Fix a residual representationρ0corresponding toM2.
Lemma VI.6. ρ0has a crystalline lift with parallel Hodge–Tate weights(0,1).
Proof. LetD be a rank two filtered φ-module with an ordered basis η of D with respect to [φ]η = diag(pλ⃗ ,⃗λ)for someλ ∈E×and
FiljD=
( D if j≤0
(Qpf ⊗QpE)·(η1+⃗y·η2) if j=1
0 if j≥2.
Since the filteredφ-module satisfies the condition 5.2.1 in Dousmanis [1]:
e f vp(pλ2) =f, e f vp(pλ)≥
∑
{i∈I0|yi=0}
1, e f vp(λ)≥0, then it is admissible. Then we have another admissible filteredφ-module
D:=SE·(e1,e2) with
Fil1D≡SE·(e1+⃗ye2) (mod Fil1SE(e1,e2)), [φ]e=
pλ⃗ 0 0 ⃗λ
! .
By Case 2, we can easily checkM=SOE·(pe1,yke2) =SOE·(E1,E2)is a strongly divisible module of D, and its Breuil module isM2.
We fix a liftψ:GQ
p f →OE×of detρ0, which is equivalent to fixingη:=λ2∈OE.Thenλ =√ ηor λ =−√
η. For each suchλ, the admissible filtered(φ,N)-modulesD(λ,⃗y)forvp(yk)<1 corresponds to the characteristic 0 closed points ofR(0,1),ψρ
0 .
Lemma VI.7. Let ζi=0 ifvp(y−1k yi)>0, and fixζi to a(pni−1)-th root of unity such thatζi is a preimage ofy−1k yjinOE for someni∈Nifvp(y−1k yi) =0.
(1) Forλ =√
η, denoting ζi+Xi =y−1k yi in the strongly divisible modules determines a strongly divisible module overSOE[[X0,···,Xf−1]];
(2) Forλ =−√
η, denotingζi+Xi=y−1k yi in the strongly divisible modules determines a strongly divisible module overSOE[[X0,···,Xf−1]];
DenoteM+(X0,· · ·,Xf−1)(resp.M−(X0,· · ·,Xf−1)) for the strongly divisible module overOE[[X0,· · ·,Xf−1]]
in (1) (resp. in (2)).
Proof. Prove only (1). We have
M+(X0,· · ·,Xf−1):=SOE[[X0,···,Xf−1]](E1,E2) with
Fil1M+(X0,· · ·,Xf−1) =SOE[[X0,···,Xf−1]](E1+p(ζi+Xi)iE2) +Fil1SOE[[X0,···,Xf−1]]M+(X0,· · ·,Xf−1), Mate,f(φ1) = λ·⃗1 ⃗0
λ(ζi+1+Xi+1)i −λ·⃗1
!
, [N]e=
⃗0 ⃗0 py−1k ·⃗1 ⃗0
! .
Indeed,M+(X0,· · ·,Xf−1)become a strongly divisible module if and only if specializingXiinmgives rise to a strongly divisible module overSOE. PuttingXi=mi∈m, we obviously have a strongly divisible moduleM+(m0,· · ·,mf−1).
Since T1st(M•(X0,· · ·,Xf−1)) for • ∈ {±} is a OE[[X0,· · ·,Xf−1]]-representation of GQ
p f, by the universal property of the deformation ring, then we haveOE-algebra morphisms
π1•:Rρ0→OE[[X0,· · ·,Xf−1]].
Lemma VI.8. For each• ∈ {±}, the morphismπ1•:Rρ0 →OE[[X0,· · ·,Xf−1]]is surjective.
Proof. LetR0be the quotient ring
OE[[X0,· · ·,Xf−1]]/(mE+ (X0,· · ·,Xf−1)2)∼=F[X0,· · ·,Xf−1]/(X0,· · ·,Xf−1)2.
By Nakayama’s lemma and by Lemma 6.1(2), it suffices to show that the induced morphismπ1•:Rρ0 → R0is surjective for each• ∈ {±}. Prove only the case•= +.Describe the strongly divisible module:
M0:=M+(X0,· · ·,Xf−1)/(mE+ (X0,· · ·,Xf−1)2) =SR0(E1,E2) with
• Fil1R0is generated byE1modulo Fil1SR0M0;
• Mate,f(φ1) = λ·⃗1 ⃗0 λ·(ζi+1+Xi+1)i −λ·⃗1
!
;
• N=0.
SinceM0is not defined over any proper subring ofR0, soπ1+ is surjective.
Lemma VI.9. For each• ∈ {±}, the morphismπ1•:Rρ
0 →OE[[X0,· · ·,Xf−1]]factors through the quo- tientq:Rρ
0→R(0,1),ψρ
0 .
Proof. The proof is the same as the proof of Lemma 6.3.
Proposition VI.10.
R(0,1)ρ
0 ∼′OE[[D0,· · ·,Df]]×OE[[D,X0,· · ·,Xf−1]]×OE[[D,X0,· · ·,Xf−1]], whereR∼′∏Ri meansRi’s are the irreducible components ofR.
Proof. By Theorem 4.12, we know that the relative dimension ofR(0,1)ρ
0 overOE is f+1. So there are at least two irreducible components by Lemma 6.4, and both of which are isomorphic toOE[[D,X0,· · ·,Xf−1]]
by Theorem 4.9.
By Lemma 6.6, ρ0 has a crystalline lift with parallel Hodge–Tate weights (0,1). By a result of [14], we know it is formally smooth. Since it also has a relative dimension f+1 overOE, we denote this crystalline deformation ring byOE[[D0,· · ·,Df]].
Note that the characteristic 0 closed points of these irreducible components exhaust all the 2-dimensional semi-stable lifts ofρ0with parallel Hodge–Tate weights(0,1).
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Acknowledgements
First of all, I would like to express my sincere appreciation to my advisor, Chol Park, for all I have learned from him. He gave ideas and assisted me in all steps of this thesis. He was the best advisor I could ever meet and the best mentor for study and research in mathematics. Through him, I was able to gain confidence and interest in mathematics.
Besides my advisor, I would like to thank the rest of my thesis committee: Hae-sang Sun and Peter Jaehyun Cho, for their insightful comments and questions. They were good math lecturers for me and good helpers for my graduation and study abroad preparation. I was happy to learn mathematics from the great mathematicians.
I also thank the UNIST Number Theory Group, especially the graduate students in the Arithmetic Geometry Lab: Jeonghyo Park, Seongjae Han, and Euntaek Lee, for the fun we have had. They always answered my mathematical questions for me.
Not only at UNIST, but also at Pusan National University, where I studied as an undergraduate, there are people to whom I should be grateful. I was able to become interested in Algebra and Number Theory through Mitsugu Hirasaka and Donghi Lee. Also, Donghoon Jang is helping me prepare for studying abroad. I express my deep gratitude to them.
There are people who must be included in these acknowledgements. I thank my father, Oik Cheon, and my mother, Eunjin An, for supporting me financially and spiritually throughout my life. I was able to become interested in mathematics because they taught me mathematics directly until high school. I also wish to extend my thanks to my younger brother, Hyeonwoo Cheon, for becoming my best friend in my life. Finally, I would like to mention a important person: my girlfriend and future wife Hyejin Noh. She has always supported my study and research in various ways. Had it not been for her help, I might have given up. She has always made me laugh, helped me not to get tired, and prayed for me.
Ulsan, January 2, 2023