A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
D AN H ARAN
M OSHE J ARDEN
The absolute Galois group of a pseudo real closed field
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 12, n
o3 (1985), p. 449-489
<http://www.numdam.org/item?id=ASNSP_1985_4_12_3_449_0>
© Scuola Normale Superiore, Pisa, 1985, tous droits réservés.
L’accès aux archives de la revue « Annali della Scuola Normale Superiore di Pisa, Classe di Scienze » (http://www.sns.it/it/edizioni/riviste/annaliscienze/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/conditions). Toute utilisa- tion commerciale ou impression systématique est constitutive d’une infraction pénale.
Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
The Group
DAN
HARAN
(*) -
MOSHE JARDEN(**)
Introduction.
The main
problem
in Galoistheory
is to describe the absolute Galois groupG(K)
of a field K. The mostinteresting
case,namely
that of the field of rationalnumbers,
is still very far frombeing
accessible.Nevertheless,
a few other
interesting
cases have been resolved.Among
them there are the finitefields, with 2
as the absolute Galois groups, real closed fields 1~ withG(R) ~ Z/2Z,
thep-adic
fieldsQ,,
with adescription
ofG(%) by generators
and relations
(Jakovlev [12], Jansen-Winberg [13]
andWinberg [25])
andthe field
C(t)
withG(C(t)) being
free.Finally
we mention PAC fields withprojective
groups as their absolute Galois groups. The lastexample
moti-vates the
present work,
we thereforeexplain
it in more detail.Recall that a field K is said to be
pseudo algebraically
closed(PAC)
ifevery
absolutely
irreduciblevariety
defined over .K has a K-rationalpoint.
On the other
hand,
aprofinite
group G is said to beprojective
if every finiteembedding problem
for G issolvable;
in otherwords, given
adiagram
where oc is an
epimorphism
of finite groupsand q a homomorphism,
thereexists a
homomorphism y : G
-B such that aoy= cpo (Gruenberg [9]).
(*)
This work constitutes a part of the Ph. D. dissertation of the first author done at Tel-AvivUniversity
under thesupervision
of the second author.(**) Partially supported by
the Israel NationalAcademy
of Sciences.Pervenuto alla Redazione il 23
Agosto
1984.It is now well known that if .K is a PAC
field,
then isprojective
(Ax [1,
p.269]). Conversely,
if G is aprojective
group, then there exists aPAC field g such that
G(K) ~ G (Lubotzky-v.d.
Dries[20,
p.44]).
An
attempt
to enrich the structure of the PAC fields has led to the de- finition of PRC fields:A field g is said to be
pseudo
real if everyabsolutely
irre-ducible
variety
V defined overK,
which has a-rational point
in every real closed fieldI~ containing K,
has a K-rationalpoint (Prestel [22]).
In
particular,
if .K has noorderings,
then K is a PAC field. Thestudy
ofPRC fields has
already
attracted a lot of attention([16], [17],
Prestel[22],
Ershov
[7]
andothers).
In order to extend theseinvestigations
it has be-come necessary to
give
a group theoretic characterization of the absolute Galois group of a PRCfield,
in otherwords,
one has to find the «right »
definition for real
projective
group ». Here is oursuggestion :
We considerthe
embedding problem (1)
and call it real if for every involution g E G such that99(g)
=A 1 there exists an involution b E B such thatoe(b)
=cp(g).
A pro- finite group G is said to be realprojective
if the subsetI(G)
of all involutions of G is closed and for every finite realembedding problem (1)
there existsa
homomorphism y : G - B
such that aoy = 99. We prove:THEOREM.
If K is a
PRCfield,
thenG(K)
is realprojective. Conversely, if
G is a realprojective
group, then’there
exists a PRCfield
K such thatG.
Unfortunately
we have to go along
way in order to prove the Theorem.Nevertheless there is a bonus for the
effort, namely
the introduction of Artin-Schreier structures. In the same way that PRC fieldsgeneralize
PACfields,
Artin-Schreier structures enrich Galois groupsby taking
into account theorderings. Indeed,
to every Galois extension.L/.K
withV -
1 E L weattach the space
of orderings X(LIK), consisting
of allpairs (L(8), P)
where cis an involution of
L(e)
is its fixed field in Land P is anordering
of
L(B).
Thecorresponding
Artin-Schreier structure is =In
particular
the absolute Artin-Schreier structure of .K is
C~ (.g)
=In the
category
of Artin-Schreier structures there areprojective objects (Section 7),
theunderlying
groups of which areexactly
the realprojective
groups
(Proposition 7.7).
If K is a PRCfield,
then(b(K)
is aprojective
Artin-Schreier structure
(Theorem 10.1). Conversely,
for every Artin-Schreier structure 0152 there exists a PRC field .K such that 0152(Theorem 10.2).
This
completes
the mainresult,
Theorem10.4,
mentioned above.Notation.
X(K)
= the set oforderings
of a field K.Ks
= theseparable
closure of a field K.If is a Galois
extension, F
is an extension of Land J is an auto-morphism
of F’ overL,
thena(x)
=x}
is the fixed field of J in L.1. - Profinite
topological
transformation groups.Sets of
orderings
offields, profinite
groups, etc. areprojective
limits offinite sets. The next Definition-Theorem characterizes these
objects
astopological
spaces.DEFINITION 1.1. A
topological
space X is said to be a Boolean space,if
itsatisfies
oneof
thefollowing equivalent
eondit2ons :(i)
X is atotally
disconnectedcompact Hausdorff
space.(ii)
X iscompact
and every x E X has a basisof closed-open neighborur- hoods,
whose intersection is(iii)
X is an inverse limitof finite
discrete spaces.(iv)
X ishomeomorphic
to a closed subsetof ~-1, for
some set I.The conditions are indeed
equivalent:
(ii)
=>(ii):
Hewitt and Ross[11,
p.12].
(ii)
~(iiii): Clearly X
is Hausdorff. Since therequired proof
is aspecial
case of a
part
of theproof
ofProposition 1.5,
we shall notbring
it here.(iii) (iv):
Assume X = limXj,
withXj
finite. Then X is a closedEJ
subset of
Also,
we may assume that{::l:
for some finiteJ6J
set Then X is closed
in {± 1~ I,
where I is thedisjoint
union of the setsh .
(iv)
~(i):
Clear.f f
Examples
of Boolean spaces :profinite
groups, sets oforderings
of fields(Prestel[21, § 6]).
Throughout
this paper wetacitly
use the fact that a continuous map betweencompact
Hausdorff spaces isclosed;
inparticular,
a continuousbijection
is ahomeomorphism.
LEMMA 1.2.
Let p : X -
Y be a continuous closed and open mapfrom
aBoolean space X onto a
topological
space Y. Then Y is also a Boolean space.PROOF. It suffices to show that Y is
Hausdorff,
since theimage
of acompact
set iscompact,
henceby
1.1(i),
Y is also a Boolean space. Thusour Lemma
follows,
e.g.,by [6,
Ch.2, § 4,
Theorem 4 and Theorem5]. j/
Let us consider the
category
of(topological) transformation
groups, i.e.pairs (X, G),
where X is atopological
space and G is atopological
groupacting continuously
on X(the
action X X G -~ X denoted henceforthby (x, a) (cf.
Bredon[2, Chapter 1]).
Amorphism
in thiscategory,
say(Y, H) - (X, G),
is apair ( f, p) consisting
of a continuous mapf :
Y -+ Xand a continuous
homomorphism
p:H - G,
such thatIf
f ( Y)
= X andq;(H)
=G,
we call[f, p)
enepimorphism.
A transformation group
(X, G)
is calledfinite,
if both X and G are finiteand discrete. A transformation group is
profinite,
it is an inverse limit of finite transformation groups.Our first aim is to characterize the
profinite
transformation groups.Let
(X, G)
be a transformation group. Apartition
Y = ... ,of X is a finite collection of
disjoint non-empty closed-open
subsets ofX,
such tha We say that Y is a
G-partition,
if for every a E G and every there is aI jn
such thatV’=
For two
partitions Y,
Y’ of X we writeY> Y,
if Y’ isfiner
thanY, i.e.,
for every V’ E Y’ there is(a unique)
V E Y such that V’ C V. Thefamily
of
partitions (resp. G-partitions)
is thuspartially
ordered.REMARK 1.3. Let Y be a finite collection of
closed-open
subsets of X.Then there is a
partition
Y’ of X such that for every V’ E Y’ and V E Y either V c V or Y’ r1 V =0.
In
particular,
every twopartitions
of .X have a common refinement.If Y is a
G-partition
ofX,
there is an obvious way to consider(Y, G)
as a transformation group
and, furthermore,
to define anempimorphism (py, idG) : (X, G)
?(Y, G) (i.e., by py(z)
= VV).
IfY’ ~
Y is anotherG-partition
ofX,
there is an obviousepimorphism (py,y, idG): (Y’, G)
--~ ( Y, G),
such thatLEMMA 1.4. Let
(X, G)
be atransformation
group. ,Assume that X is aBoolean space and G is a
profinite
group, and let Y be apartition of
X.(i)
There exists an open normalsubgroup
Nof
G such that Va = Vfor
every V E Y and every a E N.(ii)
There exists aG-partition
Y’of
Xf-iner
than Y.PROOF.
(i)
It suffices to find for aclosed-open
subset TT of X an open normalsubgroup
N of G such that Va = Y for every 6 E N. Let x E V.By
thecontinuity
of the action X X G -~ X there is aclosed-open neigh-
bourhood
Ux
of x and an open normalsubgroup Nx
of G such that v for every J EN,,.
Since V iscompact,
there are xl, ..., xk E V such thatI 1ae
Put then N has the
required property.
’
(ii)
Assume Y =~Y1, ..., i V-1
and choose N which satisfies{i).
Ifd - u’
(mod N),
thenV~
=Vf
for every a, a’ E G and1 c 2 c n.
Let ..., an berepresentatives
ofG/N.
For every function cx:{ly...y~} -~{1,...~}
,denote
vall
r1V:(m).
It iseasily
checked that Y==0}
is a
G-partition,
finer than Y.//
PROPOSITION 1.5. A
transformation
group(X, G)
isprofinite if
andonly if
Boolean space and G is aprofinite
group.PROOF. The
necessity
is obvious. To show thesufhciency,
assume that Xis a Boolean space and G is a
profinite
group. Let (T be thefamily
ofG-partitions
of X. The maps definean enimorpbism (p, ida) : (X, G) (Ribes [23,
Lemma2.5]).
But p is alsoinjective:
ifYEJ
x, XI E X are
distinct,
y there is aclosed-open
set U CX,
such that xE U, x’ 0 U; by
Lemma 1.4(ii)
there exists aG-partition
Y of X finer than{U,
X -U}.
Thus whencep(x) =,4 p(x).
Therefore(X, G)
r-v lim
( Y, G) (i) ,
since both X andlim
Y are Hausdorff andcompact
spacesY J
Thus we may assume that X is finite.
By
thecontinuity
of the action X X G ~X,
there is an opensubgroup No
of G such that for every x E X and Let X be the
family
of open normal
subgroups
of G contained inNo. Then, clearly, (X, G) /j
As an
application
of the material accumulated in this Section we con-struct a
quotient
of aprofinite
transformation group.(1)
If G = 1, thispart
proves(ii)
~(iii)
in Definition 1.1.Let
(X, G)
be aprofinite
transformation group, and let N be a closed normalsubgroup
of G. Define anequivalence
relation - on Xby :
x, if there is a orE N,
suchthat x’
= x2, and let be thequotient
space.The
quotient
map p :X - X/N
is open(if
is open, then= U
" and closed(if
’ .F’ isclosed,
then= U
ac-X.F’6 is theimage
of the
compact
set .F’ X N under the actionX X G --~ X,
hencecompact).
By
Lemma1.2,
is a Boolean space.G - G/N
be the naturalepimorphism.
It iseasily
verified that the action of G on X induces a continuous action ofG/N
onfor x E X and
ac-G).
Thus we have shown:
CLAIM 1.6.
(XIN, GIN)
is aprofinite
transformation group and(p, n):
(X, G) GIN)
is anepimorphism. Moreover,
p : is anopen map.
2. - The space of
orderings
of a Galois extension.Every
Galois extensionL/K
isnaturally accompanied by
its Galois group Another natural structure associated with.L/.K
is the set.X(L/.K)
of the maximal ordered subfields of L
containing
K. In this section we in-vestigate
this set and its relations to To ensure agood
behaviourwe assume that
V -
1 E L. It turns out thatX(LIK)
is a Boolean space andf3(L/K)
acts on it. To attain fullgenerality
we do notrequire
that Kbe of characteristic zero and
formally
real.Nevertheless,
theinteresting
case arises when K can be ordered.
We
begin by summing
up some relevant facts from the Artin-Schreiertheory.
Recall that an orderedfield
is apair (K, P),
where .K is a field andthe
ordering,
satisfies andPROPOSITION 2.1. Let be a Galois extension such that
(i)
Let 6 EG(K)
= be an involution(i.e., ~2 =1, ~ ~ 1) .
Then
.g$(~)
is realclosed,
hence has aunique ordering.
(ii)
Let P be anordering of K,
and let(L’, Q)
be a maximal ordered ex-tension
of (K, P)
such that .L’ ~ L. Then there ex2sts an involution e Esuch that .L’ _
L(e).
( iii )
Let P be anordering of
K and let(L ( el) , Q,,)
and(L ( E2 ) , Q2)
be two amaximal ordered extension
of (K, P)
contained in L. Then there exists aunique
such that in
particular
PROOF.
(i)
This follows from the fact that..gs(~)]
= 2by Lang [18,
Cor. 2 on p. 223 andProp.
3 on p.274].
(ii)
There exists([18,
Theorem 1 on p.274])
and involution EG(L’)
such that
Q
extends to the real closed fieldK,,(b).
Let 8 =ResL 6;
then~~2 = 1.
By
themaximality
ofL’,
Z’ _ r1 L _-.-L(e).
(iii) By (ii),
there are involutions6111 ð2
EG(K)
such that(.L(Ei), Qi)
c where
Qi
is the(unique) ordering
of i =1, 2.
ThusResL ð1 = ResL ð2
=E_2. By [18,
Theorem 3 on p.277]
there is aunique K-isomorphism (.Ks(~1), Q.)
-(~s(~2), (2).
Its restriction toL(81)’ ii: (L(81)’
~
~L(E~), Q2~,
is aunique K-isomorphism
between(L(81)’ QI)
andby
Prestel[21,
p.42].
N( , hence a canbe extended to a
unique
elementLet
LIK
be a Galoisextension, Ý - 1
E L. An involution e Eg(LIK)
is
real,
ifL(e)
is aformally
real field. The setX(LIK)
of the maximal ordered fields in Lcontaining
g is called the spaceof orderings of LIK. Proposition
2.1implies
that these fields are of the form(L(e), Q),
where 8 E is a realinvolution. The map d: defined
by d(L(e), Q) =
s, iscalled the
forgetful
map.If is another Galois
extension,
such thatL,
thenthe restriction map Res:
X (Lo j.g), given by (L(E), Q) - (Lo(e),
Q
r1Lo(e)),
issurjective, by
Zorn’s Lemma. Note that theforgetful
mapcommutes with the restriction of the spaces of
orderings
and the restriction of the Galois groups.Consider the Harrison
topology
on defined via the subbaseE
L*},
where =~(.L(E), Q) I
a EQ}.
The sets are~ closed-open. Indeed,
let finite Galois extension such that a,Ý- 1
ELo C L, and pick
upbl, ..., bn
ELo
such that.g(bl),
...,K(bn)
are allmaximal
formally
real extensions of .K in.Lo
which do not contain a. Thenclearly
whence is closed.
From
this one may prove as an exercise that ifLIK
isfinite,
thenX(LfK)
is a Boolean space
(see
Prestel[21,
Theorem6.5]
for a similarproof).
Foran
arbitrary
Galois extensionL jg
such thatV - 1
E L it may be verified thatX(L/K)
=lim X(Li/g),
where{Li:
i EI}
is thefamily
of finite GaloistëÏ
_extensions of K contained in L and
containing 1/-
1. ThusX(LfK)
is aBoolean space.
The restriction map Res: defined above and the
forgetful
map d : ~ areclearly
continuous in the Harrisontopology.
The group acts on in an obvious
(and continuous)
way.By Proposition
2.1(iiii)
we have thatfor every
Finally
note that , the space oforderings
of K(see [21,
p.88]).
3. - Artin-Schreier structures.
The discussion in Section 2 motivates
(see Example
3.2below)
the fol-lo-Nving
abstract definition.DEFINITION 3.1. An Artin-Schreier structure G is a
system
where
(i) (X, G)
is aprofinite topological transformation
group(the
caseX = 99 is not
(ii)
G’ is an opensubgroup of
Gof
index2 ;
(iii)
d is a continuous map such thatd(x)
is an involution inG, d(x) 0 G’,
and
(d(x))6 for
every x c- Xand GEG;
andIf a
system
Qj satisfiesonly (i)-(iii),
we call it a weakArtin-Schreier
structure.
The Boolean space ~’ is called the space
of orderings of G;
the map d iscalled the
forgetful
map ; itsimage d(X)
is called the setof
real involutions.Note that
(iv)
isequivalent
to the conditionAlso note that G = G’
implies
~~i’ = ~.EXAMPLE 3.2. If
L/K
is a Galois extension andL,
thenis, according
to Section2,
an Artin-Schreier structure.Let be the set of real involutions in and let i : be the inclusion. Then
is a weak Artin-Schreier structure.
If not
explicitely
statedotherwise,
theunderlying
group, theunderlying subgroup,
the space oforderings
and theforgetful
map of an Artin-Schreier structure @ will be henceforth denotedby G, G’, X(()
andd, respectively.
Analogously
forSJ, 9t, 113,
etc.DEFINITION 3.3. Let
§,
(S be(weak)
Artin-Schreier structures. Amorphism of (weak)
Artin-Schreier structures~: ~ --~ C~
is apair of
continuous maps(both
denotedby
abuseof
notationby q~)
such that(i)
-for
every x Ex(%)
i(ii) (gy, gy): (X { ~ ), ~) --~ (X ( Ci ), G) is a morphism of profinite trans f or-
mation groups,
i.e.,
=for
all x EA
morphism
99: C~ is called anepimorphism if
= G and(hence
alsocp(.8’’)
=G’).
An
epimorphism of
Aartin-Schreier structures g~ : Ci is said to be acover,
if
(iv) for
all xl, x2 E such that = there exists a 6 E G such that0153~
= x2 .(Then a
can be chosen to be an elementof
KerNote that if
(i) holds,
then(ii)
isequivalent
to(ii’ ) q;(x1:)
=g~(x)z~~~
for all x E and í E H’.Indeed,
if x E and cr E H -H’,
then there is a í E .g’ such thata =
d(x)í,
since(H: H’)
~ 2 and ButXà(x)==
x and=
q;(x),
hencegg(xl)
=q;(x1:)
=gg(x)9’(’r)
= + ==Also observe that
(iii)
isequivalent
toin
particular
we have Ker99 c
H’.Finally let 99: I~ -> 0
be amorphism
of Artin-Schreier structures. Then the map spaces oforderings ~: x(~) -~ X‘(C~)
induces a continuous mapNote
that qJ
is a cover if andonly
if(iv’) §5
is abijection, i.e.,
ahomeomorphism,
andqJ(H)
= G.EXAMPKE 3.4. Let
Zo
C L be two Galois extensions of .K such that Then the restriction map Res:C~(L/I~) --~
is a cover(see Prop.
2.1(iii)).
(b)
The restriction map of thecorresponding
weak Artin-Achreier structures(Example 3.2)
is anepimorphism,
but need notsatisfy
con-dition
(iv)
of Definition 3.1.Indeed,
there may exist two real involutions such that =ResLo 8’,
but noordering
ofLo(E)
extendsboth to and
L(e’).
Thus 8 and e’ are notconjugate.
(c)
Let t be transcendental overQ.
Then the map Res : ~I -
is an
epimorphism
of Artin-Schreier structures but not a cover.Examples
3.4(a)
and(c)
may begeneralized
as follows:LEMMA 3.5. Let and
FIE
be Galois extensions such that K CE, V-1
E L C F. Then the restrict ~on map Res : -QJ(LIK)
is amorphism of
Artin-Schreier structures. It is anepimorphism if
andonly if
is atotally
realextension, linearly disjoint from LIK. Here ElK
is said to betotally
realif
everyordering of
K extends to anordering of
E.PROOF.
By Example
3.4(a)
we may assume that F = LE. The Lemma follows from v.d. Dries[4, Chapter II,
Lemma2.5]. [1
4. - More about Artin-Schreier structures.
In this Section we
develop
comeconcepts
andproperties
of(weak)
Artin-Schreier structures needed later on.First a few remarks :
4.1. Let Ci be a
(weak)
Artin-Schreier structure and let N C G’ be a closed normalsubgroup
of G. Definewhere
GIN)
is thequotient profinite
transformation group(see 1.6)
and d
is the map inducedby
d :X(@) -
G.We leave to the reader the
straightforward
check that@jN
is a weakArtin-Schreier structure and that if @ is an Artin-Schreier
structure,
thenso is In the latter case the
quotient
maps and G - define a cover.Moreover,
every cover may be obtained this way.4.2. An inverse limit of
(weak)
Artin-Schreier structures is a(weak)
Artin-Schreier structure.
4.3. Let @ be a weak Artin-Schreier structure. Then C~ =
lim
where N runs
through
thefamily
of open normalsubgroups
of G con-tained in G’.
In
particular,
ifL/.K
is a Galois extension andÝ -
1E L,
thenlim (M(LijK),
where thefamily
of finite Galois extensions of .giEI
__containing vi - 1
and contained in L.LEMMA 4.4.
Every (weak)
Artin-Schreier structure 0152 is an inverse limitof finite (weak)
Artin-Schreierstructures,
which areepimorphic images of
C3.PROOF.
By
4.3 we may assume that the group G is finite. Let S be thefamily
ofG-partitions
Y of which(i)
are finer than~d-1(~) :
8 E 9i.e.,
a mapd,:
Y - G maybe defined
by
=d(x)
for all x E U with U EY;
(ii)
UZ n U = p for all U E Y and E G’ -~1~,
9 if @ is an Artin-Schreier structure.
Every
Ye S defines a finite(weak)
Artin-Schreier structure _~G, G’, Y~ G~.
Now to show that weproceed exactly
as in thefirst
part
of theproof
ofProp. 1.5,
but instead ofusing
Lemma 1.4(ii)
weapply
thefollowing
CLAIM. Let Y’ be a
partition
of X =X(C~).
Then there is finer than Y’.PROOF OF THE CLAIM. If G is a weak Artin-Schreier
structure,
thisfollows from Lemma 1.4
(ii). Assume, therefore,
that C~ is an Artin-Schreier structure. Let x E X and let V E Y’ such that x E V. There is aclosed-open neighbourhood Ux
of x suchthat xIr ft Ux
for all i E G’ -{1} .
We may as-sume that V r1
and Ux
nUx == 0
for all i E G’ -{1} (other-
wise take instead of Since X is
’
com-
pact, finitely
many of theseneighbourhoods
cover X.By
remark 1.3 there is apartition Yo
of X such that every VeYo
is contained inZTx
for somehence
Yo
satisfies(i)
and(ii)
above. Our claim therefore followsby
Lemma 1.4
(ii). //
Finite weak Artin-Scllreier structures appear
naturally, but undesirably
in the course of
proofs
in Section 7. Nevertheless we show in the next Lemma that such a structure 9f is anepimorphic image
of a minimal Artin-Schreier structure~,
whicheventually replaces %
in the above mentionedproofs.
LEMMA 4.5. be a
finite
weak Artin-Schreier structure. Then there exists afinite
Artin-Schreier and anepimorphism --> 9f,
suchthat
for
every Artin-Schreier structure 113 andfor
every(epi-)morph2sm of
weakArtin-Schreier structures there exists an
(epi-)morphism
o2:0 -->ff
such that
po&
= a.PROOF. Let xl, ... , xn be
representatives
of all the A-orbits inand denote s2 =
d(xi) (recall
thatx2 ~ =
=1,
..., n. Let Z be the set of formalexpressions ZZ,
where1 : i
n, The group A acts on each of the subsetsZi
=by (el)’
= and = for T, a EA’, (recall
that whence A acts onZ == Z, U- ... W Z,,.
Themap defined
by p(z2)
=x’
iscompatible
with the action of A It iseasily
verifiedA, A’, Z *l A~
is an Artin-Schreier struc-ture,
and ptogether
withidA
define anepimorphism
-?-~.Now let 0 be an Artin-Schreier structure and a: ~ a
morphism
We may assume that 0 is
finite,
otherwisereplace 39 by
a suitableepimorphic image, using
Lemma 4.4. For every let be a maxi- mal subset of such that yil, ...,Yini represent
distinct orbits inX(Q3) .
Then
X ( ~ ) _ ~y ~ ~ 1 c j c na , l ~ z ~ n, z~ E B’} .
Define&(yij) z’(’r).
Thena
together
with « : B -?- A is the desiredmorphism 9~ 2013~. Moreover,
,x(X(93))
= if andonly
if&(X(58) = //
Recall that if ni :
Yi -->- Y, i
=1, 2,
are two continuous maps oftopo- logical
spaces, we denoteby Yi
xYY2
the(closed) subspace
ofYI
XY2 consisting
ofpairs (yl, Y2)
such that7£1(YI)
=~2(y2)·
Let ni :
9f’
and ~2 :~2
be twomorphisms
of(weak)
Artin-Schreier structures. Define
and let
B1 XAB2
act oncomponentwise.
It is an in-structive exercise to check that the
fibred product ~¡ XellJz
is a(weak)
Artin-Schreier
structure,
and that the coordinateprojections
pa :ZI Xgll3z3
- =
1, 2,
aremorphism (cf.
also Bredon[2, Chapter 1, 6 (B)]).
To use fibred
products
we need thefollowing
characterization:LEMMA 4.6.: Consider a commutative
diagram of (weak)
Artin-Schreier structuresThe
following
statements areequivalent :
(a)
~ isisomorphic
to thefibred product ~l i.e.,
there is an iso-morphism
0: Z -*~l
such thatpio0-1
andP2°f)-1
are the coordinateprojections.
(b)
~ with pl, P2 is apullback o f
thepair (nl, n2), i.e., given
a weak Artin-Schreier structure (9 with
morphisms
0152 -+~,~
and 1Jl2: 0152 -+~2
such thatthere is a
unique
such that = 1f1 and =:: "P2.(e)
1.If
0 is aprofinite
group and and 1Jl2:0 -+B2
arecontinuous
homomorphisms,
then there exists aunique
continuous homomor-phism
1Jl: 0 -+ B such that = and = y~2 .2.
If
X is atopological
space andX(Zl)
and 1jJ2: ~ ~-X(b )
are continuous maps, there exists a
unique
continuous map ’ljJ: X ->such that = ~1 and = "P2.
(d)
1.If b2EB2
andnl(bl) _ ~2(b2),
then there is aunique
b E .B’such that
PI(b)
=bl, p2( b )
=b2 (if
PI and P2 aresurjective
this isequivalent
to Ker ; and :
2.
If
0153l E x2 E and =~~(x2),
then there is aunique
x E such thatpl(x)
= 0153l,p2 (x)
= x2 .PROOF. An
analogue
of[10,
Lemma1.1].
See also Bredon[2, Chap-
ter
I,
6(B)]. j/
We call a
diagram (1)
a cartesian square, if it satisfies one of theequi-
valent conditions of Lemma 4.6.
The
following
Lemmagives
a very usefulexample
of a cartesian square.LEMMA 4.7. Let px : Z -->
B1
be anepimorphism of
Artin-Sohreier struc-tures,
K C: B’ a closed normalsubgroup of
B such that .~ r1 Ker(Pl)
= 1.Let and nl: be the
quotient
maps. Then there exists aunique epimorphism
n2:ZIK
-~ such thatcommutes. Moreover, (2 )
is a cartesian square.PROOF. The map ~2 is defined
by
the universalproperty
of thequotient .
To show that(2)
is a cartesian square we have toverify
conditions 1.(which
istrivial)
and 2. of Lemma 4.6(d).
Indeed,
let 0153l E and x2 E with =~z2(x2).
Then there exists an x E such thatp2(x)
= x2. We have = since(2) commutes,
hence there exists such thatpx(xa)
xlFinally,
y the elementX(Z)
satisfies alsoP2(Xa)
=p2(x)
= x2 .If an element also satisfies = xs =
pz(x),
for i =1,
2then there is a T e K such xz. Therefore = I hence
pl(z)
= 1. Thisimplies T
=l,
since E r) Ker(PI)
=1;
hencex’ = x. .
//
.
5. - On PRCe fields.
A
consisting
of a field E and eorderings Q,, ..., Qe
of E is called an e-orderedfield.
If E is PRC andQl,
...,Qe
are allits distinct
orderings,
then 8 is said to be a PRCefield.
Anequivalent ([14,
Lemmas 2.2 and
2.3]
and Prestel[22,
Theorems2.1,1.2
andProposition 1.6])
definition is the
following:
An e-ordered is
PRCe,
if itsatisfies;
(i)
Let be anabsolutely
irreduciblepolynomial,
let such that
/(a,, 7 X) changes sign
on E withrespect
to each of theQ’,,s,
and letUi
be aQi-neighbourhood
of ao for i =1, ... ,
e. Then there exists an(a, b)
E Er+l such that a E~i
n ...r1 Ue
andf (a, b)
= o.(ii)
Theorderings 9i?’"?9e
induce distincttopologies
on E.Let K be a countable Hilbertian field and let 3~ _
(K, P1,
... ,Pe)
beand e-ordered
field,
fixed for this Section. Forintegers
0 e m we denoteby
the freeproduct (in
thecategory
ofprofinite groups)
of ecopies
ofZ/2Z
and m - ecopies
of2. Generalizing
results of[16]
and ofGeyer [8]
we show that there is an abundance of PRCe fields 6 that extend X such that E is
algebraic
over .K andDe,m.
To do
this,
fix involutions~l, ... , ~e E G(g)
such that the real closed fieldsgi
=.Ks(~i),
i =1,
..., e, inducePl, ... , P,
on.K, respectively.
For(~1, ... ,
EG(.K)m
letand denote
by .... , .PQe
theorderings
of inducedby~ . respectively.
Then3~= (JS~ PQl , ... , Pae)
extends X andLEMMA 5.1
(cf. [16,
Lemma6.4]): finite
extension
of X.
Letf
EL[TI’
...,Tf, X]
be anabsolutely
irreduciblepolynomial
and let
Suppose
that there exists an such thatf ( ao , X) changes sign
on L withrespect
to eachof
theQ i’s.
LetUi
s be aQi- neighbourhood of
ao in Lr. Thenfor
almost all or EG(K)m for Xa
there exists an
(a, b)
E such that a EU,
r1 ... r1Ue , f ( a, b)
= 0 andg(a) =1=
0.PROOF. Let and let
Li
be a real closure of L that inducesQ~ ;
then there exists a T, E
G(K)
such thatL
= If (1i EG(K)
is an ad-ditional element such that
K~’
inducesQi
onL,
then there exists a  eG(L)
such that
_ .KZ;~ 2
yI ,e . ,
=Li.
· Thus i EG(Li),
sinceLi
has no
L-automorphisms
besides theidentity ([21,
Cor.3.11]),
hence(1i E
i;G(L) . Conversely Kri).
induces theordering Qi
on L forevery A
EG(L).
Put ~’e+1= ... = and i =
(i1’ ..., im);
it follows that is the set of allm-tuples
(1 inG(K)m
for which £ CK,,,.
Withoug
loss ofgenerality
we may assume thatf (a, X) changes sign
on Lwith
respect
toQi
for everya E Ui
fori = 1,..., e. By
Lemma 8.4 ofGeyer [8],
and since L isHilbertian,
the set .g nU,
n ...rl Ue
is notempty
for every Hilbertian set H in
Using
the fact thatf
isabsolutely irreducible,
one can find a sequence a,,., a2, ... of elements in
.Lr,
and a sequencebi, b2, ...
of elements in
Ls
such that:a) Uln
...r1 Ue
andX)
is an irreduciblepolynomial
over Lof
degree n
=degx f
andchanges sign
on L withrespect
toQi
for every andevery j ;
b) b;)
= 0 andg(aj)
0 0 forevery 1;
c) denoting Lj==L(bj),
we have thatL1,L2,...
is alinearly disjoint
sequence of extensions of L of
degree n (cf.
theproof
of Lemma 2.2 of[14]).
Condition
a) implies
that each of theQi’s
can be extended to anordering Q¡,;
ofLj. Let Lj
=(Lj, Qi; , ... ,
As in the firstparagraph
of thisproof
there is a
T(j) E G(K)m
such that is the set of allm-tuples
or infor which then
(aj, bj)
hence o-hasthe
required property.
Thus it suffices to show that7:G(L)m- U
is a zero
set,
or,equivalently,
that i2
is a zero set.
Now observe that C 9
Cj
for everyj,
henceiG(L)m;
inparti- cular,
EG(.L)m.
Hence our result followsby
Lemma 6.3 of[16]. //
COROLLARY 5.2. Almost all a E
G(K)m
have thefollowing property : If f E
...,T,, X]
is anabsolutely
irreduciblepolynomial for
which there exists an 00 Esuch
thatf (ao, X) changes sign
onKa
withrespect
to eachof
the
porderings i f Ui
is aPac-neighbourhood of for
i =1,
..., e andif 0 ~ g
E ...,T,],
then there exists an(0, b)
EK*" 1
such that r1 ... nf(a, b)
= 0 andg(a) 00.
PROOF. Use the
countability
of .K and the fact that an intersection ofcountably
many sets of measure 1 has also measure 1. Alsoobserve,
thatif
f, Ul,
...,Ue
are asabove,
there exists a finite extension L ofK,
overwhich
they
are defined.Compare
theproof
of Theorem2, 5
of[14]. //
LEMMA 5.3. The
orderings Pal’...’ P ae
induce distincttopologies
on K-for
almost all a EG(K)m.
PROOF. It suffices to prove that for every 1 1~ l e, every finite ex- tension £ of
J{"
every6,
E L such that 0k ~l k 1,
01 61 1
1 and foralmost all a E
G(.K)m
such that £. 9Xa
there exists a b EKa
such that :With no loss let k =
1, Z
= 2.Let
f (T, X)
= X2 -T, g(T) == 1, ao
=1, Ui
=(1 - ~/3,
1+ ~J3),
fori =
1, 2, (0, 2),
f or 2 =3,...,
e. With these data go overthrough
theproof
of Lemma 5.1 and note(at
the instance ofchoosing Qii)
that each of theQi’s
can be extended inexactly
two ways toLj
= =Assume, therefore,
that we have chosenQlj, Q2j
such thatbj >10, bj
2 0.Then
b, clearly
satisfies(1 ), since bi
=U1 r1 U2, by
the construction.Now continue in the
proof
of Lemma 5.1 andget
therequired
result./ /
-
An
m-tuple
7 a. of elements ofDe,m
is called abasis,
if, and I
By
itsdefinition, ÎJe,m
has a basis with thefollowing
exten-sion
property :
if G is aprofinite
group,ai,
...,am
eG, and #(
= ... =ae
=1,
then the map =
1~
..., m, can be extended to ahomomorphism
G. In
particular,
ifo~ ... (1~
is another basis ofÎJe,m,
then the mapcan be extended to an
epimorphism By [23,
Cor.7.7]
this is an
isomorphism.
Therefore every basis ofDe,m
has the extensionproperty.
The
following
Lemmagives
a useful characterization ofLEMMA 5.4. Let G be a
profinite
groupgenerated by m elements, e of
whichare involutions. Then G is
isomorphic to
and everyfinite
groupgenerated by m elements, e of
which areinvolutions, is a homomorphic image of
G.PROOF. The
isomorphism
class of afinitely generated
group is deter- minedby
its finitehomomorphic images. Moreover
y ifG = ~arl, ... , 6~~
is a finite group such
that aî
= ...a;
=1
one mayeasily
construct another finite groupG’ = (J§ , ... , J£)
such thatu§ , ... , J§
areinvolutions,
and Gis a
homomorphic image
of G’.//
LEMMA 5.5. For almost all
for
is a basis
PROOF.
By
thepreceding
Lemma it suffices to show that ifG= 1, ... , mj
is a finite group and are
involutions,
then G is a homomorhicimage.
of
(ð~l,
..., ... ,o~m~,
for almost all 1.With no loss
Also,
we may assume that G is of even order(other-
wise e =
0, and, moreover, G
may bereplaced by
the group G which is alsogenerated by
mgenerators),
say~G)
= 2n.Let
1 : i 2.
Thepolynomial go(X )
=(X2 -~-12)
...(X2 + n2)
has no rootsin the real closed field
By Sturm,s
Theorem(cf. [8,
Lemma8.2]),
if is close
enough
to go, withrespect
toPi,
then g also has no roots in Thereforeby
Lemma 8.4 of[8]
we can construct a sequence ofpolynomials
gl, g2, ... in thatsatisfy:
a) deg g;
=2n,
andg(g,, K),
as the group ofpermutations
on the rootsof gj, is the full
symmetric
groupS2n;
b)
gj is closeenough to go
withrepsect
toPi , ... , Pe,
inparticular gj
has no roots in
K~(61),
... ,KS( ~e) ;
c) denoting by .L~
thesplitting
fieldof gj
over K we have thatL1, L2,
...is a
linearly disjoint
sequence of extensions of K ofdegree (2m) ! :
and denote