If oS belongs to the subgroup GK of GIr., then PE,N(S) is defined via the natural action of GK on P E[P] .
2. If oS E Gle  GK, then the image of S in OK/It is a. Hence s gives an isomorphism PE[P] + PE'[P], and p(s) : PE[P] + PE[P] is defined by composing this isomorphism with � l : P E'[P] + P E[P]. describes the action of G, o n the pairs of points ( z , y ) in P E[P] X P E' [P] which correspond to each other under the isogeny �. ) One checks easily that the map (In other words,
p E,N
PE,N
�
Olr.
+
PGL(PE[PD
5.3. Proof of Shih'. theorem
49
so defined is a homomorphism.
As before, the projective representation p ai N of Gle EBJI
:
Gle
to
gives
a
character
±1.
Let EK denote the character Gle to ±1 corresponding to the quadratic exten sion K/le. Recall that fp : Gle to ±l corresponds t o k{H)/k.
Proposition 5 .2.1 We have :
(�) if (�) and (;: ) ;::: if
Corollary 5.2.2 1/ K image of PIl,N
u
=
le( #),
=
1
= 1
lJ then EIl.N
contained in PSL2 (Fp) .
=
I, i. e.,
tAe
Proof of 5.1. 1 : If s E GK, then PB.N(S) acts on A2 E[P] � ".." by Xes). Hence = f,t.( .. ) . If .J f/. GK, then using the fact that the homomor EB,N(S) phism t/I l : Ecr [pl + Efp] induces multiplication by N1 on =
(�)
( where the group operation on IJ.p is written additively), one finds that
det PE.N(of)
Hence EB,N(.!)
5 .3
=
=
N1X(of).
(�) Est(s). This completes the proof.
P roof of Shih's theorem
Let KN denote the function field of the modular curve
Q,
Xo(N) of level N
over
KN =: Q[jh j 21! FN(it , j1),
where FN is the (normalized) polynomial relating the iinvariants of Niso. genous elliptic curves (cf. IF]). The Fricke involution WN acts on KN by interchanging ;l and ; 2 . Let Xo(N}P be the twist of Xo(N) by the quadratic character Ep t using the involution WN• The function field of Xo(N)P can be de scribed as follows: let leN be the field of invariants of WN in KN t i.e., the £UDc�
ti on field of Xo(N)+. Then KN and I:N( H) are disjoint quadratic extensions
50 of
Chapter
leN .
The third quadratic extension, Ie, of
t.he function field of Xo (N)"
leN contained
in
5. Elliptic
K
=
curves
KN ( H) is
Now, let E denote an elliptic curve over K wit h jinvariant. jl ' Then Ell' has = I, j invari ant j2, and hence E and Ell' are Nisogenous. Assuming cor. 5.2.2 gives a projective representat.ion Gle t PSL2(Fp l . It is not hard to see that. this representation is surjective) and gives a regular extension of le.
(�)
Remark: Instead of taking E ove r K one can take it over KN. One thus gets projective representa.tion Gle N + PGL2 (F,,) and prop 5.2. 1 shows that it is surj ective and gives a regular extens ion if C� ) =  1. Since kN is known to be isomorphic to Q(T} if N belongs to the set P a
.
,
s=
{2 . 3 � 5 , 7. 11, 13, 1 7, 19 , 23, 29, 31, 41 , 471 59, 71},
this shows that PGL2(Fp ) has property GalT provided there exists N E S with =  1 (a computer search shows tha.t this is true for all p < 5329271).
(�)
We
are interested i n values o f N for
which le
such th at :
Xo(N) has genus O.
1. 2.
Xo(N)P has
Assuming
that

a Q rat ional N
is
not a
i s i s omor p hic t o
Q (T),
i.e. ,
point .
square,
condition 1
implies that N = 21 3 , 5 , 6, = 2 , 3 , 7 ( and also for cf. {Shihl] ) . More precisely:
7, 8, 1 0, 12, 13, 18. Condition 2 is satisfied when N N
=
6,
(: )
=
1 and
N
=
1 0,
Proposition 5.3 . 1 For N a re
=
(; )
::;; I ,
2, 3, or 7, the two JUed points of WN in Xo (N)
rational over Q.
(Hence, these points stay rational on Xo ( N )P, which is thus isomorphic to
over Qj this concludes the proof of tho 5.1 . 1 . ) Let uS give t wo proofs of p rop
. 5.3.1.
P1
First proof: Let 0 b e an order of class number 1 in a quadra.tic imaginary field, a.nd Eo an elli ptic curve with endomorphism ring O. Assume that the unique prime ramified in the field is N I and that 0 contains an element '1r with '1r7r � N. This is indeed possible fo r N = 2, 3. 7:
5. 3 . Proof of Shih's theorem
51
N
0
?r
2
Z[iJ
1 +i
Z [v'2]
v'2
Z [ 1 +0] Z[v'=3]
3
Z [�]
7
Z [v'=71
vC3 v=a
.;=7 v'=7
By the theory of complex multiplication, t he pair ( Eo , ?r ) then defines a ratio po int of Xo (N) which is fixed under WH.
nal
Second
prool: Let 6. be the discriminant modular form ) q ;;::: e2ft· , A( q) = q II(l  qft 24,
and let f be the f
power series
= =
defined by
(6(z)/A(Nz) lf.::r ql II ( 1  q" )2m ,
ft� 1
".O(H)
where m =
12

N l
.
One can prove t hat I generates KN over Q, and that fWN = Nm / f. For N = 2, 3, and 7 we have m = 12, 6, 2, and 1r' is a square. This shows that the fixed p oi nt s of WH are rational. One can also show t hat the quat erni on algebra corresponding to Xo (N}P is given by (N , p.). More generally!
Exercise: Let F be
a.
field, and let
X
be the variety obtained by twisting P I
by t he quadrati c character attached to t he exten6ion F( v'b) and the involution
P I = � 4/= . Then the quaternion invariant of this curve Br2 (F).
is
of
the element (4. b) of
The ramification in the Shih covering constructed above is as fo llows :
N==2:
is of type (2, p, p). One ramifica.tion p oi nt is rational over Q , with inertia group of order 2 generated by the element The two ot hers are defined over Q( H) and are conj u gate to each other. Their i nerti a group s, of order , . correspo nd to the two different conj u ga. cy classes of unipotent elements in PSL2( Z ), and where a is a nonquadratic residue mod p. The ramification
(�l 5)'
(A i )
(6 1)'
C hapter 5.
52
Elliptic curves
N =3: The situation is simi lar ; we have ramification of type (3J P, p ) , inertia group of or d er
N =1:
The
being generated by the element ( i (01 ) .
3
ramification is of ty pe
(3, 3, p, P) i
nates, the ramification points are located at
Exercise:
that
S how
the
replaced by PSL2 (Z/P"Z).
statement of tho 5. 1 .1
(�)
Assume as before that +
Xo(N'f
= 1.
and
±H.
remains true when
is
PSL2 (Fp)
the case where
r a t ional
Then there is a regular Galois covering
with Galois group G
theorem, one takes N such that
over
wi t h a suitable choice of coordi·
±�
A complement
5 .4 C
th e
Xo( N) is
PSL2 (Fp) O.
=
Xo( N)
h as genus
by the above.
of genus I j for if the twisted curve
point . an d is of rank > 0 (where
Xo(N)P
In Shih's exp loi t
One might try to
Xo(N)P
ha.s a Q
is viewed as an elliptic curve
Q by fixing this rational point as origin) th en the following variant of
HilbertJs irreducibility theorem allows us to deduce the existence of infinitely
many extensions of Q with Galois group G:
Pr oposition 5.4.1 Let C
+ E be a regula.r Galois covering with group G, where E is an elliptic curve over a number field K. A.uume tha.t for ev
ery proper subgroup H of 0 conta.ining the commuta.tor 6ubgroup ( Ot G) J the corresponding G / H covering of E is ramified IJ.t letlSt one point. Then a.ll P E E(K) ezcept a finite number ha.ve property Irr(P) 0/ §9.3.
lit
The hypothesis in the proposition
t he covering
C /H
coverings of E
are
Falt i ngs ' theorem,
+
implies
abelian. Hence the genus of
CH
proposition.
P
s
CH
H
:I G,
ince the only unramified =
C/ H
is at least 2; by
has finitely many rational points . Let SK C E(K) be
the union of the ramification points and the
finite s e t , and if
that for all subgroups
E is rami fie d somewhere}
rt SK, then property
images of the CH(K). This is a is s at i s fied. This proves th e
Irr( P)
The above result was first obt ained by Neron (Proc. Int. Congo 1954 ) in
we ak e r
form, since Mordell's conjecture was still unproved at that t ime .
Corollary 5.4.2 Assume E( K) is infinite.
a
Then, there are infinitely many
linearly disjoint Galois extensions of K with Galois group G.
Proof' If L is any finite extension of K, we can find P E E(K ) , P rt SL (where SL is defined as above) . Th e property Irr(P) is then satisfied both over K an d over
L.
The corresponding Gextension Kp is then linearly disjoint from
The corollary follows.
L.
53
5 .5. Furths relults
For N = 1 1 , p = 47, the twisted elliptic curve XD(Nr has rank Q. Hence there are infinitely many extensions of Q wit h Galois group PSL:.(F4T)' ( An explici t example has been written down by N. Elkies. )
Example:
2 over
Problem:
Is it possible to generalize prop.
5 .4.1 to abelian varieties? More
pre
cilely, let 11' : C  A be a. finite covering of an abelia.iJ. variety A. Aalume that a.) ACE) il Zarlw denle in Ai b) 11" is ramified (Le., C is not an abelian variety). Is it
trUe that 1I"(C( K» it "xnuch Ima.ller" than A(K), i.e., that the numbs of points of logarithmic height :::;; N in 1I'(C(K» is o( NPI2), where p =:: rank A{K)? There il a pi.4tial relult in this direction in the pa.per of A. N eron referred to above.
5.5
Further results on as Galois groups
Concerning the group. results: 1.
PSL2(F I )' for q
(b)
q
( d) 2.
a. prime, is
8, by a result
=
q = 25, b y
q = p1 ,
S L2 (F q
)
is isomo rphic to As and �
of Matzat «(Ma.3J ) .
a result of Pryzwara ([Pr]).
for p p rime ,
( al q = 2 , 4, 8,
and
finite fieldlt there are the following
PS�(F,)
p :; ±2
(mod 5), d.
is known to have property
SL:.(F,)
)
known to ha.ve property GalT when:
= 4 and q = 9, because respectively in these cases.
(a) (e)
over
PSL:a and SL2
q not
P S L 2 (F q
GalT
[Mel].
for:
for then
SL, ( F9) � P S L.. ( F, ). (b) q = 3, 5 or 9, for then SL:l (F'I) is ilomorphic: to .,4", As, or �, which are known to have property G alT1
( c)
q
=
7, Mestre (unpublished).
It seems that the other values of q case q ::: 11 ).
d.
ha.ve
[Me2}, or §9.3.
not been treated (not even the
of Galoil extenlions of Q with Galois group 1 , 2, . . . , 1 6 . Their conltruction is due to Mestre (unpub lilhed), who uses the representations of GQ given by modular forms mo d 2 of prime level :5 600. There are a. few exam p les
S�(FII)
for
m. =
Chapter
6
Galois extensions of C (T) 6.1 Our
The GAGA principle
goal in this chapter is to construct G alois extensions of C(T),
us
ing the need
tools of topology and analytic g�metry. To carry out this program, we
a bridge between analysis and alge bra .
Theorem 6. 1 . 1 ( GAGA principle) Let X I Y he projective algehrtlie vtlneties over C J and let Xaa, yaa be tAe corresponding comple: tln4lytic spaces. Then. 1 . Every I1n41ytic mal" xaa + y'" is algebf'tJic. t. Every coherent antilytic sheaf over Xaa
mology coincides with W tJn4lytic one. For
a
is tdgebrtlic, tJnd its a/gebrtlic cohD
proof, see [Sc4] or [SGAl] , expo l e XII. In what follows , we will allow
ourselves to write X instead of xaa .
Remarks: 1. The functor X
t+
Xaa is the "forgetful" functor which embeds the category
of complex projective varieties into the category of complex analytic s paces .
Th. 6. 1 .1 implies that i t is fully faithful. 2 . By t he above, there is at most one algebraic structure on a compact analytic space which is compatible with it .
3.
Th. 6. 1 . 1 implies Chow's theorem:
every closed analytic subspace of
p roj ec ti ve algebraic variety is algebraic. 4. The analytic ma.p exp : G. + G., where
is not al geb raic; hence the hypothesis that X is projective is essential. Exercise;
If
isomorphism
a
X and Y are reduced varieties of dimension 1. prove tha.t any analytic of X on Y is algebraic; disprove this for nonreduced varieties. 55
Chapter 6. Galois
56 Theorem 6. 1 . 2 (Riemann) Any mension 1 is algebraic.
extensions of
C(T)
compact complez analytic manifold of
di
The proof is easy, once one knows the finiteness of the (analytic) cohomology 2 . A generalization for a
groups of coherent sheaves, see e.g. [Fo] , chap.
broader class of varieties is given by a theorem of Kodaira:
Theorem 6. 1.3 (see e.g. [GH] , ch. I, §4) Every compact Kahler mani.fold X wh.o3'! Ki.hler cl4S3 is integral ( as an element of H2 (X, R)) is a projective algebraic variety. In the above theorems, the compactness assumpti on is essential. For coverings, no such assumption is necessary:
Theorem 6.1.4 Let X be an algebraic variety over C, and let 11'" : Y ' X be a finite unramifietl. analytic covering of X . Th.en there is a unique algebraic 3tructU?"e on Y compatible with its analytic 3tructure and with 11'" . The proof i s given in [SGAl], expose XII. Let us explain the idea i n the case
X
=
X
 Zl
where X is an irreducible projective normal variety, and Z a
closed subspace. One first uses a theorem of Grauert and Remmert [G R] to
extend Y
+
X to a ramified analytic covering Y
+
X. Such a covering
corresponds to a coherent analytic sheaf of algebras over X. Since X is pro
jective, one can apply tho 6 . 1 . 1 , and one then finds that this sheaf is algebraic,
hence so are Y and Y.
(The extension theorem of Grauert and Remmert used i n the proof is a rather delicate one  however, it is quit e easy to prove when dim X = 1 , which is the only case we shall need.)
The GAGA principle applies t o real projective algebraic varieties, i n the we may associate to any such variety X the pair (XAIl, s ) ,
following way: where
X'"
=
X (C) is the complex analytic space underlying X I and
is the antiholomorphic involution given by complex conjugation on X. The real variety X can be recovered from the data (X, s ) up to a unique iso morphism. Furthermore, any complex projective variety X together with an &
R: for by X to X I the conj ugate variety of X, and hen ce gives descent data which determines X as a variety over R.
anti.holomorphic involution GAG A ,
�
determines a projective variety over
is an algebraic isomorphism from
Simi la.r remarks apply to coherent sheaves, and coverings: for example, giving a finite co'Vering of
X
over
complex analytic space
R is equivalent to giving a finite covering of the
X(C),
together with an antiholomorphic involution
which is compatible with the complex conjugation on
X(C).
57
6.2. Coveringl of Riemann lurfa.cel
6.2
Coverings of Riemann surfaces
Recall that for all 9 � 0, there exilts, up to homeomorphism, a unique c;om pad , mnnected, oriented lurface Xg of genus g (i.e., with first Betti number 2g ) . This can be proved either by "cutting and pasting" or by Morse theory (in the differentiable c;ategory). The surfac;e Xg has a standard description as 1 " 4" b" 4; 1 , b;l , and a polygon with 4g edges labelled 41. bt , 4 1 , bIl, identified in the appropriate manner. From this desaiption one can see that the fundamental group 1I'1(X,) relative to any base point has a presentation given by 2g generators at , 6 I , . . . , "" bg, and a single relation • •
1 41 6141l bl 1 ' " all bga,l bg =
1.
(To show this, one may express Xg as the union of two open sets U and V, where U is the polygon punctured at one point, which is homotopic to a wedge of 2g circle.and has fundamental group the free group F on 29 generators, and V is a disk in the center of the polygon. Applying the van Kampen theorem, One finds that 1I'1 (X,) = 1 *8 F F/(R), 1 01 L . R where IS 41"1 41 I • • • a, bg4g1 61 , ' ) Let noW Pl . ' . . , Pit be distinct point. of X"� and let WI be the fundamental group of X,  {PI , . . " PI;}, relative to a base point x . E ach 11 defines in '1['1 a conj ugacy class Oi corresponding to "turning around p. in the positive direction" (choose a disk D. containing � and no other Pi ' and use the fact that 2I'"1 (Di P,;) can be identified with Z). A similar argument as above proves that the group 1rl has a presentation given by 29 + Ie generators 41 1 bt I • • 'J 4" btl , Ch " CAl and the single relation: =
• •
where the c; belong to 01. for all i. We denote this group by 1rl (g, Ie). Observe that, if k � I , t hen 1I'1 (g, Ie) is the free group on 2 g + k  1 generators. In the applications, we shall be mainly interested in the case 9 = 0, 1c � 3.
6.3
From C to
Q
Let X be an algebraically cloled field of charaderistic 0, X a projective smooth curve of genus 9 over X, and let Ph , . . , P, be distinct points in X(K). The algebraic fundament al group of X  {PI " ' . ' Pi} may be defined as follows. Let X(X) be an algebraic closure of the function field X(X) of X over K, and let n c X(X) be the maximal extension of K(X) unramified outside the
Chapter 6. Galois extensions of C(T)
58
points Ph " ' , Pic  The algebraic fundamental group is the Galois group of n over K(X), namely,
r;ta(X

It is a profinite group. Let
{Ph ' . . , Pic } ) us
=
Gal( n / K (X)).
denote it by 71" .
Theorem 6.3.1 The algebraic fundamental group 1f' is isomorphic to the profi nite completion of the topological fundamental group :
(Recall that the profinite completion r of a discrete group r is the inverse
limit of the finite quotients of
r. )
Proof: Let first E be a finite extension of K(X) with Galois group GJ let
Qi
be a place of E above Pi, and let I( Qi) denote the inertia subgroup of G at Q.. Since the ramification is tame, we ha.ve a canonical isomorphism:
where e!i is the ramification index at Q,; this isomorphism sends an element " of I(Qi) to .s7l"/'If (mod 71" ) , where 71" is any uniformizing element at Qi. From now on, let Qi be a place of {} above Pi . The isomorphism a.bove can be generalized to such infinite extensions, by passing to the limit: write 51 as a union of finite extensions E; with ramification indices ej at Qi.. Then one has:
I( Q i)
�
�P.ei '
Choose a coherent system of roots of unity in K , i.e. , a fixed generator for each �, such that z:n � for all n, m. This provides I ( Qa) with a canonical generator c( Q .: ) . Changing Q. above P.: merely affects ce Q ,, ) by inner conjugation. Hence, the conjugacy class Ci of c( Q.:) depends only on Pi. (Changing the coherent system {� }  i.e. , replacing it by { z: } , where .. a E Z· Aut Z replaces c(Q, ) by C(Qi)4.) Th. 6.3.1 is a consequence of the following more precise statement:
Zn
=
=
Theorem 6.3.2 There eZ'ists a choice of element3 :I;j, Yj, Ci in 71" I with Cw E Ci for All i, such thAt: 1 1  1  1 1 1 . XIYtXt !It x"YgXg Yg Ct ' " c} !. The map 7r't (gl k) + 71" mapping the generators fI!;, !ljJ Cw Of 7l" 1 (g, k) to the element3 X ii Yi, C. extends to an isomorphism • • •
=
.
6.3. From C to Q
59
In other words, r is presented as a profinite group by the generators
Ci
Zi, Yj,
with the relations above. The proof will be done in two stages: first, for K = Cl then, for arbitrary algebraically closed fields K of characteristic 0.
= C, with the standard choice of roots of unity, {%on} = {e'1ri/,, }. In this case, the result follows from the GAGA dictionary between algebraic and topological coverings (see tho 6.1.3 and [SGA1]).
Step 1: Proof when K Step
2:
K arbitrary. One has the following general result
Theorem 6.3.3 Let V be an algebraic variety o'Uer an alge.braically closed field K of characteristic 0, and let K' be an alge braically closed e:tension of K . Th.e.n any covering 0/ V over K' comes uniquely from a covering defined over K .
(Note that this is not true in characteristic p, unless the coverings are tame. For example, the ArtinSchreier covering yP  Y = aT of the affine line with a E K' » a f/. K, does not come from a covering defined over K.) Two coverings of V defined over K and K� isomorphic are K�isomorphic; this is clear, e.g. by using a specialization argument. Next, one has to show that any covering which is defined over X' can be defined over K. II K' is of finite transcendence degree over K, then by using induction on tr.deg(K' I K), it is enough to show this for tr .deg( K' / K) = 1. In this case, a covering W + V over K' corresponds to a covering W x C + V x C over K, where C is the curve over K corresponding to the extension K' I K. This can be viewed as a family of coverings W + V over K parameterized by C. But it is a general fact that in characteristic 0, such families of coverings are constant. (One can deduce this from the similar geometrical statement over C, where it follows from the fact that rl (X X Y) = 71'1 (X) X rl (Y). ) By choosing a Krational point c of C, on then gets a covering We + V defined over K which is K' .isomorphic to W.
Step 3: Using tho 6.3.3, we may replace ( K, {z.}) by (Kl , {z,.}), where tr.degKt /Q < 00. One can then find an embedding Kl + C which trans forms the {z.. } into the {e21ri/"n (irreducibility of the cyclotomic polynomials). Another application of tho 6.3.3 concludes the proof. Remark:
Let
r
be a discrete group, r its profinite completion
t
=
lim r/N 
where N runs over the normal subgroups of r of finite index. The canonical homomorphism r + r is universal for maps of r into profinite groups. It is not always true that this map is inje.ctive. There are examples of finitely p� sented r with r infinite but r {I}. e.g. the group defined by four generators Xi (i E Zj4Z) with the four relations XiZi+1X, 1 = z�+l (G. Higman). =
Chap te r
60
Problem:
Can IUch
6. Galoil extensions of C(T)
a r be the fundamental group of an algebraic variety over C?
In the case of curvel, which il the case we are mostly interested in, the map 7I' l ( g . k) + 1i\(g, k) is injective. This follows from the fact that 7I'l(g, k) is isomorphic to a subgroup of a linear group (e.g., S�(R) or even better, SL2(R)); one then uses a wellknown result of Minkowski and Selberg (see e . g. [Bor}, p. 1 19).
6 .4
Appendix:
universal ramified coverings
of Riemann surfaces wit h signature Let X be a Riemann surface, S e X a finite set of points in X. Let us assign to each point P E 5 an integer np, with 2 ::;; np < 00. Such a set 5 along with a set of integers np is called a signature on X. One defines a ramified covering subordinate to the signature (5j np) to be a holomorphic map f : Y + X. with the following properties: 1 . If Sy ;;;: /1 (5), the map ! : Y  Sy + X  S is a topological covering.
{Pl. Then f l (D) splits into connected components Da , and the restriction fa of f to DQ is isomorphic to the map z ..... z..... , for some na dividing np.
2. Let P E S and choose a disk D in X with D n s
=
(When Y + X is a finite map, condition 2 just means that the ramification above P divides np .) For every P E S, let Cp be the conjugacy class "turning around P" in the fundamental group 'Kl (X  5). If Y + X is as above, Y  f1(5) + X  S is a covering, and the action of c;r is trivial. Conversely, given a covering of X  5 with that property, one shows (by an easy local argument ) that it extends uniquely to a map Y + X of the type above.
Theorem
6. 4 . 1 Let X, 5, and np b e given. Then there is a universal co ver ing Y subordinate to th.e signature (S i np) with. !a.ithju.l action of
where N is th.e normal subgroup genera.ted by the Y is simply connected} and one has Ylr X.
s"P I
s E CPI P E S. This
=
Sketch of proof. Let YN be the covering of X  5 associated to N. This is a Galois covering. with Galois group 'K1(X  S)jN. As above, one can complete it to a covering Y + X. One checks that Y is universal, simply connected, and that r acts properly on Y, with X = Y/r.
61
6.4. Appendix
We apply this to the case X = X T, where X is a compact Riemann surface of genus g, and T is a finite subset of X. Letting S = {Ph  . . , p. } , p. +th one associates to X a "generalized signature'll on and T = {p.+1 , X, where the indices are allowed to take to value 00 , by set.ting 
• • •
{ ?Ii
=
nPt
if i � a , otherwise.
7l.i = oo
The corresponding group r is then defined by generat.ors C'H and relations:
Cl ,
{
. • .
1 b1 1
41 bl(1t
c!tl
=
1,
. . .
•
• •
1b1 " CI '
Ie tlgl'IgG.
, c:.
=
1.
• •
'1+,
=
41 ,
bt ,
. . .
, Gil' b;,
1
Example: In the case 9 = 0 and a + t = 3, the group r and the corresponding universal covering X + P l (e) can be constructed. geometrically as follows. Set � = lint + 1 /n2 + 1/n3 (with the convention that 1/00 = 0). If � > 1, let X be the Riemann sphere S2 equipped with its metric of curvature 1; if l. = 1 , let X be C with the euclidean metric; and if � < 1, let X be the Poincare upper half plane, equipped with the metric of curvature 1. In each case, by the GaussBonnet theorem there is a geodesic triangle on X with angles 1r/nh 1r/n2' and 1rln3 ' (The case f&.i = 00 gives an angle equal to 0, and the corresponding vertex is a "cusp" at infinity.) Denoting by 8; the reflection about the ith side (i = 1, 2, 3), the elements 01 = 8283. C2 = 838h and 03 = 8182 satisfy the relation
and the 0, are of order ni . Hence the group r' generated by the C.. is a quotient of r, by the map sending ca to C, (i = 1, 2, 3). In fact, one can show that the projection r + r' is an isomorphism, and that X with this action of r is the universal covering of tho 6.4.1. A fundamental domain for the action of r is given by the union of the geodesic triangle with one of its reftections about a side. Theorem 6.4.2 cases :
1. !.
9 =
9
=
OJ OJ
The element ca of r has order ?Ii, e2;cept in the follo1l1ing two
8 + t = 1; a
+t
=
2, "' 1 :f:.
n1 ·
Consider first the case 9 = O. This case can itself be divided into three subcases: 1. 8 + t = 2, and "' 1 = n2 . In this case r is isomorphic to a cyclic group of order Rl (if "'1 00, then r = Z), and the statement follows. =
62
Chapter 6. Galois extensionl of
2. & + t order n•.
=
3.
The explicit const ru ction of
r ab ove shows tha.t the
C(T)
Cs are of
4. One reduces this to case 2 by adding the rela.tions Ci = 1 for a of in dic es with III = & + t  3. One thus get s a surjective homomorphism onto a group of the type treated in case 2, and the Ci with i rI. I are therefore of order 71i. By varying I, one obtains the result. 3.
a
s et I
+t�
To treat the case 9 � 1 , it suffices to ex hib it a group containing elements Al , . . . , A., Bl t . . . , B., C1, , C.+& satisfying the given rela.tion for r with the C, of order "i , (i = 1, . . . a + t). For this, one may choose elements Ci of order ns in the special orthogonal group SO J. Every element z of S02 can be written as a commutator (y, % ) in the orthogonal group OJ ; for, choosing y E S0 2 such that y2 =: :1:, and z E O2  S 02, one has yzy 1 %1 Y . 'II = � . Hence, i t suffices t o take Aj, Bi = 1 if i > 1 , an d A I , Bl E O J such that • • •
I
=
that the signature (nt l . . . , n.H ) is not one of the exceptional cases above. We then have a "universal" ramified covering Y which is simply connected. By the uniformization t heorem (d. [Fo] ) , such a Y is isomorphic to either P l (O), Ot or the upper half plane ?t. To t ell which of these three cases occur, let us introduce the "EulerPoincare characteri sti c" E of the signature: .+& 1 E = 2  2g  L(1   ) , Assume
in the theorem
with the convention
that
.1. l:1l:I
=
1ti
1=1
O.
Theorem 6.4.3 1. If E > O J !. If E = 0 ) 3. If E < O J
This can
1.
Case
1
then Y is isomorphic to Pl (e). then Y is isomorphic to C . then Y is isomorphic to the. upper half plane 11..
be proved using t he occurs only
if 9
G auss Bo nn et
= 0 and the
•
empty, corresponding to the
•
(n, n), n
< 00, corresponding
•
(2, 2, n),
corresponding to
•
(2, 3, 3),
corresponding
•
formula.
signature is;
identity t o the
covering Pl (e)
cyclic
+
Pl (C) .
covering Z 1+ zft.
r = Dft , the dihedral group of order 2n.
to r
=
A ..
( tetrahedral
group).
(2, 3, 4), corresponding to r :::; 84 (octahedral group ).
63
6.4. Appendix •
(2, 3, 5), corresponding to r
=
A5 (icosahedral group).
2. Case 2 occurs for •
9
•
g =
=: 1 and empty signature, corresponding to the covering of a complex torus by the complex plane, r = a lattice.
0 and signature (00 , 00 ) , corresponding to the map exp : C
•
.
r :: z.
C,
signature (2, 4, 4), corresponding to r ing the Ia.t tice Z + Zi.
=
group of affi ne motions preserv
•
signature ( 2 , 3 , 6 ) , corresponding to r = group of affine motions preserv ing the lattice Z + C+p)Z.
•
signature (3, 3, 3), corresponding to a subgroup of index 2 of the previous one.
•
signature (2, 2, 2, 2), corresponding to coverings
where T is an elliptic cutve and C covering map.
�
T is the universal unramified
3. Case 3 corresponds to the case where r is a discrete subgroup of PSL2(R) which is a "Fuchsian group of the first kind" , cf. e.g. Shimura [Shim] , Chap. 1; one then has X :: 11./r and X is obtained by adding to X the set of cusps (mod r). (By a. t heore m of Siegel (S ieJ , such groups r can be cha.racterized as the discrete subgroups of PSL2 (R) with finite covolume.) The genera.tors Ci with n. < 00 are elliptic elements of the Fuchsian group r, the t!.j with 1J.i = 00 correspond to parabolic elements, and the tlj and b; to hyperbolic elements.
Chapt er 7 Rigidity and rat ionality on finit e groups 7.1
Rationality
Let G be a finite group, CI( G) the set of its conjugacy classes. Choose N such that every element of G has order dividing N; the group rN = (Z/NZ) · acts on G by sending s to s� , for a E (Z/NZ )" and acts similarly on Cl( G) Let X( G) be the set of irreducible chara.cters of G. They take values in Q(J&N) , and hence there is a natural action of Gal(Q(PN)) � rN on X(G). The actions of rN o n CI(G) and on X(G) are related by the formula .
O"Q (X)(S )
=
X ( sQ),
where U'Q E Gal(Q(J&N )/Q ) is the element sending the Nth roots of unity to their ath powers.
De6nition 7. 1 . 1 A class c in CI(G) is called Qrntional if the following equivalent properties hold: 1. c is ji%ed under rN A !. Every X E X(G) ta.lces valu es in Q (and hence in Z) on c. (The equivalence of these two conditions follows from the formula above.) The rationality condition means that, if s E c, then all of the generators of the cyclic group < s > generated by s are in c, i.e., are conjugate to s. For instance, in the symmetric group Sn , every conjugacy class is rational. More generall y, let K be any field of characteristic zero. The class c of an element s E G is called Krational if X( s) E K for all X E X( G), or equivalently, if c« = c for all a such that O'Q E Gal(K(PN)/ K). For example, the altern ating group As has five conjugacy classes of order 1, 2, 3, 5, and 5 respectively. Let us call 5A and 5B the two conjugacy classes of exponent 5. (They can be distinguished as follows: given a fivecycle 8, let 65
Chapter
66
7.
Rigidity and rationality on finite groups
a(s) be the permu t ation of { 1 , 2, 3, 4, 5} sending i to si(1). Then s is in the conjugacy diUS 5A if and only if a(s ) is even. ) If " E 5A, then S 1 E 5A and s2 , 83 E 5B, 80 the classes 5A and 5B a.re not rational over Q. However , they are rational over the qu adratic field Q( v'S).
Remark: The formula O'a (X)( .t ) = XC,CI) has an analogue for supercuspidal admis. sible representations of semisimple padic groups. This can be deduced from the fact, proved by Deligne (d. [Del]), that the character of such a representation is supported by the elliptic elements (note that these are the only elements .s for which the notation "a makes sense). Rationality oC in ertia
Let K be a local field with residue field k of ch ara cterist i c zero, and let M be a Galois extension of K with group G. There is a unique maximal unramified ext ension L of K in Mj it is Galois over K. Let I be the inertia subgroup of G, i.e. , I = Gal(M/ L). Si nce the ramification is tame, I is cyclic. In fact] there is a canoni cal i somorphism
whe re e denotes the ramification index of L I K , and l is the residue field of L. (This isomorphism sends 0' E I to U1f 171' ( mo d 71' ) , where 1f is a u nifor mizing element of M.)
Proposition 7.1.2 Th.e ddSS in. G of an element of I is rational over /C . Indeed, the group H = Gal(LI K) = Gale'/ k) acts on P e e l) in a natural waYJ and on I by conjugation. These actions are compatible with the isomorp hi sm above. If now (1C1 is an element of Gal(k(p. )lk), there exists t E H such th at t acts on J'e (l) by z 1+ zQ . If 8 E I, then sa and tst 1 have the same image in Pe(l ). Hence SCi = ta r l . This shows that the class c of s is such that tf& = c, q.e.d. Corollary 7.1 .3 If Ie
=
Q, then. the cldS8es in I are rational in G .
Remark on the action oC rN on CleG) and X(C) The rNsets C1(G) and X(G) are the character sets of the etale Q alg eb ras Q ® R(G) and ZQ[G), where R(G) is the repre s ent ation ring of G (over Q), and ZQ[G] is the center of the group al ge br a Q[G]. These rNsets are easily proved to be "weakly isomorphic" in the sen se of exercise I below . However, they are not always isomorphic (see [ThID. Exercise: 1.
Let X and Y be finite sets on which acts a finite group r. (a) Show the equivalence of the following properties:
7.2 . Counting solutions of equations in finite groups
67
::: Iye l for every cyclic subgroup C of r; jXjCI = IYICI for every cyclic: subgroup C of rj
i. I XC I
ii.
iii . l X /H I
==
IYIHl for every subgroup
H or r ;
iv. The Qlinear representations of r defined by X aJ).d Y are isomorphic.
If those properties are true, t he
(b)
Show that
r·sets X and Y are said to be weakly isomorphic. weak isomorphism is equivalent to isomorphism wh en r is cyclic
( "Brauer's Lemmaft ). Give an example where r is a noncyclic group of order I X I = I YI ;::: 6, and X and Y are weakly isomorphic but not isomorphic. 2. Use exerc. l o(b) to prove that, isomorphic to Z Q[ G] . 3. Extend prop. 7. 1.2 to
MIX 7.2
is tame.
if G is
the case where
Ie
a
pgroup, p
F 2,
has cha.racteristic
then
l' >
Q ® R(G)
4, is
0, assuming that
Counting solutions of equations in finite
groups
G be a compact grouPi e quipped with its Haar measure of total mas s 1. Let be an i rred ucible cha.racter of G, wi t h p : G + GL( E) the c orresp ondi n g X linea.r representation. By Schur's lemma , IG p(tzt I ) dt is & scal ar multiple l · I E of the identity in GL(E). Taking traces gives xCz) = lX ( l ), hence
Let
X ( p(tztl)dt = ( x ) lB' JG XCI ) Multiplying on the right by
p(y ), we get :
k p(tz, 'y)dt ����p(y). =
Taking traces
give s :
X( ) (y ) .
I X(txt1y) dt == 2: X ,,( 1 ) Je This formula extends by induction to k elements Xl ,
• •
'J 2: 1: :
(7. 1)
where ex
is the inne r pro d uct of ,; with the irreduci ble character X , ex =
fc ,;(x )X(x ) tb:,
68 and
Chapter assume that
E l ex lx( 1 )
summi ng over all chara.cters
is equal
to
<
7.
Rigidity and rationality on finite groups
so t ha.t
00
the
sum is normally convergent. By see that the i ntegral
in formula 7 . 1 J we
'"
't Cx
X(Zl) " . X(Zk)X(Y) . X( I )1&
Let us compute 1(¢) in the cue where G is a finite gro up and rP is the Dirac is 1 a.t the identity element J and 0 elsewhere. One has
function which
1
t/J and hence
ex
=
I(¢) is N/IG ,",
x(l)/ tGI.
where
=
I GI
� X(I)X,
, ZJu and 11 in G, the value of number of so lut ions (th . . . I t.) of the equation
Given elements Z1 ,
N is
the
• • .
Hence
(7.2) Let 01 , , 0" denote the conjugacy c1a.sses of the elements % 1 t d efine nCOl, " . , e.) to be the number of solutions (91 , . . . , 9,,) of the • • •
• • •
9192 ' " 9" Letting Z, be the order
=
I,
and
9i E Ca.
of the centralizer of an element of Ci t one has
By applying formula 7.2 and ob serving that Z, Theorem 7.2 . 1 The number n
91 ' " g. = 1 , 9" E Oil
I %. ,
equation
is given by
=
= nCGl , . . . 1 C� )
IGI/ IO,:! , one therefore get s : of solution.! of the equation
I I C1 , . . . ' C I � X( z d · · · x(z .) n= _ I GI
" �
X( I ) I:2
'
'Where :1::, is a representative of the conjugacy class Gil (Inti the ine4'Ucible chara.cters of G.
X
"'118
through all
7.2.
Counting solution. of equations in finite groups
69
Exercises: 1. Let G be
a compact group and let p be an irreducible representation of G with character X. Show that 1 p(t:tt Z 1 )= tit = 1/x(�)2.
JJ
that the left hand side is an intertwining operator for p, and hence a sc:ala.r by Schur's lemma.. Then c:ompute its trace.) Conclude tha.t
(Hint: Show
J J p(tzr1z1 y)dz p(Y)/x(1)2, for J J x(tzt1 z1 . y)ck dt X(y)/x( 1)2 . d1
•
and that
'Y E
all
=
is
G.
=
Hence show by induction that:
J . . . J x(tlzltl1z11 . . . t,z,t;1z;1 . y)dz1 dtl
• .
· dz, dt,
=
X{Y}I'X,<.1)2'.
2. S uppose now that G is finite. We denote the commutator "'11U 1 .,,  1 by (u, 'U ) . Let 9 � 0, and let 2; be fixed elements of G, for ; = 1, , k.
let N
=
Neg, �;. y ) denote the number of tuples of elements of G,
• . .
IUch that Using
( U,1 . 11t) . . . (U" 'U, )t l�lti'l . . . t 1 2.t. 1 exerc. 1, show that :
N
=
=
of u and 11 For y e G ,
,1 ,
IGI2.+1l 1 E x(Zl ) " ·x(J.)X(y)/x(lr �·+.It  I . x
3. Show that an element y of G is sum
a
produc:t of 9 commutators
if
and only if the
y / ( ) 1fL x( ) x 1 2 1 X
is nonzero. In partkular, 'JI is a. commut ator if and only if E X( y) / x(l ) :1 O. It is a knQwn Call sequence of the "classifica.tion theorem" that every element in a finite non·abelian simple group is a commutator. The reader ma.y wish to verify this (for one of the sporadic groups, say) by using the formula above and the character t ables in [ATLAS].
Remark: Th. 7.2.1 can be used to co mpute the number of subgroups in a. fin ite group which are isomorp hic to the alternating group As . Indeed, As has a prese nt a.tion given by three generator s z, y, z, and relations 2:2
=
1l
=
Z5 =
zyz =
1.
The problem therefore amounts to finding the number of solutions of the equa tion zy z = 1 , where z, y % belong to conjugacy classcs of exponent 2, 3, and 5 respectively. The same rem ark ap plies to Sol . At , and the dihedral groups, which have similar present at ions , d. tho 6.4.3, case 1.
Chapter 7. lligidity and rationality on finite groups
70
7.3 Let
Rigidity of a family of conj ugacy clas ses
G
be a fini te group, and fix co nj ug acy classes
Cl ,
.."
Ok in G.
Let
E = E ( Cl , . . . , CII:) be the set of all (gl J . . . , gll:), with g, E Ci, such that gl • • • gil: == 1. ( Hence, by the previo us section, n(el ) • • • J C,) = Itl . ) Let E = E(C1, , C,,) be the set of (gl � " " , gle) in E s uch that 91 , gle gener at e the group G. The gro up G acts by conjugation on E and t. • • •
' " "J
AI8ume tha.t the center of G is t rivial ( as is the case, for example, when G = S,,, n � 3, or when G is a nonabelian simple group). Then tkt: action of G on E is free: for, if 9 E G fixes (91 , . . . , g,,), it commutes with the g/s, and hence with all of G, since the gi 'a gener ate G; but then 9 = 1 , because G is assumed to have trivial center. One says that a ktuple of conju gacy classes (C1 , " , GIe) is rigid if E # 0 • •
and
G
acts transitively on E, i.e. ,
One says that
(C1 ,
The order of
E
• • •
I
C. )
lEI
=
is strictly
\ G \.
rigid if it
is rigid, and
E ::;;; t.
can be comput ed usin g the formula of the p revi ous sec ti o n :
where Ci E Ci for i = 1, . . . , Ie, and X runs thro u gh the irr ed uci b le characters G. If Zi is the order of the centralizer Z(Ci ), t his can also be wri tten as:
of
(7.3) Rigi d ity is often pro ved in two steps:
1 . Compute the order of E, by us ing formula 7.3 and t he character table of C. 2. Compute the order of E  EJ by finding Ietuples i n E wh i ch do not gen· er ate all of G: for this, one uses a knowledge of the maximal subgro ups of G
(whenever possible).
Remarks: 1. Let
t7
be an outer automorphism
ktuple of conjugacy cl as ses . Then that
(l
preserves each
follows that
Ci.
(Ugl , . ' " ug. )
inner conj ugation,
Then,
9
E
• • •
letting (g1 l ' " ' I gle )
b elo ngs to
there exists
of G. S upp ose (G1 1 , Ck ) is a ri gid for some i. Indeed, suppose
u(C.) I Ci
�. Sinc e
G such that
be an elem ent of E,
G acts transi tively on
for all i.
E
it
by
7.3.
Rigidity
of a family of conjugacy classes
71
But
the 9i genera.te G and hence u is an inner automorphismt which contradicts the assumption. 2. In many cases, the term E" x(Cl ) · · · X(C.. )/X(1 )1I 2 in formula 7.3 is not
very large, the main contribution to the sum being given by the unit character X = 1. For a. rigid family of conjugacy classes, one might therefore expect that the order of magnitUde of I C1 1 · .  I Cil is clo se to I GI. As in §7. 1 , let a E (Z/N Z ) · , where N is a multiple of the exponent of G, so that (Z/NZ)· acts on Cl( G). Then one has: Proposition 7.3. 1 1 . It(Cf, ·  . , Or) 1 = I E ( Cl , . . . J Ck ) !; 2 . I E( O�., . . . , 0:)1 := IE(C!, . . . , C. )l.
The first identity follows from formula 1.3 above, combined with the formula X(c") = O"m(X)(C) of §7.1 . The second is proved by i ndu cti on on the order of G, as follows: if H is a subgroup of G, let E{H)(C1 n H" " , 0", n H) denote the set of (g1 , ' " , 9i), with 9, E Oi n H for all it s uch that 91 ' " 91c = 1 and the 9i generate H. (In general, the C, n H are not conjugacy classes in H, but unions of conjugacy classes.) The formula

E  E( 01 ,
0 0 0
, 0",) 
U
HCG, H';:G
rJH) (0
L'
1 n H, . . . C. n H) J
supplies the induction step. Corollary 1.3. 2 11 (01 , ( Oi, . 0 , 0: ) .
• • •
, 011) iI rigid (respo
strictly rigiJ)1 so is
the ja.milll
0
Remark: Here is another way to prove prop. 1.3.1. Let F be the group with presentation given by generators Zl , 0, X. and rela.tion X l ' x. = 1 , and let F be its profinite completion. Then one has: . •
Proposition 1.3.3 For that
8( Xi )
is an automorphism (J 01 F 8Uch of xi . (Equiva.lently, there exist conjugate to zr, satisfying Yl . . . Y'" = 1, and gen
each
Q
E
Z· J
• •
there
belongs to the conjugacy clo.ss
elements Yl, . . . , Y. with Yi erating F. )
Let us use the interpreta.tion of F as the algebraic fundamental group 7 of the projective line Pi with k points removed. By choosing a. coherent system {z,.} of roo ts of unity in Q, one has an identification (d. §6.3)
Chapter 7. Rigidity
72 Replacing the system
{z,.}
on finite g roups
and rationality
by {.;: } gives a different is omorphism, and com
posing the two yields the desired automorphism 8 of
F.
1.3.3 implies that 1 � (Oi, . . . , 0:) 1 I E(OI, . . . , CJ: ) I . For, the el E(Ol , . . . , Ok) are in oneone c;orrespondence with the surj ective: homomorphisms F + G sending each 2:, to an element of Ci" while the ele ments of � ( ai , . . . , Or) correspond to the surjective homomorphisms sending each :1:, to an element in Cr. The automorphism (J : F + F induces a map Hom(F, G) + Hom( P, G) which gives a bijection between the two sets �(Cl1 " " 01:) and E(Oi, . . . , Of). Prop.
=
ements of
, C1:) i s rigid (resp. IItrictly Exercise: Show that , if (C1, (0.,.(1), . . . , O"'(J:» for any permutation q of { l , . . . , k}. • • •
family
7.4
rigid),
then so i s the
Examples of rigidity
We give only
a few such examples. For more, the reader should consult [Ma3}
and the papers quoted there.
7.4.1
The
symmetric group Sn Sn (n � 3) has cycles of order n, 2,
The symmetric group corresponding to
(nA, 2A, C(l ) is strictly rigid.
a cyclic arrangement of
conjugacy classes and
n 1
respectively.
For, giving an n cycle
{ l , . . . , n}
nA, 2A,
:t
E
nA
and
C( l )
The triple
determines
(i.e. , an oriented ngon). Composi ng this
(n  i ) cycle if and only if the two vertices which are permuted are consecutive. Hence, the solutions
permutation of order
y, .;
.2: ,
n
of the equation
with a transposition gives an
:1:y,;
=
1,
with
:1:, Y, .;
cycles of order
n, 2
and
n 1
respectively, are in one to one correspondence with the oriented ngons with a
distinguished edge. Any two such configurations can be tranformed into one
another by a unique permutation in
(12)
and
( 1 2 . . . n)
strictly rigid.
Sn i
are known to generate
hence
Sn;
lEI
=
IGI.
More generally, c;onsider the conj ugacy dasses
nA, 2A ,
is the class of the permutation
(1 . . . As
b
efore,
an element
( :1:, y, .; )
E
k)( k
But E
this shows that
and
=
E, since
(nA, 2A, 0( 1 )
a(lr),
where
is
C (k )
+ 1 . . . n).
E c;orresponds to
two distinguished vertices separated by k edges.
an
oriented ngon with
1£ Ie # n/2,
any two such configurations can be transformed into one another by a uni que permutation in Sn, and hence l E I = IGI. However, (nA, 2A, a(l!l) is not rigid in generalj to get rigidity, one must assume that (k, n )
=
1 . In that case, any triple (z, Y J .; )
7.4. Examples of rilidity
73
in t generates 871: by relabelling if necessary, we may write :z: = (1 n) , and 11 = (1 1 k + 1 ) . Since ( Te, n ) I, the permutation zIc is still an ncycle. By rel abelling again, the group generated by :z:1I and y is isomorphic to the group generated by the permutations ( 1 n) and ( 12). This in turn contains 1 , n) , which i s equal to 871 by a the group generated by (12), (23) , " ' 1 (n . . .
=
. . .
wellknown result

(d. §4.4).
Exercise: Check th a.t It I = nl for the conjugacy classes (2A. nAt e(l» by applying formula 7.3 of §7.3 ( prove that the oD1y nonzero terms came from the two characters of degree 1 of 571), 7.4 2 ..
The alternating group As
The alterna.ting group As has unique conjugacy classes of order 2 and 3, de noted by 2A and 3A respectively. It has two conjugacy classes of order 5 which are ra.tional over Q( VS) and conjugate to each other, denoted by SA and SB.
P roposition 7.4.1 The fDllowing triples 0/ conjugacy cla..sses are strictly n... git! : (2A, 3At 5A) . (2A, 5A, 5B), a.nJ. (3A, 5At 5B). To prove this, one c an compute the order of E i n each case from the character ta.ble of AI ' In ATLAS style, it is: chara.cters
!
Xl )(2
)(3
X4 XS
60
4
lA 1
2A 1
3 3A 1
3
1
0
3 4 5
1
0
1
1
0
1
5 SA 1 z' •
orders of fcentralizers 4 classes
5 5B 1 �
}I
I
1
1
0
0
One then gets:
1I:( 2A, 3A, 5A)1 ;;;

IE(3A , 5A , 5B) 1
+
602

IE( 2A , 5A , 5B) 1
60:1
. . (1 + 0 4 3 5
=
=
0 + 0 + 0) ;;; 60
1 1 . . (1 + 3 + 3 4 5 5 6 02
. . (1 + 0 3 5 5
+
+
0 + 0)
1 0 + 4 + 0)
=
=
60
60.
One checks easily that any triple in any of these E genera.tes follows.
7.4. 1
As, and prop.
Chapter 7.
74
Rigidity
and
rat ionality on finite groups
Exercises:
1 . S how that the t riples (3A, 3A, SA), (3A, 5A , 5A ) , and (5A, 5A, 5A) are strictly rigid. 2. ShoW' t h at (2A, 2A, 5A ) is not rigid, even though l E I 60 in that case ( the triples in E generate dihedral subgroups of order 10). =
7.4.3
The
groups
PSL2(Fp)
The group PSL2 (F,, ) , with p > 2, co nt ai ns unique conj uga.cy classes of ele ments of order 2 and 3, denoted by 2A and 3A respectively. There are two classes pA and pB of elements of order p, which are represented by u nip ot ent ma.trices, 1 and where 1.
(� l)
Proposition
(5 1),
7.4.2
One checks that
(�)
=
The triple (2A, 3A, pA) is strictly rigid.
(�oI Yo, zo)
E
E , where �o ,
matrices: yo =
1/0 . and Zo
(0 1 ) 1 1
are represented by
Zo 
_
'
(10 11 )
the
Conversely, let (�, Y, Z) be in E. We lift �, Y , Z to i, y, z in SL2(F,,), with i of order 4, y of order 3 and z of order p, So that we have zyz ± l . We vie'll these element s iUI automorphisms of the vector space V = F" E9 Fp . Let D b e the line of V fixed by z and let D ' iD be its tr ansfor m by i. O n e has D' :1 D (otherwise1 ±ii would not b e of order 3). After conj ugat in g by an ' element of SL2(F p ) , we may assume that D (res p . D ) is the first (resp. the second ) axis of coordina.tes in V. This mean s that we have =
=

:t =
for some A, I'
in
F;.
( 0 A ) AI 0 '
By assumption,
and
i
=
(A i ) ,
z belongs to the
clus
(� l )i this ( () JI�l ) ,
pA of
we that p. i s a square. If we write p. = Jl 2 , and conjugate by see that we can further iUl sume that p. :;; 1, i.e., that i :;; zo o Moreover, since ii is of order 3 or 6, we have Tr (ii) = ± 1 . T his gives A = ± 1 , hence � = Xo, Y = 1/0 , and z = Zo, which proves the r es u lt .
impli es
Proposition 7.4.3 The triple (2A, pA, pB ) is strictly rigid if One checks that
(�Ol YO t ZO) is in E, where
(1 1 )
�o = 2 1 '
Yo  ( 01 11 ) ' _
(�)
=
1 .
7 .4.
Examples of rigidity
The element
%0
75
is conjugate to
( A i) ; since (!) =

1 , it belongs to the class
pB . ConverselYt let (:tt Y, z) E t. We lift (:tt 1/ , z) &S a.bove to (f, y, z) with z of order 4 and j, j of order p. Let D be the line fixed by y and D' be the line fixed by i. We may again assume tha.t these lines are the standard coordinate lines, and tha.t
(1 1)
( )
z = .\1 01 . Writing that jz has order 4, one gets Tr (ji) 0, i.e . .\ 
y == 0 1
I
=
=
 2, q.e.d.
Proposition 7.4.4 The triple (3A, pA, pB) is strictly rigid if
(; )
=
1 .
One checks tha.t (:to, Yo , zo) is in �t where
(
)
1 :t o = 31 2 t
Yo =
(A 1 ) ,
The assumption tha.t 3 is not a quadra.tic residue mod p ensures that Zo is in the class pB. Conversely, let (:t, y, z) be in t. Using liftings (i, j, i) as above, one may &ssume that
( )
 (1 )
11 z = ).. 01 • 1/ == 0 1 ' Writing that jz has order 3 or 6, one gets Tr (ji) = ±1 , i.e. , ).. =  l or ).. = 3. However, ).. == 1 is impossible (it would imply that i belongs to the class pAl; hence .\ 3, q.e.d. =
7.4.4
The group SL2(Fa)
The simple group G = SL2(F s ), of order 504, h as three distinct conjug;u:y classes of order g, denoted 9A, 9B, and g O which are rational over the cubic field Q(cos 2;) and conjugate to each other (d. [ATLAS], p 6 )
. .
Prop osition 7.4.5 The triple (9A, 9B, 90) is strictly rigid. The character table for G is ([ATLAS], loco cit. ):
76
7.
Chapter
Xl X2 Xa X4 X.
504 lA 1 7 7 7 7
X7 Xa X9
9 9 9
X6
m
11
2A 1 1 1 1 0
=
1
1 1
2 1 1 1 1
1
0 0 0
Y 11 Y :z:'
2 cos 2; , 2. 2 cos 7 '
11'
7
7
7A 1 0 0 0 0
1
1
8
=
9 3A
8
=
=
Rlgidity and rationality
7
7C 1
7B 1 0 0 0 0 1 Y Y.. y
0 0 0 0
9 9A
9 9B
1 1
1 1
:z:
z
:z:
:z:
x
:c"
1
1 0 0 0
1 0 0 0
Y Y
y
2 cos � 9 , 2 coa !!' 1 ,
11 "
m" =
9 9C 1
on finite
groups
orders of + centralizers + cl as ses
1
z
"
z
Z
1 0 0 0 ' " mm m
2 cos a. 9, 2 cos l a,.. '
=
111J '11
"
=
=
1; 1.
Using formula (7.3) of §7.3, one gets: 1 �(9A , 9B , 9C) 1
==
5042
1
""93 ( 1 + "7
+
1
1
;; + "7
+
1
"7
1

"8 + 0 + 0 + 0)
=
504
=
IGI·
Hence it suffices to show that any ( m, 1/, .z ) E E generates G to prove rigidity. The only maximal subgroups of G containing an element of order 9 are the normalizers of the non�8plit Cartan subgroups, which ate isomorphic to a semi�direct product C:z(F64 )1 , where (F64)1 denotes the multiplicative group of elements of F64 of norm l over F s , and the non�trlvial element of C'I. acts on (F;" )l by z � Z l j they are dihedral groups of order 1 8. If ( x,y, .2 ) E E does not generate G, then m, y, .z are contained in such a normalizer. It follows that (by interchanging y and z. if necessary) : z =
But then :z:yz. proof.
=
:cl±H4
is
not equal to
1.
m%4.
This contra.diction completes the
Exercises:
1. Show that t he triples (7 A, 7 A, 7A), (2A, 3A , 7A ) , (2A, 3A, 9A) and that (7 A, 7 B I 7C) is not rigid. 2.
Let
G·3
be the automorphism
group of SL2 (FB), cr.
are s t
ri c tly
rigid ,
[ATLAS}, p. 6. Show t hat
(9A, 3B t 3C) is strictly rigidi t he class 9A is r at ional (as a class of G · 3); the classes 3B and 3C are rational over Q( �), and c onj uga.t e to ea.ch ot her.
the
triple
7.4. Examples of rigidity
7.4.5
77
The Janko group J1
The sporadic simple group
JI
discovered by Janko is of order
175560
=
� . 3 · 5 · 7 · 1 1 · 19.
2A, 5At and 5B of orders 2.. 5, and 5; the classes SA and 5B are rational over Q( v'S) and conjuga.te to ea.ch other. If :J: E SA, then Z I E SAt but Z2 , z3 E S B ; these conjugacy classes behave like the ones of the same order in As.
It contains conjugacy classes
P roposition 7. 4.6 (d.
[Ho)) The trip le
rigid.
(2A, SA, 5B)
The relevant part of the character ta.ble of 175560
1A
1
Xl
""
120 2A
30
SA
30 5B
1
1
1
XS
76
4
X7
S
1 1 2
77
3
z' z
76
X6
4
77
Xa
77
3
Xu
133
5
133
3
%
Xu
133 209
3 1
1
Xu
Xa
2 %
G
=
JI
is
rigid but nDt strictly
is (d. [ATLAS] . p.36):
orders of +centralisers
of
classes
1
1
2
z=
z  z'
z'
(1
+
20
+
=
\1'5)/2
(1  ,;5)/2
2 Z
%
1
Using formula. (7.3) of § 7.3, one obtains:
I�I � Hence
::;:;
'i2Q.iiJ" (1
17!5602
438900
:;;
(2A, SA, 5B)
+ "
=
" + 20 +
7i  7i
tiGI.
77
3
77
+
3
77
+
iii
3
133
+
3
i3i
+
I
20i
)
i s not strictly rigid. One can check that the triples in
E  E genera.te subroups of JI isomorphic to As. It is known that J1 contains
2 conjugacy
classes of such subgroups:
1. There are
IJ1 1/(21As l)
conjugate subgroups isomorphic to
contained in the centralizer of an involution in
J1•
As which are J) was first
(Indeed,
defined abstractly by Janko as a. simple group ha.ving the properly th at
it contains
an
involution whose centralizer
is isomorphic to
2. There is a conjuga.cy class of As·subgroups which are
there
are
J JI I/I As t
such subgroups.
{±l} X
As.)
selfnormalising:
Chapter
78 In all, one has
� 1 J1 1/ 1A5 1
7.
Rigidity and rationality on finite
subgroup s of J1 which are isomorphic to
the conjugacy classes (2A, SA, SB) solutions in E. This shows that
are

rigid in
As,
each
subgroup
groups
As . Since gives I A5 1
3
I E  E I = "2 1 J1 1, and hence I � I
7.4.6
I J1 r,
=
The
i.e. ,
(2A, SA, SB) is rigid.
HallJanko group J2
604800 = 2733537. It has a r at io nal class 7A of order 7, and t wo conjugate classes SA and SB of order S, rational over Q( v'5), see e.g. [ATLAS), pp.4243.
This sporadic simple group has order
Prop osition
7.4.7 (d. [HoD The triple (SA, SB J 7 A)
is
No proper s ub group of J'l has order divisible by 35, the other hand, formula (7.3) of j7.3 and the character loc.cit.) give:
strictly rigid..
lEI l E I . On of J:z ( [ATLAS],
hence table
=
Hence the result .
7.4 .7
The FischerGriess Monster
M
The FischerGriess group M, known as the "Monster" , sporadic simple groups. Its order is
is
the largest
of the
2 46 . 3 30 • S9 . 'f'  112 . 133 · 1 7 · 19 · 23 · 29 · 31 · 41 · 47 · S9 · 71 , yet it has only 194 conjuga.cy classes. Its character table is therefore of man ageable size (in fa.ct , it was computed before M had been sh own to exist ) . The group M contains rational conjugacy classes 2A, 3B, and 29A of exponent 2,
3, and 29 (ATLAS
nota.tion).
7.4.8 (Thompson, d. [Hunt], [ Ma3] , [Th2]) triple (2A, 3B, 29A) is strictly rigid.
Proposition The
It can be verified by computer that
lEI
=
I M I . To prove
rigidity,
one must
no (z, Y J .z ) E E gene ra.te a proper subgroup of M, the maxima.l subgroups of M are not completely known at present . Hence, one must take the following indirect approach: suppose there
show that E Unfortunately,
=
f:, i.e
.
•
79
7.4. Examples of rigidity
a proper s ubgroup G in M which is generated by (z, 11, .c:: ) E :E. Let S be a simple quoti en t of G. Clearly the eleme nts z, 11, z have nontrivial image in 8, and hence the order of S is divisible by 2 · 3 · 29. Hence, it suffices to check that there are no simple groups S with 2 3 · 29 dividing l S I and l SI dividing IM I , such that S is generated by elements z, y, z conUng from the conj ugacy classes 2A, 3 E , and 29A in the Monster. This is done by checking that no group in the list. of finite simple groups satisfies these properties. One is thus forced to invoke the classification theorem for the finite simple groups to prove rigidity in this case. [Although the proof of the classification theorem has been announced, described} and a.dvertised since 1980, it is not yet clear whether it is complete or not� the part is
.
on "quasithin" groupa has never been published.]
Chapter 8 C onstruction of G alois ext ensions
of Q { T)
by the rigidity Ill e thod 8.1
The main theorem
Let K be a field of characteristic zero, let PI , " " Pk be distinct K r ational points of P1 , and let OJ , . ' . , Ole be a family of conju gacy classes of a finite group G with tri v ial center. The following result is due to Belyi, Fried, Matzat, Shih, and Th ompson ( lee [Ma3] and [Se8] for references). Theorem 8.1.1 Assume th4t the f4mily (01 , . . . , 0.) is rigid and tlut.t eAch. Ci is rational. Then. there is a. regulAr Gcovenn.g 0 . PI defin e d over K which
is
un.ramified outside {PI " "
t
Pk} 4nd such that the inertia. group over
each P, is gen.eroted by 41'1 element of Oi. . uniqu e, up to a u1'lique Guomorphism. By
taking K
=
Q,
one
Furthermore, such a coverin.g is
has:
Corollary 8. 1.2 G has property GalT (and hence is a Galois group over Q).
Proof of tho 8. 1 .1 Let L be the maximal extension of
K(T) unramified outside { Ph " " Pc } , the Galois group of L over K(T). This is the algebraic fundamental group of PI  {PI , . . . , PIli:} over fe (It is also called the geometric fundamental group b ecause the ground field is algebraically closed. ) Since L is iii. Galois extenlion of K ( T ) , one has an exact sequence:
and let
71'"
denote
1
+ 11' + 1I'K +
81
r
+
1,
( 8 .1)
Chapter
82
8.
The rigidity m.et hod
where 1fK is the Galois group of L over K(T), and r Gal(K / K). Let I, be the inertia group of "II" at Pi. . As a profinite group, 11' has a presen tation given by Ie generators :1;1 , " x� and a single relation :1:1 ' :1:1: :::: 1, ct. tho 6.3 . 1 . More precilely. choose a coherent system { Za} of roota of unity in Q. This choice determines an element x, in each 1, up to conjugacy in 11' . One can then choose the Xi so that they satilfy the relation Xl ' " ZI: = 1 (cf. §6.3) . The set Hom( 11', G) of continuous homomorphisms 11' + G il equipped with natural G· and lI'Kactions. The G act i on is defined (on the left) by =
• •
(g . f)( x )
=
.
9 E
gf( X )gl ,
Gf
and the 11' xaction is defined (on the right) by (/ . a ) ( x )
=
f( a:z:a. 1 ) ,
E
a
•
/ E Hom(1f , G),
/ E Hom(lI', G ) .
WK,
The two actions commute, i.e.t (g • f) • Q Consider the set H
C
=9
• (f • Q) .
Hom( 'If , G) defined by:
H = {,preP is surjective and ,p(:z:,)
E
Oi for all i} .
This set is stable under both the G and wKadions: •
The action of G on itself by inner a.utomorphisms stabilizes the ai, and hence G preserves H , which is isomorphic to E(Oh ' ' ' ' Ok ) as a Gset (d. §7.3). By the rigidity assumption, G acts freely and transitively on H.
•
Conjugation by an element u E 7rK sends an inertia. group Ii at � to an inertia group at p{ . Since the Pi are Krationa.I, Wx permutes the inertia groups above � . Hence, u sends each of t he Xi E Ii to a conjugate of :z:i, for some a E Z (namely, a = X ( u ) , where X is the cyclotomic character) . By the rationality of the C. , it follows that J . u maps each z, to an element of Cit and hence f . u E H.
Any � E H defines a. Galois extension E of K ( T) with Galois group G. To descend from K (T) to K(T), it suffices to show that ,p can be extended to '1CK. This is an i mmedi ate consequence of the following: Lemma 8 . 1 .3 Let 1
+
A
+
B
+
a
+
1 be
411. exact sequenet:
0/ Hom( A, G) as above ( i. e .J (g . f)(:z:) == gJ(x)gl if 9 E G , :z: E A, and (! • b)(z) f ( b:z: b  1 ) i/ b E B � :z: E A) . 1/ H is a non. empty subset 0/ Hom(A, G) on which G acts groups, let
G
b e 4 finite group, 4nd let
G
and B act on
=
freely and transitively, then the follo'Wing are equiv41ent :
,p E H tzttnds u.niquely to a homomorphism B H is stable u.nder the action of B .
1. A ny
I.
+
G.
8.2. Two variants
83
1 holds , i.e. , any ,p E H ext en ds uniquely to a homomorphism G. If b E B, then (,p . b)(%) = �(b:tbl) = tP(lnb 1 ) (�(b) * �)(z). Hence � 'II b E H, since by hypothesis H is preserved by the action of G. Conversely, if property 2 is satisfied, then given � : A + G, one may define ..p : B + G by: ,p * b = ,pCb) * ,p.
Proof: Suppose
..p : B
+
=
SuCh a 1/1 exists (G acts transitively on H) and is unique (G acts freely on H). One verifies that ", defines a homomorphism B + G w hi ch extends t/J; this follows from the compatibility of t he G and Bactions. This completes the proof of the lemma, and hence of tho 8.1.l.
Alte",4de method of proof lor tho 8. 1 . 1:
(a) Prove that the required Gcovering 0 to Pi exis t s over K and is unique up to a unique illI omorphism. (b) Use Weil's descent. criterion ( [Se3] , chap. V, no. 20) to prove that 0, together with the action of G} can be defined over K. (General principle: every "problemlt over K which has a unique solution  up to a unique isomorphism over K has a solution over K . )
•
Remark: When Ie
= 3 , one can suppo.se without loss of generality that (0, 1, 00). In this way, a rigid triple ( 0I , C2, 03) of rational conjugacy classes of G determines a canonical extension of K(T). Several natural questions arise in thilt\context:
(Pl, P2 , P3)
=
1 . C an one describe what hap'Rens when K is a local field? We will do this (in a special case) for K
=
R1n §8 .4.
2. Can one describe the decomposit'ion group above �? For example, if G is the Monster, and (01 1 Oz, 03) = (aA, 3 B, 29A) as in §7. 4. 7, then the decomposition group Di above Pi must be contained in the normalizer of an element of the class 2A. This normalizer is 2 x B, where B is the "Baby Monster" sporadic group. Aside from this, nothing seems to be known about Di . Exercise: Show that the alternating group A,. (11.
any rigid family (Ch
8.2 8.2.1
. . •
,
=
4, 5, 6, 7, 8) doe. not contain
CAl) of ra.tional. conjugacy classes (use remark 1 of §7.3).
Two variants First variant
Th. 8 . 1 . 1 can be generalized to the case where the classes are only Krat�ona1. More precisely, let US fix a choice of primitive Nth rootl of unity over It. i.e. ,
84
Chapter
8. The rigidity
method
,�rbit under Gal ( R"/ K) of primitive Nth roots of unity, T h is amounts to a K irredu ci bl e factor of the nth cyclotomic polynomial tP"", ( For example, if K = Q( V5) , N = 5, t he cyclotomic polynomial q,s factors aa an
choosing
=
q,s(X )
X" + X3 + X'J + X + 1
=
(X
2
+ 1
1
+2v5 X + l )(X:Z + 2 v's X + 1 ).
A choi ce of 5th roots of unity is just the same as choosing a s quare root of 5 in Q( v'5).) Such a choice determines a generator :cs for each inertia group at Pi (which is well defined up to conjuga.tion in 'irK ) ' After such a. choi ce haa been made, one has:
I;.
8.2. 1 If 01 , , Ole is a rigid family of K TCltional classes of G, and PI ) . . OJ p.. are KrAtiontd points of PI , then there is (I regular Gco1Jering o + PI de.fine.d over K 'UJhich is unramified outside {Pi t . . . , FA:} anti such. that the %sgenerator of the. inertia group above. Pi belongs to the class Ct . This c01Jering is un.iqu.e.ly defined up to a unique Gisomorphism.
Theorem
• • •
The proof is e&sentiall y the same as that of
tho 8.1 . 1 .
Since conjugacy classes are always ra.tional over the maximal cy clot orm c extension Q C1c1 of Q, on e only n eeds the rigidity condition to ensure th at a group can be realized as a Galois group over QC)'d(T). This property is known for : •
Most of the classical Cheva.lley groups over finite fields (Belyi [Be2) ) i
•
All th e sporadic groupSj
•
Most of the exceptional groups t wi st ed fOrIns
8. 2.2
'lG'll J D. ,
and
G'l . F.. , ESt E7 and E. ( and also the 'l E6 ) over fini t e fields (Malle [MUD.
Second variant
The assumption that the conj ugacy cl as ses are ration al is often too restrictive for applications. The fol lowi ng variant of t h o 8. 1 . 1 is useful in practice:
8.2.2 Let (G1 , G'l l G3) be a rigid triple of conjugacy classes of 0, with GI rAtional and G2 and G3 conjugate to each. other over a quadratic fie ld Q( v'D) . Let PI E PI(Q) , P2, PJ E PI (Q( v'D», with. P2 ani Fa conjugate to ea.ch other. Then there is a regular Oeztension of Q(T) 'UJhich is ramified only a.t PI P:z and PJ I and such that the canonical generator oj the inertia group at Pi ( which i s welldefined after a choi ce of roots of unity over Q( VD») is in G, .
Theorem
�
C or ollary
8.2.3 The group 0 has prope.rty GalT .
8.3. Examples
85
The proof of tho 8.2.2 is similar to that of tho 8.1.1. The set H C Hom('lf , G) is defined in the same way. The key point is to prove that H is Itill preserved under the action of 'lfQ. •
H
a
E
1fQ is trivial on Q( ¥D), then
fixes the choice of roots of unity over before. •
a
fixes
Q( ../D).
PI . P2
and
Pl,
and also
Hence a preserves
H, as
f.1 is not trivial on Q( v'D), then u interchanges Pz and Pa, and hence lz and 13• But a alao changes the choice of roots of unitYt and these effects compensate each other.
H
Remarks: l. The assumptions
on the number of classes and the field of rationality are only put to simplify the proof, and b ecause this is the principal case which occurs in practice. In fact , the same conclusion holds in greater generality, e.g. , if { CI , . . . , C�:} is stable under the action of Ga1( Q/Q), and the map { Ph " " P.} + {Ci t . . . , e.. } defined by � � et. is an antiisomorphism of Gal( Q/ Q )sets. For other variants of tho
2.
8.3
8.1.1
using the braid groUPJ see [Fr 1) and [Mal1 .
Examples see
Here also, we only give a few examples. For more,
8.3.1
The symmetric group
[Ma3] .
S"
Sn has a rigid triple of conjuga.cy when (k, n) == 1. The covering PI + PI given by 1 )nok has ramification of this type, namely:
Recall from §7.4,1 that the symmetric group classes
X
1+
(nA, 2A, C(I» ,
X l( X .
{ !:� t
Hence, by tho
8.1 .1 ,
=
��)
kiCk  n )",1I11n
2A
the polynomial
Xl( X  I ),,1  T haa Galois group S" over Q(T) when (1:t n) the splitting field of the equation
Q { T 1/C J I'r).
=
1.
=
0
Note that when
(k, n)
::::
I #:
1,
Xl ( X _ 1 )"'·  T = 0
contains Its Galois group is strictly smaller than S" (and the extension is not regular when I > 2 ) .
86
Chapter 8. The rigidity method
8.3 .2
The alternating group As
Recall from
§ 7. 4 . 2
that the triple
of conj ugacy classes in A5 is
( 2 A , 3A, 5A)
Q( v's) (but not over Q ) . By tho 8.2.1 , there is a regular extension of Q( v'5)(T) with Galois group AS J and ramification of type ( 2 A, 3A, 5 A) . The corresponding curve C has genus 0 (but is not isomorphic to Pl over Q( .J5), d. [ Se5 » . rigid . The conjugacy class SA is rational over
The action of As on
C
can be realized geometrically
Consider the variety i n P 4 defined by the equations:
{
Xl + . . . +
X5
==
u.s
follows (loc. ci t. ) .
0
X: + . . . + X; := O.
Since the first equation is linear, this variety can be viewed as a quadric hy.
The variety of lines on thi s quadric is a curve over Q which Q(.J5) to the disj oint union of two curves of genus 0 which are conjugate over Q( v's). The obvious action of 55 on V (permuting coordinates) induces an action of 55 on this curve. The extension of Q (T)
persurface in
P3.
becomes isomorphic over
corresponding to this curve is a nonregular extension with Galois group 55, which contains
Q( v'5);
Q(v'5)(T).
it can also be viewed as a regular Asextension of
An Ascovering of PI with ramification (2A, 5 A , 58) can b e realized by
taking an Sscovering with ramification of typ e
X
t+
(2, 4, 5) (e.g . , the covering §4.5). This defines a
X5  X· ) and using the double group trick (cr.
regular As�covering of Pl over
Q,
with two ramification points conjugate over
this is the situation of tho 8.2 .2 . This covering can b e shown to be isomorphic to the Bring curve, defined in p. by the homogeneous equations
Q ( V5);
{
Xl + . . . + Xs = 0, X; + . . . + Xi = 0, X� + . . . + X3
=
0,
cf. §4.4, exercise.
Exercise: Let C be the Bring cu.rve in p .. (see above) . a) Let E be the quotient of C by the group of order 2 generated by the transposition ( 12) in S5  Show that E is isomorphic to the elliptic curve defined in P by the 2 homogeneouB equation; (:t3 + (put
x
:=
:t 3, y
y3 + ,z 3 ) + (:t2y + :t2,z + y2,z =
:t4.
=
,z
=
+
y2z + ,z 2 Z + .z2y) + zy.z
0,
)
Show tha.t this curve is Q�isomorphic to the curve
+
xy
:s .
50E of [ANVERS), p.86, with minimal equation
y2
=
+
y
=
X3  X 
2
8.3. Examples
87
 52 /2 of 55 on C product of 4 copies 01 E.
and iinvariant
.
b) Use the action
to show tha.t the
Ja.cobian of C
is Q�isogenous to the
C by A.. i s Qisomorphic t o the elliptic c u rve 50H of iinvariant 2 1 5 . 5 . 2113; this curve is 15.. isogenou a to E. d) Show tha.t C has good reduction (mod p) for p ¢ 2, S. If Np( C) (resp Np( E)) de�otes the number of point t on C (resp. E) modulo Pt deduce from b ) that c) Show that the quotient o f
[ANVERS],
lac. cit. , with
Np(C)
Use
[ANVERS]. p.1l7, to
=
4 Np( E) 
3  3".
construct a table giving
3 , I 7 1 11 1 1 1 " ' 1
Np(C) for
P<
100:
3 120 30 1 97 0 1 8 1 89 9 Check these values by determining the polynomials X 5 + aX + b over FP I wit h ( CI, b) #: (0, 0) , which have all their roots rational over Fp' For instance, if p = 3, there is only one such polynomial. (up t o replacing « by CIt" an d b by W 5 , with 8 Np( C)
t e
F;), namely: Xi
This fits with
+
l lX + 11
0
==
24
30
.,.
(X + 33)(X + 13)(X  4)(X  20)(X  22).
Np( C) 120. C has semi.. stable =
(over F. ) with the union of two co pies of PI intersecting each other at the five points of P I(F4) ; describe the action of S" on this curve, using the fact tha.t As is
e) Show that
isomorphi c to
8.3_3
reduction at
2, the reduced
curve being isomorphic
SL2(F4 ) '
The group
PSL2(Fp)
Th. 8. 2.2 applied to the rigid triples of conjug acy classes (2A, pA, pB) (when �  1 ) and (3A, pA, pB ) ( when ( � ) = 1) .hows that there are regular PSL2(Fp)exteDlions of Q(T) with ramification of t hi s shape. These corre spond to the Shih coverings with N :::: 2 and N = 3 (c!. §5. 1 ). The rigidity method do cs not predict the existence of the Shi h coverings relat ed to N == 7, which are ramified at four p oint s . A coveri ng of PI h aving ramification type (2A, 3A, pAl is gi ven by t he cov ering of modular curves defined over Q( H), X (p) . XCI). T hi s covering has G alois group P SL2(Fp) over Q( #); t he rigidity property shows that it is the only PSL2(Fp ) coverin g with this ramification type. This was first
(�)
pointed out by Heeke [He] . In partiCUlar, t h e rigid t riple (2A, 3A, 7A) in covering of
functi on field E is which
is not
a.
regular.
PSL2(F7) gives rise to the Klein
Q( v:::? ) and having Galois group P SL2(F.,). Its Galo is extension of Q(T ) with Galois group PGL2 (F7),
PI defined over
Chapter
88 8.3.4
8.
The rigidity method
The GalT prop erty for the smallest simple groups
The following table lists the smallest ten non abelia.n simple groups. All of them, except the last one SL2(F 1 S), arc known to have the GalT property. The
last
column of the table indicates why.
�
GalT property
order
group =
SL2(F.) = PSL2(F,.)
60 = 2 · 3 · 5 ==
23 3 7
SL3(F2) = PSL2( F1) As = PSL2(F9) SL:z(F8) PSL2(Fu )
360 = 23 32 5 504 ::;; 23 3 2 7 660 = 22 3 . 5 . 1 1
PSL,(F13)
1 092
=
PSL2(F 1 7)
2448
=
AT
PSL2(F 19)
2520 = 23 . 32 5 . 7 3420 = 22 3 2 · 5 · 19
S L2 (F1 8)
4080
1 68
•
.
•
•
•
•
.
22 · 3 . 7 · 13 24 a 2 1 7 •
24
. 3 5 17 .
(§§5. 1 , 8.3.3); Shih with N = 2 (§ §5. 1, 8 a 3)i Shih with N 3 (§5. 1 , 8.3.3); Hilbert (§4.5); Shih with N = 2, 3 (§§5. 1, 8 . a.a)i .
=
•
=
Hilbert (§4.5);
Th. 8.2.2 and 17.4.4; Shih wi th N = 2
.
•
•
Hilbert (§§4.5, 8.3.2) Shih with N = 2, 3 (§§5.1, 8.3.3); Shih wi t h N = 3 (§§5. 1 , 8.3.3)i
.
?
In addition, it has been shown that all the ap oradic simple groups satisfy the GalT property, with the possible exception of the Mathieu group M23 1 cf.
[Hunt] ,
8.4 8 �4. 1
[Ma3] ,
{Pal .
Local
properties
P reliminaries
?f' : C + P 1 be a Gcovering defined over a. field K which is complete with respect to a. real valuation. Let z be a. point of Pl(K) S, where S is the ramification locus of the covering. The fiber A;,: at :z: is an etale K algebra. wit h action of G; its structure 18 defined by a. continuous homomorphism
Let

.
"'� : Gal ( R / K)
which
is well·defined up
to inner
+
G,
H the by conjugation;
conj ugation in G. Let us denote by
quotient of the set of such homomorphisms we endow H wi t h the discrete topology.
by the
action of G
8.4. Local
properties
89
Proposition 8.4.1 The map from Pl (K) to H defined by :c .... .. is contan uous with respect to the topology 0/ PI (K) induced by the l1cUu4tion on K . (In words, for every :c E P1 (K)t there is a neigbourhood U of :c Galois algebras A. and A. are isomorphic, for all y E U. )
other the
This
can be deduced (with some care)
P(X )
==
II(X

at)
from Krasner's
an d Q(X)
=
such
that
lemma: if
II(X

B�
as
polynomials over K, with a.a and fJi Jufficiently close for each then K(al ) == K(f31 ). See [Sal]. Another way to prove prop. 8.4. 1 is as follows: the map C + Pl gives an extension O. c O., where O. is the local ring at :c, and O� i. the semilocal ring abo\re :c. Let Mz be the maximal ideal of O�. The quotient Oz/M.O. is A One has a natural inclusion O2 C O�ol, where O� denotes the ring of power series with coefficients in K which converge in a neighbourhood of :c. Since O� is Henselian (d. [Raj , p. 79), and the extension is etale, one has: are irreducible
i,
••
This means that, for its analytic st ruct ure , product. The proposition follows. By the discussion above, we of open sets:
may write
Pl (K)  S
=
where U. = {t E P1 (K)  SI;, = ;}. the universal Z/2Zcovering defined by
P1 (Qp)

{O, oo} int o four
nonisomorphic
Remark: naud:
covering
Pi (K)  S as
U
�H
0
PI is locally a
a finite disj oint
union
U.,
For example. if
:c
+
t+
:c 2
K
=
QPJ
P
�
2,
gives a decomposition of open and dosed pieces, corresponding to the four
Galois algebras of rank 2 over Qp.
A propos of the U� , let us mention an unpublished result
if the
order of G is
Gcovering of
local field K.
8.4.2
the
a
of Ray prime to the residue ch aracteristi c t every unramified
rigi d polydisk becomes trivial after a finite extension of the
A problem on good reduct ion
Assume that 11" ; 0 � Pi li a Gcovering obtained by the rigidity metho d (th. 8. 1 . 1 ) from rati on al points Ph ' " J Pic and conj ugacy class 01 , · . . , Cle • Let p be a prime number. Assume the reductions of the � are distinct i n
Pl (Fp).
90
Chapter 8. The rigidity metho d
Pro blem: A ssum e tha.t p does not divide the oTders of the elements of C, jor 1 , , k. Is it true tha.t the CUnlf C has 900d reduction at ]l?
i
=
. . .
This is true
at
least when l' does not divide the order of G, cf. [Bcl.
Exercise: Check that given in
the a.bove problem
8.3. 1 J 8.3.2, 8.3.3.
8 .4.3
has a positive answer
for t he examples
The real case
now restrict our attention to the case when 1f : Y + P1 comes from rigid triple of conj ugacy classes (Ol t 02 , Oa), and where K = R. By p rop 8.4. 1, to ea.ch connected component of PI(R) S there is a.ttached a unique conj ugacy class of involutions in G, corresponding to complex conj ugation. There are two cases, depending on the number of such connected components:
We a
.

Case 1 : We suppose tha.t the three conjugacy classes are rational OVer R.
The corresponding covering is ramified at exactly three real points Pl J P2 , P3, which divide the circle P1(R) into three connected components. Let us choose Zi E Oi, with Zl ZZZ3 = 1 and G = (Zit Zh :1;3) ' We assume G :F {I}, hence z, :F 1 for i = 1, 2, 3. Since the Oi are rational Over R, we have
=
=
The element S 3 thus defined is such that s �
1
;1: 1 � :1: , :;
1.
Th eorem 8.4. 2 The comple:;c conjugation attached to (I point :;c in the con nected co mp on ent (Ph P2 ) 0/ P I C R) between P1 and P2 is in the con.jugacy
clus of S3 '
81 and S 2 , which are defined in the same way as 8s, and correspond to the complex conj ugations attached to the components ( P2 t Ps) and (PI t P3) respect ively. This gives the relations:
Remark: An analogous statement holds for the involutions
=
Hence, if
31
and 33
are
nontrivial, the group generated by 81 J S3 and Z 2 is 2n. where n is the order of ::r:2. There is a single
a dihedral group of order
8.4.
Local properties
91
2n
(!onjuga.cy class of involutions in a dihedral group of order Hence, one has :
with n odd.
Corollary 8.4.3
If u"e order of Z2 is odd� then Sl and S3 tin! conjugate in G.
Corollary 8.4.4
II two of the three cla.sses 01 , O2 tlnd 03 h.al1e odd ezponent,
then the s, 'oS tire conjugate.
Example: The rigid triple of conjugacy classes ster
M
satisfies the hypotheses of cor.
8.4.4.
(2A, 3B, 29A)
corresponds to a unique class of involutions in M (the das s
notation).
Remark:
The case where
oS 3
=
1
2B,
in ATLAS =
occurs only in the case G
D"
and
odd. For, if "3 = 1 , then :t:l , 02:2 are of order 2, equal to .52 and .t 1 respectively. The group generated by Zl , 02:2, Z3 is thus a dihedral group of order 2n, where n is odd (since the group has no center).
( 01 , O2 , 03)
=
(2, 2, n ) , n
in the Mon
Hence, complex conjugation
satisfying the rationality condition over 8.1.1, when applied G :# S3 J never gives totally real eztensiofU of Q.
The only dihedraJ group of this type
Q is G = S3. This
means that the rigidity method of tho
with three classes to a group
Case 2:
01 is ra.tional over R, and O2 and O;s
are
complex conjugate to each
other.
Then,
P l (R)  {Ph P2, P3} is connected, since P2 and P3
do not lie on P1 (R).
Hence there is a single conjugacy class of involutions in G corresponding to complex conjugation. Let (Z1 , Z2, Z3) in G such that:
E �(01 t O2 • 03)'
By rigidity, there is a unique involution
.t
Theorem 8.4.5 Complez conjugation belongs to the conjugacy clGS.! ol .t . Observe that i n this case
Proof of th o 8.4.2
.t
:#
and tho
1 . (Otherwise G would be cyclic. )
8.4.5
Choose a base point
on the (!onnected component (Ph Pz) in case geometric fundamental group of
X
=
01 , 02, Q3 denote the generators of
an d
P3
1)
and let
'I"
z
PI  { PI , P2, P3}, with
1r
on
P1(R),
(lying
denote as before the base point
02:.
corresponding to paths around PI ,
respectively. The complex conjugation CTx on
X
acts as
a.
Let
P2 1
symmetry
around the equator on the Riemann sphere. Hence ax a.cts on the generators ai , a2, 03 of 1r by: Case 1 : 01 1+ ail , 02 1+ a; l
Case 2:
02
1+
Oi l .
a3
1+
ai l
92
1f
Ch ap ter 8. Let
:
Y
Uy
+
denote the complex conj ugation acting on Y. Since the G coverin g
X is defi ne d over R, the following diag ram com m utes :
�
Y
'11'" 1 X Since
ux x
The rigidity metho d
=
crXJ
Y
1 .,..
X
x, t h e fiber Yz of Y over :r: is preserved by ay . The a.ction of ay
on Y:t commutes with the action of G.
The group G acts freely and transitively on Y:t, so a choice of 11
m ines a surjective map fJl : ""l(X; x )
+
with that of G, the following di agram
is commutative:
""1
�
G
""l( X ; x )
�
G
(X; x ) ax 1
On the other h an d ,
we
E
G. Because the action of u
Yz deter
com mute s
II
also have:
Hence the diagra.m below commutes:
""l(Xj x )
�
1fl(X; x )
�
ax !
G
l Inn(uy,,) G.
This proves the theorem.
Exel"Cise: Z
l
] "
Let
. ]
ZJ:
; , . '1
% ,
G be
Let
a finite group, genera.ted by elements 9 1 1 = 0 for i = I , . . ' k, and
(s ay ) R.e(Zi)
with
Z�
be the complex conjugates of .2'1 ,
point :r: < OJ let .,.. denote
{ Zi t ,
• •
�
t ZA: . Z t
• •
' . %�} .
. , " glc. Choose
points
%J: an d choose a real base the fundamental group ""1 ( P 1 ( C )  Sj x ) . where S "
'1
:=
It is genera.ted by element s :t 1 . " " :til . :t� , . , . , :t�] where :t,
of p aths going in a straight line from :t to Zi (resp , z� ) and going around this point in the positive direction. These generators, together with the rela.tion :t I ' :tJc:t� , :::: 1 t give a presentation for .,.. . Let
(resp . :t� ) denotes the homotopy class
, •
. •
z�
8.4. Local properties 8 .4.4
93
The padic case: a theorem of Harbat er
Let p be
& p ri me number.
Theorem 8.4.6 ( [Harb] ) E'Very finite
e:tension of Q,,(T).
group
is �e Ga.lois
group
of a regular
proof shows more, namely tha.t for every finite group G, there exists an a.bsolutely irreducible Gcovering X + P1 over Qp having a. "base poi nt" , i.e., an unramified point P E Pl ( Qp) which is the image of a Qppoint of X. Call R,. this property of G . The theorem clearly follows from the following two as sert ions � (i) Every cyclic group has property Rp. (ii) If G is generated by two sub groups G1 and G2 having property Jl"J then G has property Rp. Assertion (i) is easy (and true over Q, as the con struction of §4.2 shows). Assertion (ii) is proved by a gluin g process which uses rigid analytic geometry. Namely, let X. t P I (i = 1, 2) be a G.covering as above with a. base point Pt . By removing a small neighborhood of .l1 one gets a rj gi d analytic Gicovering Yi + U. where U, is a padic disk; thjs covering has the further property t hat it is trivial on Ui  U; where U; is a smaller disk contained in Ui Let Wt + Ui be the (nonconnected ) Gcovering of Ui defined by induction from Gt to G. (It is a d isjoi nt union of IG/Gi l copies of \1 .) One then embeds Ul and U2 as disjoint disks in P I , and defines a rigid analytic Gcovering of P 1 by gluing together Wi on U1 ) W� on U2 , and t he trivial Gcovering on PI U; u;. H thi s is done properly, the resulting G covering W is absolutely irreducible. By the "GAGAu t heorem in the rigid analytic se t ting (d. [Ki] , [Ko])� this covering is algebraic, and (ii) follows.
The


Remark: Harbater's original proof uses "formal GAGAn instead of "rigid GAGAn ; the idea is the same.
Chapter 9 The for:rn Tr(x2) and its applicat ions 9.1
Preliminaries
Galois cohomology (mod 2)
9.1.1 Let
K
be a field
of characteristic I 2, and let GK be Gal( K./ K), where K. K. We will be interested. in the Galois cohomology of
is a separable closure of
K
by
modulo 2; for brevity, let us denote the cohomology groups
Ir( GK ).
In the case where
interpretation of
Hi(GK ):
i
=
BI ( GK )
=
K/Kd,
H2 (GK )
=
B r2 ( K ),
where B r 2 ( K) is the 2torsion in the Brauer group of K. If a E
by (a) the corresponding element
to the
9 . 1 .2 Let
f
E
= 0,
we denote
(a. b ) i n Br2 (K)
defined by
H2(GK )
class of the quaternion algebra
i2
K· .
of Hl (GK). The cupproduct
(a)( b) corresponds
Ir( GK , Z/2Z)
1 , 2, Kummer theory provides the following
?
=
b,
ij
=
ii.
Quadratic forms
be a nondegenerate quadratic form over
K of rank
an appropriate basis, we may write
f=
"
L tl,X!,
with tli E
1=1
95
K· .
n
� 1. B y 'choosing
Chapter 9. The form Tr (:2) and its appli c ations
96
The element (1 + (a l ) ) ( 1 + (a2)) ' " (1 + (an)) in the cohomology ring H*(GK) depends only o n f . One defines the i t h StiefelWhitney class Wi of f by:
In
particular I we have: Wl
=
L(ai) 'II fti) =
W2
:::::
=
(Disc(f)
2: (a,)(a;). i.
The cohomology class W 2 is known as the Hasse (or Witt) invariant of the quadratic form f. If K is a number field, then f is completely chara.cterized by ita rank, signature, and the invari an t s WI and W2' (The same is true for ar bitrary K, when n $ 3.) The following results can be found in [Sch] , pp . 211216.
(Springer) If two quadratic forms over K become equivalent over an odddegffe =tension K' of K, then they are already equivalent over K.
Theorem 9. 1.1
(For
a generaliza.tion
to hermitian forms, see (BL] . )
Let us now consider K(T), where T is an indeterminate. If v i s a pla.ce of K(T) which is trivial on K and ¥ 00, there is a unique uniformizing par ameter '1'" at 'fJ which is monic and irreducible in K[T] . Let K(v) = K[T] /(w,,) den ot e the residue field a.t Vt and let 4 1+ a be the reduction map K[T] � K (v). If ! � E�l a.Xl is a quadrat i c form over K(T), we may assume (since the ai can be modified by squares) tha.t v ( a,:) = 0 or 1. The K(v)quadratic forms:
a1 ( !) ;::::
L
"(0.)=0
a;,
X:,
02(/) =
L (a,/'If,,) X:
.(Cli )=l
are called the first and second residues of f. One shows ( cf . e .g. [S ch}, p . 209) tha.t their images in the Witt group W (K ( v)) do not depend on the chosen representation of f as E a,Xl  (Recall that the Witt group WeLl of a field L is the Grothendieck group of the set of qu a.d rati c forms over L, with the hyperbolic forms identified to 0. ) The orem 9. 1.2 ( Milnor) If a quadratic form l over K(T) has second residue o Ilt all plilces oj K(T ) e:cept 00, then 1 is equivalent to 4 qUlldratic form over
K.
/
9.1.
Preliminaries
97
(More precisely, one has an exact o
+
W CK)
t
A qua.dratic form over a
sequence:
W( K(T))
ring
discriminant is invert ible in R.
R is
Theorem 9 . 1 . 3 (Harder) A strictly
ri""9 K [T) comes from K.
This over
theorem can be formulated. more
+
said
U W( K (v)
v�oo
+
0 .)
t o be strictly nondegenerate if it s
nondegenerAte quadratic form over the suggestively:
a quadratic vector bundle
Al is conatant. (For a generalization to ot her type! of bundles, see
9.1.3
We need
[RRJ . )
Cohomology of Sn only
H'(Sn) for i
=
1 , 2. These
groups
are
well·known:
Z/2Z, n � 2, Z/2 ED Z/2Z, n � 4 The non t rivial
(Schur) .
H1 ( S,, ) is the signature homomorphism
element in
The cohomology group H2($,, ), n � 4, has a Z/2Zbasis given by � (cupproduct) and an element 3n corresponding to the central extension
1
........
O2
+
5" + S" +
which is charact erized by the propert ies : 1 . A transposition in 8ft lift s to an element of order
2. A pro duct of two disjoint
transpositions lifts to
an
.
Et&
1
2 in Sn . elem ent of order 4 in 5",.
(The element e,.. . � correspond s to the extension s:. of Sn. ob t ai ned by taking the pullback: 04 s'" !
$"
!
{±l }.
Thia extension is characterized by t he property that a transposition lifts to an elem en t of order 4, while a p rodu ct of two disjoint transpositions lifts to an element of order 2.) The im age of Eon by the restriction map Hl (Sn) . Hl (An) i s zero. The cohomology group H2 (A,,) is isomorphic to Z/2Z for n � 4 and is generat ed
Chapter 9. The form
98 4n
by
( or
Tr (:z:2) and its
applications
Res ( " n) . The corresponding central extension of An is denoted by Aft 2 . � in ATLAS' style); it is a subgroup of index 2 of Sn '
=
by
The
En and Sn can be gi ven the following topological . On(R) gives a map of classifyi ng spaces
co homology classes
interpret ations:
the ma.p
and corres ponding
wh ere Wi
Sn
maps on
the
cohomology rings.
is the ith StiefelWhitney
element En
Exercises:
in B l (BS".)
=
class.
HI (Sn )� and
W:z
But
The class gives Sn .
WI corresponds to
the
SL:z(F3), As � SL2(FIi ), and As � SL:z(Fg). of Sft by {±1} corresponding to the element 8ft + 4a 4a. of H2(Sn ). ( a ) Show that 5;. � GL 2 (F3). (b) Show tha.t an y outer automorphism o f Se lifts to an isomorphism of 56 onto 1. Show that
2. Let Sn be
A4
::::
the
centra.l extenaion
•
56.
(c)
Show that the groups
9 .2
Sn and Sn are not
isomorphic if n � 4, n ':f:. 6.
The quadrat ic form Tr
(x2)
an etale Kalgebra of finite rank
n over Kj it is a. prod uc t of sepa. of K. There is a dictionary b etween such algebras and conjugacy classes of homo mor p hi sm s e : GK t Sft, which works as fol lows : given E, let �(E) be the set of Kalgebra homomorphisms E � K The set tI>( E) is of cardinality "" and t he natural a.ction of GK on ¢(E) gives the desired homo m orphism e : GK + Sn , after i dent ifying �(E) with {l" , . , n}. C onvers ely . E can be co nstructed as the twist of the split algebra K x . . . x K n by the lcocycle e : GK + Sn = Aut ( K ). The i m age of e is the G alois group of the smallest extension of K over which the algebra E splits; if E is a. field, it is Gal( E�eJ. / K).
Let
E
be
rable field extensions
•.
The func ti on 3: � Tr (x 2 ) defines a non degener at e quadratic form QE of rank n over K. This invariant was st u d ied extensively by 19th century ma.th ematici ans such as Jacobi and Hermite. �
Theorem 9.2.1 Let E/ K ( T) be a. finite s epa.rcble extension of degree
n.
L�t
G c Sn be the COfTesponding Ga.lois group . As.'J1,me tha.t, for a.ll places tI of K ( T ) not equa.l to 00 , the order of the inertia. group at v is odd. Then QE is consta.nt,
i. e.,
comes from
(I
form over
K.
9.3.
Application to extensions with Galois group k
99
Let A be the in tegral closure of K[T] in E, and denote by 'D the different. Using the fact that the inertia groups are of odd order, one can show that V 1 is the square of a fractional ideal,
One checks that 0 B induces a strictly Dondegenerate quadratic form over KIT ] on the K{T)module A. By tho 9 . 1 .3 the result follows. ( O ne may also show that the second residues of Q B are 0, and app ly Milnor's theorem 9 . 1 .2.)
a regular Galois extension of Q(T) with Ga.lois group the Monster, obtained from the rigid family (2A, 3B, 29A), cf. §7.4.7. Since two of the conjugacy classes in this family have odd exponent, the form QE comes from Q , and does not depend on T. One can prove that it is hyperbolic.
Example: Let E be
The fol lowin g theorem 18 proved in [Se6] .
Theorem 9.2.2
tlSsociated to
a
form oj E . Then : 1.
1lJl ( Q S) :; W2( QS) =
!.
Let E be tln ittll e K algebra of ranJc n tlnd discrimintlnt d e : GK + SnJ an.d let Os denote the trace
homomorphism
e · En., e · sft + (2)(d)
Suppose that G = e ( GK ) is contained in A ' B i.e. , that d is a square. By tho 9.2.2, we have tVl (QB) = 0 and w,(QE') ;;; e· o.,.. The element e·o.,. is the obstruction to lifting the homomorphism G K + A n to a homomorphism
GK
.
An. Hence, we have
Corollary 9.2.3
GK
+
9.3
Aft if and
The homomorphism e � GK only if
the
Witt
t An lifta to a homomorphism invariant 1.U3 ( Q B ) is O.
Applicat ion t o extensions with Galois group
An
The previous corollary applies when e is surjective: an extension E of K of degree n with Galois group A.,. can be embedded in an A,,extension if and only if W2(Q S) ::; O. This will be used to show the following result: Theorem 9 . 3. 1 (Mestre, {Me2] )
oJl
n.
The group
.4"
hD.S
the GalTproperty for
Chapter 9. The
100
( This was already shown for
n ==
0,
form
Tr (:r 2 ) and its applica.ti on s
1 (mod 8), a.nd for some other
n,
by Vila
[Vi ] . ) The pr oof constructs a covering C + P I of degree n w hose G alois closure has Galois group A,.. , and which has the following addi ti on al properties :
1.
PI
There is a point of
whose inverse i m age is a set
t he a.; a.re rational and distinct.
2. The nontrivial inertia. groups a.re all Let E 1 K(T) be the
degree n
{al "
.
. , 4n h
gener ate d by cycles of order
extension corresponding
to this
where
3.
covering. Con
2 i mpli es that the quadratic form QE comes from Q , by tho 9.2. 1 . But by cODdition 1, t her e is on e rationa.l point where this quadratic form is equiv� dition
alent to the standa.rd one, L Xl . It is easy to see that this implies that Q E is
L Xl over Q(T), and hence has trivial Wit t invariant j by can thus solve the embedding problem for this extension.
equivalent to 9.2.3, we
a.
Let us say that a. property E of
p
=
n x +
is generally true if there exists P has the
od d)
prop e rty E
for all
relies on the foll owing:
a
polynom ial
SlX" 1
+
...
+ 8n
Zariski open dense subset U
(S 1 1
.
•
cor.
•
, Sn)
of An
such that
in U. Mestre's construction ( for
n
Prop osition 9.3. 2 It is generally true th4t there e:rist polynomials Q and R in Q [T ] of degree n  1 with the following properties :
a) Q'p  P ' Q = R2 ( i . e . , ( Q I P)' = ( RI P)2 ) . b ) P, Q, R 4re p4irwise rel4ti1Jely prime.
The zero3 b1 , , bn1 of R in Q are distinct. d) The 1J4lues t, of Q / P at the Oi are distinct.
c)
. . •
Skdch
for
i
==
of proof: The matrix M
j is a skew
with ij  e nt ry
symmetric m atr ix
determinant and there exis t s a nODzerO have:
t.. � = o
j
l fli  fl,
#1.
( C. , . . . J en) in for all
Now, let Q , R be the polynomials such tha.t :
RI P
ft
=
1 /(4..  4;)
of odd dimension.
Ci E x:::'
i= J
t:t.
I.
for
i =I j
and 0
Hence it has zero
the kernel of M. We
9.3.
Application to extensions with Galoi. gTOUp A,.
One
has: =
= =
Moreover, one can
"
'�1 (X
CiC "
a,) (�  aj) t _...; c'f� ;.4; )2 + E (X 
101
"(_ 1
_ "C'_ " _Ci
i;f:; CI.i
i=l
A;
_
X  a.
_
1_ )
_
X
a;
(Q/P),
check ( by looking at a wellchosen example) that it is gen rank n  1 (so that the Ci are essentially unique) and
erally true that M haa
C), d ) hold. If (P, Q, R) are chosen as in prop. 9.3.2, with P = n(X 4i being dis ti nct , then the map PI + P1 given by X .......
that b) ,
ail with
haa degree n., and is ramified at
the being gen erated by 3 cy cles . Let G c
zeros
bi
of
T
=
4i E Q, the
P(X)/Q(X) R, the ramifica.tion groups
8ft (resp. (J c S,,) be the Galoia group
Galois extension of Q(T) (resp. of Q(T)). The group G by 3. cyclea (cf. prop. 4.4.6). By lemma 4.4.4, it is equal to A..a . Hence G is equal to An or Sft. However, we have seen that the Tr (:z: 2 ) form attached to the extension is the standard form E Xl. In particular, its discriminant is a square. This shows that G is contained in Aft , hen ce G = G = Aftt QED. There is a. similar, but more complicated , construction when n is even. Exercise 2 below proves that L haa the GalT property for n even by reducing to the cue of odd n.. of the correBponding
is transitive and generated
Remarks:
1. One may also prove Mestre's theorem by showing that the Witt i nvari ant W 2 E Br2(Q(T» of the trace form haa " no polesu (because all the ramification is odd ) , and hence is constant, i.e. , belongs to Br2 (Q). Since it is 0 at the bue point, it is zero. Another possibility is to prove by a direct con stru ctio n that the trace form of E / K(T) is con st ant , cf. exerc. 1. 2. For explicit formulas related t o the above constructions , s ee [Or) , {Schn] . 3. For n. = 6, 7, the Sch ur multiplier of � is cyclic of order 6. The correspond ing groups 6 · As and 6 . AT also have property GalT. Thi s haa been proved by JF. Mestre (unpublished). Exercises : 1.
Assume P, tJ, R are aa in prop.
9.3.2
(n odd).
Put
E :::: X(X ) and T :::: P(X)/Q(X}. so
that [E ; K(T)] :::: n..
a) Let A be the integral closure of K[T) in E. Show that A = K[X, l/Q{X)] and
that the different of A over
K[T] is the principal ideal generated by R(X ) 2 .
Chapter 9. The form
1 02 b) Let be the trace map. If
f
that:
E
Tr : E
K[X],
sh ow
degT Tr (Je X )/R( X ):l)
�
and its application6
K(T)
+
that
Tr ( z:l )
Tr (f(X)/ R(X) 2)
5Up( O, deg(f)

belongs to K[TJ a.nd
271. + 2).
(J(X)/ R(X) 2 ) belongs to K if deg( f) :5 271.  2. c) Let V be the ndimenlional Ksublpa.ce of E spanned by
In particular, Tr
i/R(X ) , X/ R(X), . . . , x
n l  / R(X) .
One hu E = V ®K K(X). Show, using b), that if lIl t V, E V, then Tr ( 1II 1I2 ) E K. Conclude that the trace form of E / K (T) comea from K. 2.
Let
n
be even, let
f
:
construction above , a.nd let
and 9
:
CJ
......
PI
CJ
.
PI
be
an
n
+ i covering given by Meltre's
be it Galois closure. One haa maps
PI is a regular An� covering.
Show tha.t tml covering lifts to a regular
An covering of Pl (hence An also haa property GalT ) .
Chapter 1 0 App endix: the large sieve inequality Statement of the t heorem
10.1
Let N be an integer � 1, and, for each prime p, let lip be a real number with o < lip � 1. Let A be a sub set of A = Z, such that for all primes p,
where
(:1:1,
• •
•
Ap C AI ph denotes the reduction of A mod p. Given a vector z := , :z:,,) E Rft and N E R, we denote by A( x, N) the set of points in A t
which are contained in the cube of side length N centered at
:1: ,
i.e.,
Then:
Theorem 10. 1 . 1 (Large sieve inequality) For every D � 1/ we h.a.lIe
Taking
D
=
Nt :
Corollary 10. 1 . 2 IA(z, N}1 'S
(2N)ft/L(Nt ).
103
104
Chapter 10. Appendix:the la.rge sieve
inequ ality
Examples! 1. H vp = � for eve ry P, then
Hence IA(z, N)I « 2.
Assume there is
p E S, with 0
primes .s D:
<
C
<
(
=
L(D)
L 1) tl ���free
.....
6  D. 7r 3
,N" L This is a typical "large sieve" si tuation. a
set S
1.
of primes of density >
0,
such that
V" =
C for
O ne may estimate L(D) from below by summing
L(D) � 1
+
L
1
.
 v" v"
::2>
Over
�. lo g D
Hence j A(x, N)I <:: ,N" l log N. A more careful estimate of LCD) by summing over all squarefree d � D allows one to r eplace the factor log N by (lo g N )'l, with 'Y < 1 , under a mild extra condition on S, d. [Seg} , chap. 13.
3. Suppose n = 1 , and v" == 1  ; . Then one can show that L(D) ;:::" log D, and hence IA(z, N) I <: lo:N I a weak form of the prime number theorem: howe ver, the method also allows one to conclude that in any interval of len gth N I there are at most O( lo:N ) primes . (More precisely, their number is � 2N/ lo g N, cf. [MoV] . ) Historically, a weaker form of the sjeve inequality was discovered first, where the sum giving L(D) was taken over the primes � D; this only gave interesting results in lar g e sieve situations (hence the name "lar g e sieve inequality" ). The poss ibility of using square�free d's was pointed out by Mont gomery, (Mol]. Exercise:
that p + 2
Use t h o 1 0 . 1 . 1 t o show that t h e number of "twin primes" (primes p such also prime) � N is sympt oticaJ1y < (iosNN )f ' Conclude that a
is
1
L
twin
"
prune

p
< 00 .
P ro of of tho 10. 1 . 1 : preliminaries
Let us assume without loss of generality that A = A(z, N). a = ( al , . . . I �), t = (tit . . . , tn ) belonging to Rn , put
Xm(t)
=
exp
G iven
vectors
(27rit ) ,= 1
a;t; .
Rn /zn by We identify A = Z N wi th the character group of the torus T a 1+ Xa l and associate to A ACz, N) the function t/J whose Fourier expansion =
=
105
10.2. A lemma. on finite groups is:
; = LA Xca. Cle
d ition on the reduction of A mod p and the fact that A is contained in a cub e of side length N give rise to inequalities satisfied by ¢; combining these will give the sieve inequality. The
co n
A lemma on finit e groups
10.2
Let C. for' l
:5 i � It. be finite
abelian group.
=
(written a.dditively), OJ
their
character groups, t/J a function on C n Ci . Suppose there are subseh 0,: of Ci with Ind :S lIi l C, I, such that the Fourier coefficient of t; rela.tive to the character X = (X. ) E t == n 6; is 0 outside n (li . Let us call :l: E C primitive if its image
in each C,
is =f. O. Then:
Lemma 10.2.1 We have :
11:
�
primitive
1;("'1 1" � 1;(0) 1"
IJ C : V;) . .
We give the proof in the case of a single group C: the general case follows by induction on the number of facton. Write tI> = E Cx;X, the sum bei ng taken over all characters X E n. Then:
Applying the
CauchySchwarz ine qual i ty, we ItI>(OW'
and hence
=
get
I E ex ' 11 2 s. E l ex l2 E 1 ,
1q,(O) 12 �
"eO
IIl CE ItI>( :t ) 12 + 1 <1>(0)12 ). z:#o
The lemma. follow s by rearranging terms in this inequality.
10.3
The D avenp ortHalberstam t heorem
Define a distance on Rn by {:l:1 = sup l :t. l i thi s defines a distance o n the torus T = Rn /zn , which we aho denote by I I_ Let 6 > 0; a set of points {til in T
is
called
5spa.ced if [ t,  ti l � 6' for all i =f:. i.
Chapter 10. Appendix:the
106
la.rge sieve inequality
Theorem 10.3. 1 (DavenportHalberstam) Let rP = L C�Xl be a continuous func tion on T who.!e Fourier coefficients Cl vAn ish when .\ is outside .!ome cube E of .!ize N. Le t � E T be 6spaced points for .!ome 6 > O. Then
�, I rP( ti)12 � 2ft sup(N, i t l i rP l l�, where 1 1 ';1 1 2 is th e L2n o rm of rP . If 6 > 1 /2, there is at most one t, and the inequality follows from the Cauchy
Schwa.rz i neq uality applied to the Fourier expansion of
�.
6�
that
One constructs
an
auxil iary function (J on
rP .
Let
us
Rn t such
1 . (J is continuous a.nd vanishes outside the cube 1%/ to view (J as a function on T.
<
��.
now suppose
tha.t
Th i s allows us
2. The Fourier transform of (J has a.bsolute value ;:: 1 on the cube
3. 1 1 91 1� � 2 n M'\ where M Let ..\
E
Rn
=
sup ( N,
E.
� ).
be the center of the cube E. Then one checks , by an elementary
compu ta.tion, tha.t the function 8 defined by
elsewhere. has the required properties . coefficient of
We are:
rP i
defi ne
For each ..\ E A, let
simila.rly cl ( B ) . We
ma.y thus define a continuous
Since Cl( '; )
B*9
of
B
=
and
fun ction
cl ( B )Cl(g) for every .\ g. Therefore:
rP (t,) where B, is the s et of
=
E A,
Cl( '; )
be the .\.th Fourier
have:
9 on T whose Fourier coefficients
if ..\
E
if ).
�E
E
rP is equal to the convolution product
( B ( t,  t )g ( t ) dt = r B( ti  t )g(t) dt, JT lB,
t such that Ittal
<
�.
By the CauchySchwarz inequa.lity;
10.4. Combining the information t he t,
Since
are 6�lIpa.ced,
the
107
B, are disjoint. Summing over i
then
gives
E 1 t/>(t.;) 12 � 2" M" l Ig l l� � 2" M" 1 It/>1 I�, i
becaus e
I l g lI; = This
1:
co
mpletes the proof.
Remark:
In the
(Selberg, see
n.
1 0 .4
case n
=
1, the
e.g. [M02] ); it
I �:�:�r
�
"","�.
can be impraved to N + i similar improvement holds for any
factor 2 sup(N, � )
is likely that
a.
Comb ining the information
Let D be given; the set {ttl of all ddivision points of T, where d ranges over positive squarefree integers ::; D, is 6spaced, for 6 = 11 D2. Applying tho 10.3.1 to t/> ::: E"E A X,,, we have
E 1t/>(tiW' 5 2" sup( N, D2 )" I A I · •
On the
splits
other hand, for ea.ch d �
as
D squarefree, the kernel T(d) of d : T
T [dJ
=
( 1 0.1) +
T
TI T[Pl Jlld
and its character group is AIdA ::: npl.r AlpA. Hypothesis (2) on A allows us to apply lemma 1 0.2. 1 to the restriction of t/> to T{d) . We thus obtain
Hence,
by summing over all squarefree d � D,
L 1 ,p(t.) 12 i
Combining equations 10. 1
and
we
;;:: IAI2 L(D).
obtain:
(10 .2)
10.2 and cancelling a factor of IAI on both sides I AI = 0 does not pose any problem.)
gives the large sieve inequality. (The case
QED.
Remark: A similar sta.tement holds for a number field K; A is replaced by
OK x · · · X OK, where OK
denotes the ring of integers of K; the corresponding
108
Chapter 10. Appendix:the large sieve inequ ality
torus
T is
then equipped with a natural a.ction of OK . The technique of the
proof is essentia.lly the same as in the case
Exercises:
Pi (i E J)
1.
Let
A
m od
be integers � 1 such tha.t
K
=
(Pi,P, )
of zn contained in a cube of side length N. Let
Q,
=
IIi
see
[Se91 , ch. 1 2.
1 if i � j. Let A be
a
subset
be BUch that the reduction of
Pi has at most ViYi' elements. Show (by the same method as for tho 10. 1 . 1 )
that
with
L{D) ;; E II{l  Vi )ll/i, J iEJ
where the sum runs through a.ll subsets J of 1 such that
ITiE J Pi :$
D. (This ap plies
for instance when the Pi'S are the squares or the cubes of the prime numbers.) 2. Let H be the set of pairs (:2:, y) of integers 1= 0 BUch that the Hilbert symbol (z,
y)
is trivial (Le. , the conic Z2

zX 2

yy 2
=
(by using exerc . 1 . ) that the number of points of <<
N 2 / log N .
0 has
H
in
a
a.
r ational point ).
(Whether or not this bound is sharp is not known. )
For generaliz ations , see [Se lO] .
Show
cube of side length N
is
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I,
Index Abhyuwll
�aUedure.
permu tatioD �oru., 36
42
abloJut.ely irreducible 'lariet.y. 20
Briq cune,
86
chvact.er pup
(of Il tonn)t e
profinite completion (of' a diac:rete property (WA), 28
property (WWA), 28
Chowt• theorem , 55
property GalT, 35
DaveDpoft..BaJbcntun theorem, lOS
double pup trick. 43
FiacherGri .. MODlter )I. 11
QratioDal (coaj ulacy
QratiODal. ziii
claII) .
ratioulity (of' a coDjulacy
fundameDt.al sroup (Illebraic). 51 GAGA principle, 55
rqulu
(extenlion). 35
raidu..
dua),
85
(of' a quadratic (orm), 8S
rntridiOil of' IC&lan (CuDdor). 21
GaJr property, 35
rigid «(amDy of' CODjUlacy claua). 10
pDerally true (property), 100 poeric (coveriq map). ziv
rilidity, 70
HIlI.Janko croup J2 , 18
Schol... Reicbardt. iheorem. 8
GraueRRemmert ut.maiOll theorem, 56
Harder'l theorem. e1
SN
•
property. 10
_parable (profizU�e lfouP). 15
leXtic raolveDt.. 25
B...... Witt iDvuiut
(of' a quadratic Corm). IS
(of a Driety).
85
ra�ioul over Q (variety), ziii
Ftattini Clubpoup), 16
Hilbert property
l1'0up). 58
Shafarevicb'l theorem, 8 18
Bilbert'. irreducibility theorem, 25
Shih'l iheorem, 47
ailllature (OD a RiemaDn .udace). 60
Hilhertian (field), 18
�lit (toru.), 6
Irr(P) property. 23
Irreducibility iheorem (HUb ert) , 25
StiefelWhitDey cluIa
Jordan'. lemma, 45
drietly rilid (family oC eODjUluy daueI). 70
KuDirational (variety). 30
torUl, 6
Mestre'. theorem, gg
t.ype (el), 18
Janko &!,ou p Jlt 77
Krat.ional (cODjulacy cl ..) , 85
IUle aiew iDequalit.y. 103 Milt"tt '. t.heorem, IiU'
Monlte lfouP. 78
MotM C1llIctiOD. 3IiI nonnalilcrt 1 1
SpriDler'. iheorem, 88
(of a quadratic f'ann ) , e6
thin (Iublet of' V(K», 1 8 trace Conn (of' aD etale allebra), 88
type
(C, ).
18
Yena.I (eoverinl map). ziv weak approximation property, 29 weak weak approzimatioD property. 28
117