C OMPOSITIO M ATHEMATICA
D AVID A. C OX W ALTER R. P ARRY
Torsion in elliptic curves over k(t)
Compositio Mathematica, tome 41, n
o3 (1980), p. 337-354
<http://www.numdam.org/item?id=CM_1980__41_3_337_0>
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337
TORSION IN ELLIPTIC CURVES OVER
k(t)
David A. Cox and Walter R.
Parry
Let k be a field of characteristic
p > o, P ¥: 2, 3,
and let t be transcendental over k. The purpose of this paper is tostudy
thegroups
E(k(t»t’.,
=(x
EE(k(t»tor:
p does not divide the order ofxl
where E is an
elliptic
curve overk(t)
with nonconstantj-invariant.
Since
E(k(t»
isfinitely generated (the
Mordell-Weiltheorem), E(k(t»,’O,
isisomorphic
toZlnZ EB Z/mZ
where n and m arepositive integers with p,r n
andm , n.
Acomplete description
of thepossible
groups is
given
in Theorem 5.1.We
approach
thisproblem
in a classical way,using
thesubgroups Tm ( n ), ml n,
ofSL(2, Z)
definedby
Tm (n )
acts on the upper halfplane b
asusual,
and thequotient Ym (n )
=rm(n)BSJ
is related to moduliproblems
ofelliptic
curves con-taining
asubgroup isomorphic
toZ /n Z ® Z /m Z (we
make this pre- cisein § 1).
The basic idea is that thepossible
groupsE(C(t))tor
are thoseZlnZ (D ZlmZ
for whichXm(n)
=rm(n))J*
has genus 0. The methods ofDeligne
andRapoport
then allow us togeneralize
to anarbitrary
field k of characteristic not 2 or 3.The first section defines a level
(n, m)
structure on anelliptic
curveover a
base,
and uses[2]
to solve theresulting (coarse)
moduliproblem
and relate it to-r,,,(n).
§2 ispreliminary
to§3,
which is a 0010-437X/80/060337-18 $00.2010© 1980 Sijthoff & Noordhoff International Publishers - Alphen aan den Rijn Printed in the Netherlands
catalog
of theproperties
ofTm(n),
and §4applies
this to the finemoduli
problem, studying
the universal curves for level(n, m)
struc-tures
(these
exist in mostcases).
Then §5 puts this alltogether
toprove the classification theorem. An
appendix
contains atheorem,
used in§4,
which is a nice extension of arepresentability
result in[2, VI.2].
The usual ways of
writing Tn(n), Yn(n)
andXn(n)
arer(n), Y(n)
andX (n ),
and we will use the latter. Note that when m =1,
our notation agrees with standard notation.We would like to thank
Barry
Mazur for several usefulsuggestions.
§1
In this section we make extensive use of
[2].
Let n and m bepositive integers
with n > 2 and mn.
A level(n, m)
structure on ageneralized elliptic
curve E - S(see [2, II.1.12]
for adefinition)
is an S-inclusion of groupssuch that:
1.
Cm
islocally,
in the étaletopology, isomorphic
to(Z/ m Z)s,
and2. the
image
of a meets every irreducible component of everygeometric
fiber of E ---> S.To relate this to
[2],
let H be thesubgroup
ofGL(2, Z/nZ)
definedby:
Then,
from[2, IV.3],
we get thealgebraic
stackJlt%[1/n]
and itscompactification (relative
toZ[I/n]) .JlH[I/n].
Theseobjects
have thefollowing interpretation:
PROPOSITION 1.1:
Jlt%[1/n] (resp..JlH[I/n])
is thealgebraic
stackclassifying equivalence
classesof
level(n, m)
structures onelliptic
curves E over S
(resp. generalized elliptic
curves E overS),
where twolevel
(n, m )
structures a :Z /n Z
xCm
E and a’ :Z /n Z
xC2 -
E areequivalent if
there is anS-isomorphism q: Cm ----> C’ m
such that a =a’- (1
xq).
PROOF: We first treat
JK%[1/n].
From[2, IV.3.2],
a level H struc-ture on E is an element a of
FH(S),
whereFH
is the étale sheafHBIsos(En, (Z/nZ)}).
Such an a thus consists of an étale cover(S; - S);i
of S andisomorphisms ai:(En)s,---->(ZInZ)’.
1 such that fori, j
E I, there isan hij
EHom(Sij, H) (S;j
=Si xs sj)
and a commutativediagram:
Let C be the
subgroup
of(Z/nZ)2 generated by (0, n/ m ).
Sincethe
ai’(1,O) (resp.
theai’(C» patch
togive
us a map(Z/nZ)s E (resp.
anS-group
schemeCm
and a mapCn--->E). Together,
thesedefine a level
(n, m)
structure whoseequivalence
class is well defined.Then, using (1),
one sees thatFH(S)
is the set ofequivalence
classesof level
(n, m )
structures onE,
as desired.With this
interpretation
of«’ H [Iln],
thetechnique
used in theproof
of Construction 4.13 of
[2, IV.4] easily gives
us the desired inter-pretation
ofJK [ 1/n]..
Let
M’ H [Iln] ]
andMH[I/n]
denote theunderlying algebraic
spaces of,«’[Iln]
andAlH[I/n] (i.e., they
are coarse moduli spaces for theunderlying
functors ofM[I/n]
andAlH[I/n]).
As we will most oftenbe
working
over afield k,
we introduce the notation:(We
areassuming
that the characteristic of k does not dividen.)
If there is anydanger
ofconfusion,
we will writen .Ãln,m,h
etc.We can say the
following
aboutM’ k
andMk :
PROPOSITION 1.2:
If
k is afield
whose characteristic does not dividen, then :
1.
Mk
is a smooth,geometrically
connected curve whose genus isindependent of
k.2. When k = C, there are
isomorphisms:
(rm(n)
isdefined
in theintroduction.)
PROOF: The map
is smooth and proper
by [2, VI.6.7]
and has connectedgeometric
fibers
by [2, IV.5.5] (note
thatdet: H (Z/nZ)*
issurjective).
So weneed
only
show thatMQ = rm(n)B.
But this follows from[2, IV.5.3]
since
det : H - (Z/nZ)*
issurjective
andrm(n)
is the inverseimage
ofH n
SL(2, Z/nZ)
inSL(2, Z). a
A level
(n, m)
structure has asimple
form when the base has aprimitive
mth root ofunity:
PROPOSITION 1.3: Let
a:(Z/nZ)xCmE be
a level(n, m)
struc-ture over
S,
where n is invertible on S and S has aprimitive
m th rootof unity.
Then there is anisomorphism CX (Z/mZ)s
over S.PROOF: Pick an étale cover
{Si
-S’},e/
so that(Cm)s
isisomorphic
to(Z/mZ)St
Let en :En
XEn -
ILn be the usualpairing,
and letCm
be aprimitive
mth root ofunity
on S. Over eachSi
there is aunique
section ui of
Cm
such that ui generatesCm
anden(a(l, 0), a (0, ui»
=(m.
Then the ui
patch
togive
anisomorphism (Z/mZ)s - Cm
over S..Thus,
whenever S has aprimitive
mth root ofunity,
we will write alevel
(n, m )
structure as§2
For every
prime
p, let7p
be the intersection ofSL(2, Z)
with anopen
subgroup
ofSL(2, Zp)
such that7p
=SL(2, Z)
for almost all p.Let
which will be fixed
throughout
this section. r is a congruence sub- group ofSL(2, Z),
and let n be its level(so
thatr(n)
Gn.
We also fix thesubgroup
For our purposes, the way to understand T is to reduce modulo
T(n).
We will use the well-known
isomorphisms:
induced
by
the natural maps, where vp is the usualp-adic valuation. 4>
will denote map
SL(2, Z) - SL(2, ZINZ),
andOp
will denote the mapSL(2, Z) ---> SL(2, Z/p vp(n)z).
We see that
r(p vP") ç Tp,
and it then follows thatis an
isomorphism.
We also have anisomorphism:
Combining
all ofthis,
we get abijection
which,
combined with thebijection
shows that the natural map
is
bijective.
The cusps of T can be identified with the set
Every
cusp has one or twopreimages
inmSL(2, )/N,
and this leadsus to define the sets
Since a double coset
Fu(-N)
is inC-(n
if andonly
if -u is inruN,
we see that:
This proves the first two assertions of the
following:
LEMMA 2.1: l
1.
If - 1 £ r,
thenC+(I)
is the setof regular
cusps andC-(I)
is theset
of irregular
cusps(see [3,
p.29)].
2.
If -1
Er,
thenC+(I) = W.
3.
If - 1 £
r and the levelof
r isodd,
thenC-(n = 0.
PROOF: To prove
3,
assume thatC-(n 0 0. By (4),
there existu E
SL(2, Z)
and y E r sothat _u-Iynu
Er(n),
and thisimplies
_yn
Er(n)
C r. It follows that -1 Er. -Let
vl(F = =ttC+(I)
andv-.(n = *C-(I).
Thenv.(F) P.’(n
+v-.(n
is the number of cusps ofr,
andWe can now compute
lIoo(r)
in terms of theTp’s:
THEOREM 2.2: For each odd
prime
p,definee(p)
so thatThen
PROOF: . then we have an
isomorphism
which then
gives
abijection:
Interpreting
this in terms of cusps(via
theanalog
of(2)
for ±r and:trp) gives
the desired formula.When-iiÉnp dd Fp,
the last assertion of Lemma 2.1implies
thatC-(Fp) = 0
for some oddprime
p, and thenC-(n = 0.
The result nowfollows from
(3), (5)
and the last two assertions of Lemma 2.1. Il§3
We will
study
the groupTm(n)
of the introduction. We will assumen - 2.
LEMMA 3.1: The index
g (I- (n» = [SL(2, Z): :trm(n)] is given by:
PROOF: The index of
r(n)
inrm(n)
isn/m,
and the index of-F(n)
inSL(2, Z)
is well-known(see [3,
p.22]). ·
We next want to determine the number of cusps of
rm(n).
The first step is to prove:PROPOSITION 3.2:
PROOF:
Identify SL(2, Z)/ N
with the setvia the map
sending (f ))
to(a). Using [3,
Lemma1.41],
it is easy to see that(a)
and(fJ)
as above represent the same double coset ofrm(n))SL(2, Z)/N
if andonly
ifNote that
Tm(n)
has the form of the r in§2, i.e.,
1
where each
7p equals Tp·(ps)
for some r and s, r:5 s.By (3),
we arereduced to the case n =
pS,
m =p’.
For every i between 0 and s there are
cp(ps-i)
different c’s between 1 andpS
withgcd(c, pS) = pi
‘(cp
is the Eulercp-function). By (6),
forevery such c there are
double cosets
represented by (f)
for some a.Forming
theappropriate
sum over i and
simplif ying yields
the formula:The next step is to determine
v.-(F. (n»:
PROPOSITION 3.3:
and
in ail other cases.
PROOF: Let
(’), gcd(a, c)
=1,
represent a cusp inC-(rm(n».
Then(a)
and_ (a C)
represent the same double coset inFn(n)BSL(2, Z)/ N,
sothat
by (6). Using
the second congruence tosimplify
thefirst,
we see thatn = 2 or 4. It is easy to
compute v+
and v- in these cases tocomplete
the
proof. a
Propositions
3.2 and3.3, together
with the results of§2, give
animmediate
proof
of:PROPOSITION 3.4: The number
of
cuspsof rm(n)
isgiven by
and in all other cases,
Next, we consider elements of finite order in
Tm(n):
PROPOSITION 3.5:
1.
FI(2)
hasexactly
twoconjugacy
classesof elliptic elements,
allof
which have order 4.
2.
FI(3)
hasexactly
twoconjugacy
classesof elliptic elements,
allof
which have order 3.
3. For
(n, m) -:;é (2, 1)
or(3, 1), rm(n)
has noelliptic
elements.Since -1 is in
Tm (n )
if andonly if n
= 2, we get:COROLLARY 3.6:
rm(n)
istorsion-free if
andonly if (n, m) 0 (2, 1),
(2, 2)
or(3, 1).
PROOF oF PROPOSITION 3.5:
By [3, § 1.4],
everyelliptic
element ofSL(2, Z)
isconjugate
to one of thefollowing:
Trace considerations now show that
Tm(n)
has noelliptic
elements forn > 3.
Also,
none of the above elements is congruent to(oo)
modulo 2or
3,
so thatF(2)
andT(3)
have noelliptic
elements. This leavesonly FI(2)
andT,(3),
and it is easy to determine theirelliptic
elements..Knowing
li, v. and theelliptic
elements forrm(n)
enables us to compute the genus ofXm (n ) by applying
the formula of[3,
Pro-position 1.40].
Then we can prove:PROPOSITION 3.7:
Xm(n)
has genus 0if
andonly if (n, m)
is oneof
the 18
following
orderedpairs :
PROOF: The genus formula referred to above shows that
Xm ( n )
does have genus 0 for the
pairs
listed in(7). Conversely,
assume thatXm(n)
has genus 0. The mapsXn(n)---.>Xl(n)
andX,,(n)--->X(m)
showthat both
XI(n)
andX(m)
have genus 0. As iswell-known,
thisimplies
2 s n s 10 or n = 12 and 1:5 m 5. Thepairs (n, m)
withm 1 n
satisfying
theseinequalities
consist of the 18 listed in(7)
and 7 more:(10, 2), (12, 2), (9, 3), (12, 3), (8, 4), ( 12, 4)
and( 10, S).
In each of these 7cases, one computes that
Xm(n)
has genus >_ 1. Il We nowstudy
the ramification of the natural map fromX ( n )
toXm(n):
PROPOSITION 3.8: The
ramification
indexof
the mapabove a cusp
of Xm(n) represented by (a) is gcd(n/m, c),
except that when(n, m) = (4, 1),
theramification
index above(2)
is 4.PROOF: From
(6)
it is evident that the number of double cosets inr(n)BSL(2, Z)/ N
which are contained in the double coset ofrm(n)BSL(2, Z)/N represented by (’)
isn/gcd(n, mc). Proposition
3.3shows but for the case
(n, m) = (4, 1 )
that this isequal
to the number of cusps ofX(n) mapping
to(a )
inXm(n).
Becauser(n)
is normal inrm(n),
thedegree n/m
of the map is theproduct
of the ramification index and the number ofpreimages.
Thistogether
with an examination of theexceptional
casegives
the result. aProposition
1.2 in § 1 shows thatXm (n)
can beregarded
as thecomplex points
of avariety Ma
defined over Q. We want to deter- mine the field ofrationality
of each cusp.Using [2, VI.5],
the action ofGal(C/Q)
on the cusps can be described as follows. The cusps arerational over
O(Cn).
Let(f)
represent a cusp, and take o- EGal(Q(l’n)/Q).
If u is aninteger relatively prime
to c and n whoseimage
in(ZInZ)* corresponds
to u, then u takes(f)
to(acu). Using this,
we can prove:
PROPOSITION 3.9: Let
(f)
represent a cuspof Xm(n),
and letThe
field of rationality of (f) is
the maximal realsubfield of Q(£r) if
c = 0 or
n/2
mod n, andQ(Çr)
otherwise.PROOF: Lift the above action of
Gal(Q(Çn)/Q)
on the cusps ofXm(n)
torm(n)BSL(2, Z)/N.
Let u be aninteger relatively prime
to cand n. Then
(6) implies
that(acu)
represents the same double coset as0
if andonly
ifau --- a mod
gcd( n, m c ), equivalently,
Since
a/gcd(m, a )
and r arerelatively prime,
the last congruence isequivalent
tou = 1 mod r.
The passage from the above double cosets to cusps is
straightforward
and concludes the
proof
of theproposition.
Il§4
Let n and m be as usual. In this section k will denote a field of characteristic
p >_ 0,
where:1.
p § n
andp 4 2, 3
2. k contains a
primitive
mth root ofunity.
Assume that
(n, m) 0 (2, 1), (2, 2), (3, 1 )
or(4, 1 ).
ThenF,, (n)
is torsion-free(Corollary 3.6)
and all of its cusps areregular (Pro- position 3.3),
soby
TheoremA.1, Mk
representsJUb i.e.,
there is a universal level(n, m)
structureon some
generalized elliptic
curveEk
overMk (note
that we’reusing Proposition 1.3).
The part of
Ek lying
overM’ k
is a smoothelliptic
curveEk
overM’ k -
The
complement Mk - M’,
when k isalgebraically closed,
can beidentified with the set
We can prove the
following:
PROPOSITION 4.1: Let n and m be as above.
1.
If
k isalgebraically closed,
thefiber of Ek ---> Mk
over the cusprepresented by (a)
isof
typeIb,
where b =n/gcd(n/m, c).
2.
Ek - Mk
is the Néron modelof Ei - Mk.
3. The group
of
sectionsof Ek ---> Mk having finite
order is isomor-phic
toZINZ EB ZlmZ.
4.
Ec - Mc
isisomorphic
to theelliptic
modularsurface for Fn (n) (see [4, §41).
Since all sections of an
elliptic
modular surface are torsion(see [4,
Theorem5. 1]
or[ 1 , 3.20]),
we get an immediatecorollary:
COROLLARY 4.2:
If
k has characteristic 0(and
aprimitive
m th rootof unity),
then the groupof
sectionsof Ek - Mk
isisomorphic
toZlnZ (D ZlmZ.
PROOF OF PROPOSITION 4.1: We will use the notation of the ap-
pendix (in particular,
the groupcP(N)
of §2 is writtenU).
The cusprepresented by (f)
is a double cosetThe fiber of
Ek ---> Mk
over this cusp is of typeIb,
where b is theunique positive integer dividing n
such that(see (10)
in theappendix).
It is easy to calculate that b =n/gcd(n/m, c),
as desired.To prove the second
assertion,
we first compute the order ofj : Mk --* P’
at(a). Using [2, VI.5.3]
and theproof
ofProposition
3.8, wesee
that j: Mn,n,k ----> P k
1 has apole
of order n at every cusp. ThenProposition
3.8 showsthat j
has apole
of order b =n/gcd(n/m, c)
at(a).
SinceEk
has sections which hit every irreducible component of the fiber over(a),
it follows thatEk
is the Néron model ofE’ --- > M’
at(a).
The third assertion is now easy to prove. We can assume that k is
algebraically closed,
and let G be the group of sections ofEk -- Mk.
The map ak
(see (8)) gives
aninjection
The fiber over the cusp
represented by (ô)
is of typelm by
1, so that[4,
Remark1.10] gives
aninjection
It also follows from 1 that for a- in
G,
nu hits the zero component ofevery fiber.
Thus, by [4, Proposition 1.6], nGtor
= 0. From this weimmediately
see thatGtor:::::: ZlnZ EB ZlmZ.
To prove the last
assertion,
letX--->Xn(n)
be theelliptic
modularsurface for
rm(n).
Note that X is analgebraic
surface. Letf : X ° -
Ym(n)
be the restriction of this overYm(n).
For T EJ ,
the fiber off
over[T]
EYn(n)
is theelliptic
curveXT
=C/(Z
+Z T),
and the mapssending [T]
to[1 ln]
and[T/m ]
inX, give holomorphic
sectionsof f
which define aholomorphic injection:
We want to show that â is
algebraic.
X’,
the kernel ofmultiplication by n,
is analgebraic
curve(being
étale over
Ym(n»,
and soXn,
its closure inX,
is finite overXm(n).
Thus we have a finite map
Xn ---> Xn (n)
andholomorphic
sections(given by à)
overYn(n).
Theseclearly
extend and hence are al-gebraic.
Then, using
the factthat.,Uo c
isrepresented by Ym ( n ),
there is a map(3: Ym(n) Ym(n)
and a cartesiandiagram:
Suppose (3([ ’Tt]) = (3([ ’T2]), ’Ti ES).
ThenaTj: Z/nZ 0153 Z/mZ 4 XTj (i =
1, 2)
areisomorphic
level(n, m)
structures. From this it is easy to findy E rm(n)
withy(T,)
= T2. Thus13
isinjective
and hence an isomor-phism.
This proves ourassertion.
Let us
briefly
discuss the cases(n, m) = (4, 1)
and(3, 1).
FI(4)
istorsion-free,
so thatMk
represents.1l2 by [2, VI.2.7]. Thus,
there is a universal level(4, 1)
structure on anelliptic
curveFor k
algebraically closed,
the Néron model ofE’ k -->MO k
has fibers oftypes
7t, 14
andIt
over the cusps(ô), (0)
and(2) (the irregular cusp),
and for
any k,
its group of sections of finite order isisomorphic
toZ/4Z. Also,
theelliptic
modular surface forFI(4)
is the Néron model ofE’--->M’,
andE’--->M’
hasonly
torsion sections when k has characteristic zero.FI(3)
has anelliptic element,
so thatAl2
is notrepresentable.
Thiscorresponds
to the fact thatover k
there is aunique
level(3, 1)
structure ao with a nontrivial
automorphism (over C,
it isgiven by [(1- W)/3]
EC/Z
+ Zw, w =e2Tri/3).
But the functoril2
definedby
; a never
equals
ao overil
is
representable by M2 ç M2.
Thus there is a universal level(3, 1)
structure on an
elliptic
curveIf k is
algebraically closed,
thenùi
=Mo - laol,
and the Néron modelof
Ê’---> ll’
has bad fibers of typesIl, 13
and IV* over(1), (0)
and ao.For any
k,
the group of sections isisomorphic
toZ/3Z. Also,
the Néron model ofE" c ---> M’ c
is theelliptic
modular surface forFI(3),
andall sections are torsion when k has characteristic zero.
§5
Now we come to the main result of the paper. For any field
k, k(t)
will denote the field of rational functions in a variable t.THEOREM 5.1: Let k be a
field of
characteristic pa 0,
and assume that p 02,
3. Let n and m bepositive integers
withm n,
and setG =
Z/nZ E9 Z/mZ.
Then thefollowing
areequivalent:
1. There is an
elliptic
curve E overk(t)
with nonconstantj-invariant
such that G::::::
E(k(t»,o,,
the rationalpoints of finite
order notdivisible
by
p.2. p does not divide n, k contains a
primitive
mth rootof unity,
andG is one
of
thefollowing
19 groups :For k = Q or C, this means:
COROLLARY 5.2:
If
E is anelliptic
curve overC(t) (resp. Q(t»
withnonconstant
j-invariant,
thenE(C(t»tor (resp. E(Q(t»tor)
must be oneof the
19 groupsof (9) (resp.
oneof
the 15 groups on thefirst
twolines
of (9)). Furthermore,
allof
these do occur.PROOF oF THEOREM 5.1 :
1 => 2. Certainly p 4’ n,
and since(ZlmZ)’C G C E(k(t)), k
must have aprimitive
mth root ofunity (this
is a well-known consequence of the existence of the
pairing
em :Em
xE,,, --> By Propositions
3.7 and1.2,
weonly
have to prove thatMk
has genus 0.But G C
E(k(t» gives
a level(n, m)
structure onE,
so that we get a commutativediagram:
where j
=j(E)
is thej-invariant
of E.Since j
isdominating,
u must bedominating.
Thus the function field ofMk injects
intok(t),
which shows thatMk
has genus 0.2# 1. First,
assume thatG40, Z12Z
or(Z/2Z)2.
Let K be the function field ofM,,,,,,k.
ThenProposition
4.1 and the discussion of level(3, 1)
and level(4, 1)
structuresgive
us anelliptic
curve E overK with nonconstant
j-invariant
such thatE(K)tor
= G.In §3 we described the Galois action on the cusps of
Xn(n).
Construction 5.3 of
[2, VI.5]
shows that thisdescription
alsoapplies
to the cusps of
Mk.
It is then easy to see that the cusprepresented by
(°)
is rational over k. SinceMk
has genus 0(Propositions
3.7 and1.2),
we see that
Mk = P’.
Thus K =k(t).
To show that the groups
0, Z/2Z
and(Z/2Z)2
can occur, considerthe
following elliptic
curves overk(t),
definedby
theequations:
Each of these
equations
has a Néron model overPk.
The bad fibersare of types
Il, h
and II * for the firstequation, I,, I2
and III * for the secondand I2, I2
andI 2
for the third.Then, working over k
andusing [4, Proposition 1.6]
as in§4,
oneeasily
sees that the group oftorsion solutions is
0, Z/2Z
and(Z/2Z)2 respectively. ·
Appendix
Let H be a
subgroup
ofGL(2, ZINZ).
Thealgebraic
stack«’ H [lin]
has a
compactification «H[Iln]
relativeto Z[Iln] (see [2, IV.3]),
andwe set
.JlH[I/n] = «H[Iln] -.,«’ H [Iln].
Let r be the inverse
image
of H nSL(2, ZINZ)
inSL(2, Z).
Thepurpose of this
appendix
is to relate therepresentability
of.ÁlH [1/6n]
and
.l H [ 1 /n ]
to some well-knownproperties
of r.Specifically,
we willprove:
THEOREM A.l:
AtlH[1/6n]
is analgebraic
spaceif
andonly if
r istorsion-free
and allof
its cusps areregular.
THEOREM A.2:
«Ir[l/n] is
analgebraic
spaceif
andonly if C-(D = 0 (see §2).
The first theorem follows from the second
using
Lemma 2.1 and[2, VI.2.7].
To prove thesecond,
we use theinterpretation
of,«Z[I/n]
given
in[2, IV.6].
Let k be analgebraically
closed field whosecharacteristic does not divide n, and let C be a Néron
polygon
with bsides, b 1 n,
over k.A level H structure on C is described as follows. Let C° = creg =
Gm,k
xZlbZ,
and letÛo
=Gm,k
xZINZ.
There is a natural inclusion C° CÛ’.
Anisomorphism
JLn,k =Z / n Z
defines anisomorphism s : C° ->
(ZInZ)’,
and let B be theimage
ofCn
under s. InGL(2, ZINZ),
definethe
subgroups:
Then,
from[2, IV.6],
a level H structure on C is a double cosetHaU(B), a
EGL(2, ZlnZ),
such thatNext we describe how
automorphisms
of(C, +) (see [2, II.I])
act on level H structures on C. LetUo
be theimage
of the mapAut(C, +) --->
Aut(C" . Every automorphism
ofCn
extends to anautomorphism
ofCn,
andusing [2, II.1.10],
we get an exact sequenceThen an
automorphism 4>
of(C, +)
takes a level H structureHa U(B)
to the level H structure
HauU(B),
where u in ± Uand 0
map to thesame
thing
inUo.
LEMMA A.3: A level H structure
Ha U(B)
on C has a nontrivialautomorphism if
andonly if
PROOF: The case n = 2 is trivial. When n
3,
the lemma is an easy consequence of(10)
and the fact thatUo
isisomorphic
toAut(C, +).
Il Now Theorem A.2 follows
easily.
An element inC-(n gives
us, via(4)
in§2,
an element ESL(2, ZlnZ)
such thatBy
the abovelemma, HaU(B)
has a nontrivialautomorphism.
Con-versely, again using
LemmaA.3,
suppose we havea-Iha
E - U forsome a
E GL(2, Z/nZ)
and h E H. Note that h lies in HnSL(2, Z/nZ).
Let16 = (o " ,
wherer = det(a)-’. Then u=a{3
is inSL(2, ZINZ)
and satisfies(11).
From(4)
in§2,
we get an element ofC-(r).
aREFERENCES
[1] D. Cox and S. ZUCKER: Intersection numbers of sections of elliptic surfaces.
Inventiones math., 53 (1979) 1-44.
[2] P. DELIGNE and M. RAPOPORT: Les schemas de modules de courbes elliptiques, in Modular functions of one variable II. Lecture Notes in Math. 349, Springer- Verlag, Berlin-Heidelberg-New York, 1973.
[3] G. SHIMURA: Arithmetic theory of automorphic functions. Princeton U. Press, Princeton, 1971.
[4] T. SHIODA: Elliptic modular surfaces. J. Math. Soc. Japan, 24 (1972) 20-59.
(Oblatum 12-XII-1978) Department of Mathematics Rutgers University
New Brunswick New Jersey 08903 Current Addresses:
D. Cox
Department of Mathematics Amherst College
Amherst, Mass. 01002
W. Parry
Department of Mathematics SUNY at Stony Brook Stony Brook, N.Y. 11794