C OMPOSITIO M ATHEMATICA
R AVI R AMAKRISHNA
On a variation of Mazur’s deformation functor
Compositio Mathematica, tome 87, n
o3 (1993), p. 269-286
<http://www.numdam.org/item?id=CM_1993__87_3_269_0>
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On
avariation of Mazur’s deformation functor
RAVI RAMAKRISHNA*
Math Dept. Princeton University, Princeton, NJ08544 Received 20 February 1992; accepted 16 July 1992
List of notations
W(Fq):
Thering
of Witt vectors of the fieldof q
elementsFq where q
=p’.
rco:
Thecategory
whoseobjects
arecomplete
Artinian localrings
with residue fieldFq
and whosemorphisms
are localhomomorphisms
that are theidentity
onthe residue field.
K : A finite extension of
Q,
withring
ofintegers
A.G :
Gal(KIK).
p :
A residualrepresentation, p :
G ~GLn(Fq).
REMARKS. All
homomorphisms, including p above,
are assumed continuous.We take p > 2
throughout
this paper. All group schemes are commutative.Introduction
Let R be in
rcO
and mR be the maximal ideal of R. Let0393n(R)
be the kernel of thereduction map
GLn(R) ~ GL"(Fq).
Let p: G ~GLn(R)
be ahomomorphism
suchthat 7c°p = p
where 03C0 is the canonicalprojection
R ~RIMR
=Fq.
We call p 1 and p2
strictly equivalent
if pi =YP2 y-l
for some Y in0393n(R).
Astrict
equivalence
class of lifts ofp
to R is called a deformation ofp
to R.DEFINITION. Let
p
begiven.
For R inrcO
we define Mazur’s functor F:W° -
Setsby F(R) = {the
set of deformations ofp
toRI.
Note that F is afunctor.
*The author held NSF graduate and Department of Education fellowships while this research was conducted.
Mazur has shown that F satisfies the first three
Schlessinger
criteriagiven
below. He has also shown that when
p
isabsolutely irreducible,
F satisfies the fourth of these criteria. Infact,
one needsonly
that theendomorphism ring
of thegalois
module associated top
beFq
to ensure that F satisfies the fourth criterion.The
argument
used in[B]
works with this weakerhypothesis.
We letC(p)
denote this
endomorphism ring. Schlessinger
showed that a functorsatisfying
these four criteria is
pro-representable.
Thus for suchp
there exists a universal deformationring, R(p).
For more on thesetopics
see[B], [Ml]
and[Sch].
We want to define a modified version of Mazur’s functor. We restrict our
attention to those elements of
F(R)
such that thegalois
modules determinedby
the deformation to R are the
generic
fibers of finite flat group schemes over A.The aim of this paper is to do this
functorially
and in some casescompute
the(uni)versal
flat deformationrings, Rfi(p).
Mazur has considered in[Ml]
arestriction that is similar in the
ordinary
case. The results hereapply
in thesupersingular
case. We find that if K =Qp, C(p)
=Fq
andp
comes from thegeneric
fibre of finite flat group scheme overZp
thenRfl(03C1)
=W(Fq)[[Ti, T2]].
Section 1
We let X be a
property
for finiteW(Fq)[G]-modules
such that the set of finiteW(Fq)[G]-modules
withproperty
X is closed under direct sums,subobjects
andquotients.
We define a subfunctorFx
of F.DEFINITION. We define
Fx(R)
to be those p inF(R)
such that when viewed asan
W(Fq)[G]-module,
p hasproperty
X. Note that R has finitecardinality
andwe assume that
p
hasproperty
X.PROPOSITION 1.1.
Fx is a functor.
Proof.
Let R and S beobjects
in0 and q5:
R ~ S amorphism
in0.
Then itsuffices to show that
03C1 ~ FX(R) implies ~ 03C1 ~ FX(S)
Let B = R" and D = S" as
rings
with G-actions. Themap 0
induces a map0":
B ~ D ofG-rings.
As D is a finitering,
it isfinitely generated
as a B-module.Let Xl, x2,..., xm
generate
D over B. Then we have asurjection
ofW(Fq)[G]-
modules 03A8: Bm ~ D
given by 03A8(ei)
= xi, where the ei are canonical basis elements of Bm. Asproperty
X is closed under direct sums, Bm hasproperty
X.As it is closed under
quotients,
D hasproperty X,
the desired result. Note thatfor 0 -
p to haveproperty
X we do not need R and S to be inrco.
All wereally
require
is ahomomorphism
R ~ S of finiterings.
This willcome
up in Section 4.DEFINITION. Let E: ~ Sets be a functor such that
E(Fq)
is apoint.
LetR°,
RI,
andR2 ~ 0
withmorphisms Ri ~ Ro
for i =1, 2.
We have the map:The
Schlessinger
criteria are:Hl.
R2 ~ Ro
smallimplies (*)
issurjective.
H2.
Ro = Fq, R2
=Fq [81 implies (*)
isbijective.
H3.
dim tE
=dim(E(Fq[81»
oo.H4.
Ri
=R2, R2 ~ Ro
smallimplies (*)
isbijective.
A map R - S is small if it is
surjective
and has kernel aprincipal
idealannihilated
by
the maximal ideal of R. The dual numbersFq [91 of Fq
is thering
with elements a + b03B5 where a and b are in
Fq
andg2
= 0.Schlessinger
shows thattE is a vector space over
Fq .
THEOREM 1.1. The
functor F x satisfies
thefirst
threeSchlessinger
criteria.If C(p)
=Fq
thenFX
alsosatisfies
thefourth
criterion.Proof.
We make use of the fact that Mazur’sfunctor, F,
satisfies the criteria.We first note that if we have Hl for
FX
then the rest follow.H2. As
Fq[03B5] ~ Fq
issmall, (*)
issurjective
forFX by
Hl forFX.
We know(*)
isbijective
for F so(*)
must beinjective
forFX.
H3. From Lemma 2.10 of
[Sch]
weknow tF x and tF
areFq
spaces. We havetFX ~ tF
which is finite dimensional and we are done.H4. Here we suppose
C(p)
=Fq. By
Hl forFX, (*)
issurjective
forFX. By
H4 forF, (*)
isbijective
for F. Thus(*)
isbijective
forFX.
Thus we
only
needverify
Hl forFX.
LetRo, R1,
andR2
be as in Hl andput
R3 = R1 R0 R2.
Let03C11 03C10 03C12 ~ FX(R1) FX(R0) FX(R2). By
Hl for F we canchoose
p E F(R3)
such that p - pi03C10 03C12
under the map:We show p has
property
X. Weeasily
see that the mapR3 ~ R1
xR2
isinjective
so the map
is an
injective
map ofW(Fq)[G]-modules.
As theW(Fq)[G]-modules
determinedby
pl and P2 haveproperty X
and theW(Fq)[G]-module
determinedby
p is asubmodule of their direct sum we are done.
PROPOSITION 1.2. Let
p
haveproperty X
and supposeC(p)
=Fq.
LetR(p)
andRx(p)
be thepro-representing objects of
F andFX respectively.
Then the mapR(03C1) ~ Rx(p)
issurjective,
i.e.FX
is a closedsubfunctor of
F.Proof.
Let S be theimage
ofR(p)
inRx(p).
Then the universalproperty
X deformation toRx(p)
factorsthrough
S. As S is asubring
ofRx(p)
thisdeformation to S lies in lim
FX(S/mlS).
Thusby universality
there is aunique
mapRX(03C1) ~
S.Composing
this with theinjection
S -Rx(p)
we recover the deformation toRx(p). By universality
thiscomposed
map is theidentity
so themap S -
Rx(p)
issurjective
and we have S =RX(03C1)
the desired result.Section 2
We now
provide
twoexamples
ofproperty
X which are closed under direct sums,subobjects
andquotients.
Recall that K is a finite extension ofQ p
withring
of
integers
A. We need thefollowing
lemma.LEMMA 2.1. Let 0 ~ T ~ U - V ~ 0 be an exact sequence
of
G-modules.Suppose
U is thegeneric fibre of
afinite flat
group scheme U over A. Then thereare
unique finite fiat
group schemes f7 and 1/ over A such that the above sequence is thegeneric fibre of
the exact sequenceof finite flat
group schemes over A.Proof.
The idea is to take for f7 the schematic closure of T in U. One then takes0/11f7
for "1/. See[S]
and[R]
for details.DEFINITION. Let
Ffl
be the subfunctor of Fconsisting
of those p EF(R)
suchthat the
galois
module determinedby
p is thegeneric
fiber of a finite flat group scheme over A.From the above lemma and the fact that a direct sum of finite flat group schemes over A is
again
a finite flat group scheme over A we see thatFfl is
afunctor
satisfying
theSchlessinger
criteria. Thus whenC(p)
=Fq
we have thatFfl
ispro-representable.
Note that there is noassumption
on the ramification of A.For the second
example
ofproperty
X we need to review the modules of[FoLa].
We areonly
concerned with aspecial
case of their modules and weassume, in their
notation,
that -9 =Zp throughout.
Here we insist that K beunramified over
Qp.
Let 6 be the absolute Frobenius.A filtered Dieudonné A-module is an A-module furnished with a
decreasing, exhaustive, separated
filtration(M’)icz
of sub-A-modules such that for eachinteger i,
we have a 6-semi-linear map~i: Mi ~
M. Furthermore it isrequired
that for
x ~ Mi+1, ~i+1(x)
=p~i(x).
These filtered modules form aZp-linear
additive
category,
MF.We denote
by MFt
f the fullsubcategory
of MF whoseobjects
M haveunderlying
spaces that are A-modules of finitelength
andsatisfy 1 Im ~i
= M.The
category MF f j
is a fullsubcategory of MFfor whose objects satisfy M°
= Mand.M’
= 0. We callRepfZp
thecategory
of finiteZp[G]-modules.
In[FoLa] they
construct a faithful exact contravariant
functor, Us : MFf,qtor ~ RepfZp. They
showthat when restricted to certain
subcategories
ofMFf,", US
isfully
faithful. All wewill need is their result that
Us
isfully
faithful onMFf,jtor for j
p.Fontaine and Lafaille have shown in Section 5 of
[FoLa]
that when theresidue field of A is
algebraically
closed the G-modules associated to these filtered modules have the closureproperties
ofproperty
X of theprevious
section.
They
do not state this result in theexplicit
form we want, butthey give
the
explicit galois
action forsimple
filtered modules. In the case ofarbitrary
residue fields
Faltings
has shown in[Fa]
Theorem 2.6 that the G-modules associated to these filtered modules have therequisite
closureproperties.
DEFINITION.
Let j
p. LetFi
be the subfunctor of F such that each p comes from a Fontaine-Lafaille module of filtrationlength equal
toj.
From the
previous paragraph
we know thatFi
is a functorsatisfying
theSchlessinger
criteria.Fontaine and Lafaille have shown in Section 9 of
[FoLa]
that thecategory
MFf,2tor
isantiequivalent
to thecategory
of finite flat group schemes over A. For M anobject
ofMFfc;r2, US(M)
is thegeneric
fiber of thecorresponding
group scheme.(As Us
isfully faithful, they
recover the result ofRaynaud
that a finitegroup scheme over an unramified extension of
Qp
extends to a finite flat groupscheme over the
ring
ofintegers
in at most oneway.)
We say that arepresentation
hasweight j
if it comes from anobject
M ofMFfc;/.
Suppose
now that K is unramified overQ,
and we aregiven
arepresentation p : Gal(K/K) ~ GL2(Fp)
such that thegalois
moduleF 2
is thegeneric
fiber of afinite flat group scheme over A. In this situation the functors
F fl
andF2
are thesame. The
problem
ofcalculating
therepresenting object
of these functors is addressed in thefollowing
sections.Section 3
We are now
ready
to do acomputation.
We restrict our attention to twodimensional
representations
as thosearising
from modular forms are two dimensional. Whendoing computations
in this paper weonly
mean that therepresenting object
istopologically isomorphic
to thering given.
Our results arenot
explicit
as those in[B].
Let K =
Q p
andp : G ~ GL2(Fp)
be therepresentation coming
from thegalois
action on the
p-division points
of anelliptic
curve overQp
withgood supersingular
reduction. From the Weilpairing
we know detp
= x the cy- clotomic character and thegood
reductionimplies
thatp
is ofweight
2. Weknow from Section 1.11
of [Se1]
thatC( p)
=Fp.
Thus in the sense of Section 2 of this paper and Section 2.8[Se2] p
isweight
2. We show that in this caseRfl(p)
=R2(03C1)
=Zp[[Tl, T2]].
In fact all we need tocompute Rfl(03C1)
is thatp
isweight
2 and thatC(p)
=Fp.
First wecompute tF2.
LEMMA 3.1.
F2(Fp[E])
=Ext12(H, H)
where H is thegalois
module associated top.
The extensions on theright
are in the categoryof weight
2representations
andthe tilde indicates that we
only
consider those extensions killedb y
p.Proof.
We view(Fp[03B5])2
as a 4-dimensionalFp
space. Thenp ~ F2(Fp[03B5])
canbe written as
03C1(03C3) = (03C1(03C3)Z03B1 003C1(03C3)).
Such areprésentation clearly ives
rise to anelement
of Ext12(H, H).
An elementof Ext12(H, H) gives
such arepresentation
andhence
gives
an elementof F2(Fp[03B5]).
It is asimple
calculation toverify
that strictequivalence
of liftscorresponds
toequivalent
extensions.As H =
U s(M) for
M anobject of MFf,2tor, Ext12(H, H)
=Ext12(M, M)
where theextensions on the
right
are inMFf,2tor
and are also killedby
p. From[FoLa]
weknow that
lgAM = lgZp US(M)
so M hasunderlying
space a 2 dimensionalFp
space, and
Fp
=EndZp[G](H)
=End(M) by
full faithfulness. We will see thatonly
the size of the
steps
in the filtration andknowledge
ofEnd(M)
are needed todetermine
Ext12(M, M).
To
simplify computation,
we consolidatethe ~i
attached to M into asingle
notation. Recall that in this case the
object
M is killedby
p and thus a vectorspace over
Fp.
We will soon show thatdimFpM1 = 1.
Thus we can choose a basis(el, e2)
overFp
ofM° where e2
spansM1.
We havewhere a,
03B2, 03B3, 03B4
EFp.
As~1: M 1 --+ M,
the *’s are to indicate that~1
is not definedat el. We combine thèse to
get
asingle
matrix:Xm
=(03B103B2|03B303B4).
Thé dashesgive
the filtration structure. The number of columns to the
right
of the dashes is the dimension ofMl.
So here we have thatM’
is 1-dimensional. Weget
that anobject
M ofMFf,2tor
is determinedby
asingle
dashed matrixXM,
where the dashes indicate the size of thesteps
in the filtration. When wemultiply X M by
amatrix R which
respects
the filtration ofM, RX M
denotes the dashed matrix formedby R(po
andR~1. Similarly XMR
denotes the dashed matrix formed from(POR
and~1R.
We sayX M
commutes with R ifX MR
=RXM.
LEMMA 3.2.
Ext12(M, M) = Hom(M, M)/([R, XM]| R in M2(Fp)
respects thefiltration of M).
Proof. Let 0 ~ M ~ N ~ M ~ 0, N ~ Ext|2(M, M).
Thenwhere CE
M2(Fp)
is a dashedmatrix.
Note thatXN
has two sets of dashes. Thesecond and fourth elements of an ordered basis for the
underlying
space of N form a basis forNI.
The set of allpossible
C’scorresponds
to theHom(M, M).
We mod out
by equivalent
extensions. An elementP ~ Ext12(M, M)
withcorresponding
matrix D ~M2(Fp)
isequivalent
to N if there is amatrix 0
such that
We
get
the relation D = C +[R, XM].
Here R must preserve the filtration of M.This is because the
isomorphism
of N to P must preserve the filtrations. So C and Dgive equivalent
extensions wheneverC - D = [R, XM]
for some Rrespecting
the filtration asabove,
and we are done.We now use some of our
knowledge of p.
From[Sel]
weknow
thatC(p)
=Fp
so H and thus M have
endomorphism ring Fp.
We know thatdimFp M’
= 2 anddimFp M2 =
0. All that remains isdimFp M1.
Thisequals
1 because otherwise anendomorphism
of M would not have torespect
any filtration structure and any element ofM2(Fp)
that commuted withXM
would do. As the centralizer of every element ofM2(Fp)
is at least 2 dimensional we concludediMF.(M’)
= 1. So tocompute tF2
we noteHom(M, M)
is 4dimensional,
the set of R that preserve the filtration is 3dimensional,
andkernel(R ~ [R, XM])
issimply End(M)
which is 1dimensional.
Thus tF2
is4-(3 -1)
= 2 dimensional. One of these dimensions iseasily
seen to come fromtwisting p by
the étale character 1 + eX. The other dimension is not so clear.We now have that
R2(p)
=Zp[[T1, T2]]II
for some ideal I. We show that 1 iszero. We do this
by counting
theZp/(pj)-valued points
ofR2(p).
We will showthat for every
1,
there arep2(1- 1)
of them. This is the number ofZ pl(pl)-valued
points of Zp [[Tl, T2]]. So if f ~ I and (x, y)
E(pZpl(pl»2
thenf(x, y)
~ 0 modpl.
Lifting
to characteristic zero, we see that if(x, y) ~(pZp)2
thenf(x, y)
= 0.Without
difficulty
one can show such anf
must be zero. Thus I = 0 andR2(P)
=Zp[[Tll T2]].
Let us now count
Zp/(p’)-valued points
ofR2(p).
Wetry
to findobjects
N ofMFf,2tor
whoseunderlying
spaces are freeZp/(pl)
modules of rank 2. We denotethe kernel
of p
on Nby Np.
So we seek N such thatNp ~
M. This is the same asrequiring XN ~ XM mod p,
whereX N
has a dashed structure likeXM.
AsXN E M2(Zp/(pl))
and is determinedmod p
there arep4(l-1) possibilities.
We nowshow that there are
only p2(l-
1)possibilities
up toisomorphism.
We see thatXN1 ~ XN2
if andonly
if there is a matrixR ~ M2(Zp/(pl))
such that:We
require
R to be lowertriangular
to preserve the filtration. There arep’(1 - 1)
such R and
pl-1
lie in the centerof M2(Zp/pl)
and thus commute with allXN.
Weknow that no others commute because
C(03C1)
=Fp.
Thus up toisomorphism
thereare
p4(l-1)/(p3(l-1)/p(l-1))
=p2(l - 1)
lifts of M toZp/(pl).
Thus we have
proved:
THEOREM 3.1. Let p > 2 and G =
Gal(QpIQp),
andp :
G ~GL2(Fp)
beweight
2and suppose
C(p)
=Fp.
ThenRf’(P)
=R2(p)
=Zp[[T1, T2]].
Section 4
We now
generalize
the results of theprevious
section. Let G =Gal(Qp/6p)
Weconsider
p :
G ~GL2(Fq),
an irreducible flatrepresentation
with detplI
= x and q =p".
This is therepresentation
attached to aneigenform
ofweight
2 onho(N)
where
p 1
N at anon-ordinary prime.
Thefollowing proposition
is due to Serre.See
[Se2].
PROPOSITION 4
(Serre).
Let G =Gal(QpIQp).
Letp :
G ~GL2(Fq).
We assumeF 2 c F .1 is irreducible and at with det - -
then 03C1|I=(03C80 0 03C8p
whereFp2
cFq. If 03C1 is
irreducibleand flat
with detp 1X
thenp 1 ( n 0 where
is
a fundamental
characterof
level 2. FurthermoreC(p)
=Fq.
Proof. Let It = I/Ip
the tame inertia group. Recall that a fundamental character of level s is ahomomorphism It ~ F*ps ~ F*p
which extends to anembedding ofFp.
intoFp
as fields. Let V be the 2 dimensionalFq
space on whichG acts. As
p
is irreducible it follows fromProposition
4 of[Sel]
that the wild inertia group1 p
actstrivially
on EThus I
t acts on V and this action issemisimple. By Proposition
1 of[Se2]
this action is via two characters of level 1or 2. As det
03C1|I
= x it followsthey
are both level 1 or both level 2.If they
are bothlevel
1,
Serre shows in Sections 2.3 and 2.4 of[Se2] that 03C1|I = x a 0 b
where0 a b p - 2. As
detplI
= x andp
is flat we must haveby
Theorem 3.4.3 of[R]a=0
and b = 1. Thusp )j = (10 0~).
Thenby
local class fieldtheory
theimage of p
is abelian and one seesby considering
centralizers that the fullimage of 03C1
would be contained in( 0
andp
would not be irreducible. Thus we have that03C1|I=(03C8a0 b where
is a fundamental character of level 2 anda + b = p + 1 as
detp I r
= x. Butby
the flatness and Theorem 3.4.3 of[R]
we musthave,
without lossof generality, a
=0,
and b = p, the desired result. It remains to showC(p)
=Fq.
Oneeasily
sees that those elements ofGL2(Fq)
that commutewith the
image
ofp II
are of the form(*0 0*).
If in fact all such elementscommute with the full
image
ofp
then we find withoutdifficulty
that the fullimage of 03C1
is contained in the set of matrices of the form(*0 0*).
Inparticular
itis abelian so
by
local class fieldtheory
the order of theimage
of tame inertiadivides
p -1.
But this cannot be true asplI
acts via fundamental characters of level 2 so not alldiagonal
elements commute with the fullimage
ofp.
ThusEndFQ(H)
is 1dimensional,
and we are done.We want to
compute R(p),
the universal deformationring
with no restrictions.We recall from
[Ml]
and[B]
that thetangent
space isgiven by H 1 (G,
Adp).
Furthermore if
H2(G,
Adp)
is trivial thenR(p)
is a power seriesring
indimFq H1(G,
Adp)
variables. We show that for thep above, H2(G,
Adp)
is trivial.It will then follow from the Euler characteristic of local
galois cohomology
thatdimFqH1(G,
Adp)
= 5. See[Se3]
for the necessary results ingalois cohomology.
LEMMA 4.1. For the
p given above, H2(G,
Adp)
is trivial.Proof. By
Tateduality, H2(G,
Adp)
is dual toH’(G, (Ad p)*).
We show thatH0(G,(Ad03C1*) = ((Ad P)*)’
is trivial.Let ~~((Ad p)*)G.
Weshow 0
has 4dimensional kernel
so 0
= 0 and we will be done. We havethat 0: Ad p - Fp
where the G-action on Ad
p
isconjugation by p
and the G-action onFp
isby
thecyclotomic
character which isdet 03C1|I.
ForZ E Ad p, ~(g. Z) = g.~(Z).
For allg E I
we see03C1(g)Z03C1(g)-1-(det p(g))Z
E kernel0.
It suffices to show for some g e Ithat the
image
of the mapTg:
Z03C1(g)Z03C1(g)-1-(det p(g))Z
is 4 dimensional. We finda g ~ I
such that the kernel of this map istrivial,
anequivalent
result. ChoosegEl
so that03C8(g)
= a, an element oforder p2 -1
inFp2. (Recall 03C8
is a fundamental character of level2).
If weput Z = ( t
then03C1(g)Z03C1(g)-1-(det 03C1(g))Z simplifies
toOne
easily
sees that this last matrixequals
0only
if Z = 0. So we have that kernel~
is 4 dimensionalso ~
=0,
the desired result.LEMMA 4.2. For the
p given, H’(G, Ad p) is
5 dimensional.Proof.
We havedim H0 - dim H1 + dim H2 = - dim Ad 03C1 by
the Euler charac-teristic of Ad
p.
From the above lemma thatH2
is trivial.H°
is 1 dimensional becauseC(p)
=Fq
and Adp
is 4 dimensional.We now obtain similar results for the case
Fp2 ~ Fq.
PROPOSITION 4.2. Let G =
Gal(QpIQp).
Letp :
G ~GL2(Fq). Ifp
isirreducible, flat
and det03C1|I
= x thecyclotomic character,
thenC(p)
=Fq.
Proof.
Letp : G -4 GL2(Fq) GL2(Fq2)
where i is theinjection
i :Fq ~ Fq2.
Wehave
Fp2 c Fq2
anddet |I
= x.Furthermore p
is flatby
theargument
used inProposition
1.1. Weshow p
is irreducible. Then we canapply Proposition
4.1.As
p
is irreducible the wild inertia actstrivially through p
as before. Thusp(7p)
is trivial. We
suppose p
is reducible and arrive at a contradiction.If
p is
reducible then theimage of p
is contained in matrices of the form( * .
AsIp
actstrivially
we seethat |I = ( Y2
wherethe
arecharacters.
Since p
is flat we haveby
Theorem 3.4.3 of[R]
that y, and y2 areproducts
of distinct fundamental characters of level 2r. Weshow 71
=1 03B32.Let 03C8
be a fundamental character of level 2r and let a =
1 + p + p2 + ··· + p2r-1.
Asdet
plI
= x, and x= t/Ja
we see that if yi = y2 thenthey
bothequal 03C8a/2
or -03C8a/2.
But
by
Theorem 3.4.3 of[R], 03C8a/2 and -03C8a/2
are not flat. Thus03B31 ~
72- We nowshow p
must besemisimple.
If it is not, there is a 03C3 E G such thatAs
03B31 ~
72, there is a 03C4 ~ I such thatWe see that
But this must lie in the
Image p II
as inertia is normal. Asy(s - t)z-1 ~
0 we havea contradiction so no such u exists
and p
issemisimple
and thus abelian.As p
is abelian we see that the order of theimage
of inertia must dividep - 1.
By
flatness we seefi)j = ().
ThusplI = () and
asp is
abelian weeasily
see that theimage
ofp
is contained in thediagonal
matrices. This contradicts the fact thatp
is irreducible.Thus p
is irreducible.By Proposition
4.1 we seeC(p)
=Fq2.
As themap C(03C1)~Fq Fq2 ~ C(p)
isinjective
we seeC(p)
=Fq .
PROPOSITION 4.3. For
p as in Proposition 4.2, H2(G, Ad p)
= 0.Proof.
We knowC(p)
=Fq
and this isequivalent
toH°(G,
Adp) being
1dimensional. From the Euler characteristic we see
It is thus
enough
to showH’(G,
Adp)
is 5 dimensional.We know
H 1 (G, Ad p)
is at least 5 dimensional and we show that themap H1(G, Ad 03C1)~FqFq2 ~ H1(G, Ad )
isinjective, where p
is as before.Applying
Lemma 4.2 will finish theproof.
We show that no nontrivial deformation p
of p
toFq[03B5]
becomes trivial when considered as a deformationof p
toFq2 [9] .
Suppose
p is such a deformation. Thenp(6)
=A03C3+03B5B03C3.
We claim there is anX E M2(Fq2)
such that(I+03B5X)03C1(03C3)(I-03B5X)=A03C3. Simplifying,
we see that[A(1’ X]
=B(1.
We show that there exists a YEM2(Fq)
such that[A(1’ Y]
=B(1’
i.e.that p is the trivial deformation to
Fq[03B5].
Let i be the
automorphism
ofFq2 given by raising
to theqth
power. Then i fixesFq.
We haveas
A03C3
andB(1
lie inM2(Fq). Subtracting
one from the other we seeA03C3(X-X03C4) = (X-X03C4)A03C3.
AsC(p)
=Fq2
we see that X - XT lies in the center ofM2(Fq2).
Thus if X= s t
we have t =tt,
u = u’ and(s - v) = (s
-v)’.
Thus wehave
Choosing
Y =( t we
are done.From the
preceding
discussion we haveproved
thefollowing
theorem.THEOREM 4.1. Let p >
2,
G =Gal(Qp/Qp)
andp:
G ~GL2(Fq)
be an irreducibleweight
2representation
such that det03C1|I
=~. Then
the universaldeformation ring R(p)
=W(Fq)[[T1, T2, T3, T4, T5]].
We now compare
R(p)
toRfi(p).
The ideas are the same as in Section 3 but alittle more care must be taken. We
again
let H be thegalois
module and M the Fontaine-Lafaille module associated to H. NowdimFp M’
= 2r and we have amap
Fq ~ End(M).
This mapgives
M anFq
structure. Theunderlying
space ofM can now be realized as a 2-dimensional
Fq
space. Asendomorphisms
of aFontaine-Lafaille module preserve the filtration we see that
M 1
is also anFq
space. Since
C(p) = Fq
the sameargument
as in Section 3 shows thatdimFq M’
= 1. We also note that if N is the Fontaine-Lafaille module associated to aW(Fq)/(p’)-valued point
ofRfl(03C1)
then theunderlying
space of N is a freeW(Fq)/(pl)-module
of rank 2 andN1
is direct summand free of rank 1 overW(Fq)/(P’).
Tofind tF2
we cannotmerely compute É 2(M, M).
Instead we haveto
compute Ext12,Fq (M, M)
the extensions of Mby
M killedby
phaving
anFq
structure.
LEMMA 4.3.
F2(Fq[03B5])
=EXt12,Fq(H, H)
where H is thegalois
module associated top
and the extensions on theright
are in the categoryof weight
2representations
with an
Fq
structure.Proof.
As before.LEMMA 4.4.
Ext12,Fq(M, M)
=HomFq(M, M)/([R, XM]|R in M2(Fq)
respects thefiltration).
Proof.
Note that the Fontaine-Lafaille modules here areZp-modules,
but wegive
them anFq-structure.
We can do this as the a semilinear structure vanishes for modules overZp
so there is nodifficulty
inmaking
these two structurescompatible.
ForN~Ext12,Fq(M, M )
we haveXN= ()
where C is adashed matrix in
M2(Fq).
Thatis,
the set of allowable C’scorresponds
toHomFq(M, M).
An elementP E EXt 2,Fq (M, M)
withcorresponding
matrix D iséquivalent
to N if there is a matrix() R
such thatHere the matrices
I, R, C,
D andX M
are all inM2(Fq).
Furthermore notonly
does R préserve the filtration as
before,
but to ensure that( /7 R )
is anisomorphism respecting
theFq
structure, we see R must commute with theFq
structure. We
again get
D =C+[R,XM],
and the resultfollows.
So the set of R that preserve the filtration is the set of R e
M2(Fq)
that preservea 1 dimensional
subspace
of a 2 dimensionalFq
space. The set of such R is a 3dimensional Fq
space. Weget
thatjust
as before.So we have
Rfl(03C1)
=W(Fq)[[Tl, T2]]/I
for some ideal I. The samecounting
argument
as before shows I=0. Theonly
difference is that here we findRfl(03C1)
has
q2(l-1) nonisomorphic W(Fq)/(pl)-valued points.
We haveproved
thefollowing
theorem.THEOREM 4.2. Let
p : Gal(Qp/Qp) ~ GL2(Fq)
be an irreducibleweight
2 repre- sentation wheredet 03C1|I = ~
thecyclotomic
character. ThenRfl(03C1) = R2(p) = W(Fq)[[Ti, T2]1-
The
following corollary,
which is similar to Mazur’s result in theordinary
case,is immediate.
COROLLARY 4.1. For
p as above,
there is asurjective
mapR(03C1) ~ Rfl(03C1).
Thekernel
of
this map is an idealgenerated by
three elements.Section 5
We now make a
straightforward generalization
of theprevious theory.
We concern ourselves first with
computing weight j
universal deformationrings for j
p. Thekey points
here is thatUs : MFf,jtor ~ Zp [G]-modules
isfully
faithful. We assume then that
K = Q p
andp:G--+GL2(Fq)
is aweight j
representation
whose associatedgalois
module H hasendomorphism ring Fq.
To H there
corresponds
to a Fontaine-Lafaille moduleM ~ MFf,jtor.
Here weassume the filtration
length
of M isexactly j.
We have a mapFq ~ End(M).
Theunderlying
space of M is a 2r dimensionalFp
space. AsFq
acts onM,
we havedimFp M’is
one of0,
r or 2r. It cannot be 0 or 2r for the same reason as before.Here
X M
has aslightly
different dashed structure. There are infact j -
1 dashedlines. The basis with
respect
to whichX M
is written is foundby choosing
a basisof Mi - ’
andaugmenting
it to a basisof Mj- 2
and thenaugmenting
to a basis ofMj-3
and so on. The number of columns between the ith and i + lst dashed linesof X M
is the dimensionof Mi/Mi+1.
For theproblem here,
eachMi is
anFq
space so most of the dashed lines are
adjacent
with no columns between them.We
again
have thefollowing
lemma fordetermining tFj
=ËXfiFq(M, M).
LEMMA 5.1.
Ext1j,Fq(M, M) = HomFq(M, M)/([R,XM]|R in M2r(Fp)
respectsthe
filtration
andFq structure).
As before we see
tF2
is 2 dimensional. The sameargument
ofcounting W(Fq)/(pl)
valued
points
ofRj(03C1)
carriesthrough
and we have thefollowing:
THEOREM 5.1. Let p >
2, G = Gal(Qp/Qp)
and03C1: G ~ GL2(Fq)
beweight j.
Suppose
also thatC(p)
=Fq.
ThenRip)
=W(Fq)[[Tl’ T2]].
Section 6
Finally
we do acomputation
for anordinary
flat residualFp representation. By ordinary
we mean therepresentation
hassemisimplification
a direct sum of 1dimensional
representations,
one of which is unramified.Assuming
detp = x,
from Section 2.8 of[Se2]
we see that thisrepresentation
restricted to inertia isgiven by
We consider the latter case. We see
that 03C1 = ( ) where 03C81
andare
étale characters. One
easily
sees thatC(p)
=Fp,
soby
the methods of Section 4we see that
Rfi(p)
=Zp[[T1, T2]1-
We now insist that
03C81 ~ 03C82.
That such arepresentation
exists is an easy exercise incomputing
extensions of Fontaine-Laffaille modules. AZp-valued point
x ofRfl(03C1) gives
thereprésentation
03C1x =( 0 2 )
where X is thecyclotomic
character and theWi
are unramified characters with values inZ*
such that
03A8i ~ 03C8i
mod p. This follows as thep-divisible
group associated to px has non-trivial connected-étale sequence onp-torsion and
thus itself has non- trivial connected-étale sequence. The characters on thediagonal of px
must be asdescribed because there are
only
two 1-dimensionalp-divisible
groups overW(Fp), 03BCp~
and the constantp-divisible
groupQp/Zp,
i.e.03C1x|I = (). We
willshow that for any
pair
of étale characters(03A81, 03A82) lifting (03C81, 03C82)
there is aunique lifting
p ofp
toZ p
such that p =( )
. Asimple counting argument
will then show that thislifting
is then aZp-valued point
ofRfl(03C1).
Thus we will have shown that any
ordinary
lift ofp
that "looks"p-divisible,
i.e.has
semisimplification
the Tate module of ap-divisible
group overZp,
is in factthe Tate module of a
p-divisible
group overZp.
All that follows can be done more
generally,
but we treatonly
thespecific
casefor the
p
described above.Let 03A81 and ’P2
be étale liftsof 03C81 and 03C82
toZp*
We want to consider for R e0,
liftsof p
to R suchthat 03C1 =(
0Y2)
Thecharacters XIF 1 and ’y 2
acton R via the
composite
of their action onZp
and the mapZp ~
R. We call twolifts pi and P2 as above very