C OMPOSITIO M ATHEMATICA
R AYMOND R OSS
K
2of Fermat curves with divisorial support at infinity
Compositio Mathematica, tome 91, n
o3 (1994), p. 223-240
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K2 of Fermat
curveswith divisorial support
atinfinity
RAYMOND ROSS*
(Ç) 1994
Louisiana State University, Baton Rouge, LA 70803-4918 Received 2 October 1991; accepted in final form 29 April 1993
Introduction
The Beilinson
conjectures ([2], [3]) interpret
thespecial
values of the L-functions associated to a
projective variety X
defined over a number field tothe
K-theory
ofX,
via aregulator
whichgeneralizes
that of Dirichlet. In this paper, we examine part of theseconjectures
in the case where X is the Fermatcurve
FN:xN+yN=zN
of exponentN,
where N 3. Ourinvestigation
isinspired partly by
Beilinson’s theorem on modular curves([3],
also see[17]),
and
by
a result of Rohrlich on the divisor class group ofFN [12].
Let us recall what Beilinson’s theorem says,
following
theexposition
in[17].
Let
X/Q
be a modular curve, andlet g
be the genus of X. For each modulatorcurve
Y/Q
which covers X via amorphism 0y:
Y ~ X defined overQ,
there isa
homomorphism 03B8Y*: K2Y ~ K2X.
Letfly
be thesubspace K2
Y 0Q
withdivisorial support at the cusps of Y
(this
is madeprecise
in Section 1below),
and let
&x
denote thesubgroup
ofK2X
QQ spanned by 0y*(2Y)
for all suchY Then
&x
cH2H(X, Q(2))z
and theimage
ofYx
inH22(X, R(2))
under theregulator homomorphism
regx is aQ-structure
ofH2D(X, R(2)),
andwith c ~
L(g)(0, X)
modQ*.
In
general,
one must look at&x
andnot just Jx.
Forexample, if p
isprime,
then
Xo(p)
hasexactly
two cusps, and thereforeonly
one(up
to scalarmultiple)
modular unit
f
It then follows thatLXo(p)
is trivial.For
FN,
ananalogue
of the group of modular units isprovided by
thosefunctions whose divisorial support is contained in the
points
atinfinity,
whichare those
points
P onFN
such thatxyz(P)
= 0. Inpartial analogy
withBeilinson’s
construction,
weinvestigate
thesubgroup
ofK2FN generated by
theimages
under the transfer mapsK2FdN ~ K2FN of
those elements whose divisorial support is contained in thepoints
atinfinity.
We show that thissubgroup
is ofpositive rank,
andthat,
overQ(p2N), it is
acyclic Aut FN-
module.
Using this,
we descend toQ
and obtain a bound for the rank of thecorresponding
group overQ.
In contrast to the modularsituation,
this*Research partially supported by NSF grant DMS-9102786.
subgroup
isusually
of rank smaller than the rank ofK2FN
aspredicted by
theconjecture, namely, (N - 1)(N - 2)/2,
the genus ofFN.
The rank of thissubgroup,
and a set of generators for the vector space obtainedby tensoring
with
Q,
will be determined in[14].
We now
briefly
indicate theorganization
of this paper. Webegin by summarizing
what we need fromK-theory.
In Section2,
we exhibit anexplicit
element in
K2FN
which we show is of infinite order. In Section3,
we describethe
subgroup
ofK2FN
which isgenerated by
the K-theoretic transfers of those elements inK2FaN arising
from functions with divisorial support on thepoints
at
infinity.
We will see that this issimply
the group of such elements inK2FN.
Section 4 is an
interlude,
where we look at theexample
ofF4.
In the finalsection,
we show that thesubgroup
ofK2FN
underinvestigation
is aprincipal Aut FN-module,
and use this to determine an upper bound on its rank as anabelian group in case N is an odd
prime.
This work is based upon part of the author’s
Rutgers
Ph.D. thesis. 1 would like to thank myadvisor,
DavidRohrlich,
both forsuggesting
thistopic
forresearch and for the
generosity
with which he offeredadvice, assistance,
and encouragement. 1 also thank Chuck Weibel forpointing
out that in an earlierdraft,
1 missed a fewsymbols
in theproof
of Theorem 3.Finally,
1 wouldlike to thank the referee for many
helpful
comments andsuggestions
forimprovement.
1. The
regulator,
Bloch’strick,
andsymbols
with supportWe
begin by recalling
what theregulator
of Beilinson and Bloch is in thespecial
case that X is a smoothprojective
curve defined overQ.
We fix analgebraic
closureQ
c C ofQ.
Allalgebraic
extensions ofQ
which arise aretacitly
understood to lie in this fixed choice ofQ.
Let F be a field.
By
Matsumoto’s theorem[10], K2F ~ (F*
QF*)/R,
whereR is the
subgroup
of F*0 F* generated by
the tensors of the formf
Q(1 - f),
withf ~ 0, 1.
Theimage
off
Q h inK2F
isdenoted {f, hl,
andis referred to as a
symbol.
Let X be a smooth
projective
curve of genus g defined overQ.
Thelocalization sequence
provides
us with thefollowing
exact sequence:where
with Tp
being
the tamesymbol
at P:By
Garland’s theorem[9], K2Q(P)
istorsion,
whenceK2X
and ker 03C4 agree,up to torsion. If k is any finite extension of
Q,
thenby viewing
X as a curveover k,
theseremarks,
withQ replaced with k,
remain valid.In
[2],
Beilinson defines aregulator
forK2X,
which agrees with that of Bloch[6].
This may be viewed as ahomomorphism
which is functorial in X. The
superscript
denotesthe -1-eigenspace
ofH1(X(C), R(1))
under the actionof complex conjugation
on bothX(C)
andR(l) =
2niR. We remark thatH1(X(C), R(1))-
isisomorphic
to theDeligne cohomology
groupH2D(X, R(2))
mentioned above in the Introduction[18].
To describe regx, we assume for
simplicity that {f, hl
E ker 1:. Fordetails,
werefer the reader to
[8]
or[11].
We viewH1(X (C), C)
as the space of C-valued functionals onH1(X(C), Z).
Let y be a closedpath
onX(C),
and fix abasepoint Po
E y; assume thatf
and h are bothregular
and nonzero on y. ThenHere,
we choose fixed branches oflog f
andlog
h on someneighborhood
ofy, and the
integrals
are takenstarting
atPo.
One may show that this definitiondepends only
upon the class of03B3~H1(X(C), Z).
Let
03A91+X
dénote thesubspace
ofH0(X(C), 03A91X/C)
which is invariant under the actionof complex conjugation
on bothX(C)
and03A91X/C.
Under the identifica- tionH1(X(C), R(1))-
withHomR(03A91+X R),
we viewregX({f,h))
as a functionalon real 1-forms. One then obtains the
following
formula for theregulator:
One can check that this
integral
converges for anypair
of functions onX,
andwe
thereby
obtain ahomomorphism
which is functorial in X and extends the
regulator
as defined above. As an immediate consequence of thisfunctoriality,
we have thefollowing:
It follows from
this,
ordirectly
from(2),
that theregulator
vanishes on thosesymbols
of which one entry is constant.We now turn to the transfer map in
K-theory,
and describe those facts whichwe will need in the
sequel.
Let
F/L
be a finite Galois extension of fields with Galois groupG,
and denoteby 0
the inclusion L c+ F. The two mapsand
are then related to the action of Galois on
KnF
in thefollowing
manner:This follows from the fact that
0*
is induced from the extension of scalars functorP(L) ~ P(F),
andcP*
is induced from the restriction of scalars functorP(F) - P(L),
whereP(A)
is the category offinitely generated projective
A-modules for any commutative
ring
A.We will also need to know how the transfer behaves with respect to the
regulator
map, as follows. Letk/Q
be a finite Galoisextension,
and lettf¡: Q(X) 4 k(X)
be the field inclusion.LEMMA 2. Let
cxEK2k(X).
Thenwhere
Proof.
Since theregulator
is defined over C and03C8*
is inducedby
the fieldinclusion,
we haveregx(fl)
=regX(03C8r*03B2)
for allfi
EK2Q(X). Letting fi
=03C8*03B1,
the lemma then follows from
(3).
DIn
particular,
ifregx(a)
=0,
thenregX(03C8*03B1)
= 0.We now recall Bloch’s trick
[4].
Let S ={P0,...,Pn}
cX(Q).
Choose anembedding
X 4 Jac X such thatPo corresponds
to theidentity
element ofJac X. Assume that the
images
of thePi
under thisembedding
are torsionpoints.
Let
f and g
be functions inQ(X)
whose divisors consistentirely
ofpoints
in S. One version of Bloch’s trick asserts that there is an
integer N,
functionscPi E Q(X)*,
and constantsci~Q*
such that a ={f,g}N 03A0i {~i, cil
is in the kernel of the tamesymbol.
We will refer to such an oc as a normalization of{f,g}.
Note that03B1~K2L(X)
for some finite extension L ofQ; by taking
theK-theoretic transfer from
L(X)
toQ(X)
we obtain an element inK2Q(X)
n kerr. We remark that if k is a finite extension ofQ,
S cX(k),
andf, g
Ek(X)*,
thenOi
and ci can be chosen to be defined over k.Utilizing
this version of Bloch’strick,
we nowconstruct
asubgroup
ofK2X
0Q
which is associated to S.Choose k so that S c
X(k),
letWs = {f~Q(X)*: ordQ( f )
= 0 for allQ ~ SI,
and
let fnl
be generators foruS/Q*
defined over k. Let{uS, uS}
denote the
subgroup
ofK20(X) generated by symbols {f,g}
such thatf,g~uS.
Then{uS, uS} is generated by symbols
of thefollowing
three types:The
symbols
in(5)
aretorsion, being
inKlo
= limK2L,
with the limitbeing
taken over all finite extensions L of
Q
inQ.
Thesymbols
in(6)
arepullbacks
from
K20(p’);
we may thereforeapply
Bloch’s trick on P’ andthereby
obtaina
symbol
inK2Q(X)
which is apullback
fromK2p’/O.
This all occurs over some finite extension ofQ,
whence thissymbol
is torsion.Let
NXS(k)
be the groupgenerated by
a normalization of each of thosesymbols
in(7).
This groupdepends
on the normalizationschosen;
see theremarks below for more about this
dependence. Observe, however,
that ifJVx,s
denotes the group
generated by
all thesymbols
in(5)-(7),
thenJVx,s(k)
0Q
and
JVx,s
0Q
have the sameimage
under theregulator
map.We now put
9-x,s(Q)
=tr JVx,s(k),
where tr is the transfer map fromK2k(X)
toK2Q(X). By (3), JX,S(Q)
ckerT,
and we may thereforeidentify 9-x,s(Q) 0Q
with asubspace
ofK2X~Q,
which we shall refer to as thesubspace of K2X
0Q
with divisorial support on S.REMARKS. Bloch’s
trick,
as stated in[4],
does notspecify
how to choose Nand ci, or even that the modifications
to {f,g}
toyield
an element ofkerr areof the type that we have selected.
Any
two normalizationsof {f, gl
would differby
a rationalmultiple
and an element of r = ker i n ker regx.Very little,
ifanything,
seems to be known aboutr;
Beilinsonconjectures
that(r
~Q)~H2H(X, Q(2))Z
= 0. If X is such thatH2H(X, Q(2))
Z =H2H(X, Q(2)),
then
conjecturally
our constructionyields
the same vector space as any version of Bloch’s trick. Inparticular,
the Fermat curvessatisfy
thiscondition,
sincetheir Jacobians have CM:
Letting 5N
denote aregular
proper model forFN
over
Z,
we have the localization exact sequence inK-theory:
By ([19],
Theorem3)
the Euler factorLp(FN,S)
has nopole
at s =0; therefore,
the
right-most
vector space is zero([11], Proposition 4.7.9). Finally,
theleft-most vector space is zero
([1]).
2. An element of infinité order
Let 03B6
=e203C0i/N, and
letAi,j
denote theautomorphism
of FN.
Let t1/N denote theprincipal
branch of the Nth rootfunction,
and let y denote thefollowing path
from(1,0)
to(0, 1)
onFN(C):
For
integers
mand n,
let ym,n denote thefollowing
closedpath
onFN(C):
By
aslight
abuse ofnotation,
we will also denoteby
03B3m,n thecorresponding
element
of H1(FN(C), Z).
We will need the beta function
B(u, v),
which is defined forpositive
realnumbers u and v
by
THEOREM 1. The
symbol {1 - x,
1- y}2N belongs
ker i and has non-zeroimage
under regFN. Inparticular, {1 -
x, 1 -yl2N
represents an element inK2FN of infinite
order.Proof.
An easy calculation showsthat {1 -
x, 1 -y}2N ~
ker L. We show that{1 -
x, 1 -y}2N
has nontrivialimage
under theregulator homomorphism by computing
its value on the1-cycle
y 1,1 + y 1, -1. Let e > 0 besmall,
and choosea
representative path
Ym,n,f: for ym,n such that the initialpoint
of y"t,n,E is((1 - 03B5)1/N, 03B51/N),
and such that both 1 - x and 1 - y are nonzero andregular
on 03B3m,n,03B5. Let Q
= {1 -
x, 1 -yl2N. Choosing
the branches oflog(l - x)
andlog(l - y)
which are real-valued on y, we find thatregFN(03C3)(03B3 m,n)
isequal
to:Note that as 03B5 tends toward zero, the
imaginary
part of the secondintegral
tends toward zero. Since the value of this
integral depends only
on thehomology
class of ym,n and not on thebasepoint chosen,
we conclude thatFor any
a, b
EZ,
letThen
The
interchange
of the two sums and theintegral
maybe justified by
a doubleapplication
of theLebesgue
dominated convergence theorem.Moreover,
the double sum convergesabsolutely,
which may be seenby considering
the casea = b = 0.
Noting
that the lastintegral
above isB(j/N, klN
+1)
andusing
theidentity
we obtain
Returning
to 03B3m,n, we find thatTherefore
Define bk by
and for 1
KN, let
Note
that bk
> 0 for all k and that the sequence{bk}
ismonotonically decreasing,
so03B21 > 03B22
>...> fiN
> 0. Let N’ be definedby
Then
which is nonzero, since each summand is
positive.
D3.
Symbols
with support atinfinity
and the Fermât towerWe now turn our attention to the elements in
K2FN
which have divisorialsupport on the
points
atinfinity. By
thepoints
atinfinity
we mean the setSN
={P~FN(Q):xyz(P) = 01.
There are 3N suchpoints, given
inprojective
coordinates
by [(j, 0, 1], [0,’ j, 1],
and[03BE03BEj, 1, 0],
for 1j N, where (
is aprimitive
Nth root ofunity and j
is aprimitive
2Nth root ofunity,
which weshall
always
take tosatisfy 03BE2 = 03B6.
Let k =
Q(P 2N), and,
to fix matters, choose anembedding l:FN Jac FN
defined over
Q
andsending
thepoint [0,1,1]
to theorigin.
A theorem ofRohrlich
[12]
asserts thatI(S N)
is torsionand,
in the notation of Section1,
thegroup
%s/Q* is generated by
thefollowing functions,
all of which are definedover k:
with
E(N) =
N if N is odd andE(N) = N/2
if N is even.We are interested in the space
JFN,SN(Q)
0Q,
the space ofsymbols
inK2FN Q Q
with divisorial support atinfinity.
Recall that this space is03C8*(NFN,SN(k))
QQ,
where,41’F,,S,(k)
is the groupgenerated by
normalizations of thesymbols
obtained from the functions in(9)-(12),
and03C8*
is the transfermap induced
by
the fieldinclusion 03C8~:Q(FN) k(FN).
We will see below
(Theorem 3)
that for a suitable choice ofnormalizations,
where
rN
is theautomorphism
groupof FN.
LetyFN,SN(k)
denote theQ-vector
space
Q[0393N]{1 - x,
1 -yl,
and letflFN,SN(Q) = 03C8*yFN,SN(k).
It follows from Theorem 3 thatJFN,SN(Q)
0Q
=flFN,SN(Q).
We will assume these facts in thissection,
since the results below do notdepend
on those discussed here.For each
positive integer d,
letbe the
morphism given by
Let u and u be the standard coordinate functions on
FN,
such that u = Xd andv =
ya. Put k.
=Q(P2dN)’
soka(FaN)
=kd(FN)(x, y)
is an abelian extension ofkd(FN)
with Galois groupGd~03BCd
x itd-For each
integer n 1,
letAi,j(n) (i and j
read modulon), a(n),
and~(n)
denote the
following
elements ofrn
=Aut Fn :
where (n
is aprimitive
nth root ofunity and 03BE2n = 03BE, 03BEnn
= -1. ThenTn
isgenerated by Ai,j(n), a(n),
andtl(n),
and isisomorphic
to a semidirectproduct
of Pn 03BCn and the
symmetric
groupS3.
Now fix
N,
let03B6N and 03BEN
be asabove,
and for eachinteger d 1,
choose03B6dN
and
ÇdN
such thatÇjN
=(N and ÇjN
=03BEN.
Then the map0393dN ~ rN given by
is a
homomorphism,
the kernel of which is the relativeautomorphism
groupAut(FdN/FN).
We viewyFN,SN(k)
as ardN-module
via this map. Letjd:k1(FN) ka(FN)
be the field inclusion.PROPOSITION 1.
~d*yFdN,SdN(kd) = j*dyFN,SN(k).
Proof.
A direct calculationusing (3)
shows thatis
raN-equivariant,
and thatTherefore
It follows from
(4)
that ker0*
is trivial(recall
that we have tensored withQ).
Therefore
~d*yFdN,SdN(kd) = j*YF,,S,(k).
~We now descend to
Q.
Let03C8d: Q(FdN) kd(FdN)
be the field inclusion.Proof.
What we must show is thatFrom the
following diagram
of field inclusions:we obtain the
following
commutativediagram
inK-theory:
Note that
by (4), (jd*·j*d)yFN,SN(k) = !/ FN,SN(k).
Thereforeas desired. D
4. The Fermât curve of exponent 4
In this
section,
we examine in some detail the curveF4.
Inparticular,
we willexhibit a
symbol {f, 91 EflF4SJQ)
such that the divisor off
containspoints
which are of infinite order under the canonical
embedding F4 ~
JacF4;
compare with
[15].
In some contrast to[15],
we can relate thissymbol
to theL-value
L(3)(F4, 0)
in the mannerpredicted by
the Beilinsonconjecture.
We
begin by showing
that thesymbols
generate a rank 3
subgroup
ofK2F4.
Note that thesesymbols
lie inflF 4,S4(Q).
Let ym,n be as in Section 2.
Calculating
as in theproof
of Theorem1,
we findthat
where
Note that
/31
>03B22 > 03B23 > 03B24 > 0;
thus in the first two cases theregulator
value is non-zero.
A similar calculation for a2
yields:
Define
Boaa
andBeven by
Note that
Boaa
>Beven
> 0.Calculating
as in theproof of Theorem 1,
one finds thatwhich is nonzero.
Finally,
one mayverify
that these elements arelinearly independent by considering
their values on thepaths
03B32,1, 03B31,3 and 03B33,3.It is not difficult to write down an
explicit isomorphism X0(64) ~ F4
whichis defined over
Q,
and is such that the cusps ofXo(64) correspond precisely
with the
points
atinfinity
onF4.
Thus 03B11, a2, and a3 generate thesubspace 19X.(64)
ofK2X0(64) ~ Q
described in the Introduction. Inparticular,
theimages
of al, a2, and 03B13 under regF4 define aQ-structure
onH;(F 4’ R(2)),
withvolume
equal
to a rationalmultiple
ofL(3)(F 4’ 0).
On the other
hand,
we canapproach K2F4
"from below". The JacobianJo(64)
ofX0(64) is isogenous
overQ
to E x E’ xE’,
where E and E’ are theelliptic
curvesand we have
morphisms
p:F4 ~
E,03B2: F4 ~ E’,
and03B2’: F4 ~
E’given by:
One computes that
Noting
that the divisors of thepullbacks
toF4
of8X/Y’
and 1 -2X/Y by fl and /3’
consist ofpoints
atinfinity,
we conclude that these functions have torsion divisorial support on E’. A calculationusing
Rohrlich’sexplicit
version[13]
of Bloch’s theorem[5]
shows thatdefines a
Q-structure
onH’(E, R(2))
with volumeequal
toc’L’(0, E’),
wherec’~Q*.
Theconjectures
of Beilinsonimply
thatis a generator for
K2E’
QQ.
Note that
while functions on
F4, actually
define functions onE,
which we also denoteby f
and g. The divisors off and g
on E consist of torsionpoints,
andcomputing
as above we find that
regE({f, gl)
determines aQ-structure
onH2D(E, R(2))
with volume
equal
tocL’(0, E)
withc~Q*.
The Beilinsonconjectures imply that {f,g}
is a generator forK2E
QQ.
Puta3
=03C1*{f,g}8.
There is a
decomposition
ofcohomology
inducedby 03C1, 03B2,
and03B2’:
H1(F4(C), C) ~ H1(E(C), C)
(f)H 1 (E’(C), C)
(DH 1 (E’(C), C)
which is
orthogonal
with respect to thepairing
where we are
using
DeRhamcohomology.
Therefore 03B11, a2, andâ3
span a3-dimensional
subspace V
ofK2F4 Q Q,
and theimage
of V under theregulator
is aQ-structure
ofH 2(F,, R(2))
with volumeequal
to a rationalmultiple
ofL(3)(F 4’ 0).
If the Beilinson
conjectures
are true, we should have V =LF4,S4(Q) (and
bothshould be
equal
toK2F4
QQ).
We nowverify
that it is indeed the case that V =flF4,S4(Q).
Note that it suffices to check thatâ3
EJF4,S4(Q)
OQ.
A
computation using (3)
showsthat,
up totorsion,
Using
thealgorithm
in[16],
we computethat,
up totorsion,
Therefore, â3
Ef2p 4,S (Q),
as desired.In
closing,
wenoee
that the divisor of 1 -xly
containspoints
which are notpoints
atinfinity
onF4; by
a result of Coleman[7],
thesepoints
are of infiniteorder in
Jac F4.
5. The group of
symbols
with support atinfinity
In this
section,
we fix our attention onFN,
and let k =k1
=Q(J.l2N)’
JV =
NFN,SN(k), y = yFN,SN(k), J = JFN,SN(Q), 03C8
=1 ’ Q(FN) k(FN),
andG =
G
1 =Gal(k/Q).
In
general,
Y QQ
cannot beequal
toK2FN
~Q, provided
that one accepts the Beilinsonconjectures.
For thepredicted
rank ofK2FN
is the genus ofFN,
which
equals (N - 1XN - 2)/2,
and we havePROPOSITION 2.
Let p be an
oddprime.
ThendimQ JFp,Sp(Q)
0Q
3(p - 1).
1 nparticular, for
p >7,
we havedimQ JFp,Sp(Q)
0Q genus(Fp).
Proposition
2 is a consequence of thefollowing:
THEOREM 3. With an
appropriate
choiceof normalizations,
N 0Q
= y =Q[0393N]{1 - x, 1 - y}.
Proof. By (9)-(12),
it is clear that X QQ
isgenerated
overQ by
normaliz-ations of the
following symbols:
for
0 j, k
N - 1. Note that we do not need to includesymbols
for whichone of the entries is a function listed in
(10)-(12),
because we have tensored withQ.
The
symbols
in(16)
are torsion inK2k(FN).
Although
nottorsion, appropriate
powers of thesymbols
in(15)
arepullbacks
fromK2k(P’),
and therefore have trivial normalizations.The
symbols
in(14)
are alsopullbacks
fromK2k(P1). Indeed, letting
q =
q(N),
one finds that:Thus the
symbols
in(14)-(16)
all have trivial normalizations. We now show that thesymbols
in(13)
have normalizations which lie inE7,
andexplicitly
determine these normalizations. Given a
symbol {f, gl,
we will denote anormalization of
{f, g} by v( f, g).
In
examining
the othersymbols
in(13),
we will need thefollowing.
Since{1 -
X,y}2N = {1 -
x, 1 -xN}2~K2k(x),
weapply
Bloch’s trick and select b =ni {fi(x), cil
withf
Ek(x)*
and ci E k* such thatTherefore {1 -
x,y}2N·03B4
istorsion;
let M denote its order. We thus haveWe note that the
image
of 03B4 under anyautomorphism
ofFN
is aproduct
ofsymbols
in which one entry is constant.A routine calculation shows that we may select normalizations of the
remaining symbols
in(13)
as follows:and
From this it is clear that X Q
Q
cy;
a routine verificationusing (17)-(20)
yields
the reverse inclusion. DProof of Proposition
2. Note that asQ-vector
spaces, !Y 0Q ~ 03C8*J
0Q ~ 03C8*03C8*N
~Q.
Let E =03A303C3~G
or. Thenby (3)
we see that J QQ --
%1: QQ;
inlight
of theprevious result,
we consider y03A3.We first make a convenient choice of
spanning
set for y. Since p isodd,
x -
ç,ky
= x+ 03B6k’y,
where k’ =(2k
+ 1 +p)/2.
In what amounts to a re-indexing,
letThen ô3° is
spanned by
the normalizationsv«Tj,k), v(Uj,k),
andv(Vj,j
of thesesymbols.
Therefore
y03A3
isgenerated by v(Tj,k)03A3, v(Uj,k)03A3,
andv(Vj,k)03A3.
Onefinds, using (18), (19),
and(20)
thatV(1j,k)1:. = (TI,k)2N, v(U j,k)1:. = (U03A3j,k)2MN,
andv(Vj,k)03A3 = (V03A3j,k)2MN .
Assume that
( j, k) ~ (0, 0).
LetBj,k
denote any one ofT,k, Uj,k,
andVj,k.
Ifa is an
integer relatively prime
to p, thenB03A3j,k
=B03A3aj,ak,
where thesubscripts
areread modulo p. This follows from the fact that
B03C3aj,k
=Baj,ak,
where03C3a:03B6~03B6a
isthe Frobenius
automorphism
at a.It follows that y03A3 is
generated by
thefollowing 3(p
+2)
elements:for
0 j
p - 1. Note that we have thefollowing
relations:So we may discard the elements
T6.o, T03A30,1,
andTl o.
The same remarkholds,
mutatis
mutandis,
for theU03A3j,k
and theJtj:k,
sincethey
are obtained from theT03A3j,k
by automorphisms.
So J isgenerated
overQ by 3(p - 1) symbols,
as claimed.D
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