C OMPOSITIO M ATHEMATICA
J IU -K ANG Y U
On the moduli of quasi-canonical liftings
Compositio Mathematica, tome 96, n
o3 (1995), p. 293-321
<http://www.numdam.org/item?id=CM_1995__96_3_293_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
On the moduli of quasi-canonical liftings
JIU-KANG YU
© 1995 Kluwer Academic Publishers. Printed in the Netherlands.
Department of Mathematics, Harvard University
Received 11 March 1993; accepted in final form 27 May 1994
0. Introduction
In
[11],
Lubin and Tate established a versionof explicit
class fieldtheory
over alocal field
K0:
the ray class fields aregenerated by
torsionpoints
of the Lubin-Tateformal
Ao-module,
whereAo
is thering of integers
inh’o.
This is ananalogue
ofthe Kronecker-Weber theorem that the ray class fields of
Q
aregenerated by
rootsof
unity.
In this paper, we will derive an
analogue
of another classical theorem[ 15,
Theo-rem
5.7]
which asserts that thering
class field of an order in animaginary quadratic
field is
generated by
the moduli ofelliptic
curves withcomplex multiplications by
that order. Infact,
ourtheory (Section 4)
describes thering
class field of anyAo-order
0 in anarbitrary separable
extension K ofKo.
The moduli space here is the Lubin-Tate moduli space(Section 2)
thatparametrizes liftings
of a formalAo-module.
The role ofelliptic
curves withcomplex multiplications
will beplayed by liftings
withendomorphism
actionby
0. Theseobjects
are first introducedby
Gross
[4]
in case Il is aquadratic
extension ofKo. Following
histerminology,
such
liftings
are calledquasi-canonical liftings.
Some
properties
of thesequasi-canonical liftings
will be studied. Inparticular,
we will
compute
the Newtonpolygon
of themultiplication-by-03C00
map(03C00
is aprime
element of
K0)
of aquasi-canonical lifting (Section 6).
From thiscomputation,
thevaluations of the moduli
(in
asuitably
normalized coordinatesystem)
can bederived. In
particular,
we characterize the situations where the moduli areprime
elements of the
ring
class field. WhenK/K0
isquadratic,
this has been doneby
Gross
[4]
andFujiwara [3].
The result is fundamental in the non-archimedean localheight computation of Heegner points by Gross-Zagier.
For otherapplications,
see[3],[8].
In Sections
11-13,
westudy
theGross-Hopkins period
map(see [5], [6],
andSect.
11 )
from the Lubin-Tate moduli space to aprojective
space. Wepresent
anexplicit
form of theperiod
map andstudy
its valuationalproperties,
which arecontained in its valuation function
(Section 7).
It tums out(Section 13)
that theexceptional
set of its valuation function looks like walls in(part of)
theapartment
of the Bruhat-Tits
building
ofSLn(K0),
and the vertices areexactly
the valuations of moduli ofquasi-canonical liftings.
Finally,
in Section 14 we determine theendomorphisms
of thecanonical lifting,
which is defined to be a
lifting
withendomorphism
actionby
the maximal order.Thus all results in Gross
[4]
in the case ofheight
2 aregeneralized
to the case ofarbitrary
finiteheight.
1. Notations and conventions
Let
K0
be a non-archimedean localfield, Ao
itsring
ofintegers,
iro a fixedprime element,
andk0 ~ Fq
its residue field. We useK0 to denote
thecompletion
of thealgebraic
closureof K0,
andAo
itsring
ofintegers, ko
its residue field.Let Il be a finite
separable
extension of.Ko
and letA,
7r, k -rF y
be the corre-sponding
constructs for K. The valuation onK0
is normalized so thatord(K*)
= Z.Therefore, ord(K*0)
=eZ,
where e is the ramification index ofK/K0. Finally,
weuse n =
e f
to denote the extensiondegree [K : K0].
An
Ao-lattice
in a finite-dimensionalh’o-vector
space V is a freeAo-submodule
of V of rank
equal
to dim V. AnAo-order
in K is anAo-subalgebra
of K whichis an
Ao-lattice.
In whatfollows,
we will saysimply
"lattice" and "order" instead of"Ao-lattice"
and"Ao-order".
All formal modules are assumed to be
1-dimensional, commutative,
and offinite
height.
If G is a formalAo-module,
then themultiplication-by-a
map on G(a
EA0)
is denotedby [03B1]G.
The group of a-torsionpoints
is denotedby G[a].
If x, m are
integers,
m >0,
we use(x
modm)
to denote theunique integer
ksuch that
0 k
m - 1 and k m z(mod m).
2.
Liftings
and the Lubin-Tate moduli spaceLet k’ be an extension of
ko, regarded
as anAo-algebra.
We denoteby Ck’
the cate-gory of
complete, noetherian,
localAo-algebras
whose residue fields are extensions of k’. This category has an initialobject WAo( k’) (see [2]
and also[14,
II. Sect.5]).
In
particular, every e Ck,
has a canonical structure of aW Ao( k’)-algebra.
Let
G
be a formalAo-module
over k’ and let{R, mR ) ~ Ck’.
We define alifting
of
G
over .R to be a formalAo-module
G over R such that G o(R/mR)
=G.
Twoliftings G,
G’ are consideredisomorphic
if there is anisomorphism f :
G ~ G’ offormal
Ao-modules
over R such thatf
0(R/mR)
is theidentity automorphism
ofG.
Let
XG(R)
be the set ofisomorphism
classes ofliftings
ofG
over R. Thisis a co-variant functor on the category
Ck’.
A fundamental theorem due to Lubin and Tate[12,
Theorem3.1] (see
also[1, Proposition 4.2], [7,
Theorem22.4.4])
asserts that when
G
is of finiteheight
n, the functorXo
isrepresentable by
WA0{k’)[[t1,..., tn-1]],
i.e.XG ~ Spf WA0(k’)[[t1,..., tn-1]].
Moreprecisely,
THEOREM. Assume that G is
a formal Ao-module offinite height
n over an exten-sion
k’ofko.
There is alifting g of G over WA0(k’)[[t1,..., tn-1]] with the following property : for any li, fting
G over(R, mR)
EC,
there is aunique
element m EmR 1
such that the
lifting
obtainedfrom 9 by specializing
ti to mi isisomorphic
to G.
The
lifting g
is called auniversal lifting
ofG.
It is notunique though
any two such areisomorphic.
To fix a choiceof 9
is to choose a coordinatesystem
onXG,
i. e. to attach an element m E
m"R-1
to everylifting
G EXG(R).
We call m themodulus of G.
When
(R, mR)
is a discrete valuationring,
it isinteresting
tostudy
the valuations of the moduli ofliftings.
Forexample,
ifG,
G’ areliftings
with moduli m,m’,
then
has the
following meaning:
theidentity automorphism
ofG 0 (R/mR)
=G’ 0 (R/mR)
can be lifted to anisomorphism
G ~G’ over R/mR
but cannot be liftedto one over
R/mN+1R.
3. The
isogeny
constructionNow asume that R is the
ring
ofintegers
in some finite extension ofKo
and that Gis a
lifting
over R ofG,
which is defined overR/mR
and isof height
n. The Tatemodule
T(G)
is defined to bewhere the map
G[03C0m+10]
~G[03C0m0]
is[7rO]G-
It is a freeAo-module
of rank n.Specifying
a lattice L inT(G) ~ Ko containing T(G)
isequivalent
tospecifying
a finite
Ao-invariant subgroup
schemeCL
of G. We then can form thequotient G/CL.
This formalAo-module
can begiven
a canonical structure of alifting
of0(qN) (qN = #CL(A0), 0(qN ) being
the formalAo-module
obtainedby applying
the base
change x ~ xqN to G),
which we now describe:We can find a formal
Ao-module G’
and anisogeny
g: G ~G’
such that the kemelof g is CL and g 0 (R/mR)
is the Nth power of Frobenius: X -XqN.
Infact,
such apair (g, G’)
is fumishedby
Serre’s formula[10,
Theorem1.4]:
and G’ is such that G’(g(X), g (Y»
=g(G(X, Y)) and [a]G’(g(X))
=g([a]G(X))
for all a E
Ao.
The conditionthat g ~ (R/mR)
is the Nth power of Frobeniusimplies
that the reduction of G’ isC(qN) . Therefore,
G’ is alifting
of0(qN) .
Theisomorphism
class of thislifting
does notdepend
on the choice of(g, G’).
We willdenote this
lifting by GL and
denote theisogeny
g: G ~GL by
gL. It is immediate from Serre’s formula thatGL
can be defined overR[CL(A0)].
Let K’ =
End(G) ~ K0.
This is an extension ofKo
ofdegree
n. For any two latticesLI, L2 2 T(G),
we canidentify Hom( GL, , GL2) ~ Ko
with K’by
thecorrespondence
Here9L21 means h~b-1
EHom(GL2’ G)~K0, if b
EA0, and h:GL2 ~
G is anisogeny such
that h o 9L2 =[b]G ([10], 1.6).
We claim: under thiscorrespondence, Hom(GL1, GL2)
is identified withTo see
this,
assumethat 0
is anisogeny
andput 03C8 = g-1L2 o ~
o 9LI- Choose b EAo
such that
[b]G
og-1L2
is anisogeny.
It iseasily
checkedL2 that e 0
b is anisogeny
and(03C8 ~ b)([b]-1GCL1) ~ CL2.
Thatis, (03C8 ~ b). (b-1L1) C L2.
Conversely, if lb = 9£21 0 cjJo 9LI
is suchthat 0 . L
1g L2,
we can choose b EAo
such
that 1b ~ b, ~ ~
b are bothisogenies.
For any y eG LI [b],
write y = 9LI(x)
with x E
[b]Õl(CLl).
We then have(~ ~ b)(y)
= gL2 0(’ljJ 0 b)(x)
= 0 since(1b
0b)(x) E CL2 by assumption.
Thus we haveproved
thatGL1 [b] 9 ker(o 0 b).
Therefore, ~
is anisogeny ([10], 1.5).
4.
Quasi-canonicalliftings
Let F be the Lubin-Tate A-module associated to
(K, 7r) (i.e.
theunique
formalA-module over A such that
[03C0]F~k(X0 = Xqf,
see[12]),
considered as a formalAo-module
ofheight n
over A. LetF
=F 0 k,
so that F can beregarded
as alifting
ofF.
We call F thecanonical lifting
ofF.
Alifting
of the formFL (for
some lattice L in
T(F) 0 1(0 containing T(F))
is called aquasi-canonical lifting
(of F(qN), where qN = [L : T(F)]).
The modulus ofFL
will be denotedby
mL.Since the
endomorphism [1r]F
of F satisfies([03C0]F ~ k)(X)
=Xqf
andqf
=#F[03C0],
we see thatFL
=F1r-IL
for any L. Thus we are led to the action of7rz
onthe set of all lattices in
T(F) ~ Ii7o.
The orbits of this action are called lattice classes and the lattice class of a lattice L is denotedby [L].
Frompreceding discussions, FL depends
on[L] only.
Thus we can defineFL
for any Lby FL
=F1r-m L
form » 0.
By
Lubin-Tatetheory [11], FL
can be defined over an abelian extension of K.We shall determine the smallest field of definition of
FL.
First ofall,
we note thatthe group A* acts on lattice classes in an obvious way. For every
N,
it also acts onXF(qN)(R)
=mR through
the Artinsymbol
whenever R is an abelian extension of A. The relation of these actions is:Fa-1L = F(a,K)L,
orequivalently
Ma-IL= m(a,K)L,
for any a ~ A*.Indeed, letting
gL : F ~FL
be themorphism given by
Serre’sformula,
we thenhave
g(a,K)L
= ga -1 Lby
Lubin-Tatetheory [11]
and the above relation followsimmediately.
Now it is easy to see thatFL
can be defined over the class field of03C0Z
xOr,
whereThis class field is called the
ring class field
of the orderVL (Note:
thisdepends
on
03C0).
We aregoing
to show that this isactually
the smallest field of definition ofFL.
As we have seen in Section
3, End(FL) 0 K0
can be identified with K.PROPOSITION 1. Let
Li, L2
be lattices inT(F) 0 Ko. The following
threestatements are
equivalent:
(i) FLI is isomorphic to FL2 as formal Ao-modules ; (ii)
The lattice classes[L1], [L2]
are in the sameA*-orbit;
(iii) End(FL1), End(FL2)
are the same orderO in K,
andLi, L2
areisomorphic
O-modules.
Proof. (i)
~(iii):
First we observe thefollowing:
For any L and any non-zero x ET(G), Hom(FOL.x, FL)
is anOL-module isomorphic
to Lby
the discussion in the end of Section 2. Now if(i)
isfulfilled, clearly End(FL1)
andEnd( FL2)
areidentified with the same order O in
K,
andL 1 and L2
areisomorphic
as 0-modulesby
the observation wejust
made.(ii)
~(i):
ifLi
=a-1L2,
theautomorphism [a]:
F ~ F induces anisomorphism FL1 ~ FL2. (iii)
~(ii):
an (9-moduleisomorphism
fromL1
toL2
ismultiplication by
some element u03C003BD of K*.Clearly
u sends
[Li]
to[L2].
~PROPOSITION 2. Let
Li, L2
be lattices inT(F) 0 K0.
ThenFL1
andFL2
areisomorphic
asliftings (of some F(qN)) if and only if[L1] = [L2].
Proof.
Without loss ofgenerality,
we may assume thatLi, L2
are latticescontaining T(G).
LetgL:F ~ FL be
theisogeny defining FL (say given by
Serre’s
formula). By
thepreceding proposition,
we may assume thatL1
=aL2
forsome a E A*. Then an
isomorphism
fromFL1
toFL2
isgiven by
9L2 o[a]F
o9£11
(see
the discussion at the end of Section3). Any
otherisomorphism
is obtainedby composing
anautomorphism
ofFL1,
therefore is of the form 9L2 o[ab]F
o9£11,
with b E
OL1 = OL2.
It is easy to see that the reduction of thisisomorphism
is[ab]F(qN).
The latter isidentity
if andonly
if ab = 1 since x E A ~[x]F(qN)
isinjective. Therefore,
theisomorphism
is anisomorphism
ofliftings
if andonly
ifab = 1 and
Li
=L2.
~THEOREM 1. The
smallest field containing
K over whichFL is defined is
theclass field of 03C0Z
xVL.
Inother words, K(mL) is
thering class field of OL.
Proof.
This is immediate from the relation ma-IL =m(a,K)L
and thepreceding
proposition.
~COROLLARY.
For any N
>0, K (ML, FL[03C0N]) is the class field of 03C0Z
x(1
+7rN OL).
Proof. Indeed,
if u eO*L, x
EFL[1rN],
we have(u, K). x
= gL o[u-1]F
og-1L(x) by
Lubin-Tatetheory.
Thus(u, K)
fixesK(mL, FL[03C0N])
if andonly
if uy m y
(mod L)
for all y E1r-N L. Clearly
this condition isequivalent
tou ~
1 (mod 03C0NOL). ~
Let O be an order in K. It is
customary
to call a lattice in K a proper 0-lattice ifEnd( L )
= O. Two proper O-latticesL1, L2
areequivalent
ifL
1 =x L2
forsome x e K*.
COROLLARY Let G ~
A*/O*
be the Galois groupof the ring class field of
0. Then G permutes the
quasi-canonical liftings FL
withOL =
O. The action issimply
transitive on eachorbit,
and the orbitscorrespond bijectively
toequivalence
classes
ofproper
0-lattices. 0PROPOSITION 3. Let G be a
lifting of F 0k ko
overAo
such thatEnd(G) ~
Ois an order in A. Then there is some L such that G is
isomorphic
toFL
overAo
asformal Ao-modules.
Proof.
It isenough
to show that G and F areisogenous.
For any lattice M inT(G)~K0 containing T(G), the endomorphism ring of GM is {a
EK aM
CM}
(note
thatT(G)
~Ko
is a 1-dimensional K-vector spaceby
theassumption
onEnd(G)).
Therefore we can choose M so thatGM
has the structure of aheight
1formal A-module. As a
height
1 formal A-module isunique
overAo, GM
must beisomorphic
toF. o
With the
hypotheses
of the lastproposition,
choose anisomorphism f :
G ~FL.
Then g
=f
~k0
eAutk0(F).
Let thering homomorphism
l: A =EndA0(F) ~
D =
Endk0 (F)
be definedby
reduction. Then G is aquasi-canonical lifting
of thepair (F,
g o t og-1)
in the sense of Gross[4],
cf. Section 14.5. Some numerical invariants of
quasi-canonical liftings
Let T =
T(F)
and let L be a lattice in T 0K0.
Form thefiltration {Li =
L n
03C0-iT}i~Z
andlet li
=dimk0Li+1/Li.
Wehave li li-e
because the map 03C00:Li+1 ~ Li+1-e
induces aninjection Li+1/Li ~ Li+1-e/Li-e.
On the other
hand,
weclearly have li
= 0 for all isufficiently large, and l i
=f
for all i
sufficiently
small.Therefore,
there is aunique nondecreasing
sequence ofintegers,
oflength
n,a0(L) ··· an-1(L),
with thefollowing property:
forevery
integer
i and every natural number m, i appears m times in the sequence if andonly
if m =li-e - lie
EXAMPLE 1. If L =
T,
then the set{a0(T),
... , an-l1(T)}
consistsof 0, 1,
... , e-1,
eachoccurring
withmultiplicity f.
It is easy to see that in
general,
for each r EZ/eZ,
there areexactly f
indicesjl, ... , j f
such that aji(L)
~ r(mod e).
EXAMPLE 2. If
K / K 0
is unramified andL 2 T,
then thenumbers {a0(L),..., an-1(L)}
are theexponents
of theelementary
divisors ofL/T,
i.e.L/T ~
~n-1j=0A0/03C0aj(L)0A0.
In
general,
the structure ofL /T
as anAo-module
isL/T ~ ~n-1j=0A0/
03C0~aj(L)/e~0 Ao if L D
T. Moreprecisely,
there are elements xo, ..., xn-1 E T 0Ko
such that
Axj = 03C0-(aj(L)-e+1T
foreach j
and L =~n-1j=0A0xj.
Indeed,
if mi =li-e - li
>0,
we let yi1,...,yimi ELi-e+1
be such thatthey generate
the cokemel of 03C00:Li+1/Li ~ Li-e+1/Li-e.
Let L’ be theAo-
submodule of L
generated by
the nelements {yij}i,1jmi.
Form the filtration{L’i}i~Z
for L’and define thenumbers {l’i}.
Weclearly have l’i = li
- 0 forlarge
i. We also have
l’i-e - l’i li-e - li
= mi, since the cokemel of TTo:L’i+1/L’i ~ L’i-e+1/L’i-e
contains thelinearly independent
elements yi1,...,yimi. It followsthat lengthA0(L’/T) lengthA0(L/T).Since L’
CL, we must have L’
= L. Nowwe can
suitably
re-label the nelements (yij}i,1jmi
as xo, ... , xn-1 to concludethe
proof
of the above claim.EXAMPLE 3. If we consider the lattice
03C0-t L,
we will obtainaj(03C0-tL) = aj(L)+t.
EXAMPLE 4.
Let (, )
be anon-degenerate symmetric K0-bilinear
form onT ~
K0 satisfying ~ax, y) = ~x, ay)
for all a EK,
x, y E T ~Ko
DefineLV = {x
E T ~K01 (x, L~ ç A0}.
This is a lattice withOLV
=OL.
Thenaj(LV)
= d + e - 1 -an-1-j(L),
where d is such thatTV
=03C0-dT.
This is clear from the
following computation:
Now let 0 r0(L) ··· rn-1(L) be integers such that aj(L) ~ rj(L)( mod e)
and such that
rn-1(L)
is minimal withrespect
to this property. Forexample,
ife =
3, f
=1,
then(ro,
r1,r2)
is one of thefollowing: (o,1, 2), (o, 2, 4), (1, 3, 5),
(1, 2, 3), (2, 4, 6), (2, 3, 4).
It is easy to see that the sequence
r(L) = (ro(L),..., rn-1(L))
is determinedby f(L) = (r(L)
mode). Therefore,
there aren!/(f!)e possible
values ofr(L).
The sequence
of integers ao(L) - ro(L),
... ,a,,, - 1 (L) - rn-1(L)
is still a non-decreasing
sequence, with all elements divisibleby
e.Therefore,
we can writewhere
so(L),
... , sn-1(L)
arenon-negative integers.
In the lattice class
[L],
we can choose aunique representative Lo satisfying Lo D T, Lo 1; 1r-IT.
ThisLo
is also characterizedby ao(Lo)
= 0. The numbersa(Lo), r(Lo), r(Lo), s(Lo)
arecanonically
associated to the lattice class[L].
Wedenote them
by a[L], r[L], r[L], s[L] respectively.
The above constructions can be
applied
to any rank 1 free A-module T. For the rest of thissection,
take T = A and(x, y)
=TrK/K0(xy).
Let V be an orderin K.
PROPOSITION 4. The
following
statements areequivalent:
(i) Every
proper 0-latticeis free
over 0.(ii)
The Galois groupof
thering
classfield of
O actssimply transitively
onquasi-canonical liftings FL
withVL
= V.(iii) 0’ is a free
0-module.(iv) ai(O)
+a, - 1 - i (0) is a
numberindependent of i.
Proof.
Theequivalence
of(i)
and(ii)
is clear from the secondcorollary
toTheorem 1. It is obvious that
(iii) implies (iv). Conversely,
if(iv) holds, put Ov
=xL,
where x eOV/{0}
is such thatord(x)
is minimal. Then 1 E L and o C L C A. But one caneasily
see thata(O)
=a(L) by assumption
and therefore#(A/O) = #(A/L).
Thus O = L andVV
is free.Since
(i) clearly implies (iii),
it remains to show that(iii) implies (i).
Let L bea proper 0-lattice. We claim that
LV L
=Vv.
One direction is easy:(LV L, 0) = Tr(LVLO) g Ao, so LV L
C0’.
To show the otherinclusion,
it isenough
to showthat
(LVL)V
C 0. We have x E(LVL)V ~ Tr(xLV L) g A0 ~ xLv
C(LV)V =
L => x E
OL
= 0. Now sinceOv
is a free0-module, LV L
=Ov implies
that Lis an invertible module over the local
ring
0.Therefore,
L is free. 0Let 4v =
1r - d A
be the inverse different ofK/K0.
LEMMA 1.
Suppose
that theresidue field of
O isof
orderqf 1,
thenProof.
We have#(A/O) = #(OV/AV) = #(OV/A)q-df,
andao(GV) =
d, a[OV]
=a(OV) -
d. So the first formula follows from the discussion afterExample
2. The second formula followseasily
from the first. 06. The Newton
polygon
of[7rO]FL
As a shorthand we use qj to denote
(qi - 1)1(q - 1).
THEOREM 2. Let
FL
be aquasi-canonical lifting
and let the Newtonpoly-
gon
of [7rO]FL
be thepolygonal
linejoining (1, e), (q, w1),...,(qn-1, Wn-1),
to
(qn,
Wn =0).
Thenwhere
fj,r[L](q)
EZ[q, q- 1 is
apolynomial depending on j
andr[L] only.
Note that in the statement of the
theorem,
we do not assume that(qi, wj)
is abreak on the Newton
polygon.
Weonly
assume that it lies on the Newtonpolygon.
The
proof
of the theorem will begiven
in Sections 9 and 10. Anexplicit
formulafor
fi,r[L]
will begiven
in Section10,
and we will see:COROLLARY The
point (qj, W j)
is a breakof
the Newtonpolygon of [03C00]FL
if
andonly if aj[L] ~ aj-1[L].
Inparticular, if K/K0
isunramified, (qj, wj)
isa break
if
andonly if sj ~ 0; if K/K0
istotally ramified, (qj, wj)
isalways
abreak.
Hazewinkel
[7] (cf.
Sect.Il)
shows that a universallifting
F overWA0(k)[[t1,..., tn-1]]
can be chosen in a way such thatChoosing
thisparticular universal lifting
meanschoosing
aparticular
coordinatesystem on the Lubin-Tate moduli space
XF.
Now we can reformulate Theorem 2as follows:
THEOREM 2’. Let wl, ... , Wn-l, wn = 0 be
given by the formula
in Theorem 2.Assume that
aj[L] ~
aj-i[L].
Thenord((mL)j)
Wi- ~COROLLARY. Let O be an order in K such that
an-1[OV] ~ a,,-2[0’1,
Putr =
r[Ov].
Then(mov )n-l
is aprime
element in thering class field of
(1if and only if
thefollowing
two conditions aresatisfied: (i)
the residuefield of
(1 isko;
(ii)
we have the relationREMARK. The
proof will
begiven
in Section 10. The condition on r is satisfied inparticular when ri
+ rn-1-iis independent
of i.EXAMPLE. The Newton
polygon
of[03C00]F,
where F is thecanonical lifting,
iseasily
seen to be thepolygonal line joining (1, e), (qf,
e -1),..., (q(e-1)f, 1 ),
to
(qn, 0).
This is the case r = +can =(0,..., 0, 1,..., 1,...,
e -1,...,
e -1)
(each
number occursf times).
So we obtainfor j = if + k, 0 k f - 1
We can use this to
compute
the valuations of the moduli of anyquasi-canonical liftings
with r = +can(e.g. when K/K0
isunramified,
wealways
have r = r-c. = 0and fj
= 1 forall j).
Thisgeneralizes Keating [8, Prop. 7],
where the caser =
rcan,
s =(0,
... ,0, Sn-1),
O =Ao
+ 03C0sn-1A is treated.If 0 is an order with
r[O]
= fcan and residue fieldko,
then( mov )n- i
is aprime
element of its
ring
class field.Let n = 2. We obtain Gross’ formula
[4, Prop. 5.3]:
Note that in this case, every
quasi-canonical lifting
has f = r-,. so thisgives
thevaluations of the moduli of all
quasi-canonical liftings,
and itdepends only
on thesingle
number s =s1[L],
called the "level" of thequasi-canonical lifting.
In thiscase mL is
always
aprime
element of thering
class field. See alsoFujiwara [3, Prop. 2].
7. Valuation functions
Let
f
be a non-zerorigid analytic
function ontu ~ K | ord(u)
>0}.
There is aunique piecewise
linear continuous functionVf: R>0
--+ R such thatV j (ord(u))
=ord(f(u))
for all u whose valuationord(u)
lies in a dense open subset ofR>0.
Thefunction
Vf
is called thevaluation function
off.
It followsimmediately
from thedefinition that
Vfg
=Vf
-I-Vg,
and we expect to have the relationVfog
=Vf
oVg
in
general.
Explicitly,
take coordinates and writef(X)
=EakX k,
thenVf(x)
=min{ord(ak)
+kx}.
The set
Ef
of all x such thatmink f ord(ak)
+kx}
is achieved at more than onevalue of k is called the
exceptional
set off.
For any u such thatord(u) e E f
wehave
Vf(ord(u))
=ord(f( u)).
The formula
Vf(x)
=mink{ord(ak)
+kx}
enables us to define valuation func- tions for moregeneral
power series(non-convergent,
withnegative
or fractionalpowers,
etc)
and extend the domain ofVf
to alarger
subset of R. Theidentity Vfog
=Vj
oVg
holdsprovided
thatV-1g(Ef)
contains no open set.It is well known that
Irf
and the Newtonpolygon
off
contain the same infor- mation aboutf.
The latter is defined to be theboundary
of the convex hull ofThe
graph
ofVf
is sometimes called the Newtoncopolygon
off.
All these can be
generalized
to power series of several variables. For moredetails,
see Lubin[9,
Sect.3].
8.
Spécial subgroups
Now let G be an
arbitrary
formalAo-module
overAo
ofheight
n. Let the Newtonpolygon of [03C00]G
be thepolygonal line joining (1, wo(G)), (q,
w(G)),
...,@ (q n-1 wn-1(G)),
to(qn, wn(G)).
We havewo(G)
= e andwn(G)
= 0.Again,
we donot assume that every
(q2, wi(G))
is a break of the Newtonpolygon.
The
following
dataclearly
all convey the same amount of information about G:2022 The Newton
polygon of [03C00]G;
2022 The valuation function of
[7rO]G;
2022 The
decreasing
sequencew( G) = (w1(G),..., Wn-t (G));
. The sequence
u(G) = (u1(G),..., un-1(G)),
where(recall
that qj =(qj - 1)/(q - 1))
2022 The
decreasing
sequencev(G) = (vi(G),
...,Vn( G)),
whereThe numbers
vj(G)
aresimply
theslopes
of the Newtonpolygon,
and we havethe
following interpretation: vj (G)
is thelargest
number thaving
thefollowing
property:
there is an
Ao-invariant subgroup
scheme C ofG[03C00],
of orderq3,
such thatord(x) t
for all x E C(here
andelsewhere,
weidentify
a finite group scheme with itsAo-points).
An
Ao-invariant subgroup
scheme C ofG[7rol
oforder qd is
calledspecial
ifand
Clearly,
aspecial subgroup
of orderqd
exists for d =0, 1,..., n.
And it isunique
if and
only
if(qd, wd(G))
is a break on the Newtonpolygon
of[7ro]G.
When(q, wl(G))
is abreak,
theunique special subgroup
of order q is studied in[9],
where it is called the canonical
subgroup.
As
suggested by
K.Keating
to me, when(qd, wd(G))
is abreak,
thesubgroup {x
eG[03C00]| ord(x) wd(G)}
can be called ageneralized
canonicalsubgroup.
So the
generalized
canonicalsubgroups
form a filtration ofG[7ro],
indexedby
thebreaks. Then a
special subgroup
can be characterized as anAo-invariant subgroup
that lies between two successive
generalized
canonicalsubgroups.
We do not have an
interpretation
of the numberUj( G).
But it tums out thatfor G =
FL, uj(G)
can beexpressed by
aparticularly
nice formula(Theorem 2).
Proposition
5 belowpartially explains why:
whenG’
isisogenous
to G and thekemel of the
isogeny
is aspecial subgroup, u(G’)
is related tou(G)
is a veryelegant
way.The
following
lemma can be verifiedby straightforward computations.
LEMMA 2. For
any positive integer d
n,the following
areequivalent:
(i) The polygonal line joining (1, w’0), (q, w’1),...,(qn-1, w’n-1) to (qn, w’n) is
a
convex polygonal line, where wi
=Wi( G)
+(q2 - 1)wd(G) for i d, wi
=PROPOSITION 5. For any
positive integer
d n,if
oneof
theequivalent
con-ditions in the
preceding
lemmaholds,
and C is anyspecial subgroup of G[03C00]
of order qd,
G’ =GjC,
then we havewhere
w’i is given
in(i) of the preceding lemma;
Proof.
It is routine toverify
theequivalence
of the statements(i), (ii), (iii),
so itis
enough
togive
aproof
of(i).
This can be doneby
the same method forproving
Theorem B of Lubin
[9],
which is thespecial
case d =1,
q = p of ourproposition.
The idea is to calculate
V[03C00]G’
asVg
oV[03C00]G
oV-1g,
where g : G ~ G’ is anisogeny
with kemel C. Because C is
special,
we can determineVg by
Serre’s formula. Weomit the details. 0
This
proposition
is ofindependent
interest. But it is notabsolutely
necessary in thesequel.
9. Proof of Theorem 2
Now consider the
quasi-canonical lifting
G =FL.
We assume that L is such thata(L)
=a[L],
i.e. L D T andL 1J 7r-’T.
We can find elements x0,...,xn-1 ~ T 0K0
such thatAxj = 03C0-(aj(L)-e+1)T
foreach j
and L =~n-1j=0A0xj.
Let gL :
F ~FL
be theisogeny defining FL,
with kemelCL,
which weidentify
with
LIT.
LetqC(i)
=#(CL
nF[7r’]).
Thefollowing
lemma is immediate from Serre’s formula and Lubin-Tatetheory.
LEMMA 3. For any torsion
point
x EFL(Ao),
leto(x)
be the smallestinteger
tsuch that x E
F[7r’]
+CL.
The valuationord(gL(x)) depends
ono(x) only
and isa
strictly decreasing function of o(x). Explicitly,
let t =o(x),
thenLemma 3 is the
key
to compute the Newtonpolygon
of[03C00]FL.
From it we seethat the valuations of the torsion
points
ofFL
arecomputable,
and these arejust
the
slopes
of the Newtonpolygon (see
Lemma 4(i) below).
All we need is somepatience
to unravel the formula.Let ~ be the function such that
ord(9L( x))
=~(o(x)).
LEMMA. 4.
(i) vj(FL) = ~(aj-1(L)+1).
(ii)
LetLd
=~d-1j=0A003C0-10xj
+ L. ThenSd, the subgroup of FL[03C00] identified with Ld/L, is special of order qd.
(iii) FLd is isomorphic to FL/Sd.
(iv) If moreover we
havesd(L)
> 0, thenProof. (i)
is immediate from Lemma 3 because the numbersvj(G)
are the val-uations of elements in
FL[03C00]
=gL([03C00]-1F(CL)). (ii)
and(iii)
are obvious. In(iv),
we need the
assumption
thatsd(L)
> 0only
to ensure that(ao(L), ... , ad- 1 (L), ad(L) - e,..., an-l (L) - e)
isnon-decreasing.
0Fix some d and assume that
sd(L)
> 0. LetG’
=F03C00Ld, qc’(i)
=#(C03C00Ld ~ F[03C0i]).
LEMMA 5.
Proof. (i)
Observe thatc(i
+1) - c(i)
=li, where li
is defined in Section 5.By definition, li-e-li
=#{j| aj(L)
=i}.
Now the formula can beproved by
an easy induction.(ii)
is obvious and(iii)
is an easy consequence of(ii). (iv)
is deducedusing (i), (ii)
and the factad[Ld]
=ad [L] -
e.(v)
iseasily
seen from the remarkafter
Example
2 of Section 5. cLEMMA 6. Assume that
sd(L)
> 0. We have(i) vd(G’)
=vd(G) - wd(G), (ii)
vd+1(G’)
=qdVd+1(G). Consequently, Vd(G) - wd(G) qdvd+1(G).
Proof. (ii)
can beeasily
deducedusing
Lemma5(iv).
In fact we see thatvj(G’) =
qdvj(G) for all j d+1.Similarly, using Lemma 5(iii) and the fact that ad(L)-e
ad- 1
(L),
we can show thatvj(G’) -
vj(G)
= x is the samefor j
=1,...,
d. Sincewe have
03A3nj=1(qj - q j-1 ) vj (G)
= wo - wn = e, we cancompute
andget
X = -wd. 0
By
Lemma3,
Lemma4(i),
and Lemma5(i), u(FL) depends
ona[L] only.
Thuswe can define
fj,r(q)
to be such thatuj(FL)
=fj,r(q)qj/(qjqj+1)
for anyFL
such that
a[L]
= r. We will compute/y,f(?)
in the next section and show that it is apolynomial
function eZ[q, q-1]. Assuming this,
we can prove Theorem 2now. Use an induction on
03A3sj[L].
We have assumed that the theorem is true whens [L]
=0,
i.e.03A3sj[L]
= 0. If03A3sj[L]
>0,
choose any d such thatsd[L]
> 0. Note thatr[Ld]
=r [L]
andsd[Ld]
=Sd[L] - 1, sj[Ld]
=sj[L] if j ~
d.By
Lemma 6and
Proposition
5(or rather,
theproof
of Lemma6),
we haveNow we can
apply
the inductionhypothesis
toFLd
to conclude theproof.
10.
Computation
offj,f (L)
We maintain the notations and
assumptions
of thepreceding
section.Moreover,
until the end of this
section,
we assume thata(L)
=a[L]
=r(L),
sos(L)
= 0.We are
going
to derive a formula foruj(FL)
=fj,r[L](q)qj/(qjqj+1).
We shallomit the reference to
FL
or L in notations because this causes noambiguity.
Forexample, rj
isrj(L); uj
isuj(FL).
One more definition is in order. For x E
Z/eZ,
we use x to denote theunique representative x
E Z of x such that0 x
e - 1. If xo, ..., xk EZ/eZ,
wewrite
if 0 (XI - x0) ··· (Xk - xo) .
If x, y, z EZ/eZ and x ~
y ~ z, we saythat y
is between x and z. Forexample,
if e =10,
we have2 ~ 5 ~ 6 ~ 8,
and7 ~ 9 ~ 0 ~ 4.
LEMMA 7.