Proof of the main result

4.1 ±Construction of H¹_r(K; )

LEMMA4.1.1. Let T be a crystallineZp-representation of GK. The inclusion H¹_f(K;T)H¹(K;T)induces, for every integer n1, an injection

H¹_f(K;T)=pⁿH¹_f(K;T),!H¹(K;T=pⁿT):

(4:1:1:1)

PROOF. The mapH¹_f(K;T)=pⁿH¹_f(K;T)!H¹(K;T=pⁿT) factors through H¹_f(K;T)=pⁿH¹_f(K;T)!H¹(K;T)=pⁿH¹(K;T)!H¹(K;T=pⁿT):

The second morphism is injective since it is part of the long exact sequence asso-ciated to 0!T!^pn T!T=pⁿT!0.

To show that the first one is injective we have to prove that H¹_f(K;T)\ pⁿH¹(K;T)pⁿH¹_f(K;T). Lethbe a class of a cocycle inH¹_f(K;T)\pⁿH¹(K;T).

By definition the image of h in H¹(K;T_ZpB_cris) is trivial and since H¹(K;TZpBcris) is aQp-vector space the same is true for the image ofh=pⁿ. This

exactly means thathbelongs topⁿH¹_f(K;T). p

DEFINITION4.1.2. For every integersn1,i2 f0;1;2g and 0rp 2 we define functors

Hⁱ_r(K; ):Rep_Zp…G_K0†^[0;r]_crys !Mod(Z_p);

Hⁱ_r(K; ):Rep_Z=pnZ…G_K0†^[0;r]_crys !Mod(Z=pⁿZ);

by putting, for everyTin Rep_Zp…G_K0†^[0;r]_crysandTin Rep_Z=pnZ…G_K0†^[0;r]_crys,

Hⁱ_r(K;T):ˆHⁱ(Fil^rC(M(Dcris(T)))); Hⁱ_r(K;T):ˆHⁱ(Fil^rC(M(D¹_cris(T))));

where M( ),D_cris( ) andD¹_cris( ) are the functors defined in 3.1.3 and 3.2.1, and Mod(Zp) denotes the category ofZp-modules.

REMARK4.1.3. For everyr_maxrp 2 we haveHⁱ_r(K;T)ˆHⁱ_rmax(K;T) and Hⁱ_r(K;T)ˆHⁱ_rmax(K;T). Indeed by Remark 3.1.4, we have

Fil^rM(D^[0;r]_cris(T))ˆFil^r(D^[0;r]_cris(T)_WS)

ˆX^r

jˆ0

Fil^j(D^[0;r]_cris(T))_WFil^{r j}Sˆ

ˆX^r

jˆ0

Fil^{j‡rmax r}(D^[0;r_cris^max](T))_WFil^{r j}Sˆ

ˆ X^rmax

jˆrmax r

Fil^j(D^[0;r_cris^max](T))WFil^{rmax j}Sˆ

ˆX^rmax

jˆ0

Fil^j(D^[0;r_cris^max](T))_WFil^{rmax j}Sˆ

ˆFil^rmaxM(D^[0;r_cris^max](T));

and soH¹_r(K;T)ˆHⁱ(Fil^rC(M(D^[0;r]_cris(T))))ˆHⁱ(Fil^rmaxM(D^[0;r_cris^max](T)))ˆH¹_rmax(K;T).

The case ofTis the same.

PROPOSITION4.1.4. Let T be inRep_Zp…G_K0†^[0;r]_crys, with0rp 2. The short exact sequence

0!T ^p!ⁿ T!T=pⁿT!0

induces two long exact sequences of abelian groups connected as in the commutative diagram below.

(4:1:4:1)

PROOF. The right vertical sequence is the long exact sequence of continous Galois cohomology. Set TˆT=pⁿT, DˆD_crys(T), DˆD¹_cris(T), MˆM(D_crys(T)) and MˆM(D¹_cris(T)). We have a short exact sequence 0!D!^pn D!D!0, cf. (3.1.3.1), and by applyingM( ) another short exact sequence 0!M!^pn M!M!0, cf. 3.2.1.

Therefore, by using Lemma 3.3.2, we have a short exact sequence of complexes 0!Fil^rC(M)!^pn Fil^rC(M)!Fil^rC(M)!0:

(4:1:4:2)

The left vertical sequence in (4.1.4.1) is the long exact sequence of cohomology as-sociated to the short exact sequence (4.1.4.2).

To constructt¹_st:H¹_r(K;T)!H¹(K;T=pⁿT) we proceed as follow.

H¹_r(K;T)ˆH¹(Fil^rC(M))

Ext¹_ModFI(S=pnS)^r(1;M) (by Prop:3:3:4)

!Ext¹_Rep

Z=pnZ…GK†^[0;r]_crys(Z=pⁿZ;T_st¹(M)) (by exactness of T¹_st; cf:2:3:7)

ˆExt¹_Rep

Z=pnZ…GK†^[0;r]_crys(Z=pⁿZ;T_jGK) (by Prop:3:2:2) Ext¹_{RepZ=pnZ…GK†}(Z=pⁿZ;T_jGK)

ˆH¹(K;T=pⁿT)

To construct the morphismtstwe proceed similarly.

H¹_r(K;T)ˆH¹(Fil^rC(M))

ˆExt¹_Mod(S)^r(1;M) (by Prop:3:3:4)

! Ext¹_Rep

Zp…GK†^[0;r]_crys(Z_p;T_st(M)) (by Cor:2:3:5)

ˆExt¹_Rep

Zp…GK†^[0;r]_crys(Z_p;T_jGK) (by Prop:3:2:2)

ˆH¹_f(K;T) (by definition)

Ext¹_RepZp…GK†(Zp;TjGK)ˆH¹(K;T) (by definition)

We get a morphism ofZ-modulestst:H¹_r(K;T)!H¹(K;T) with imageH¹_f(K;T).

Finally the commutativity for the upper left square in the diagram follows from the additivity ofTst, for the upper right square is evident and for the lower square it follows from the the commutativity of the diagram (2.3.7.1). p COROLLARY 4.1.5. The injection H¹_f(K;T)=pⁿH¹_f(K;T),!H¹(K;T=pⁿT), cf.(4.1.1.1), factors canonically through t¹_st:H¹_r(K;T=pⁿT)!H¹(K;T=pⁿT). We denote by

e:H¹_f(K;T)=pⁿH¹_f(K;T),!H¹_r(K;T=pⁿT) the monomorphism obtained.

PROOF. The morphism e is induced by the compositionp⁽¹⁾_r t_st¹. The factor-isation throught¹_st follows by the commutativity of the diagram (4.1.4.1). p

We now state and prove the main result of this article.

THEOREM 4.1.6. Let p3 be a prime integer and we fix an integer r with 0rp 2. We consider a crystallineZp-representations of GK0, T, with Hodge-Tate weights in[0;r]and assume e(rmax 1)p 1. Then for every integer n1, H¹_f(K;T)=pⁿH¹_f(K;T)is isomorphic to H¹_r(K;T=pⁿT)viae.

PROOF. LetTbe as in Theorem 4.1.6. To show thate:H¹_f(K;T)=pⁿH¹_f(K;T)! H¹_r(K;T=pⁿT) is an isomorphism we have to prove thatp⁽¹⁾_r in the diagram (4.1.4.1) is surjective. It is clear that ifH²_r(K;T)ˆ0 (or more generally ifH²_r(K;T) had not p-torsion) thenp⁽¹⁾_r would be surjective. We will prove below thatH²_r(K;T)ˆ0 ifeˆ1 orr1 (cf. Remark 4.1.11) but in general this does not seem to be true. What we do instead is:

(1) we prove that if H²_r(K;T=pⁿT)ˆ0 thenp⁽¹⁾_r is surjective, see Proposition 4.1.7;

(2) we prove H²_r(K;T=pⁿT)ˆ0 under the hypothesis e(r_max(T) 1)p 1, see Proposition 4.1.10.

PROPOSITION4.1.7. Let T be inRep_Zp…GK0†^[0;r]_cryswith rp 2. If H²_r(K;T=pⁿT)ˆ0 then p⁽¹⁾_r :H¹_r(K;T)!H¹_r(K;T=pⁿT)

is surjective.

PROOF. Set MˆM(D_cris(T)) andMˆM=pⁿM. It is enough to show that the reduction modpⁿ mapZ¹(Fil^rC(M))!Z¹(Fil^rC(M)) on 1-cocycles is surjective.

Leta2Mandg2Fil^{r 1}MV¹be such that r_M(a)ˆC_M(g);

where C_M(g)ˆ(W^M_{r 1}W₁)(g) g. Let a02M, (resp. g₀2Fil^{r 1}MV¹) be any lifting of a (resp. g). We have rM(a0) CM(g₀)ˆpⁿj₁, for some j₁2MV¹. Denote byj₁the class ofj₁inMV¹. SinceH²_r(K;T=pⁿT)ˆ0, there exist z₁2M andv₁2Fil^{r 1}MV¹, such that

r_M(z₁) C_M(v₁)ˆ j₁:

Take any lifting z12M (resp. v12Fil^{r 1}MV¹) of z1 (resp. v1) and set a₁ˆa₀‡pⁿz₁ and g₁ˆg₀‡pⁿv₁. We have r_M(z₁) C_M(v₁)ˆ j₁‡pⁿj₂ for somej₂2MV¹and

r_M(a₁) C_M(g₁)ˆpⁿj₁ pⁿj₁‡p²ⁿj₂ˆp²ⁿj₂:

SinceMand Fil^{r 1}MV¹arep-adically complete and separated, it is clear that we

can finish by induction. p

4.1.8

-Let us introduce some notations on the ring S=pS. We have E(u)u^emodp.

For any integersj0 and 0de 1 we denote byu^hej‡dithee-partial divided power of u(it satisfies j!u^hej‡diˆu^ej‡d). We set V¹_S=pSˆS=pS_SV¹_SˆS=pSdu.

The canonical differential d:S!V¹_S composed with the projection V¹_S!V¹_S=pS factors through a derivation S=pS!V¹_S=pS that we denote still by d. For any f 2S=pSwe will writedf ˆ_du^d (f)du.

LEMMA 4.1.9. Let f :ˆau^hej‡di be in S=pS, with j0, 0de 1 and a2k. Then f is integrable in S=pS, i.e. there is g2S=pS such that_du^d (g)ˆf , if and only if:

- ford6ˆe 1, p does not divide ej‡d‡1;

- fordˆe 1, p does not divide e.

PROOF. It is obvious that the conditions are sufficient. Indeed ifd6ˆe 1 andp does not divideej‡d‡1, a primitive offis _ej‡d‡1^a u^hej‡d‡1i; whendˆe 1 andp6 je, a primitive is^a_eu^he(j‡1)i.

The conditions of the lemma are also necessary because if they are not satisfied,

we have_du^d(u^hej‡d‡1i)ˆ0. p

PROPOSITION4.1.10. Let T be inRep_Zp…G_K0†^[0;r]_crys, w ith0rp 2. For every n1, theZ=pⁿZ-module H²_r(K;T=pⁿT) is of finite type and if e(rmax(T) 1) p 1, then H²_r(K;T=pⁿT)ˆ0.

PROOF. Thanks to Remark 4.1.3 we may assumerˆr_max(T). Thenˆ1 case implies the casen2 by induction onnanddeÂvissageon the exact sequence

0!pⁿT=p^n‡1T!T=p^n‡1T!T=pⁿT!0:

Let us compute H²_r(K;T=pT). Set TˆT=pT, DˆD_cris(T) and DˆD¹_cris(T)ˆ D=pD, cf. 3.1.3. By construction the smallest jump in the filtration FilⁱD is r_min(D)ˆ0 and the biggest jump is r_max(D)ˆr r_min(T), where r_min(T) is the smallest Hodge-Tate weight of T. Put M:ˆM(D)=pM(D)2ModFI(S=pS)^r and C:ˆC_M:ˆ(W_r^M₁W₁) Id. The idea is to show first that

C:ˆCoker(C:Fil^{r 1}M_SV¹!M_SV¹)

is a finite dimensionalFp-vector space, which implies of course thatH²_r(K;T=pT) is

also finite. Then, we show that, under the hypothesis e(r 1)p 1, every ele-ment in this cokernelCis in the image ofr_M(see the diagram below), which means H²_r(K;T=pT)ˆ0.

In the particular case r1, the cokernel C is zero. Indeed, in this case, Fil^{r 1}MˆM so that the operatorW_{r 1}W₁:MV¹!MV¹ can be iterated.

The divided FrobeniusW₁:V¹!V¹is nilpotent:

W^m₁(du)ˆu^{pm 1}duˆh^pm_e¹i

!^upm^pm1¹

‰ e Š^!du; thereforeW_{r 1}W₁is also nilpotent,Cis invertible andCis zero.

For the rest of the proof we supposer2. Let us prove dimFpCe(r 1)dimFp(T=pT):

(4:1:10:1) We write

M_SV¹ˆD_WS=pS_SSdu:

Anyg inM_SV¹ is a finite sum of elements of the formvu^hej‡didu, where v2D,j0 and 0de 1. We want to find conditions such thatgbelongs to the image ofC. It is enough to treat the casegˆvu^hej‡didu, andv6ˆ0. Letsbe the weight such thatv2Fil^sDnFil^s‡1D; thengbelongs to Fil^{r 1}M_SV¹if and only if jr 1 s.

We compute

(W_{r 1}W₁)(g)ˆW_s(v)W_{r 1 s}(u^hej‡di)u^{p 1}du

ˆW_s(v)1W_{r 1 s}…^uej‡d_j! †u^{p 1}du

ˆW_s(v)1 u^p(ej‡d)

j!p^{r 1 s}u^{p 1}du

ˆW_s(v)u^hp(ej‡d)i

p(ej‡d) e

h i

!

j!p^{r 1 s} u^{p 1}du:

Sinceh^p(ej‡d)_e i

where, for every m,sm denotes the sum ofp-adic digits ofm. Clearly, we have s_pjˆs_jand so conclusion the element gˆvu^hej‡didu belongs to the image ofC for every jr 1 and the inequality (4.1.10.1) follows.

To proveH²_r(K;T=pT)ˆ0 it is enough to show that everygˆvu^hej‡didu inMV¹, withv2D,j(r 2) and 0de 1, is in the image ofr_M. Recall that everyv2Dis a horizontal section (by definition ofr_M) andeis coprime withp (because ofr2 ande(r 1)p 1). By Lemma 4.1.9 we may assume moreover that 0de 2 andej‡d‡1ˆpmfor some integerm1. We have

pmˆej‡d‡1e(r 2)‡d‡1(p 1) e‡d‡1p 2;

which shows that there are no such elements. p

REMARK4.1.11. It is clear by the proof of Proposition 4.1.10 that if eˆ1 or r1, then H²_r(K;T)ˆ0. Indeed, if eˆ1, every element in S is integrable, and so rM is surjective. If r1, then Fil^{r 1}MˆM, which implies that CM:ˆW_{r 1}W₁ Id is an isomorphism: the series P_‡1

iˆ0 (W_{r 1}W₁)ⁱ converges to an inverse of C_M. This was the idea of the proof in [13]. We do not know if under the hypothesis of Proposition 4.1.10, we have H²_r(K;T)ˆ0 or not.

End of the proof of Theorem 4.1.6. p COROLLARY4.1.12. Let p3 and let T₁ and T₂ be two crystalline Z_p -representations of G_K0with Hodge-Tate weights in[0;r][0;p 2]and assume e(r 1)p 1. Then for every morphism (resp. isomorphism) i:T1=pⁿT1! T2=pⁿT2ofZ=pⁿZ[GK0]-modules, there exists a morphism (resp. isomorphism)~i ofZ=pⁿZ-modules making the following diagram commutative.

(4:1:12:1)

PROOF. By the constructions above (cf. (4.1.4.1) and Corollary 4.1.5) we have the following commutative diagram

wheree₁ande₂are isomorphisms by Theorem 4.1.6. Set~iˆe₂¹H¹_r(K;i)e₁. p

4.2 ±Some complements

Let us finish this section with a variant of Corollary 4.1.12 and some ex-amples.

PROPOSITION4.2.1. Let T1and T2be two crystallineZp-representations of GK0

with Hodge-Tate weights in [0;r][0;p 2] and assume erp 2. Then for every morphism (resp. isomorphism)i:T₁=pⁿT₁!T₂=pⁿT₂ ofZ=pⁿZ[G_K ]-mod-ules, there exists a morphism (resp. isomorphism)~iofZ=pⁿZ-modules making the following diagram commutative.

(4:2:1:1)

REMARK4.2.2.

(1) The hypothesiserp 2 is more restrictive than that of Corollary 4.1.12 (it implies (e;p)ˆ1 even forrˆ1), but Proposition 4.2.1 gives functoriality forG_K-morphismes.

(2) The hypothesis erp 2 is necessary. Indeed consider the following counterexample for eˆp 1 and rˆ1. Take KˆQ_p(m_p(Q_p)), T₁ˆZ_p(1) andT₂ˆZ_p. Clearly T₁ and T₂ are congruent modulop but H¹_f(K;T1)ZpFp and H¹_f(K;T2)ZpFp do not even have the same di-mension over Fp.

PROOF. The proof is similar but simpler than that of Corollary 4.1.12 so we only sketch it. Foriˆ1;2 setDiˆDcris(Ti) andMiˆM(Di); we haveTi=pⁿTiˆ T_st¹(M_i=pⁿM_i). Under the hypothesiserp 2, the functorT¹_st is fully faithful (cf. [8, TheeÂoreÁme 1.0.4]) therefore the morphism i is induced by a unique

morphism i⁰:M1=pⁿM1!M2=pⁿM2 in ModFI(S=pⁿS)^r and we have a commu-tative diagram

wheree1ande2are isomorphisms by Theorem 4.1.6. Set~iˆe₂¹H¹(Fil^rC(i⁰))e1. p For an abelian groupAwe denote byA[n] then-torsion subgroup ofA.

PROPOSITION4.2.3. Let2rp 2be an integer and assume ep, then:

(1) The groups H²_r(K;Zp(r))[pⁿ]and H²_r(K;Z=pⁿZ(r))are non-zero.

(2) The monomorphism e:H¹_f(K;Z_p(r))=pⁿH¹_f(K;Z_p(r)),!H¹_r(K;T=pⁿT) de-fined in Corollary4:1:5is not surjective.

PROOF. The statement (2) follows from (1) by using (4.1.4.1). Let us prove (1):

it is enough to treat the case nˆ1. Set DˆDcris(Zp(r)). We have Dˆ D_cris(Z_p( r)(r))ˆHom_Zp[GK0] Z_p;A_cris

ˆW, Fil⁰DˆD, Fil¹Dˆ0, andW^D₀ ˆs.

HenceM(D)ˆD_WSˆS, Fil^rM(D)ˆFil^rSasS-modules, andM(D)ˆ1(r), cf.

2.3.6. We haveH¹_r(K;Zp(r))ˆH¹(Fil^rC(1(r))) andH¹_r(K;Fp(r))ˆH¹(Fil^rC(1(r))), where1(r)ˆ1(r)=pⁿ1(r). Denote byjthe class ofu^{p 1}_SduinH²(Fil^rC(1(r))).

We haved(u^p)ˆp(u^{p 1}Sdu), sopjˆ0. We claim that the class ofu^{p 1}duin H²(Fil^rC(1(r))) is not zero, which implies at the same timej6ˆ0,H²_r(K;Fp(r))6ˆ0 and H²_r(K;Zp(r))[p]6ˆ0. Let us prove this claim by contradiction: let a be in 1(r)ˆS=pSandgdube in Fil^{r 1}1(r)V¹ˆFil^{r 1}(S=pS)SV¹, such that

d

du(a)du (W_{r 1}W₁)(gdu)‡gduˆu^{p 1}du;

or equivalently

d

du(a) u^{p 1}W_{r 1}(g)‡gˆu^{p 1}: Since r 11 and ep, by writing gˆP

jr 1g_ju^hej‡di (with g_j2k and 0de 1), and by taking reduction modulo Fil^p(S=pS), we get that there exists a2S=(pS‡Fil^pS)ˆk[u]=u^p, such that

d

du(a)ˆu^{p 1}

which is impossible. p

REMARK4.2.4. Let 2rp 2 be an integer, assumeep and that the residue field ofK is finite. We have

H¹_f(K;Z_p(r))=pH¹_f(K;Z_p(r)) ,!^6e H¹_r(K;F_p(r)) ^t!^1st H¹(K;F_p(r));

and by [1, Example 3.9, pg. 359]H¹_f(K;Z_p(r))ˆH¹(K;Z_p(r)); in particulart¹_st is surjective but not injective and H¹_f(K;Zp(r))=pH¹_f(K;Zp(r)) can be identified to t¹_st(H¹_r(K;Fp(r))). We suspect that this phenomenon happens more generally.

PROPOSITION 4.2.5. Let T be a crystalline Z_p-representation of G_K0 with Hodge-Tate weights in[0;r],0rp 2. We have an isomorphism

t¹_st:H⁰_r(K;T=pⁿT)! (T=pⁿT)^GK:

PROOF. SetDˆDcris(T) andMˆM(D). A direct computation shows H⁰_r(K;T=pⁿT)ˆHom⁰_Mod(S)^r 1=pⁿ1;M=pⁿM

ˆFil^r((S=pⁿS)^rˆ0_WnD=pⁿD)^WrˆId Fil^r(D=pⁿD)^WrˆId(T=pⁿT)^GK0:

The last inclusion is strict in general. Indeed we have the exact sequence 0!H⁰_r(K;T) ^p!ⁿ H⁰_r(K;T)!H⁰_r(K;T=pⁿT)!^d H¹_r(K;T) ^p!ⁿ H¹_r(K;T)! induced by the short exact sequence 0!M ^p!ⁿ M!M=pⁿM!0:By the theo-rem of Liu (2.3.4)Tstis an equivalence of categories and therefore we have

H⁰_r(K;T)ˆHom⁰_Mod(S)^r 1;M

ˆHom_Mod(S)^r 1;M

! T^GK

andH¹_r(M)H¹_f(K;T), cf. (4.1.4.1). By definitionH¹_f(K;T) contains all the torsion of H¹(K;T), so the kernel ofH¹_r(K;T) ^p!ⁿ H¹_r(K;T) is equal to thepⁿ-torsion subgroup H¹(K;T)[pⁿ] ofH¹(K;T). As in the proof of (4.1.4.1), the exact functorsT_standT_st¹ induce the commutative diagram

where the rows are exact. This proves the claim. p

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