The cuspidal torsion packet
onhyperelliptic
Fermat quotients
par DAVID GRANT et DELPHY SHAULIS
RÉSUMÉ. 7 un nombre premier, soit C la courbe projec-
tive lisse définie
sur Q
par le modèle affiney(1- y)
= soit ~ le point à l’infini de ce modèle deC,
soit J lajacobienne
de Cet
soit~ :
C ~ J lemorphisme
d’Abel-Jacobi associé à ~.Soit Q
uneclôture
algébrique
deQ.
Noustraitons
ici un cas non couvert dans[12],
en montrant que~(C)
nJtors(Q)
estcomposé
del’image
par~
des points de Weierstrass de C ainsi que les points(x, y) = (0,0)
et
(0, 1)
de C. Ici, Jtorsdésigne
les points de torsion de J.ABSTRACT. 7 be a prime, C be the
non-singular
projec-tive curve defined
over Q by
the affine modely (1- y) =
~the point of C at
infinity
on thismodel,
J the Jacobian ofC, and
~:
C ~ J the albaneseembedding
with ~ as base point.Let Q
be an
algebraic
closure ofQ. Taking
care of a case not covered in[12],
we show that consistsonly
of the image under~
of the Weierstrass points of C and the points(x, y) = (0, 0)
and(0, 1),
where Jtors denotes the torsion points of J.1. Introduction
En route to
calculating
thecuspidal
torsionpackets
on Fermat curves,Coleman, Tamagawa,
and Tzermias studied thecuspidal
torsionpackets
on the
quotients
of Fermat curves[12]. Specifically,
let £ be an oddprime,
and for any 1 C a .~ -
2,
letCa
be thenon-singular projective
curvedefined
over Q by
the afhnemodel xl =
which is a curve of genus g =(~20131)/2.
Let oo denote the lonepoint
onCa
which is atinfinity
on thismodel, and 0 : Ca ~ Ja
the Albaneseembedding
ofCa
into its JacobianJa
with oo as base
point. By definition,
thecuspidal
torsionpacket Ta
is theset of torsion
points
ofJa
which lie ono(Ca).
LetRo
andR1 respectively
denote the
points (x, y) = (o, 0)
and(o,1 )
on C. Thefollowing,
and moregeneral results,
wereproved
in[12].
Manuscrit regu le 14 février 2003.
Theorem
(Coleman, Tamagawa, Tzermias). 11,
then there is an a, with 1 a ~ -2,
such thatCa
is nothyperelliptic,
and such thatThis leaves open the
question
ofdetermining Ta
for any fixed aand
and in
particular,
for the a for whichCa
ishyperelliptic,
which we addresshere. It is known that
Ca
ishyperelliptic precisely
when a =1, (~ -1)~2,
or £ - 2. There are
isomorphisms
fromC(i-I)/2
andCt-2
toCi,
whichinduce
bijections
fromT(i-I)/2
andT,~_2
toTl,
so we will lose nogenerality
in
concentrating
onCl.
Let C =Cl,
J =Jl,
and T =Tl.
Note that on Jthe
cuspidal
torsionpacket
coincides with thehyperelliptic
torsionpacket.
Let (
denote aprimitive
ofunity, let,
be a fixedtth-root
of1/4,
and let
Wi,
1 i~ ~
denote thepoints 1/2)
on C.Theorem 1.1. 7. Then
Theorem 1.1 is due to Shaulis
[20].
Theexposition
hereincorporates
some
simplifications
of theoriginal argument.
The case = 5 does not ex-actly
fit thepattern
of Theorem1.1,
and in that case T isgiven explicitly
in[9]
and[5]. Indeed, Proposition 4.3(d) depends crucially
on theassumption that g
> 3.The
proof
here owesheavily
to the work of[12],
but differs in somekey respects.
Inparticular,
it does notrely
on thep-adic integration theory
ofColeman,
which he used in[10]
to show thatTa
consists ofjust £-power
torsion if
Ca
is nothyperelliptic 11,
and that T consists of(2t)-
power torsion for . > 5. Our
proof
combinesexplicit geometry
with theaction of Galois groups on torsion
points,
anapproach
to Manin-Mumfordproblems
that goes back toLang [17],
and has been usedby
many authors since. We refer the reader to the survey article of Tzermias for thehistory
of the work on the Manin-Mumford
Conjecture [23],
and mentiononly
somemore recent work
~2J, [3], [4], [6], [7], [19], [22].
For results on the torsion ofJa
that lies on a thetadivisor,
see~1J, ~13J,
and[21].
The next two sections of the paper are
preliminary.
In the first we detailwhat we need about the
explicit geometry
of C andJ,
and in the secondwe derive the necessary results on Galois actions on torsion
points
fromthe
theory
ofcomplex multiplication.
Theproof
of Theorem 1.1 isgiven
insection 4.
We would like to thank the referee for a careful
reading
of this paper anda number of useful
suggestions.
2.
Geometry
of C and JWe assume that > 7. Let K =
Q((),
let K be analgebraic
closureof
K,
and let O = denote thering
ofintegers
of K. Let À = 1 -~,
so tO = Note that
Gal(K/Q)
consists of theautomorphisms
Qi,i E
(71/f71)*,
such that(i.
The
automorphism ~ : (x, y)
-y)
of C extends to anautomorphism
of
J,
so we can endow J withcomplex multiplication by
Oby defining
anembedding
p : (~ ~End(J)
such thatp(() =
E. Sincexidx/y,
0 ig -
forms a basis for the
holomorphic
differentials ofC,
weget immediately (and
it is wellknown)
that theCM-type
of J is ~ _~Ql, ..., ~9~.
We write[a]
forp(a),
and let + denote the groupmorphism
on J. For any a E 0we let
J[a]
denote the kernel of[a]
inJ(K)
andJ[a°°]
= andfor any ideal a C
(7,
we letJ[a] =
Weidentify points
on J withthe
corresponding
divisor class inPic°(C).
Let 0 denote theorigin
on J.The
hyperelliptic
involution t : C ~ C isgiven by ~((x, y)) _ (~,1 - y),
and its fixed
points
are the2g + 2
Weierstrasspoints
oo andWi,
1 i ~.The
following
results are standard.Lemma 2.1.
a)
For any P onC, ~-1~~(P) _ 0(t(P)),
so[-1]*
pre-serves
~(C).
b)
For any P onC, [(]O(P) = ~(~’(P)), so [(]*
preservesO(C).
c) Every point
on J isrepresented by
aunique
divisorof
theform for
sorrte 0 r g,and Pi
onC,
1 ~ i _ _ r,where Pi 54
oo, andd)
Thepoints
inJ[2]
areprecisely
thoserepresented by
a divisorof
theform (1),
where thePi,
1 i r, are distinct Weierstrasspoints.
Lemma 2.1
immediately gives
thefollowing.
Lemma 2.2. Let
Q
E J berepresented by
a divisor as in(1),
and letdenote the
image of
under thetranslation-by-Q
map. Then isempty if r > 3,
islo(Pl), ~(P2)~ if r
=2,
and isOf ifr=1.
Indeed,
if~(P)
=Q
+Ø( PI),
then P +i (P’) -
20o andPi
+ ...+ Pr -
roorepresent
the samepoint
inJ,
whichby
Lemma 2.1 isimpossible
for r >3,
which
implies
that P isPI
orP2
for r =2,
and that P isPl
or 0 for r = 1.Recall that a
point Q
EJ(K)
is called almost rational overK,
in thesense of Ribet
[4], ~19~,
if wheneverT(Q)
+v (Q) = [2]Q
for some EGal(K/K),
then we haveT(Q)
=v(Q)
=Q.
Weget
thefollowing
fromLemma 2.1 and the fact that oo is K-rational.
Lemma 2.3.
a)
The curveO(C),
and inparticular
the torsionpacket T,
arepreserved
under the actionof Gal(K/K).
b) Every Q
E§(C) - J[2]
is almost rational over K.We will need the
following,
which isProposition
C in[6].
Lemma 2.4. Let k be a
perfect field,
G be a commutativealgebraic
groupover
k,
and Q be afinite
setof primes.
LetGn
denote the torsionpoints of
G whose order is divisible
only by prirraes
inS2,
denote the subsetof
almost rational torsion
points of
G over k inGn.
For an oddprime
p, set6(p)
=1,
and setb(2)
= 2.Define A
Letki
=k (G [A]),
andsuppose that there exist an
integer B,
divisibleonly by primes
inS2,
suchthat
G(kl)
9G[B].
Then E CG[B].
Since the orders of the
poles
of xand y
at oo arerespectively
2and
we see that for P on
C, §(P)
EJ[l] - 0
if andonly
if there is a function onC of the form y +
f (x),
wheref
is apolynomial
ofdegree
at most g, whose divisor is£(P - oo).
If this is the case, then the divisor of1 - y
+ isl( ¿( P) - oo),
socomparing
divisors andmatching
up coefficients ofxl gives
Lemma 2.5.
#(¢(C)
n(j[t] - O)) ~
+ 2.Proof.
Theonly points
P on C withx(P)
= 0 areRo
andRl,
and con-sidering
the divisorsof y
and 1 - y shows thatJ~~~ -
O.Furthermore,
if((z, w))
EJ[l] -
0with z # 0,
weget
that((z, w))
aredistinct
points
inJ(~~ -
0 for i E So it suffices to show that thereare not two
points (z, w)
and(r, s)
onC,
with 0r~,
suchthat
~((z, w)), ~((r, s))
EJ[~] -
0.If such
(z, w)
and(r, s) exist,
then there arepolynomials f
and h ofdegree
atmost g
such thatSimplifying
thesegives
Substituting x
= zX anddividing by z~
in the formerequation
of(2),
andsubstituting x
= rX anddividing by r’
in the latterequation
of(2), gives
two
expressions
for(X - 1)~. Equating
theseyields
Letting
u and v be square roots ofzf
andr’, (3)
can be rewritten as(4)
By assumption ze - rl =f 0,
so weget
from(4)
thatu(2h(rX)
+1) ~ v(2 f (zX)
+1)
are both constantpolynomials,
hence thatf (x)
andhex)
are both constants. But
by (2)
this violates theassumption
thatzr ~
0. 03. Galois actions on torsion
points
Let H be the Hilbert class field of K. Recall that
by using cyclotomic units,
one can see that everyprincipal
ideal of 0prime
to(A)
has a gen-erator
congruent
to 1 modÀ2. Therefore, by
Artinreciprocity,
H is alsothe ray class field of K of conductor
(À)2 (so
also the ray class field of K of conductor(a)).
Let E =
K(7)
and L = Recall that C hasgood
reduction atevery rational
prime ~,
and it is shown in[11]
thatC,
and henceJ,
achieves
everywhere good
reduction over L. For any extension of number fieldsFl/F2,
letD(Fl/F2)
denote the discriminant ofFlIF2,
andNF’
denote the norm from
Fl
toF2.
IfFl/F2
is a Galoisextension,
letdenote the Galois group of
Fl/F2. Finally,
ifFlIF2
is an abelianextension,
let
f(Fl/F2)
denote the conductor ofLemma 3.1. We have 2.
Proof.
This is a well-knownargument
that webriefly
recall. Theanalogue
to
elliptic
curves withcomplex multiplication plays
akey
role in theproof
of the Coates-Wiles theorem
(see [8], [15]).
If
E/K
is unramified over(À)
there isnothing
to prove, so suppose it is ramified. Then it istotally
ramified over(.~),
soQ(7)/Q
istotally
ramifiedover t. Hence ~. But for any r E
Q,
thepolynomial
discriminant of r is so t.
Calculating
the discriminant of
E/Q
two ways shows thatThe order at t of the
right-hand
side is(t - 2)
+f(i - 1),
so the order at i ofN~ (D(E/K))
is2(t - 1).
Henceord(À)(D(E/K))
=2(f -1).
Since all non-trivial characters onGEIK
havethe same
conductor,
the conductor-discriminant formulaapplied
toE/K
now
gives
the result. 0Lemma 3.2. For any m and n
coprime
inZ, H(J[moo])
andare
linearly disjoint
over H.Proof.
Let M =By
thetheory
ofcomplex multipli- cation,
M is abelian overK,
andby
the criterion ofNéron-Ogg-Shafarevich,
since m and n are
coprime, M / K
is ramifiedonly
over(~),
soM / H
is to-tally
ramified at everyprime
over(A).
For the same reason,ML/L
isunramified,
so the ramification index over K of anyprime
of ML above(A)
divides[L : K]
= 2~. Hence[M : H]
divides 2~. Ourgoal
is to show[M:H]=1.
Suppose
for a contradiction that2~~M : H],
soM/H
has aquadratic
subextension
M’/H.
Then since M’ istotally
ramified at everyprime
ofH above
(A), D(M’ / H) = (A).
SinceH/K
isunramified,
if h is the classnumber of
K, D(M’ / K) = (A)h.
So the conductor-discriminant formulaapplied
toM’/K gives
thatwhere
f x
is the conductor of the fixed fieldof X
over K. Notethat f x
istrivial if and
only
if the fixed fieldof X
is contained inH,
whichhappens
if and
only if X
factorsthrough Therefore fx
isnon-trivial for
precisely h
=[H : K]
choices of x, andby (5),
for each ofthese, fx
=(.~).
Hence the conductor ofM’ / K
is also(.~).
Since the rayclass field of K of conductor
(A)
isH,
no such M’exists,
and[M : H]
divides ~.
Finally
suppose for a contradiction that[M : H]
= ~. Then sinceML/L
is
unramified,
the same must be true ofME/E.
Therefore ME =EI,
where I is the inertia subfield in ME of any
prime
of ME above(A).
Since
(A)
does not divide the formula for conductors ofcomposite
extensions and Lemma 3.1
imply
that 2. Hence M iscontained in the ray class field of K of conductor
(.~)2,
which is H. 0If p is a rational
prime,
for every x in thep-adic integers Zp
and u Ewe define
~x)u
=[xn]u,
where zn E Z is such modPn.
Lemma 3.3. Let
p 0 1
be a rationalprime.
Thenfor
every x E there is a 7~ EGal(K/H)
such[x]u for
every u EProof.
Since J hasgood
reduction at p, it follows from Theorem 2.8 ofChapter
4 of[18]
that isisomorphic
to whereOp
is the inverse limit as n goes toinfinity
ofO/pn,
andNp,
I denotes thereflex norm
(that is,
for any a EK, N4,,(a) =
If c =~’ ~- ~-1,
then is thering
ofintegers
ofKo
=K(e),
and since $ is aset of
representatives
for the orbits of under the action ofcomplex conjugation,
the reflex norm restricted toKo
isjust
the normNKO
ThenN§°
induces a norm map N :Z[e]J
so it suffices to show that there is a y E such that x =N(y).
But via the Chinese RemainderTheorem,
if pl,...,pr are the
primes
ofKo
above p, then we have anisomorphism
: - where is the valuation
ring
in(Ko)p2,
thecompletion of Ko
at pi.Furthermore, (Ko) p, /Qp
is an unramified extension ofcomplete
local fields and x is a unit inZp,
so there is a z E whosenorm from
(Ko)pl
toQp
is x. We can then take y such that(z,1, ...,1),
and we will haveN(y)
= x. DLemma 3.4. There is a T E
Gal(K/K)
such thatT(u) = [1
+f]u for
allu E
J[tM].
Indeed,
it is stated in theproof
ofProposition
2 in[12]
that there is sucha T e
Gal(K /Q),
but the construction there shows that T fixes K. It is also stated in[12] only
for ~ >11,
but the sameproof
holds for £ = 7.4. Proof of Theorem 1.1
Suppose t
E T. If t =0,
then t =~(oo),
so from now on we assumet
~
0. We canuniquely
writewhere t2 E
J(2°°~, ta E J~~°°~,
and t’ is of orderprime
to 2A.Theorem 1.1 will follow
directly
fromPropositions 4.1, 4.2,
and 4.3.Proposition
4.1. We have t’ = O.Proof.
Let t’ have order r.By applying
Lemmas 3.2 and 3.3 to everyprime dividing
r, there exist a T EGal(K~H)
such~2~t~. Furthermore, applying
Lemma 3.2 with m = 2t and n = r, we can further assume that Tfixes t2 and
ta.
Hence12(t) + t + t = let) + let) + I( t),
andt,,r (t), -r2(t)
areall in
O(C) by
Lemma 2.3.By
Lemma2.1,
eitherr(t)
EJ~2~,
or t =T(t).
In either
case, t’
= O. 0Proposition
4.2.0,
=0,
and t = t2 =for
some1 ~ i ~ £.
Proof.
Sincet2 ~ 0,
we claim that there is a T EGal(K/H)
such thatT(t2)
= t2 + v, for some v EJ [2] -
O.Indeed,
let 2 n be the smallest power of 2 such that[2 n] t2
=0,
so n > 1.If n > 2, from Lemma 3.3
we can let r be an element such that
T(t2) _ [1
~-2~"~]~2* Suppose
nowthat n = 1. Lemma 2.1 shows that
H(J[2])
=H(y),
which is a non-trivial extension of H since(2)
is unramified in H. is any non-trivial element in thenj3(t2) =
for some1 :::; j:::; £ -1.
Hence in thiscase we can take T =
j3.
From Lemma
3.2,
we can further assume that T fixest)...
ThenT(t) =
t+v,
soT(t)
Eo (c) n o (c) v -
Lemmas 2.1 and 2.2 nowimply
that t =~(Wi)
for some 1 i ~. D
Proposition
4.3. then t= ta
and, , i - - , , - - - -
Proof. a) Suppose
y EJ(K(J[f]))
nJ[too]. By
Lemma3.4,
there is aT E
Gal(K/K)
such that-r(x) = [1
+t]x
for every x EJ~~°°~.
HenceT fixes
K(J~~~),
soT(y)
= y. Therefore y EJ [t] -
b)
Lemma 2.3 shows that t EJ[t’]
is almost rational over K.Taking
Q = G =
J,
and k = K in Lemma2.4,
thenpart (a) gives
thatt E
J[f].
c) Making
use of results ofGreenberg [14]
and Kurihara~16~,
Lemma 2of
[12] gives
that if t EJ[~] - J~~3~,
then#(§(C)
nJ[~]) ~ ~2.
Theproof
there is stated for ~ > 11 but worksequally
well for f = 7. This violates Lemma2.5,
so t EJ~~3~.
d)
This isProposition
4 of[12]
for ~ >11,
and in thehyperelliptic
case is easy to do
directly
forany ~ >
7.Indeed,
if t EJ~~3~,
then[(1
+~) (1 - ~)3~t
= O.Multiplying
this outgives
t +~~3~t
+~~3~t
=+ +
(~4~t,
soby
Lemma2.1,
either(~3~t
EJ[2], t
=~-~3~t,
t =
[(]t,
or t =[(4]t.
The first twopossibilities give
acontradiction,
and the other
possibilities imply
that t EJ[A] .
e)
Note thatJ[A]
=[i]O(Rl)l.
The result now followsfrom Lemma 2.1.
D
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David GRANT
Department of Mathematics
University of Colorado at Boulder Boulder, CO 80309-0395 USA E-maid : grant0boulder. colorado. edu
Delphy SHAULIS
Department of Mathematics
University of Colorado at Boulder Boulder, CO 80309-0395 USA
E-mail : shaulis~euclid. colorado. edu