aa r X i v : . [ m a t h . N T ] M a y The higher Hilbert pairing via ( φ, G )-modules
Sarah Livia ZerbesOctober 30, 2018
Abstract
Following the strategy in [6], we prove a Tate duality for higher dimen-sional local fields of mixed characteristic (0 , p ), p = 2. The main tool isthe theory of higher fields of norms as developed in [1] and [7]. Assumingthat p is not ramified in the basefield, we then use this construction todefine the higher Hilbert pairing. In particular, we show that the Hilbertpairing is non-degenerate, and we re-discover the formulae of Br¨ucknerand Vostokov. Contents ( φ, G ) -modules 3 A cris . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.4 Differentials, residues and duality . . . . . . . . . . . . . . . . . . 52.5 The equivalence of categories . . . . . . . . . . . . . . . . . . . . 72.6 The module Z p (2) . . . . . . . . . . . . . . . . . . . . . . . . . . 72.7 Calculation of the Galois cohomology . . . . . . . . . . . . . . . . 82.8 Construction of the pairing . . . . . . . . . . . . . . . . . . . . . 10 Introduction
Let p be an odd prime, and let K be a d -dimensional local field of mixedcharacteristic (0 , p ) Denote by G K the absolute Galois group Gal( ¯ K/K ). For n ≥
1, denote by µ p n the group of p n th roots of unity.This paper consists of two parts. In the first part, we prove a higher Tateduality for the G K -module µ p n :- Theorem 1.1.
Let K be a d -dimensional local field of mixed characteristic (0 , p ) , and denote by G K its absolute Galois group. Let F be the maximal al-gebraic extension of Q p contained in K , and assume that O K /O F is formallysmooth. Then for i ∈ { , . . . , d + 1 } and all n ∈ N we have a perfect pairing H i ( G K , µ ⊗ ip n ) × H d +1 − i ( G K , µ ⊗ d − ip n ) → Q p / Z p . Remark.
The above pairing should certainly be the same as the cup productpairings, but this is not that easy to show.To prove Theorem 1.1, we follow the strategy of Herr in [6] and express theGalois cohomology groups in terms of the ( φ, G )-module of µ p n . We will alsoprove a higher dimensional Tate isomorphim:- Theorem 1.2.
Let K be a d -dimensional local field of mixed characteristic (0 , p ) . Let F be the maximal algebraic extension of Q p contained in K , andassume that O K /O F is formally smooth. Then there is a canonical isomorphism H d +1 ( G K , µ ⊗ dp ∞ ) ∼ = Q p / Z p . In the second part of the paper, we assume that p is prime in K , and we useTheorem 1.1 to define a pairing K d ( K n ) × K ( K n ) → µ p n . Composing it with the natural multiplication map K ( K n ) × d → K d ( K n ), weobtain a pairing V n : K ( K n ) × ( d +1) → µ p n which factors through V n : (cid:0) K ( K n ) /p n (cid:1) × ( d +1) → µ p n . In Section 4 we give a an explicit description of V n :- For 1 ≤ i ≤ d + 1, let α i ∈ O × K such that α i ∼ = 1 mod ¯ π n , and let F i ( X ) ∈ A + K such that h n ( F i ) = α i .Let f i ( X ) = (1 − φp ) log F ( X ). Theorem 1.3.
The pairing V n is non-degenerate. Moreover, we have V n ( α , . . . , α d +1 ) = µ Tr Res πn,T ,...,Td (Φ) p n , here Φ is given by the formula Φ = 1 π d +1 X i =1 ( − d +1 − i p d +1 − i f i ( π n ) d log F ( π n ) ∧ · · · ∧ d log F i − ( π n ) ∧ d log F φi +1 ( π n ) ∧ · · · ∧ d log F φd +1 ( π n ) . Comparing these formulae with the explicit descriptions of the higher Hilberpairing of Br¨uckner and Vostokov (c.f. [3] and [8]), we get the following result:-
Corollary 1.4.
The pairing V n is the higher Hilbert pairing. For proving Theorem 1.3, we follow the strategy of Benois in [2].
Remarks. (1) To keep the notation as simple as possible, we will prove theabove results for local fields of dimension 2. However, the proofs generalizewithout problems to local fields of arbitrarily high dimension.(2) Theorem 1.1 can certainly be generalized to an arbitrary Z p - of G K . We willdeal with this in a different paper. ∗ For a 2-dimensional local field K with ring of integers O K , let k K ∼ = F p (( T ))be the residue field. ∗ For a ( φ, G )-module M , we sometimes denote the action of the Frobeniusoperator on M by φ M . ∗ For a 2-dimensional local field K , let G K = Gal( ¯ K/K ). I am very grateful to John Coates for his interest and encouragement. Also, Iwould like to thank Otmar Venjakob, Kay Wingberg and Ivan Fesenko for somehelpful comments. Finally, I would like to very warmly thank Guido Kings forhis invitation to Regensburg in Spring 2007 when part of this paper was written. ( φ, G ) -modules Let K be a 2-dimensional local field of mixed characteristic (0 , p ), and let F bethe maximal algebraic extension of Q p contained in K . Let k F be the residuefield of F and let ω F be a uniformizer of F . Assume that O K /O F is formallysmooth, i.e. ω F is a uniformizer of K . Let X be a unit in K whose reduction¯ X is a p -basis for the residue field k K of K , so k K ∼ = k b (( X )) for some finiteextension k b of k F . Fix an algebraic closure ¯ K of K . Let ( ξ i ) i ≥ be compatiblesystem of primitive p i th roots of unity, and let ( X i ) i ≥ be a compatible systemof p i th roots of X . Denote by µ p i the group of p i th roots of unity.Let K i = K ( µ p i , X i ) and K ∞ = S K i . Also, let F i = F ( µ p i ).3 emma 2.1. The extension K ∞ is a -dimensional p -adic Lie extension of K .More precisely, we have Gal( K ∞ /K ) ∼ = Γ ⋊ Γ , where Γ = Gal( K ∞ /K ( µ p ∞ )) ∼ = Z p . and Γ = Gal( K ( µ p ∞ ) /K ) is isomorphic (via the cyclotomic character χ ) to anopen subgroup of Z × p . Let γ and γ be topological generators of Γ and Γ , respectively. Let a ∈ Z p such that γ γ = γ a γ . (1) Note.
We have a = χ ( γ ) ∈ Z × p . It follows that in particular we have γ γ a = γ γ . (2)Let E F be the field of norms of the tower ( F i ) i ≥ , and let k F be its residuefield. Let ¯ π F be a uniformizer of E F , so E F ∼ = k F ((¯ π F )). Let E K be the field ofnorms of the tower ( K i ). Let ǫ = (1 , ξ , ξ , . . . ) and T = ( X i ) i ≥ ∈ E K . Define¯ π = ǫ −
1. Let k = k b k F . Lemma 2.2.
The field E K is given by E K ∼ = k F ((¯ π F )) ˆ ⊗ k F k b (( T )) ∼ = k (( T ))((¯ π F )) . Proof.
See the section on Kummer towers in [7]. Let A F be a lift of E F to characteristic 0, so A F ∼ = W ( k F )[[ π F ]][ π − F ] ∨ , where π F is a lift of ¯ π F . Let φ be a lift to A F of the Frobenius operator commutingwith the action of Γ . Let T = [ T ]. Define A K = W ( k )[[ T ]][ T − ] ∨ [[ π F ]][ π − F ] ∨ . Then A K is a lift of E K to characteristic 0. Let B K = A K [ p ] be its field offractions. Note that A F ⊂ A K . Define a lift of Frobenius to A K by φ ( T ) = T p . Note that φ commutes with the action of G on A K . Define N ∈ Z p by γ ( π F ) = π F ,γ ( T ) = ( π + 1) N T. One can show that N ∈ Z × p since X is a p -basis of k K .Note that A K is a free finitely generated module over φ ( A K ) of degree p .It follows that we can define a left inverse ψ of φ by the formula φ ( ψ ( x )) = 1 p Tr A K /φ ( A K ) ( x ) . roposition 2.3. Let E be an algebraic closure of E K . Then we have anisomorphism of Galois groups Gal( E / E K ) ∼ = Gal( ¯ K/K ∞ ) . Proof.
See [7].Let A be a lift of E containing A K . Then the actions of φ and ψ can beextended uniquely to A (c.f. [7]). A cris Let C K be the p -adic completion of ¯ K , and let O C K be its ring of integers.Let ˜ E be the set of sequence x = ( x (0) , x (1) , . . . ) of elements in O C K satisfying( x ( i +1) ) p = x ( i ) . Then ˜ E has a natural structure as a ring of charactristic p . For n ≥
1, let ǫ n = ( ζ ( n ) , ζ ( n +1) , . . . ) be the p n the root of ǫ in ˜ E . Let A inf = W (˜ E )be the ring of Witt vectors of ˜ E , φ the Froenius of A inf , and if x ∈ ˜ E , thendenote by [ x ] its Teichm¨uller representative in A inf . Then the homomorphism θ : A inf → O C K X p n [ x n ] → X p n x (0) n is surjective and its kernel is a principal ideal with generator ω = [ ǫ ] − ǫ ] − . Let B inf = A inf ( p − ). Note that θ extends to a homomorphism B inf → C K . De-fine B ∇ +dR = lim ←− B +inf / (ker θ ) n , and extend θ by continutiy to a homomorphism B ∇ +dR → C p . This makes B ∇ +dR into a discrete valuation ring with maximal idealker( θ ) and residue field C K . The action of G K on B +inf extends by continuity toa continuous action on B ∇ + dR .Let A cris be the subring of ( B ∇ dR ) + consisting of the elements of the form P ∞ n =0 a n ω n p ! , where a n is a sequence of elements in A inf tending to 0 as n → + ∞ . Let Ω A K be the module of continuous Z p -linear 1-differentials of A K . Note thatwe have an isomorphism of A K -modulesΩ A K ∼ = A K dπ F ⊕ A K dT. Let Ω A K be the module of continuous Z p -linear 1-differentials of A K , and letΩ A K = ^ Ω A K . Lemma 2.4.
We have an isomorphism of A K -modules Ω A K ∼ = A K dπ F ∧ dT. orollary 2.5. If K ′ is a finite separable extension of K , then the natural map A K ′ ⊗ A K Ω A K → Ω A K ′ is an isomorphism.Proof. Clear.
Corollary 2.6. If K ′ = K ⊗ F F ′ for some finite unramified extension F ′ of F ,then the natural map W ( k F ′ ) ⊗ W ( k F ) Ω A K → Ω A K ′ is a W ( k F ) -linear isomorphism. It follows from Lemma 2.4 that if ω ∈ Ω A K , then there exist a i,j ∈ Z p suchthat ω = ( P a i,j T i π jF ) dπ F ∧ dT . Definition.
Define the residue mapRes : Ω A K → Z p , Res( ω ) = a − , − . Since φ is a lift of the Frobenius operator, we have φ ( π F ) = uπ pF for some u ∈ A × K satisfying u ∼ = 1 mod p . Lemma 2.7.
For any λ ∈ A K , we have Res( φ ( λ ) dφ ( π F ) ∧ dφ ( T )) = p φ (Res( λdπ F ∧ dT )) . Proof.
It is sufficient to prove the formula for λ = π F T . Write u = 1 + pa forsome a ∈ A K . We have φ ( T ) = T p , so φ ( λ ) dφ ( π F ) ∧ dφ ( T ) = p π F T dπ F ∧ dT + puT du ∧ dT. But uT du ∧ dT = T − P d ( ( − r +1 p r r a r ) ∧ dT . It is easy to see that the coefficientof π − F in P d ( ( − r +1 p r r a r ) is 0, which finishes the proof. Definition.
Let M be a torsion ( φ, G )-module. Define˜ M = Hom A K ( M, B K / A K ⊗ A K Ω A K ) . Lemma 2.8.
The residue map induces an isomorphism
TR : ˜ M → M ∨ . roof. We imitate the proof of Lemma 1.3 in [6]. By continuity, the residuegives a homomorphism from ˜ M to M ∨ . Now Ω A K is a free A K -module of rank1. The ring A K is principal and the A K -module B K / A K ⊗ A K Ω A K is divisibleand hence injective, and the functor which associates ˜ M to M is exact. Also,the functor which to M associates M ∨ is exact, so by the snake lemma we canassume without loss of generality that M is an A K -module of length 1 and hencea 1-dimensional vector space over E K . By choosing a basis, we can thereforeassume that M = E K . Note that we can identify E ∨ K with Hom Z p ( E K , F p ) and f E K with Hom E K ( E K , Ω A K /p Ω A K ). We need to show that the natural map f E K = Hom E K ( E K , Ω A K /p Ω A K ) → Hom Z p ( E K , F p )induced by the residue is bijective. Now if f ∈ f E K is non-zero, then sincethe image of f is an E K -vector space of dimension 1 and the residue map issurjective, it is clear that TR( f ) is non-zero. Now let a ∈ Hom Z p ( E K , F p ). Forall m, n ∈ Z , let α m,n = a ( T m ¯ π nF ). Since a is continuous, there exists N ∈ N such that α m,n = 0 for all n ≥ N and all m ∈ Z . Define ω = ( X n ≥− N X m ∈ Z α − m, − n T m − π n − F ) dT ∧ dπ. Then for all m, n ∈ Z , the class mod p of Res( T m π n ω ) is α m,n . It follows thatif we define f ∈ f E K by f (1) = ω mod p , then TR( f ) = a . Denote by G K the absolute Galois group Gal( ¯ K/K ). Definition. A Z p -representation of G K is a Z p -module V of finite type equippedwith a continuous linear action of G K . If V is annihilated by a power of p , thenit is called a p -torsion module. Theorem 2.9.
The functor V → D ( V ) = ( V ⊗ Z p A ) H K gives an equivalence ofcategories ( Z p -representations of G K ) → ( ´etale ( φ, G ) -modules over A K ) , and an essential inverse is given by D → ( A ⊗ A K D ) φ =1 .Proof. See [1] or [7]. Z p (2) For an element a ∈ E K , denote by [ a ] its Teichm¨uller representative. Recallthat if K contains a primitive p th root of unity, then [ ǫ ] = π + 1 is an elementof A K . To simplify notation, let Ω( K ) = Ω A K .7 emma 2.10. Let K ′ = K ( µ p ) , and define an A -linear map ρ K ′ : A ⊗ Z p Z p (1) → A ⊗ A K ′ Ω( K ′ ) ,λ ⊗ ǫ → λ ⊗ d log[ ǫ ] ∧ d log T. Then ρ K ′ is an isomorphism of A -modules which does not depend on the choiceof the generator ǫ of Z p (1) .Proof. Imitate the proof of Lemma 3.6 in [6].Note that we can give Ω( K ) the structure of a ( φ, G )-module by defining φ Ω( K ) ( xdy ∧ dz ) = 1 p φ ( x ) dφ ( y ) ∧ φ ( z ) ,g ( xdy ∧ dz ) = χ ( g ) g ( x ) dg ( y ) ∧ dg ( z )for g ∈ G . Proposition 2.11.
With this stucture as a ( φ, G ) -module, Ω( K ) is isomorphicto D ( Z p (2)) .Proof. Let K ′ = K ( µ p ). By Corollary 2.5 the natural map Ω A K → Ω A K ′ is in-jective. Composing it with the natural Gal( ¯ K/K )-equivariant injection Ω A K ′ → A ⊗ A K ′ Ω A K ′ gives an A K -linear G K -equivariant map Ω A K → ( A K ⊗ A K ′ Ω A K ′ ) H K .Explicit calculation shows that this map is also surjective. Composing it withthe restriction of ρ − K ′ to the points fixed by H K gives an isomorphism of A K -modules between Ω A K and D K ( Z p (2)) which commutes with the action of G K .It is easy to see that it also commutes with the action of φ , which finishes theproof. Let V be a Z p -representation of G K , and denote by D the corresponding ( φ, G )-module D K ( V ). Define the following complex:- C φ,γ ,γ ( D ) : 0 ✲ D f ✲ D ⊕ f ✲ D ⊕ f ✲ D ✲ , (3)where the maps f i are defined as follows:- f : x → [( φ − x, ( γ − x, ( γ − x ] ,f : ( x, y, z ) → [( φ − y − ( γ − x, ( φ − z − ( γ − x, ( γ − y − ( γ γ a − γ − − z ] ,f : ( x, y, z ) → ( γ − x − ( γ γ a − γ − − y − ( φ − z. C ψ,γ ,γ ( D ) : 0 ✲ D g ✲ D ⊕ g ✲ D ⊕ g ✲ D ✲ , (4)where the maps g i are defined as follows:- g : x → [( ψ − x, ( γ − x, ( γ − x ] ,g : ( x, y, z ) → [( ψ − y − ( γ − x, ( ψ − z − ( γ − x, ( γ − y − ( γ γ a − γ − − z ] ,g : ( x, y, z ) → ( γ − x − ( γ γ a − γ − − y − ( ψ − z. Definition.
Denote by H iφ,γ ,γ ( D ) (resp. H iψ,γ ,γ ( D )) the cohomology groupsof the complex C φ,γ ,γ ( D ) (resp. C ψ,γ ,γ ( D )). Proposition 2.12.
Let V = Z p (1) or µ p n , and let D be its ( φ, G ) -module.Then for all ≤ i ≤ , we have isomorphisms H i ( G K , V ) ∼ = H iφ,γ ,γ ( D ) ∼ = H iψ,γ ,γ ( D ) . Remark.
This result can certainly also be shown to be true for a general Z p -representation V of G K . However, in the general case the argument is technicallymuch more complicated, and since our main interest is the construction of theHilbert pairing, we restrict ourselves to the case above. We will prove the generalcase in [9]. Proof.
Scholl [7] and Andreatta [1] have shown that we have isomorphisms H i ( G K , V ) ∼ = H iφ,γ ,γ ( D ). It is therefore sufficient to explicitely give isomor-phisms ι i : H iψ,γ ,γ ( D ) ∼ = H iφ,γ ,γ ( D ). i = 3:- Let x ∈ D . Since ψ is surjective, we can choose u ∈ D such that ψ ( u ) = x . Define ι ( x ) = u . This is well-defined:- If u ′ also satisfies ψ ( u ′ ) = x ,then a = u − u ′ ∈ D ψ =0 . Write u = P i ∈ Z f i ( π F ) T i . Then in particular ψ ( f ( π )) = 0. In Proposition I.5.1 in [4], it is shown that γ − D F ( V ) ψ =0 . Let h = ( γ − − ( f ). For j = 0, let α j = ( γ − T j T j ∈ A ∗ F .Let v = P j =0 α j f j ( π ) T j . Then u = ( γ − h + ( γ − v and hence is zero in H φ D ,γ ,γ ( D ). i = 2:- Let [ x, y, z ] ∈ D ⊕ satisfy ( γ − x − ( γ − y − ( ψ − z = 0. Denote ι ([ x, y, z ]) by [ u, v, w ]. Write ( φ − w = P i ∈ Z a i f i ( π ) T i . Note that f ( π F ) ∈ D f ( Z p (1)) ψ =0 , so by Proposition I.5.1 in [4], there exists h ∈ D F ( Z p (1)) suchthat ( γ − h = f . Also, when i = 0, then α i = ( γ − T i T i is invertible in A F .Define u = x + h ( π ), v = y + P j =0 f j ( π ) α − j T j and w = z .9 = 1:- Let y, z ∈ D satisfy( γ − y = ( γ γ a − γ − − z (5)and ( ψ − y = ( ψ − z = 0. We need to show that there exists x ∈ D suchthat ( φ − y = ( γ − x and ( φ − z = ( γ − x . How do we construct this x ? Write ( φ − z = P i ∈ Z f i ( π F ) T i . By Proposition I.5.1 in [4], ( γ − − f iswell-defined. When i = 0, then α i = ( γ − T i T i ∈ A × F . Define x = ( γ − − f ( π F ) + X i ∈ Z α − i f i ( π F ) T i . Using (5) it is not difficult to see that x has the required properties. i = 0:- ι = id. When x ∈ D satisfies ψx = x , γ γ n − γ − x = x and γ x = x , thenit is easy to see (using again the result from [4] mentioned above) that φx = x .It is not difficult to see that the above maps are indeed isomorphisms. Remark. If γ ′ and γ ′ is a different pair of topological generators of Γ and Γ ,then the complexes C φ,γ ,γ and C φ,γ ′ ,γ ′ are quasi-isomorphic. More generally,we can replace γ and γ by ω γ and ω γ for any ω , ω ∈ Λ( G ) × . Definition.
The Pontryagin dual M ∨ of an ´etale ( φ, G )-torsion module M isdefined as the continuous homomorphisms M ∨ = Hom Z p ( M, Q p / Z p ) . As shown in Lemma 2.8, the map TR induces an isomorphism ˜ M → M ∨ . Wecan therefore give M ∨ the structure of a ( φ, G )-module. Denote the operationof Frobenius on it by φ M ∨ . We quote the following result from [6]:- Proposition 2.13.
Let C = ( M i , d i : M i → M i +1 ) be a cochain comlex ofabelian groups which are compact and locally separated (with M i in degree i ).Suppose that the d i are strict homomorphisms with closed images. Then C ∨ =( N i := ( M i ) ∨ , d i := t d i − : N i → N i − ) is a chain complex of abelian groupswhich are compact and locally separated (with N i in degree i ). The d i are stricthomomorphisms with closed image, and for all i , we have natural isomorphisms α i : H i ( C ∨ ) ∼ = ( H i ( C )) ∨ . In order to be able to apply Proposition 2.13 to the complex C φ,γ ,γ ( M ),we need the following result:- Lemma 2.14. If M is an ´etale ( φ, G ) -torsion module over A K , then the imageof φ − contains a compact neighbourhood of on which φ − induces a bijectionand hence a homeomorphism by compactness. roof. Reduce to the 1-dimensional case, using Proposition 2.4 in [5].For the rest of this section, let M = D ( µ p n ). Note that as a φ -module, we have M ∼ = A K mod p n . In particular, this implies that ψ M is defined. Proposition 2.15.
The map ψ : M ∨ → M ∨ : f → f ◦ φ M agrees with ψ M ∨ .Proof. Imitate the argument in Section 5.5.1 in [6].Let 0 ≤ i ≤
3, and define an isomorphism v i : ( M ∨ ) ⊕ ( i ) → ( M ⊕ ( i )) ∨ , ( g j ) ≤ j ≤ ( i ) → ⊕ ≤ j ≤ ( i ) g j . Lemma 2.16.
For all ≤ i ≤ , the v i induce isomorphisms (which we willalso denote by v i ) H i ( C ψ M ∨ ,γ − ,γ − ( M ∨ )) → H − i ([ C φ M ,γ ,γ ( M )] ∨ ) . Proof.
We only have to show that the actions of ψ M ∨ and φ M are compatiblewith the maps v i . But this is shown in Proposition 2.15.Combining the isomorphisms of Proposition 2.12 and Lemmas 2.13 and 2.16,we therefore get for all 0 ≤ i ≤ u i ( M ) : H i ( C φ M ∨ ,γ ,γ ( M ∨ )) → H i ( C ψ M ∨ ,γ − ,γ − ( M ∨ )) → v i H − i ([ C φ M ,γ ,γ ( M )] ∨ ) → α i [ H − i ( C φ M ,γ ,γ ( M ))] ∨ Using these isomorphisms, we get the following proposition:-
Proposition 2.17.
For all ≤ i ≤ , we have a perfect pairing H i ( C φ M ,γ ,γ ( M )) × H − i ( C φ M ∨ ,γ ,γ ( M ∨ )) → Q p / Z p , ( x, y ) → [ u − i ( M )( y )]( x ) . Making the above isomorphisms explicit (which is very messy, so we omitthe details), one can show that H φ M ,γ ,γ ( M ) × H φ M ∨ ,γ ,γ ( M ∨ ) → Q p / Z p , (6)( x, y, z ) × ( f, g, h ) → h ( γ γ x ) − g ( γ φ M ( y )) (7)+ f ( γ γ a − γ − φ M ( z )) + g (˜ ωφ M ( z )) , (8) H φ M ,γ ,γ ( M ) × H φ M ∨ ,γ ,γ ( M ∨ ) → Q p / Z p , (9)( x, y, z ) × ( f, g, h ) → − h ( γ x ) − g ( γ γ a − γ − y ) − h (˜ ωy ) (10)+ f ( φ M ( z )) , (11)11here ˜ ω ∈ Λ( G ) is the element satisfying˜ ω ( γ − −
1) = n γ a − γ − − . Definition.
Define the map TR K : Ω A K → Z p to be the composition Tr W ( k ) / Q p ◦ Res K . Note that TR K is Z p -linear. Proposition 3.1.
For all ω ∈ Ω( K ) , we have TR K (( γ γ n − γ − − ω ) = 0 , (12)TR K (( γ − ω ) = 0 , (13)TR K (( φ Ω( K ) − ω ) = 0 . (14) Proof.
Since γ ( T ) = (1 + π ) N T from some N ∈ Z × p , it is clear that (13) holds.Write ω = P i ∈ Z a i T i dT ∧ dπ F , where a i ∈ A F for all i ∈ Z . As shown in [6],we have TR K (( φ − a dπ F ) = 0, which (by the compatibilities of the actionsof φ ) implies (14). It remains to show (12). Let x = γ −
1. Expanding γ a − γ − in terms of x gives γ a − γ − a + higher order terms . Since γ is trivial on A F , the operator γ γ a − γ − − γ − a . Equa-tion (12) therefore follows from the 1-dimensional case as treated in [6].For j ≥
1, let Ω j ( K ) = Ω( K ) mod p j , which is an ´etale ( φ, G )-torsionmodule over A K isomorphic to D K ( µ p j ). By reduction mod p j , Res induces acanonical Z /p j Z -linear mapTR K,j : Ω j ( K ) → Z /p j Z . By Proposition 3.1, TR
K,j factorizes through the quotient of Ω j ( K ) by ( τ (Ω j ( K )+ τ (Ω j ( K )+( φ − i ( K )), where τ i = γ i −
1. We therefore get a homomorphismTR
K,j : H φ,γ ,γ (Ω i ( K )) → Z /p j Z . Passing to the direct limit gives a map H φ,γ ,γ ( B K / A K ⊗ A K Ω( K )) → Q p / Z p . v p be the p -adic valuation of Z p normalized by v p ( p ) = 1. Let n p ( γ ,n ) = v p (log χ ( γ )) and n p ( γ ,n ) = v p ( η ( γ ,n )). Let c = p np ( γ ,n ) log χ ( γ ,n ) p np ( γ ,n ) η ( γ ,n ) . Proposition 3.2.
The map − c TR gives a canonical isomorphism between thegroups H φ,γ ,γ ( B K / A K ⊗ A K Ω( K )) and Q p / Z p . Remark.
The factor c may seem bizarre, but we will see its use in Section 4. Proof.
It is sufficient to show that TR
K,j gives an isomorphism for all j . Sincewe can expand any element in Ω j ( K ) as a power series in T with coefficients in A F mod p j , it is sufficient to show that for all k = 0, γ − T k A F . The proposition follows from Herr’s proof of the Tate isomorphism in the1-dimensional case (c.f. Theorem 5.2 in [6]). Expanding the power series showsthat ( γ − T k ) = T k f k ( π ), where f k ( π ) ∈ A × F , which finishes the proof.Combining this result with Proposition 2.11 and the main result of Sec-tion 2.7 proves Theorem 1.2. Let V be a torsion Z p -representation of G K , and let M = D ( V ). Recall that˜ V = Hom Z p ( V, µ p ∞ ) . Lemma 3.3.
The ( φ, G ) -module D K ( ˜ V ) is isomorphic to Hom A K ( M, B K / A K ⊗ A K Ω( K )) .Proof. It follows from the equivalence of categories (Theorem 2.9) that D K ( ˜ V )is isomorphic to Hom A K ( M, D ( µ ⊗ np ∞ )). It is now sufficient to observe that D K ( µ ⊗ np ∞ ) = lim −→ D K ( µ ⊗ np j ) ∼ = lim −→ Ω( K ) p j Ω( K ) ∼ = B K / A K ⊗ A K Ω( K ) . Theorem 1.1 is therefore a consequence of Lemma 2.8 and Theorem 2.17.
Let F be the maximal algebraic extension of Q p contained in K . Throughoutthis section, we assume that the extension of F over Q p is unramified. As inthe previous sections, let ǫ = (1 , ξ p , ξ p , . . . ) ∈ E K and π = [ ǫ ] −
1, where [ ǫ ] isthe Teichm¨uller representative of ǫ . Also, let T = ( X, X p , X p , . . . ) ∈ E K , andlet T = [ T ] ∈ A K be its Teichm¨uller representative.Fix n ≥
1. Let π n = φ − n ( π ) = [( ξ p n , ξ p n +1 . . . )] − T n = φ − n ( T ) =[( X pn , X pn +1 , . . . )]. 13 roposition 4.1. Let n ≥ . Then for all i ≥ , taking cup product with ξ p n gives an isomorphism of Gal( K n /K ) -modules ∪ ξ p n : H i ( G K n , Z /p n Z ) → H i ( G K n , µ p n ) . Proof.
Let D = D ( Z /p n Z ) ∼ = A K n mod p n . Let Γ ( n )1 = Gal( K ( µ p ∞ , X pn ) /K n ) and Γ ( n )2 = Gal( K ∞ /K ( µ p ∞ , X pn )). Let γ ,n and γ ,n be topologicalgenerators of Γ ( n )1 and Γ ( n )2 , respectively. Recall that the G K n -cohomology of Z /p n Z is given by the complex C φ,γ ,n ,γ ,n ( D ) : 0 ✲ D f ✲ D ⊕ f ✲ D ⊕ f ✲ D ✲ , where f : x → [( φ − x, ( γ ,n − x, ( γ ,n − x ] ,f : ( x, y, z ) → [( φ − y − ( γ ,n − x, ( φ − z − ( γ ,n − x, ( γ ,n − y − ( γ ,n γ a ,n − γ ,n − − z ] ,f : ( x, y, z ) → ( γ ,n − x − ( γ ,n γ a ,n − γ ,n − − y − ( φ − z. Since µ p n = < ξ p n > , it is easy to see from this description of the cohomologygroups that cup product (which is the same as multiplication) with ξ p n gives aGal( K n /K )-equivariant isomorphims H i ( G K n , Z /p n Z ) → H i ( G K n , ξ p n ) for all0 ≤ i ≤ H ( G K n , µ ⊗ p n ) × H ( G K n , Z /p n Z ) → Q p / Z p . Taking cup product with ξ p n and using Proposition 4.1 gives a perfect pairing H ( G K n , µ ⊗ p n ) × H ( G K n , µ p n ) → µ p n . (15) Definition.
The Hilbert pairing K ( K n ) × K ( K n ) → µ p n . is the composition of (15) with the Galois symbol map δ × δ : K ( K n ) × K ( K n ) → H ( G K n , µ ⊗ p n ) × H ( G K n , µ p n )Note that we have a natural (surjective) multiplication map K ( K n ) × K ( K n ) → K ( K n ). We therefore interprete the Hilbert pairing as a pairing V n : K ( K n ) × K ( K n ) × K ( K n ) → µ p n . (16)14rom the definition of the Galois symbol it is clear that V n factors through V n : (cid:0) K ( K n ) /p n (cid:1) × → µ p n . (17)Since the pairing (15) is perfect, the pairing V n in (17) is non-degenerate. Lemma 4.2.
We have a commutative diagram K ( K n ) × K ( K n ) δ × δ ✲ H ( K n , µ p n ) × H ( K n , µ p n ) K ( K n ) ❄ δ ✲ H ( K n , µ p n ) ∪ ❄ Let F by the maximal extension of Q p contained in K , and let R = O F [[ T n ]][ T − n ].Note that we can identifiy A + K n with the abstract power series ring R [[ Y ]] (where Y = π n , but we forget this for the time being). Let m = ( p, Y ) be the maximalideal of A + K n , and let A = 1 + m . For F ( Y ) ∈ A , define l ( F ( Y )) = (1 − φp ) log F ( Y ) . To shorten notation, let f ( Y ) = l ( F ( Y )). Define the differential operators D = ( Y + 1) ddY and D = T n ddT n . Fix n ≥
1, and let S n = R [[ π n ]]. Note. (1) The action of γ on A + K n is trivial mod π .(2) The element ǫ is a generator of Z p (1), so ǫ mod p n is a generator of µ p n and can be identified with ξ p n .Let ǫ ( n ) = ǫ mod p n and τ = π − . Proposition 4.3.
Let F ( Y ) ∈ A . Then there exist unique a γ ,n ( π n ) , b γ ,n ( π n ) ∈ S n such that ( φ − a γ ,n ( π n ) ⊗ ǫ ( n ) ) = ( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) , ( φ − b γ ,n ( π n ) ⊗ ǫ ( n ) ) = ( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) . Moreover, we have a γ ,n ( π n ) = 1 − χ ( γ ,n ) p n D log F ( π n ) mod π,b γ ,n ( π n ) = η n ( γ ,n ) D log F ( π n ) mod π. Proof.
Arguing as in the proof of Lemma 2.1.3 in [2] we have( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) = − − χ ( γ ,n ) p n D log f ( π n ) ⊗ ǫ ( n ) mod π. D φ = pφD , we can write( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) = ( φ − − χ ( γ ,n ) p n D log F ( π n ) ⊗ ǫ ( n ) ) mod π. Let ˜ a γ ,n ( π n ) = − χ ( γ ,n ) p n D log F ( π n ), so( φ − a γ ,n ( π n ) ⊗ ǫ ( n ) ) = ( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) . Since φ − πS n , we deduce that there exists a unique a γ ,n ( π n ) ∈ S n such that a γ ,n ( π n ) = ˜ a γ ,n ( π n ) mod π and ( φ − a γ ,n ( π n ) ⊗ ǫ ( n ) ) =( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) . The existence of b γ ,n ( π n ) follows from similar arguments:- Let η n ( γ ,n ) = η ( γ ,n ) p n . Note that γ ,n ( T n ) = (1 + π ) η n ( γ ,n ) T n , (18)so γ ,n ( T n ) = T n + η n ( γ ,n ) T n π mod π , and hence γ ,n f ( π n ) = f ( π n ) + η n ( γ ,n ) D f ( π n ) π mod π , ( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) = η n ( γ ,n ) D f ( π n ) ⊗ ǫ ( n ) mod π. By assumption we have f ( π n ) = (1 − φp ) log F ( π n ) . Since D φ = pφD , it follows that( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) = (1 − φ ) η n ( γ ,n ) D log F ( π n ) ⊗ ǫ ( n ) mod π. Let ˜ b γ ,n ( π n ) = η n ( γ ,n ) D log F ( π n ) ⊗ ǫ ( n ) . It follows that (1 − φ )˜ b γ ,n ( π n ) =( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) . Since φ − πS n , there exists a unique b γ ,n ( π n ) ∈ S n such that b γ ,n = ˜ b γ ,n mod π and( φ − b γ ,n ( π n ) ⊗ ǫ ( n ) = ( γ ,n − f ( π n ) τ ⊗ ǫ ( n ) ) . Definition.
Let ι n : A → H φ,γ ,n ,γ ,n ( C n ) be the homomorphism F ( X ) → [ f ( π n ) τ ⊗ ǫ ( n ) , a γ ,n ( π n ) ⊗ ǫ ( n ) , b γ ,n ⊗ ǫ ( n ) ] . (19) Lemma 4.4.
The map ι n is well-defined. roof. Explicit calculation shows that( γ ,n − a γ ,n ( π n ) ⊗ ǫ ( n ) ) = ( γ ,n γ N − γ ,n − − b γ ,n ( π n ) ⊗ ǫ ( n ) ) . (20)It follows that ι n is really a map into H γ ,n ,γ ,n ,φ . Proposition 4.5.
Let δ n : K × n → H ( G K n , Z p (1)) be the Kummer map. Wehave a commutative diagram A ι n ✲ H φ,γ ,n ,γ ,n ( C n ) K × n h n ❄ δ n ✲ H ( G K n , Z p (1)) ∼ = ❄ To prove the proposition, we follow the strategy of Benois in the proof ofProposition 2.1.5 in [2]. We split the proof of the proposition into a sequence oflemmas.Note that the action of G K on A K factors through G K = Gal( K ∞ /K ).Recall that G K ∼ = Γ ⋊ Γ , where Γ is congruent (via the cyclotomic character χ K ) to an open subgroup of Z × p and Γ is congruent (via a character η K ) to Z p . Lemma 4.6.
Let [ x, y, z ] ∈ H γ ,n ,γ ,n ,φ ( C n ) , and let u ∈ A be a solution of ( φ − u = x . Then h ([ x, y, z ]) is given by the cocycle σ → c ( σ ) which isdefined as follows:- Let ˜ σ be the image of σ in G K under the projection map.Let k = χ (˜ σ ) and l = η (˜ σ ) , so ˜ σ = γ k ,n γ l ,n . Then c ( σ ) = ( σ − u − γ k ,n − γ ,n − y − γ k ,n γ l ,n − γ ,n − z. Proof.
Let N x,y,z = D ( µ p n ) ⊕ A K n e , where the action of φ, γ ,n and γ ,n on e is given by φ ( e ) = e + x , γ ,n ( e ) = e + y , γ ,n ( e ) = e + z . Then the long exactsequence associated to the short exact sequence of G K n -modules0 → D ( µ p n ) → N x,y,z → A K n → δ : H ( A K n ) → H γ ,n ,γ ,n ,φ ( C n ), and aneasy diagram search shows that δ (1) = [ x, y, z ]. Applying ( φ −
1) to (21) givesa short exact sequence 0 → µ p n → T x,y,z → Z p → δ Gal : Z p → H ( G K , µ p n ). We have u + e ∈ N x,y,z ⊗ A Kn A and ( φ − u + e ) = 0, so u + e ∈ T x,y,z . So δ Gal (1) can berepresented by the cocycle σ → σ ( u + e ) − ( u − e )=( σ − u + ( σ − e. G K acts on A K n via the quotient G K n . Since e ∈ A K n , we have(˜ σ − e = ( γ k ,n γ l ,n − e = γ k ,n ( γ l ,n − e + ( γ k ,n − e = γ k ,n γ l ,n − γ ,n − z + γ k ,n − γ ,n − y The lemma now follows from the commutativity of the diagram H ( A K n ) δ ✲ H γ ,n ,γ ,n ,φ ( C n ) Z p h ❄ δ Gal ✲ H ( G K n , µ p n ) h ❄ In particular, h ( ι n ( F )) is given by σ → ( σ − u − γ χ ( σ )1 ,n − γ ,n − a γ ,n ( π n ) ⊗ ǫ ( n ) − γ χ ( σ )1 ,n γ η ( σ )2 ,n − γ ,n − b γ ,n ( π n ) ⊗ ǫ ( n ) , where (1 − φ ) u = f ( π n ) τ . Since γ ,n ( π n ) = π n mod π and γ ,n ( T n ) = T n mod π , we have γ χ ( σ )1 ,n − γ ,n − a γ ,n ( π n ) ⊗ ǫ ( n ) ∼ = χ ( σ ) 1 − χ ( σ ) p n D log F ( π n ) ⊗ ǫ ( n ) mod π,γ χ ( σ )1 ,n γ η ( σ )2 ,n − γ ,n − b γ ,n ( π n ) ⊗ ǫ ( n ) ∼ = η ( σ ) η n ( γ ,n ) D log F ( π n ) mod π. These congruences imply that c ( σ ) ∼ = ( χ ( σ ) σ − u + 1 − χ ( σ ) p n D log F ( π n )+ η ( σ ) χ ( σ ) η n ( γ ,n ) D log F ( π n ) mod π. We now interprete c ( σ ) in terms of A cris . Denote by I the ideal of A cris generatedby π and π p − p . Lemma 4.7.
There exists a unique x ∈ Fil A cris such that x = u ( π − π )mod I and (1 − φp ) x = f ( π n ) . roof. Imitate the proof of Lemma 2.1.6.2 in [2].Define the element µ ( σ ) = ( σ − x − log( σ ( F ( π n )) F ( π n ) ) , which belongs to 1 + Fil A cris for all σ ∈ G K n , and it is easy to check that themap µ : G K n → Fil A cris is a cocycle. Lemma 4.8.
We have µ ( σ ) = c ( σ ) .Proof. We have (1 − φp ) µ ( σ ) = 0, so µ ( σ ) has the form a ( σ ) t for some a ( σ ) ∈ Q p .On the other hand, from the congruences˜ σF ( π n ) = F ( π n ) + χ ( σ ) 1 − χ ( σ ) p n D log F ( π n ) π + η ( σ ) η n ( γ ,n ) D log F ( π n ) π mod π and ( σ − x = ( χ ( σ ) σ − uπ mod I it follows that µ ( σ ) =( χ ( σ ) σ − uπ + 1 − χ ( σ ) p n D log F ( π n ) π + η ( σ ) χ ( σ ) η n ( γ ,n ) D log F ( π n ) π mod I = c ( σ ) T mod I, which implies that µ ( σ ) = tc ( σ ). Corollary 4.9.
One has [ ǫ ] c ( σ ) = exp( µ ( σ )) = σ exp( x )exp( x ) F ( π n ) σF ( π n ) . Proof of Proposition 4.5.
Let y = exp( x ). Then the equation (1 − φp ) x = f ( π n )can be written of the form y p φ ( y ) = exp( pf ( π n )) . Consider the short exact sequence1 → [ ǫ ] Z p → W ( R ) → ν pW ( R ) → , where ν ( a ) = a p φ ( a ) . It shows that the inclusion W ( R ) ⊂ A cris gives a 1 − Y of Y p φ ( Y ) = exp( pf ( π n )) and solutions X =log Y of (1 − φp ) X = f ( π n ). Hence y ∈ W ( R ), and it is easy to see byinduction that y p n φ n ( y ) = F ( π n ) p n φ n ( F ( π n )) . z = φ − n ( yF ( π n ) − ). Applying the map θ : W ( R ) → O C p to both sides ofthis equation, we obtain that θ ( z ) p n = h n ( F ) − . Hence the connecting map δ n sends h n ( F ) to the class of the cocycle σ → θ ( z/σ ( z )). On the other hand, one has θ ( zσ ( z ) ) = θφ − n ( yσF ( π n ) σ ( y ) F ( π n ) ) = θφ − n ([ ǫ ] − c ( σ ) ) = ξ − c ( σ ) p n , which finishes the proof. In this section we reprove Vostokov’s formulae for (16). More precisely, we provethe following result:- For 1 ≤ i ≤
3, let α i ∈ O × K such that α i ∼ = 1 mod ¯ π n , andlet F i ( X ) ∈ A + K such that h n ( F i ) = α i . Let f i ( X ) = (1 − φp ) log F ( X ). Theorem 4.10.
We have V n ( α , α , α ) = µ Tr Res πn,T (Φ) p n , (22) where Φ is given by the formula Φ = − π ( 1 p f ( π n ) d log F φ ( π n ) ∧ d log F φ ( π n ) − p f ( π n ) d log F ( π n ) ∧ d log F φ ( π n )+ f ( π n ) d log F ( π n ) ∧ d log F ( π n )) . We will prove the theorem in the rest of this section.
Lemma 4.11.
Let [ x, y, z ] , [ x ′ , y ′ , z ′ ] ∈ H φ,γ ,n ,γ ,n ( C n ) . If [ x , y , z ] representsthe cohomology class of [ x, y, z ] ∪ [ x ′ , y ′ , z ′ ] , then x = y ⊗ γ ,n x ′ − x ⊗ φy ′ , y = z ⊗ γ ,n x ′ − x ⊗ φz ′ . Moreover, if z, z ′ ∈ S n , then z = x ⊗ γ ,n y ′ − y ⊗ γ ,n x ′ mod π. Proof.
Since Γ ( n )1 (resp. Γ ( n )2 ) is isomorphic to an open subgroup of Z × p (resp. Z p ), the formulae for x and y follow from [6], using that the cup product iscompatible with restriction. The formula for z follows from the observation that γ and γ commute on S n mod π .For 1 ≤ i ≤
3, let ι n ( F i ( X )) = [ f i ( π n ) τ ⊗ ǫ, a ( i ) γ ,n ( π n ) ⊗ ǫ, b ( i ) γ ,n ⊗ ǫ ].20 orollary 4.12. If [ x , y , z ] = ι n ( F ( X )) ∪ ι n ( F ( X )) , then x = 1 − χ ( γ ,n ) p n ( D log F ( π n ) ⊗ f ( π n ) − p − f ( π n ) ⊗ D log F φ ( π n )) ⊗ τ ⊗ ǫ mod S n , y = η n ( γ ,n )( D log F ( π n ) ⊗ f ( π n ) − p − f ( π n ) ⊗ D log F φ ( π n )) ⊗ τ ⊗ ǫ mod S n , z = 1 − χ ( γ ,n ) p n η n ( γ ,n )( D log F ( π n ) D log F ( π n ) − D log F ( π n ) D log F ( π n )) ⊗ ǫ mod π Proof.
Observe that γ ,n ( f i ( π n ) τ ) = χ − ( γ ,n ) f i ( π n ) τ mod S n ,γ ,n ( f i ( π n ) τ ) = f i ( π n ) τ mod S n . The lemma now follows from the previous lemma and Proposition 4.3.
Proof of Theorem 1.3.
We prove the Theorem using the formulae (6). Let H α ,α ,α = ι n ( F ( X )) ∪ [ x , y , z ] , where (to simplify the notation) we write[ x, y, z ] = [ f ( π n ) τ, a (3) γ ,n ( π n ) , b (3) γ ,n ] . Recall that γ ,n and γ ,n commute on S n mod π . It follows that the formu-lae (6) simplify to ι n ( F ( X )) ∪ [ x , y , z ] = z ⊗ γ γ ( x ) − y ⊗ γ φ M ( y ) + x ⊗ γ φ M ( z ) mod S n . Using the formulae in Corollary 4.12 it follows that H α ,α ,α = γ φ M ( b (3) γ ,n ( π n ))( 1 − χ ( γ ,n ) p n ( D log F ( π n ) ⊗ f ( π n ) − p − f ( π n ) ⊗ D log F φ ( π n )) ⊗ τ ⊗ ( ǫ ( n ) ) ) − γ φ M ( a (3) γ ,n ( π n ))( η n ( γ ,n )( D log F ( π n ) ⊗ f ( π n ) − p − f ( π n ) ⊗ D log F φ ( π n )) ⊗ τ ⊗ ( ǫ ( n ) ) )+ γ γ ( f ( π n ) τ )( 1 − χ ( γ ,n ) p n η n ( γ ,n )( D log F ( π n ) D log F ( π n ) − D log F ( π n ) D log F ( π n )) ⊗ ( ǫ ( n ) ) ) mod S n As before, we have γ ,n ( f i ( π n ) τ ) = χ − ( γ ,n ) f i ( π n ) τ mod S n ,γ ,n ( f i ( π n ) τ ) = f i ( π n ) τ mod S n ,
21o (rearranging the terms) the above formula simplifies to H α ,α ,α = 1 − χ ( γ ,n ) p n η n ( γ ,n ) π − ×− p − f ( π n )( D log F φ ( π n ) D log F φ ( π n )+ D log F φ ( π n ) D log F φ ( π n )))+ p − f ( π n )( D log F ( π n ) D log F φ ( π n )+ D log F ( π n ) D log F φ ( π n )) − f ( π n )( D log F ( π n ) D log F ( π n )+ D log F ( π n ) D log F ( π n )) ⊗ ( ǫ ( n ) ) mod S n . Recall that D = ( π n + 1) ddπ n and D = T n ddT n . It follows that the image inΩ( K ) of the above expression (which we also denote by H α ,α α ) with respectto the map in Lemma 2.10 is1 − χ ( γ ,n ) p n η n ( γ ,n ) π − × ( − p − f ( π n ) d log F φ ( π n ) ∧ d log F φ ( π n )+ p − f ( π n ) d log F ( π n ) ∧ d log F φ ( π n ) − f ( π n ) d log F ( π n ) ∧ d log F ( π n )) mod S n . Taking into account that p − n log χ ( γ ,n ) = p − n ( χ ( γ ,n ) −
1) mod p n , we obtainthat − c TR( H α ,α ,α ) = − Tr F/ Q p Res π n ,T π − ( p − f ( π n ) d log F φ ( π n ) ∧ d log F φ ( π n ) − p − f ( π n ) d log F ( π n ) ∧ d log F φ ( π n )+ f ( π n ) d log F ( π n ) ∧ d log F ( π n )) , which finishes the proof. References [1] F. Andreatta,
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Higher exponential maps and explicit reciprocity laws I , preprint.Saral Livia ZerbesDepartment of MathematicsImperial College LondonLondon SW7 2AZUnited Kingdom email:email: