Scaffolds and (Generalised) Galois Module Structure Nigel Byott University of Exeter 24 June 2015
Scaffolds give a new approach to Galois module structure in local fields. When they exist, they give a lot of information in purely numerical form, but interpreting this to get explicit module-theoretic statements requires further effort. This is joint work with Griff Elder and Lindsay Childs. Main reference: NB + L. Childs + G. Elder: Scaffolds and Generalized Integral Galois Module Structure arXiv:1308.2088 Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 2 / 24
Outline of Talk: Motivation I: Inseparable Extensions Motivation II: Galois Module Structure in Prime Degree What is a Scaffold? When do Galois Scaffolds Exist? Consequences of Having a Scaffold Example: Weakly Ramified Extensions Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 3 / 24
Motivation I: Inseparable Extensions Can we ”do Galois theory” for inseparable extensions? Take: K a field of characteristic p > 0; L a primitive, purely inseparable extension of K of degree p n : L = K ( x ) with x p n = α ∈ K × \ K × p . The only K -automorphism of L is the identity, but another familiar sort of K -linear operator is given by (formal) differentiation. Let δ : L − → L be the K -linear map given by δ ( x j ) = jx j − 1 . This makes sense as δ ( x p n ) = 0 = δ ( α ), but depends on the choice of generator x . We have δ ( x j ) = 0 if p | j ; δ p = 0 . Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 4 / 24
Motivation I: Inseparable Extensions d s We want to introduce operators δ ( s ) that “behave like” 1 dx s . s ! For 0 ≤ s ≤ p n − 1, write s = s (0) + ps (1) + · · · + p n − 1 s ( n − 1) with 0 ≤ s ( i ) ≤ p − 1 . Then define a K -linear map δ ( s ) : L − → L by � n − 1 �� � j � � j ( i ) x j − s = � δ ( s ) ( x j ) = x j − s . s s ( i ) i =0 (Think of δ ( p i ) as differentiation with respect to x p i , where we pretend that x , x p , x p 2 , . . . , x p n − 1 are independent variables.) Notation: For 0 ≤ s , j ≤ p n − 1, s � j means s ( i ) ≤ j ( i ) for 0 ≤ i ≤ n − 1 . Then δ ( s ) ( x j ) = 0 unless s � j . Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 5 / 24
Motivation I: Inseparable Extensions We have � s + t � δ ( s ) δ ( t ) = δ ( s + t ) . s (This is 0 if s + t ≥ p n .) The commutative K -algebra A with basis ( δ ( s ) ) 0 ≤ s ≤ p n − 1 acts on L . This is analogous to action of the group algebra in standard Galois theory. The group algebra is a Hopf algebra, and its action is compatible with the comultiplication. In the same way, if we make A into a Hopf algebra with comultiplication s δ ( s ) �→ δ ( r ) ⊗ δ ( s − r ) , � r =0 then L is an A -Hopf-Galois extension of K . A is the divided power Hopf algebra of dimension p n . Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 6 / 24
Motivation I: Inseparable Extensions Now bring in ramification. Say K is the local field F p f (( T )) with valuation v K ( T ) = 1. Suppose v K ( α ) = − b with p ∤ b . So L / K is totally ramified and v L ( x ) = − b . ( x j ) 0 ≤ j ≤ p n − 1 is a K -basis of L with valulations distinct modulo p n , and � v L ( x j ) + bs if s � j , v L ( δ ( s ) · x j ) = ∞ otherwise. Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 7 / 24
Motivation I: Inseparable Extensions The action becomes even more transparent if we adjust our bases by suitable units: set � − 1 � n − 1 � n − 1 � Ψ ( s ) = y ( j ) = � � δ ( s ) , x j . s ( i ) ! j ( i ) ! i =0 i =0 Then v L ( y ( j ) ) = v L ( x j ) = − jb and � y ( j − s ) if s � j , Ψ ( s ) · y ( j ) = otherwise. . 0 The elements Ψ ( p i ) and y ( j ) form a prototypical example of a scaffold . Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 8 / 24
Motivation II: Galois Module Structure in Prime Degree Let K be a finite extension of Q p with absolute ramification index v K ( p ) = e . Let L / K be a totally ramified Galois extension of degree p . Let G = � σ � = Gal ( L / K ). We want to study the valuation ring O L of L as a Galois module. As L / K is wildly ramified, O L cannot be free over O K [ G ], so consider the associated order A := { α ∈ K [ G ] : α · O L ⊆ O L } . This is the largest order in K [ G ] over which O L is a module. Basic Question: When is O L a free module over A ? Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 9 / 24
Motivation II: Galois Module Structure in Prime Degree L / K has ramification break b characterised by ∀ x ∈ L \{ 0 } , v L (( σ − 1) · x ) ≥ v L ( x ) + b , with equality unless p | v L ( x ) . Then ep ep 1 ≤ b ≤ p − 1 , p ∤ b unless b = p − 1 . ep We assume b ≤ p − 1 − 1. Bertrandias, Bertrandias and Ferton (1972) showed that O L is free over A ⇔ ( b mod p ) | p − 1 . Ferton (1972) determined when a given power P h of the maximal ideal P of O L is free over its associated order, in terms of the continued fraction expansion of b / p . Analogous results in characteristic p (so K = F p f (( T )) and e = ∞ ) were given by Aiba (2003), de Smit & Thomas (2007) and Huynh (2014). Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 10 / 24
Motivation II: Galois Module Structure in Prime Degree These results all depend on the following idea: Let Ψ = σ − 1 and choose x ∈ L with v L ( x ) = b . For 0 ≤ j ≤ p − 1 set y j = Ψ j · x , so v L ( y j ) = ( j + 1) b . Then, for 0 ≤ s ≤ p − 1, � = y s + j if s + j ≤ p − 1; Ψ s · y j (mod x s + j P T ) ≡ 0 otherwise where T = ep − ( p − 1) b . Then Ψ and the y j form a scaffold. Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 11 / 24
What is a Scaffold? Let K be a local field of residue characteristic p > 0, let π ∈ K with v K ( π ) = 1, and let L / K be a totally ramified extension of degree p n . Fix b ∈ Z with p ∤ b and for each t ∈ Z define a ( t ) = a ( t ) (0) + pa ( t ) (1) + · · · + p n − 1 a ( t ) ( n − 1) := ( − b − 1 t ) mod p n . A scaffold of shift b and infinite tolerance on L consists of elements λ t ∈ L with v L ( λ t ) = t for each t ∈ Z ; K -linear maps Ψ 1 , Ψ 2 , . . . , Ψ n : L − → L such that � λ t + p n − i b if a ( t ) ( n − i ) ≥ 1 , Ψ i · λ t = 0 if a ( t ) ( n − i ) = 0 , and Ψ i · K = 0. Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 12 / 24
What is a Scaffold? For 0 ≤ s ≤ p n − 1 , set Ψ ( s ) = Ψ s (0) s (1) s ( n − 1) n Ψ n − 1 · · · Ψ . 1 Then � λ t + sb if s � a ( t ) , Ψ ( s ) · λ t = 0 otherwise. Moreover L is a free module over the commutative K -algebra A = K [Ψ 1 , . . . , Ψ n ] on the generator λ b . Example: K = F p f (( T )) and L = K ( x ) purely inseparable of degree p n , b = − v L ( x ), λ cp n − bj = T c y ( j ) and Ψ i = δ ( p n − i ) . Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 13 / 24
What is a Scaffold? Now fix T > 0. A scaffold of tolerance T is similar except that the formula for the action of A on L only holds “up to an error”: � λ t + sb if s � a ( t ) , Ψ ( s ) · λ t ≡ 0 otherwise. where the congruence is modulo terms of valuation ≥ t + sb + T . (Then A no longer need be commutative.) Example: L / K totally ramified Galois extension of degree p . Ψ 1 = σ − 1, and λ cp n + b ( j +1) = π c Ψ j 1 · x where v L ( x ) = b ; here T = ep − ( p − 1) b . Remark: In the BCE paper, we allowed a slightly more general definition of scaffold. Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 14 / 24
When do Galois Scaffolds Exist? Suppose L / K is a totally ramified Galois extension of degree p n . Take a generating set σ 1 , . . . , σ n of G = Gal ( L / K ) so that the subgroups H i = � σ n − i +1 , . . . , σ n � , 0 ≤ i ≤ n satisfy | H i | = p i and refine the ramification filtration. Then we have (lower) ramification breaks b 1 ≤ b 2 ≤ · · · ≤ b n , characterised by ∀ y ∈ L × , v L (( σ i − 1) · y ) ≥ v L ( y ) + b i , with equality if and only if p ∤ v L ( y ). Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 15 / 24
When do Galois Scaffolds Exist? Now if x is any element of the intermediate field K i = L H n − i of degree p i over K , then p n − i | v L ( x ), and if p n − i +1 ∤ v L ( x ) then v L (( σ i − 1) · x ) = v L ( x ) + p n − i b i . We now make 3 assumptions: Assumption 1 (very weak): p ∤ b 1 . Assumption 2 (fairly weak): b i ≡ b n (mod p i ) for each i . If G is abelian, this holds by the Hasse-Arf Theorem. Now set Ψ n = σ n − 1. Assumption 3 (pretty strong): For 1 ≤ i ≤ n − 1, we can replace σ i − 1 with Θ i ∈ K [ H n +1 − i ] so that ∀ y ∈ L × with v L ( y ) ( n − i +1) � = 0 . v L (Θ i · y ) = v L ( y ) + p n − i b i Nigel Byott (University of Exeter) Scaffolds and Galois Module Structure 24 June 2015 16 / 24
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