Taylor Wiles Small

8/13/2019 Taylor Wiles Small

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Ring Theoretic Properties of Certain Hecke

Algebras

Richard Taylor

D.P.M.M.S.,

Cambridge University,

16 Mill Lane,Cambridge,

CB2 1SB,

U.K.

Andrew Wiles

Department of Mathematics,

Princeton University,

Washington Road,Princeton,

NJ 08544,

U.S.A.

7 October 1994

Introduction

In the work of one of us (A.W.) on the conjecture that all elliptic curves definedover Q are modular, the importance of knowing that certain Hecke algebrasare complete intersections was established. The purpose of this article is toprovide the missing ingredient in [W2] by establishing that the Hecke algebrasconsidered there are complete intersections. As is recorded in [W2], a methodgoing back to Mazur [M] allows one to show that these algebras are Gorenstein,but this seems to be too weak for the purposes of that paper. The methods ofthis paper are related to those of chapter 3 of [W2].

We would like to thank Henri Darmon, Fred Diamond and Gerd Faltingsfor carefully reading the first version of this article. Gerd Faltings has also

suggested a simplification of our argument and we would like to thank him forallowing us to reproduce this in the appendix to this paper. R.T. would like tothank A.W. for his invitation to collaborate on these problems and for sharinghis many insights into the questions considered. R.T. would also like to thankPrinceton University, Universite de Paris 7 and Harvard University for theirhospitality during some of the work on this paper. A.W. was supported by anNSF grant.

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1 Notation

Letpdenote an odd prime, let O denote the ring of integers of a finite extensionK/Qp, let denote its maximal ideal and let k = O/.

IfL is a perfect field GL will denote its absolute Galois group and if thecharacteristic ofL is notp then : GL Zp will denote the p-adic cyclotomiccharacter. If L is a number field and a prime of its ring of integers thenG will denote a decomposition group at and I the corresponding inertiagroup. We will denote by Frob the arithmetic Frobenius element ofG/I.

IfG is a group and M aG-module we will let MG andMG denote respec-tively the invariants and coinvariants ofG on M. If is a representation of

G into the automorphisms of some abelian group we shall let V denote theunderlying G-module. IfH is a normal subgroup ofG then we shall let H

andHdenote the representation ofG/H on respectively VH andV,H.

We shall also fix a continuous representation

: GQ

GL2(k)

with the following properties.

is modular in the sense that it is a modprepresentation associated tosome modular newform of some weight and level.

The restriction ofto the group Gal (Q/Q(

(1)(p1)/2p)) is absolutelyirreducible.

Ifc denotes complex conjugation then det (c) =1.

The restriction of to the decomposition group atp either has the form

1

0 2

with1and2 distinct characters and with 2unramified; or is inducedfrom a character of the unramified quadratic extension of Qp whoserestriction to the inertia group is the fundamental character of level 2,IpF

p2.

Ifl =p then

either |Il

00 1

,

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or|Il 1

0 1

,

or |Gl is absolutely irreducible and in the case |Il is absolutelyreducible we have l 1 modp.

(This implies that|Glis either unramified or of type A, B or C as definedin chapter 1 of [W2]. On the other hand if|Gl is of type A, B or C thensome twist satisfies the condition above.)

In the case that |Gp

1

0 2

we will fix the pair of characters 1, 2.

Note that in some cases this may involve making a choice.

We will let Q denote a finite set of primes qwith the properties

is unramified atq,

q 1 modp,

(Frob q) has distinct eigenvalues, which we shall denote q andq.

Much of our notation will involve a subscript Q to denote dependence on Q,wheneverQ= we may simply drop it from the notation.

For q Q we shall let q denote the Sylow p-subgroup of (Z/qZ). Weshall let q denote a generator. We will write Q for the product of the q

with q Q. We will let aQ denote the kernel of the map O[Q] O whichsends every element of Q to 1. Letq denote the character

GQ

Gal(Q(q)/Q)=(Z/qZ) q,

and letQ=qQ q.

We will denote by NQ the product of the following quantities:

the conductor of;

the primes inQ;

p, if is not flat (i.e. does not arise from the action of Gp on theQp-points of some finite flat group scheme over Zp) or if det |Ip = .(We remark that if|Gp is flat but det |Ip = then |Gp arises from anetale group scheme over Zp. We also note that in [W2] the term flat isnot used when the group scheme is ordinary.)

We will let Q denote the inverse image under 0(NQ) (Z/NQZ) of theproduct of the following subgroups:

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the Sylow p-subgroup of (Z/MZ), where Mdenotes the conductor of;

for each q Q the unique maximal subgroup of (Z/qZ) of order primetop.

Let T(Q) denote the Z-subalgebra of the complex endomorphisms of thespace of weight 2 cusp forms on Qwhich is generated by the Hecke operatorsTl and l for l|pNQ, byUq forq Q and by Up ifp|NQ. Let m denote theideal ofT(Q) Z Ogenerated by , by tr (Frob l) Tl and det (Frob l) llforl|pNQ, byUq q forq Q and by Up 2(Frobp) ifp|NQ. Note that ifQ= this definition only makes sense ifO is sufficiently large thatk contains

the eigenvalues of(Frob q) for all q Q. It is a deep result following fromthe work of many mathematicians that m is a proper ideal (see [D]), and somaximal. We let TQ denote the localisation of T(Q)

Z

O at m. Note thatTQis reduced because the operatorsTl forl|NQ act semi-simply on the spaceof cusp forms for Q and the Uq for q Q act semi-simply on all commoneigenspaces for the Tl for which the corresponding p-adic representation iseither ramified at q or for which (Frob q) has distinct eigenvalues. Thereis a natural map O[Q] TQ, which sends x Q to y where y Z,y x mod qfor all q Q and y 1 modN.

It follows from the discussion after theorem 2.1 of [W2] or from the work

of Carayol [C2] that there is a continuous representation

modQ :G Q GL2(TQ);

such that ifl|NQpthenmodQ is unramified atland we have tr modQ (Frob l) = Tl

and det modQ (Frob l) = ll. In particular the reduction of modQ modulo the

maximal ideal ofTQ is. From [C1] we can deduce the following.

Ifq Q thenmodQ |Gq =1 2where1 is unramified and1(Frob q) =Uq, and where 2|Iq =q|Iq .

Ifl =p and |Il is non-trivial but unipotent then modQ |Il is unipotent.

If l Q {p} and either |Il = 1 or |Gl is absolutely irreduciblethenmodQ (Il)

(Il).

det modQ =Qwhere is a character of order prime to p.

Moreover if |Gp is flat and if det |Ip = then p|NQ so modQ |Gp is flat (i.e.

the reduction modulo every ideal of finite index is flat). If |Gp is not flat

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or if det |Ip = then p|NQ, |Gp 1

0 2

and Up is a unit in TQ. Itfollows from theorem 2 of [W1] (or more directly in the case 1|Iq = from

proposition 12.9 of [G]) that modQ |Gp

1

0 2

, where 2 is unramified

and 1(Ip) has order prime to p. In the case that 1 is unramified we knowfurther that1= 2 (see proposition 1.1 of [W2]) and that this character has

finite order. It will be convenient to introduce the twist Q= modQ

1/2Q of

modQ . In particular we see that det Q is valued in O

.The main theorem of this paper is as follows. Recall that we may write T

for T.

Theorem 1 The ringT is a complete intersection.

We note that ifO is the ring of integers of a finite extension K/K thenthe ring constructed using O in place ofO is just TQO O. Also T is acomplete intersection if and only ifT O O is (using for instance corollary 2.8on page 209 of [K2]). Thus we may and we shall assume that O is sufficientlylarge that the eigenvalues of every element of are rational over k and thatthere is a homomorphism : T O . In particular the definition ofTQmakessense for all Q. There is an induced map Q : TQ T O. The mapTQ T takes the operators Tl and l to themselves and the operator Uq to

the unique root of U2

TqU+qq in T above q. We will let Q denotethe kernel of Q and will let Q denote the ideal Q(Ann

T Q(Q)). Then it

is known that > #Q/2Q #O/Q with equality if and only if TQ is acomplete intersection (see the appendix of [W2] or [L], we are using the factthat TQ is reduced).

2 Generalisation of a Result of de Shalit

In this section we shall use the methods of de Shalit (see [dS]) to prove thefollowing theorem.

Theorem 2 The ringTQ is a freeO[Q] module ofO[Q]-rank equal to theO-rank ofT.

By lemma 3 of [DT] we may choose a prime Rwith the following properties:

R|6NQp;

R1 modp;

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As R 1 modp and det is unramified at R, no component of TQnor ofTQ can correspond to an eigenform with a non-trivial action of(Z/RZ).

AsR/R =R1 ink, no component ofTQnor ofT

Qcan correspond to

an eigenform which is special atR (i.e. an eigenform which correspondsto a cuspidal automorphic representation ofGL2(A) whose componentatR is special).

As for each prime q Q, q/q = q1 in k, no component ofTQ cancorrespond to an eigenform which is special at q.

The ring generated by the Hecke operators Tl and l forl|pNQR, byUqforq Q {R}and by Up ifp|NQ onS2(Q)2 is isomorphic to T(Q)[uR]/(u2R TRuR+ RR). In fact UR acts by the matrix

TR 1RR 0

onS2(Q)2. Similarly the ring generated by the Hecke operators Tl and lfor

l|pNQR, byUq forqQ {R}and byUpifp|NQon S2()2#Q+1 is isomorphic

to T()[uq : q Q {R}]/(u2q Tquq + qq : q Q {R}). Tensoringwith O and localising at the appropriate maximal ideal we get the desired

isomorphism. We have to use the fact thatu

2

R TRuR+RR has two rootsin TQwith distinct reductions modulo the maximal ideal and the similar factsover Tforu2q Tquq+ qq with qQ {R}.

Because is irreducible we see thatH1(YQ, O) m Q =H1(XQ, O) m Q and that

H1(YQ, O)m Q =H1(XQ, O) m Q. By corollary 1 of theorem 2.1 of [W2] we

see thatH1(XQ, O)m

Qare free rank one TQ-modules and thatH

1(XQ, O)m

Q

are free rank one TQ-modules. Hence it will suffice to prove the followingproposition.

Proposition 1 H1(YQ, O) is a freeO[Q]-module, withO[Q]-rank equal

to theO-rank ofH1(YQ, O).

BecauseH1(YQ, K) = H1(YQ, K)

Q we need only show that H1(YQ, O)

is a free O[Q]-module. Because R 5, Q acts freely on the upper halfcomplex plane and so may be identified with the fundamental group ofYQ.In particular we see that Q is a free group. Similarly

Q acts freely on the

upper half complex plane and we get identifications

H1(YQ, O)=H1(Q, O)

=H1(Q, O[Q]),

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the latter arising from Shapiros lemma. Under these identifications com-plex conjugation goes over to the involution induced by conjugation by =

1 00 1

and trivial action on the coefficients. (This follows because the

action ofc on YQ is induced by the map z z of the upper half complexplane to itself.)

Because Qis a free group, the cocyclesZ1(Q, O[Q]) are a free O[Q]-

module. (If1,...,a are free generators of Q then we have an isomorphism

Z1(Q, O[Q]) O[Q]

a

((1),...,(a)).)

On the other handacts trivially on Qand so the coboundaries are containedin Z1(Q, O[Q])

+. Thus H1(YQ, O) = Z1(Q, O[Q])

is a free O[Q]-module, as desired.

Before leaving this section we remark the following corollaries of theorem2.

Corollary 1 Ifq Q then TQ{q}/(q 1) TQ. Moreover (q 1)TQ{q}

and(1 + q+ ...+ #q1q )TQ{q} are annihilators of each other in TQ{q}.

Corollary 2 If for someQ the ringTQ is a complete intersection then so isT

.The proof is by showing that under the assumption that TQ{q} is a com-

plete intersection, so is TQ. The argument in section 2 of [K1] shows that ifa complete local noetherian ring Sis a complete intersection, iff Sand ifHomS(S/(f), S) = HomS(S/(f), Ann S(f)) is a free S/(f) module thenS/(f)is a complete intersection. We apply this result withS=TQ{q}andf=q1.The final condition is met because the annihilator ofq 1 in TQ{q} is a freerank one TQ module by the last corollary.

Corollary 3 Q= #Q.

We remind the reader thatQis defined at the end of section one and that = . The proof of the corollary is by showing that Q{q} = Q#q ifqQ. WriteQ =Q {q}and let denote the natural surjection TQ TQ.It suffices to prove that in TQ we have the equation Ann Q = (1 + q +

...+ #q1q )1(Ann

T Q(Q)). However

1(AnnT Q

(Q)) is just the set ofelements t of TQ for which tQ (q 1)TQ. The inclusion Ann Q

(1 + q+ ...+ #q1q )1(Ann

T Q(Q)) is now clear. Conversly ift Ann Q

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then t annihilates ker and hence t = (1 +q+ ...+#q1q )s. Then we see

that sQ (q 1)TQ, i.e. thats 1(AnnT Q

(Q)). The other inclusionnow follows.

Corollary 4 #(aQTQ/QaQTQ)#(O/) = #(O/Q).

To prove this corollary note that aQ/a2Q

=qQO/#q. Thus it follows

from theorem 2 that aQTQ/a2QTQ

=TQ O[Q] aQ/a2Q

=qQ T/#q and so

we deduce that aQTQ/QaQTQ =qQO/#q. This corollary now follows

from the last one.

3 Some Algebra

In this section we shall establish certain criteria for rings to be complete inter-sections. We shall rely on the numerical criterion established in the appendixof [W2]. In that appendix there is a Gorenstein hypothesis which can bechecked in the cases where we will apply the results of this section. Howeverin order to state the results of this section in somewhat greater generality weshall reference the paper [L], where the Gorenstein hypothesis in the appendixof [W2] is removed, rather than the appendix of [W2] directly.

Fix a finite flat reduced local O algebra T with a section : T O.

We will consider complete local noetherian O algebras R together with mapsR T. We will denote by JR the kernel of the map R T, by R theinduced map R O, byR the kernel ofR and by R the image under Rof the annihilator in R ofR. We will let R= (

2R JR)/RJR.

IfSRTthen2S2R,JSis the pre-image ofJRand so SR.

We have an exact sequence

(0)R JR/RJR R/2R T/

2T(0).

From this we deduce the following facts.

#R < . (To see this it suffices to show that (R)R = (0). How-

ever as T is reduced (JR)R = (R)R and so the map (JR/RJR)R (R/

2R)R is an isomorphism. The result follows.)

#R#(R/2R) = #(T/2T)#(JR/RJR).

Lemma 2 Suppose we have the inequalities#(T/2T) #(O/T)#R and

#(O/T)#(JR/RJR) #(O/R) < and suppose R is a finite flat O-algebra. ThenR is a complete intersection.

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To show this first note that we have the inequalities

#(R/2R) = #(T/

2T)#(JR/RJR)/#R

#(T/2T)#(O/R)/(#R#(O/T)) #(O/R).

Now applying the criterion of [L] we see thatR is a complete intersection.

Lemma 3 We have the inequality:

#(O/R) #(JR/RJR)#(O/T).

As Fitt R(JR) Ann R(JR) we see that Fitt O(JR/RJR)RAnnR(JR).On the other hand it is easy to see that

Ann R(R) {s R|sR JR}AnnR(JR).

ApplyingR we see that

R TFittO(JR/RJR),

and the lemma follows.

Lemma 4 IfR is a complete intersection which is finite and flat overO and

R = (0) then #(T/2T) #(O/T)#R. If R is a power series ring thesame result is true without the assumptions that it is finite overO and thatR= (0).

For the first part we see that, as R is a complete intersection, #(R/2R) =

#(O/R) (see [L]). Thus we see that

#R#(O/R) = #(T/2T)#(JR/RJR) #(T/

2T)#(O/R)/#(O/T).

The first result follows. For the second result note that we can factor R Tas R R T with R a complete intersection which is finite and flat

over O and for which R/2R

T/2T (by the proof of lemma 9 of [L]).

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1. In is contained inm2T

Qn

andTQn/In has finite cardinality.

2. In+1T InT andn InT= (0).

3. There is a surjective map ofO-algebrasTQn+1/In+1TQn/In such thatthe diagram

TQn+1/In+1 TQn/In

T/In+1T T/InT

commutes. (Note that this map is not assumed to take a given Heckeoperator to itself.)

4. lim TQn/In is a power series ring.

Then forn sufficiently largeTQn is a complete intersection, and henceT is acomplete intersection (by corollary 2 of theorem 2).

LetPdenote lim TQn/In. We get a natural mapPT and can choosemaps P TQn compatible with the maps TQn T and TQn TQn/In.Because In m

2T Qn

we see that the map P TQn is surjective. We have asequence

P Qn ((JQn+ In) (2Qn+ In))/(JQnQn+ In).

(Note that although the mapsPTQn and P Qn are not compatible asnvaries the composite map above is.) Moreover P= lim((JQn+In)(

2Qn+

In))/(JQnQn+ In) (using the fact that TQn/Inis finite for all n) and so as Pis finite we have that the map P ((JQn+ In) (

2Qn+ In))/(JQnQn+ In)

is injective for n sufficiently large. Thus for n sufficiently large P

Qn.We deduce the inequality

#(/2) #(O/)#P = #(O/)#Qn,

where the first inequality follows from lemma 4. The proposition follows on

applying corollary 4 of theorem 2 and lemma 2.

Corollary 1 Suppose that we have an integerr and a series of setsQm withthe following properties:

1. ifq Qm thenq 1 modpm;

2. is unramified atqand(Frob q) has distinct eigenvalues;

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3. #Qm=r;

4. TQm can be generated as anO-algebra byr elements.

ThenT is a complete intersection.

To prove this corollary it is useful to have the following definition. By aleveln structure we shall mean a quadruple B= (A,,,), where

Ais anO-algebra,

: O[[T1,...,Tr]]A,

:O[[S1,...,Sr]]/(pn, (S1+ 1)pn

1, ..., (Sr+ 1)pn

1) A makes A afree module overO[[S1,...,Sr]]/(pn, (S1+ 1)p

n

1, ..., (Sr+ 1)pn

1),

and: A/(S1,...,Sr) T/pn.

IfB is a structure of level nand n n then it induces a structure of level n

by reducing mod(pn

, (S1+ 1)pn

1,..., (Sr+ 1)pn

1).Let Am = TQm/(p

m, pm

q 1|q Qm). This extends to a level mstructurethat we will denote Bm. For n m we will let Bm,n denote the level nstructure induced byBm. There are only finitely many isomorphism classes ofstructures of level n and so we may choose recursively integers m(n) with the

following two properties.

1. Bm(n),n1=Bm(n1),n1.

2. Bm(n),n=Bm,n for infinitely many integers m.

LetIndenote the kernel of the map from TQm(n) to the ring underlyingBm(n),n.We claim that the pairs (Qm(n), In) for n 2 satisfy the requirements ofproposition 2 (we use n 2 to ensure that In m2Qn). We need only checkthat lim Bm(n),n is a power series ring. On the one hand it is a finite freeO[[S1,...,Sr]]-module, and so has Krull dimension r + 1. On the other hand itis a quotient ofO [[T

1,...,Tr]] and so must in fact equalO[[T

1,...,Tr]].

4 Galois Cohomology

It remains to find a sequence of sets Qm with the properties of corollary 1of proposition 2. We must recall some definitions in Galois cohomology. Wedefine H1f(Ql, ad

0).

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Forq Qwe have thathq = #H0(Qq, ad0(1)) = #k. It remains to check that

hph 1. In the case that|Gp is not flat or det |Ip = this is proved in parts(iii) and (iv) of proposition 1.9 of [W2]. Thus suppose that|Gp is flat anddet |Ip =. We must show that dimkH

1f(Qp, ad

0) 1 + dimkH0(Qp, ad

0)(= 1 if|Gp is indecomposable and = 2 otherwise).

Following [FL] let M denote the abelian category ofkvector spacesMwitha distinguished subspace M1 and a k-linear isomorphism : M/M1 M1

M. Then there are equivalences of categories between:

Mop;

finite flat group schemes A/Zp with an action ofk ;

k[Gp]-modules which are isomorphic as modules over Fp[Gp] to the Qppoints of some finite flat group scheme over Zp.

See section 9 of [FL] for details. The only point here is that an action ofk onthe generic fibre of a finite flat group scheme over Zp extends uniquely to anaction on the whole scheme. Let M() denote the object ofM correspondingto . Then dimkM() = 2 and dimkM()

1 = 1 (since det |Ip = ). We getan embedding Ext1M(M(), M())H

1(Qp, ad). We will show that

1. dimkExt1M(M(), M()) = 2 if|Gp is indecomposable and = 3 other-

wise;

2. the composite map Ext1M(M(), M())H1(Qp, ad)

trH1(Qp, k) is

non-trivial, where tr denotes the map induced by the trace.

The lemma will then follow.For the first point it is explained in lemma 4.4 of [R] how to calculate

Ext1M(M(), M()). Let {e0, e1} be a basis ofM() with e1 M()1. Let

(e0, 0) = e0+ e1 and (0, e1) = e0+ e1. Then Ext1M(M(), M()) can

be identified as a k-vector space with M2(k) modulo the subspace of matricesof the form

r 0s t

r 00 t

=

0 (r t)s+ (t r) s

,

for any r, s, t k. Thus dimkExt1M(M(), M()) = 2 if = 0 and = 3 if

= 0. However = 0 if and only ifM()1 is a subobject ofM() inM. Thisis true if and only if has a one dimensional quotient on which inertia acts bywhich itself is true if and only if|Gp is decomposable.

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For the second point consider the k[Gp]-module where is the un-ramified representation

Frobp

1 10 1

.

Then is an extension of by itself. Moreover its extension class mapsto the element ofH1(Qp, k) = Hom(Qp, k) which is trivial on Z

p and takes

p2. Finally it is isomorphic to the action ofGp on the Qp points of a finiteflat group scheme over Zp, because this is true over an unramified extension.

Lemma 6 TQ can be generated as an O algebra by dimkH1Q(Q, ad

0) ele-ments.

Let mQ denote the maximal ideal ofTQ. It will suffice to show that thereis an embedding ofk-vector spaces

: Homk(mQ/(m2Q, ), k)H

1Q(Q, ad

0).

We first define

: Homk(mQ/(m2Q, ), k)H

1(Q, ad).

If is a non-zero element of the left hand group we may extend it uniquely toa map of local O-algebras : TQ k [] where 2 = 0. Let =

Q

. Weget an exact sequence

(0)VV V(0),

and hence a class() in Ext1k[GQ

](, )=H1(Q, ad). Because det is valued

in k k[] we see that () actually lies H1(Q, ad0).We claim that resl() lies inH

1f(Ql, ad

0) forl Q. This computation isvery similar to some in [W2], but is not actually carried out there, so we givean argument here. First suppose that l = p and that either p|#(Il) or |Glis absolutely irreducible. In this caseQ(Il)

(Il) and det Q|Il has order

prime to l. Because eitherp|#(Il) orp= 3 and ad(Il)

=A4 we have thatH1((Il), ad0) = (0) and so|Il =|Il k k[]. The result follows in this case.

Secondly suppose that|Il is unipotent and nontrivial. Then the same is trueforQ|Il and then also for. However the Sylowp-subgroup ofIl is pro-cyclic

and so |Il must also be of the form kk [] and resl() H1(Il, ad

0())must vanish. In the casel = p, is flat and det |Ip = the claim is immediate

from the definitions. In the case l= p and|Gp

1

0 2

with1|Ip =

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use the fact that Q|Ip

1 0 1

where 1 denotes the Teichmuller lifting

of1|Ip. Finally in the case l = p, |Gp

1

0 2

with 1|Ip = but

not flat use the fact that

Q|Gp

0

where is an unramified character of order prime to p.It remains to show that is injective. Suppose it were not. Then we could

find a non-zero such that kk[]. Thus tr Q is valued in O+ kerand in particular TQ is not generated as an O-algebra by tr Q. We will show

this is not the case. Ifq Qand q then we can find Gq such that(Uq)

2 (tr Q())(Uq)+det Q() = 0. (will in fact lie above Frob q.) This

polynomial has distinct roots in k and so both its roots in TQ lie in the sub-O-algebraTgenerated by the image of tr Q. Thus for all q, Uq T.Hence Uq T, and as Uq is a unit, T. Moreover for all l Q for which is unramified we see that TlQ(Frob l)

1/2 T and hence Tl T. Ifp|NQthenUpQ(Frobp)

1/2 is a root of the polynomial X2(tr Q())X+det Q()

for any element ofGp which lies above Frobp. For some over Frobp thispolynomial has two distinct roots in k and so Up T. ThusT = TQ as werequired.

Finally we turn to the proof of the main theorem. As in [W2] (after equa-tion (3.8)) we may find a set of primes Qm with the following properties:

1. ifq Qm then q 1 modpm;

2. ifq Qmthenis unramified atqand(Frob q) has distinct eigenvalues;

3. H1(Q, ad0(1))

qQm

H1(Fq, ad0(1)).

As for each such q, H1(Fq, ad0) = k we see that by shrinking Qm we may

suppose that the latter map is an isomorphism. Then we have that #Qm =dimkH

1(Q, ad

0(1)). Also H1Qm(Q, ad0(1)) is the kernel of the map in 3.

above and so is trivial. Thus by lemma 5 we see that dimkH1Qm(Q, ad

0)

#Qmand so TQm can be generated by #Qm = dimkH1(Q, ad0(1)) elements.The main theorem now follows from corollary 1.

References

[C1] H.Carayol,Sur les representationsp-adiques associees aux formes mod-ulaires de Hilbert, Ann. Sci. Ec. Norm. Super. 19 (1986) 409-468.

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[C2] H.Carayol,Formes modulaires et representations Galoisiennes a valeursdans un anneau local complet, in p-adic monodromy and the Birch-Swinnerton-Dyer conjecture (eds. B.Mazur and G.Stevens), Contem-porary Math. 165 (1994).

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[DT] F.Diamond and R.Taylor,Lifting modular mod l representations, DukeMath. J. 74 (1994) 253-269.

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Ann. Sci. Ec. Norm. Super. 15 (1982) 547-608.

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the proceedings of the 1993 Hong Kong conference on modular formsand elliptic curves.

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Appendix

The purpose of this appendix is to explain certain simplifications to the argu-ments of chapter 3 of [W2] and to section 3 of this paper. These simplificationswere found by G.Faltings and we would like to thank him for allowing us toinclude them here. We should make it clear that the arguments of this ap-pendix (just as those of chapter 3 of [W2] and section 3 of this paper) applyonly to proving conjecture 2.16 of [W2] for the minimal Hecke ring and mini-mal deformation problem. In order to prove theorem 3.3 of [W2] one needs toinvoke theorem 2.17 and the arguments of chapter 2 of [W2].

We will keep the notation and assumptions of the main body of this paper.

LetQ denote a finite set of primes as described in section 1 of this paper. Bya deformation of of type Q we shall mean a complete noetherian local O-algebraA with residue fieldk together with an equivalence class of continuousrepresentations: G

Q

GL2(A) with the following properties:

mod mA = ;

1 det is a character of finite order prime to p;

ifl Q {p}and |Il is semi-simple then (Il) (Il);

ifl Q {p}and |Il 1 0 1

then|Il 1 0 1

; if is flat and det |Ip = then is flat;

if either is not flat or if det |Ip = then |Gp

1

0 2

where 2

is unramified and2mod mA= 2.

As in chapter 1 of [W2] there is a universal liftunivQ :G Q GL2(RQ) of typeQ. Recall that the universal property is for lifts up to conjugation. Moreoverone checks (c.f. the second paragraph of the proof of lemma 6) that there is anatural isomorphism

Homk(mRQ/(,m2RQ

), k)=H1Q(Q, ad0).

There is also a natural map RQ TQ so that univQ pushes forward to a

conjugate ofQ.Recall that if Q = we shall often drop it from the notation. In this

appendix we shall reprove the following result.

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Theorem 3 R T and these rings are complete intersections.

We note that ifO is the ring of integers of a finite extension K/K thenthe rings TQ and R

Q constructed using O

in place ofO are just TQOO

andRQOO . Also TQ is a complete intersection if and only ifTQOO is(using for instance corollary 2.8 on page 209 of [K2]). Thus we may and weshall assume that O is sufficiently large that the eigenvalues of every elementof(G

Q

) are rational over k.We recall that in the penultimate paragraph of section 4 of this paper we

showed that TQ is generated as an O-algebra by tr Q(GQ ). Thus we see thatthe mapRQ TQ is a surjection.

We will need the following result.

Lemma 7 Ifq Q thenunivQ |Gq

1 0

0 2

where1|Iq =2|

1Iq

and both

these characters factor throughq :Iq q.

It suffices to check the first assertion. As is unramified at q, univQ |Gqfactors through Z Zp(1), where Z is topologically generated by some liftf of Frob q, Zp(1) is topologically generated by some element and wheref f1 = q. As (Frob q) has distinct eigenvalues it is easy to see that after

conjugation we may assume that univQ

(f) = a 00 b

wherea b mod mRQ

.

We will show that univQ () is a diagonal matrix with entries congruent to1 mod mRQ. We will in fact prove this mod m

nRQ

for all n by induction on n.Forn = 1 there is nothing to prove. So suppose this is true modulo mnRQ withn >0. Then

univQ ()

1 0

0 2

(12+ N) mod m

n+1RQ

,

where each i 1 mod mRQ and where N 0 mod mnRQ

. We see that mod

mn+1RQ

we have

1 + a 0

0 b

N a 0

0 b1

q11 0

0 q12

(1 + N)q

q11 0

0 q12

(1 + qN)

q11 0

0 q12

+ N,

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and asa b mod mRQ we deduce that N is diagonal modmn+1

RQas required.

We can choose 2 so that 2(f) q mod mRQ. Then we can define amap q R

Q to be 2|

2Iq . This makes RQ into an O[Q]-algebra. Using the

last lemma and the universal properties ofRQ and ofR it is easy to see thatRQ/aQ

R. It moreover follows from the discussion preceding theorem 1 of

this paper that the map RQTQ is a map ofO [Q]-algebras.The key observation is the following ring theoretic proposition. Theorem 3

follows on applying it to the ringsRQn andTQn for the setsQn constructed insection 4 of this paper. Note that there is a map O[[S1,...,Sr]] O [Qn] withkernel the ideal ((1+ S1)

#q1 1, ..., (1 + Sr)#qr 1), whereQn = {q1,...,qr}.Note also that by the displayed isomorphism a couple of lines before theorem

3 there exists a surjection ofO-algebrasO[[X1,...,Xr]]RQn.

Proposition 3 Supposer is a non-negative integer and that we have a mapofO-algebrasR T withTfinite and flat overO. Suppose for each positiveintegern we have a map ofO-algebrasRnTn and a commutative diagramofO-algebras

O[[S1,...,Sr]] Rn R

Tn T,

where

1. there is a surjection ofO-algebrasO[[X1,...,Xr]]Rn,

2. (S1,...,Sr)Rnker (RnR),

3. (S1,...,Sr)Tn = ker (TnT),

4. if bn denotes the kernel ofO[[S1,...,Sr]] Tn then bn ((1 +S1)pn

1,..., (1 + Sr)

pn 1)andTn is a finite freeO[[S1,...,Sr]]/bn-module.

ThenR Tand these rings are complete intersections.

Reducing mod we see that it suffices to prove this result withk replacing

O everywhere. In this case we see that the last condition becomes bn (Sp

n

1 ,...,Spn

r ). Further we may replace R by its reduction modulo mRker(R T), and so we may assume that R is finite over k. We may replace Tnby Tn/(S

pn

1 ,...,Spn

r ) and so assume that bn = (Spn

1 ,...,Spn

r ). Finally we mayreplaceRn by its image in R Tn.

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Now define ann-structure to be a pair ofk-algebrasB A together witha commutative diagram ofk-algebras

k[[S1,...,Sr]]

k[[X1,...,Xr]] B R A T,

such that

1. B R A,

2. (S1,...,Sr)B ker (BR),

3. (S1,...,Sr)A= ker (AT),

4. A is a finite free k[[S1,...,Sr]]/(Spn

1 ,...,Spn

r )-module.

Note that #B (#T)pnr

#R and so we see that there are only finitely manyisomorphism classes ofn-structures. IfS is an n-structure and ifm n thenwe may obtain anm-structureS(m) by replacingA by A/(Sp

m

1 ,...,Spm

r ) andBby its image in R (A/(Sp

m

1 ,...,Spm

r )).As explained above, it follows from the hypotheses of the proposition that

ann-structureSn exists for each n. We next claim that we can find, for eachn, n-structuresSn such that form n we have S

m

=(Sn)(m). To prove this

observe that we can find recursively integersn(m) with the following properties

S(m)n(m)=S

(m)n for infinitely manyn

and form >1, S(m1)n(m)

=S(m1)n(m1).

Then setSm = S(m)n(m).

Thus we obtain a commutative diagram

... R

n ... R

2 R

1 R ... T n ... T

2 T

1 T

ofk [[X1,...,Xr, S1,...,Sr]]-algebras. Moreover we have that

k[[X1,...,Xr]]RnTn,

Tn is a finite free k [[S1,...,Sr]]/(Spn

1 ,...,Spn

r )-module,

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Rn/(S1,...,Sr)R andTn/(S1,...,Sr)

T.

Let R denote the quotient ofk[[X1,...,Xr]] by the intersection of the idealsker(k[[X1,...,Xr]] Rn) and let T

denote the quotient of k[[X1,...,Xr]]

by the intersection of the ideals ker (k[[X1,...,Xr]] Tn). Then we have acommutative diagram

k[[S1,...,Sr]]

k[[X1,...,Xr]] R R

T T,

such that

RT,

T is a finite free k[[S1,...,Sr]]-module,

R/(S1,...,Sr)R andT/(S1,...,Sr)

T.

We deduce that T has Krull dimension r and hence we deduce that themap k[[X1,...,Xr]] T has trivial kernel. That is we have isomorphismsk[[X1,...,Xr]]

R

T. ThusR

T. As Thas Krull dimension 0 and

T = k[[X1,...,Xr]]/(S1,...,Sr) we see that T is a complete intersection, andthe proposition is proved.

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