C OMPOSITIO M ATHEMATICA
R ALPH G REENBERG
On the jacobian variety of some algebraic curves
Compositio Mathematica, tome 42, n
o3 (1980), p. 345-359
<http://www.numdam.org/item?id=CM_1980__42_3_345_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
345
ON THE JACOBIAN VARIETY OF SOME ALGEBRAIC CURVES
Ralph Greenberg*
@ 1981 Sijthoff & Noordhoff International Publishers - Alphen aan den Rijn
Printed in the Netherlands
1. Introduction
Let p
be aprime
and let F be a field of characteristic different from p. We will assumethroughout
this paperthat p ~
5 and that F contains aprimitive p-th
root ofunity.
For anyinteger a
such that1 :5 a ::5 p -
2,
letJa
denote the Jacobianvariety
of the curveyP
=xa(1- x)
over F. In this paper we will prove several results about thepoints
of p-power order onJa.
Our mostinteresting
results willconcern the case when F is
Q(p1)
orQp(p1),
whereQ
denotes the rational numbers andQp
thep-adic
numbers. However, we willbegin by stating
twogeneral
results. We letF
denote a fixedalgebraic
closure of F.
THEOREM 1: The group
Ja(F) of
F-rationalpoints
onJa
contains asubgroup isomorphic
to(Z/(p ))3.
THEOREM 2: Let
F(a)~
denote thefield generated by
allpoints of
p -power order on
Ja(F).
ThenGal(F(a)~/F)
~Zdap,
where 0 ~da ~ p+1 2
and
Zp
denotes the additive groupof p -adic integers.
We have a number of remarks to make about these theorems. If F is a finite field of characteristic
~(~p),
then the fact thatJa(F)
contains asubgroup
of orderp3
was noticedby
Iwasawa. It followsquite simply
from the fact that the roots of the zeta function ofyP
= xa(1- x)
are certain Jacobi sums over Fby using
anelementary
congruence for these Jacobi sums which Iwasawa proves in
[6].
Our* Supported in part by a National Science Foundation grant.
0010-437X/81/03345-15$00.20/0
proof
of theorem 1provides another,
somewhat moreconceptual proof
of this congruence.By
a similarapproach,
congruences for other Jacobi sums can also be derived.Also,
as our arguments willshow,
the result thatGal(F(a)~/F)
is torsion-free isquite closely
relatedto Iwasawa’s congruence.
The value of
da depends
on the nature of the field F. LetÊ
denote thecomposite
of allZp-extensions (contained
inF)
of F. Theorem 2of course shows that
F(a)~
CF.
If F is either a finite field or a finite extension of the fieldQe
of e-adicnumbers,
thenGal(F/F) ~ Zp
andone can see
easily
that we must haveF(a)~
=Ê
and henceda
= 1 in thiscase. On the other
hand,
if F =Qp(p1),
then it follows from local class fieldtheory
thatGal(F/F) ~ ZP
and thereforeF(a)~
is consider-ably
smaller thanF.
The mostinteresting
case to consider is when F is the field K =Q(p1).
Then one can showby
class fieldtheory (using Leopoldt’s conjecture
which is valid forK)
thatGal(K/K) ~ Z(p+1)/2p. (See [2]
and[4].)
It turns out that"usually" K(a)~
=K.
How- ever, for certain values of a and p,K!:)
is a proper subfield ofK.
Thisoccurs for
example
whenever p ~ 1(mod 3)
and a is either of the two solutions to the congruencex2 + x + 1 ~ 0 (mod p). (This
can beexplained by
the factthat,
for these values of a,Ja
is not asimple
abelianvariety.
See[8].)
There are also other cases whereK(a)~ ~ K
which wewill describe at the end of Section 3.
In contrast, it is conceivable that the group
Ja(K)
never containsmore
than p3 points
of p -power order. In section5,
we willgive
anecessary and sufficient condition
(involving cyclotomic units)
for thep-primary subgroup
ofJa(K)
to bejust (Z/(p))3.
Our condition is satisfiedif p
does not divide the class number of the maximal real subfield of K.Despite
extensivecalculations,
there are noknown p
that do divide this class number.
(See [7]
and[10].)
Our
approach
toproving
the above theorems is tostudy
thep -adic representation
ofGal(F/F)
on the Tate modulefor la
for theprime
p,using
the factthat Ja
is ofCM-type.
In the case F =K,
thisp-adic representation
can be describedexplicitly
in terms of certain Jacobisums. Our final results
depend
on thisexplicit description
of theabove
p-adic representation. By following
some of the arguments in Iwasawa’s paper[6] (which depend
onStickelberger’s
theorem on thefactorization of Jacobi
sums),
we can determineprecisely
the fieldgenerated by points
of order p onJa
overQp
and overQ. (For Q,
wealso need to use an
explicit reciprocity
lawproved by
Artin andHasse.)
Forsimplicity,
we willjust
state our results for J =03A0p-1a=1 Ja.
LetL and
Lp
denote the fieldsgenerated
overQ
andQp respectively by
the
points
of order p on J. Then it is well-known that K C L and thatKp
=Qp(p1),
thecompletion
of K at theprime
over p. It turns outthat
Gal(L/K)
andGal(Lp/Kp )
are bothelementary
abelian p-groups and therefore one can consider them asrepresentation
spaces overZ/(p)
for thegroup 0394
=Gal(K/Q) ~ Gal(Kp/Qp). (If
x EGal(L/K)
orGal(Lp/Kp)
and 5E L1,
the action of 8 is x ~03B4x03B4-1,
where9
is anyautomorphism
of L orLp extending 8.)
The irreducible represen- tations of 0394 overZ/(p)
are one-dimensional andcorrespond
to thepowers
03C9i(i
=1, ...,
p -1)
of thecharacter 03C9:0394 ~ (Z/(p))x
definedby w(5)
= cmod p if 03B4(03B6p) = 03B6cp
for aprimitive p -th
root ofunity Cp
inK(and Kp).
If V is anyrepresentation
space for d overZ/(p),
we letWe will also
regard cv
and its powers as characters of 0394 with values inZ’ by using
the canonicalisomorphism
of(ZI(p»’
intoZ’.
We canthen define
Vi
whenever V is aZp-module
on which 0394 acts. We can now state our main results.THEOREM 3: Let
i + j = p.
ThenGal(Lp/Kp)i fOl
orZ/(p).
It isisomorphic
toZ/(p) if
andonly if
i isodd, ~ 1,
and p does not divide the numeratorof
thej-th
Bernoulli numberBj.
THEOREM 4: Let C denote the group
of cyclotomic
units in themaximal real
subfield
K+of
K. Then L =K({p~|~
ECI).
We want to add a few remarks
concerning
these theorems. First ofall,
it is not hard to see from local class fieldtheory
that theorem 3actually
does determineLp uniquely.
One needs to knowonly
theso-called "indices of
irregularity." Also,
theorem 4implies
im-mediately
thatGal(L/K) ~ (Z/(p))t
wheret ~ p-3 2,
since this is the rank of C and the torsionsubgroup
of C has orderprime
to p. If we let E+ denote the full unit group ofK+,
then it is well-known that[E+ : C]
is
equal
to the class number h+ of K+ and from this one finds that t= p 2 3 p recisel y
whenp
h+. Now the action of 0394 on C is suchthat
(C/Cp)j
isisomorphic
toZ/(p) when j
is even but 0 p - 1 and is otherwise trivial. Thus one can find a set of generatorsfqjl, where j
iseven and 2
~ j ~
p -3,
such thatqjCP
generates(C/ CP )j.
It is not hardto show that à acts on
Gal(K(p~j)/K) by
thecharacter (,)’ (where
i
+ j
=p).
ThusGal(L/K);
isisomorphic
toZ/(p)
when~j~ (Kx)p
andis trivial otherwise. If we let
(E+/C)p
denote thep-primary subgroup
ofE+/C, then 71j
is ap-th
power in K’ if andonly
if thewi-th
component
(E+ / C)p,j
is non-trivial. Thus the structure of(E+/C)p
as a0394-module
completely
determines the structure ofGal(L/K).
The fieldL itself is
actually
determinedalso,
as we can seeby using
the factthat L C
K.
We also want to
point
out that theorem 3 could be derived from theorem 4by using
the classical resultthat qj
is ap-th
power inKp
ifand
only
if p divides the numeratorof Bj together
with the obser- vation thatLp
is thecompletion
of L for anyprime dividing
p.However,
we willgive
asimpler,
directproof
of theorem 3.2. Generalities
Let Ep denote a fixed
primitive p -th
root ofunity
in F. Webegin by considering
the curveyP
=f (x),
wheref (x)
is a non-constantpoly-
nomial over F whose irreducible factors occur withmultiplicity
notdivisible
by
p. In much of thissection,
we won’t need to assume that p 0 2 or 3. We willstudy
the divisor classes on the above curve fromtwo
points
ofview - namely by using
genustheory
for thecyclic
extension
F(x, y)/F(x)
and alsoby using
the fact that the Jacobianvariety Jp
of the above curve has A =Z[Cp] ]
as aring
of endomor-phisms,
whereep
is a fixedprimitive p -th
root ofunity
in K.We can determine the group of divisor classes for the above curve
which are defined over F and invariant under the action of
Gal(F(x, y)/F(x)) by
thefollowing
standardgenus-theoretic
arguments. To avoid aslight complication,
we assume thatf (x)
has àtleast one linear factor over F.
Let ~
EGal(F(x, y)/F(x))
be definedby ~(y)
= Epy. Assume that D is a divisor ofdegree
zero forF(x, y)
such that§(D)-D
is aprincipal divisor, ~(D) - D
=(z),
say.Clearly, N(z)
EFX,
where N is the norm map for the abovecyclic
extension. In
fact,
ourassumption
aboutf(x) implies
thatN(z)
E(Fx)p
as one can see
by just noting
that every residue class for a ramifiedprime
ofF(x, y) corresponding
to a linear factor off (x)
has arepresentative
from F and that z must be a unit at the ramifiedprimes. Hence,
we may assume thatN(z)
= 1. Thus z =~(w)/w
forsome non-zero element w of
F(x, y)
and therefore D -(w)
is in- variantunder 0
and in the same divisor class as D. LetPo P2, ..., Pt
denote theprimes
ofF(x, y)
ramified in the extensionF(x, y)/F(x).
Itfollows that every divisor class of
degree
zero invariantunder 0
contains a sum of the
P;’s.
These divisor classes have orderdividing
p. The
only
non-trivialprincipal
divisors formed from thePi’s
aremultiples
of(y) (up
to divisors fromF(x)).
Thus we see that thegroup of invariant divisor classes of
degree
zero forF(x, y)/F(x)
isisomorphic
to(Z/(p )1-2.
If e denotes the number of distinct irreduci- ble factors off (x)
over F and if d =deg(f(x)), t
= e + 1if p t
d sincein this case the infinite
prime
is ramified. Ifp ) d,
then t = e.The curve
yP = f (x)
has a birationalautomorphism (x, y) ~ (x, Epy) (corresponding
to0)
which induces anendomorphism
Zof Jf
definedover F. Since 1+Z+Z2+...+Zp-1 is the
zero-endomorphism,
weobtain a
homomorphism
of A =Z[Cp]
intoEndF(Jf) by mapping ep
toZ. If the genus of the above curve is
positive,
then this is anisomorphism and,
tosimplify notation,
we willidentify
A with itsimage.
The Tate module
Tf of Jf
for theprime
p can be viewed as amodule over A and therefore over
Ap
=A~ZZp.
Of course,Ap
isjust
the
ring
ofintegers
inKp
=Qp(03B6p).
NowAp
is aprincipal
ideal domain and henceT
must be a freeAp-module
of rank r= 2g , where g
isthe genus of
yP
=f (x).
One couldeasily determine g by
Hurwitz’formula but one can also do this as follows. Let 1T =
Cp - 1,
a genera- tor of the maximal ideal ofAp.
Then r isequal
to the dimension ofTf/03C0Tf
overZ/(p).
If a EA,
we will letJf[03B1]
denote the kernel of theendomorphism
ofJf corresponding
to a. The groupJf[p]
ofp-division points
onJf
defined overF
isisomorphic
toTf/p T f
as anAp-module
and contains
Jf[03C0].
Hence we can see thatpr = |Jf[03C0]|.
ButJf[03C0]
isisomorphic
to the group of divisor classes ofdegree
zero ofF(x, y)
invariant underGal(F(x, y)/F(x))
and thisjust depends
on the numberof distinct roots of
f (x)
and itsdegree
d. Inparticular, if p
is odd and iff(x)
=xa(1 - x)
which 1 ~ a ::5 p -2,
theng = p - 1 2
and the cor-responding
Tate moduleTa
is a freeAp-module
of rank 1.From now on we will assume that p is odd and that
f (x)
is such thatg
= p - 1 2.
If cr EGal(F/F),
then the action of cr onT
commutes withthat of
Ap.
SinceTf
is a freeAp-module
of rank1,
we have03C3(t) = p(o,)t
for all t ETf,
where p is ahomomorphism
ofGal(F/F)
intoU =
Axp.
Let Fm denote the subfield ofF generated by
the coordinates of all p-power divisionpoints
onJf.
Then Gal(F~/F)
is of courseisomorphic
to theimage
of p. Theimage
of p also determines the structure of thep-primary subgroup
ofJf(F);
it must beprecisely Jf[03C0sf], where sf
is thelargest integer
such that03C1(03C3)
~ 1(mod 03C0sf)
forall (T E Gal(F/ F).
Onesimple
consequence of this is thatJf(F)
willcontain
Jf[p"]
if andonly
ifsf ~ n(p - 1),
since7TP-lA = pA.
Thiswould occur for
example
ifJf(F)
contained apoint
of orderpn+1.
We will assume from here on that
f(x)
factors into linear factorsover F. The genus theoretic arguments described above make it clear that
Jf(F)
contains at leastJf[03C0].
It follows thatIm(p)
ÇU1,
where weare
using
the notationUS
for{u E V u ---
1(mod 03C0s)}
when s ~ 1. Wecan say more. Let T ~ 0394 be defined
by 7-(Cp) 03B6-1p.
If a EAp,
we willusually
write03C1(03B1)
as â. We let U* ={u
EU1|
uù EZ’I.
Then we willshow that
Im(p) Ç
U*. This in fact followseasily
from Weil’spairing.
LetM = lim 03BCpn
as n ~ ~, where lip- is the group ofpn -th
roots of
unity
in F and the inverse limit is definedby
means of thep-th
power map from 03BCpn+1 to gpn. ThenM == Zp
and the action ofGal(F/F)
on M isgiven by
ahomomorphism
K :Gal(F/F) ~ Zp.
Nowif
Cl, C2
are divisor classes forF(x, y)
of orderdividing p",
chooseDi
and
fi(i = 1, 2)
such thatDi
ECi, Dt
andD2
aredisjoint,
andpnDi = (fi)
wherefi
EF(x, y).
Then(Ch C2) ~ f1(D2)/f2(D1)
defines a non-degenerate pairing Jf[pn] Jf[pn]~03BCpn.
Heref(D)
is defined forf
eF(x, y)
and for aprime
divisor Ddisjoint
from(f))
in the obviousway and extended
multiplicatively
tonon-prime
divisors. It is clearthat for a E Gal(F/ F), (03C3(C1), 03C3(C2)) = (T« Ch C2)). Also, if 0
EGal(F(x, y)/F(x))
is the generator definedby 0(y)
=,Epy, then(~(Ci),
~(C2)) = (Cl, C2).
The Weilpairing Tf
xTf ~ M,
which we denoteby , .),
will then have theproperties: (a(t1), t2> = (th 03BA(03C3)03C3-1(t2)>
andZ(t1), t2>
=(tl, Z-1(t2)>
for all tl, t2 ETf. By using Zp-bilinearity,
wefind that
(ati, t2) = t1, ât2)
for a EAp.
If a =p(a),
it follows that a =03BA(03C3)03C1(03C3)-1,
i.e.03C1(03C3)03C1(03C3)
=03BA(03C3).
This proves thatIm(p)
C U*.We also mention in
passing
that if F is a finite field of order e’ and if03C3 is the Frobenius
automorphism over F,
then03BA(03C3) = ~’
and there-fore,
since03C1(03C3)
and itsconjugates
overQp
are the roots of the zetafunction for
YP
=f(x)
overF,
the above statement isjust
the RiemannHypothesis
in this veryspecial
case.The fact that
Im(p)
Ç U* has a few consequences that we want to mention.First,
ifJf(F)
contains apoint
of orderp n,
then sf >(n - 1)(p - 1)
and so03BA(03C3) ~
1(mod pn-l7T)
and therefore ~ 1(mod pn)
for all o- EGal(F/F).
It follows that lipn C F.Secondly,
it is easy tosee that d acts on
Us/Us+1 by
the character 03C9s. Inparticular,
J acts as(-1)S.
Now if s ~ 0(mod p - 1)
and aE US
~ U*but ~ Us+h
thenclearly
aâ EUs+1
and hence s must be odd. Therefore we see thateither s f
is odd or s f = 0(mod
p -1).
Forexample,
if we knew thatJf(F)
did not contain all thep-division points of Jf,
then it wouldfollow that the
p-primary subgroup
ofJf(F)
isisomorphic
to(Z/(p)S,
where s = sf is odd. We have one final
general
remark to make. LetFI
denote the field
generated by
allpoints
of order p onJf.
As a Galoismodule and as an
Ap -module,
we haveJf[p] ~ Tf/pTf.
It follows thatGal(F/F1)
=p-’(Up-1)
and hence thatGal(F1/F)
isisomorphic
to asubgroup
ofUllUp-1.
We seeeasily
from this thatGal(F1/F)
is anelementary
abelian p -group, as we mentioned in the introduction.3. The curves
yP
= xa(1 - x)We
just
assume p is odd at first and that 1 :5 a ~ p - 2. Let po, p l, and px denote theprimes
ofF(x) corresponding
to x, 1 - x, and1/x.
These
primes
are ramified in the extensionF(x, y)/F(x).
LetPo, Pl
and P.be the
primes
ofF(x, y) lying
above po, pl, and p~. Let c denote the divisor classcontaining
D =Po -
P~. Then c is invariant under the actionof 0
and thecorresponding point
onJa(F)
is a generator ofJa[03C0].
We will show thatJa(F)
containsJa[03C02].
This isequivalent
tofinding
a divisor class CI ofF(x, y)
such that0(cl) -
cl = c. Webegin by showing
that such a c, exists if andonly
if one can find az e
F(x, y)
such thatN(z)
= x, where N is the norm map for thecyclic
extensionF(x, y)/F(x).
The existence of c, amounts to the existence of a divisor D’ E c withN(D’)
= 0. IfN(z)
= x, then D’ =D - (z)
has this property sinceN(D) = p0 - p~ = (x). Conversely
ifsuch a D’
exists,
then we must at least be able to find a z’ EF(x, y)
withN(z’)
= ax where a E Fx. Butby considering
residue classes moduloPl,
weeasily
see that03B1 = 03B2p
where03B2 ~ Fx
and hencez
= 03B2-1 z’
has therequired property.
We will now show that x is in fact a norm for the extension
F(x, y)/F(x).
This is a consequence of thefollowing
rather well-known lemma since both x and
1-x
and thereforexa(1-x)
areclearly
norms for the extensionF(p1x)/F(x).
LEMMA 1: Let F be any
field containing
aprimitive p-th
rootof unity.
Let u, v E Fx. Then u is anorm for
the extensionF(pv)/F if
and
only if v is a
normfor F(pu)/F.
PROOF: Let Ep be a
primitive p -th
root ofunity
in F. If either u or v is ap -th
power inF,
thelemma
is trivial.Assuming that
this isnot
the case, let
Ou
EGal(F(pu)F)
be definedby ~u(pu) = ~ppu.
Define ov similarly.
Consider thecyclic aglebra A
=[F(pu), ou, v].
Let
V = pu.
The fieldF(U)
can be considered as asubalgebra
ofA. As a vector space over
F(U), A
has a basis 1, V,V2, ...,
Vp-’where V03B1V-1 =
0,,(a)
for all a EF(U)
and VP = v.Furthermore, v
isa norm for the extension
F(U)/F
if andonly
if A isisomorphic
tothe
algebra of p
x p matrices over F. But we canclearly identify F(pv)
with thesubalgebra F(V).
AlsoUVU-1
=~-1v(V)
so that theinner
automorphism
of dby
V restrictsto ~-1v
onF(V).
Since1, U, U2, ..., Up-1
is a basis forA_over F(V),
we see that A isisomorphic
to the
cyclic algebra [F(pv), ~-1v, u]. Since
this is the p p matrixalgebra
if andonly
if u is a norm forF(pv)/F,
lemma 1 isproved.
The
following proposition
followsimmediately
from what we havesaid above and at the end of section 2. Theorem 1 of course follows from this.
Theorem 2 also follows
quickly. Letting
pa denote thep-adic representation
ofGal(F/F)
on the Tate moduleTa
forJ,,,
we haveGal(F(a)~/F) == Im(Pa).
But U* ~Z(p+1)/2p X Z/(p)
as atopological
groupand U*~U2 ~ Z(p+1)/2p.
SinceIm(03C1a) ~ U*~U2,
we obtain theorem 2.In the case where F is a finite field of order
er(with er
~ 1(mod p)
since we areassuming
that F contains aprimitive p-th
root ofunity), proposition
1immediately gives
a congruence for the roots of the zeta function ofyp = xa(1-x)
over F. If cr denotes the Frobeniusautomorphisms
ofF/F,
then these roots areprecisely
theconjugates
of
pa(o’)
overQp
andthey
mustbe 1 (mod )
and even mod7T3
ifp > 3. The
fact
that uactually gives
anendomorphism of Ja
whichcommutes with Z
together
with the fact that A is a maximal com-mutative
subring
ofEnd(Ja)
shows thatp,, (o,)
is in A and notjust Ap.
Let us assume that F is the residue class field
A/F,
where 9 is aprime
ideal
dividinge.
We also assumethat Ep
isCp
mod 9. Now the roots ofthe zeta function are known to be the
conjugates
overQ
of the Jacobisum
where 0
is thep-th
power residue character mod 9 andg(t)
is theGaussian sum for F
corresponding
to the character~t.
It is notessential for our arguments to know which
conjugate
of a, isPa(u).
But this could be found as follows. Because of the above congruence
03C1a(03C3)
is determinedby
its factorization. TheCM-type of Ja
is knownto be
Ka = {03B4 ~ 0394 |[03C9(03B4))
+[a03C9(03B4)] p},
where[ ]
denotes the firstdigit
in thep-adic expansion,
and thus we have(03C1a(03C3)) = II8EKa 03A603B4-1.
(See [3].)
On the otherhand, by
a theorem ofStickelberger
we have(g(1)p)=JR
where R= 03A3p-1c=1 c03B4-1-c,
where5c
~ 0394 is determinedby
03B4c(03B6p) = 03B6cp.
It follows that(03B1F) = FSa, with Sa = 1 p (03B41
+ 8a -03B41+a)R. By comparing,
one finds that03C1a(03C3)
= a,.Now we come to the case
F = K = Q(03B6p).
Let G= Ga
=Gal(K(a)~/K)
and let H =Ha
be the inertiasubgroup
of Gcorrespond- ing
to theprime (1T)
ofK,
which is theonly
ramifiedprime
inK(a)~/K
since
K(a)~
CK. Letting
I denote the group of fractional ideals of Kprime
to p, the Artinsymbol
o-a =Q
defines ahomomorphism
o-,, of I to G. The
image
of I is dense in Gby
the Tchebotarevdensity
theorem. If 0 is aprime
ideal ofK(~ (03C0)),
then o- ,0 isjust
thecorresponding
Frobeniusautomorphisms
and we have03C1a(03C3F)
= a,j.The
image 03C1a(G)
isclearly just
the closure of thesubgroup generated by
the03B1’s
for all 9. We will follow Iwasawa[6]
tostudy
thisimage.
Let
Io
be thesubgroup
of I ofprincipal
ideals. Theimage
ofIo
in G under the Artin map is a densesubgroup
of H. If a EI,
thenobviously (Pa«(Ta))
= asa and if a EIo,
then Pa(03C3a)
= a Sa where a is any generator of a such that z == 1(mod 1T2).
It followseasily
thatpa(H)
=U2a.
We will conclude this section
by determining da da(K)
rankZp(G).
Since H has finite index inG,
this amounts tofinding
rankzp ( U 2 Sa
It is easy to determine this rank
by decomposing U2 by
the action of 0394.If X
=Wi,
1 ~ i :5 p -1,
then theX-component
ofU2
is isomor-phic
toZp
andSa
acts asmultiplication by
Now
B1,~
is non-zeroprecisely when X
is odd or X is theprincipal
character.If ea = p+1 2 - da,
then it is clearthat ea
is the number of oddX’s
such thaty(a + 1) = 1 + X(a).
It is easy to see that this holdsonly
wheny(a)
is aprimitive
6-th root ofunity
andX(a
+1)
=X(a)2. Obviously,
if p = 2(mod 3), then ea
= 0. On the otherhand,
if p ---1 (mod 3)
and ifa2 + a
+ 1~ 0(mod p),
thenX(a
+1)
=1
+ ~(a)
wheneverX(a) 0
1. Thusda = p - 1 6
+ 1 and one can see thatK(a)~
=KK’,
where K’ is the subfield of K fixedby {1, Sa, 03B42a}.
For sucha,
la
turns out to beisogenous
over K to(J’a)3
whereJ’
is an abelianvariety
ofdimension p - 1 6
with thering
ofintegers
of K’ as aring
ofendomorphisms. (See [8].)
There are many other
pairs (a, p)
forwhich ea
is non-zero. The firstexceptional pairs
occur for p = 67. In addition to the values of adiscussed in the last
paragraph (a
= 29 and 37 for p =67),
wehave ea
= 2 for a =6, 10, 19, 47, 56,
and 60.(Note
theeasily explained
symmetrya H p - 1 -
a.)
Theexceptional
characters in these cases are the two characters of 0394 of order 6. Thefollowing
remark due to Lenstra shows that similarexceptional pairs (a, p)
occur for allsufficiently large
p such that p --- 7(mod 12).
Acharacter X
of order 6 of 0394 will thencorrespond
toan odd character mod p.
Let g
be aprimitive
root mod p. We have~(a + 1) = 1 + ~(a) if a --- gx6 and a + 1 ~ g2y6 (mod p)
with(x, p) = (y, p)
= 1. But the RiemannHypothesis
for the curvegx6
+ 1 =g2y
overZ/(p ) provides
many such values of a.(The
genus of this curve is10.)
More
generally,
if p == 6 r + 1(mod 12 r)
andif y
is any character of 0394 of order6r, then X
willagain
be odd and the existence of a’ssatisfying
~(a
+1)
= 1+ ~(a)
forsufficiently large
p followsby considering
thecurve
grx6r
+ 1 =g2ry6r
overZ/(p).
One canmake ea arbitrarily large
in this way. We also want to remark that for any fixed value of a >1,
onecan prove the existence of
infinitely
manyprimes
p suchthat ea
=ea(p)
is non-zero
by using
the Tchebotarevdensity
theorem. We will omit the details.Finally,
we want topoint
out here that the fieldK(a)~
isactually completely
determinedby
theexceptional X’s. By
class fieldtheory,
itis not difficult to show that
if X
is either an odd character or theprincipal
character ofd,
then there exists aunique ZP-extension KX
of K such that
Kx/Q
is a Galois extension and such that 0394 acts onGal(KX/K) by
means of X.By noting
that pa induces a0394-isomorphism
of G with
Im(pa),
one can see thatK(a)~
isprecisely
thecomposite
ofthe
Kx’s
forthose y
such thatX(a
+1) ~
1 +X(a).
4. The local field of
p-division points
Let
Kp
=Qp(03B6p)
and letL(a)p
denote the extension ofKp generated by
thepoints
oforder p
onJa.
NowHa
can be identified withGal(K(a)p,~/Kp).
We will then haveGal(K(a)p,~/Lp) = {h ~ Ha|03C1a(h) = 1
(mod pAp)}
and henceGal(L(a)p/Kp)
will beisomorphic
to the cor-responding quotient
group ofHa. Using
the fact thatpa (Ha )
=USa2,
wefind that as à -modules
Gal(L(a)p/Kp) ~ Vsa,
where V =U2/Up-l.
Now asmentioned earlier
U;/ Ui+1
isisomorphic
toZ/(p)
with L1acting by
(ùi.Thus, considering
V as arepresentation
space for 0394 overZ/(p),
eachcharacter X
of 0394 occurs withmultiplicity
one in V except forwp-’
= cvo and03C91,
which don’t occur in V at all. If X=,w’,
2:5 i ~ p -2,
thenSa
actson
Vx
asmultiplication by X(Sa) = ±(1
+X(a) - X(a
+1)) B1,x
-i. Forthese X’s,
thex-component
ofGal(L(a)p/Kp)
is non-trivialprecisely
when~(Sa) ~
0(mod pZp).
It is well-knownthat p
dividesB1,~-1
if andonly if p
divides the Bernoulli number
B;,
where i+ j
= p.(Up
top-adic units,
both occur as values of the same
p-adic
L-functionLp(s, 03C9j).)
There-fore, if p ) B;
thenGal(L(a)p/Kp)i
will be trivial for every a. It is alsopossible
that 1 +X(a) - X(a
+1)
is divisibleby p
for certain values of a.Obviously
this cannot occur for every a since then one would have X = w which has been excluded. Also we note thatonly
oddX’s
canoccur
non-trivially
inGal(L(a)p/Kp).
The above remarks allow us to determine
L(a)p completely. By
localclass field
theory, Kp
has aunique cyclic
extensionLx
ofdegree p
such that
Lx/Qp
is Galois and such that à acts onGal(Lx/Kp) by
thecharacter X, where
X = 03C9i
with2 ~ i ~ p - 2. (This
is not true forX = w° and
03C91!)
We see thatL(a)p
isjust
thecomposite
of thoseL,,’s
for which
~(Sa) ~
0(mod pZp). Obviously L~
will be contained inL(a)p
for some a if and
only
ifp f Bj.
Theorem 3 followsimmediately
fromthese considerations.
It is
interesting
to examine the structure of thep-primary subgroup
of
Ja(Kp).
Of course,Ja[1T3]
is contained inJa(Kp). (We
areassuming
here
that p ~ 5.)
In order to have more thanthis,
we would have to have03C1a(Ha) ~ U4.
Hence a necessary and sufficient condition forhaving j a [lr4l C Ja(Kp)
is that(V3(Sa)
~ 0(mod pZp).
Now it is easy tosee that
1 + ~(a) ~ ~(a+1) (mod p)
for~ = 03C93
and1 s a S p - 2.
Therefore
03C93(Sa) ~ 0 (mod pZp)
if andonly
if p dividesBp-3.
The firstsuch
prime (and
theonly
one less than125,000) is p
= 16843.(See [7]
and
[10].)
Of course, forthis p
weactually
haveJa[03C05]
çJa(Kp)
forevery
a(1 ~ a ~ p - 2).
NowpBp-5
forthis p
and hence the p -primary subgroup
ofJa(Kp)
will beprecisely Ja[1T5]
unless03C95(a
+1)
~03C95(a)
+ 1(mod p).
There arejust
two values of a within the range 1 ::5 a ~ p - 2 for which this congruence holds - the two solutions of a2 + a + 1 0(mod p).
Infact,
for these two values of a, we haveX(a
+1)
=X(a)
+ 1whenever X
=03C9i
with i ~ 0(mod 3). Since p f Bp-9
for the above p, we can see
that,
for these two values of a, thep-primary subgroup
ofJa(Kp)
isJa[1T9],
anelementary
abelian p- group of orderp9.
Ingeneral,
forp ~ 5,
one can at least say thatJa(Kp)
never contains all thep-division points
onJa.
This followsby
an easy argument
using
the trivial fact thatB2
andB4
are not divisibleby p.
5. The
global
field ofp -division points
Let L(a) denote the extension of K =
Q(ep) generated by
thepoints
of order p on
Ja.
The field Lgenerated by
thep-division points
on Jis of course