A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
P IERRE D ÈBES
Galois covers with prescribed fibers : the Beckmann-Black problem
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 28, n
o2 (1999), p. 273-286
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Galois Covers with Prescribed Fibers:
the Beckmann-Black
ProblemPIERRE
DÈBES
(1999), pp. 273-286
Abstract. The Beckmann-Black problem asks whether every Galois extension
E/K is the specialization of a Galois branched cover of I~1 defined over K with the same Galois group. E. Black conjectures this is indeed always possible. We give three results about this conjecture. The first one is that it implies the Regular
Inverse Galois Problem. The second one considers a "mere" form of the problem
which requires that the realizing Galois cover be defined over K only as mere
cover (i.e., without the Galois action). This mere form is shown to hold if K contains an ample field, e.g. K contains a complete valued field. Our last result is a proof of the original Beckmann-Black condition in the case K is a PAC field.
Mathematics Subject Classification (1991): 12F12 (primary), 12E25, 14H30, 11 Gxx (secondary).
Introduction
This paper is devoted to the
problem
ofrealizing
groups as Galois groupsover
K(T)
with the extra constraint that thespecialization
at somepoint
isgiven
in advance. Atypical
result is thefollowing.
THEOREM
(Theorem 3.1).
Let K beafield containing
acomplete valued field K ((x))).
Thengiven a finite
group G and a Galois extensionE / K of
group G, there exists anabsolutely
irreduciblepolynomial P (T, Y)
EK [T, Y]
such that
(i) Y)) is a
Galois extensionof K(T) of
group G, and(ii)
thesplitting field
over Kof the polynomial P (0, Y)
is the extensionE / K.
The
proof
usesprevious
arithmetic structure results of ours on the set of models of agiven
cover of the line[De3];
the main idea consists intwisting
covers
(Section 2).
Theorem 3.1 isactually proved
in the moregeneral
case Kcontains an
ample
field k. Recall a field k is calledample
if every smooth k-curve hasinfinitely
many k-rationalpoints provided
there is at least onePervenuto alla Redazione il 15 maggio 1998 e in forma definitiva il 8 febbraio 1999.
(e.g. [DeDes]). Complete
valued fields areample.
From results ofPop [Po;
Appendix I] (who
also introduced the notion ofample fields),
the fieldsQ’
and
~tp
of alltotally real,
andrespectively totally p-adic algebraic
numbers areother
typical ample
fields.P(seudo) A(lgebraically) C(losed)
fields areample
too:
by definition,
over a PACfield,
every curve has at least one(in
factinfinitely many)
K-rationalpoints [FrJa; Chapt. 10].
Thus the conclusion of Theorem 3.1 also holds for PAC fields. Infact,
astronger
conclusion holds for PACfields (Theorem 3.2):
one mayrequire
that(i)
holds over K itself(and
notonly
overK),
thatis,
that the fieldK(t)[Y]/(P(T, Y))
is a Galois extension ofK ( T )
of group G. We note(Remark 3.3)
that an argument for the PACcase is more or less
implicit
in theproof
of a related result from[FrVo].
This solves the so-called Beckmann-Black
problem
for PAC fields. In moregeometric
terms, thequestion, originally
statedby
S. Beckmann[Be] (over
K =
Q),
iswhether, given
a field K and a groupG,
thefollowing
arithmeticlifting property
holds: every finite Galois extensionE / K
of group G is thespecialization
of a G-cover ofI~1
defined over K(the prefix
"G" in "G-cover"indicates that the Galois action is part of the
data;
see[DeDo]
for aprecise definition).
S. Beckmannproved
thelifting
property when G is an abelian group or asymmetric
group(over
numberfields).
Theproblem
has also beeninvestigated by
E. Black who introduced acohomological
method. She obtained that over a hilbertian field K, a semi-directproduct
of a finitecyclic
group A with agroup H having
thelifting
property also has thelifting
property if and(char(K), IAI)
= 1[B13].
That includes the case of abelian groups andprovides
newexamples
of groups with thelifting
property overarbitrary fields, e.g.the
dihedral groupsDn
of order 2n when n is odd([Bll],
see also
[B13]
for a detailed account of theproblem).
Sheconjectures
that thelifting property
holdsunconditionally, i.e.,
for every group G and over every field K[B13].
Our initial statementsrephrase
as follows. Theorem 3.2 shows the Blackconjecture
holds over PACfields, i.e.,
that over thesefields,
the arithmeticlifting property
holds for all groupsG ~ 1 ~ .
Theorem 3.1 is that overfields
containing
anample field,
aslightly
weaker form of thelifting property,
called the mereform,
holds.Namely,
this mere formrequires
that therealizing
cover be defined
over K only
as mere cover(that is,
without the Galoisaction)
and be Galois over K.
E. Black also established some connections between the arithmetic
lifting
property and other classical Galois realization
properties
of groups. Forexample
she showed that if a group G has a
generic
extension over K(e.g.
if Gsatisfies the classical Noether’s
problem [Sa]),
then G has thelifting property
over K
[B12].
In Section 1 we show one furtherconnection, namely
that thearithmetic
lifting property
for all groups and fields(i.e.,
the Blackconjecture) implies
theRegular
Inverse Galois Problem(Proposition 1.2) (we
note this (’)Note that unlike algebraically closed fields which have no non-trivial algebraic extensions (andso for which the Beckmann-Black problem is trivial), PAC fields may have many fields extensions:
in fact, each finite group is a Galois group over PAC hilbertian fields K (like the field
more precisely, the absolute Galois group G(K) of such fields is pro-free of infinite rank.
was observed
independently by
A.Tamagawa).
This suggests that the whole Blackconjecture, though plausible,
isprobably
out of reach at the moment.The
Regular
Inverse Galois Problem is known to hold overample
fields(and
so over PAC fields
too). Solving
theoriginal
Beckmann-Blackproblem
overample
fields(and
not thejust
the mere form ashere)
would bequite interesting.
ACKNOWLEDGEMENTS. The author wishes to thank E.
Black,
M. Emsalem and F.Pop
for mathematical discussions and valuable comments.1. - Beckmann-Black and the
regular
inverse Galoisproblem
1.1. - The
conjectures
Throughout
the paper the Galois group of a Galoisextension E / k
is denotedby
Given a field K, we denoteby KS [resp. by K]
aseparable [resp. algebraic]
closure of K andby G(K)
the absolute Galois groupG(K) =
K. Recall a field extension
ElK
isregular
ifE / K
isseparable
and E n K = K. As
usual,
we sometimes shorten thephrase
"Galois extension withElK regular" to just
say"regular Galois
extensionfor such
extensions,
we have =(which
is in factequivalent
to theregularity condition). By
cover ofI~1
wealways
mean asmooth
projective
model(of
the associated function fieldextension).
We usethe
phrase
"G-cover" for Galois coversgiven
with theirautomorphisms;
non-necessarily
Galois coversgiven
without theirautomorphisms
are referred to asmere covers.
Given a
degree
d mere coverfK : XK - I~1
defined over K and anunramified K-rational
point to E
denote thecompositum
of all fields ofdefinition over K of
points
in the fiberby equivalently, KfK,to
isthe
compositum
of all residue fields at to of the Galois closure of the extension We call the fieldKfK,to
thesplitting field
or thespecialization
of
fK
at to. With thisnotation,
the Beckmann-Black arithmeticlifting
conditioncan be reformulated as follows. We
distinguish
a G-form(the original one)
anda mere form. The
former,
which isconjectured
to hold for all fields K andgroups G
by
E.Black, implies
the latter.BECKMANN-BLACK ARITHMETIC LIFTING CONDITION. Let K be a
field,
G bea finite group and
E / K
be a Galois extension of group G.G-form (G-BB):
There exists a G-coverf : I~1
of group G definedover K and some unramified
point to E
such that thesplitting
fieldextension of
fK
at to isK-isomorphic
toElK.
Mere form (mere BB)_:
There exists a mere coverfK :
X -*plover
K,Galois of group G over K, and some unramified
point
to E such that thesplitting
field extension offK
at to isK-isomorphic
toElK.
The
following conjecture
is the centralproblem
of Inverse GaloisTheory.
REGULAR INVERSE GALOIS PROBLEM
(RIGP).
Given a field K and a finite groupG,
there exists a Galois extension withE/A" regular
such that=
G ;
or,equivalently,
there exists a G-coverf ’ :
X -Pl
1 of group G defined over K.REMARK 1.1.
(a)
A G-form and a mere form of the RIGP could also bedistinguished.
The mere form would be:given
a fieldK,
each finite group G is the Galois group of a Galois cover defined over K as mere cover(but
notnecessarily
Galois overK).
To myknowledge,
this mere form is not knownto hold over more fields than the classical form
(which
would be the G-form of theRIGP).
(b)
Given a field K and a groupG,
G-BB for G andK(x) (with
xtranscendental over
K) implies
G-BB for G and I~: this follows from Bertini’s theorem. The converse is unclear. So are theanalogous questions
foralgebraic
extensions.
1.2. - G-BB
implies
RIGPPROPOSITION 1.2. The Beckmann-Black
lifting
conditionimplies
theRegular
Inverse Galois Problem. More
precisely, given
a group G andafield
K,G-BB for
G and over every
regular
extensionof
Kimplies
RIGP over K. Inparticular,
G-BBfor
G and overevery field of
characteristic 0implies
RIGP overQ.
PROOF.
Suppose given
a group G and a field I~ and assume that G-BB holds for G over everyregular
extension of K.Regard G,
via itsregular representation,
as asubgroup
of thesymmetric group Sd (with
d=
SetE -
K (T )
where T ={ Tl , ... , Td }
and let G act on E via its action on{ Tl , ... , Td } .
The fixed fieldE~ is
aregular
extension of K, hence the function fieldK (B)
of a d-dimensionalK-variety
B, whichby
construction is unirational.We will now use the G-BB
lifting
property to construct aGalois extension of
group G withregular
and such that the extensionsand are
linearly disjoint.
Thefollowing diagram
summarizes the argument.More
specifically,
fromG-BB,
there exists a Galois extension of group G withregular
whichspecializes
to at T = 0.Apply
then the hilbertian
property
to the hilbertian field(dim(B)
>0)):
thereexists t E
K(B)(2)
such that thespecialized
is Galois ofgroup G and is
linearly disjoint
from the extension Set E’ =Et ;
the extension has the
required properties:
inparticular,
theregularity
of the extension follows from
The
following
lemma shows that E’ E is a Galois extension of E =K (T)
of group G and is
regular
over K. Tocomplete
theproof,
we note thatregular
realization over
K (T)
isequivalent
toregular
realization overK (T ) .
0LEMMA 1.3. Let
ElK (B)
andEl K (B)
betwo finite
extensions with bothElK
and
E’/K regu_lar. Assume
thatE’IK(B)
is Galoisof group
G.Assum_e further
thatthe extensions and
KEIK(B)
arelinearly disjoint
overK (B).
ThenE’E/E
is a Galois extensionof
group G and isregular
over K.PROOF. Consider the
diagrams
We have a
priori
G =G(E’IK(B)) D G(E’E/E)
But itfollows from the linear
disjointness assumption over K
and theregularity
ofthe extension
E’IK
thatG(KE_’E/KE_)
=G(KEIIK(B))
=G(E’IK(B))
= G.Conclude that
G(E’E/E)
=G(KE’E/KE)
= G. D1.3. - The linear
disjoint
realization conditionThe G-BB condition was used in the
proof
ofProposition
1.2 to showthat the unirational
K-variety
B has thefollowing
LinearDisjoint
Realizationproperty:
(LDR)
Given a Galois extension of group G withE / K regular,
there exists a Galois extension
E’IK(B)
of group G withE’IK regular
andsuch that the extensions and are
linearly disjoint.
Lemma 1.3
actually
shows that LDR(for
all K-varieties B of dimension>
0) implies
RIGP. Thusshowing
LDR canactually
be an alternate way ofusing
Noether’soriginal
idea to prove the inverse Galoisproblem.
What is tobe shown is that every G-cover can be somehow "deformed" into a
G-cover (of
the same base and with the same
group)
that islinearly disjoint (over K).
Notethat if B is a rational
variety (over K),
then the LDR condition holds(use
the (2) That t can be picked in K ( B ) follows for example from Theorem 2.2 of [De2].argument
in theproof below). Furthermore,
unlike Noether’soriginal
program, thisapproach
wouldyield
theregular
form of the inverse Galoisproblem.
Infact,
as we nowshow,
for infinitefields,
LDR(for
allK-varieties)
isequivalent
to RIGP.
PROPOSITION 1.4. Let K be an
infinite field
and assume the RIGP holds over K.Then every
K-variety
B(of dimension > 0) satisfies
condition LDR.PROOF. Fix a
K-variety
B of dimension > 0 and a finite extensionwith
ElK regular.
Let T ={Ti,... , Tr }
be a transcendence basis ofK(B)
over K. It follows from the RIGP over K that there exists a Galois extension of group G with
E*/A" regular: namely
realize G overregularly
and extend
K(T) to_K (T).
Denote thesupport of the
ramification divisor of the extension[resp.
the extensionby ram(E) [resp.
ram(E*)].
Thenpick
anautomorphism X
ofK(T)
such that the intersection(ram(E)) n ram(E*)
consistsonly
of components of codimension > 1(this
is
possible
since K isinfinite;
x can bepicked
inGLr(K)).
Denote thepull-
back of the extension
along
theautomorphi sm x by
Thesituation is summarized
by
thediagram
at the end of theproof.
_
It follows
from theconstruction
and thepurity theorem
thatthe
intersection is not ramified aboveK (T).
ThereforeKENKE*
=Y(T). Indeed,
otherwise,
would be a proper unramified extension ofK (T),
whichcanonically corresponds
to a proper etale cover of P . The contradiction follows since the fundamental group ofPr is trivial.
_ _Conclude that the extensions and are
linearly
dis-joint (the
former isGalois).
Set E’ =K(B)E*.
It follows[La; Proposition
1p.
262]
that the extensions and arelinearly disjoint.
Theextension has the
required properties.
2. -
Twisting
coversThis section is the technical core of the paper. In
particular
weexplain
how to twist a G-cover
fK :
X -I~1
defined over K of group Gby
anhomomorphism : G (k ) ---->
G. In order to do so we will view covers ofPl
1as
representations
of K-arithmetic fundamental groups; we refer to[DeDo]
formore details on the
dictionary
between these twocategories.
Denote the affine subset
I~1
with the reduced ramification divisor offx
removed
by (I~ 1 ) *
and the K-arithmetic fundamendal group of(I~ 1 ) * by
Let
ØK :
G be therepresentation corresponding
to the G-coverfK.
The
representation corresponding
to the mere cover associated withfK (by dropping
the Galoisaction)
is obtainedby composing q5K
with theleft-regular representation
y : G ~Sd
of G(where d = Identify
G andy(G).
The
representation corresponding
to the mere coverf = fK
Q9KKS
is the restrictionØKs :
Gc Sd
of theprevious
one to theK, -fundamental
group of
(P~)*.
Each unramified K-rationalpoint
to E(P’)* provides
asection
sto : G(K) ~ IIK
of the canonicalsurjection nK - G(K).
Recall this result from[Del].
PROPOSITION 2.1
[Del; Proposition 2.1].
For each t EG(K),
the element(q5 K conjugate
inSd
to the actionof
í on thefiber fi 1 (to). Consequently,
the
splitting field K fx,to of fK
at tocorresponds
via Galoistheory
to the homo-morphism G(K) -*
G; that is, it is thefixed field
inKs ofker(ØKsto)
and the Galois group
of
the extension is theimage
groupof sto.
Thehomomorphism OK sto is
called the arithmetic actionof G(K)
on thefiber fx 1 (to).
Fix an unramified rational
point
to E Denote the mapØK Sto :
G(K) -
Gby
Denote theright-regular representation
of Gby 3 :
G -~Sd.
Definecp* : G(K) -
Gby ~p*(g) -
Consider then the map~x : Sd
definedby
The
homomorphism 6w*
EHom(G(K), Censd G) (3)
can be viewed as a 1-cochain in
Censd G) (with
trivialaction).
It follows fromProposition
2.3of
[De3]
that themap Øk : Sd
induces arepresentation
ofII K
and thatthe associated mere cover, denoted
by fK - jl"
-P~,
is a K-model of themere cover
f.
The mere coverfx
isby
definition the twisted cover offix by
thehomomorphism
cp. Thefollowing
statement contains theproperties
ofthe twisted cover that we will use. In the case of PAC
fields,
conclusion(b) corresponds
to the "fieldcrossing argument"
from[FrVo;
Lemma1].
~3~Note that is an actual homomorphism because 8 and cp* are both anti-isomorphisms.
PROPOSITION 2.2. The twisted cover
fx :
-~I~1
has thefollowing
proper-ties :
(a)
The arithmetic actionof G(K)
on thefiber ( (%)~~ (to)
is the map8~p* :
G(K)
-+GCensd
G(by "~po ~ 8~p* "
we mean theproduct
map and8~p*).
(b)
Let x, EXCfJ(K)
be anunramified
K-rationalpoint.
Set t, = Then the arithmetic actionof G(K)
onthe fiber
in theoriginal
coverfK
isconjugate
in G to ~p :G(K) -
G.(c)
Let t, EPl (K)
not a branchpoint.
Denote the map~x Stl : G(K) --+
Gby
~pl
(the
arithmetic actionof G(K)
onthe fiber
Then there is a K -rationalpoint
on the twisted coverI~1
above ti.PROOF.
(a) immediately
follows from the definition ofi;.
(b)
Consider the sectionssto
andStl (from G(K)
tonK).
FromProposi-
tion
2.1,
for each T EG(K),
the action of T on the fiber isgiven by
In
Sd
the elementØK(St¡ (i))
E G shouldreally
be viewed as themultiplication
on the left
by q5K (St¡ (r))
in G while the element8~p* (t)
is themultiplication
on the
right by By assumption,
the elements(i)) (r
EG(K))
have a common fixed
point,
say a) E G. It follows that(c) By definition,
we have~~pl (i)
whereWhence
-UL ,
°
That
is,
for each i EG(K), (StI (r))
is theconjugation by (r)
and so fixes 1. Thecorresponding point
above t, is K-rational. 03. - The Beckmann-Black
problem
overlarge
fields3.1. - The mere Beckmann-Black
problem
overample
fieldsThe
following
result is thegoal
of this section. Aspecial
case is thetheorem stated in the introduction.
THEOREM 3.1. The
mere form of the
Beckmann-Blacklifting
property(mere BB)
holds
if
K contains anample field.
That is,given
a group G, each Galois extensionE / K
is thesplitting field
extension at someunramified point
to EpI (K) of
somemere cover Y -
I~1 defined
over K and Galois over Kof
group G.PROOF. Let k be an
ample
field contained in K. The first stage uses theRegular
Inverse Galois Problem over k. Morespecifically
there exists a G-cover fk :
X -I~1
of group G defined overk ; furthermore,
thecover fk
can be
required
to have atotally
k-rational fiber above some unramifiedpoint
to E
(that is,
the fiberfC’(to)
consistsonly
of k-rationalpoints
onX).
This result over
ample
fields is due to F.Pop [Po]
in its final form. It had first beenproved
in the case k is acomplete
valued fieldby
D. Harbater[Har],
andin other various
special
cases(e.g. [DeFr], [Del];
see[DeDes]
for acomplete bibliography).
The condition that therealizing
cover has atotally
k-rationalunramified fiber is not
explicitly
stated in[Har]
and[Po]
but can be deduced from theproofs.
For more detailsconcerning
thispoint
see[Del]
and[Li]
forthe
complete
valued field case and[DeDes;
Section4.2]
for thegeneral
caseof an
ample
field.Consider then the G-cover
fx
obtainedfrom fk by
extension of scalars from k to K.Properties
offk
carry over tofK over
K, thatis, fK
is a G-coverof group G defined over K and has a
totally
K-rational fiber above to.The last
stage
usesProposition
2.2. Let cp :G(K) -
G be asurjective homomorphism
thatcorresponds
to thegiven
extensionE/K (i.e., Kker(CfJ)
=E).
By construction,
the arithmetic action on the fiber above to in the coverfK
istrivial. Consider the mere cover
" - X~ 2013~ I~1
obtained fromfK by twisting by
cpo FromProposition
2.2(a),
the arithmetic action ofG ( K )
on the fiberis the map
~~p* : G(K)
--GCensd G.
This map has the same kernel as Conclude that thesplitting
field extension offx
at to isK-isomorphic
to the extensionEIK;
the coverfk - I~l
is the desiredcover Y ~
I~1
of the statement. D3.2. - Related comments 3.2.1. - Pointed RIGP
The
proof
shows moregenerally
that the mere form of the Beckmann-Blacklifting
property holds over every field K for which a certain"pointed"
formof the
Regular
Inverse Galois Problem holds. Thispointed
formrequires
thateach group G be realized as the
automorphism
group of a G-coverf :
X -I~1
of group G defined over K and with the additional
property
that there is anunramified
totally
K-rationalfiber,
orequivalently,
that X has an unramifiedK-rational
point.
To myknowledge,
no field is known for which the RIGP is known to hold but not thepointed
RIGP. In fact no field is known forwhich the RIGP is known to hold and which does not contain an
ample
field(in
which case thepointed
RIGPholds). Producing examples
of such fieldsseems to be a difficult
question, firstly, because,
such fields may not exist(if
the RIGP
only
held over fieldscontaining ample fields), secondly, because, it
happens
to be difficult toproduce non-ample
fields at all(at
leastinside Q
and apart from number
fields).
There is a natural candidate for thatquestion
though:
the field but that the RIGP holds over~ab
and that~ab
is notample
are both still openconjectures.
Incidentally
we notethat,
even for numberfields,
theexisting
argumentsthat
they
are notample
are notelementary:
one usesFaltings’s
theorem and another uses Merel’s result on rationalpoints
on modular curves[Me].
In fact a natural way to prove that no number field isample
is to find a curve C definedover Q
such thatC(K)
is finite for all number fields K(and C(Q) # 0).
That
is,
to prove Mordell’sconjecture
for one curve over all number fields.However some works of
Szpiro
andMoret-Bailly [Mo]
showthat,
for someeffective version of
it,
the Mordellconjecture
for asingle
curveimplies
theabc
conjecture
andtherefore, by
a result of Elkies[El],
the Mordellconjecture
for all curves. We
finally
mention that anexample
of an infinitealgebraic
extension
of Q
that is notample
was communicated to usby
P.Corvaja;
hisconstruction is not
elementary
either since it uses theLang conjecture
aboutrational
points
on subvarieties of abelianvarieties,
alsoproved by Faltings (and
which
implies
the Mordellconjecture).
It is unclear whether the RIGP holdsover this field.
3.2.2. - G-BB for abelian extensions
The strategy of Theorem 3.1 can also be used to prove the G-form of the BB
lifting property
for abelian groups G over anarbitrary
field K. This showsin
particular
that the G-form of the BB condition holds if K is a finite field.The
argument
wasalready developed
in[De3].
Wegive
below a brief sketch of theproof.
The first step is to realize the
given
abelian group G as the Galois group ofa G-cover X -->
P1 over K
with at least one unramifiedpoint
to EThen, just
asabove,
it ispossible
to "twist" the G-coverfK by
any element ofHom(G(K), Censd (G)), which,
since G isabelian, equals Hom(G(K), G).
The
resulting
cover is still defined over K as G-cover and thesplitting
fieldextension at to can be any Galois extension of group G
given
in advance. TheG-BB
lifting
propertyclearly
follows. As to therequired preliminary regular
realization of
G,
it is a classical resultexcept possibly
for the existence of at least one unramifiedpoint
to EObviously
this is adifficulty only
when K is a finite
field,
and moreparticularly,
when the characteristic divides the order of G. For this technicalpoint
we refer to[De4].
3.3. - The case of PAC fields
3.3.1. - The Beckmann-Black
problem
over PAC fieldsOver PAC
fields,
theoriginal
Beckmann-Blackproblem
can be solved.THEOREM 3.2. The
G-form of the
Beckmann-Blacklifting
property(G-BB)
holds
if K is a PAC field.
That is,given
a group G, each Galois extensionElK of
group G is the
splitting field
extension at someunramifzed point
to EPl (K) of some
G-cover Y --+
Pl of
group Gdefined
over K.PROOF. PAC fields are
ample.
So as in theproof
of Theorem3.1,
we can realize G as theautomorphism
group of G-coverfK :
X -I~l
of group G defined over K. The extra condition about the existence of atotally
K-rationalfiber will not be needed here. Also we fix an unramified K-rational
point
to, welet cp :
G ( K ) -~ G be a surjective homomorphism
thatcorresponds
to thegiven
extension
E/K
and consider the mere cover1; : jl"
-~I~1
obtained fromf by twisting by
cpo Since the field K isPAC,
the curvek"
hasinfinitely
many K-rationalpoints.
Let xl be one of them that is unramified and set tl =fx(xl).
From
Proposition
2.2(b),
the arithmetic action ofG(K)
on the fiber in theoriginal
coverfK
isconjugate
in GG(K) -
G.Consequently,
the
specialization
offK
at to is the field extensionElK.
0REMARK 3.3. The
proof
shows moregenerally
that if K is PAC(*) any given G-cover fK :
X -pI of group
Gdefined
over K has the property that every Galois extensionElK of group
G is thespecialization of fx
atinfinitely
many
unramified points
t EI~1 (K).
This
specialization property actually implicitly
appears in theproof
of The-orem B of
[FrVo]
as a consequence of the fieldcrossing argument, which,
as we said
earlier, corresponds
toProposition
2.2(b)
in the context of PACfields. Fried and Volklein work over fields of characteristic 0 but this
part
of theirargument
remains valid inarbitrary
characteristic. Thus[FrVo] implicitly
contains a solution to the Beckmann-Black
problem
for PAC fields.3.3.2. - Some
specialization properties
The
specialization
property(*)
is used in[FrVo]
to showthat, given
a PACfield K, the
following
assertions areequivalent:
(i)
Each group is a Galois group overK,
(ii)
Each G-coverfK :
X -~I~1
defined over K can bespecialized
to a Galoisextension
ElK
with the same Galois group.Fried and Volklein call RG-hilbertian a field
satisfying
condition(ii). They
also
produce
anexample
of a non-hilbertian RG-hilbertianfield;
otherexamples
are
given
in[DeHa].
Using Proposition
2.2(b) (instead
of the fieldcrossing argument)
leads tothe
following
characterization of RG-hilbertian fields. Let K be anarbitrary field, i.e.,
wedrop
theassumption
"K PAC" of[FrVo].
Given a group G anda G-cover
fK :
X -l~l
defined over K of groupG,
consider the collection of all twisted coversfK : jl"
-~P~
where w is anysurjective homomorphism GK -
G. Denote then thedisjoint
union of all curves k"by xfK .
PROPOSITION 3.4. The
field
K is RG-hilbertianif
(**) for
each G-coverfK :
X -I~l defined
overK, the set XfK
containsinfinitely
many K -rationalpoints.
Furthermore, if the
RIGP is known to hold over K, then the converse holds as well.Note that condition
(i)
above - each group is a Galois group over K - isimplicitly
contained in(**)
since it assures thatXfK
is not the empty set.Thus there are two
aspects
in the characterization of RG-hilbertian fieldsgiven
in
Proposition
3.4: condition(i)
is the Galois-theoretic part and the rest of con- dition(**)
is a purediophantine
condition(in
that itonly
consists insolving polynomial equations). Proposition
3.4 followsstraightforwardly
fromPropos-
tion 2.2
(b)
and(c):
these indeed assert that K-rationalpoints
on some twistedcurve
k" exactly correspond
tospecializations
of the coverfK
of Galois group G.Condition
(*),
which is satisfiedby
PACfields,
is stronger than condi- tion(**)
since then each of the twisted curves containsinfinitely
many rationalpoints.
We end this paperby
some further observations on these twoconditions. We note in
particular (Section
3.3.3 and Section3.3.4)
that the stronger condition(*)
does not hold ingeneral
overample
fields and so theproof
of the Beckmann-Blacklifting
property for PAC fields does not carry over, as itstands,
to the moregeneral
case ofample
fields.Ample
fields how-ever
satisfy
thefollowing specialization
property, whichreadily
follows fromProposition
2.2 and the definition ofample
fields. This was first observedby
M. Fried in the
special
case K =~tr.
(***)
any G-coverfK :
X --->Pl of
group Gdefined
over K has the property thatif a
Galois extensionE / K of
group G is thespecialization of fK
at someunramified point
t EI~1 (K),
then it is thespecialization of fK
atinfinitely
manyunramified points t
E3.3.3. - The local field case
Take K = R and consider the G-cover
fR
of groupZ/2
associated with the extensionR(T, Clearly
the extensionC/R
is not aspecializa-
tion of
fR.
There are similarexamples
overQp.
Consider the G-cover of groupZ/2
associated with the extension wherey 2-y- pT =
P 0.It is
straightforwardly
checked that for each t EQp, p)
EZp;
hencethe
polynomial y2 -
Y - reduces modulo p toY2 -
Y. Conclude fromt2-p
Hensel’s lemma that for each t E
Qp, Y2 -
Y - t2-p istotally split
overQp.
Thus no
degree
2 extension ofQp
is aspecialization
of3.3.4. -
Using
anargument
of Saltman[Sa;
Section5]
Denote the
cyclic
group of order 8by Cg.
Thisexample
shows that theunique
unramified extensionL2/Q2
of groupCg
is thespecialization
of no G-cover
fQ2 :
Xp1 over Q2
with groupCg
and that is definable overQ.
Indeedlet
fQ
be a G-cover definedover Q
of groupCg
and letfQ2
=/Q0QQ2.
Assumethat
L2/Q2
is thespecialization
offQ2
at some unramifiedpoint t
EpI (~2).
From Krasner’s
lemma,
there exists ti EI~1 (Q)
with the same property. But then thespecialization
off~
at t, is a Galois extensionof Q
of groupCg
andit remains Galois of group
C8
after extension of scalars toQ2.
This contradictsWang’s counter-example
to the Grunwald theorem.Equivalently,
thisexample
shows that there is no
Q2-rational point
on any of the coversIÓ2 : X
-->P,
obtainedby twisting by
asurjective homomorphism
cp :G(Q2) -~ C8
ofkernel
G(L2).
3.3.5. - The number field case
Assume K is a number field and that the G-cover
fK (of
groupG)
is of genus > 2. Then fromFaltings’s theorem,
agiven
Galois extension of group G is thespecialization
offK
at at mostfinitely
manypoints t
EI~1 (K).
Furthermore,
there may be nopoints
at all. Forexample
take the coverf~
associated with the extension
Q(T,
Thespecializations
offQ
are
only
ramified atprimes
p - 1 modulo 4. Therefore an extensionE/Q
ramified for
example
at 7 cannot be aspecialization
off~.
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