J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
L
UIS
V. D
IEULEFAIT
Newforms, inner twists, and the inverse Galois
problem for projective linear groups
Journal de Théorie des Nombres de Bordeaux, tome
13, n
o2 (2001),
p. 395-411
<http://www.numdam.org/item?id=JTNB_2001__13_2_395_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Newforms,
inner
twists,
and the inverse Galois
problem
for
projective
linear
groups
par LUIS. V. DIEULEFAIT
RÉSUMÉ. Nous reformulons de manière
plus explicite
les résultats deMomose,
Ribet etPapier
sur lesimages
desreprésentations
ga-loisiennes attachées à des newforms sans
multiplication complexe,
en nous
restreignant
aux formes depoids
2 et de caractèretri-vial. Nous calculons deux tels
exemples
denewforms, possédant
une
unique
tordueconjuguée
à laforme,
et nous démontrons quepour tout nombre >
3,
l’image
est aussi grosse quepossible.
Nous utilisons ce résultat pour prouver que les groupes~ 3,5
(mod
>3)
et~
0, ± 1(mod
>3)
sont groupes de Galois surQ.
ABSTRACT. We reformulate more
explicitly
the results ofMo-mose, Ribet and
Papier concerning
theimages
of the Galoisrepre-sentations attached to newforms without
complex multiplication,
restricted to the case ofweight
2 and trivialnebentypus.
Wecom-pute two
examples
of thesenewforms,
with asingle
inner twist, and we prove that for every inertprime greater
than 3 the im-age is aslarge
aspossible.
As a consequence, we prove that thegroups
PGL(2,
for every 3, 5(mod 8),
> 3 ,and
PGL(2,
for everyprime
0, ±1
(mod
> 3, areGalois groups over
Q.
1. Introduction Let
f
=¿~1
anqn
be aweight
2 newform onro(N).
Inparticular
f is
a normalizedeigenform
for the whole Heckealgebra.
Let Qj
be the number fieldgenerated by
its coefficients an and 0 itsring
ofintegers.
For everyprime £
put:
Ot
= 0 0z7Ge
=Qj
0Q
Qt
-By Deligne’s
theorem(see
[D71]),
there exists a continuous Galoisrepre-sentation :
Manuscrit reçu le 8 septembre 1999.
unramified outside iN
satisfying,
for everyprime p f
2N :Let G =
Gal(Q/Q)
andGi
= closedsubgroup
ofGL(2,
Ot).
From thedecompositions:
Qj,t
=fl xjt
andOi
=fl xjt
Ox
we obtain ade-composition
of Pi
as direct sum of therepresentations:
From now on, we will assume that the newform
f
does not havecomplex
multiplication
(CM);
see[R80],
page48,
or[S71b]
for a definition.Let A be a
prime
inQj .
Consider the reductionpa
of pA, obtainedby
composing
pA with the reduction map:GL(2, Ox) - GL(2,1FÀ)’
where1Fa
is the residue field of A. Let Q =
f(A)
be the rationalprime
such thatA I t
and letGa
be theimage
ofpa.
In order to determine theimages
of therepresentations pa
we will need thefollowing
result of Ribet([R85],
theorem
2.1):
Theorem 1.1. Let H be an open
subgroup of
G. Thenfor
almost every Ame have:
a)
Therepresentations
PÀIH is
an irreducible 2-dimensionalrepresentation
of
H overb)
The orderof
the groupsHa
= is divisibleby
e.For every
automorphism -y
of the fieldQj
consider the newform:Suppose
that there exists a Dirichletcharacter X
such that:for almost every p. This character must be
unique
(because
we areassuming
that
f
does not haveCM)
and we will call it xq. Ingeneral,
for a cuspform
f
Es2(ro(N),,-)
it holds: =,(E).
In ourparticular
case(E
=1)
this
implies
that xq is aquadratic
character. When such a character exists we saythat, I
is an inner twist off .
Let r be the set of
those y giving
an inner twist. We recall someproperties
ofr:
~ For every, E r the conductor of xq is divisible
only by
theprimes
dividing
N.~ r is an abelian
elementary
2-group.
. LetFf
=Qj,
the fixed field of r. ThenFf
is the fieldgenerated by
f a~} ,
with p~’
Nranging
over adensity
one set ofprimes.
o Whenever N is
square-free,
nof
ES2(ro(N))
has CM or innertwists,
so that r ={1}.
For all but the first of these
properties,
see[R80],
page 49. The firstprop-erty
follows from theequality
q(p£)
= that will beproved
later.Using
the fact that the Galoisrepresentation
q(p£)
is unramified outside iN it follows that x,y canonly ramify
atprimes
dividing
N.Every
character Xy may bethought
of as a character on the Galois groupG,
so its kernelH.y
is an opensubgroup
of G. Let H be the intersection ofthe E
r ,
and let K be its fixed field(cf.
[R85],
page190).
LetHe
=pE(H)
and letwhere R is the
ring
ofintegers
ofFf
andRI
= R 0z7Ge.
The
following
theorem of Momose([M81];
[R85],
theorem3.1)
determines theimage
of these modular f-adic Galoisrepresentations:
Theorem 1.2. For almost every
prime f
we haveHt
=At.
What is known for every
prime £
isthat pt
restricted to Hgives
a map:where D is a
quaternion
divisionalgebra
overFf
(this
quaternion algebra
is related to a central
simple
algebra
overFf
definedby
a certain2-cocycle
on r with values inQ*f ).
Thisimplies
that,
first ofall,
we should restrictourselves to those
primes
such that:For these
primes,
HE
is an opensubgroup
ofAt
(condition (0)
holds foralmost every
E,
theexceptions
are theprimes
dividing
the discriminant ofD).
Then the theorem is
proved
afterapplying
a result of[R75]
(cf.
[R75],
corollary
2.2 and theorem3.1,
and[R85],
page192)
stating
thatHt = At
if the
following
hold:(1)
The determinant mapHt -
7Le
issurjective.
(2) ~ >
5.(3)
Ht
contains anelement xi
such that:(trace
Xl)2
generates Rt
as aZ,,algebra.
(4)
For each A1£,
the grouppx (H)
is an irreduciblesubgroup
ofGL(2, JFÀ)
whose order is divisible
by
t.To see that these conditions hold for almost every
prime,
theorem 1.1 isapplied
(to
deal with condition(4))
together
with thefollowing
facts:o There exists an
integer v
such that v isrelatively
prime
toN,
Frob v E Ho The determinant
of pt
coincides on H with the f-adiccyclotomic
characterXe
(condition
(1)).
We are interested in the
image
ofG, Gi
=pe(G). pP
is a continuoushomomorphism
whose restriction to H takes values in
At
9GL(2, Ri),
if i satisfies condi-tion(0).
For y E r let us consider qp£
and Pi ø
Xy. These tworepresentations
areisomorphic
(semisimple
with the samecharacter).
Thus there is a matrixX E such that:
X-t
being
trivial on Hand -y
onHi
CAi,
X commutes withHt.
Then Xis a scalar matrix and we obtain the
equality:
For g
E G leta(g)
EQj
be an element such that:-y(a(g))
=xy(g)a(g)
for
all 7
Er,
whose existence isguaranteed by
Hilbert’s Satz90,
anda(g)
does notdepend
on i.Moreover,
itonly depends
on theimage g of g
inG/H.
Therefore,
we haveonly
a finite number ofa(g),
andthey belong
to0e
for almost every E.The matrices
pe(g)a(g)-1
are r-invariants(because
7 E r acts on both piand a
by multiplication by
X,y ),
sothey
are inGL(2, Ri).
In theequality:
, . - , ,
-let us call
Bt
thesubgroup generated by
theproducts
incurly
brackets,
when g
ranges over G. It is contained inAt
because it is inGL(2,
Rt)
andhas determinant Xi. Note that
a2 (g)
ERi
because it is r-invariant:For an
element g
E H we can takea(g)
=1,
so thatHi
CBt.
An
application
of theorem 1.2 proves thefollowing
theorem of E.Papier
([R85],
theorem4.1):
Theorem 1.3. For almost every
prime
E,
theimage Gt
of pt is
thesub-group
of
GL(2,
generated by
the groupAi
and thefinite
setof
matrices:Now,
let us introduce a variant of this theorem that will allow us tocompute
theexceptional
primes
in someexamples.
The
key
observation is that we canreplace
the conditionby
condition theorem 1.3
applies.
To check whichprimes
verify
thisequality,
we will use the conditions
(0)
to(4)
of theorem1.2,
this timeapplied
toBi.
Condition
(0)
is needed apriori
to ensure thatHe
ç Ai
and(see
theproof
of theorem
1.3)
it alsoimplies
Be
C Ae .
In[Q98] J.
Quer gives
a formulafor the
quaternion
algebra
D(see
also[C92])
and condition(0)
holds for allprimes i
such that:Besides,
aspointed
out in theproof
of theorem1.3,
theprime
mustverify:
For ease of
notation,
let us write:IX, y]
for thediagonal
matrix:( ).
(0
yAs a consequence of
(1.1),
condition(1)
forBe
is satisfiedby
everyprimes
because the matrices
[a(g), lla(g)]
have determinant 1 and so all we needis the determinant map of
Gt
to besurjective;
this isequivalent
to thesurjectivity
of thecyclotomic
character xe : G -7Ge_
Again,
we look for the elementspecified by
condition(3)
inHe,
this is morethan
enough
becauseHQ
C Be.
Then if v is suchN,
Frob v E H anda 2generates
Ff
over mustverify:
Condition
(4)
is the hard one. Inspite
of theorem 1.1 the determination ofthe finite
exceptional
set ofprimes
is nottrivial,
even for G itself(see
the nextsection).
Taking
Be
instead ofHt
we can at least reduce theproblem
to
verifying
the condition onGi, avoiding
the restriction i >(G : H) (see
the
proof
of theorem 2.1 of[Ri85]),
not to mention thegain
in condition(1).
_Let us call the reduction mod A’ of the
A’-component
ofB~,
whereA’
1£
is aprime
inR ,
and let ...,Ài
be theprimes
dividing
A’ in O.Assume that for the
prime i
theorem 1.1applied
to G holds. Then from theirreducibility
of all the and formula(1.1)
it follows thatB)..’
isan irreducible
subgroup
ofGL(2, 1FA’).
In the same way we see that if thegroups
(G)
have ordersmultiple
of £ the same holds for because in(l.l)
the matrices[a(g),
a-’
(g)],
afterreducing
their coefficients moduloany
prime
dividing 2,
have ordersrelatively
prime
to £.Thus,
we haveproved:
Theorem 1.4. Let ~ > 5 be a
prime
such thatfor
everyplace A
lying
above
i,
thefollowing
holds:Gx
is an irreduciblesubgroup of
GL(2, 1F).)
whose order is a
multiple
of
t.If, furthermore,
fsatisfies
conditions(i),
Gi is
thesubgroup of
GL(2,
generated by
the groupAl
and thefinite
set
of
matrices:Remark. This result
applies
to almost everyprime.
Let A be a
prime
in 0dividing t
for aprime i
satisfying
thehypothesis
of theorem 1.4.
Let us call
P(pa(G))
theimage
of inPGL(2,1FÀ).
It is well known(cf.
[RV95],
lemma2.2)
that the inclusionAt
CG~ implies:
where
A’ = AnR,
and PXL = PSL if r =[Fx, : F£]
is even and PXL = PGLif r is odd.
__ ____
Now observe that as elements
of PGL(2,
the matrices(a(g),
a-1
(g)~
and[a2(g), 1]
can beidentified,
and this is useful becausea2 (g)
ERt
implies
[a2(g), 1]
E From this and formula(1.2)
we see that if r isodd we
already
have:If r is even, we have to determine if all the matrices
~a2(g),1~
are inPSL(2, FA,),
in which case we will have:or if any one of them is not in
PSL(2, F~),
in which case we willagain
have(1.3).
___The determinant of these matrices
being
a2 (g) ,
weeasily distinguish
be-tween these two cases, and we obtain:Proposition
1.5.Let £, A, A’
and r be as in theprevious
discussion. Then2.
Exceptional
Primes for theorem 1.1We will review the
proof
of theorem 1.1given
in[R85],
finding
conditions2.1. Reducible
Representations.
Let us suppose that for aprime A
E0
dividing t,
PÀ(G)
is a reduciblesubgroup
ofGL(2,lFa).
Thenwith l/Ji
= for i =l, 2,
El, E2 Dirichlet characters unramified outsideN with
image
inFx
and Ti
thecyclotomic
character mod .~.By Deligne’s
theorem,
for everyprime
p~’
iN we have:From the last formula we see that ml+m2 - 1
(mod
e-1) andel (P)IE2 (P)
1 for every
p f N,
so that we can choose0 :::; ml m2 £-1.
Generalizing
results of
[S73]
to the case ofgeneral
level,
Faltings
and Jordanproved
in[FJ95]
thefollowing
result,
which we will state forgeneral
weight,
i.e.,
when therepresentation
comes from a newformf
ESk(N):
Theorem 2.1.
Suppose
that therepresentation
pA
is reducible. Thenif
i >N, pa
= 61 s3 with the characters Eiunramified
outside N.Carayol
and Livn6([C89] , [L89])
havegiven
bounds for theconduc-tors of modular Galois
representations.
Using
thisresult,
together
with theorem2.1,
we have(cf.
[FJ95],
pages 13 and46) :
Corollary
2.2. Letf
be anewform of weight
2 and level N andÀ 1£
aprime
in C~ such thatpa is
reducible. Thenif e
>2, 1 f
N,
wehave,
for
every p
~’
fN:with E a character
unramified
outside N whose conductor cverifies:
c2
N.
In
particular
ifp # £, p m
1(mod
c) ,
then: ap z 1 + p(mod A)
and ifp # ~,P - -1
(mod c) ,
then:ap -=
~(1 +p) (mod A).
These congruences cannot be
equalities,
because:[
so thatonly
finitely
many A cansatisfy
them.2.2. The second condition. We now turn to the second condition in
theorem 1.1.
Let A and E be as before and suppose that the order of
Ga
is not divisibleby
~. Itsimage
P(Gx)
inPGL(2, IF,X)
has to be :1)
cyclic,
2)
dihedral,
or1)
In this case,using
the fact that therepresentation px
isodd,
we concludethat it would be
reducible,
so it is coveredby
the above results.2)
In this case there exists a Cartansubgroup
CA
ofGL(2,lFa)
such thatGA
is contained in the normalizerNa
ofCA,
but not inCA.
Let px : G - be thecomposition:
.. - -- , -- , ,
The kernel of p x is then an open
subgroup
of G of index2,
so its fixed fieldKA
is aquadratic
field unramified outside IN. Weimpose: t
f
N .Suppose
thatKa
ramifies at t. Let~3
= (2t) -
a be the Dirichlet charactercorresponding
toKÀ,
with cx unramified outside N.We
have,
for every p~’
INAs in
[S73],
page17,
we will use these congruences to restrict thepossible
values of t. Theproblem
is that we need(2.1)
to hold also for theprimes
piN.
They
obviously
because ap = 0 for thesep
([AL70]).
The trick for
dealing
with the otherprime
factors of N is to raise the levelto reduce the situation
again
to the case ap = 0. Moreprecisely,
for aprime
p 11
N wereplace f
by :
f ’ =
anqn. Applying
theorem 3.64of
[S71]
we conclude thatf’
has levelplV. Repeating
theseprocedure
forevery such
prime
factor of N we see that we can suppose that:For every
prime
piN,
ap = 0 andp2 ~
N.
Therefore,
we can(and will)
assume that(2.1)
holds for every ~.Let 8 = be the derivation for mod t modular forms introduced
by
Serreand
Swinnerton-Dyer.
The existence of anoperator
satisfying
the sameproperties
in thegeneral
case of level N mod i modular forms wasproved
by
Katz in[K77] ,
provided
thate ~
N.Applying
0 to both sides of(2.1)
we obtain theequality,
as modular forms overFa :
for f t
N.Comparing
the filtration of both sides of thisequality
we concludethat t 3
(see ~573)).
Lemma 2.3.
Suppose
that A is such thatP(Ga)
is dihedral. Thenif ~
> 3and f t
N,
we have:I I , 11 " I
for
everyp f
£N;
where a is aquadratic
characterunramified
outside N.In
[R85]
it isproved
that this can holdonly
forfinitely
many i. For us,formula
(2.2)
will suffice in theexamples
to detect thefinitely
many inertprimes
that fall in this case.3. The
Examples
We will
apply
the last two results of section 1 to the realization ofprojec-tive linear groups over finite fields as Galois groups over
Q.
We aregoing
tostudy
twoexamples
of newforms without CMhaving
asingle
inner twist.Our
goal
is to check the conditions of theorem 1.4 for the inertprimes
inthe extension In our
examples
Ff
will be aquadratic
or a realcyclotomic field,
in order to obtain anexpression
of these inertprimes
in terms of congruences. In the firstexample,
the existence of the inner twistgiven by
an oddcharacter,
and the fact that the inertprimes
inFf
remaininert in will allow us to
apply corollary
2.2 withoutusing
the bound for the conductor of e.The same method could have been
applied
in the secondexample,
butonly
to those
primes
that remain inert inQf
-The
examples
below have beencomputed
with analgorithm implemented
by
W. Stein[St]
based on ideas of J. Cremona.3.1. First
Example. Computing
the characteristicpolynomials
of the Heckeoperators
we found inS2 ew (1024)
a newformf
withand
a2 = 0
(because
4N,
see[AL70])
and with thefollowing
coefficient,
given
by
their minimalpolynomials:
The level
being
a power of2,
iff
had CM it should begiven by
aquadratic
character a unramified outside 2 so that either
a(3) =
-1 ora(5) =
-1.But a3, a5 are both non-zero, so we have a contradiction.
The fact that in the case of a newform of level N with CM the
quadratic
character a is unramified outside N can be
proved
as in the case of thecharacters
giving
an inner twist(see
the discussion after theorem1.1),
andis immediate
taking
the definition of modular forms with CM in terms of Grossencharactersgiven
in[S71b] .
When we
computed
the characteristicpolynomial
of the Heckeoperator
T3
acting
onS2ew(1024)
we observed also that thepolynomial defining
a3appears in this characteristic
polynomial
withmultiplicity
2.Galois
conjugates
are each 2-dimensional.The list of coefficients of
f
suggests
thatf
has the inner twistgiven by
the involution Q EGal(Q¡/Q(ý’2))
and with XQ =X the mod 4 character
corresponding
toQ(i).
Infact, ifp ~
13,
a~
=X(p)ap,
and this fails for the other characters unramified outside2,
namely: ?p
corresponding
toand cp
toQ( yC2) .
We remark that
f
has at most one innertwist,
because:To prove the
equality:
10’
=X f
(we
equate
coefficients even for p = 2because a2 =
0)
we use the fact thatXf
ES2(1024),
because the conductorof X
divides1024,
the conductor off
(see
theorem 3.64 of[S71]).
Note that for the same reason alsoof
ES2(1024).
The coefficients a2jbeing
allequal
to0,
we have: =f
and =f .
It follows that bothX f
and of
are new of level1024,
because if not anotherapplication
of[S71],
theorem3.64,
wouldimply
that(or
is an oldform inS2 ( 1024),
contradicting
the aboveequalities.
The modular forms:
f ~,
have,
withrespect
to the action of the Heckeoperator
T3,
the commoneigenvalue
-a3, whosecorresponding eigenspace
in
S2ew(1024)
is 2-dimensional.Thus,
two of these modular forms mustbe
equal. Finally,
X(5)
=1,
~(5) _ -1
andas
= a5imply
that theonly
possible equality
isand this proves that
f
has a(single)
innertwist,
so thatFf
=Q(B/2).
Now we will
apply
theorem 1.4 to thisnewform,
but restricted to theinert
primes
in which are: 2 -3, 5
(mod 8).
We will show thatthey
remain inert in the extension
Qf /Q.
Let [ be one of theseprimes, A
I
t aprime
in 0 and A’ = a n R.We know that
Fy
=Ft2 ,
and that afterreducing
mod A’ thefollowing
holds:
Z!32
What we want to know is whether in the
quadratic
extensionQj
=Q(a3 )
over
Ff
=Q(a5)
the element a3 is inFJ2
ornot,
orequivalently,
whethera3
EF¡2
or not.Suppose
thata3
EF¡2.
Thentaking
norms withrespect
toFt2
IFt
we shouldhave:
N(a3)
= 8 EPi,
and then:(î)
= 1. This is false for theprimes t =-
3, 5
(mod 8),
therefore:IFl2
and then:(3.1)
forevery t
3,5
(mod 8),
t is inert inQf /Q.
We start
by checking
for which of theseGa
is irreducible of order amultiple
of
t,
with A aprime
in 0dividing
t.to use the bound for the conductor c nor to
compute
a coefficient ap for aprime
p(mod c) .
In thepresent
case, if therepresentation
werereducible the character e defined in section 2.1 would be unramified outside
2,
thus:with u ~
Z,u >
0.Take a
prime
p = 20131
(mod 2~),p ~ 20131?0 (mod ~)
(in
case u =1,
impose
also p == -1
(mod 4) ).
Then,
6(p)
= ±1 andcorollary
2.2implies:
Because of the inner
twist,
aP
=x(p)ap
=-aP so that ap E
Thus,
its minimal
polynomial
hasdegree
4.Using
the fact that both ap and a3 are square roots of elements ofFf
thatgenerate
the samefield,
we see that: ap = z. a3 , z EFf.
Replacing
in(3.2)
we obtain:The
prime t being
inert inQj ,
I and the minimalpolynomial
of a3having
discriminant a power of 2 we conclude
that,
afterreducing
modA,
a3 is aprimitive
elementof Fx
=Ft4,
and this contradicts(3.2’)
becausez-1
EFf.
So
far,
we haveproved:
Lemma 3.1. There exists a
newform f
ES2(1024)
with:such that
for
everyprime
f inert in(f
-3, 5
(mod 8),
and f remainsinert in
Qj )
and A E 0 theprime
dividing f,
therepresentation
pa(G)
isirreducible.
Let us now
compute
the inertprimes ~ >
5 such thatpa
has order notdivisible
by £, by
means of lemma 2.3 and formula(2.2).
Applying
thelemma we see that if the residual
representation
is in the dihedral case:ap *
a(p)ap
(mod A),
for everyp t
2t ,
with a aquadratic
characterunramified outside 2.
Independently
of which of the 3possible
charactersa is we know that either
a(3) =
-1 ora(7) _ -1 (observe
3,7).
So that for p = 3 or 7 we have:
ap - -ap and then ap m 0
(mod ~).
This is false for every inertprime £
because the norms of a3 and a7 arepowers of 2.
We have to find the inert
primes
falling
in case(3)
of section 2.2. Take£ > 3 an inert
prime
in andput
p = 3 in formula(2.2).
Because £ remains inert in we know that:is the minimal
polynomial
for a3 EFA
-This is
incompatible
with(2.2)
unless thispolynomial
agrees with:on
Fx;
but this is never the case. This concludes theproof
of thefollowing
Lemma 3.2. For the
newform
in lemma 3.1 it also holds:for every i
> 3 insert in the orderof p~(G) is
multiple oft;
where Ais the
prime
in 0dividing
~.Therefore,
itonly
remains to check the technical conditions(i),(ii)
and(iii)
on theseprimes
to conclude thatthey satisfy
thehypothesis
of theo-rem 1.4.Applying
the results of[Q98]
to deal with condition(i)
we see thatD,
thequaternion
divisionalgebra
overFf,
isgiven by
(~3, -1) .
The -1 comesfrom the fact that the inner twist is
given by
thequadratic
charactercor-responding
toQ(i).
Wehave a 2
= 4 + so thatusing
Hilbertsymbols
we see that condition(i)
for an inertprime t
isequivalent
to the existence of a non-trivial solution of theequation:
But this
equation
has theglobal
solution(~, y, z) _ (l,1 +
.J2, 1),
so that(i)
holds for every ~. Infact,
in this case D is itself a matrixalgebra
(be-cause noprime
ramifies):
D ’7--M(2,~).
Condition
(ii)
is easy to check because there is asingle
inner twist. Fromthe definition of the elements
a(g)
we see that we can take asa(g)
for9 tt
H the coefficient a3.Then a
prime A
E (~ over aprime t
for which(ii)
fails must divide a3, but this isimpossible
for an inertprime.
Finally,
it remains to check condition(iii).
We have to choose v such that Frob EH, i.e.,
Q(i)
must be fixedby
Frob v.Thus,
we can take as v anyodd
prime
thatdecomposes
in~(i),
i.e.,
such that = 1 orequivalently:
v - 1(mod 4).
Choosing v
= 5 the other conditions of(iii)
are satisfied because5 t N
=1024 and
a 2=
6 +generates
Besides,
forevery l =1=
5 inert ina2
generates Rt
as aZealgebra,
so that(iii)
is satisfied for theseprimes.
To rescue the
prime t
= 5 observe that we can also take v = 13 becausea13
isequal
orconjugate
toa2
From this discussion and the
previous
twolemmas,
we canapply
theo-rem 1.4 and conclude:such that
for
-3, 5
(mod 8),
2 >3,
theimage Ge
thesubgroup of
GL(2,
generated by
the groupAt
defined
in section 1 and the matrix:/- n B
Now we
apply proposition
1.5 to theseprimes.
For 9 ft H
(there
isonly
one
coset)
we have takena(g)
=a3. In the
proof
of formula(3.1)
it wasshown that
~2
so thatproposition
1.5gives:
Theorem 3.4. For the
newform of
theprevious theorem,
for
every t =3, 5
(mod 8),
~ > 3 and A E0, À
~,
we have:Corollary
3.5. Forevery £
=3,5
(mod
8),
t >3,
PGL(2, IFez ) is
aGalois group over
Q.
3.2. Second
Example. Looking
in the spaceS2 e"’ ( 1331 )
we found anewform
f
with Qj
aquadratic
extension of the realcyclotomic
fieldLl
=Q((
+ (-1),
where (
is aprimitive
11-th root ofunity.
We will prove below that
Ff
=L11
(and
this isprecisely
the field where wewanted to
work),
the choice of the level 1331= 113
was motivatedby
thefollowing
result of Brumer([B95]):
Theorem 3.6. Let
f E
S2(N)
be anewform,
without CM.Suppose
thatand (
aprimitive pSp-root
of
unity.
Remark. This theorem also
helped
us to find theexample
in section 3.1. The coefficient a2 isgiven by
thepolynomial:
and the
following
holds:Q(a 2)
=L11,
in fact:And a3 =
((
+~-1) ,
so that a3 ELll.
We also know that all = 0 because
112 ~
N
(see
[AL70]).
If
f
hadCM,
it should begiven by
the charactercorresponding
tobut this is ruled out
by
the fact that:a2 #
0,
2
-1.The
computation
of the characteristicpolynomial
of the Heckeoperator
T2
on the spaceS2eT"(1331)
shows that the factorcorresponding
to a2 is asimple
factor. As in the section3.1,
we prove that ESZ eT" ( 1331)
and if we call Q the involution inGal(Q¡ / LI1),
we observe that the modularforms:
f a
andw f
have the commoneigenvalue
-a2 withrespect
to theTherefore, f
has an inner twist and weeasily
see that it isunique,
so that:=
Ll1.
oThe
application
of theorem 1.4 to this newformf
is as in section3.1,
except
for the fact that we willapply corollary
2.2 in its fullstrength.
Recall that we are
only
interested in the inertprimes
in which arethe
primitive
roots mod 11 and their squares, i.e:When
applying
corollary
2.2 to deal with the reducible case, the characterE must be unramified outside
11,
provided 1
>2,
and its conductor c mustverify:
c2 ~
1
1331,
thus c1
11. So the values of E are 10-roots ofunity,
contained in
Filo -
We know that if 101 ilo -
1 then 10I £2 -
1,
so that theimage
of E is contained inThen, applying
the formula incorollary
2.2 to p = 2 we conclude that U2 EF12.
But we also know that the residue class
degree
of £in Qj
is 5 or10 ;
a2generates Qj
and the discriminant of the minimalpolynomial
of a2(
=2~ . 118 -
199)
is not divisibleby
i.Therefore a2 has
degree
at least 5 over and we have a contradiction.We conclude that the
representations
are irreducible for every odd i inertin
Let us now treat the dihedral case. If t > 3 is in this case, we have:
for every p
f
l ll .Taking
p = 2 we havew(p) _ -1
and then Aa2 .
The norm of a2over Q
is-199,
so that weget A
199,
and 199 is not inert inF
The next
step
is theapplication
of formula(2.2)
to find the odd inertprimes
such that case(3)
of section 2.2 holds.Applying
formula(2.2)
to p =2,
from the fact thata2 mod A has a minimal
polynomial
ofdegree
atleast
5,
we obtain a contradiction.Therefore,
forevery ~
> 3 inert in and A E 0dividing i,
we concludethat the order is divisible
by
E.It remains to check conditions
(i), (ii)
and(iii)
in order toapply
theo-rem 1.4. In condition
(i)
thequaternion algebra
D isgiven by
(see
[Q98]):
(a2,
-11).
It is known that condition(i)
is satisfied if t is such that botha2 and -11
are units inF¡,À’
(here
A’ = AnR).
But from theequation
of
a2
and the fact that i is inert inFf,
we conclude that this holds for allthese E.
In condition
(ii),
let us take asa(g) for 9 fj.
H the coefficientsa2.Then
con-dition
(ii)
holds forevery i
inert inFf,
because none of these can divide a2.Finally,
for condition(iii)
we take v = 3 because:(-31)
=1,
and then Frob 3 fixesa3
= 2 +(~
+(-2
generates
Ff
overQ.
Moreover,
for 3 inert ina3
generates
Re
as aZtooalgebra.
Applying
theorem 1.4 we obtainTheorem 3.7. There is a
newform f
in52(1331)
with~f
aquadratic
extension
of
(C
aprimitive
11-th rootof
unity)
andFf
=£11
such thatfor every 10
:1:1(mod 11),
1
2, 3,11,
theimage
GI
of
pt is thesubgroup of
GL(2, Ot)
generated by
the groupAt defined
in section 1 and the matrix:To
apply
proposition
1.5 observe thatFf
is adegree
5 extensionof Q ,
so that r = 5 is odd and then we haveTheorem 3.8. For the
newform of
theprevious
theorem, for
everyt 0
:f:1(mod 11),
~ ~
2, 3,11
and A E0,
1£,
we have:Corollary
3.9. For(mod
11),
t =,4
2, 3,11,
PGL(2,
a Galois group over
Q.
4.
Concluding
RemarksThe conditions of 1.4 can be
effectively
verified for any newform withoutCM
having
asingle
inner twistusing
the results of section 2.The method used to deal with the reducible case in the first
example only
works if we have an odd inner
twist,
andapplies
only
to thoseprimes
inert but it has the
advantage
ofavoiding
thecomputation
of acoefficient ap with p as described after
corollary
2.2(p
grows as a functionof q
andw).
Though
these conditions may seemhighly
restrictive,
we have found otherexamples
verifying
them. Forinstance,
there is a newformf
ESZ (4096)
whose first coefficients have minimal
polynomials
(a2
=0):
This newform has an inner
twist,
given
by
the mod 4 character X, so thatFf
=Q(a5) =
~( 2 -i- ~).
Let us consider theprimes £
with residueclass
degree
2 ini.e., i =-
7,9
(mod 16).
It can be shown that the2
places
ofFf
lying
above £ are inert inQf IFf.
The verification of theconditions of theorem 1.4 for these
primes
isexactly
as in section3.1,
anddiscriminants,
norms and resultantsinvolving only
the coefficients a3 anda5 . It follows that for every
.~ ._--_ 7, 9
(mod 16),
and A EC~, ~
( .~,
This, together
withcorollary
3.5,
proves thatPGL(2, IFe2 ) is
a Galois groupover Q ,
for every(mod 16), l
> 3 .The same conditions are also satisfied
by
thefollowing
twoexamples:
of
ESZ(192)
with an inner twist andFf
=Applying
our methodto this
newform,
we find noexceptional
inertprime
so that the groupsPGL(2,
are Galois groupsover Q
forevery i =-
2,3
(mod 5), ~
> 3.9
f E
S2(333)
with an inner twist andFf
=Q(Jl) .
Applying
our methodto this newform and
combining
with the result obtainedusing
theexamples
of level 1024 and 4096 we conclude that the groupsPGL(2,
are Galoisgroup
over Q ,
forevery t $
±1(mod 48), 1
> 3 . References[AL70] A. ATKIN, J. LEHNER, Hecke operators on Math. Ann. 185 (1970), 134-160.
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Luis V. DIEULEFAIT
Dept. d’Algebra i Geometria Universitat de Barcelona
Gran Via de les Corts Catalanes 585 08007 - Barcelona
Spain