ABSTRACT
Title of dissertation: UNIVERSAL DEFORMATIONS AND P-ADIC
L-FUNCTIONS
Arijit Sehanobish, Doctor of Philosophy, 2019
Dissertation directed by: Professor Lawrence Washington
Department of Mathematics
In this thesis we study deformations of certain 2-dimensional reducible representations
whose image is in the Borel subgroup of GL2(F). Our method of understanding the uni-
versal deformation ring is via the Jordan-Ho?lder factors of the residual representation. Us-
ing the vanishing of cup products of appropriate cohomology classes we can compute the
tangent space of the universal deformation ring and some obstruction classes to lifting
representations. In this process, we can also explicitly construct certain big meta-abelian
extensions inside the fixed field of the kernel of the universal representation. We give
an explicit example of our construction of an unramified extension in the case of elliptic
curves of conductor 11. We also give an Iwasawa theoretic description of various fields
that are cut out by the universal representation. The Galois theoretic description of the
constructed meta-abelian unramified extension is then later used as an ingredient for the
isomorphism criterion in the modularity lifting results. When the isomorphism criterion
is satisfied, we could prove some modularity lifting results allowing us to recover some
results of Skinner-Wiles and prove a conjecture of Wake in this special case. We also show
that the representations considered by Skinner-Wiles have big image inside the universal
deformation ring.
UNIVERSAL DEFORMATIONS AND P-ADIC L-FUNCTIONS
by
Arijit Sehanobish
Dissertation submitted to the Faculty of the Graduate School of the
University of Maryland, College Park in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
2019
Advisory Committee:
Professor Lawrence Washington, Chair
Professor Patrick Brosnan
Professor Thomas Haines
Professor Niranjan Ramachandran
Professor William Gasarch
?c Copyright by
Arijit Sehanobish
2019
Dedication
To my parents and to the wonderful Leah A. Drew.
ii
Acknowledgments
It is nearly impossible to thank everyone that I have met in this long journey and had
a hand in my successful completion of this dissertation. In fact, some of the people men-
tioned below might not remember their contribution, but their impact was substantial and
will not be forgotten.
First and foremost, I would like to thank my adviser Lawrence Washington for being
a truly great adviser. He has spent countless hours discussing various aspects of Iwasawa
theory and carefully going through my thesis. Without his guidance, encouragement and
perseverance, this thesis would not have been possible.
Next I would like to thank Professor Tom Haines for giving me the opportunity to spend
a year in France. I will forever be grateful to the Hadamard Fondation for the generous
support that allowed me to do research at Universite Paris Sud and to Universite Paris Sud
for being a wonderful host. I would like to Professor Laurent Clozel for facilitating my
visit there and also for patiently answering many questions about automorphic forms and
Ihara?s lemma. The main idea for this thesis came from long conversations with Olivier
Fouquet who asked me to look at Skinner-Wiles and explaining the difficulties that the
authors had to overcome. I thank him for that insight. Finally I thank Jacques Tilouine
and Ariane Mezard for answering various questions about Eisenstein Hecke algebras and
reducible representations.
I would finally like to express my gratitude to University of Maryland and to the many
faculty members (including my thesis committee members) for making this a truly won-
derful place to work.
I would also like to thank the close friends who were with me through this period
of my life. Special thanks to (in no particular order) Giovanni, Arthur, Andrew, Arno,
Przemyslaw, Richard, Ran, Alex, Sam, Jonathan, Catie, Amy, Tim and the list actually
goes on. You know who you are and I appreciate your support.
iii
My family: my parents. Well without them, I would not have been born. And to my
girlfriend Leah Drew who has brightened up many a dreary day with her smile, love and
encouragement.
iv
Table of Contents
Dedication ii
Acknowledgements iii
1 Introduction 1
2 Galois cohomology and Selmer groups 11
2.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Local duality theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Restricted ramification and Global duality . . . . . . . . . . . . . . . . . . 17
3 Universal Deformation Rings 26
3.1 Basic Definitions and preliminaries on representations . . . . . . . . . . . . 26
3.2 Preliminaries on deformation theory . . . . . . . . . . . . . . . . . . . . . 33
3.3 Ordinary deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.4 Computations of some tangent spaces and obstruction classes . . . . . . . . 43
3.5 Understanding extension classes and cup-products . . . . . . . . . . . . . . 60
3.6 An explicit construction of a meta-abelian extension . . . . . . . . . . . . . 82
3.7 Pseudo-deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4 Hida theory of ordinary modular forms and Hecke algebras 92
4.1 Galois representations attached to classical modular forms . . . . . . . . . 94
4.2 Hida theory of ?-adic modular forms . . . . . . . . . . . . . . . . . . . . . 96
4.3 Structure of Hida Hecke algebras and big modular Galois representations . 99
5 Images of Galois representations 102
5.1 History of related results . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
5.2 A new big image theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
v
6 Modularity lifting and Wake?s conjectures 109
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
6.2 Congruence Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
6.3 Modularity lifting and Wake?s conjecture . . . . . . . . . . . . . . . . . . . 129
7 Selmer groups 138
7.1 Review of some basic definitions . . . . . . . . . . . . . . . . . . . . . . . 138
7.2 Some calculations on Selmer groups . . . . . . . . . . . . . . . . . . . . . 140
Bibliography 145
vi
Chapter 1: Introduction
One of the central themes in modern number theory is to understand Gal(Q/Q). The
standard way to understand an abstract group is via representations which in this case are
called Galois representations. A large class of naturally occurring representations comes
from the etale cohomology of smooth projective varieties defined over Q.
Definition 1.0.1. We call a Galois representation geometric if it occurs in a subquotient of
the cohomology of a smooth projective variety.
Definition 1.0.2. Let c be a complex conjugation in Gal(Q?/Q). A representation ? : GQ?
GL(V ) is called odd if det(?(c)) =?1
Conjecture 1.0.3. (Fontaine-Mazur): now a theorem by Breuil, Emerton, Kisin, Paskunas,
Colmez et al (for n = 2): Any odd Galois representation ? : GQ ? GLn(Qp) which is
unramified outside a finite set of primes and De Rham at p is geometric.
In view of this theorem, it is natural to study Gal(QS/Q), i.e. the maximal extension
unramified outside a finite set S. The first case of understanding this group is via characters
(or 1-dimensional representations) or to understand Gal(QabS /Q) and this is achieved by
Class Field Theory.
To prove the conjecture, one proves a R = T theorem and one needs a local-global com-
patibility result supplied by the p-adic local Langlands conjecture. In this thesis, we revisit
R = T theorem of [57] and reinterpret and reprove some of their results. To be precise:
1
Let p be an odd prime number. Let ??: Gal(Q?/Q)? Gl2(F) be an odd continuous rep-
resentation unramified outside a finite set S of rational primes, where F is a finite field of
characteristic p. In that situation, ?? factors through GS = Gal(KS,p/Q) where K denotes
the field fixed by Ker ?? in Q? and KS,p is the maximal pro-p-extension of K, unramified
outside S. We will throughout assume that all our representations have conductor N which
is a squarefree integer.
Mazur?s deformation theory of Galois representations shows the existence of a (uni)versal
deformation ring Runiv?? and the associated (uni)versal representation ?
univ which allow us
to parametrize all deformations ?? : GS? Gl2(R) of ?? : GS? Gl2(F) where R stands for
any complete noetherian local ring with residual field F. By application of Schlessinger?s
criterion, Mazur ([37] subsection 1.2) shows that Runiv?? is a quotient ring of a formal power
series ring whose minimal number of variables is d = dimFH1(GS,Ad(??)). In this paper,
we consider ?? whose image is contained in upper triangular matrices and whose diagonal
characters ?1 and ?2 satisfy a particular assumption. In this case, we would expect to un-
derstand quite a bit of the structure just by studying the one-dimensional pieces. Indeed,
we give a simple formula for d, where all the terms in the formula are given by class field
theory.
Definition 1.0?.4. We c?all a prime l difficult if l ? 1 (mod p) and ?? : GQl ? GL2(F) be
1 ?
such that ?? ?? ?= ?, where ?|Il 6= 0.
1
In fact a main result of this thesis is the following theorem. Assume Ext1(?2,?1) = F
(we call this Hypothesis 1) and ?1 =6 ?2 and we do not have any difficult primes in the
deformation problem, then
2
Theorem 1.0.5. There exists an exact sequence
f g
0?? Ext1(? 1 11,?1)?Ext (?2,?2)?? tRuniv ?? Ext (?1,?2)?? 0??
There is an exactly similar result by [2] for the pseudo-deformation ring. Our guiding
principle in this thesis is to understand various arithmetic objects (which we will make pre-
cise) by one-dimensional objects. We can also handle the case of difficult primes, but we
have to tweak the deformation problem by adding a Steinberg condition. See remark 3.4.3.
Going back to the structure of deformation rings, Mazur shows Runiv?? = O[[Y1, ...,Yd]]/J
where J will be called the ideal of relations (sometimes also called obstruction classes)
so now the determination of Runiv?? amounts to that of J. But finding explicit obstruction
classes is quite difficult and they relate to non-vanishing of cup products or more generally
Massey products. But nonetheless, we manage to explicitly compute an obstruction class
in a very specific situation and we can relate it to some cohomology class coming from
one-dimensional representations. We set up the following problem.
Given a surjective ring homomorphism A1  A0 between complete local Noetherian O-
algebras with kernel I, generated by a single element with I.mA1 = 0, let ? be a represen-
tation to A0 lifting ?? and ?? be any lift of ? to A1. Now assume ? is an upper triangular
lift of ?? to A0. We will find the obstruction to lift ? to a not necessarily upper triangular
representation to A1. We know that this obstruction class is independent of the choice of lift
and will only depend on ? . We call that class O(?) ? Z2(??, ??)? I. We assume a technical
condition called Neben, in which we fix the non-p part of the diagonal characters in our
deformation problems. This allows us to rule out p-power non-trivial characters ?l which
are congruent to 1 mod p.
Theorem 1.0.6. Assume Neben and Hypothesis 1 and b? c = 0 where b ? Ext1(?2,?1)
3
and c ? Ext1(?1,?2). Then
O(?) = 0 iff f21 ? B2(a,d)? I
We then try to understand the fields that are cut out by our representation, since this is
ultimately our object of study. And surprisingly we can construct some big meta-abelian
extensions via our Galois representations. To state the next theorem we have to prepare
some notations.
Definition 1.0.7. The ideal of reducibility is the smallest ideal Ired of Runiv?? such that ?
univ
mod I is reducible.
To simplify notation, we will use I instead of Ired , when there?is no chance?of confusion.
?? ?
To be consistent with the notations of the later sections, let ?? ?? ?= ?.
?
Let F be a field that is cut out by the kernels of the characters ? and ? where ??(?1) =
1, and they are of finite order and their orders are prime to p and their conductors are
squarefree and prime to each other. Assume F ?Q(?p) = Q. Let F? be the cyclotomic
Zp extension of F(?p). Now let K? be the maximal abelian p extension of F?, unramified
outside N p, with ???1?-action and let L? be the maximal abelian p extension of F?,
unramified outside N, with ???1??1-action. Then we have the following theorem:
Theorem 1.0.8. Let M? be the maximal extension abelian p-extension of K?L?, unramified
everywhere with trivial ???-action. Then
Gal(M?/L?K?) =? I/I2
Moreover, we have a much stronger statement in the case where ? = ? = 1, then
4
Theorem 1.0.9. Assume Hypothesis 1. Let M? be the maximal extension abelian p-
extension of K?L?, unramified everywhere with trivial ???-action. Then
Gal(M 2?/L?K?) ?= I/I =? (IGGal(K?/K 2?)/IGGal(K?/K?))???
where I is the ideal of reducibility of Runiv?? , G=Gal (K?/Q(?p?)), and K? is the maximal
extension of K? unramified outside N and ? = Gal(Q(?p)/Q).
Similar identification is possible in our general case but one has to assume some extra
hypothesis on Gal(K?/F?) and Gal(L?/F?). This identification was first done by Sharifi,
[54]. But our result is different than his theorem as we allow ramification at auxiliary
primes in our intermediate fields and thus it applies in greater generality. Finally we give
a description of these methods in constructing these fields explicitly in the case of elliptic
curves of conductor 11.
Continuing on our theme of understanding these Galois representations, we asked Chris
Skinner at the Arizona Winter School in 2017, if he knew if the Galois representations
used in his paper [57] have big image, i.e. contains an open subgroup of SL2(F[[T ]]). Big
image questions are particularly important to cut down the size of Selmer groups and recent
works of Kato have shown that they are a crucial ingredient in his Euler systems. We have
a positive answer to that question.
Theorem 1.0.10. Let ? : GQ? GL2(F[[T ]]) be such that
i) ? is irreducible, mod p distinguished and ordinary. (cf. Definition 3.3.1 and Defini-
tion 3.3.3)
ii) Determinant is of infinite order. ? ?
1 ?
iii) There exists ? ? Ip such that ?(?
?
) =? ?? where ? 6= 0.
0 1
Then Im(?) contains an open subgroup of SL2(F[[T ]]).
5
On the other side of the picture, we have the rich theory of ordinary modular forms
which is primarily developed by Hida. In our special case, we have a complete description
of these Hecke algebras. Let Hm and hm be the Hida ordinary Hecke algebras acting on
spaces of modular forms and cusp forms respectively of fixed tame level N and Nebentypus
?? . Let us assume Neben. Then we have the following theorem:
Theorem 1.0.11. Hm = hm??/(A? ? ) ?, where h is the cuspidal Hecke algebra and H is,
the Hecke algebra of modular forms. We define the annihilator of the unique Eisenstein
series E? (? ,?) by I(? ,?) and we denote by I? ,? the image of I(? ,?) under the canonical
projection H h. And
( )
A? ,? := ? ((1+X)s(l)????1??2(l)l?2) G(T,???1?2)
l|N
l-cond(???1)
where G(T,???1?2) is a twist of the Deligne-Ribet p-adic L-function as follows:
G(?(u)us?1,???1?2) = Lp(?1? s,???1?2?)
where ? is a Dirichlet character of the second kind.
An important corollary of this theorem is the following:
Corollary 1.0.12. Assume p|Lp(?1,???1?2) but p2 - Lp(?1,???1?2), then (Hord2 )m is
Gorenstein iff (hord) =? O or some ramified DVR over O, where (Hord2 m 2 )m and (hord2 )m are
the ordinary Hida Hecke algebras acting on weight 2 modular and cusp forms with level N
and Nebentypus ?? respectively. In particular, (hord2 )m is a regular local ring of dimension
1.
Now going back to our problem at hand: we want to show that there is an isomorphism
Runiv ??? = T , where T is some suitable Hecke algebra. But this is where one faces serious
6
difficulties in proving modularity lifting results as there may not even be a map ? : R? T .
This is where we again need Hypothesis 1 to construct the map. The ? is shown to be an
isomorphism using a numerical criterion.
The second problem in proving an ?R = T ? theorem in this reducible case is that T is no
longer always Gorenstein and thus the numerical criterion fails. Then one can still hope
to prove R = T by some other method bypassing the numerical criterion. But in view
of Fontaine-Mazur conjecture, we should recover these representations in cohomology of
some geometric object. Indeed by the results of Hida, the cohomology of the towers of
modular curves ?lim
1 r
?H (Y (N p ),Zp)m is our natural candidate. But results of Tilouine-
Mazur show that?lim
1 r
?H (Y (N p ),Zp)m is free over Tm iff Tm is Gorenstein and thus we lose
the crucial geometric input to prove such a result. Thus we need new ideas to prove ?R= T ?
results where some appropriate cohomology group is not free over the Hecke algebra. See
the results of Erickson-Wake in this direction, where they work with pseudo-deformation
rings. But one can ask for the next best case scenario.
Question 1.0.13. Is 1?lim?H (Y (N p
r),Zp)m representable by a perfect complex of Tm-modules
when the residual Galois representation is mod-p distinguished?
In that case one can use Nekovar?s machinery of Selmer complexes to construct a 2-
variable p-adic L-function, the existence of which is only known in the free case.
Even though the Hecke algebras are not Gorenstein, we have the following conjecture by
P. Wake.
Conjecture 1.0.14. (Wake): Let Ih and IH be the Eisenstein ideals in hord and Hord. Then
for all height 1 prime ideals p and q such that Ih ? p and IH ? q, hord ordp and Hq are Goren-
stein. In that case we say Hord and hord are weakly Gorenstein.
Using our characterization of I/I2 and the structure of the Hecke algebras, we prove
Wake?s conjecture in our special case.
7
Theorem 1.0.15. (Wake?s conjecture) Hordm is weakly Gorenstein where m is an Eisenstein
maximal ideal of Hord . (cf. Definition 6.2.16)
And finally we reprove some results of Skinner-Wiles in [56] and [57]. Our proof also
shows that the Hecke algebras considered by Ohta in [43] are Gorenstein. To prove the
next theorem, let us assume the following additional hypothesis:
? p - ?(N) or Neben.
? (? ,?) 6= (??2,1)
? p|B1,???1??1
? (Vandiver type conjecture) Gal(L?/F?) is cyclic as a ? module, cf. section 3.5.
Theorem 1.0.16. We have an isomorphism Runiv??? =Hm of complete local intersections over
?.
Definition 1.0.17. (Skinner-Wiles) A prime p in Runiv?? is nice if it is height 1, contains p
and p does not contain the ideal of reducibility and is an inverse image of a prime q in hm.
Theorem 1.0.18. (a) There is an isomorphism (Runiv ??? )p = (Hm)q of complete local inter-
sections over ? , where p ? Runivp ?? and q ? Hm are any height 1 prime ideal (sometimes
referred to prime divisors) over (p)? ? such that ??1(q) = p
(b) (Skinner-Wiles) Under the isomorphism above, (Runiv?? ) ?p = (hm)q for all nice primes p
and both the rings are complete local intersections.
The organization of the thesis is as follows:
Chapter 2 : Sections 1 and 2 deal with basic definitions. In section 3, we prove some tech-
nical Galois cohomology results and give some criteria for the cup products. These results
are then used throughout Chapter 3,6 and 7. We also provide examples where Hypothesis 1
8
is satisfied in the form of proposition 2.3.8.
Chapter 3 : Sections 1 and 2 are respectively about basic definitions and properties of Ga-
lois representations and Mazur?s results about the structure of universal deformation ring
and in section 3, we recall some properties of ordinary deformation rings. In section 4,
we compute the tangent space of the universal deformation ring in theorem 3.4.1 which
is one of the main results of the section. We then carry our analysis further by trying to
find explicit obstructions to lifting Galois representations. In some special cases, we man-
age to find our obstruction classes as in theorem 3.4.19. Some of these computations give
alternate proofs of the smoothness of local deformation rings. At the end of the section,
we show how Hypothesis 1 can be used to simplify some arguments in [57]. In section
5, we make a deeper look at the fields that are cut out by the Galois representations con-
structed in the previous sections. We construct unramified extensions in theorem 3.5.15
and 3.5.25. We also prove various Iwasawa theoretic properties of some intermediate fields
which will be quite important for us in Chapter 6. In section 6, we use our ideas to explic-
itly construct these fields in the case of elliptic curves of conductor 11. Section 7 is about
pseudo-representations and gives us a link between pseudo-representations and representa-
tions and this will be important for us to construct the map ? : R? T .
Chapter 4 : Chapter 4 is basically a crash course on Hida theory and the properties of the
Galois representations attached to modular forms.
Chapter 5 : We start with some history and motivation behind big image questions and in
section 2, we prove our big image theorem ( theorem 5.2.5).
Chapter 6 : In Section 2, we introduce the Eisenstein series and compute the congruence
modules attached to the Hecke algebras and prove the structure theorem for Hida Hecke
algebras in proposition 6.2.26. We also construct a cusp form which is congruent to our
Eisenstein series. In section 3, we state Wake?s conjecture and state the various isomor-
phism criteria. We then use these criteria to prove the modularity lifting results and Wake?s
9
conjecture.
Chapter 7 : We revisit Selmer groups and we show a factorization property for Selmer
groups.
10
Chapter 2: Galois cohomology and Selmer groups
In this chapter, we recall some main results on Galois cohomology and use them to
compute tangent spaces of deformation rings and study Selmer groups. Most of the expo-
sition can be found in Milne, Tate, Rubin, Neukirch-Schmidt-Wingberg and Wiles in the
references.
2.1 Basic definitions
Let G be a group and M be a module with an action of G. The cases of interest are
when both G and M are discrete or G is profinite and M is discrete and finally both G and
M are profinite. In any case, we will always require the action of G on M to be continuous.
For a topological group G and a module M, the i-th group of continuous cochains Ci(G,M)
is the group of continuous maps Gi?M. There is a differential d : Ci(G,M)?Ci+1(G,M)
satisfying
(d f )(g1, ...gi+1) = g1. f (g2, ...gi+1)+?(?1) j f (g1, ..g jg j+1, ..gi+1)+(?1)i+1 f (g1, ...gi)
It is easy to check d2 = 0, so we have a complex C?(G,M). Then we define H i(G,M) :=
ker d/im d.
In the sequel, we will drop the term continuous, where all our cocycles will be continuous
maps. Let ?1,?2,?3 be 3 representations of G in Gld1(A),Gld2(A) and Gld3(A) respectively.
Then one can define the 4 modules:
11
? Z1G(?2,?1) = {c : G?Md1,d2(A) : ?g1,g2 ? G,cg1g2 = ?1(g1)cg2 + cg1?2(g2)}
? B1G(?2,?1) = {c : G?Md1,d2(A) : ?M ?Md1,d2 ,?g ? G,cg = ?1(g)M+M?2(g)}
? Z2G(?2,?1)= {c : G?G?Md1,d2(A) :?g1,g2,g3 ?G,?1(g1)cg2,g3?cg1g2,g3 +cg1,g2g3?
cg1,g2?2(g3) = 0}
? B2G(?2,?1)= {c : G?G?Md1,d2(A),? f : G?Md1,d2(A),?g1,g2 ?G,cg1,g2 = f (g1g2)?
?1(g1) f (g2)? f (g1)?2(g2)}
The elements in Zi and Bi are called i-homogeneous cocycles and i-homogeneous cobound-
aries. It is easy to see that BiG(?2,?1)?ZiG(?2,?1) and the quotient is denoted as Ext iG(?2,?1).
We will drop G from the notation whenever the group G is clear from the context. The
case that is important for us is when ?i are characters then it is easy to see Ext i(?2,?1) =
H i(G,? ??11 2 ). We will use this identification throughout this thesis.
Finally the Yoneda product is defined as follows:
Ext1(?2,?1)?Ext1(?3,?2)?? Ext2(?3,?1)
given by
(c1,c2)? (g1,g2 7? c1(g1)c2(g2))
It is easy to see that it maps
Z1(?2,? 1 21)?Z (?3,?2)? Z (?3,?1)
and maps
Z1(?2,?1)?B1(?3,?2)+B1(?2,?1)?Z1(?3,?2) to B2(?3,?1).
12
Thus it induces a bilinear map on
Ext1(?2,?1)?Ext1(?3,?2)?? Ext2(?3,?1).
In the context of inhomogeneous cocycles, the Yoneda product is also called the cup prod-
uct.
Proposition 2.1.1. (Tate) (a) Let T =?lim?Tn and assume the Tn?s are finite. If i > 0 and
H i(G,Tn) is finite for all n, then H i(G,T ) =?lim?H
i(G,Tn).
(b) If T is a finitely generated Zp-module and i? 0, then H i(G,T ) has no divisible elements
and H i(G,T )?Qp ?= H i(G,T ?Qp).
Proposition 2.1.2. Suppose H is a closed normal subgroup of G, and let M be a dis-
crete, finitely generated Zp-module or a finite dimensional Qp-vector space. There is a
Hochschild-Serre (inflation-restriction) exact sequence
0? H1(G/H,MH)? H1(G,M)? H1(H,M)G/H ? H2(G/H,MH)? H2(G,M)
13
2.2 Local duality theorems
Let K be a finite extension of Qp and let ?n be the n-th roots of unity inside K? , where
K? is an algebraic closure of K. We will write H i(K,?) to denote H i(Gal(K?/K),?)
Proposition 2.2.1. (a) H0(K,?n) = ?n?K
(b) H1(K,? ) = K?/(K?)nn
(c) H2(K,?n) = Z/nZ
(d) H2(K,Gm) =Q/Z
(e) H i(K,?n) = 0 for i? 3.
Corollary 2.2.2. If M is a finite GK module, then H i(K,M) is finite.
Now we can state Tate?s local duality theorem.
Theorem 2.2.3. Let M be a finite GK-module and let M? = Hom(M,Gm). Then for 0? i?
2, the cup-product
H i(K,M) ? H2?i(K,M?)?? H2(K,Gm)?=Q/Z
is a perfect pairing.
For a finite GK-module M, we define the local Euler-Poincare characteristic to be
#H0(K,M) ?#H2(K,M)
?(M) = 1 .#H (K,M)
We can extend the concept to the case where M is a finite free Zp-module or a finite-
dimensional Qp-vector space by making the more familiar definition
?add(M) = h0(M)?h1(M)+h2(M)
14
where hi(M) = rank H i(K,M). We have the following useful formula for the Euler charac-
teristic.
Proposition 2.2.4. ?(M) = p?vp(#M).N = 1[O:#MO]
where N = [K : Qp] and O is the ring of integers in K. In particular, ?(M) = 1 if order of
M is coprime to p.
Corollary 2.2.5. Let ? be the mod p cyclotomic character. If l 6? 1 mod p, then the deg
p extension of Ql(?p) given by the fixed field of the kernel of c, where c is any non trivial
cocycle in H1(Ql,?), is ramified (tamely) at l.
Proof. This is a straightforward application of the Hochschild-Serre exact sequence 2.1.2
by noting that the restriction H1(Q 1l,?)? H (Il,?) is an injection.
We will end this section by recalling some facts about unramified cohomology.
We define H inr(K,M) := H
i(Knr/K,MI) to be the unramified classes, i.e. classes vanishing
on inertia.
By inflation-restriction exact sequence we get
H1nr(K,M) = ker(H
1(K,M)? H1(Knr/K,M)) (2.1)
and since H2(Z?,M) = 0, we get the following exact sequence
0??MGK ??MI F??ro?b?-?id MI ??MI/(Frob? id)MI ?? 0 (2.2)
In particular, if M is finite, we get
#H1nr(K,M) = #H
0(K,M) (2.3)
15
and
H2nr(K,M) = 0 (2.4)
Proposition 2.2.6. If #M is relatively prime to p, then H1nr(K,M) and H1nr(K,M?) exactly
annihilate each other under the Tate pairing.
Proof. Note that the inclusion H1nr(K,M) ??H1(K,M) is compatible with cup-products so
the map
H1nr(K,M) ? H1nr(K,M?)?? H2(K,Gm)
factors through H2nr(K,Gm) which is 0. The only thing left to show is that
#H1nr(K,M) ?#H1 ? 1nr(K,M ) = #H (K,M).
Now, #H1nr(K,M) = #H
0(K,M) and #H1 (K,M?nr ) = #H
0(K,M?) and H0(K,M?) is dual to
H2(K,M) via Tate pairing. And finally since (#M, p) = 1, ?(M) = 1 and this gives the
desired result.
16
2.3 Restricted ramification and Global duality
In this section we will be concerned with the cohomology of GQ,S where S is a finite set
of primes containing p and ?. One of the main reasons for doing so is that the cohomology
of GQ is not well-behaved.
For a finite module M, we define M? = Hom(M,Gm). We will start with a few lemmas, the
results of which are used later.
Lemma 2.3.1. h1(GQ,S,F) = #{q ? S : q? 1 mod p}+1, where F= Fpr and r ? 1.
Proof. H1(GQ,S,F) = Hom(GQ,S,F), so this reduces to finding cyclic extensions of Q of
degree p unramified outside of S. Now take any prime q ? S such that q? 1 mod p. Then
the field of q-th roots of unity has a cyclic subfield of degree p. And finally we can construct
a degree p extension from Q(?p2). And it?s easy to see that these are the only ones.
Lemma 2.3.2. h2(GQ,S,F) = #{q ? S : q? 1 mod p}
Proof. This is corollary 8.7.5 in [50]. It is a trivial consequence of the global Euler
Poincare characteristic formula (2.9).
Recall we have the restriction maps:
res : H i(GQS ,M)? H
i(GQq,M) (2.5)
which gives rise to a map called localization
loc : H i(GQS ,M)??H i(GQq ,M) (2.6)
The next lemma is a well-known local-global principle in algebraic number theory.
?
Lemma 2.3.3. loc: H2(G 2QS ,F)? q?S:q?1 (mod p)H (GQq ,F) is an isomorphism.
17
Proof. By local Tate duality (theorem 2.2.3), H2(GQq,F) 6= 0 iff l ? 1 (mod p). Thus both
the sides have the same F-dimension. So it suffices to show that the map is an injection.
But the the injectivity is (i) of Corollary 9.1.10 in [50].
?
It is customary to write H2(Q ,M) := P2q?S q (GQ,S,M). See theorem 2.3.12 for more
details. This lemma is quite useful as it allows us to check if some 2-cocycle is 0, by
checking it locally.
The next lemma also allows us to detect vanishing of cup products.
Lemma 2.3.4. If a ? Ext1(?2,?1) and b?? Ext1(?1,?2), t?hen a?b = 0 is equivalent to the?? ?1 a ? ??
existence of a representation of the form??? 0 ?2 b ???
0 0 ?1
Proof. a? b = 0 is equivalent to the existence of a function f such tha?t d f = a? b, i.e.
??? ?1 a f? ?
?
?
f (gh) = ?1(g) f (h)+ a(g)b(h)+ ?2(h) f (g). Now construct the matrix ? 0 ? b ?2 ??
0 0 ?1
and one can easily see it?s a representation. We remark that the choice of f is not unique
but can differ by an element in Z1(?1,?1).
Since our goal in the next chapter is to construct these meta-abelian extensions, we
would like to come up with some criteria for the vanishing of the above cup-products.
Note that the cup product trivially vanishes if there are no primes congruent to 1 in S.
To formulate our problem: let ? be a non-trivial character of order prime to p and let
Q(?p) ? K? , where K? is the fixed field of the kernel of ? . In view of lemma 2.3.3,
we can check the vanishing of the cup products locally since cup products are compatible
with restriction. The cases of interest are the primes congruent to 1 (mod p). Let q be
such a prime and let ? be ramified at q, then by theorem 2.2.3 and proposition 2.2.4,
18
H1(Qq,??1) is trivial. Again note that by the same argument if ? is unramified but non-
trivial, H1(Q ?1q,? ) is trivial. Thus we have nothing to prove. Thus the only case of interest
is when q? 1 (mod p) and ?|GQ = 1. In that case, H1(Qq,F) is a 2 dimensional F vectorq
space spanned by a ramified cocycle and an unramified cocycle. The cup product of any
two unramified cocycles is trivially zero and since the Tate pairing is non degenerate, the
cup product of two ramified cocycles is zero iff they lie on the same line. We summarize
this discussion in the form of the following proposition:
Proposition 2.3.5. Let ? be any finite order non-trivial character of order prime to p. Let
H1(GQ,S,?)?H1(GQ,S,??1)??
? H2(GQ,S,F)
be the usual cup-product. Then ?= 0 if one of the following conditions are satisfied:
? There are no primes congruent to 1 (mod p) in S.
? ? is ramified or non-trivial at all primes q ? S which are congruent to 1 (mod p)
Moreover let H1(GQ,S,?) be one-dimensional as a F-vector space and let b be a basis for
this space. Let c ? H1(G ?1Q,S,? ), then b? c = 0, iff the following hold.
b and c are both unramified at q, or if b is ramified at q, then c must be ramified at
q and c|GQ must be a multiple of the basis element of H1(Qq,??1), i.e. when q ? 1q
(mod p),c|GQ = kb|GQ for some k ? F.q q
Remark 2.3.6. We will only use this proposition in the case where h1(GQ,S,?) = 1. Then
we can drop the assumption that ? is non-trivial. If ? is trivial, then in that case b and c are
F multiples of each other and since the cup product is anti-symmetric b? c = 0
In fact we can try to see how to find obstructions in constructing upper triangular rep-
19
resentations of the form ?? ??? ?1 a f1 ??? ?
?
? 0 ?2 b f ?2?? ?
?
0 0 ? ?1 a ??
0 0 0 ?2
where a?b= 0= b?a. So define d f2 = b?a and f1 = f in the previous lemma (lemma 2.3.4).
If a representation were to exist, the top right corner, call it ? , should satisfy
?gh = ?1(g)?(h)+a(g) f2(h)+ f1(g)a(h)+?(g)?2(h) (2.7)
A brute force computation shows that
a(g) f2(h)+ f1(g)a(h) ? Z2(?2,?1)
So the same calculation from the previous corollary shows that existence of the representa-
tion is equivalent to the fact that a(g) f2(h)+ f1(g)a(h) is a coboundary.
Remark 2.3.7. a(g) f2(h) + f1(g)a(h) is called a triple Massey product and the higher
Massey products measure obstructions to constructing higher dimensional representations.
But we do not know of any necessary or sufficient conditions to make the above triple
Massey product a coboundary in such a general situation, except when the target group is
trivial, nor can we compute it explicitly in any generality.
Now we will give some conditions and examples where our Hypothesis 1 is satisfied. Let
? be a character of order prime to p and let K? be the fixed field of the kernel of ? . Let
K =K?(?p), so p - [K :Q]. The characters associated to K are ? i? j. Assume the values of ?
lie in F and we will use ? to denote the one-dimensional space over F on which Gal(K/Q)
20
acts via ? . Let S be the set of primes that divide the conductor of K. By inflation-restriction
(proposition 2.1.2), we get the following exact sequence:
0? H1(K/Q,?)? H1(QS/Q,?)? H1(Q /K,?)Gal(K/Q)S ? H2(K/Q,?).
Since p - [K : Q], H1(K/Q,?) = H2(K/Q,?) = 0. Therefore,
H1(QS/Q,?)?= H1(Q /K,?)Gal(K/Q)S
Since Gal(QS/K) acts trivially on ? , we get
H1(Q /K,?)Gal(K/Q)S ?= Hom(Gal(QS/K),?)Gal(K/Q)
Let ? ?Hom(Gal(Q?/K),?)Gal(K/Q), h ?Gal(K/Q) and let h lift to h? ?Gal(QS/Q). Then
?(h?gh??1) = ?(h)?(g) for all h,g.
By Kummer theory,
g(b1/p)
?(g) = = ?g,b?
b1/p
where b ? K?/(K?)p and ?,? is the Kummer pairing that takes values in ?p. Therefore,
?g,b??(h) = ?gh,b?= ?g,h?1b?h = ?g,h?1b??(h).
The non-degeneracy of the Kummer pairing implies that
h?1b? b?(h)?
?1(h) mod(K?)p.
Changing h to h?1,
bh ? b?
?1(h)?(h) mod(K?)p.
21
Now K(b1/p)/K is unramified outside S and assume the ??1?-component of the class
group of K is prime to p. It follows that b is an S-unit (times a p-th power). Let ES be the
S-units in K. Then we have shown the following proposition.
Proposition 2.3.8. Let ? be a character whose order is prime to p. Assume ??1? compo-
nent of the p-part of the class group of K is trivial. Then
H1
?1
(QS/Q,?)?= (E /(E )p)? ?S S
Finally, we need to decompose ES/(ES)p into irreducible components under the action
of Galois. Since p - [K : Q], the representation is semi-simple. Let E be the group of
units in K. The units E+ in K+ have an unit whose Galois conjugates generate a finite
index subgroup L of E+/{?1} (cf. 5.27 in [67]). Therefore L/Lp decomposes into one-
dimensional components corresponding to the non-trivial even characters of Gal(K/Q). By
the Brauer-Nesbitt theorem 3.1.8, the same is true for E+/(E+)p. Since E+ and ?p gener-
ate a subgroup of index 1 or 2, we find the characters that occur in E/E p are the non-trivial
even characters and ? .
For simplicity, assume the conductor of ? is q where q ? 1 mod p and q is a prime. So q
splits completely in Q(?p)/Q. Therefore the primes over q contribute to one-dimensional
F vector spaces on which Gal(Q(?p)/Q) acts via ? i, for 0? i< p?1. Let ? be a character
of conductor q in the group generated by ? and ? . Then ? is a character of Gal(K/Q(?p))
of order dividing the degree of the extension. Let f be the order of the root of unity ?(p).
Then the prime of Q(?p) over p splits into g = [K : Q(?p)]/ f primes in K. Therefore, the
primes of K above p contribute to the characters 1,? f ,?2 f , ...?(g?1) f . Note that they all
have conductors q or 1. Putting all of the above arguments together, we obtain exactly what
characters occur (with multiplicities) in ES/(E pS) .
If ? =6 ? is odd, then ???1 is even and non-trivial and so occurs at least once. If ???1
22
has conductor pq, then it does not occur in the representations coming from p or q. There-
fore, it occurs exactly once. This implies that if both ? and ???1 have conductor pq, then
H1(QS/Q,?) is one-dimensional.
In the final part of this section, we introduce cohomology classes that have special local
behavior. Recall we have the map
res : H i(G iK,M)? H (GKv ,M) (2.8)
Definition 2.3.9. A Greenberg-Wiles Selmer system is a collection L = {Lv} of subgroups
L 1v ? H (Gv,M) such that for almost all primes v 6= p,
Lv = H1nr(Kv,M) = ker(H
1(Kv,M)? H1(Knrv /Kv,M))
The condition at p is subtle and one generally uses the Greenberg condition. And gen-
erally we will require some conditions on the primes where the representation is ramified.
For a more complete list of local conditions see [60].
Definition 2.3.10. The Selmer group associated to a set of local conditions is given by
( )
H1 1L (Q,M) = ker H (Q,M)??H1(Qv,M)/Lv
v
Definition 2.3.11. Define L? ? H1v (G ?v,M (1)) to be the annihilator of Lv under the local
Tate pairing. Then we call L ? the dual Selmer system for L and H1 ?(Q,M?(1)) the dualL
Selmer group.
Now we state the main results of this section, which allow us to compute various Selmer
23
groups.
Theorem 2.3.12. (Poitou-Tate) Let K be a number field, S be a finite set of primes con-
taining the archimedean primes and all places v such that v(#M) =6 0. Then we have the
following 9-term exact sequence.
0? H0(GK,S,M)? P0(GK,S,M)? H2(G ,M?)?K,S ?
? H1(G ,M)? P1(G ,M)? H1(G ,M?)?K,S K,S K,S ?
? H2(G 2 0K,S,M)? P (GK,S,M)? H (G ,M?)?K,S ? 0,
where A? = Hom(A,Q/Z), and
Pi(GK,S,M) = ??H i(Kp,M), i? 0
p?S
where the restricted product is taken with respect to the subgroups H inr(Kp,M) of H
i(Kp,M).
Note that we take H?0 at the Archimedean primes.
We define the global Euler-Poincare characteristic to be
#H0(GK,S,M) ?#H2(GK,S,M)?(M) :=
#H1
(2.9)
(GK,S,M)
We have the following useful formula to compute the Euler-Poincare characteristic.
#H0(K ,M) #H?0(K ,M)
Proposition 2.3.13. ?(M v v) = ? = ? 0 ?
v?S ?#M? v?S #H (K ,M )? ? v
Theorem 2.3.14. (Greenberg-Wiles) Let M be a finite GQ module and let L be a set of
Selmer conditions. Then H1L (Q,M) and H
1 (Q,M?) are finite and
L ?
#H1L (Q,M) #H0(Q,M) #Lv
#H1 (Q,M?
=
? ) #H0(Q,M?)
? #H0(Q
L v
,M)
24
Note that this formula makes sense as #Lv = #H0(Qv,M) for all unramified primes.
25
Chapter 3: Universal Deformation Rings
In this chapter we will study deformations of various Galois representations and study
the tangent spaces of these rings.
3.1 Basic Definitions and preliminaries on representations
Let G be a profinite (compact), Hausdorff topological group. In our applications, G
arises from one of the following situations.
? Local Fields: Let L be a local field, i.e. a finite extension of Qp, and L? be an algebraic
closure. Then G= Gal(L?/L)
? Global fields: Let K be an extension Q and let S be a finite set of primes in K. Let KS
be the maximal extension of K unramified outside S. Then G= Gal(KS/K)
Definition 3.1.1. A Galois representation is a continuous group homomorphism from G to
GLn(R) where R is a topological ring and G is as above. We call a Galois representation
a local (Galois) representation if G is the Galois group of the local field and we call the
representation global if G is the Galois group of the global field.
Now we state a few well-known lemmas without proofs.
Lemma 3.1.2. Let G be as above. And let
? : G?? GLn(Q?p)
26
be a continuous representation. Then there exists L, a finite extension of Qp such that
?(G)? GLn(L).
Lemma 3.1.3. Let ? : G ?? GLn(L) be a continuous representation. Then there exists
M ? GLn(L) such that M?(G)M?1 ? GLn(OL).
The above lemma shows that there is at least one lattice stable under the action of G.
Proposition 3.1.4. The number of stable lattices (up to homothety) is finite iff ? is irre-
ducible.
In fact, in our situation the number of stable lattices will not be 1. As an example, take 2
elliptic curves E1 and E2 over Q. If they have Q-rational 2-isogeny between them, then they
give rise to non-isomorphic Galois stable lattices for their associated 2-adic representations.
This motivates the following proposition. But we need some notations and definitions
before we can state it.
Definition 3.1.5. The semi-simplification of a representation G on a finite dimensional
vector space V over a field k is the direct sum of all the Jordan-Ho?lder constituents of the
k[G]-module V . We usually denote it by V ss. In this definition, we take multiplicities into
account so that dimk V = dim V ssk .
Definition 3.1.6. The representation V is semi-simple iff V ?= V ss as k[G]-modules. Con-
cretely, if (Vi) is an increasing filtration of sub-representations of V such that Vi+1/Vi is
irreducible, then V ss ?=?iVi+1/Vi.
Example 3.1.7. Let k be any field and let ? : G?? GL2(k) be given by
?? ?? a(g) b(g)g 7? ??
0 d(g)
then ?ss = a?d
27
Let ?? Ln be a stable lattice under the action of G.
Proposition 3.1.8. (Brauer-Nesbitt) The semi-simplification of the representation of G on
?/?? is independent of the choice of ?.
Corollary 3.1.9. G has a unique stable lattice in Ln (up to homotheties) iff G acts irre-
ducibly on knL, where kL is the residue field.
Proof. This is exercise 4 in [52] page 3. We give a quick sketch. Let G act irreducibly on
the residue field and assume we have 2 stable lattices call them L1 and L2 and moving them
by homothety, assume L2 ? L1 and L2 6? ?L1. Now, L2/(L2 ??L1) ?? L1/?L1. Since G
acts irreducibly on the residue field, we get L2/(L2 ? ?L1) = 0 or L1/?L1. If L2 = ?L1,
then L1 and L2 are homothetic. So assume L2/(L2 ? ?L1) = L1/?L1. Then we can see
that L2/?L2  L1/?L1 and by irreducibility L2/?L2 ?= L1/?L1. Thus there is a matrix
M? ? GL2(kL) that sends L1/?L1 to L2/?L2. By Nakayama?s lemma, we can lift M? to
GL2(O) that sends L1 to L2. Thus the two lattices are isomorphic.
The converse is easy.
The above proposition motivates the following definition.
Definition 3.1.10. Let L be a finite extension of Qp and ? : G ?? GLn(L) be a Galois
representation. We denote by ??ss : G ?? GLn(kL) the semi-simple representation defined
above. It is called the residual representation of ? .
Definition 3.1.11. We call ?? absolutely irreducible if ?? ? k?L is irreducible where k?L is an
algebraic closure of kL.
Recall the following well-known lemma by Ribet.
Lemma 3.1.12. (Ribet) If ? : G? GL2(L) is irreducible but ??ss is reducible, then there
exists a G-stable lattice where ?? is indecomposable.
28
Proof. This is Prop. 2.1 in [48]. We recall some key aspects of the proof as those ideas
will be important for us. Ribet considers the graph of stable lattices (up to scaling by L?),
where 2 lattices [L1] and [L2] are connected by an edge if ?L1 ? L2 ? L1. Let us call the
graph X. He shows X is a tree. Note that X is bounded by Prop 2.1.4 and note that ?? is
indecomposable iff it has one neighbor. Such a vertex is called a leaf. We recall a theorem
from [53] which say trees have leaves, and this gives us the desired lattice.
In this thesis, we will be dealing with reducible, indecomposable representations. In
view of Ribet?s lemma, we would like to pin down our choice of lattice. Note that the proof
of Ribet?s lemma gives us not one but two lattices sitting on opposite sides of our tree.
L1 L2 . . . Ln
indecomposable at p
And ?? is indecomposable for exactly 2 lattic?es, and let?us denote th?e 2 represe?ntations
?1 ? ?2 ?
by ??1 and ?? . If ??ss ? ? ? , then ?? ? ?? ?? and ?? ? ?? ??2 = 1 2 1 = 2 = Since
0 ?2 0 ?1
we are dealing with odd representations, i.e det(?(c)) = ?1 for any choice of complex
conjugation c, the order the diagonal characters is determined by fixing a basis for the
complex conjugation. So one still manages to get a unique lattice, even if the ?? is not
irreducible. In fact one has the following corollary.
Corollary 3.1.13. At least one of the indecomposable ?? constructed above has the added
property that ?? restricted to inertia group at p is indecomposable.
Proof. This is corollary 5 on page 190 in [38].
Example : Let p = 691, Deligne constructs a Galois representation attached to the Ra-
manujan cusp form ?. The following is an unpublished result of Greenberg-Monsky, the
29
proof of which can recently be found in [70].
Proposition 3.1.14. (Greenberg-Monsky, Yan) There are exactly 2 lattices (up to homoth-
ety) for the Galois representation associated to ? such that the mod 691 representation is
reducible and indecomposable with the diagonal characters 1 and ?11 occurring in oppo-
site orders. So the lattice chain looks like:
L1 L2
indecomposable at I691 split at I691
Remark 3.1.15. Our previous discussion shows that at least one of the ?? is no?t semi-simp?le
1 ?
at I691. Note that the ??2 which is the reduction of the lattice L2 has the shape?? ??.
0 ?11
Since Q(?691) has a cyclic 691 degree extension, unramified everywhere with ??11-action,
we see that ? on I691 is trivial. In fact, exercise 8.9 in [67] shows how to find a Kummer
generator for the extension. So this implies that ??1 contains a wildly ramified 691 degree
extension over Q(?691). This can also be verified by checking that the ?11 component of
the 691 class group of Q(?691) is trivial.
Remark 3.1.16. Serre in [51] observed that the number of lattices is tied to the fact that ??
is not reducible mod 6912. Greenberg and Monsky used that idea and showed that the piece
of the 691-Hilbert class field of Q(?691) corresponding to the character ??11 is contained
in the field extension of Q cut out by Deligne?s 691-adic representation of GQ associated
with ?. We use Serre?s observation as our motivation to carry out these investigations for
the elliptic curves of conductor 11 and more general reducible representations in Section
3.5 and 3.6.
Lemma 3.1.17. (Schur) If ?? is absolutely irreducible, then EndkL[G](??) = kL.
30
Note: The converse is not true. For example the centralizer of the Borel in Gl2(kL) is
kL.
Definition 3.1.18. We call a representation ?? Schur if EndkL[G](??) = kL.
From now on, let F be a finite extension of Fp (with discrete topology) and let O be
a complete local Noetherian algebra with maximal ideal mO with residue field F and we
fix an isomorphism O/m ?O = F. Let CO be the category of complete, local Noetherian
O-algebras with residue field F. The objects are A with maximal ideals mA such that
the structural map O ? A induces an isomorphism O/m ?O = A/mA. Maps are local ring
homomorphisms compatible with the identification of residue fields with F.
We will now prove a version of Schur?s lemma for objects in CO .
Lemma 3.1.19. Let A ? CO , ? : G?GL (A) and End (??) = F. Then End (?) = A?n F[G] F[G] .
Proof. This is essentially the proof of Mazur in [37]. So we only give a sketch. The idea is
to use completeness and reduce to the case of Artinian local O-algebras and then induct on
the length of A. The base case is when length of A is 0, i.e. A = F and the induction step is
F[?]/(?2). Assume now A is local Artinian:
0 =me+1A (m
e
A ( ...mA ( A
with each quotient an F-vector space. Choose a minimal non-zero ideal I ?meA which is a 1-
dimensional F-vector space, and by choosing an F basis, we identify I with F. Now let M ?
GLn(A) commute with ? . The induction hypothesis implies M mod I ? EndA/I(? mod I) =?
(A/I)?. So M is of the form M = ?In + M0, where M0 ? Mn(I). Now for all g ? G,
(?In +M0)?(g) = ?(g)(?In +M0), which implies M0?(g) = ?(g)M0. And by the above
identification, we can view the above equation in Mn(F). And Schur?s lemma holds in this
case. So M0 is also a scalar and this proves the lemma.
31
Lemma 3.1.20. (Carayol, Serre) Let A,B ? CO and B ? A closed. Let ? : G? GLn(A).
Assume ?? is irreducible and tr(?) lies in B. Then there exists M ? ker(GLn(A)? GLn(F))
such that M?(G)M?1 ? GLn(B).
Proof. This is proposition 2.13 in [24].
Remark 3.1.21. This lemma is the main ingredient to show that Runiv?? is generated by the
traces of the Frobenii.
Remark 3.1.22. One can also replace the finite field F by any other other field k with trivial
Brauer group, i.e. H2(k, k??) = 0.
However there is a generalization of Carayol?s lemma due to Kisin which will be used
throughout the thesis. The main application of this lemma will appear in Chapter 3. See
the section on Pseudorepresentations for more details.
32
3.2 Preliminaries on deformation theory
Definition 3.2.1. Let ?? : G? GLn(F) be Schur (definition 3.1.18) and define the defor-
mation problem R?? from CO to SETS to be given by:
A 7? {? : G? GLn(A) : ? mod mA = ??}/?=
Note:
?1 =? ?2??M ? ker(GLn(A)? GLn(F)),M? M?11 = ?2
Theorem 3.2.2. (Mazur):(a) (Existence) R?? is representable by a ring, say Runiv?? .
(b) (Twisting by character) If ?? ? is another representation equivalent to ???? where ? is a
character, then there is a canonical isomorphism between Runiv?? and R
univ
?? ?
Proof. See [37].
Before we delve into deeper properties of this ring and define interesting deformation
problems, let us state the n = 1 case. This will be important to us later on.
Proposition 3.2.3. The universal deformation ring for a character ?? : G ?F? is W F Gab,(p)Q ( )([[ Q ]])
where W (F) is the ring of Witt vectors, Gab,(p)Q is the abelianization of its pro-p completion.
e
In fact R?=W (F)[[X1, ...Xk,Y ]]/((1+Xi)pi ?1)), where k is the number of primes in S con-
gruent to 1 mod p and ei = valp(li? 1). We will call it the Iwasawa algebra. The reason
for this will be clear at the end of the next example.
Remark 3.2.4. Even though the above ring is formally smooth, its special fiber is com-
plicated. To work with the special fiber, Mazur-Wiles introduced sheets. We will restrict
ourselves to a very special case to avoid dealing with sheets.
Example 3.2.5. Let p be an odd prime, S = {p}, F = Fp and G =Gal (QS/Q). In that
case, the deformation ring is Zp[[1+ pZp]], which is the usual Iwasawa algebra.
33
Remark 3.2.6. Given a residual representation ?? , det(??) is a 1-dimensional representation.
If ? is a deformation to a ring R, then clearly det(?) is a deformation of det(??). If ?univ
is the universal deformation, then it follows det(?univ) is a deformation of det(??) to Runiv?? .
By the universal property, there exists a unique map, which we will call ?det?, from the
Iwasawa algebra to Runiv?? which allows us to view R
univ
?? as an algebra over the Iwasawa
algebra.
In certain situations, we will demand our deformations to have a fixed determinant. One
can then easily show the existence of a universal deformation ring Runiv,det?? as parametrizing
deformations of ?? to CO with a fixed determinant.
We now define and compute tangent spaces of some deformation rings.
Definition 3.2.7. For a ring R ? CO , its tangent space is defined as
tR := HomO(R,F[?]/?2)?= Hom 2F(mR/(mR +mOR),F)
Remark 3.2.8. Given two deformations of ?? to F[?]/(?2) given by the matrices
?? ? ? ?? a+a?? b+b?i i? ?? ?? a bA ?i = = ?[ ]1+Mi?
c?? d +d?i i? 0 d
then one can define the sum A as?follows:
a b
A ?? ?
?
?[ ]= 1+(M1 +M2)? .
0 d
This will be used in our calculation of tangent spaces.
Proposition 3.2.9. a) t ? 1Runiv = H (G, ad??), where ad ?? is the space of 2?2 matrices over??
F on which ?? acts by conjugation.
34
b) t univ,det =? H1(G, ad0??) where ad0?? is the space of 2?2 matrices with trace 0 over FR??
on which ?? acts by conjugation.
c) Runiv has a presentation Runiv??? ?? =O[[T1,T2, ..,Td]]/J where d is the dimension of H1(G, ad??)
as an F-vector space, and there exists a surjective homomorphism
H2(G, ad??)? J/mO[[T1,..T Jd ]]
where H2(G,ad??)? is the F-dual of the F-vector space H2(G,ad??).
d) Runiv,det has a presentation Runiv,det ??? ?? = O[[T1,T2, ..,Td1]]/J1 where d1 is the dimension
of H1(G, ad0??) as an F-vector space, and there exists a surjective homomorphism
H2(G, ad0??)? J1/mO[[T1,..T ]]Jd 1
Proof. See [18]
The previous discussions show
Runiv ?Runiv,det= ab,(p)?? ?? ??W (F)W (F)[[G ]]
So we don?t really lose any information or restrict the problem if we fix a determinant. On
the automorphic side, this corresponds to fixing the central character (in general) or the
weight of the modular form (in our particular case).
Instead of fixing the determinant, or rather allowing the determinant to arbitrarily vary, we
will impose the following condition on the determinant:
Definition 3.2.10. Define the deformation problem from CO to SETS
A 7? {? : G? GL2(A) : ? mod mA = ?,
det? = ?1?,where ? is trivial on Gal(Q(?p?)/Q) and
35
?1 = [det?|Gal(Q /Q(? ?))],where [ ] denotes the Teichmu?ller lift.}S p
We will refer to such a deformation problem as a deformation with a fixed Neben char-
acter, i.e. we have fixed the non-p and tame p-part of the determinant, or in short Neben.
Remark 3.2.11. About the notation regarding ?1, these are the characters that are referred
to as characters of the first kind by Iwasawa. Recall that a character is of first kind if p2
does not divide its conductor.
Lemma 3.2.12. The above is a deformation problem and is given by a quotient of Runiv.
We will denote it by Runiv,det=?1 . Furthermore, Runiv,det=?1 is a ? =W (F)[[Y ]] algebra via
the determinant map.
Proof. (Sketch) Since this condition looks a bit non-standard, we give a quick sketch to
show what it means for a collection D of liftings (R,?) of (F,?) to be a deformation
problem. Recall that one needs to check the following conditions:
1. (F,?) ?D
2. If f : R? S is a morphism in CO and (R,?) ?D , then (S, f ??) ?D . The converse
holds if f is injective.
3. Suppose R1,R2 ? CO , I1, I2 are ideals of R1,R2 respectively, such that there is an
isomorphism f : R1/I1 ?
?? R2/I2. Suppose (R1,?1) and (R2,?2) ? D and f ? (?1
mod I1) = ?2 mod I2.
Then ({(a,b) ? R1?R1/I1 R2;?1?R1/I1 ?2}) ?D .
4. If (R,?) is any lifting and I1 ? I2 ? ... is a sequence of ideals in R with ? jI j = 0 and
(R/I j,? mod I j) ?D for all j, then (R,?) ?D .
36
5. If (R,?) ?D and x ? ker(Gl (A)? GL (F)), then (R,x?x?12 2 ) ?D .
Checking these conditions is straightforward. One just needs that Teichm?uller lifts are
n
given by ?(x) = lim xp and f is a continuous homomorphism. Then [20] tells us that
there exist a closed ideal in Runiv?? such that the deformation problem is represented by
Runiv?? /I.
Remark 3.2.13. This condition on determinants is satisfied by Hida families. See the next
chapter. So it?s a fairly natural condition to impose on our deformations.
We will finish this section by constraining our deformations to have certain local con-
ditions. These local deformations are well understood. For the remainder of the section, let
l =6 p. The references are [60] and [12].
Choose an embedding GQl ?? GQS .
Minimal deformations : Let p - ??(Il). Take Cl to be the class of lifts of ??|GQ that factorl
through Gl/Il ? ker(??) with fixed determinant. Then Taylor shows that these deformations
correspond to the subspace L ? H1(G 0 1 0 Il Ql ,ad ??) given by H (Gl/Il,(ad ??) l). The cor-
responding deformation problem is smooth and the universal deformation ring is given by
W (F)[[Td]] where d = H0(GQl ,ad
0??).
?? ???? ?Steinberg deformations : Suppose l?6 1 (mod p) or p | ?? ?(Il). And let ?? =? ?.
??
Following Taylor in [60], we?define Cl to be?the class of deformations where ? (with respect
?cyc? ?
to some basis) is of the form?? ?? where ?cyc is the p-adic cyclotomic character.
0 ?
Sometimes, in literature, authors twist the representation by ? and only define Steinberg
deformations with cyclotomic determinant. Note that if l 6? 1 mod p, ? is non-trivial and in
this case, if ? (mod p) = 0, we have to use versal deformation rings or one has to choose
an inertia fixed line to rigidify the problem. The second method was used by Dickinson
37
and Calegari-Emerton [8]. In any case, we see that the deformation problem is smooth as
well and the (uni)versal ring is given by W (F)[[Td]] where d = H0(Gl,ad0??). Finally if
l ? 1 (mod p) and ? (mod p) = 0, one follows Kisin?s method. However we will not be
dealing with that case. Before we move to the next section, let us make a few remarks.
?? may be unramified but ? can be ramified and this can only happen if l ? 1 (mod p).
However Neben prevents such a situation to happen. Finally we will take ? (mod p) 6= 0,
then the same computations as before will show that the dimension of the tangent space is
h0(G ,ad0Ql ??) = 0.
Since we will be dealing with square free level N, we shall take ? = 1, otherwise f 2? |N.
The Steinberg condition will be particularly important for us later so we will study it in
some detail.
Now let us consider the case when l ? 1 (mod p) and ?|Il (mod p) 6= 0, thus ? (mod p)
correspon?ds to a ra?mified cocycle in H1(Qq,F), call it b?. Let a characteristic 0 lift ? be of
a b
the form?? ??. We show that if c(?) = 0 for all ? ? Il , then c = 0 on GQl .
c d
Let ? be the Frobenius at l, then we have the following relation:
????1 = ? l. (3.1)
? ? ? ?
a b
Let ? ? ?(?) =? ? and ?(?) =?? 1 x ?? and using the above equation, we immedi-
c d 0 1
ately get c = 0.
In fact we can say more. Note by [60], we know the universal deformation ring for Stein-
berg representations is W (F) := O . Let ? be an uniformizer of O . Then factoring out an
appropriate power of ? , we can consider that c takes values in O . Now
c (mod ? ) is a ramified 1-cocycle with values in F, i.e. c (mod ? ) ? H1(Qq,F). We
38
know by Tate duality (theorem 2.2.3) and the Local Euler-Poincare characteristic formula
(proposition 2.2.4) that H1(Qq,F) is a 2 dimensional F vector space and is spanned by
two cocycles b? and b??, where b? is the ramified cocycle and b?? is the unramified one. We
now show that c? := c (mod ? ) is an F multiple of b. If not, let c? = xb?+ yb?? Since by
definition of Steinberg deformations, any lift of ?? can be conjugated into an upper tri-
a?ngular matrix, we can?now construct a non-trivial deformation to F[?] by the following?? 1 b?+ ?(xb?+ yb??) ??. But that contradicts the fact the the dimension of the tangent
1
space is 0, i.e., there are no non-trivial upper triangular deformations. We record this dis-
cussion in the form of a proposition.
?? ?? 1 bProposition 3.2.14. Let ?? : G ?GL F be of the form ?Ql 2( ) ?, where b|Il=6 0. Con-
1
sider the problem of Steinberg deformations. Then c? constructed above is either split at l,
i.e. c?(GQl) = 0 or c? is totally and tamely ramified at l and c? = kb for some k ? F.
This idea will be used later on in our section on pseudo-deformations.
D?efinitio?n 3.2.15. We call a prime l difficult if l ? 1 (mod p) and ?? : GQl ? GL2(F) =
?? 1 ? ??, where ?|Il 6= 0.
1
In fact we can also pin down when c? is ramified and when it is split.
Let ? be any lift of ?? to A then ?(Il) ? SL2(A), since det?(?) is un?ramified at l. Since Il
is pro-cyclic, the image of Il will be a cyclic p group. Let
?? a b ?? ? SL2(A) be a non-
c d
identity matrix whose order is a power of p. Since the minimal polynomial of the matrix
divides the characteristic polynomial, we immediately get the following equations
? a+d = 2
39
? ad?bc = 1
Thus if a or d = 1, this forces bc = 0, and since b is an unit, c = 0. Thus we get an easy
criteria to check if c? is non-zero. We record th?is result i?n the form of an easy proposition.
a b
Proposition 3.2.16. c? 6= 0 iff there exist a lift?? ?? with a or d =6 1.
c d
For l = p, we choose the ordinary condition. This will be discussed in some detail in
the next section. Morally speaking, we would like our deformations to look similar to ??
when restricted to various decomposition subgroups. In this thesis, we will not consider
the case where ??|GQ is unramified but it?s lift is of Steinberg type.l
40
3.3 Ordinary deformations
Definition 3.3.1. We call ?? : GQp ? GL2(F) ordinary if
? ?
??? ?? ?? ?= ?
0 ??
where ?? is an unramified character.
Remark 3.3.2. The choice of the stable line will depend on the embedding GQp ?GQ,S, but
any two embeddings are conjugate by some g ? GQ,S and that g will transport one stable
line in one embedding to a stable line to the other embedding. If GQp stabilizes more than
one line, we will choose a line and we will call it a special line. Mazur calls this a choice
of p-stabilization.
Definition 3.3.3. We call an ordinary representation mod-p distinguished if ?? 6= ?? .
Remark 3.3.4. This definition implies that there are at most two special lines.
Assumption (Non?CM) We will assume the * in the above definition is non-zero.
Remark 3.3.5. The above assumption along with the previous definitions and comments
imply that there is an unique special line.
Remark 3.3.6. The non-vanishing of ? is strongly related to the fact that if this ? is attached
to a modular form f , then f is not a CM form. For more precise statement, we refer the
reader to the papers on local indecomposability of non-CM forms by E. Ghate.
Under the above assumption we have the following theorem:
Theorem 3.3.7. There exist an universal ordinary deformation?of ?? cal?l it ?ord and an
? ?
universal deformation ring which we call Rord such that ?ord ?? ?=? ? where ? is an
0 ?
unramified character and * 6= 0.
41
Proof. This is Theorem 3.30 in [24]
There is a related concept called nearly ordinary deformation rings.
Definition 3.3.8. A representation ? : GQp ? GL2(A) is called nearly ordinary if
? ?
??? ?? i1cyc? ?? ?= ?
0 ?? i2cyc?
and ? and nd ? are unramified characters and i1 > i2.
It is not hard to see that this a representable deformation problem and the universal de-
formation ring is denoted by Rn.ord and ordinary deformations form a subfunctor of nearly
ordinary deformations. The technical reason for introducing these objects are that they are
twist invariant. In fact, if V is an ordinary deformation then V ?(1) in general will not be
ordinary but will be nearly ordinary. Nonetheless V ?(k) will be ordinary for some k ? 1.
Remark 3.3.9. If we assumed ? mod p = 0, we will modify the deformation problem by
the following: We will consider pairs (?VA,LA), w?here A ? CO where ? : Gp ? GL(VA)
? ?
with fixed determinant and is given by ?? ?? and LA is a free unramified rank 1 A-
0 ?
submodule and LA reduces to ?? . One can show that this deformation problem is now
representable, call this Rord,det,L.
Theorem 3.3.10. Rord ?= O[[T ]].
Proof. This is well-known to due to works of Wiles and Mazur. For example one can
see section 2 of [12] or Proposition 3.6 in [30] or example 6.5 Case-2, subcase (ii) in [5].
Of all the references, the computations in [5] with upper triangular lifts are similar to our
computations on obstructions in the next section.
42
3.4 Computations of some tangent spaces and obstruction classes
For the rest of the section, fix a ?? : Gal(Q?/Q)? GL2(F) where F is a finite extension
of Fp and ? ?
??? ?1 ??? ?= ? with ?1 6= ?2
0 ?2
such that ?? is unramified outside of a finite set of primes, i.e. ?? factors through GQ,S where
GQ,S is the Galois group of the maximal extension of Q unramified outside S and S is a
finite set of primes including p and ?.
Under the above hypothesis, Theorem 3.2.2 guarantees the existence of the universal de-
formation ring, which we call Runiv?? .
Before stating the main theorem, let us make the following assumption.
Hypothesis 1 : dimF Ext1(?2,?1) = 1.
Let us now fix a non-zero element in Ext1(?2,?1) and let us call it b.
The main result of this section gives a complete description of the tangent space of this ring
in terms of the Jordan-Ho?lder factors of ?? and assume we are in any one of the conditions
satisfied by proposition 2.3.5.
Theorem 3.4.1. Assume we do not have any difficult primes in our deformations. Then
there exists an exact sequence
0?? Ext1 ? f ? ?( 1,? )?Ext11 (? 12,?2)?? tRuniv ?? Ext (?1,?2)?? 0??
? ?
a+a?? b+b??
Proof. Let ? be any lift of ?? to GL ? ?2(F[?]/?2) given by the matrix? ?.
c?? d +d??
43
For the above lift to be a group homomorphism, the coefficients satisfy the following rela-
tions.
1. a = ?1
2. a? ? ? ?gh = agah+ahag or in other words a /a is a continuous group homomorphism from
GQ,S? F. Lemma 2.3.1 counts the number of such homomorphisms.
3. d = ?2
4. d?gh = d d
?
g h +d d
?
h g
5. c?gh = dgc
?
h + c
?
gah or in other words c
? ? Ext1(?1,?2).
6. bgh = agbh +bgdh
7. b? = a? b +a b? ? ?gh g h g h +bgdh +?b?gdh ??
a+a?? b+b??
The map ? is defined by sending ???? ???? to c?.
c?? d +d??
First we show that the map is well-defin?ed. Suppose we are?given two deformations which
1+ x? y?
are equivalent, i.e. there exists a matrix?? ?? that conjugates a deformation
z? 1+w?
?1 to ?2. Let ci be the bottom left corner of ?i. Then
c2 = c1 +(a?d)z
To show that the map is well-defined, w?e want to?show ci?s give?rise to th?e sam?e extensio?ns
1 z d c1 d c2
up to isomorphism. Indeed, the matrix ?? ?? conjugates ?? ?? to ?? ??.
0 1 0 a 0 a
Thus the map is well defined.
44
To show that it is a group homomorphism: Note that by remark 3.2.8, the bottom left cor-
ner of the sum of two deformations is given by (c1 + c2)? and so the fact that g is a group
homomorphism is obvious.
Then we show that the map is surjective. Given an extension class c ? Ext1(?1,?2), the
cup products b c and c b are a priori in Z2(a,a) and Z2g h g h (d,d) respectively but by propo-
sition 2.3.5, we know that these cocycles are actually in B2(a,a) and B2(d,d) respectively.
Thus there exist functions ?,? : G? F, such that
? bgch = ?(gh)?ag?(h)?ah?(g)
? cgbh = ? (gh)?dg? (h)?dh? (g)
Consider the map r : G?G? F, given by r(g,h) = ?(g)bh + bg? (h). An easy but long
and tedious calculation shows that r ? Z2(?2,?1). Again, by the global Euler character-
istic formula 2.3.13, we see that Z2(?2,?1) = B2(?2,?1). Thus there exists a function
?? such that ?(gh) =?r(g,h)? ag?(h)? dh?(g). Finally, we can construct a lift of ?? as?? a+?? b??? ??. The kernel of the map ? is the group of all upper triangular lifts.
c? d +??
To defi?ne the map f , let us observe the following facts.
(a) If?
?
? a b+b?? ?? is a lift ?? , then an easy calculation shows that
0 d
b?gh = agb
?
h + b
?
gdh and by taking strict equiva?lence into acco?unt,?b? ? Ext?1(?2,?1) so by
? ?? a b+ kb? a bour assumption b = kb. And finally note that ????=? ?? as the matrix
? 0 d 0 d? ? ? ? ? ?? 1+ k? 0 ?? a b a b+ kb?conjugates?? ?? to?? ??.
0 1 0 d 0 d
(b) Finally given a,a?,b,d,d?, we can construct a unique upper triangular lift by hand. Note
that a?gbh and bgd
?
h lie in Z
2(?2,?1). By our assumption and by the Global Euler-Poincare
45
characteristic formula (cf. proposition 2.3.13), we get Z2(?2,? 21) is B (?2,?1). So there
exist functions ? and ? such that
1. b a?h g = ag?(h)??(gh)+dh?(g)
2. bgd?h = ag? (h)?? (gh)+dh??(g) ?
a+a?? b+(???? + kb)?
Our desired lift is then of the form?? ??
?? ? ?
0 d +d???
? a+a?? b+b?1? ?? ?? a+a?? b+b?Finally , if and 2? ?? are 2 lifts, then
0 d +d?? 0 d +d??
b? ? ? ? ? ? ? ?1,gh?b2,gh = ag(b1,h?b2,h)+dh(b1,g?b2,g), i.e. b1?b2 = kb. And the same calculation
as before shows that these matrices are conjugate. The same calculations also show that
changing our functions ? and ? do not change the strict equivalence class of our lift. Thus
in our lift, we can take k = 0.
We are now in a position to define the ma?p f . ?
a+a?? b???
Given a? in Ext1(? ? ?,? ), we map it to ? ? and d? ? Ext1(? ,? ), we
? 1 1? ?
2 2
0 d
? a b???map it to ??. By looking at the conjugacy classes to these matrices just as
0 d +d??
before we see that these maps are well defined, so that gives a map f from Ext1(?1,?1)?
Ext1(?2,?2) to tRuniv??
From the discussion above, we see that f sends (0,0) to the identity. To check that the map
is a?n injection, note tha?t if M ? ke?r(GL2(F?[?]? GL2(F)) and?? a+a?? b+b??M ??M?1 ? a b ?=? ?, then by a tedious matrix calculation one can
0 d +d?? 0 d
see that a? and d? are constant multiples of a and d as desired. But this is precisely the
1-coboundary relation. Thus the map is injective. The homomorphism is clear by matrix
46
multiplication and formulae satisfied by a? and d?. And this gives us the desired theorem.
An important consequence is that we can now readily compute the dimension of the
tangent space. Since we will be needing this later, we record it in the form of the following
corollary.
Corollary 3.4.2. dim tRuniv = 2#{q? 1 (mod p)+1} + dim Ext1(?1,?2).??
Proof. Follows easily from the previous theorem and Lemma 2.3.1.
Remark 3.4.3. If we have difficult primes in our deformation problem, then we take Runiv,st??
to be the universal deformation ring parametrizing Steinberg deformations. Then proposi-
tion 3.2.14 shows that the cocycles appearing the lower left corner must be either ramified
or completely split at l. Let Ext1 1ram(?1,?2) ? Ext (?1,?2) be the subspace of all cocycles
that are split or ramified at the difficult primes. Then if c ? Ext1ram(?1,?2), then b? c = 0
by proposition 2.3.5. Then we have an exact sequence just like in theorem 3.4.1
0?? fExt1(?1,?1)?Ext1 ? ? ?? t
? 1 ?( 2, 2) univ,st ?? Extram(?1,?2)?? 0R??
The proof of this statement is exactly the same as the proof of theorem 3.4.1 since all the
relevant cup products are now 0.
Remark 3.4.4. These above exact sequences also appear in the work of Chenevier-Bellaiche,
and our proof, even though similar in spirit, is slightly different.
Remark 3.4.5. Define a function f : G?G? F given by
f (g,h) = b?gh?a?gbh?agb?h?b d?g h?b?gdh (3.2)
47
where a,a?,d,d? are as before but b? is any set theoretic map. An extremely tedious but sim-
ple ca?lculation will?show f ? Z2(?2,?1). Or in other words if one twists ?? to the following
?? ? ??11 2 ?form ??. The above calculation and previous observations immediately show
0 1
us that the obstruction to lifting ? lies in H2(G ?1Q,S,?1?2 ) which is 0. So there is no ob-
struction to lifting the top right corner. This observation is actually used in computing the
tangent spaces of ordinary deformations and Steinberg deformations.
We define Ired to be the smallest ideal such that every lift ? mod Ired is reducible. We
call Ired the ideal of reducibility and Rred the quotient of Runiv?? by I
red . One can easily see
that Rred is the universal ring parameterizing upper triangular lifts. An easy upshot of the
above discussion is the following lemma.
Lemma 3.4.6. Ired 6= 0 iff Ext1(?1,?2) 6= 0.
Proof. Suppose Ired =6 0, then by our setup, we have an irreducible lift to F[?]. The pre-
vious calculations then imply the bottom left corner of the lift is a non-trivial element
in Ext1(?1,?2). Conversely, if Ext1(?1,?2) =6 0, then we can construct a non-trivial irre-
ducible lift to F[?], thus Ired 6= 0.
In fact one can do better.
Proposition 3.4.7. If n is the minimal number of generators of Ired , then
dim Ext1(?1,?2)? n.
Proof. The proof follows from noticing that the representation is reducible when bc = 0,
but one can identify b and c with the appropriate extension classes. Up to scaling by F, we
have an unique extension class. For more details see [2], Prop: 1.7.1.
The following proposition is well-known and gives a complete description of Ired .
48
Proposition 3.4.8. Ired is generated by any of the following sets:
{trace(?univ(Frobl)? [?1](Frobl)? [?2](Frobl)) : l ?/ S}
{a(?)? [?1](?)}
{d(?)? [?2](?)}
{b(?)c(?)} ? ?
a b
where ?univ ? ?=? ? and by abuse of notation, we denote [?i] as the Teichmu?ller lift
c d
of ?i to W (F) which is then mapped to Runiv?? via the structure map.
Proof. See [24] or [2].
Proposition 3.4.9. dimFtRred = 2#{q? 1 mod p}+2.
Proof. Rred parametrizes upper triangular lifts. Now from the exact sequence in Theo-
rem 3.4.1, the subspace of all lifts (up to equivalence) where c?= 0 is given by Ext1(?1,?1)?
Ext1(?2,?2) and we have already seen b? is uniquely determined once these above quanti-
ties are fixed. The result follows immediately.
Remark 3.4.10. It is really important for us that dimRuniv?? > dimR
red .
We will now try to find the obstruction to lifting to an upper triangular representation.
Given a surjective ring homomorphism A1  A0 with kernel I, generated by a single ele-
ment with I ?mA1 = 0, let ? be a representation to A0 lifting ?? and ?? be any lift of ? to A1.
To measure if this is a representation we calculate
?? ?a?gh? a?ga?h b?gh? a?gb?h? b?gd?hM ?gh = ??(gh)? ??(g)??(h) =? ? (3.3)
0 d?gh? d?gd?h
49
One can easily see that the above matrix is in Z2,up(??, ??)? I ? Z2(??, ??)? I?, where Z2,up?is
f11 f12
the set of matrices with the lower left entry is 0. Write an element in Z2,up as?? ??.
0 f22
One can see the following relations:
? f 211 ? Z (a,a).
? f22 ? Z2(d,d).
? a ?g f12(g ,g??)+b f ? ?? ? ?? ? ?? ? ?g 22(g ,g )? f12(gg ,g )+ f12(g,g g )? f11(g,g )bg??? f12(g,g )dg?? =
0.
?? ? ? ?? f11 f12Finally if an element ?? ? ?is in B2,up 11 12(??, ??), and we let ?? ?? be the
0 f22 0 ?22
matrix that realizes that coboundary, then we have the following relations:
? f11 ? B2(a,a).
? f 222 ? B (d,d).
? f12(g,g?) = ? ? ? ?12(gg )?ag?12(g )?bg?22(g )??11(g)bg???12(g)dg? .
Finally we note Mgh is independent (up to a coboundary, i.e. in B2(??, ??), of the choice of
a lift ??). Now let us start by assuming that there exists a lift of the diagonal characters to
A1, i.e. we assume that f11 and f22 are coboundaries. In fact we now choose a? and d? to be
those diagonal characters. This forces f11 = f22 = 0. And the upper right corner takes a
very simple form, i.e.
a f (g?,g??)? f (gg?,g??)+ f (g,g?g?? ?g 12 12 12 )? f12(g,g )dg?? = 0,
i.e. f 212 ? Z (d,a). Now let us change b? by any function, call it b?(1), keeping a? and d? fixed.
Then define f (1)12 via
f (1) : b?(1)? a? b?(1)12 = gh g h ? b?
(1)
g d?h.
50
Now b?(1)? b? ? I and
f (1) g h ? f g h b?(1)? b? ? a? b?(1)12 ( , ) 12( , ) = gh gh g( h ? b? ? d? b?
(1)
h) h( g ? b?g)
b?(1)= gh ? b?gh?ag(b?
(1)
h ? b?
(1)
h)?dh(b?g ? b?g)
changes f by an element in B212 (d,a). Let us now summarize the above discussion in the
form of two propositions.
Proposition 3.4.11. There exists an injection Ext2?(d,a) ?? E?xt2(??, ??).
0 f12
Proof. Let f ? Ext212 (d,a), then define a class ?? ?? in Ext2(??, ??). The above
0 0
discussion shows that this map is well-define?d. To show? ?
injectivity: Let f ,g ? Ext2(d,a)
0 f ?g
map to the same element in Ext2(??, ?? ?), i.e ? ? in B2(??, ??). By the previous
0 0
discussion f ?g ? B2(d,a). This proves the injectivity.
Proposition 3.4.12. Assuming the diagonal characters can be lifted, Rred ?= O[[X1, ...,Xt ]],
where t = 2#{q? 1 (mod p)}+2.
Proof. Under our assumption and by our previous discussion the obstruction to lifting to
an upper triangular representation is in Ext2(d,a). But our assumption (Hypothesis 1)
that Ext1(d,a) is 1-dimensional and the Euler-Poincare characteristic formula in 2.3.13
immediately tells us that dim Ext2(d,a) = 0. Thus the obstruction vanishes. And the
proposition follows immediately.
Corollary 3.4.13. Assume Hypothesis 1, Neben and let the deformations be ordinary at p,
then Rred ?= Rord .
Proof. By previous proposition, we know that the deformation problem is unobstructed
and we have also calculated the tangent space. Under Neben, the relative dimension of the
51
tangent space is 1, i.e. Rred ?=O[[X ]]. There is a canonical surjective map from Rord?Rred ,
since under Neben any reducible deformation is automatically ordinary. But a surjection
from O[[X ]]? O[[X ]] is an isomorphism.
Remark 3.4.14. The above corollary can be summarized by the following statement: There
are no non-trivial upper triangular deformations of ?? of fixed determinant to F[?].
To conclude our study of the universal deformation ring, we need to study the obstruc-
tions to lifting. Recall A1  A0 is a ma?p O-alge?bras with kernel I, generated by a single
a? b?
element with I ?mA1 = 0. Now let M :
?
=? ?? be a lift to A1 and we are measuring the
c? d?
failure for this lift to be a homomorphism. Define a function fgh := c?gh? c?ga?h? d?gc?h.
I?t?s easy to ?check f ? Ext2gh (a,d)? I. Just as before, write the matrix in Z2(??, ??) as?? f11 f12 ??.
f21 f22
Proposition 3.4.15. There exists a surjection Ext2(??, ??) Ext2(??1,?2). ?
f11 f12
Proof. First we note that given a matrix in Z2 ?? ?? written as ?? ?( , ) ?, then we
f21 f22
immediately see f21 satisfies the following relation:
d f (g?,g??)? f (gg?,g??)+ f (g,g?g?? ?g 21 21 21 )? f21(g,g )ag?? = 0, ? g,g?,g?? ? G.
? ?
f11 f12
Thus f21 is an element in Z2(?1,?2). Now, if
?? ?? ? B2(??, ??), then
f21 f22
f21(g,h) = ?21(gh)?dg?21(h)??21(g)ah,
?? ?f11 f12where ? ?21 is a function from G to F. Thus the map sending ? ?? f21 is a
f21 f22
52
well-defined map. It is clearly an F-linear homomorphism. Finally we have to show that
the map is surjective. Note that bg f ? ??21(g ,g ) ? Z3(a,a) and f21(g,g?)bg?? ? Z3(d,d) but
Z3(a,a) = B3(a,a) and Z3(d,d) = B3(d,d). So there exist functions ? and ? : G?G? F,
such that
bg f ?21(g ,g??) = ag?(g?,g??)??(gg?,g??)+?(g,g?g??)?ag???(g,g?)
f21(g,g?)b ?? = d ?(g?,g??g g )??(gg?,g??)+?(g?,g?g??)?d????(g,g?g ).
? 0
Thus one can construct an element in Z2(?,?) given by ?? ?? which maps to f21.
f21 ?
Corollary 3.4.16. Suppose there are no difficult primes then Krull dimension of Runiv?? ?
2#{q ? S : q ? 1 (mod p)}+3
Proof. We know that Krull dimension of Runiv?? ? d1?d2, where di := H i(GQ,S,ad ??) Writ-
ing down the formulae for d1 and d2, and noting that
dim H1(G ,? ??1Q,S 2 1 )?dim H
2(G ?1Q,S,?2?1 ) = 1
by the Euler-Poincare characteristic formula in 2.3.13. So we obtain the result.
In the subsequent discussion, we will assume that there is no obstruction to lifting the
lower left corner. Under that assumption, we will try to calculate other obstructions. In fact
just like before, we choose c? such that f21 = 0, i.e. c? ? Ext1(a,d)? I.
Lemma 3.4.17. If there is a c? which makes f21 = 0, then c? is independent of choices of a?
and d?.
53
Proof.
c? ?gh = (a?h +?h)c?g +(d?g +?g)c?h.
= a?hc?g + d?gc?h +(?hc?g +?gc?h).
where ?,? : G? I are arbitrary functions and since ?? mod mA1 is upper triangular, c? takes
values in mA1 . But I.mA1 = 0, so ?hc?g = ?gc?h = 0 and hence the lemma follows.
Proposition 3.4.18. Assume we have lifted the bottom left corner such that f21 = 0. Then
the diagonal entries satisfy the following properties:
i. f11 ? Z2(a,a)? I.
ii. f22 ? Z2(d,d)? I.
iii. Changing a? by any arbitrary function changes f11 by an element in B2(a,a)? I.
iv. Changing d? by any arbitrary function changes f22 by an element in B2(d,d)? I.
Proof. We will give two proofs of part i.
The first proof: i and ii are clear from the previous discussion and the proof is exactly the
same since f21 = 0.
The secon?d proof: T?his is a more direct formal computation. Let ?0 be any lift of ?? to A0
a0 b0
given by ?? ?? which is a homomorphism. Let ? be any lift of ? to A
? ? 0 1
given by
c0 d0
?? a? b? ??. The map A1? A0 is surjective with kernel I and ImA1 = 0. Recall
c? d?
f11(g,h) = a?gh? a?ga?h? b?gc?h
54
And we now want to compute
a f (g?,g??g 11 )? f ? ??11(gg ,g )+ f (g,g?g??11 )?a ?g?? f11(g,g )
=ag(a?g?g?? ? a?g? a?g?? ? b?g? c?g??)? (a?gg?g?? ? a?gg? a?g?? ? b?gg? c?g??) + (a?gg?g?? ? a?ga?g?g?? ? b?gc?g?g??)?
ag??(a?gg?? a?ga?g?? b?gc?g?)
Now, we can write b?c? = b0c?, a?c? = a0c?, d?c? = d0c?, since c? takes values in mA1 . And we
know that:
I. b0(gh) = a0(g)b0(h)+b0(g)d0(h).
II. c?gh = c?ga?h + c?hd?g.
Now in the above formula, we can replace a by a? and we get
a?g(a?g?g??? a?g? a?g??? b?g? c?g??)?(a?gg?g??? a?gg? a?g??? b?gg? c?g??)+(a?gg?g??? a?ga?g?g??? b?gc?g?g??)? a?g??(a?gg??
a?ga?g?? b?gc?g?)
After canceling out the terms and using the formula for c? in II, we get c?g??(b?gg? ? a?gb?g? ?
b?gd?g??). But by using the formula for b0, we get b?gg? ? a?gb?g? ? b?gd?g?? ? I and c?g??(b?gg? ?
a?gb?g?? b?gd?g??) ? mA1 ? I = 0 and hence the claim follows.
iii and iv are exactly similar so let us prove iii. Let ? : G? I be any function. Now
(a?gh +?gh)? (a?g +?g)(a?h +?h)? b?gc?h
= (a?gh? a?ga?h? b?gc?h)+(?gh? a?g?h? a?h?g)??g?h
Now since I2 = 0, ?g?h = 0 and ?gh? a?g?h? a?h?g = ?gh?ag?h?ah?g but this quantity
is a coboundary, i.e it belongs to B2(a,a)? I.
We are ready to state and prove one of the main theorems of this sub-section.
Let ? be an upper triangular lift of ?? to A0. We will find the obstruction to lift ? to
55
a not necessarily upper triangular representation to A1. We know that this obstruction
class is independent of the choice of lift and will only depend on ? . We call that class
O(?) ? Z2(??, ??)? I.
Theorem 3.4.19. Assume Neben, Hypothesis 1 and b? c = 0 where b ? Ext1(?2,?1) and
c ? Ext1(?1,?2). Then
O(?) = 0 iff f21 ? B2(a,d)? I
Proof. Recall: f21 = c?gh? c?ga?h? c?hd?g. We have already seen f21 is not a coboundary, and
that gives rise to an obstruction class.
Conversely, assume we have chosen a c? such that f21 = 0. So now we are reduced to show-
ing the existence of a lift of a to A1 which makes f11 = 0. We can lift a to a multiplicative
character to A1, call that character ? . Write a? = ? +? . We are looking for the existence of
? such that the following equations holds:
? a?gh = a?ga?h? b?gc?h
? ?gh +?gh = (?g +?g)(?h +?h)+ b?gc?h
So ? should satisfy :
?gh = ?g?h +?h?g + b?gc?h
And b?gc?h = bgc?h, since c? takes values in I.
Now bc?? Z1 d a ?Z1 a d ?I???( , ) ( , ) Z2(a,a)?I and by our assumption this is a coboundary.
So there exists a function that realizes this coboundary. In fact ? is a function that works.
The same proof works verbatim for f22 and the only difference is that the coboundary lies
in Z2(d,d). So, now we choose lifts such that f11 = f22 = 0. The matrix representing O(?)
56
? ?
?? 0 f12can be written as ??. Now f12 satisfies
0 0
a ?g f12(g ,g??)? f12(gg?,g??)+ f12(g,g?g??)?d ?? f (g,g?g 12 ) = 0,
i.e. f 212 ? Z (d,a). Changing the lift changes f12 by an element in B2(d,a). Now our
assumption on 1-dimensionality of Z1(d,a) (Hypothesis 1) and the Euler-Poincare char-
acteristic formula 2.3.13 shows Z2(d,a) = B2(d,a), or in other words we can find a b? such
that f12 = 0. Hence we have shown that we can construct a lift that forces the vanishing of
the relevant cohomology class.
We finish this section by talking about some special prime ideals in Runiv?? .
Lemma 3.4.20. Let p be any dimension 1 prime in Runiv?? , not containing the ideal of re-
ducibility, let A be the normalization of Runiv?? /p, let E be the fraction field of A and ? be
the induced representation ? : GQ,S? GL2(A). Then ??A E is absolutely irreducible.
Proof. This is lemma 3.33 in [62].
Definition 3.4.21. (Skinner-Wiles) The primes in the above lemma are called good primes,
if the residue characteristic is p.
Remark 3.4.22. These primes were used in [57] to do level raising. However their argu-
ment is unnecessarily complicated and the whole patching process can be simplified using
Kisin?s method. That was carried by Lue Pan [45] in their thesis. Our simplifying assump-
tion makes the patching arguments quite simple but since the arguments already exist in
literature in quite generality, we do not repeat it here.
Finally we give a sketch of the construction of such good primes under Hypothesis 1
and Neben. These are the main results in Chapter 4.3 of [57]. The point here is to impress
57
on the reader about how strong Hypothesis 1 is and how it can be used to significantly
simplify the proof of a key proposition, Proposition 4.3 in [57].
Lemma 3.4.23. Let ? : G redQS ? GL2(R /Q) be such that det ? is of finite order, where Q
is any prime ideal of Rred containing p. Then dim Rred/Q = 0.
Proof. Since Q contains (p) and we reduce to the case where Q has height 1, thus we get
Rred/Q = F[[X ]]. We assume that F is chosen large enough to contain the values of the
diagonal characters. Now since the det ? is of finite order, under Neben, we can make the
diagonal characters to be of finite order as well, wh?ich implies the cha?racters take values
? ?? a b+?b?X?in F . So now we can represent ? via the matrix ??. Since this is a
0 d
homomorphism and the target space is a domain, we get, b?(gh) = agb?(h)+ b?(g)dh.
This impl?ies b? = k?b, for som?e constant k? ? F by our Hypothesis 1. Thus the matrix
becomes ?? a b(1+?k X?? ) ??. But 1+?k ??X is an invertible power series so one
0 d ? ?
1+?k?X? 0
can conjugate the above matrix by?? ?? and one immediately gets ? ?= ?? .
0 1
Thus the result follows.
Proposition 3.4.24. There exist good primes in every component of Runiv?? .
Proof. This is Proposition 4.2 in [57]. In this case, one should take Q to be a minimal
prime in Runiv/m Runiv?? ? ?? . Then by our calculations on tangent spaces, dim Q? 1. Now if Q
contains Ired , then the induced Galois representation
?Q : GQ? GL2(Runiv?? /Q)
is reducible. By the choice of Q, det(?Q) has finite order. But the previous lemma con-
tradicts that dim Q ? 1. Thus Q does not contain Ired and hence ?Q is irreducible. For
58
complete details of the above steps and to complete the rest of the proof, one should follow
the rest of the arguments in [57].
59
3.5 Understanding extension classes and cup-products
In this section we will generalize Sharifi?s method in [54] t?o constru?ct big non-abelian
a b
extensions of Q. We set up a general framework. Let ?univ ?=? ?? be the universal
c d
deformation to Gl (Runiv2 ?? ). We are interested in understanding the kernel of the fixed field
of ?univ. To simplify notation we will use I to denote Ired . We can define two ideals of
Runiv??
B := ?b(?) : ? ? GQ? and C := ?c(?) : ? ? GQ?
Recall we have a product map
B/IB?C/IC ?? I/I2
which comes from the fact that BC = I. Since Rred is a domain we know that I is a prime
ideal and our method will give a complete description of the cotangent space at I in terms
of Iwasawa theory. Our arguments are general and do not need that I is a prime ideal.
However this description combined with the numerical criteria will be used to prove cases
of modularity lifting theorems and Wake?s conjectures in Chapter 6. However most of the
results in this chapter are quite general and are completely independent of the hypotheses
in the previous chapter.
Note that by proposition 3.4.8, I is generated by ?a(?)? ?1(?)?= ?d(?)? ?2(?)? where
?1 and ?2 are defined as before. Now one can check the following formulae:
b(?)c(?)= (a(??)??1(?)?1(?)??1(?)(a(?)??1(?))??1(?)(a(?)??1(?)) (mod I2)
60
Similar formula holds for c(?)b(?) where we use d? ?2, instead of a? ?1. Checking the
statement is routine and we leave it to the reader.
?? ?? ?1 b? a?? ??1 ??
The above formula can also be summarized via this matrix representation??? 0 ? ?2 c? ??
? ? 0 0 ?1
?? ?2 c? d?? ??2
(or equivalently by ??? 0 ? b? ?
??
1 ?? ), where b? ? B/IB, c? ?C/IC, a?? ??1, d?? ??2 ?
0 0 ?2
I/I2 or by the following diagram of fields.
61
M?
P
K?L?
K? L?
B/IB C/IC
F?
Q
Figure 3.1: A field diagram explaining the subfields cut out the above matrix
where F? is the fixed field of the kernel of the diagonal characters. In what follows,
we will work out the above picture explicitly in the case where ?2 is the trivial character,
?1 = ? , the cyclotomic character and generalize it later. Thu?s F? in this case?is just Q(?p?).?? 1 c d?1 ??
The matrix representation that we will be working with is: ??? 0 ? b ??? .
? ? 0 0 1
??? ? b a?? ??Remark 3.5.1. One can also work with?? 0 1 c ???, but one must work with homo-
0 0 ?
geneous cocycles or Ext-groups. All arguments otherwise remain unchanged.
In any case, K? is the maximal abelian pro-p extension of F?, unramified outside S with
a ?cyc action of ? and an action of ? by ? := Gal(Q(?p)/Q). L? is the maximal abelian
pro-p extension unramified outside S\{p} with a ??1cyc action of ? and an action of ??1 by
62
?, and M? is an abelian pro-p extension of K?L? with trivial action of ??? .
Before we start with the structure of the above Galois groups, let us recall some basic facts
about Iwasawa theory.
Definition 3.5.2. We say M and M? are pseudo-isomorphic if there exist a homomorphism
M?M? with finite kernel and cokernel.
The next theorem is a structure theorem for ?-modules.
Theorem 3.5.3. (Iw?asawa) Let M?be a finitely generated ? module. Then M is pseudo-
isomorphic to ?r? s ai t b ji=1 ?/p ? j=1 ?/Fj where Fj is an irreducible Weierstrass poly-
nomial, i.e., it is an irreducible polynomial of the form
F kj(T ) = T + c T k?11 + ...+ ck
where ci ? m?.
Following Iwasawa?s theorem one can define the following invariants attached to M.
? r(M) = r = ? rank of M
? ?(M) = ?si=1 ai (Iwasawa ?-invariant)
? ? (M) = ?tj=1 b jdeg(Fj) (Iwasawa ? -invariant)
? F tM,? = ? j=1(F )bj j (characteristic polynomial of M)
Note that r(M),?(M) and ? (M) are independent of the choice of a generator ? but not the
characteristic polynomial, where ? is a topological generator of ??= Zp (non-canonically).
Let F be a totally real field and let F? be the cyclotomic Zp-extension. Let H? be the max-
imal unramified abelian p-extension of F?. Let X? := Gal(H?/F?). Then X? is naturally a
?-module.
63
Conjecture 3.5.4. (Iwasawa): ?(X?) = 0
Remark 3.5.5. This is known to be true for abelian F by the work of Ferrero-Washington
and this result will be used throughout in our thesis.
Conjecture 3.5.6. (Leopoldt): Let K be any number field and S be the set containing all
infinite primes and all primes over p, then
H2(GK,S,Qp/Zp) = 0.
Remark 3.5.7. This conjecture is known for abelian number fields and this will also be used
throughout in this section.
Now let us summarize some basic Iwasawa theoretic results about the Galois groups at
the intermediate levels of this diagram. Let ? = Sp ? S?, i.e. the primes above p and ?.
Let X? and XS be the Galois groups of the maximal abelian pro-p extensions of Q(?p?)
unramified outside ? and S respectively. The following theorem gives a complete structure
of X? and XS.
p?1
Theorem 3.5.8. (Iwasawa) X? is pseudo-isomorphic to ? 2 ? (?-torsion)
Proof. This is theorem 13.31 in [67].
However we are only interested in the ?-component of X?. In that case, we have a
precise statement due to Theorem 11.3.18 in [50].
Proposition 3.5.9. X?? is isomorphic to ?.
Theorem 3.5.10. X? has no finite nontrivial ?-modules. Moreover we have an exact se-
quence of ?-modules ?
0? Zp(1)? XS? X?? 0
l?S??
l?1modp
64
In particular XS also has no finite ?-submodules.
Proof. See [50] page 750.
In fact one can say more about the ?-component of XS.
X?
?
S = ? Zp(1) (3.4)
l?1 mod(p)
And each Zp(1) extension is totally ramified at p and totally (tamely) ramified at l that
contributes to it. And the ? extension is only ramified at p. Finally we also see that our
Hypothesis 1 forces Gal(K?/F?) to be cyclic. In fact we say more. If ? in the upper right
corner of ?? is ramified only at p, then we get ? but if ? is ramified at p and another prime
q, then we get a Zp(1)-extension, just like in our case of the elliptic curve X (11A2 in
Cremona tables). See section 3.6.
Remark 3.5.11. Theorem 3.5.10 is true for any totally real base field F if one assumes
Leopoldt?s conjecture for the field F and p.
Proposition 3.5.12. Gal(L?/F?) does not have any finite ? submodules. Moreover Gal(L?/F?)
is a pseudo-cyclic ?-module, if one assumes Vandiver?s conjecture.
Proof. Let S be the set of primes dividing N. Consider the exact sequence coming from
class field theory
E? ?Fn ??q|N(O?Fn/q) ? Gal(MS(Fn)/H(Fn))? 0 (3.5)
where G? is the p-adic completion of G and EFn are the group of units of Fn = Q(?pn) and
the first map is the diagonal embedding, H(Fn) is the p-Hilbert class field and MS(Fn) is
the maximal p extension of Fn unramified outside of S. Now taking projective limits under
65
the norm maps, one obtains the following exact sequence:
?
E?? Rq? Gal(MS(F?)/H(F?))? 0 (3.6)
q|N
where R ? + ?q =?lim?(O?Fn/q) . Note that E? = E? ?E? , under the action of complex conjuga-
tion and (E?)+ ???1 = 0 and E? =Zp(1) as Galois modules so (E?)
?
??1 = 0 as well. Now one
can look at ??1- component of the exact seque?nce by taking ??1-coinvariants which is an
exact functor, so we now get an isomorphism q|N(Rq)??1 =? (Gal(MS(F?)/H(F?)))??1 .
We will write down the exact structure of (Rq)??1 but we note that this has no finite sub-
modules by lemma 3.5.13. So all we have to understand is the structure of Gal(H(F?)/F )??
as a ?-module. But it is a theorem of Iwasawa that the above module does not have any
finite submodules. This final statement is quite well known, see for example [67], Proposi-
tion 13.28.
Finally note that Vandiver implies the maximal abelian unramified p-extension of F? is
cyclic as a ?-module. Now lemma 3.5.22 shows that the characteristic ideal of Rq and that
of H(F?) are coprime. And this proves that Gal(L?/F?) is pseudo-cyclic.
Lemma 3.5.13. r r(Rq) ???1 = ?/ fq(T ) where fq(T ) = (1+ T )p ??(q)(1+ p)p , where r
is the number of primes over q in K?. In particular, Rq is non-trivial iff 1??(q) = 0
(mod p), i.e. q? 1 (mod p).
Proof. See [28] page 528.
We will use a general version of this lemma later.
The following corollary of Proposition 3.5.12 will be a key in our modularity lifting theo-
rem (theorem 6.3.6).
Corollary 3.5.14. Gal(L?/F?) is cyclic if one of the following two conditions are satisfied:
66
(a) H(F?)/F? is cyclic and fq are units or
(b) H(F?)/F? is trivial and there is only one non-unit fq.
Proof. The corollary follows from the fact if f ,g are two relatively prime distinguished
polynomials in ?, then the map ?/( f g)? ?/( f )??/(g) has finite kernel and co-kernel.
Theorem 3.5.15. M?/K?L? is unramified everywhere and under Hypothesis 1
Gal(M?/L?K?) ?= I/I2 ?= (IGGal(K?/K?)/I2GGal(K?/K?))
where K? is the maximal abelian p-extension of K?, unramified outside N with ??1-action.
Proof. First we show that it is unramified at p. But this is obvious since d = 1 on Ip by
ordinarity of ? . Now let l ? S. First let us assume that the inertia subgroup at l inside
Gal(K?/F?) or Gal(L?/F?) is non-trivial. Then by the previous results 3.5.10 and 3.5.12,
l must be infinitely ramified. Otherwise if the inertia subgroup is finite, it will generate a
finite ?-submodule. Without any loss of generality, assume that l is infinitely ramified in
Gal(K?/F?). Now, we localize our field diagram at l. We let Ml over Kl,?Ll,? be a totally
tamely ramified at l extension. This extension is given by a root ? of some irreducible
polynomial f with coefficients in Kl,?Ll,?. But since Ll,? is a compositum of Ll,n at finite
levels, we can assume that the coefficients lie in some Kl,?Ll,n. Thus we can think of Ml
as some degree pr totally tamely ramified extension of Kl,? with Galois group Hl . By the
same argument we can reduce this to some finite extension Cl given by the coefficients of
this polynomial. Now Hl injects into Aut(Cl(?)/Cl). Let Dl be the fixed field of Hl . Now
?n
Cl(?)/Dl is totally tamely ramified hence by Abhyankar?s lemma Cl(?) = Dl(? p ) for
n
some uniformizer ? of Dl . But ? = ? p u for some unit u and ? in Kl,? as l is infinitely
?n
ramified. Hence Ml = Kl,?(up ). Hence Ml is unramified at l. So this takes care of all
67
primes that are ramified in either K? or L?. To finish the proof: let l be a prime that is
unramified in K?L?/F? but ramifies in M?/K?L?. In that case, localizing the entire field
diagram at l, Kl,?Ll,? is the unique unramified Zp extension of Fl,?. Now, Zp acts on the
inertia subgroup at l inside M?/K?L? via lifting and conjugating and this action is given by
? ? ? l . But from our matrix calculations we know that this action must be trivial. Hence
M?/K?L? is unramified everywhere. This finishes the proof of the first part of the theorem.
Let us denote by P := Gal(M?/K?L?). Note that
?? ??? 1 b x? ?????
? ? ? ?? ?
?? 1 0 y ??? ??? 1 0 y ?? 1 b x? ??? ?? ?? ???????? ?
?
0 a c 0 1 0 = 0 1 0 0 a c ???
0 0 1 0 0 1 0 0 1 0 0 1
Thus P is central in Gal (M?/Q). A simple matrix multiplication shows that P is the com-
mutator subgroup of Gal(M?/F?). We will now show how to construct M? and in the
process will identify Gal(M?/K?L?). But first note that d? 1 mod I2 gives an isomor-
phism between P and I/I2. We will make the map explicit. Let ? ,? ? GF? be such that
b(?) = c(?) = 0 and let f be any function that satisfies d f = c? b. Then one computes
mod I2,
f ([? ,?]) = f (????1??1)
= f (??)+ f (??1??1)+ c(??)b(??1??1)
= f (?)+ f (?)+ c(?)b(?)+ f (??1)+ f (??1)+ c(??1)b(??1)+ c(?)b(??1)
Now note that f (x?1) = ? f (x), b(y?1) = ?b(y), c(z?1) = ?c(z), and for g,h ? GF? ,
b(gh) = b(g)+b(h) and c(gh) = c(g)+ c(h)
68
Using the above relations we get the formula:
f ([??]) = c(?)b(?).
Let K? be the maximal abelian extension of K? unramified outside S with ??1 action by
? and let K?? be the maximal subextension of K? on which G:= Gal(K?/F?) acts trivially.
Then K?? is abelian over F and Gal(K??? /K?) = Gal(K?/K?)/IG, where IG is the augmen-
tation ideal. Note that by Abhyankar?s lemma, K? is unramified everywhere. Similarly
Abhyankar?s lemma shows L?K?/K? is unramified at all l, and also at p, since L?/F? is.
Thus K? contains L?K?. Now there is a field K /F?, such that Gal(K /F?) = Gal(K??/K?).
Now ? acts on Gal(K /F?) by ??1. Furthermore K /F? is unramified outside N. By con-
struction, it is the maximal such extension. Thus K = L?. Since P is the center and the
commutator inside Gal(M?/F?), we seek to construct an extension of K??, call it K???, such
that Gal(K???/K??) is central and the commutator subgroup of Gal(K???/F?). We also de-
mand that ??? act trivially on Gal(K???/K??).
Let us draw a picture summarizing what we have so far.
69
K?
N
K??? M?
IGN/I2GN
P
K?? K?L?
N/IGN
Z K? L?
G
H
F?
???
Q
Let us focus on the K???/F?, and let Z be the Galois group. Note that Z is necessarily
non-abelian. Observe that K??/F? is abelian and Z sits inside the following exact sequence:
0? IGN/I2GN? Z? N/IGN?G? 0
and IGN/I2GN lies in the center of Z. We will make a quick sketch of that fact. Pick some g
and lift it inside Z and by abuse of notation continue to call it g. We would like to show the
conjugation by g on IGN/I2GN is trivial.
Note (g? 1)(h? 1)n = 0 for all g,h, so g(h? 1)n = (h? 1)n and similarly for any lift of
an element from N/IGN. Since G is cyclic as a ?-module (by Hypothesis 1), we get that
every element of IGN/I2GN is a commutator. By the general theory of central extensions,
70
one get a map from
?2(G?N/IGN)? [G,N/IGN] = I 2GN/IGN
given by x? y?7 [x,y], where all the commutators are taken in the group Z, where the
elements x and y are arbitrarily lifted and then taken commutators. It is trivial to check
that such an action is well-defined. Finally note that ?2(G?N/IGN) can be identified with
G?N/IGN and there is a canonical map :
G?N/IGN  IGN/I2GN
given by g?n 7? (g?1)n.
The upshot of the above discussion is that this map is given by taking commutators. Thus
we have a commutative diagram:
G?N/I N I N/I2G G GN
?=
B/IB?C/IC I/I2
And this gives us the desired isomorphism, i.e
Gal(M /K L )?= I/I2? ? ? ?= IGN/I2GN
Finally we note that the action of ??? on IGN/I2GN is necessarily trivial since g(a?b) =
?(g)?(g)a???1(g)??1(g)b = a?b
Question 3.5.16. (Comm) Are there any weaker conditions to ensure that every element of
IGN/I2GN is a commutator even if G or H is not cyclic?
Remark 3.5.17. Hypothesis 1 is only used to ensure that G is cyclic.
71
Remark 3.5.18. There is a similar picture on the other side of our diagram. Let us briefly
recall the construction. Call H:=Gal(L?/F?). let L? be the maximal unramified exten-
sion of L? outside N p with ? action. By Abhyankar?s lemma, L?/L? is unramified at
N, since H does not have any finite submodules so all primes that ramify are infinitely
ramified. Let L?? be the maximal subextension on which H acts trivially on Gal(L??/F?).
Thus L??/F? is abelian and Gal(L??/L?) = Gal(L?/L?)/IH . Thus there is a L /F? such
that Gal(L /F?) = Gal(L??/L?). Now note that L /F? is unramified outside N p and is the
maximal such extension and Gal(L /F?) has a ? action of ? . Thus L = K?.
Remark 3.5.19. Another alternate and equivalent construction will be to look at the maxi-
mal abelian p-extension of K?, unramified outside N, call it X , then look at (IGX/I2GX)???
Remark 3.5.20. Our proof also shows that M? is the maximal abelian p-extension of K?L?
with trivial ??? action.
Remark 3.5.21. Note that in the proof of unramifiedness, we do not require the Galois
modules to be cyclic. All we need is that there are no finite non-trivial ?-submodules. This
observation will be very useful to us later. See theorem 5.2.5.
Lemma 3.5.22. G does not contain any submodule isomorphic to Zp(?1) and H does not
contain any submodule isomorphic to a submodule of Zp(1).
Proof. First we show that H does not contain any submodule isomorphic to a submodule of
Zp(1). If this was the case, then the characteristic polynomial of H would have a common
k k
factor with (1+T )p ?(1+ p)p for some k. Now the characteristic polynomial for (Rq)??1
is coprime to the characteristic ideal of X??1 , where X is the Galois group of the maximal
abelian p-extension of F? unramified everywhere. Let f (T ) be the characteristic power
series of X??1 . Then
f (? s?(1+ p) ?1,?) = Lp(s,??)
72
where ? is a character of first kind and ? is a character of second kind.
If (1+T )?? (1+ p)| f (T ), where ? is some p-power root of unity, then
((??(1+ p))s?? (1+ p))|Lp(s,??)
This implies Lp(1,??) = 0 where ? is chosen so that ? = ?? . But this contradicts a result
of Brumer.
pk kAnd finally (1+T ) ? (1+ p)p is clearly coprime to fq(T ) as ?(q) 6= 1. Thus H does
not contain any submodule isomorphic to a submodule of Zp(1).
First we show that X? does not have any submodule isomorphic to Zp(?1). By a theorem
of Coates [9], there is an isomorphism
K (O )(p) =? (Cl(k(? ?)?Z (1))Gal(k(?p?)/k)2 k p p
We use the above non-trivial result of Coates by taking k = Q. The group on the left is
finite. If X? has a submodule isomorphic to Zp(?1), then (Cl(k(?p?)?Zp(1)) must be
infinite and is fixed by ???. This shows that (Cl(k(? ?)?Z (1))Gal(k(?p?)/k)p p contains a
subgroup of infinite order which is a contradiction.
Now let us handle the case where we consider the maximal abelian pro-p extension of
Q(?p?), unramified outside with ? action. Denote the Galois group of the maximal abelian
p-extension unramified outside p by Y . We recall a construction of Soule. Let En be
n
the group of p-units of Q(?pn), define E? := l p?im?E/E , under the norm map. For each
e = (en) ? E?, define
?(m) ? .??(?)
m?1?
n (e) := ? e nn , n? 1
??Gal(Q(?pn)/Q)
73
where ? is the p-adic cyclotomic character and for each a ? Zp, ?a?n is the unique integer
in the interval [0, pn) that is congruent to a modulo pn. The Kummer map associated with
the system of p-units {?(m)n (e)}n is the unique homomorphism ? : Y ? Zp determined by
the following formula:
n
?(?) := { ?(m)( 1/p ??1n (e)) } , ? ? Y,n? 1
Thus we get a pairing :
E?(m?1) ?Y ?? ? Zp
(e ???m?1n )n,? 7? ?(?)
We can see that the map factors through Y (m). We apply this pairing for m = 2. Then
Soule?s theorem says Y (m)? is finite. Applying this to our group G, we see that G(1)?
must be finite. But this contradicts that G does not have any finite submodules. Thus
G(1)? = 0. The case for the auxiliary primes dividing N are handled exactly in the same
manner as earlier.
74
A?generaliza?tion of the above picture : Let us consider the case where:
?? ? ?? ? ?=? ?, where ? is unramified at p and ?? is an even character and the con-
0 ?
ductor of ?? is squarefree. We still demand that ? 6= 0 on Ip. As before,?we consi?der the
a b
universal ordinary, minimally ramified deformation ring. Write ?univ ?= ? ??. I is
c d
the ideal of reducibility and we assume I 6= 0. Then ???1 cuts out a totally real abelian
number field F and our hypotheses on ? and ? force ?? to be of type S (in Greenberg?s
terminology), i.e. F ?Q(?p) = Q. Let B and respectively C be the ideals generated by
b(?) and c(?), then BC = I.?Just like before, we have a?n explicit function d?? that gives
?? [?] c? d?? [?] ?
rise to the following matrix: ??? 0 ?cyc[? ] b? ???? where [ ] denotes the Teichmu?ller
0 0 [?]
character, b?, c? and d? ? B/IB,C/IC,Runiv/I2?? respectively. However to be consistent with the
notations in [54] and notational convenienc?e of working with H i, rather?than homogeneous?? 1 c? d??1 ??
cocycles, we instead consider the matrix: ??? 0 ?cyc[? ][?]?1 b? ???. By abuse of no-
0 0 1
tation we still use the notation b?, c?, d?. We summarize the above matrix in the form of the
following diagram of fields:
75
M?
P
K?L?
K? L?
G H
F?
?
F(?p)
?
F
G1
Q
Figure 3.2: Description of the fields cut out by the above matrix
The following properties are straightforward and the proofs are exactly the same as
before. M?/K?L? is unramified at p. G1???? act trivially on P and P is the commutator
subgroup of Gal(M?/F?). Furthermore,
(d??)([? ,?]) = c(?)b(?) mod I2.
Note that Hypothesis 1 ensures that K?/F? is a cyclic ? module. Then Theorem 3.5.10
and the remark after the theorem shows that Gal(K?/F?)?= ? if K?/F? is only ramified at
76
p. Moreover Gal(K?/F?)?= Zp(1) as Galois modules if K?/L? is ramified at p and some
auxiliary prime q and q? 1 (mod p).
Before we state the next theorem, let us recall the p-adic L-function of Kubota-Leopoldt.
Let F be a totally real abelian field and ? be an even character of type S and let F? be the
extension attached to ? . Kubota and Leopoldt proved existence of a function Lp,S(s,? )
which satisfies the following interpolation property:
Lp,S(1?n,? ) = L(1?n,???n) ? (1????n(p)Npn?1) (3.7)
p?S?Sp
where Sp contains all primes above p. Let ? be a topological generator of Gal(F?/F) and
let u ? Z?p be such that ? ? = ? u for any ? ? ?p? . Deligne-Ribet and Wiles also proved that
there exist a unique G? ,S(T ) ? Zp[? ][[T ]]?Qp such that:
Lp,S(1? s,? ) = G s? ,S(u ?1) (3.8)
Moreover if ? is of type W then
G??(T ) = G(?(?)(1+T )?1) (3.9)
Theorem 3.5.23. char(Gal(L?/F?)) = G???1?2(u(1+T )?1?1) ? fq(T ) where
l| Ncond(???1)
r r
fq(T ) = (1+T )p ?????1(q)(1+ p)p . Moreover f (T ) is coprime to L (s,??1q p ??2).
Proof. Let S be the set of primes dividing N/conductor(???1). Recall the exact sequence
(3.5) in the previous section
E?Fn ??q?S(O?Fn/q)?? Gal(MS(Fn)/H(Fn))? 0 (3.10)
77
where EFn is the group of units of Fn and the first map is the diagonal embedding, H(Fn)
is the p-Hilbert class field and MS(Fn) is the maximal p extension of Fn unramified outside
of S. Note by our hypothesis ???1 6= ? i. Taking the ??1??1? components of the above
exact sequence we get
? ?q?S(O?Fn/q) ?? ?1??1? = Gal(MS(Fn)/H(Fn))??1??1? (3.11)
Taking the inverse limits, we get
?q?S(Rq)??1??1 ?? = Gal(MS(F?)/H?)??1??1?
where H? is the p-Hilbert class field of F?. The same proof as in 3.5.9 (this is also calcu-
k
lated in Itoh [28] section 6), shows (Rq) ?1 ?1 ?? ? ? = ?/ fq(T ), where fq(T ) = (1+T )p ?
????1 k(q)(1+ p)p . In particular Rq has no finite submodules. See the remark below to
see when fq is not an unit. Now let X? = Gal(H?/F?), then by solution of Iwasawa main
conjecture by Wiles we get
char?((X?) ?1??1??1? = G???1?2(u(1+T ) ?1) (3.12)
To finish the proof of the first part of the theorem, we note that by lemma 3.5.26, L?/F? is
unramified at primes dividing the conductor of ???1. For simplicity write ? = ????1(q).
So the only thing left for us to check is: G (u(1+T )?1???1?2 ?1) and fq(T ) are co-prime.
Plugging in ? (1+ p) for (1+ T ) in G ?1???1?2(u(1+ T ) ? 1), taking u = 1+ p we get
G ?1???1?2(? ?1). Plugging this back inside the p-adic L-function, we get Lp(1,???1?2)
which is non-zero.
Remark 3.5.24. Note that if ? = Frobq and ? ? Iq, then ????1 = ?q. Using this equation
78
we get ???1(q) = 1. Finally we see fq is not an unit iff 1??(q)???1(q)? 0 (mod p),
i.e., q? 1 (mod p).
We note that Gal(L?/K?) has no submodule isomorphic to a submodule of Zp(1).
Thus we get the following theorem which summarizes all the Iwasawa theoretic properties
of I/I2.
Theorem 3.5.25. (a) char?(I/I2) = sym(char(Gal(L?/F?)) ? char(Gal (K?/F?))), where
sym of 2 polynomials is the symmetric product of two polynomials.
(b) char?(I/I2) does not have multiple roots iff char((X?)???1?2 does not have multiple
roots.
(c) ? (I/I2) = ? (Gal(L?/F?) where ? (M) is the Iwasawa?s ? -invariant. In particular, I is
cyclic iff Gal(L?/F?) is cyclic.
Note that Iwasawa?s theorem 3.5.8 and theorem 3.5.10 also holds in this case. Thus we
get
Gal(K?/F k?)?= ?? (Zp(1)) as ? modules (3.13)
Thus under Hypothesis 1, we get the cyclicity of Gal(K?/F?).
If a prime l is infinitely ramified at either K?/F? or L?/F?, we repeat the same argument.
Now let l be a prime that divides the conductor of ? or ? , then by the squarefree assumption
l divides the conductor of both ???1 and ???1.Then,
Lemma 3.5.26. K?/F? and L?/F? are both unramified at l.
Proof. This argument appears in the work of Wiles, Sharifi, Ohta and Itoh. We merely
repeat the standard arguments in our case.
Let H? be the p-Hilbert class field of F?. and let N? be the the maximal p-extension of F?
unramified outside l. Define N? to be the maximal subextension of N? which is unramified
over F? at all primes dividing l. Then by class field theory, Gal(N ??/N ) is isomorphic
79
to a quotient of ?lim??? |l O
?
F . Since l - p and N? is a pro-p extension of F?, the aboven,?
Galois group is in fact a quotient of J := lim? O??? ? |l k , where kn,? is the residue field ofn,?
Fn,? . Now Il acts trivially on J. Now let ? = ???1 or ???1. Then ?(Il) =6 1. Thus,
(Gal(N?/N?))? = 0. This ends the proof of the lemma.
Note that theorems 11.3.5 and 11.3.18 ensure that Gal(K?/F?) does not have any finite
? modules. In this above situation, Kl,?Ll,? is the unique unramified Zp-extension of
Fl,?. Then we can repeat the above proof and the proof in the previous section to show
that Gal(M?/K?L?) is unramified everywhere. We summarize this in the form of the next
theorem. Also note that this is perhaps one of the few results that is true in this thesis even
without Hypothesis 1.
Theorem 3.5.27. M?/L?K? is unramified everywhere.
Proof. We have already proved most of this theorem. We just summarize the steps for the
convenience of the reader.
M?/L?K? is unramified outside N and unramified at p. If there is any prime l that is in-
finitely ramified at either K? or L?, then by applying Abhyankar?s lemma as in 3.5.15,
we can show M?/K?L? is unramified at l. Since any prime that ramifies in K?/F? or
L?/F? has to be infinitely ramified since Gal(K?/F?) and Gal(L?/F?) have no finite
? submodules, so this implies l is unramified in K?L?/F?. Then looking locally at a
prime l|l, K?,lL?,l is the unique unramified Zp extension of F?,l. Now if the inertia sub-
group J of M?,l/K?,lL?,l is non-trivial, then choosing a generator ? for the Galois group
Gal(K?,lL?,l/F?,l, we see that conjugation action of ? on J sends j? jl . Thus the action is
non-trivial but this contradicts that the fact that action on M?/K?L? is trivial.
To summarize the results of the last two sections, one can always construct big unram-
ified extensions over K?L? under the following condition:
80
? Gal(K?/F?) and Gal(L?/F?) do not have any finite ? submodules.
However to determine it explicitly in terms of some known Iwasawa modules, we need the
following condition, which seems extremely hard to check in general
? Every element of I/I2 is a commutator.
As remarked before, that is certainly true if Gal(K?/F?) or Gal(L?/F?) is cyclic. In that
case, the proof of theorem 3.5.15 works out verbatim and we get
Gal(M?/K?L?) =? I/I2 ?= Gal(K?/F?)?? Gal(L?/F?) (3.14)
Remark 3.5.28. One can follow the above procedure to construct big meta-abelian exten-
sions, starting with a totally real base field F , if one assumes Leopoldt?s conjecture and
Iwasawa?s ? = 0 conjecture.
Remark 3.5.29. The fq appearing in the theorem 3.5.23 are exactly the same as the factors
of B(T ;? ,?) appearing in [43], appendix A.
81
3.6 An explicit construction of a meta-abelian extension
We will give an explicit construction of the fields considered in a special case of elliptic
curves of conductor 11. These computations can also be found in [17]. We will be working
with
X 2 3 21(11) : y + y = x ? x 11A3
X : y2 + y = x3? x2?7820x?263580 11A2
and their mod 5 Galois repr?esentatio?ns (under suitable choice of basis) can respectively
1 ?
be given by the following ?? ?? and ?
?
? ? ? ?
?
?, where * in both cases is non-
0 ? 0 1
zero. The ideal of reducibility I (proposition 3.4.8) in both the cases is (25) as we have
a degree 25 isogeny between the curves but to make our calculations easier we will be
working with the maximal ideal m = (5). Call ?1 and ?2 the associated 5-adic Galois
representations attached to each curve. Then the ideal generated by the b(?) in each
case is Z5 and that of c(?) is (25). In fact, we can say more: By Kummer theory, we
can identify B/mB as the field generated by the 5-division polynomial over Q(?5). Then
B1/mB1 can be identified by Gal(K1/Q(?5)) where K1 is the splitting field of the polyno-
mial: x5+2x4+6x3?2x2+4x?1. One can easily check that 11 ramifies in this extension
and the primes above 5 split and Gal(Q(?5)/Q) acts via ??1. And B2/mB2 can be identi-
fied with Gal(K2/Q(?5)) where K2 is the splitting field of the polynomial x5?11 and in this
case both 5 and 11 ramify and Gal(Q(?5)/Q) acts via ? . One can now identify C1/mC1
with Gal(K2/Q(?5)) and C2/mC2 with Gal(K1/Q(?5)). In this case, one can come to this
conclusion by looking at the 125 division points. But we take a slight digression to explain
this phenomenon in a more theoretical context as this is not an accident but one of the main
driving forces behind the theory of modularity and p-adic L-functions.
82
Let f be a modular form such that its associated mod p Galois representation is reducible.
Then by Ribet?s lemma (proposition 3.1.12), there exists at least one lattice?such that th?e
?? ?
representation is non-semisimple. Up to a twist, the representation is given by?? ??.
0 1
Then following Ribet?s proof in 3.1.13, one can show that th?ere exists a?nother lattice in Q2p
1 ?
where the mod p Galois representation can be given by ?? ?? where * in both
0 ??
cases is non-zero. Call these lattices ?1 and ?2. In fact these 2 lattices sit at two opposite
extremes of the chain of lattices which realizes the Galois representation attached to f . One
of them has the highest ?-invariant and the other the lowest, which is called the ?-deprived
quotient.
Conjecture 3.6.1. (Greenberg): The ?-deprived quotient lattice has ?-invariant 0.
In our case, X1(11) indeed has ?-invariant 0 and X has ?-invariant 2 and they are on
two extreme edges of the isogeny graph with X0(11) sitting between them. Now let us
look at the first lattice. B/mB can be identified with H1(QS,Fp(??)) and C/mC can be
identified with H1(QS,F ?1 ?1p(? ? )). Now let us assume the Hecke algebra associated to
?1 (or ?2) is Gorenstein. Then the above cohomology groups are 1-dimensional. In fact it?s
an if and only if statement by the work of Mazur-Tilouine. Now it?s easy to see that B/mB
in the 2nd lattice is the C/mC in the first lattice and vice versa.
?? ?? 1 c ? ??
Now we turn our attention to the following matrix over F5 given by ??? 0 ? b ??? where
0 0 1
we regard c ? H1(G ,??1Q5 11 ) and b ? H1(G, Q5 11,?). Then the above computations show,
that the fixed field of the kernel of c is K1 and the fixed field of the kernel of b is K2. Since
the cup product c?b = 0, we have the above 3-dimensional representation by lemma 2.3.4.
83
? ?
a b
In fact, write the 5-adic representation of X as the matrix?? ??. Then ?= d?1 mod
c d
I/mI. And this gives rise to the following diagram of fields given by the fixed field of the
kernel of the representation. So the only unknown object in the following diagram is the
degree 5-extension M of K1K2.
M
K1K2
K1 K2
Q(?5)
Q
But before we write down what M is, we want to make some preliminary observations.
Gal(M/K1K2) has trivial action of Gal(Q(?5)/Q) and it is central and the commutator
subgroup of Gal(M/Q(?5)) and these can be easily seen by the matrix representation of
the field diagram.
Now, conversely given?any field d?iagram as above with the above properties, we claim?? 1 ? ? ??
that Gal(M/Q(?5)) =??? 0 1 ? ???. Let G := Gal(M/Q(?5)). Note that G is necessarily
0 0 1
84
non-abelian. We have an exact sequence
0? Z/5? G? Z/5?Z/5? 0.
So we have a non-abelian group of order 53. In fact we know that there are only 2 non-
abelian groups of order p3; one of them is the group of unipotent matrices with exponent
p and the other is a group of exponent p2. We now show that the latter situation can not
happen. If there is an element of order 25, consider the fixed field of the subgroup generated
by that element, call it L. Now since any subgroup of index 5 is necessarily normal, we
see that L is a degree 5 Galois extension over Q(?5). Now we need to show that L is
Galois over Q. To see this, note that the number of elements of order 25 in G is 125?25.
These give rise to 125?25?(25) = 5 distinct subgroups, where ? is the Euler phi function. Now
Gal(Q(?5)/Q) permutes these subgroups and already fixes K1 and K2. Thus Gal(Q(?5)/Q)
permutes the other 3 subgroups so there is a fixed point for this action. Call it L. Thus we
can take L to be Galois over Q and L is necessarily disjoint from both K1 and K2. In fact
we have the following diagram that explains our situation:
85
M
K1K2
K1 L K2
Q(?5)
Q
Now Gal(Q(?5)/Q) acts semi-simply on Gal(K1K2/Q(?5)), since their orders are co-prime
and there are two eigenvectors for this action, which contradicts the above picture as K1,
K2 and L correspond to 3 different eigenvectors. Thus we have shown any diagram of the
above form is in one-one correspondence with a matrix as described above. To get hold
of the group Gal(M/Q), note that order of Gal(Q(?5)/Q) is coprime to Gal(M/Q(?5), so
H2(G,(Z/p)?) = 0. Thus Gal(M/Q)?= GoGal(?Q(?5)/Q) and now it is straightforward?? 1 ? ? ??
to see that Gal(M/Q) is of the form??? 0 ? ? ??? .
0 0 1
Now let M? be another field and let f ? be the function in the top right corner of the matrix.
Let f be a function such that d f = d f ? = c?b. Thus
( f ? f ?)(??) = ( f ? f ?)(?)+( f ? f ?)(?) for all ? ,? ? GQ
Thus f ? f ? is a homomorphism, which corresponds to a degree 5 Galois extension of Q
86
with possible ramifications only at 5 and 11. Thus once we find one M, we can get hold
of all such M? by just composing with these fields. So that gives a complete description of
all such M?. So now all we have to find is one such field. Now the 5-class group of K1 is
Z/5 and is generated by any prime over 5 or 11 and the action of Gal(Q(?5)/Q) is trivial
on the 5-class group of K1. For proof see [17] Prop 6.2. Call H1 the 5- Hilbert class field.
Since [H1 : Q(?5)] = 25, H1 is abelian over Q(?5). Thus we can take M to be H1K1K2. An
alternative construction would be to see that the 5-class group for K2 is Z/5?Z/5 and is
generated by the primes above (5) and (11). One can see that the class generated by the
prime over (5) is acted on trivially by Gal(Q(?5)/Q). Call that unramified 5-extension H2.
Then H2K1K2 is our desired extension.
We think of the above picture as the base of the cyclotomic tower and thus this allows us to
find non-trivial examples.
Finally one can put X1(11) in the Hida family. For a precise definition, see chapter 4.
The Hida family in this case is just ? and one can ask the same question for this Hida
family. Since 5 is a regular prime, and the Hecke algebra is particularly nice, it is fairly
straightforward to figure out the ?-adic picture. Note the field K? is a Kummer extension
?
and is explicitly given by Q(? 1/55?,(11) ). The ?-module Gal(K?/Q(?5?)) is cyclic and
isomorphic to ?/(T ? 5). L? is given by adjoining a compatible sequence of 5?-roots of
?/?? . We also see that Gal(L?/Q(?5?)) is also a cyclic module. A theorem of Iwasawa
(Theorem 2 in [29]) and the vanishing of ? by Ferrero-Washington shows that maximal
extension of Q(?5?) unramified outside 5 and 11 is a free pro-5 group on 2 generators. Thus
the compositum of the two independent Z5 extensions that we constructed is the maximal
abelian 5-extension and the field M? is the field fixed by the commutator subgroup. Thus,
by Nakayama?s lemma, I is cyclic. For more precise results relating Gorensteinness of an
universal deformation ring and the cyclicity of the ideal of reducibility, we refer the reader
to [2].
87
Remark 3.6.2. The reader can clearly see an easy generalization of this above picture. Take
a prime q? 1 mod p. Then one can work out the entire picture, basically replacing 11 with
q, for regular primes p. However for irregular primes, one needs Vandiver?s conjecture
to ensure the cyclicity of the appropriate Galois group as a ? module. Thus Vandiver?s
conjecture implies the cyclicity of I. This should be reminiscent of the results in [36].
Remark 3.6.3. An example of the above type is worked out in [54] for the prime p = 37,
N = 1 and for ??ss = 1??32. And it seems to the author that Greenberg and Monsky in
proposition 3.1.14 used very similar ideas in their unpublished note on the Ramanujan ?
function.
Remark 3.6.4. The Galois group of the maximal abelian extension of Q(?5?) unramified
outside 5 and 11 with ? action is ??Z5. ? extension is only ramified at 5 and so we can
not recover that extension in our construction.
88
3.7 Pseudo-deformations
In this section, we make a slight digression into pseudo-deformations. All the results
in this section are fairly standard and readily available in the literature. We follow the
notations of Taylor [59].
Let G be a group and R is a commutative ring with 1. We will assume d! is invertible in R.
Definition 3.7.1. A R-valued (continuous) pseudo-character of dimension d is a R-linear
(continuous) function T : G? R such that
? T (e) = d
? T (g1g2) = T (g2g1) ?g1,g2 ? G.
? ???S ?(?)T ? (g1, ...gd+1 d+1) = 0
where T ? : Gd+1? R is given by the following.
Let x = (x1, ...x d+1d+1) ? G . Let ? be the cycle ( j1, ... jm), then T ? := T (x j1....x jm). Now
for any general ? , define T ? := ?T ?i , where ? = ??i be it?s cycle decomposition.
Of course trace of a representation satisfies the above 3 conditions. All our pseudo-
characters will be continuous so we will drop the word continuous in the sequel. Taking R
to be F, one can consider deformations of T to C . Let DpsO T (A) be the set of deformations
of T to A.
Lemma 3.7.2. The functor DpsT is (pro)represented by a complete local Noetherian O-
algebra RpsT .
Now Carayol-Serre lemma 3.1.20 immediately shows that if ?? is absolutely irreducible
and T is it?s trace, then there is an isomorphism: Runiv ? ps?? = RT . In the sequel, let d = 2. How-
ever if ?? is reducible, such an isomorphism does not hold in general as the following exam-
ple in [31] shows. Suppose that ? , ? : G? F?1 2 are characters and c 11, c2 ? Ext (?2,?1).
89
??
?
?1 c1 + c2T
Then ?
?
? is a representation of G ? GL2(F[T ]). More naturally, one
0 ?2
obtains a family of representations of G over P(Ext1(?2,?1)), the projectivization of
Ext1(?2,?1), and all have same the pseudo-character ?1 +?2. However our Hypothesis 1
will enable us to compare the two rings.
Lemma 3.7.3. Assume Hypothesis 1. Let VF[?]/?2 be a deformation of ?? . If VF[?]/?2 induces
the trivial deformation on pseudo-characters, then VF[?]/?2 is the trivial deformation.
Proof. This is lemma 1.4.3 of [32]. In fact we gave a different proof of this result in our
proof of Theorem 3.4.1, even though we did not state this result explicitly. In fact, there we
showed that if a? = d? = 0, then b? = kb and the above lemma follows from that.
Corollary 3.7.4. (a) Under Hypothesis 1, there exists a surjection ? : Rps  RunivT ?? induced
by sending a representation to it?s trace.
(b) tRps ?= tRuniv??
Proof. This is corollary 1.4.4 of [32].
(b) These modules are finite and have the same cardinality due to a result of Chenevier-
Bellaiche [2] and corollary 3.4.2. And the map induced by the trace map is surjective so
it?s an isomorphism.
Proposition 3.7.5. Assume Hypothesis 1 and Neben and fix the determinant (for simplic-
ity). There are no non-trivial upper triangular deformations to F[?] and so Runiv,det?? is
generated by the traces of the Frobenii.
Proof. We first show that there are no non-trivial upper triangular deformations. Under
Neben, we see that we have unique lifts of our diagonal characters to F?[?] and in fact w?e
?? b+b?1 ?
will take these lifts to be trivial. Thus our deformation has the shape of?? ??.
0 ??2
90
A simple matrix calculation shows that b? ? Ext1(?? ?2, ??1), thus b = kb. Calculations in 3.3
immediately show that this is a trivial deformation.
The next statement follows from the surjection ? in the previous lemma. We give a sketch
of another proof following the lines of Carayol-Serre. This is a fairly standard and straight-
forward argument. It suffices to show that any non-trivial deformation of ?? to GL2(F[?]) is
generated by traces. The proof is similar to Carayol?s proof. Let ? be a deformation of ??
to GL2(F[?]). Write the matrix entries of ? as functions a+a??,b+b??,?c and d +d?? of
Gal(Q?/Q). Let K be the fixed field of the kernels of ??1 and ??2, by Neben, det(?) factors
through K. Thus if ? ? Gal(Q?/K), then Det(?(?)) = 1 = 1+ a?? + d?? ? bc? . Since c
is non-trivial (by assumption), the Cebotarev density theorem implies there exists a ? such
that b(?)c(?) = 0. But trace(?(?)) = 1+a?? +d?? = 1, it follows that the traces of ? gen-
erate F[?]. Since Runiv,det?? is generated a ?-algebra by the generators of m 2Runiv,det/mRuniv,det?? ??
and the result follows via Nakayama?s lemma.
91
Chapter 4: Hida theory of ordinary modular forms and Hecke algebras
We will assume the reader is familiar with the definition of modular forms and Hecke
operators. In this section, we will give a summary of facts about Galois representations
attached to modular forms and Hida theory. We will end this chapter by giving an explicit
structure of the Hida Hecke algebra. We follow the exposition in Hida, Emerton-Pollack-
Weston, Ohta and Fukaya-Kato-Sharifi. Just to set up our notation, for any congruence
subgroup ? ? SL2(Z), we denote by Mk(?) (resp. by Sk(?)) the space of all modular
forms?(resp. c?usp forms) of weight k and level ?. For any Dirichlet character ? and
a b
? : ?? ?= ? ? SL2(Z), we define
c d
?(?) := ?(d)
Definition 4.0.1. We say a modular form f on ?1(N) has Nebentypus ? if ?0(M) acts on
f via the character ? .
Note that by considering q-expansions one can define
Mk(?)Z := Mk(?)?Z[[q]]
92
and for any ring R,
Mk(?)R := Mk(?)Z?Z R ?? R[[q]]
and similarly for Sk(?)Z and Sk(?)R. Now let ? = ?1(N). Then one has the double coset
operators given by the following:
For all primes l, we define Tl to be the double coset operator
T ??
? ?
? 1 0 ?l = ??
0 l
and for d coprime to N, we define ?d? to be the operator, ?d? = ??d?, where ?d ? ?0(N)
satisfies ? ?
? 0
? ??? ?d ? mod (N)
0 d
The Hecke algebra T is the commutative algebra defined over Z by the operators Tl and
?d?. The action of the Hecke operators on modular forms is well known and Mk(?) and
Sk(?) are stable under the Hecke operators. With this in mind, we define the Hecke al-
gebra (resp. cuspidal Hecke algebra) Hk(?) (resp. hk(?)) to be the image of T inside
EndZ(Mk(?)Z (resp. EndZ(Sk(?)Z). Note that these are the Z subalgebra of EndZ(Mk(?)Z
and EndZ(Sk(?)Z generated by Tl and ?d?. For any ring R, we define
H(?)R := H(?)Z
h(?)R is defined analogously.
93
4.1 Galois representations attached to classical modular forms
Let f be a normalized i.e. a1 = 1, cuspidal newform for ?1(N pr), of weight k ? 2 and
nebentypus ? which is a Hecke eigenform for all Hecke operators. Let K f denote the field
generated by the coefficients of the q-expansion of f . It is well-known that K f is a number
field.
Definition 4.1.1. We call a modular (cusp) form f ordinary if ap is a p-adic unit.
The next theorem is a landmark result in the theory of modular forms. It is the culmi-
nation of the work of Shimura, Carayol, Deligne, Serre, Mazur and Wiles.
Theorem 4.1.2. Let f be as above. Choose a prime p above p in K f and let K f ,p denote the
p-adic completion of K f . Then there exists a compatible system of absolutely irreducible
p-adic representations ? f ,p of Gal(Q?/Q) in GL2(K f ,p) satisfying
1. ? f ,p is unramified outside N p.
2. For any prime q - N p,
det(1?? f ,p(Frobq)T ) = 1?aqT +qk?1?(q)T 2
3. det(? f ,p(c)) =?1, where c is a complex conjugation.
4. Let ?p be the p-adic cyclotomic character, then det? f ,p = ??k?1p , where we view ?
as a Galois character by defining ?(Frobq) := ?(q) for all q - N p.
5. (ordinary) (Deligne, Mazur-Wiles) The restriction of ? f ,p to the decomposition sub-
94
group at p is isomorphic to an upper triangular representation of the form
?? ?? ?(g) ?g 7? ??
0 ? (g)
where ? is unramified and ? (Frobp) is the unique p-adic root of
x2?apx+?(p)pk?1 = 0
Here we take ?(p) = 0 if p|N, so then ? (Frobp) = ap.
6. (Langlands, Carayol) Let q 6= p and q|N, let C be the conductor of ? . Write N = qeN?
?
(resp. C = qe C?) so that q - N? (resp. q -C?).
(a) If e = e? >?0, then ??f ,p restricted to inertia subgroup at q is equivalent to the fol-
? 0
lowing form: ?? ??. Moreover ? f ,p restricted to the Decomposition subgroup
0 1
at q is still diagonal. Let ?q be the unique unramified character appearing in ? f ,p|Dq ,
we have ?q(Frobq) = aq.
(b) (Steinberg case) If e = 1 and e? = 0, ? f ,p restricted to th?e Decomposi?tion sub-
?(1) ?
group at q is ramified and is equivalent to the following form: ?? ??. where
0 ?
? is an unramified character that takes Frobq to aq and ?(1) is the twist of ? by the
p-adic cyclotomic character.
(Note: In this case the image of inertia is infinite)
Remark 4.1.3. The representation depends on the choice of p above p and we have abused
notation to write the representation as ? f ,p.
95
4.2 Hida theory of ?-adic modular forms
Given integers k ? 2 a prime p ? 5, and N, such that p - N and let O be the ring of
integers in some complete subring of Cp, we note the inclusion
? : Mk(?1(N p?))?Mk(?1(N p? )) for any ? ? ?
commutes with the action of Hecke operators. Thus the restriction of a Hecke opera-
tor from level N p? is an Hecke operator for level N p? . This map induces a projection
morphism Hk(?1(N p? ))  Hk(? (N p?1 )) taking Tl to Tl . Now, we define Sk(N p?,O)
and M (N p?k ,O) to be the space of weight k cusp forms and modular forms respectively
that are in ? r1(N p ) for some r ? 0 and whose q-expansion (at the cusp at ?) lie in O .
We have an action on these spaces by the groups (Z/N)? via the nebentypus character
and Z?p via the product of the nebentypus character along with the map ? 7? ?k (weight
map/action). This action makes the above spaces an O[[Z?p]]-module, which we will call
(by abuse of notation) ?. For all l - N p we have the action of the Hecke operator Tl , and
Up and Uq for primes q|N on these spaces. Thus Sk(N p?,Zp) and Mk(N p?,Zp) are natu-
rally hk(N) :=?lim?hk(?1(N p
r)) and Hk(N)-modules. One can also define the ring of p-adic
modular forms (cusp forms) via p-adic completion of the divided congruences of the ring
?kMk(N p?,O) (respectively Sk(N p?,O)). Hida has defined an ordinary projector on these
spaces
e := lim Un!p .n??
If f is an eigenform for the operator Up with eigenvalue ap, one can easily check e f = f
iff ap is a p-adic unit, otherwise e f = 0. By using the projector e, we denote the space of
ordinary modular forms (resp. cusp forms) by Mordk (N) (resp. S
ord
k (N)). We also denote
the ordinary Hecke algebra (resp. cuspidal Hecke algebra) by Hordk (N) (resp. by h
ord
k (N)).
96
Theorem 4.2.1. (Hida) The spaces Sordk (N) and M
ord
k (N) are finite free ? modules and
moreover these spaces are independent of k as long as k ? 2.
Proof. See [22] Theorem 1.
Before we state the next definition, let us introduce some special prime ideals in ?. An
arithmetic prime of ? is a prime ideal of the form
Pk,? := (1+T ? ?(1+ p)(1+ p)k) (4.1)
for k ? 2 and character ? : 1+ pZp ? O? of p power order, say pr(?). If I is a finite
extension of ?, we call a prime of I arithmetic if it lies over some Pk,? .
Definition 4.2.2. (Hida-Wiles) Fix a finite flat integral domain I over ?. A I-adic form F
of level N and character ?: (Z/NZ)?? C?p is a formal q-expansion
?
F = ? an(F)qn ? I[[q]]
n=0
such that for almost all arithmetic primes P , F mod P ? Mord(N pr(?) ?kk,? k,? k ,??? ). One
can similarly define a I-adic cusp form.
To ease notation, we will drop N when the tame level is understood. The following
theorem gives a brief summary of Hida theory.
Theorem 4.2.3. (Hida) (a) Hord(?) and hord(?) are free of finite rank over ?.
(b) We have the following specialization property: For every arithmetic prime of the form
Pk,? , k ? 2, there are isomorphisms
Hord(?)/P Hordk,? (?)?= Hordk (N p
r(?)+1,????k)
97
hord(?)/P hord(?) =? hordk,? k (N p
r(?)+1,????k)
sending Tl to Tl .
(c) Both Hord(?) and hord(?) are etale over all arithmetic points of ?.
Proof. See [24], Corollary 3.20 for (a), Corollary 3.19 for (b), Corollary 1.4 in [22] for (c)
and Theorem 1.2 in [22] for (a).
We have a duality between modular forms and Hecke algebras.
Theorem 4.2.4. (Hida-Ohta) There is a non-degenerate pairing
Sord(?)?hord(?)? ?
given by ( f ,T )?7 a1( f |T ). The pairing induces isomorphisms Hom?(hord(?),?)?= Sord(?)
and Hom ord ord?(S (?),?)?= h (?)
Define Mord(?) := { f ? Mord(?)Q(?) : an( f ) ? ?, ? n ? 1}, then Mord(?) and Hord(?)
are duals of each other via the above map.
Proof. This is theorem 3.17 in [24].
Remark 4.2.5. There seems to be no consensus in literature about how these maps should
be normalized. We have followed Hida?s notations in this chapter. The reader should make
sure that the normalizations are consistent when they are checking other references.
98
4.3 Structure of Hida Hecke algebras and big modular Galois representa-
tions
For the rest of the section, let us assume that N is squarefree.
Lemma 4.3.1. Hord and hord are reduced.
Proof. This is corollary 1.3 in [26].
Lemma 4.3.2. Hord is a complete semi-local ring, with maximal ideals m1,..,ms and the
maximal ideals correspond to a mod p system of eigenvalues of modular forms of tame
level N.
Proof. It?s a fairly standard argument in Hida theory to reduce to the weight 2 case and
then there are only finitely many mod p modular forms of weight 2 and level N. For more
details see [24] Chapters 3.1 and 3.2.
Piecing together the above arguments, we get the following well-known proposition.
This can also be found in Chapter 3 in [24].
Proposition 4.3.3. Hordm is equidimensional, reduced of dimension 2 and any minimal prime
ideal has characteristic 0. Mordm is a Cohen-Macaulay H
ord
m -module.
Proof. (Sketch) We know that Hordm is a finite free ? algebra, thus dim Hordm = dim ? = 2.
Equidimensionality follows from the fact that any finite torsion-free algebra M over ? is
necessarily equidimensional. Theorem 17.3 from [35] shows for any minimal prime p of
X , dim X/p = dim X . Since p ? Hordm can be extended into a regular sequence, p must be
characteristic 0. Finally,
depthHordM
ord
m ? depth?Mordm = dim ? = 2m
99
Thus Mordm is a Cohen Macaulay H
ord
m module.
Remark 4.3.4. This proposition is crucial for imposing various hypotheses on class groups
on the deformation theory side. We saw that the tangent space on the deformation rings can
be arbitrarily large depending on the size of the class groups or ray class groups, whereas
the Hecke algebra is quite small.
Now we come to the one of main theorems of this section.
Theorem 4.3.5. (Hida, Wiles) There exist a continuous 2-dimensional pseudo-character
T : GQ?Hordm , such that T (Frobl) = Tl , for l - N p. Or in other words one has a continuous
?2-dimensional r?epresentation ? : GQ?GL (Hord2 m ?Q(?)), such that tr(?)= T and ?|Dp=
?? ?(g) ? ??, where ? is unramified and ? (Frobp) =Up.
0 ? (g)
Remark 4.3.6. Note that there is no reason why we can choose our representation to take
values in GL2(Hm), such that ? mod m = ??. So to remedy the situation, from now on we
will be assuming Hypothesis 1 and Neben.
Let us summarize our discussion in the form of the following commutative diagram,
which we view as a special case of local-global compatibility. The complete details of the
proof of the next proposition can be found in [25] page 230 and 232.
Proposition 4.3.7. (Local-Global compatibility) We have a commutative diagram as fol-
lows:
Rord Runiv? hm
?
Proof. Before giving the proof, we explain why we call it a Local-Global compatibility, we
think of the top map coming from the Galois side and the map from ?? hm is coming from
100
the automorphic side. The top arrow is constructed via the following: Note Runiv?? is gener-
ated by the traces by Proposition 3.7.5 and by the modularity of ?? (cf. Proposition 6.2.35)
the map ? : Runiv univ??  hm is given by Tr(? (Frobl)) 7? Tl and ? (Frobp) 7?Up. The map
from Rord ? Runiv?? is given by taking a representation and restricting it to Dp. The map
?? hm is given by l 7? ?l? for l - N p. Now the map from ?? Rord is essentially given by
the ?weight? character. We refer the reader to page 232 in [25] for a detailed description of
this map and the commutativity of the two ? actions.
Corollary 4.3.8. The map Rord ?? hm is finite.
Proof. This is straightforward since ? ?= Rord by theorem 3.3.10 and ?? hm is finite by
theorem 4.2.3 and the diagram commutes.
Remark 4.3.9. The corollary basically says that there are only finitely many ordinary modu-
lar forms of a fixed tame level N with a given mod p representation at Dp. Such a statement
is extremely hard to prove in the non-ordinary case and is a result of Emerton-Breuil-
Paskunas, using their solution of the p-adic local Langlands correspondence. In fact even
in our special case, we needed an automorphic input and to get that automorphic input in
the non-ordinary case, one needs the full power of Colmez?s Montreal functor.
101
Chapter 5: Images of Galois representations
The goal of this section is to prove a big image result and deduce some consequences.
We will start the section with summarizing the motivation and known results. Then we will
then use modularity lifting results to prove the Galois representation is big in GL2(F[[T ]]).
5.1 History of related results
Understanding the images of Galois representations was first initiated by Serre who
showed non CM elliptic curves have big images. The work was then extended by Momose
and Hida-Lang in the ?-adic setting. The big image question appears naturally in control-
ling the sizes of Selmer groups which are used in the calculations of Wiles [69] and others
in the context of modularity lifting. The big image also appears in the work of Kato in
constructing an Euler system in his proof of the main conjecture for elliptic curves.
5.2 A new big image theorem
Our goal in this section is to give a very simple proof of the big image properties of the
representations studied by Skinner and Wiles. This question was inspired by a comment of
Hida in [27] and subsequent discussions with Chris Skinner at the Arizona Winter School
in 2017.
We start off with a couple of group theoretic lemmas.
102
Lemma 5.2.1. Let ? be an irreducible 2-dimensional representation of a group G. Let G?
be an finite index normal subgroup of G such that ?|G? is a sum of 2-distinct characters,
then there exist a subgroup H of G of index 2 and a character ? of H such that ? = IndGH? .
Proof. Call the characters appearing in the lemma ?i. Since G? is normal in G and ?i are
distinct characters, we claim G acts transitively on the set {?1,?2}.
Proof of claim: Suppose on the contrary, G fixes ?1. Let L1 and L2 be the two G?-lines on
which G? acts via the characters ?i. Let li be the G?-bases of Li. Then there exist a g ? G,
such that
gl1 = al1 +bl2
with b 6= 0, otherwise L1 will be G-stable contradicting the irreducibility of V . Then, for
h ? G?,
ghl1 = g?1(h)l1 = ?1(h)(al1 +bl2)
Now gh = h?g for some h? ? G?. This implies
h?gl1 = h?(al1 +bl2) = a? (h?1 )l ?1 +b?2(h )l2 = a?1(h)l1 +b?1(h)l2
Thus ? (h) = ? (h?), since ghg?11 2 = h? and ? ? ?11(h ) = ?1(ghg ) = ?1(h), since G fixes ?1.
Thus we get ?1 = ?2, which is a contradiction.
Thus G can not fix ?1. Let H be the stabilizer of ?i under this action. Then clearly H is of
index 2. Now we show that we can extend ?i to characters on H.
Let h? ? H and let h?l1 = al1 +bl . Let g ? G? and h? = g?12 hg, where h ? H. Then
gh?l1 = a?1(g)l1 +b?2(g)l2
103
But
hgl1 = h?1(g)l1.
Therefore hl1 = al1. So L1 is a H-stable 1-dimensional vector space so ?1 extends to H
i.e. HomH(?,?i) 6= 0. Call the extended character ? . By Frobenius reciprocity, we get
HomG(?, IndGH?) 6= 0. But since ? is irreducible, this proves ? = IndGH? .
Definition 5.2.2. We call a representation dihedral if it is induced from a character from a
quadratic extension. Note that the projective image of the reduction of such a representation
is a dihedral group.
We now need to show that the restriction of any irreducible representation to any finite
index normal subgroup is semi-simple. But this is Clifford?s theorem. We give a quick
sketch of the proof.
Lemma 5.2.3. (Clifford) Let W be a simple G-module. Let H be a finite index normal
subgroup of G. Then W is a direct sum of simple H-modules.
Proof. Let U be any simple H-submodule of W . Then the conjugates of U are H-submodules
since H is normal. The intersection of one with the sum of others is another H-submodule
but this intersection must be 0 as U is simple. Since the span of all G-conjugates is a
G-submodule of W , it must be all of W .
We give a short and quick exposition to Pink?s theory of Lie algebras in [46].
Let A be any complete semi-local p-profinite ring, where p > 2. In our examples, A will be
F[[T ]] where F is a finite extension of Fp which contains all the relevant eigenvalues (see
below). Pink defines a map
? : SL(2)? sl(2)
M 7?M? (tr(M)/2)Id.
104
Let G be a p-profinite subgroup of SL(2). Then define L1(G) to be the closed subgroup of
sl(2) that is topologically generated by ?(G). Let L1 ?L1 be the closed additive subgroup of
M2(A) that is topologically generated by {?(x)?(y) : x,y ?G}. Let C = Tr(L1 ?L1) which
we view as scalar matrices. In fact we can define subgroups inductively by
L2 = [L1,L1], Ln+1 = [L1,Ln] for all n? 1
Hn := {x ? SL2(A) : ?(x) ? Ln and tr(x)?2 ?C} for all n? 1.
We summarize the main results of Pink?s theory in the following theorem.
Theorem 5.2.4. (Pink) The map ? : Hn? Ln is a homeomorphism. Hn is a pro-p subgroup
of SL2(A) and Hn is normalized by H1. Conversely if G is any pro-p subgroup and L is the
closed additive group generated by ?(G), then G ? H1 and the commutator subgroup of
G is H2.
The main facts that we will use from Pink?s theory are as follows:
? C ?L? L.
? If g ? GL2(A) normalizes G, then g normalizes L(G).
With the above results in hand, we show that the images of the representations are big, i.e.
they contain an open subgroup of SL(2).
Theorem 5.2.5. Let ? : GQ? GL2(F[[T ]]) be such that
i) ? is irreducible, mod p distinguished and ordinary. (cf. Definition 3.3.1 and Defini-
tion 3.3.3)
ii) Determinant is of infinite order. ? ?
1 ?
iii) There exists ? ? Ip such that ? ?
? ?
( ) =? ? where ? 6= 0
0 1
Then Im(?) contains an open subgroup of SL2(F[[T ]]).
105
? ?
a? b?
Proof. Let ? ? Ip and let ? ?
? ?
( ) = ? ?. Since det(??) |Ip= ??cyc and p ? 5, we can
0 1
always choose a? to hav?e order g?reater than 2. Now ?(?) is upper triangular by ordinar-
a b
ity of ? . Call the lift ?? ??. Since a? =6 d?, a? d is a unit power series in F[[T ]].
0 d ? ?
1 b/(a?d)
Conjugating the image of ? by the matrix M : ?? ?= ?, we can assume that
? 0 1?? a 0 ?
?
? ? Im(?). Note that by raising ? to pn powers and taking limits, we can assume
0 d
that a,d ? F n, since lim pn?? T = 0. The action of Ad(?(?)) on L has 3 distinct eigenval-
ues, namely 1,ad?1 and a?1d. Call ? = ad?1. Thus the Lie-algebra L?decompos?es into the
a 0
corresponding eigenspaces, L ? ?= L[1]?L[? ]?L[??1], where L[1] =? ? , L[? ] =
? 0 ?a?? 0 b ?
? ? ?
? ? ?? 0 0L ? 1 ?, [ ] = ?, where L[i] is the eigenspace corresponding to eigenvalue
0 0 c 0
i. Let ? be a topological gene?rator of Gal(Q?p(?p?)/Qp(?p)), such that ?cyc(?) = 1+T ?
1+T u
F[[T ]]. Hence we get ? ?(?) =? ??. Since we are looking at the image of ? con-
0 1 ? ?
?? 1+T u? T bjugated by M, we need to conjugate ?(?) by M and thus we get a?d ?? is in
0 1
MIm(?)M?1. Finally note that M commutes with unipotent matrices. Thus conjugating the
image of ? by M does not change our underlying assumptions ab?out Im(?). Now,? ?
by Pink?s
0 b(1+T )s
result, ?(?) normalizes L. Conjugating L[? ] with ?(?)s, we get? ?? ? L[? ]
0 0
for all s ? Zp. And similar computation with L[??1] shows that both L[? ],L[??1] are
F[[T ]]-modules. Since we know that there exist a b whose constant term is non-zero, we
106
have just shown that L[? ] is isomorphic to F[[T ]]. Note that [L[? ],L[??1]]? L[1]. Thus if
L[??1] = a, where a is a non zero F[[T ]] ideal, then a? L[1]. Now we are left to show that
L[??1] is non-trivial. Let K be the fixed field of the kernel of det(?) and H= Gal(Q/K).
Thus H is a finite index normal subgroup of GQ. Now we claim that ?|H is irreducible. If
not, then it is a sum of characters by lemma 5.2.3, which implies that ? is induced from
a character by lemma 5.2.1. In that case the image of ? is contained in diagonal and anti-
diagonal matrices. By our hypothesis, we have a non-trivial unipotent element in Im(?).
Thus raising ? to an appropriate power, we can assume ? ? H and ?(?) is a non-trivial
unipotent matrix, since the index of H is prime to p. So ? is not?induced?and hence ?|H is
a b
irreducible. Thus there exist an element in Im(?| of the form??H) ?? with c 6= 0.
c d
Now det(?|H) is of the form 1+T f (T ) and so admits an unique square root. Twisting ?|H
by det(?| )?1/2H , we see that the image now lies in SL(2). But this is a p-p?ower cha?racter
a b
and so does not change the image, so by abuse of notation we assume that ?? ?? lies
c d
in the image of the twisted representation. Now applying Pink?s ? map and projecting onto
the L[??1] subspace we get a non-zero element in L[??1]. Now since Fp[[T ]] is a PID, we
can identify L[??1] with a non-zero element, call it a. Now it?s an easy exercise to see the
image contains the principal congruence subgroup defined by a. Now if the image contains
a principal congruence subgroup, then the image is open in SL(2) is lemma 2.4 in [27].
Remark 5.2.6. i) Skinner-Wiles instead of using condition (iii) impose the condition that ?
is not induced.
ii) They assume that the the image contains a diagonal matrix of infinite order with deter-
minant 1, and this is now known from the works of Hida and others.
iii) If p? Runiv?? is a good prime, then the Galois representation
? : G ? GL (RunivQ 2 ?? /p) satisfies conditions (i) and (ii).
107
iv) Hida assumes that there exists a ? such that ??(?) = diag(??, ?? ), with ?? 6= ?? . Then he
claims that one can assume that ?(?) is diagonal as well. We give a?proof of that claim.
?? x
Proof. Let A ? and let ? be a lift of ?? to A. Write ? ? ?? ?
?
CO ( ) = ?. Now the
y ??
characteristic polynomial of ?(?) has distinct roots in A, as (??? ?? )2 +4xy has a non-zero
square root in A? since xy ? mA and ?? ? ?? ? A? and the roots reduce to ?? and ?? . Call
these roots r1 and r2. Let VA be the A module on which ? acts. Thus we can use the roots of
the characteristic polynomial to decompose VA into 1-dimensional eigenspaces. Let e1,e2
be an A-basis of M lifting the basis e?1, e?2. Thus one writes the eigenvectors v = e1 +m2e2
and v2 =?e2 +m1e1, f?or?unique ele?m?ents m1,m2?? mA. Thus? 1 m1? ?? ??
?1 ? ?
r1 1 m1
(?) =? ?? ?? ?? 1 m1. Conjugating by ?? ?? does
m2 1 r2 m2 1 m2 1
not change the strict equivalence class of the lift. Thus we can assume that the lift ?
contains a diagonal element.
Remark 5.2.7. A direct application of lemma 1.5 in [27] shows that the image of ?univ is
big in Runiv?? .
Remark 5.2.8. Condition (iii) is by far the most difficult condition. It relates to non-CM
ness of the Hida family. See the papers of Bin Zhao, Hida, Ghate on local indecomposabil-
ity and non-CM forms.
108
Chapter 6: Modularity lifting and Wake?s conjectures
6.1 Introduction
In this chapter, we will reprove a weaker but a more explicit version of Skinner-Wiles
[57] and we can also treat the case where ? 6= 1. The proof uses our calculation of I/I2 and
various isomorphism criteria. To apply the criteria, we relate the congruence module of
Wiles to the congruence module of Hida. As a byproduct, under Vandiver?s conjecture, we
prove that the Hecke algebras considered in Ohta [43] are Gorenstein. Finally we apply our
ideas to prove Wake?s conjecture for the Hecke algebras considered in the Chapter 4. Since
we are only dealing with ordinary Hecke algebras, we will drop ord from the superscript.
6.2 Congruence Modules
In this section we briefly recall Hida?s formalism of congruence modules and apply the
setup to understand the congruences between cusp forms and Eisenstein series. In fact we
will show the two concepts of congruence modules, one by Hida and one by Wiles, are
the same and we will freely use the results of Hida-Ohta to get results towards modularity
lifting and Wake?s conjecture.
Setup: We follow the notations in [23]. Let A be an integral noetherian domain of charac-
teristic 0, and let R,S be A-algebras. Let F be the quotient field of A and let ? : R  S and
? : S  A be A-algebra homomorphisms. Define ? := ? ?? .
109
We assume the following:
? R,S are reduced and finite flat over A.
? ? and ? induce unique F -algebra decompositions as follows:
R?A F = F?X , S?A F = F?Y, R?A F = (S?A F)?Z (??)
Let RX (respectively SY ,RZ) be the images of R (respectively S,R) in X (respectively
Y,Z)
Definition 6.2.1. Define modules of congruence by
C0(?;A) = (A?SY )/S, C0(? ;S) = (S?RZ)/R, C0(? ;A) = (A?RX)/R
By chasing through some diagrams, one can easily prove this lemma.
Lemma 6.2.2. C0(?;A)?= S/(a)?= A/(?(a)) =? SY/b?= S/(a?b), where a= ker(S? Y )
and b=ker(?).
Proof: See lemma 5.2 in [23].
Lemma 6.2.3. C0(?;A) = SY ?S A, C0(? ;S) = S?R RZ and C0(? ;A) = RX ?R A, where A
and S are R-modules via ? and ? .
Proof. This is lemma 6.3 in [23].
Even though C0(? ;S) is defined as a module, it is actually a ring. Observe that if A1 and
A2 are A-algebras, then Spec(A1?A A2) = Spec(A1)?Spec(A)Spec(A2)., i.e. tensor products
correspond to fiber products. Then using previous lemma, we make the following remark:
Remark 6.2.4. Spec(C0(? ;S)) is the scheme theoretic intersection of Spec(S) and Spec(RZ),
inside Spec(R).
110
Consider the setup where R = Runiv?? , A = ? = R
red , ? is the canonical projection of
Runiv??  R
red . Note that Runiv?? is reduced. If a prime p is in the support of C0(? ,A), then
by the discussions above it is clear p ? I, where I is the ideal of reducibility which is the
kernel of ? . The upshot is if a prime is in the support of the congruence module, then the
Galois representation is reducible, i.e. it measures the congruences between irreducible and
reducible Galois representations. Thus if q? Runiv?? is a good prime, we get that q is not in
the support of the congruence ideal.
In fact one can define higher congruence modules Ci for i > 0.
Definition 6.2.5. Ci(?;A)= TorS(A,A), C (? ;A) = TorRi i i (A,A),
Ci(? ;S) = TorRi (S,S), where we view these modules as R-modules via the maps ? and ?
as previous lemma.
Then another diagram chasing gives the following easy lemma:
Lemma 6.2.6. 1) C1(?;A) = ?S/A?S,? A. If S is an universal deformation ring, then this
module is nothing but the dual of an adjoint Selmer group as discussed and calculated in
Chapter 3.
2) C1(? ;S) = I/I2 where I =ker(R  S). If R,S are as in the previous discussion, I is the
ideal of reducibility.
Proof. Since we will not be needing them, we refer the reader to see page 276 in [24] and
various references in the book. But nonetheless, this description gives another way to think
about the invariants that we already defined and calculated.
Since throughout this section, we will be using Ohta?s results, we would like to state
the setup in Ohta. To begin the comparison between various congruence modules, let us
recall the following definition/setup in Ohta.
111
Let
0? A??i B??? C? 0 (6.1)
be an exact sequence of finite flat reduced R-modules. We consider C as a B-module via
the map ? . Suppose we are given a B-module section over Q =Frac(R), i.e.
0? A?Q??t B?Q??? C?Q? 0 (6.2)
such that
t ? (i?1Q) = 1A?Q and (??1Q)? s = 1C?L (6.3)
Then one defines the congruence module as
COhta =C/?(B??(C))?= t(B)/A
For more details about congruence modules and Ohta?s setup, we refer the reader to section
1 in [43]. Note that the two definitions are the same. Applying lemma 6.2.2, to C0(?,C) in
Ohta?s setup, we see that
B?Q = (C?Q)? (A?Q)
and C0(?,C) = (Im(B? A?Q))/ker(?), which is exactly t(B)/A as desired.
In the introduction of [43], Ohta introduces this module as the module that measures the
failure of this section to be defined over R.
Let the setup be as before. Wiles in [69] defines a congruence module via:
CWiles =C/?(AnnB((ker(?))).
It is very easy to see that this module measures the failure of the splitting of the exact
112
sequence 6.1. The next lemma shows that these modules are in fact the same.
Lemma 6.2.7. B??(C) = AnnB(A).
Proof. Let x ? AnnB(A), then x ? i(a) = 0. Applying t to it, we get, t(x) ?a = 0. Since A is
flat over R, by clearing denominators, we can assume t(x) ? A. So i(t(x)) ? AnnB(A). But
i(t(x)) ? ker(?). So i(t(x)) ? ker(?)?AnnB(A).
Claim 1: ker(?) ? AnnB(A) = 0
Let b ? ker(?) ? AnnB(A) = 0. Since b ? ker(?), then b ? i(A). Now b ? b = 0, since
b ? AnnB(A), but B is reduced so b = 0.
This claim implies that t(x) = 0, therefore x ? ?(C), so AnnB(A)? B??(C).
Conversely, let x ? B??(C), now ?(x ? i(a)) = 0 and t(x ? i(a)) = t(x) ?a = 0, as x is in the
image of C. So x ? i(a) ? ker(?) ? ker(t).
Claim 2: ker(?)? ker(t) = 0.
Proof: Let ? ? ker(?) ? ker(t), then ? = i(a). Since t(?) = t(i(a)) = a = 0, this shows
? = 0 and thus proves the lemma.
It is useful to know all the above viewpoints when dealing with congruence modules.
Remark 6.2.8. In the original work of Wiles in [69], CWiles is controlled by the special
value of some adjoint L-function coming from the work of Doi-Hida which was further
axiomatized by Diamond-Flach-Guo. But COhta is controlled by the Kubota-Leopoldt p-
adic L-function as will be explained later. The motivation and evidence for such a strategy
for comparing the two modules comes from the work of Mazur-Wiles and Fukaya-Kato-
Sharifi that the cuspidal Hecke algebra mod the Eisenstein ideal is controlled by the Kubota-
Leopoldt p-adic L-function. In particular cases, where the Eisenstein ideal is ?nice?, one
can get hold of the full Hecke algebra.
Finally to finish the discussion about our general setup, let us prove the following easy
proposition:
113
Proposition 6.2.9. Consider a commutative diagram of rings:
f
A B
g h
C ? D
Suppose all the maps are surjective. Then the following are equivalent.
i. g induces an isomorphism ker( f )?= ker(?)
ii. f induces an isomorphism ker(g)?= ker(h).
iii. the canonical map from A? B?D C is an isomorphism.
Proof. We show that i and iii are equivalent. Let (b,c)?B?DC, then pick any lift of b in A,
call it a. Now g(a) and c have the same image in D, thus g(a)?c? ker(?). Thus there is an
unique element in ker f which maps to this element via g. Call that element a?. Then a?a?
maps to (b,c). Thus the map is surjective. To show injectivity, if there exists an a mapping
to 0 in both B and C, then a? ker( f ). But g is an isomorphism from ker( f ) to ker(?). Thus
a must be 0. Conversely suppose A is the fiber product of B and C over D, then A can be
written as {(b,c) ? B?C : h(b) = ?(c)}. So ker( f ) = {(0,c) ? B?C : ?(c) = 0}=ker(?).
The proof of the other equivalences are similar.
Since we want apply our results towards Wake?s conjectures in [63], we would like to
have a better understanding about the prime ideals in fiber products of rings. The following
proposition gives a complete description of all prime ideals in fiber products.
Proposition 6.2.10. Let the setup be as in the previous proposition, then
Spec(A) =U ?V ?W
114
where
U := { f?1(p) : p ? Spec(B) and ker(h) 6? p}
V := {g?1(p) : p ? Spec(C) and ker(?) 6? p}
W := {??1(p) : p ? Spec(D)} where ? := ? ?g = h? f}
Proof. If p ?W , then p must contain ker( f ) and ker(g). This is clear since ker( f )?= ker(?)
and ker(g) ?= ker(h). Conversely any prime ideal in A containing both ker(?) and ker(h),
must lie in W as the image of this prime ideal under the surjective homomorphism ? is a
prime ideal. In fact W is a closed subset of Spec(A). Now, let us pick an arbitrary prime
ideal q in A. We want to show that q lies in either U or V . Now without loss of generality
assume q do not contain ker( f ). Then pick an element ? ? ker( f ) \ q. Now localizing at
? , we get a commutative diagram
f
A? B?A A?
g h
C? ?A A? D?A A?
Now A? is flat over A and we can think of the above diagram as a fiber product. Now note
that B?A A? = 0 = D?A A? . Thus (by abuse of notation) g is an isomorphism from A? to
C?A A? . Thus there exists an unique prime p that maps to q, under g?1. Similarly for the
case where p does not contain ker(g).
Finally we show that there does not exist any prime p in A such that p + ker( f ) and p +
ker(g). Note that Ap = 0 by using the above argument and choosing elements from ker( f )\
p and ker(g)\p. Thus we see that U and V are disjoint subsets. The argument also shows
the open set U is isomorphic to the the open subset of Spec C defined by ker(?).
Now we will apply our above setup in the case of Hecke algebras. This work is already
115
done by Ohta in [43] and Lafferty in [33]. We will briefly recall their work in our squarefree
level case N. We warn the reader that the notation in Hida and that of Ohta-Lafferty and
Wiles differs by a twist of the p-adic cyclotomic character.
Let ? and ? be Dirichlet characters mod u, v with uv | N p, v prime to p, and ??(?1) =
1. We shall also extend our ground field to K, some extension of Qp which contains all
the values of ? and ? . Let O be its ring of integers and ? be its uniformizer. Define
Ur := 1+ prZp and fix u a topological generator of U1. Then under the identification:
O[U1] = O[[T ]],u? 1+T . The Eisenstein series we are interested in are of the following
form :
? ( )
E (? ,?;c) := ? (?)G(T,??2)+ ? ? ?(t)?(n/t)At(T ) qcn (6.4)
n=1 0<t|n
p-t
Here, c is a positive divisor of N p/uv, prime to p,
??
??
?
1/2 if ? = trivial
? (?) :=??0 otherwise
A (T ) := t(1+T )s(t)t = t?(?t?), if ?t?= us(t) (6.5)
and
G(?(u)us?1,??2) = L 2p(?1? s,?? ?) (6.6)
for every character ? of finite order on U1, which we can identify with a Dirichlet character
of the second kind. G lies in ? unless ??2 = 1, in which case it has a simple pole at s = 1.
Finally we set
E (? ,?) := E (? ,?;1) (6.7)
Then by the work of Hida [22], we know that E (? ,?) generates all Eisenstein series. For
116
the rest of the thesis, we will assume that (? ,?) 6= (??2,1).
Lemma 6.2.11. E (? ,?) is an eigenvector for Tl for l - N with eigenvalue ?(l)?(l)+?(l)
and for Up with eigenvalue ?(p).
Remark 6.2.12. But E (? ,?) is not an eigenvector for all Ul if uv 6= N. If l|N/uv, then
if one writes down the double coset operator for Tl and works out explicitly the action of
Tl on E (? ,?), one gets that Tl acts via the eigenvalue ?(l)?(l)+?(l). But one does not
get any such formula for the action of Ul on E (? ,?). But since l|N, we have the Hecke
operator Ul and not Tl . However, if l|uv, then E (? ,?) is indeed an eigenvector for Ul with
the appropriate eigenvalue.
The method of l-stabilization of a newform f for level N produces two newforms of
level Nl by the following formulae:
f1(z) = f (z)?? f (lz) (6.8)
and
f2(z) = f (z)?? f (lz) (6.9)
where ? and ? are roots of the characteristic polynomial of Ul acting on the 2-dimensional
space spanned by f (z) and f (lz). In particular, the L-function of f1 and f2 is the L-function
of f with the Euler factors (1??l?z)?1 and (1?? l?z)?1 respectively removed.
The method of l-stabilization gives us a way to remedy our problem. We diagonalize the
action of Ul on the 2-dimensional space spanned by ?E (? ,?)(z),E (? ,?)(lz)? and this
action has no eigenvalues ?(l) and ?(l)?(l). We choose our l-stabilized Eisenstein series
with the eigenvalue at l to be ?(l). Or in other words, this forces ? to be an imprimitive
character. We denote E? (? ,?) this new stabilized Eisenstein series. For ease of the reader,
we write down its eigenvalues.
117
Lemma 6.2.13. The eigenvalues of E? (? ,?) for Tl , l -N p, Ul for l|N and Up are respectively
? ?(l)?(l)+?(l) for l - N or l|uv
? ?(l) for l|N/(uv)
? ?(p) for l = p
Proof. It follows from lemma 6.2.11 and the previous remark.
Remark 6.2.14. In E? (? ,?), ? is generally chosen as the imprimitive character, whereas ?
is always a primitive character. We will write ? prim to denote the primitive character that
induces ? .
Remark 6.2.15. This method of level raising is necessary to find congruences between
Eisenstein series and cusp forms. This cusp form was constructed in proposition 6.2.35.
Definition 6.2.16. We define the Eisenstein ideal to be the ideal generated by
?Tl??(l)?(l)??(l),Ul??(l)?(l) for l|v,Ul??(l) for l|N/u,Up??(p)?
and let m? ,? be the maximal ideal containing the Eisenstein ideal. We will refer to m? ,? as
the maximal Eisenstein ideal and all our Hecke algebras will be localized at this maximal
ideal. To simplify notation, we will just use m to denote this maximal ideal.
Now we would like to write down some conditions such that E? (? ,?) is the unique
Eisenstein series in the localized Hecke algebra with the given system of eigenvalues mod
the maximal ideal.
Lemma 6.2.17. Let E (?1,?1) ? M?. Then the eigenvalues of T (l) for E (?1,?1) and
E? (? ,?) are congruent modulo (? ,T ) for all l - N p iff one of the following conditions
hold:
118
?1 ? ?2 and ?1? ? ?2? mod(?)
?1 ? ??2 and ?2 ? ?1? mod(?)
Proof. This is lemma 1.4.9 in [43] and the proof is standard and follows from independence
of characters.
The following proposition is a strengthening of the above lemma.
Proposition 6.2.18. Assume Neben then E? (? ,?) is the unique Eisenstein series in our
localized Hecke algebra with the given system of eigenvalues mod the maximal ideal.
Proof. By linear independence of characters, we get either
? ? ? ?1 (mod p) and ? ? ?1 (mod p) or
? ? ? ??1 and ?2 ? ??
But Neben forces the congruence into an equality as we do not have any non-trivial p-
power characters congruent to 1 mod p.
We now show that the second case can not happen. The same argument with Neben shows
that
? = ?1? and ?1 = ??
As the Up eigenvalues are also congruent, we get ?(p) ? ?1(p) (mod ?). Since the
conductors and orders of ?1 and ? are prime to p, we get ?(p) = ?1(p). Combining
these facts just like in lemma 1.4.9 in [43], we get ? ?1 ?1i|(Z/p)?= ? and ??? = 1, but
this forces the conductor of ? to be divisible by p, which contradicts our assumption that
p - N.
Recall the Residue map of Ohta in [43] which is essentially the constant term map
which sends an Eisenstein series to the formal sum of (cusps)? residue at each cusp.
119
Theorem 6.2.19. (Lafferty-Ohta) Suppose E (? ,?;c) ?M?, then
Res(E (? ,?;c)) = A? ,? .c? ,?;c (6.10)
and c? ,?;c ?/ m?, where
( )
A := ? ((1+X)s(l)????1??2(l)l?2? ,? ) G(T,???1?2) (6.11)
l|N
l-cond(???1)
where G(T,???1?2) is a twist of the Kubota-Leopoldt p-adic L-function as follows:
G(?(u)us?1,???1?2) = L ?1p(?1? s,?? ?2?) (6.12)
where ? is a Dirichlet character of the second kind.
Proof. This is theorem 4.2 in [33].
In fact Ohta constructs the following exact sequence:
Theorem 6.2.20.
0? S??M? ?
R??es C?? 0
and the sequence canonically splits over Frac(?) with the Hecke-equivariant splitting given
explicitly by c E (? ,?;c)? ,?;c? A .? ,?
Proof. This is the exact sequence 2.4.6 and Theorem 2.4.10 in [43].
Recall the following proposition from Lafferty.
Proposition 6.2.21. ?(c? ,?;c) is a free ? module.
Proof. This is proposition 3.3.2 in [33].
120
So now we can generalize Ohta?s exact sequence using the above results. Our strategy
is exactly the same as in Ohta?s in [43]. We will start with the maximal ideal in the Hecke
algebra and Proposition 6.2.18 allows us to isolate an unique Eisenstein series E? (? ,?) and
then localizing Ohta?s exact sequence immediately gives us:
Proposition 6.2.22. Assume either one of two the following conditions are satisfied:
1) p - ?(N) or
2) Neben
then we have a fundamental exact sequence
0? Sm?Mm?Cm? 0,
where Cm is a free ?-module of rank 1 with generator c? ,? and c? ,? ?/ m?. The sequence
canonically splits over Frac(?) and the congruence module is given by ?/A? ,? .
Proof. This is the sequence when we get after localizing the exact sequence Theorem
6.2.19 at m. This is the also the exact sequence 3.1.4 in [43], the only difference being
our proofs showing the existence of a unique Eisenstein series with the given system of
mod p eigenvalues.
We would like to apply the formalism of congruence modules to the context of Hecke
algebras. For a slightly more general formalism, we refer the reader to page 253 in [43].
Recall from theorem 4.2.4 we have the following
? Hom?(Mm,?) = Hm
? Hom?(Sm,?) = hm
Recall that Hm and hm are reduced, flat and finite over ?. Tensoring the exact sequence in
121
Proposition 6.2.22 by Frac(?), and taking Hom(?,Frac(?)), we get
Hm?Frac(?) = (hm?Frac(?))? (C?m?Frac(?)) (6.13)
Let ?1 and ?2 be the projections of Hordm in the respective components. Thus we get the two
exact sequences
0? a? ??H ?1m hm? 0 (6.14)
0? ?I(? ,?) ? H ??2m m ?? 0 (6.15)
The second exact sequence (6.15) comes from identifying Cm with ? by choosing a basis
and a and I(? ,?)m are ideals in Hm defined to make the above sequences exact.
Remark 6.2.23. I(? ,?)m is better known in the literature as AnnHm(E (? ,?)). One can also
see that ?1 is the canonical surjection induced by the inclusion of Sm ??Mm. Unraveling
the definitions, one can see that ?2 is given by Tl(a1(E? (? ,?))).
The congruence module attached to the exact sequence (6.14) is generated by a single
element as a ?-module by lemma 1.19 and then lemma 1.1.12 in [43] shows that the con-
gruence module attached to that exact sequence is isomorphic to the congruence module
from Proposition 6.2.22 and is equal to ?/A? ,? . In fact Ohta in page 254 in [43] shows the
congruence module attached to the exact sequence (6.15) is also ?/A? ,? as Hm modules.
For a more detailed description of the congruence modules attached to the Hecke algebras,
we refer the reader to section 3.2, pages 253-255 in [43]. Alternatively one can show this
directly using the definitions and the lemmas at the beginning of this section. Now using
the second exact sequence we can define CWiles = ?/?2(AnnHm(ker(?2)))
Proposition 6.2.24. CWiles = ?/(A? ,?).
122
Proof. We have shown in Lemma 6.2.7, CWiles = COhta. And the rest follows from the
previous paragraph.
Recall the main result of Mazur-Wiles in [39]:
Theorem 6.2.25. Let h be the cuspidal Hecke algebra and denote by I? ,? the image of
I(? ,?) under the canonical projection H h. Then
hm/(I? ,?)m ?= ?/(A? ,?).
In view of our abstract algebra and the deep result of Mazur-Wiles, we immediately
obtain:
Proposition 6.2.26. Hm = hm??/(A? ? ) ?.,
Proof. We already know from (6.13), Hm?Q(?) = hm?Q(?)?Q(?) and using the exact
sequences (6.14) and (6.15) above we have a commutative diagram:
pr1
Hm hm
pr2 ?1
? ?2 ?/(A? ,?)
In view of the proposition 6.2.9 we have to show ker(pr2) =? ker(?1). Now, ker(pr2) =
I(? ,?)m and the ker(?1) = (I? ,?)m. And (I? ,?)m is defined as the image of I(? ,?)m under
the projection map. So the hypotheses of the proposition 6.2.9 are satisfied.
Recall the following definition.
Definition 6.2.27. Let R be a n-dimensional local Noetherian ring. Then R is Gorenstein if
ExtiR(F,R) = 0 for all i < n and ExtnR(F,R) = F.
123
Note that this definition takes a particularly simple form if R has dimension 0. In that
case R is Gorenstein iff HomR(F,R) =? F as F-vector spaces. Gorensteinness of Hecke
algebras is important because in that case the spaces of modular forms are free of rank 1
over the Hecke algebras and one can then realize Hida?s big Galois representation inside
the cohomology of towers of modular curves.
Sometimes when checking whether the big Hecke algebra is Gorenstein, it is convenient to
check at a finite level or weight. In fact restricting ourselves to the weight 2 Hecke algebra,
we get this corollary. Let ? be an uniformizer of O . Then
Corollary 6.2.28. Assume ? |Lp,S(?1,???1?2) but ? 2 - Lp,S(?1,???1?2), then (H2)m
is Gorenstein iff (h2)m ?= O or some ramified DVR over O .
Proof. Note that from proposition 6.2.26 we have a fiber product diagram:
pr1
(H2)m (h2)m
pr2 ?1
?
O 2 O/(L 2 ?1p,S(?1,?? ? ))
The existence of such a diagram was clear in the works on Mazur-Wiles, Kurihara and
Harder-Pink. Our fiber product diagram also appears in [43], section 2.4, the exact se-
quence 2.4.1. The fact that the specialization to the weight 2 using (4.1) corresponds to
Lp,S(?1,??2??1) is clear from the calculations in [43], section 2.5, particularly the for-
mula in 2.5.3. The rest of the proof is surprisingly a commutative algebra result from [58],
corollary 2.7 which says R?F S is Gorenstein iff R and S are DVRs. The corollary follows
directly from that statement.
Remark 6.2.29. This result is an analog of a theorem of Mazur in [36]. In fact we can
say more here: let k ? 2 (mod p? 1), assume p|L (1? k,???1 2p,S ? ) but p2 - Lp,S(1?
k,???1?2), then (Hk)m is Gorenstein iff (hk)m =? O or some ramified DVR over O . This
124
is analogous to some results of P. Wake and C. Erickson in this higher weight case.
Remark 6.2.30. Calegari and Emerton in [8] used this above description of Hecke algebra in
their case to compute the ramification of the cuspidal Hecke algebra by using deformation
theory arguments. More explicitly, if e is the ramification index of the cuspidal Hecke
algebra over O , then e is given by a non-trivial deformation to F[X ]/Xe+1. However, if our
tame level N is prime and p||N?1, our Hypothesis 1 forces the cuspidal Hecke algebra to
be O by theorem 1.2 in [8].
The above corollary can be summarized by the following picture:
Cuspidal Family Zeros of Kubota-Leopoldt
p-adic L -function
Weight Map
Eisenstein Family
Weight Space
This is the picture of our space of modular forms. The weight map is etale over classical
points. We do not know in general the nature of these intersections. It is widely believed
that the zeros are simple zeros but we do not know how to prove it.
125
Remark 6.2.31. Given this description of the Hecke algebra, it is very easy to see that
CWiles = ?/(A? ,?). In fact ker(pr2) = I? ,? ? 0 and since AnnhmI? ,? = 0, we immediately
get AnnHmker(pr2) = A? ,? ?0.
We give some further properties our Hecke algebras in the form of a couple of lemmas
which will be used in the next section.
Lemma 6.2.32. AnnHm(I(? ,?)) = ker(Hm  hm)
Proof. Let x?AnnHm(I(? ,?)), then localizing the fiber product diagram in Proposition 6.2.9,
we get
pr1
(Hm)x hm?Hm (Hm)x
pr2 ?1
?? ?Hm (Hm)
2
x ?/(A? ,?)?Hm (Hm)x
Since I(? ,?) is the kernel of pr2, so localizing at x, (by abuse of notation) pr2 is an
isomorphism. Since the above is a fiber product diagram, this forces hm?Hm (Hm)x = 0 =
(hm)x?, where x? is the image of x inside hm. Since hm is reduced, this forces x?= 0, as desired.
Conversely, let x ? ker(Hm  hm), then localizing the fiber product diagram as before, we
get (Hm)x ?= ?x?, where x? is the image of x in ?. This implies I(? ,?)x = 0. Since Hm is
reduced this shows that x kills I(? ,?). This proves the claim.
Lemma 6.2.33. pr1|I(? ,?) is injective.
Proof. If pr1(x) = 0, then x ? ker(Hm  hm)? I(? ,?). But AnnHm(I(? ,?)) = ker(Hm 
hm) by lemma 6.2.32, thus x ? x = 0, but Hm is reduced, so x = 0.
The proof of this theorem also proves a familiar statement about cuspidal Hecke alge-
bras.
Corollary 6.2.34. I? ,? is a faithful hm module.
126
Proof. Note that we have a commutative square:
I(? ,?) Hm
?=
I? ,? hm
Let y ? Annhm(I? ,?), then we can lift y to some x ? Hm. If x ? AnnHm(I(? ,?)), then the
previous theorem shows y = 0 and so we are done. So assume x ?/ AnnHm(I(? ,?)), then
localizing the fiber product diagram at x, we get:
pr1
(Hm)x (hm)y
pr2 ?1
?? ?2Hm (Hm)x ?/(A? ,?)?Hm (Hm)x
where ?1 is an isomorphism but pr2 is not, which is a contradiction.
To finish off this section, we would like to sh?ow that the?? considered in Chapter 3 is
?? ?
reduction mod p of some cusp form. Recall ?? ?=? ?? where ? is unramified at p,
?
??(?1) = 1 and ? is ramified at p and all primes where ? and ? are unramified.
Proposition 6.2.35. Assume A? ,? is not an unit. Then there exists a cusp form f such that
? ?f = ? .
Proof. Sord is free of finite rank over ? and let { f1, ... fk} be a ?-basis. By Hida duality
(theorem 4.2.4), there exist a dual ? basis for h, i.e.
????0 if i =6 j
a1( fi|h j) =???1 if i = j
Let Hi be arbitrary lifts of hi under the canonical surjection H h. Since the residue map
127
is surjective, there exists a modular form g such that Res(g) = c? ,? . Now define
g? := g??a1(g|hi) fi.
Since Res(A ?? ,?g ?E ?? ,?) = 0, it follows that A? ,?g ?E? ,? =?F is a cusp form.
Obviously F ? E? ,? mod A? ,? . Thus F is a mod A? ,? eigenform with the same eigenvalues
as E? ,? . We would like to apply Deligne-Serre lifting lemma to lift these mod p system of
eigenvalues to characteristic 0. However to apply the lifting lemma, we have to work with
a DVR. To remedy the problem, we look at the specialization of F and E? ,? at weight 2,
since the prime ideal corresponding to arithmetic specializations is a height 1 prime. Then
the constant term of E(2) (2) ? (2)? ,? vanishes mod p. Thus we get F = E? ,? mod p. Now by the
Deligne-Serre lemma, we can lift this mod p-eigenform into our desired eigenform.
128
6.3 Modularity lifting and Wake?s conjecture
Now let us recall Wake?s conjecture in [63].
Conjecture 6.3.1. (Wake): Let Ih and I ord ordH be the Eisenstein ideals in h and H . Then for
all height 1 prime ideals p and q such that I ? p and I ? q, hord and Hordh H p q are Gorenstein.
Recall proposition 6.2.10 gives us a complete description of all the prime ideals in Hordm .
Since in this thesis, we are only concerned with ordinary Hecke algebras, we will drop ord
from the superscript.
Let p be a height 1 prime of Hm such that I(? ,?) ? p. Let p1 := pr1(p). Then p1 ? I? ,? ,
i.e. p ?W in our terminology, i.e. in that case p is the inverse image of some prime divisor
of A? ,? ? ?.
Conversely given a height 1 prime p in hm such that I? ,? ? p. Let p? := pr?11 (p). Then p?
is either in V or W . If it is V , then it is not in the support of the congruence module, then
Theorem 4.1 in [26] shows that hp? and Hp are Gorenstein.
Now let us recall some facts about Fitting ideals in the appendix of [39]. Let M be a R
module. Then
? FitR(M)? AnnR(M)
? If M is generated by n elements, then AnnR(M)n ? FitR(M)
? If I is an ideal of R, then FitR/I(M/IM) is the image of FitR(M) in R/I
? If M is a direct sum of cyclic R-modules, i.e. M = R/p1? ...?R/pk, then FitR(M) =
p1....pk
Now we need some isomorphism criterion to prove our modularity lifting results.
Theorem 6.3.2. (Isomorphism criterion) Let us consider the following commutative trian-
gle
129
R ? T
f
g
?
Let I be ker( f ), then the map ? is an isomorphism between complete intersections over ?
iff ?(FitR(I)) 6? m?T .
Proof. See [11]
Corollary 6.3.3. If T is Gorenstein over ? or if we replace ? by a DVR, then ? is an
isomorphism iff ?(FitR(I)) = AnnT (J), where J=ker(T ? ?).
Proof. See [11]
We use this theorem to state Wiles numerical criterion. Let J be the kernel g. Define
the ?-module ? by g(AnnT (J)).
Corollary 6.3.4. Let O be a DVR and assume we have a commutative triangle as before
where we replace ? by O . Assume further that T is free as O-module and ? 6= 0. Then ? is
an isomorphism of complete local intersections over O iff lengthO(I/I
2) = lengthO(O/?).
Proof. See [11].
Remark 6.3.5. Note that for a surjection ? as above, lengthO(I/I2)? lengthO(O/?). The
difficult part in the above corollary is to show the reverse inequality.
Now we are ready to use the various isomorphism criteria. We assume hypothesis 1
throughout the rest of the section. Let u?s recall our?deformation problem.
?? ?
Fix a squarefree tame level N. Let ?? ?=? ??, where
?
? ??(?1) = 1
? ? is unramified at p
130
? ? is ramified at all l|N/cond(???1) and ?|Ip=6 0.
? ?? is mod p distinguished.
Then we have shown in proposition 6.2.35 that ?? ?= ? f , for some cusp form f . We consider
the universal deformation ring Runiv?? parametrizing lifts which are ordinary at p, Steinberg
at all primes where ? is ramified and we take the minimal lifts at primes where the diagonal
characters are ramified. Note that Galois representation attached to the modular form f
constructed in proposition 6.2.35 has all the above properties by theorem 4.1.2. Using the
results in Chapter 3 and proposition 3.7.5, we get a commutative triangle
Runiv ??? Hm
f
pr2
?
where ? : Runiv?? ? Hm given explicitly by
tr(?univ)(Frobl) 7? (Tl,Tl(E? (? ,?))) (6.16)
pr2 is the projection onto the Eisenstein component and f is the canonical surjection
f : Runiv  Rred ??? = ?. Thus the kernel of f is the ideal of reducibility as defined before.
Assuming cyclicity conjecture we will use the isomorphism criterion to give a new simple
proof of the Gorensteinness of the Hecke algebras constructed by Ohta.
Skinner-Wiles in [56] and [57] have proved modularity lifting results for reducible repre-
sentations. However in both their works they only treat the case where ? = 1. However, we
bypass that problem by using Ohta?s residue map. Calculation of the cotangent space I/I2
was inspired by Sharifi?s results in [54] and our results generalize his results. The compu-
tation of the congruence modules is already done by Ohta and the following hypotheses are
required to ensure that the isomorphism criterion is satisfied. Moreover our proof is explicit
131
in the sense that our proof shows that the reducibility locus coincides with the Eisenstein
series which is a rather intuitive situation. However just like [56], we have to assume
Hypothesis 1, otherwise we can not get a map from Runiv?? ?Hm. This was explained in our
section on pseudo-representations. Skinner-Wiles in [57] removed Hypothesis 1, but their
proof is not explicit. In the current literature, one works with pseudodeformations and this
work is been carried by P. Wake and C. Erickson in a series of papers.
To prove the next theorem, let us assume the following additional hypothesis:
? p - ?(N) or Neben.
? (? ,?) 6= (??2,1)
? p|B1,???1??1
? (Vandiver-type cyclicity conjecture) C/IC is cyclic as a ? module, cf. section 3.5
Theorem 6.3.6. (Skinner-Wiles, Ohta) We have an isomorphism Runiv ??? = Hm of complete
local intersections over ?.
Proof. Let Ired be the ideal of reducibility, then we get by the commutative triangle that
Ired surjects onto I(? ,?) via ? . Thus using the properties of Fitting ideals, we have the
chain of inclusions:
?(Fit (IredRuniv ))? FitHm(I(? ,?))? AnnHm(I(? ,?))??
Now, AnnHm(I(? ,?)) = ker(Hm  hm) by lemma 6.2.32 and by remark 6.2.31, we know
that
AnnHm(I(? ,?)) = A? ,? ? ?, where we view ? as a Hm module via pr2. By Hida duality
(theorem 4.2.4),
ker(Hm  hm) = Image(Res)
132
But by Lafferty-Ohta (theorem 6.2.19), the image does not lie in m?. To complete the proof
of the theorem via theorem 6.3.2, we need to show
?(FitRuniv(I
red)) = AnnHm(I(? ,?)) = A?? ? ,?
Since for any R module M, and for any ideal I ? R, M?R R/I =?M/IM, and by the proper-
ties of Fitting ideals, we see that
Fit?(Ired/(Ired)2) = f (Fit red redRuniv(I )) = pr2(?(Fit?? Runiv(I )))??
and the above inclusions show that Fit?(Ired/(Ired)2) ? A? ,? . Under the Vandiver type
assumption, we get Ired is cyclic as a ?-module by theorem 3.5.15 and theorem 3.5.25.
Thus we get Fit (Ired? /(Ired)2) = Ann red red 2?(I /(I ) . Under the assumption of cyclicity,
proposition 3.5.12 and theorem 3.5.23 shows that either
C/IC ?= ?/(G ?1???1?2(u(1+T ) ?1))
or G ?1???1?2(u(1+T ) ? 1) is a unit and C/IC ?= Rq but the p-divisibility of B1,???1??1
contradicts (G ?1???1?2(u(1+ T ) ? 1)) is a unit. Thus all the fq?s are units. Note that
B/IB?= ?. Thus Ired/(Ired)2 =C/IC. Thus Ann (Ired/(Ired))2? = (G???1?2(u(1+T )?1?
1)). Now Gal(L?/F?) is unramified everywhere by lemma 3.5.26. Since all the fq?s are
units, by the formula (6.11), we see that there are no Euler factors in the definition of A? ,? .
Combining this observation with theorem A.1.13 and the displayed formula A.1.12 in [43]
shows that A? ,? is the characteristic polynomial of C/IC. Thus A? ,? ? Fit?(Ired/(Ired)2).
Thus Fit (Ired/(Ired? )2) = A? ,? which implies pr2(?(Fit redRuniv(I )) = pr2(Ann?? Hm(I(? ,?)).
Similar arguments as in lemma 6.2.33 shows that pr2|ker(Hmhm) is injective and maps
AnnHm(I(? ,?) isomorphically onto A red? ,? . Thus, ?(FitRuniv(I )) = AnnHm(I(? ,?))??
133
Remark 6.3.7. The proof actually shows that Gorensteinness of the Hecke algebra is ?al-
most? equivalent to the cyclicity conjecture.
Remark 6.3.8. We want to make a few comments regarding the hypotheses made at the
beginning of the theorem.
? Ohta used the condition p - ?(N) but we can remove that condition with a slightly
general Neben.
? (Trivial zero case) p|B1,???1??1 along with p - ?(N)(or Neben) ensures Ired 6= 0.
However if p - B , then Ired1,???1??1 is non zero iff there is a trivial zero coming from
the Euler factor at the auxiliary primes. Recall the example of X1(11) worked out
earlier where Ired 6= 0 and the trivial zero came from the Euler factor at 11. This is
the so called case of trivial zero which was first solved by Greenberg-Stevens in their
proof of Mazur-Tate-Teitelbaum. Moreover since C/IC is cyclic, we get that there
is only prime q congruent to 1 (mod p) and C/IC = Rq by proposition 3.5.13. And
A? ,? is fq. This is theorem 6.3.1 in [65]. Note that our cyclicity of B/IB ensures the
hypothesis of theorem 6.3.1 in [65] are satisfied. In this case, one also gets that Hm
is Gorenstein.
Definition 6.3.9. (Skinner-Wiles) A prime p in Runiv?? is nice if it is good (see defini-
tion 3.4.21) and is an inverse image of a prime q in hm.
Theorem 6.3.10. (a) There is an isomorphism (Runiv?? ) ?p = (Hm)q of complete local inter-
sections over ?p, where p ? Runiv?? and q ? Hm are any height 1 prime ideal (sometimes
referred to prime divisors) over (p)? ? such that ??1(q) = p
(b) (Skinner-Wiles) Under the isomorphism above, (Runiv ??? )p = (hm)q for all nice primes p
and both the rings are complete local intersections.
134
Proof. Proposition 2.1 in [26] shows that
(Hm)p = (hm)p??p
or in other words the congruence ideal vanishes when localized over any prime divisor
above (p). We will abuse notation and use the letter p to denote the prime ideals above (p)
in the following commutative diagram of rings.
(Runiv ??? )p (Hm)p
f
pr2
(?)p
Now, annihilator of ker(pr2) is clearly ?p from the formula above. Thus, ? = ?p and
so length?(?/?) = 0. To use the numerical criterion, we need to find the length of
(Ired/(Ired)2)p. But theorems 3.5.15 and 3.5.25 give us a description of the ? module
(Ired/(Ired)2)p in terms of the tensor product of two Iwasawa modules. Now support of
Supp(M?N)? Supp(M) ? Supp(N). Finally note that (C/IC)p = (Rq)p??p/(A? ,?)p but
A? ,? and the characteristic ideals of Rq all have ?-invariant 0, or in other words, localizing
at the prime ideal (p) above the prime (p) ? ?, makes them all units. Thus (C/IC)p = 0,
and thus (Ired/(Ired)2)p = 0, thus the length is also 0 and so the numerical criterion is
satisfied (corollary 6.3.4). So to summarize, we have shown an isomorphism
(Runiv?? )p
?= (Hm)p
Now, note that if p is a good prime, then the height of q is necessarily 1. Suppose not, if
the height is 0, then q is a minimal prime but minimal primes are in characteristic 0, but p
has characteristic p. Thus q has characteristic p and thus sits over (p) ? ?. In this case q
135
pulls back to a height 1 prime q? ? Hm and
(h ) =? (H ) ? ?= (Runivm q m q ?? )p
Since the map pr2 : (hm)q ? ?p is the 0 map, (by abuse of notation) ? can not factor
through Rred which implies the Galois representation with values in (Hm)q? is irreducible.
Theorem 4.1 in [26] shows that (Hm)q? is Gorenstein.
Remark 6.3.11. In theorem 3.29 in [24] Hida constructs a pseudorepresentation with values
in h, and to upgrade that pseudorepresentation into a representation, we need
Hypothesis 1.
Remark 6.3.12. Our proof also shows that the Eisenstein locus of (Hm)q coincides with the
reducibility locus in the universal deformation ring. Indeed in that case, we will get
(Runiv ??? )p = (Hm) ?q = ?p
Theorem 6.3.13. (Wake?s conjecture) Hordm is weakly Gorenstein.
Proof. From the discussions before, we are also concerned about the prime divisors of
A? ,? . Let f be a prime divisor of A? ,? , then we get a commutative triangle
(Runiv ??? )p (Hm) f
?1
pr2
? f
where ? f is a DVR and p is the prime above f . To prove the conjecture, it is enough to
use the numerical criterion. Now length?(I/I
2)p = multiplicity of f in A? ,? , by Theorem
3.4.14, since A? ,? does not share any prime factors with Rq and T ? p - A? ,? . And the
congruence module on the right is nothing but (A? ,?) f ,which has length equal to multi-
136
plicity of f in A? ,? . Thus by the numerical criteria, ? is an isomorphism of complete local
intersections over ? f .
Remark 6.3.14. The two theorems above have two different flavors. One of them deals with
primes above p and needs deep results like the vanishing of ? invariants whereas the pre-
vious theorem deals with characteristic zero primes and needs results like the coprimality
of characteristic ideals of various Iwasawa modules.
137
Chapter 7: Selmer groups
7.1 Review of some basic definitions
We start by recalling the definitions of various Selmer groups. We will be assuming
Neben th?roughout th?e section.
?? ?
Let ?? ??? ?= ?, where ??(?1) = 1 and ? and ? are both unramified at p. Let M
0 ?
be the F vector space on which ?? acts.
Definition 7.1.1. The residual Greenberg Selmer group associated to a set of local condi-
tions is given by
( )
Sel(Q,M) = ker H1(QS,M)??H1(Qv,M)/Lv
v
where
? L := ker(H1p (Qp,M)? H1(Ip,?))
? for v ? S and v - p, L 1v := Hnr(Qv,M)
where S contains all primes dividing N and p and ?. Note that since p is odd, the primes
above ? will not contribute towards our calculations of Selmer groups.
Remark 7.1.2. There is also a version of Selmer groups in which we replace Ip by Dp.
Those Selmer groups are called strict Selmer groups.
138
We recall the definition of imprimitive residual Selmer groups.
Definition 7.1.3. Let ?0 be any finite subset of S, not containing p and ?, then
( )
Sel?0(Q,M) := ker H1(Q ,M)? ? H1S (Qv,M)/Lv
v?S\?0
where Lv are defined as above.
Now we come to some other Selmer groups. We closely follow the definitions in [49].
Let V be an ordinary Galois representation and let T be any Galois stable lattice. Let
M := V/T , then we define the Selmer group. Let S be the set containing all ramified
primes and p and ?.
Definition 7.1.4.
( )
Sel(Q,M) := ker H1(Q,M)? H1(Qv,M)/Lv
Since V is ordinary, there is a Dp stable line V1 and Ip acts trivially on V/V1. Let M1 denote
the image of V1 in M and let M2 := M/M1, then we define
Lp := ker(H1(Q ,M)? H1p (Ip,M2)).
for v ? S, v 6= p, L := H1v nr(Ql,M).
We can analogously define the imprimitive Selmer group as before.
139
7.2 Some calculations on Selmer groups
Let ? =6 1. Then we have an exact sequence
0? H1(Q,??)? H1(Q,M)? H1(Q,?)? H2(Q,??) (7.1)
We want to understand what elements in H1(Q,??) land inside Lp.
0 H1(Qp,??) H1(Qp,M) H1(Qp,?)
res res res
0 H1(I ,??) H1p (Ip,M) H1(Ip,?)
from the diagram above, we see that image of H1(Qp,??) lies in Lp. In fact even if ? = 1,
we still get H1(Qp,??) lands inside Lp. This is clear from the diagram below:
F H1(Qp,??) H1(Qp,M) H1(Qp,F)
res res res
F H1(Ip,??) H1(Ip,M) H1(Ip,F)
Finally assume ? =6 1, by Tate duality (theorem 2.2.3) we get H2(Qp,??) = 0. And as a
consequence, we see that
0? H1(Qp,??)? Lp? H1nr(Qp,?)? 0
is exact.
We remark that in the case of strict Selmer groups, we replace H1nr(Qp,?) by H1(Qp,?).
Now we need to deal with other auxiliary primes. We break it down into a few cases. First
we assume that the conductor of ?? is N, and N will always be square free.
140
Let l|N, then H1nr(Ql,??) = H0(Ql,??) from (2.3). Thus H1nr(Ql,??) 6= 0 iff ?? = 1.
In particular, ? is unramified so ? is ramified. In that case by (2.3), H1nr(Ql,?) = 0. Also
note that H1nr(Ql,M) =6 0 iff ?? or ? is the trivial character.
Thus when ? is ramified, we get an isomorphism H1nr(Ql,??) ?= H1nr(Ql,M) as both of
them are 1-dimensional vector spaces over F. Now if ? is the trivial character, we have the
long exact sequence,
H0(??)? H0(M)? H0(F)? H1(??)? H1(M)? H1(F)? H2(??) (7.2)
Now ? is ramified at l, thus H0(??) = H2(??) = 0 and H0(M) = F
This shows that we have an isomorphism H1 ? 1nr(Ql,M) = Hnr(?). Combining all the cases,
we get an injection:
H1L (QS,??)? Sel(Q,M)
In the case where f? f? 6=N, we will use the imprimitive Selmer groups. Now pick a prime l
dividing N but not f? f? . Then ? is necessarily ramified at Il . In that case, we relax the local
condition at l by defining Ll = 0. Thus we get an injection of H1 ?0L (Q,??) ?? Sel (Q,M).
We summarize the above discussion in the form of the following proposition:
Proposition 7.2.1. Let f? , f? = N and ? =6 1, then we have an exact sequence of Selmer
groups,
0? Sel(Q,??)? Sel(Q,M)? Sel(Q,?)
Note the right group is 0 if ? = 1 as there are no unramified extensions of Q. It is also 0 if
Clp(K )?? = 0 where K? is the fixed field of the kernel of ? .
Finally we would like to estimate the size of H1L (Q,??). This estimation is a stan-
dard application of the Greenberg-Wiles formula. Let M = F(??). Then M? = F(??1).
We define H1 (Q,M?? ) by requiring the classes to be locally trivial at p and unramifiedL
141
everywhere else. Recall the Greenberg-Wiles formula from Chapter 2 :
#H1L (Q,M) #H0(Q,M) ? #Lv1 = .#H ?(Q,M?) #H0(Q,M?) #H0(Qv,M)L
To analyze this, we will study each of the terms separately. Under our hypotheses, we
are just left to calculate Lv and H0(Qv,M). If l| f? , then H0(Qv,M) = 0 = Ll . If l - f? ,
then M is unramified at l and so L 0l = H (Ql,M). Finally we are left to calculate l = ?.
H1 (R,M) = 0 and H0(R,M) = prnr iff ? is odd, where pr = #F.
Remark 7.2.2. In [19] and [3], Sel(Q,M) is defined by taking Lv = 0, for all v ? S and
v 6= p, i.e., ?0 contains all ramified primes, except p. In that case, a similar diagram chase
as above shows H1(Q ,??) ?? Sel?S 0(Q,M).
?1
The group H1 ??(Q,M ) can be seen as the Cl (K )?p ? by noting that under NebenL
[K? : Q] is prime to p and the rest follows from [49]. We remark that if ? is even, then
H1 ?(Q,M?) is given by the p-adic L-function by the work of Wiles [68].L
Now we relate the Selmer groups and the residual Selmer groups. Most of these results are
standard calculations in Iwasawa theory so we only provide a sketch.
Since M is ?-divisible, we have an exact sequence
0?M[?]?M ??? M? 0 (7.3)
Under the assumptions in this section we note that H0(Q,M[?]) = 0, thus we get an iso-
morphism
H1(Q,M[?]) =? H1(Q,M)[?] (7.4)
We want to show the following proposition:
Proposition 7.2.3. The exact sequence above induces a map f : Sel(Q,M[?])? Sel(Q,M)[?]
and f is an isomorphism if f? f? = N. Otherwise, let ?0 contain all primes except p, then
142
Sel?0(Q,M[?])? Sel?0(Q,M)[?] is an isomorphism.
Proof. Let c ? Sel(Q,M[?]), let v ? S and v 6= p, and if v is such that
H0(Qv,M[?]) = 0, then by (2.3), H1nr(Qv,M[?]) = 0, thus there is no local condition and
we have nothing to check. Now suppose that H1nr(Qv,M[?]) =H0(Qv,M[?]) 6= 0, by a long
exact sequence in cohomology we still get H1(Qv,M[?]) =? H1(Qv,M)[?]. We just have
to check H1nr(Qv,M[?])?H1nr(Qv,M)[?]. Taking the Iv cohomology of the exact sequence
(7.3), we get H1(Iv,M[?])?= H1(Iv,M)[?], since MIv and M[?]Iv are one dimensional K/O
and F modules respectively and M is divisible. And finally we have a commutative square
H1
?
(Qv,M[?]) = H1(Qv,M)[?]
res res
?
H1(Il,M[?])
= H1(Il,M)[?]
And this takes care of all the primes v 6= p. For the prime p, see [3].
The map f is obviously injective. Now we would like to show that the map f is surjective.
Let c ? Sel(Q,M)[?]. Now let v ? S and v 6= p, then c(?) = ?(?v)??v for some ?v ?M
and for all ? ? Iv. We want to show ?v ?M[?]. Also note that ?(??v) = ??v. From the
exact sequence (7.3), c comes from H1(Q,M[?]), which implies
?(?v)??v = tv, where tv ?M[?]. Now under some choice of basis {e1,e2},
M|Iv= 1? ? , where ? is a ramified character. Thus ??v = xve1 for some x ? K/O . Let y
be such that ?yv = xv, then combining the above relations, we get
?v = yve1 + tv
and since ?(e1) = e1, we immediately get
c(?) = ?(tv)? tv
143
Checking the condition at p is well known. For example, see [19] or [3].
When f? f? 6= N, one must work with imprimitive Selmer groups and the second part of
this proposition is essentially the crux of [19]
Remark 7.2.4. We think of the above proposition and proposition 7.2.1 as a factorization
(or congruence) statement between algebraic p-adic L-functions. For the corresponding
statement about analytic p-adic L-functions, one needs that the Hecke algebra is Gorenstein
and to prove a similar factorization statement for analytic p-adic L-functions led us to this
thesis. When the Hecke algebra is Gorenstein, this is known from the work of [19] or
recently by [3].
144
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