An Introduction to Galois Module Structure

AN INTRODUCTION TO GALOIS MODULE STRUCTURE

Nigel Byott

University of Exeter

Omaha, May 2019 Its Galois group is Gal(N/K) := {ﬁeld automorphisms σ : N → N with σ(k) = k ∀k ∈ K}. Then Gal(N/K) is a group of order n. The Fundamental Theorem of Galois Theory says there is a bijection between subgroups of Gal(N/K) and ﬁelds E with K ⊆ E ⊆ N. The Nomal Basis Theorem says there is an element α ∈ N (called a normal basis generaor) such that {σ(α): σ ∈ Gal(N/K)} is a basis for N as a K-vector space. Thus every β ∈ N can be written in a unique way as X β = cσσ(α), cσ ∈ K. σ

Galois Theory: A Quick Summary An extension N/K of ﬁelds of (ﬁnite) degree n = [N : K] is Galois if it is normal and separable.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 2 / 25 The Fundamental Theorem of Galois Theory says there is a bijection between subgroups of Gal(N/K) and ﬁelds E with K ⊆ E ⊆ N. The Nomal Basis Theorem says there is an element α ∈ N (called a normal basis generaor) such that {σ(α): σ ∈ Gal(N/K)} is a basis for N as a K-vector space. Thus every β ∈ N can be written in a unique way as X β = cσσ(α), cσ ∈ K. σ

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 2 / 25 The Nomal Basis Theorem says there is an element α ∈ N (called a normal basis generaor) such that {σ(α): σ ∈ Gal(N/K)} is a basis for N as a K-vector space. Thus every β ∈ N can be written in a unique way as X β = cσσ(α), cσ ∈ K. σ

Galois Theory: A Quick Summary An extension N/K of fields of (finite) degree n = [N : K] is Galois if it is normal and separable. Its Galois group is Gal(N/K) := {field automorphisms σ : N → N with σ(k) = k ∀k ∈ K}. Then Gal(N/K) is a group of order n. The Fundamental Theorem of Galois Theory says there is a bijection between subgroups of Gal(N/K) and fields E with K ⊆ E ⊆ N.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 2 / 25 Thus every β ∈ N can be written in a unique way as X β = cσσ(α), cσ ∈ K. σ

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 2 / 25 Galois Theory: A Quick Summary An extension N/K of fields of (finite) degree n = [N : K] is Galois if it is normal and separable. Its Galois group is Gal(N/K) := {field automorphisms σ : N → N with σ(k) = k ∀k ∈ K}. Then Gal(N/K) is a group of order n. The Fundamental Theorem of Galois Theory says there is a bijection between subgroups of Gal(N/K) and fields E with K ⊆ E ⊆ N. The Nomal Basis Theorem says there is an element α ∈ N (called a normal basis generaor) such that {σ(α): σ ∈ Gal(N/K)} is a basis for N as a K-vector space. Thus every β ∈ N can be written in a unique way as X β = cσσ(α), cσ ∈ K. σ

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 2 / 25 Then K[G] is a ring with multiplication ! ! ! X X X X X cσσ dτ τ = cσdτ στ = cσdσ−1ρ ρ. σ τ σ,τ ρ σ We call K[G] the group algebra of G over K. Then K[G] acts on N; for β ∈ N we have ! X X cσσ · β = cσσ(β) ∈ N. σ σ Thus N becomes a module over the ring K[G].

Reinterpreting the Normal Basis Theorem Write G = Gal(N/K) and let ( ) X K[G] = cσσ : cσ ∈ K , σ∈G a K-vector space over K of dimension n.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 3 / 25 We call K[G] the group algebra of G over K. Then K[G] acts on N; for β ∈ N we have ! X X cσσ · β = cσσ(β) ∈ N. σ σ Thus N becomes a module over the ring K[G].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 3 / 25 Then K[G] acts on N; for β ∈ N we have ! X X cσσ · β = cσσ(β) ∈ N. σ σ Thus N becomes a module over the ring K[G].

Reinterpreting the Normal Basis Theorem Write G = Gal(N/K) and let ( ) X K[G] = cσσ : cσ ∈ K , σ∈G a K-vector space over K of dimension n. Then K[G] is a ring with multiplication ! ! ! X X X X X cσσ dτ τ = cσdτ στ = cσdσ−1ρ ρ. σ τ σ,τ ρ σ We call K[G] the group algebra of G over K.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 3 / 25 Thus N becomes a module over the ring K[G].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 3 / 25 Reinterpreting the Normal Basis Theorem Write G = Gal(N/K) and let ( ) X K[G] = cσσ : cσ ∈ K , σ∈G a K-vector space over K of dimension n. Then K[G] is a ring with multiplication ! ! ! X X X X X cσσ dτ τ = cσdτ στ = cσdσ−1ρ ρ. σ τ σ,τ ρ σ We call K[G] the group algebra of G over K. Then K[G] acts on N; for β ∈ N we have ! X X cσσ · β = cσσ(β) ∈ N. σ σ Thus N becomes a module over the ring K[G]. Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 3 / 25 This means that N is a free K[G]-module of rank 1. [Note that, unlike vector spaces over a ﬁeld, modules over a ring do not always have a basis, i.e. are not always free. Thus the Normal Basis Theorem gives non-trivial information about the structure of N as a K[G]-module.]

The main question of Galois module structure is: Can we ﬁnd an analogue of the Normal Basis Theorem at the level of integers?

Now if α ∈ N is a normal basis generator for N/K, then, for each β ∈ N, there is a unique λ ∈ K[G] with β = λ · α.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 4 / 25 [Note that, unlike vector spaces over a ﬁeld, modules over a ring do not always have a basis, i.e. are not always free. Thus the Normal Basis Theorem gives non-trivial information about the structure of N as a K[G]-module.]

The main question of Galois module structure is: Can we ﬁnd an analogue of the Normal Basis Theorem at the level of integers?

Now if α ∈ N is a normal basis generator for N/K, then, for each β ∈ N, there is a unique λ ∈ K[G] with β = λ · α. This means that N is a free K[G]-module of rank 1.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 4 / 25 The main question of Galois module structure is: Can we ﬁnd an analogue of the Normal Basis Theorem at the level of integers?

Now if α ∈ N is a normal basis generator for N/K, then, for each β ∈ N, there is a unique λ ∈ K[G] with β = λ · α. This means that N is a free K[G]-module of rank 1. [Note that, unlike vector spaces over a ﬁeld, modules over a ring do not always have a basis, i.e. are not always free. Thus the Normal Basis Theorem gives non-trivial information about the structure of N as a K[G]-module.]

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 4 / 25 Now if α ∈ N is a normal basis generator for N/K, then, for each β ∈ N, there is a unique λ ∈ K[G] with β = λ · α. This means that N is a free K[G]-module of rank 1. [Note that, unlike vector spaces over a ﬁeld, modules over a ring do not always have a basis, i.e. are not always free. Thus the Normal Basis Theorem gives non-trivial information about the structure of N as a K[G]-module.]

The main question of Galois module structure is: Can we ﬁnd an analogue of the Normal Basis Theorem at the level of integers?

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 4 / 25 Let N is a ﬁnite extension of Q (i.e. a number ﬁeld) with [N : Q] = n. We have an analogue of Z inside N:

ON = {α ∈ N : α is the root of some monic f (X ) ∈ Z[X ]}.

ON is a ring, it contains Z, and it has a basis over Z of size n.

We call ON the ring of algebraic integers of N.

Rings of algebraic integers

Let’s take K = Q. Inside Q we have the ring of integers Z.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 5 / 25 We have an analogue of Z inside N:

ON = {α ∈ N : α is the root of some monic f (X ) ∈ Z[X ]}.

ON is a ring, it contains Z, and it has a basis over Z of size n.

We call ON the ring of algebraic integers of N.

Rings of algebraic integers

Let’s take K = Q. Inside Q we have the ring of integers Z. Let N is a ﬁnite extension of Q (i.e. a number ﬁeld) with [N : Q] = n.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 5 / 25 ON is a ring, it contains Z, and it has a basis over Z of size n.

We call ON the ring of algebraic integers of N.

Rings of algebraic integers

Let’s take K = Q. Inside Q we have the ring of integers Z. Let N is a ﬁnite extension of Q (i.e. a number ﬁeld) with [N : Q] = n. We have an analogue of Z inside N:

ON = {α ∈ N : α is the root of some monic f (X ) ∈ Z[X ]}.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 5 / 25 We call ON the ring of algebraic integers of N.

Rings of algebraic integers

Let’s take K = Q. Inside Q we have the ring of integers Z. Let N is a ﬁnite extension of Q (i.e. a number ﬁeld) with [N : Q] = n. We have an analogue of Z inside N:

ON = {α ∈ N : α is the root of some monic f (X ) ∈ Z[X ]}.

ON is a ring, it contains Z, and it has a basis over Z of size n.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 5 / 25 Rings of algebraic integers

Let’s take K = Q. Inside Q we have the ring of integers Z. Let N is a ﬁnite extension of Q (i.e. a number ﬁeld) with [N : Q] = n. We have an analogue of Z inside N:

ON = {α ∈ N : α is the root of some monic f (X ) ∈ Z[X ]}.

ON is a ring, it contains Z, and it has a basis over Z of size n.

We call ON the ring of algebraic integers of N.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 5 / 25 √ Any element α = u + v d of N with u, v ∈ Z will be an algebraic integer: it is a root of √ √ 2 2 2 X − (u + v d) X − (u − v d) = X − 2uX + (u + v d) ∈ Z[X ].

However, these might not be all the algebraic integers: √ 1 α = 2 (1 + d) is a root of 1 √ 1 √ 1 X − (1 + d) X − (1 − d) = X 2 − X + (1 − d) 2 2 4 so α ∈ ON ⇔ d ≡ 1 (mod 4).

Example: Quadratic Fields √ Let N = Q( d) with d a squarefree integer.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 6 / 25 However, these might not be all the algebraic integers: √ 1 α = 2 (1 + d) is a root of 1 √ 1 √ 1 X − (1 + d) X − (1 − d) = X 2 − X + (1 − d) 2 2 4 so α ∈ ON ⇔ d ≡ 1 (mod 4).

Example: Quadratic Fields √ Let N = Q( d) with d a squarefree integer. √ Any element α = u + v d of N with u, v ∈ Z will be an algebraic integer: it is a root of √ √ 2 2 2 X − (u + v d) X − (u − v d) = X − 2uX + (u + v d) ∈ Z[X ].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 6 / 25 √ 1 α = 2 (1 + d) is a root of 1 √ 1 √ 1 X − (1 + d) X − (1 − d) = X 2 − X + (1 − d) 2 2 4 so α ∈ ON ⇔ d ≡ 1 (mod 4).

However, these might not be all the algebraic integers:

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 6 / 25 Example: Quadratic Fields √ Let N = Q( d) with d a squarefree integer. √ Any element α = u + v d of N with u, v ∈ Z will be an algebraic integer: it is a root of √ √ 2 2 2 X − (u + v d) X − (u − v d) = X − 2uX + (u + v d) ∈ Z[X ].

However, these might not be all the algebraic integers: √ 1 α = 2 (1 + d) is a root of 1 √ 1 √ 1 X − (1 + d) X − (1 − d) = X 2 − X + (1 − d) 2 2 4 so α ∈ ON ⇔ d ≡ 1 (mod 4).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 6 / 25 ( 2Z if d ≡ 2, 3 (mod 4), Notice that TrN/Q(ON ) = Z if d ≡ 1 (mod 4).

Theorem √ Let d be a squarefree integer and N = Q( d). Then √ √ ( [ d] = {u + v d : u, v ∈ } if d ≡ 2, 3 (mod 4); Z √ √ Z ON = h 1+ d i n 1+ d o Z 2 = u + v 2 : u, v ∈ Z if d ≡ 1 (mod 4).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 7 / 25 Example (Cyclotomic Fields) 2πi/m Let N = Q(ζm) with ζm = e , a primitive nth root of unity. Then ON = Z[ζm]. We can now ask the question: If N/Q is a Galois extension, does it have a normal integral basis, i.e. does there exist α ∈ ON such that each β ∈ ON can be written uniquely as β = λ · α for some λ in the integral group ring Z[G]?

Theorem √ Let d be a squarefree integer and N = Q( d). Then √ √ ( [ d] = {u + v d : u, v ∈ } if d ≡ 2, 3 (mod 4); Z √ √ Z ON = h 1+ d i n 1+ d o Z 2 = u + v 2 : u, v ∈ Z if d ≡ 1 (mod 4). ( 2Z if d ≡ 2, 3 (mod 4), Notice that TrN/Q(ON ) = Z if d ≡ 1 (mod 4).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 7 / 25 We can now ask the question: If N/Q is a Galois extension, does it have a normal integral basis, i.e. does there exist α ∈ ON such that each β ∈ ON can be written uniquely as β = λ · α for some λ in the integral group ring Z[G]?

Example (Cyclotomic Fields) 2πi/m Let N = Q(ζm) with ζm = e , a primitive nth root of unity. Then ON = Z[ζm].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 7 / 25 If N/Q is a Galois extension, does it have a normal integral basis, i.e. does there exist α ∈ ON such that each β ∈ ON can be written uniquely as β = λ · α for some λ in the integral group ring Z[G]?

Example (Cyclotomic Fields) 2πi/m Let N = Q(ζm) with ζm = e , a primitive nth root of unity. Then ON = Z[ζm]. We can now ask the question:

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 7 / 25 Theorem √ Let d be a squarefree integer and N = Q( d). Then √ √ ( [ d] = {u + v d : u, v ∈ } if d ≡ 2, 3 (mod 4); Z √ √ Z ON = h 1+ d i n 1+ d o Z 2 = u + v 2 : u, v ∈ Z if d ≡ 1 (mod 4). ( 2Z if d ≡ 2, 3 (mod 4), Notice that TrN/Q(ON ) = Z if d ≡ 1 (mod 4).

Example (Cyclotomic Fields) 2πi/m Let N = Q(ζm) with ζm = e , a primitive nth root of unity. Then ON = Z[ζm]. We can now ask the question: If N/Q is a Galois extension, does it have a normal integral basis, i.e. does there exist α ∈ ON such that each β ∈ ON can be written uniquely as β = λ · α for some λ in the integral group ring Z[G]? Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 7 / 25 √ √ √ 1 Take α = 2 (1 + d). Here G = {id, σ} with σ( d) = − d. √ ! √ ! 1 + d 1 + d ∀a, b ∈ : a + b = (a + b + aσ) · . Z 2 2

However, if d ≡ 2, 3 (mod 4), the answer is No.

If we have α with Z[G] · α = ON , then

(1 + σ) · α = TrN/Q(α) = ±1,

but we have seen that TrN/Q(ON ) = 2Z.

Example (Cyclotomic ﬁelds with prime conductor) ∼ × If N = Q(ζp) with p prime, then G = {σ1, σ2, . . . , σp−1} = Fp , where a σa(ζp) = ζp , and we have ON = Z[G] · ζp.

Example (Quadratic Fields) √ If N = Q( d) with d ≡ 1 (mod 4), the answer is Yes.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 8 / 25 However, if d ≡ 2, 3 (mod 4), the answer is No.

If we have α with Z[G] · α = ON , then

(1 + σ) · α = TrN/Q(α) = ±1,

but we have seen that TrN/Q(ON ) = 2Z.

Example (Cyclotomic ﬁelds with prime conductor) ∼ × If N = Q(ζp) with p prime, then G = {σ1, σ2, . . . , σp−1} = Fp , where a σa(ζp) = ζp , and we have ON = Z[G] · ζp.

Example (Quadratic Fields) √ If N = Q( d) with d ≡ 1 (mod 4), the answer is Yes. √ √ √ 1 Take α = 2 (1 + d). Here G = {id, σ} with σ( d) = − d. √ ! √ ! 1 + d 1 + d ∀a, b ∈ : a + b = (a + b + aσ) · . Z 2 2

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 8 / 25 If we have α with Z[G] · α = ON , then

(1 + σ) · α = TrN/Q(α) = ±1,

but we have seen that TrN/Q(ON ) = 2Z.

Example (Cyclotomic ﬁelds with prime conductor) ∼ × If N = Q(ζp) with p prime, then G = {σ1, σ2, . . . , σp−1} = Fp , where a σa(ζp) = ζp , and we have ON = Z[G] · ζp.

However, if d ≡ 2, 3 (mod 4), the answer is No.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 8 / 25 Example (Cyclotomic ﬁelds with prime conductor) ∼ × If N = Q(ζp) with p prime, then G = {σ1, σ2, . . . , σp−1} = Fp , where a σa(ζp) = ζp , and we have ON = Z[G] · ζp.

However, if d ≡ 2, 3 (mod 4), the answer is No.

If we have α with Z[G] · α = ON , then

(1 + σ) · α = TrN/Q(α) = ±1, but we have seen that TrN/Q(ON ) = 2Z.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 8 / 25 Example (Quadratic Fields) √ If N = Q( d) with d ≡ 1 (mod 4), the answer is Yes. √ √ √ 1 Take α = 2 (1 + d). Here G = {id, σ} with σ( d) = − d. √ ! √ ! 1 + d 1 + d ∀a, b ∈ : a + b = (a + b + aσ) · . Z 2 2

However, if d ≡ 2, 3 (mod 4), the answer is No.

If we have α with Z[G] · α = ON , then

(1 + σ) · α = TrN/Q(α) = ±1, but we have seen that TrN/Q(ON ) = 2Z.

Example (Cyclotomic ﬁelds with prime conductor) ∼ × If N = Q(ζp) with p prime, then G = {σ1, σ2, . . . , σp−1} = Fp , where a σa(ζp) = ζp , and we have ON = Z[G] · ζp. Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 8 / 25 As in our quadratic example, if N/K is a wild Galois extension then there cannot be a normal integral basis for N/K. So the tame/wild distinction is fundamental for Galois module structure.

The ﬁrst general result of integral Galois module structure was:

Theorem (Hilbert 1897; Speiser 1916) If N is a tame Galois extension of Q whose Galois group is abelian, then N/Q does have a normal integral basis.

The key idea is that N ⊆ Q(ζm) for some squarefree m.

Tame and wild extensions

We call an extension N/K of number ﬁelds tame if TrN/K (ON ) = OK ; otherwise it is wild.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 9 / 25 So the tame/wild distinction is fundamental for Galois module structure.

The ﬁrst general result of integral Galois module structure was:

Theorem (Hilbert 1897; Speiser 1916) If N is a tame Galois extension of Q whose Galois group is abelian, then N/Q does have a normal integral basis.

The key idea is that N ⊆ Q(ζm) for some squarefree m.

Tame and wild extensions

We call an extension N/K of number ﬁelds tame if TrN/K (ON ) = OK ; otherwise it is wild. As in our quadratic example, if N/K is a wild Galois extension then there cannot be a normal integral basis for N/K.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 9 / 25 The ﬁrst general result of integral Galois module structure was:

Theorem (Hilbert 1897; Speiser 1916) If N is a tame Galois extension of Q whose Galois group is abelian, then N/Q does have a normal integral basis.

The key idea is that N ⊆ Q(ζm) for some squarefree m.

Tame and wild extensions

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 9 / 25 Theorem (Hilbert 1897; Speiser 1916) If N is a tame Galois extension of Q whose Galois group is abelian, then N/Q does have a normal integral basis.

The key idea is that N ⊆ Q(ζm) for some squarefree m.

Tame and wild extensions

The ﬁrst general result of integral Galois module structure was:

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 9 / 25 The key idea is that N ⊆ Q(ζm) for some squarefree m.

Tame and wild extensions

The ﬁrst general result of integral Galois module structure was:

Theorem (Hilbert 1897; Speiser 1916) If N is a tame Galois extension of Q whose Galois group is abelian, then N/Q does have a normal integral basis.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 9 / 25 Tame and wild extensions

The ﬁrst general result of integral Galois module structure was:

Theorem (Hilbert 1897; Speiser 1916) If N is a tame Galois extension of Q whose Galois group is abelian, then N/Q does have a normal integral basis.

The key idea is that N ⊆ Q(ζm) for some squarefree m.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 9 / 25 But we could try replacing Z[G] with a bigger ring.

The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module. Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

The key idea is that N ⊆ Q(ζm) for some m.

Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module. Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

The key idea is that N ⊆ Q(ζm) for some m.

Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module. But we could try replacing Z[G] with a bigger ring.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module. Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

The key idea is that N ⊆ Q(ζm) for some m.

Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module. But we could try replacing Z[G] with a bigger ring.

The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

The key idea is that N ⊆ Q(ζm) for some m.

Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module. But we could try replacing Z[G] with a bigger ring.

The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

The key idea is that N ⊆ Q(ζm) for some m.

Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module. But we could try replacing Z[G] with a bigger ring.

The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module. Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 The key idea is that N ⊆ Q(ζm) for some m.

Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module. But we could try replacing Z[G] with a bigger ring.

The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module. Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 Wild abelian extensions of Q If N/Q is a wild abelian extension, then ON cannot be a free Z[G]-module. But we could try replacing Z[G] with a bigger ring.

The associated order of ON is

A(N/Q) = {λ ∈ Q[G]: λ · ON ⊆ ON }.

Then A(N/Q) is a subring of Q[G] containing Z[G], and ON is an A(N/Q)-module. Also A(N/Q) = Z[G] ⇔ N/Q is tame.

Theorem (Leopoldt, 1959) If N is a wild Galois extension of Q whose Galois group is abelian, then ON is a free module over A(N/Q).

The key idea is that N ⊆ Q(ζm) for some m. Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 10 / 25 This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 The Hilbert-Speiser Theorem and Leopoldt’s Theorem mean we have a good understanding of Galois module structure for abelian extensions of Q. We would like to generalise in two directions: non-abelian Galois groups; base ﬁelds K ⊃ Q; (both for tame and wild extensions).

Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 We would like to generalise in two directions: non-abelian Galois groups; base ﬁelds K ⊃ Q; (both for tame and wild extensions).

Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

The Hilbert-Speiser Theorem and Leopoldt’s Theorem mean we have a good understanding of Galois module structure for abelian extensions of Q.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 non-abelian Galois groups; base ﬁelds K ⊃ Q; (both for tame and wild extensions).

Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

The Hilbert-Speiser Theorem and Leopoldt’s Theorem mean we have a good understanding of Galois module structure for abelian extensions of Q. We would like to generalise in two directions:

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 base ﬁelds K ⊃ Q; (both for tame and wild extensions).

Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 (both for tame and wild extensions).

Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 Example (Wild quadratic extensions) √ If N = Q( d) with d ≡ 2, 3 (mod 4) then 1 + σ 1 − σ A(N/ ) = · + · . Q Z 2 Z 2

This is the maximal order in Q[C2]. We have √ ON = A(N/Q) · (1 + d).

The Hilbert-Speiser Theorem and Leopoldt’s Theorem mean we have a good understanding of Galois module structure for abelian extensions of Q. We would like to generalise in two directions: non-abelian Galois groups; base ﬁelds K ⊃ Q; (both for tame and wild extensions). Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 11 / 25 So being tame does not guarantee the existence of a normal integral basis.

Question: What determines which tame Q8-extensions do have a normal integral basis?

A tame non-abelian case

∼ For tame Galois extensions N/Q with Gal(N/Q) = Q8 (the quaternion group of order 8), calculations of Martinet (1971) showed that there is a normal integral basis in some cases, but not in others.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 12 / 25 Question: What determines which tame Q8-extensions do have a normal integral basis?

A tame non-abelian case

∼ For tame Galois extensions N/Q with Gal(N/Q) = Q8 (the quaternion group of order 8), calculations of Martinet (1971) showed that there is a normal integral basis in some cases, but not in others.

So being tame does not guarantee the existence of a normal integral basis.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 12 / 25 A tame non-abelian case

∼ For tame Galois extensions N/Q with Gal(N/Q) = Q8 (the quaternion group of order 8), calculations of Martinet (1971) showed that there is a normal integral basis in some cases, but not in others.

So being tame does not guarantee the existence of a normal integral basis.

Question: What determines which tame Q8-extensions do have a normal integral basis?

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 12 / 25 There are no irreducible symplectic characters if G is abelian or |G| is odd.

The group Q8 has just one irreducible symplectic character.

If N/K is a Galois extension of number ﬁelds with Galois group G, then for each irreducible character χ of G there is a complex function L(s, N/K, χ) called the Artin L-function.

Digression: irreducible characters and L-functions

Attached to a ﬁnite group G are certain functions χ : G → C called irreducible characters. Some of these may be symplectic.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 13 / 25 The group Q8 has just one irreducible symplectic character.

If N/K is a Galois extension of number ﬁelds with Galois group G, then for each irreducible character χ of G there is a complex function L(s, N/K, χ) called the Artin L-function.

Digression: irreducible characters and L-functions

Attached to a ﬁnite group G are certain functions χ : G → C called irreducible characters. Some of these may be symplectic. There are no irreducible symplectic characters if G is abelian or |G| is odd.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 13 / 25 If N/K is a Galois extension of number ﬁelds with Galois group G, then for each irreducible character χ of G there is a complex function L(s, N/K, χ) called the Artin L-function.

Digression: irreducible characters and L-functions

The group Q8 has just one irreducible symplectic character.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 13 / 25 Digression: irreducible characters and L-functions

The group Q8 has just one irreducible symplectic character.

If N/K is a Galois extension of number ﬁelds with Galois group G, then for each irreducible character χ of G there is a complex function L(s, N/K, χ) called the Artin L-function.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 13 / 25 Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

s s log(n) where n = e for s ∈ C. This is the famous Riemann Zeta Function ζ(s). The series (and the product) converge for Re(s) > 1. However, ζ(s) is deﬁned for all s ∈ C (except for a pole at s = 1) by analytic continuation, and it has a functional equation relating ζ(s) to ζ(1 − s).

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 This is the famous Riemann Zeta Function ζ(s). The series (and the product) converge for Re(s) > 1. However, ζ(s) is deﬁned for all s ∈ C (except for a pole at s = 1) by analytic continuation, and it has a functional equation relating ζ(s) to ζ(1 − s).

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

s s log(n) where n = e for s ∈ C.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 The series (and the product) converge for Re(s) > 1. However, ζ(s) is deﬁned for all s ∈ C (except for a pole at s = 1) by analytic continuation, and it has a functional equation relating ζ(s) to ζ(1 − s).

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

s s log(n) where n = e for s ∈ C. This is the famous Riemann Zeta Function ζ(s).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 However, ζ(s) is deﬁned for all s ∈ C (except for a pole at s = 1) by analytic continuation, and it has a functional equation relating ζ(s) to ζ(1 − s).

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

s s log(n) where n = e for s ∈ C. This is the famous Riemann Zeta Function ζ(s). The series (and the product) converge for Re(s) > 1.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 In general, each Artin L-function L(s, N/K, χ) is a meromorphic function on the whole of C, and has a functional equation. The functional equation for L(s, N/K, χ) involves a number W (N/K, χ) called the root number. If χ is an irreducible symplectic character then W (N/K, χ) = ±1.

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 The functional equation for L(s, N/K, χ) involves a number W (N/K, χ) called the root number. If χ is an irreducible symplectic character then W (N/K, χ) = ±1.

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

In general, each Artin L-function L(s, N/K, χ) is a meromorphic function on the whole of C, and has a functional equation.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 If χ is an irreducible symplectic character then W (N/K, χ) = ±1.

The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 The simplest possible example is when N = K = Q, so G = {e}, with just one irreducible character, χ0. Then X 1 Y 1 L(s, / , χ ) = = , Q Q 0 ns 1 − p−s n≥1 p prime

In general, each Artin L-function L(s, N/K, χ) is a meromorphic function on the whole of C, and has a functional equation. The functional equation for L(s, N/K, χ) involves a number W (N/K, χ) called the root number. If χ is an irreducible symplectic character then W (N/K, χ) = ±1. Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 14 / 25 Theorem (Conjectured Serre 1971; proved Fr¨ohlich,1972)

Let N/Q be a tame Galois extension with Galois group Q8, and let χ be the irreducible symplectic character of Q8. Then

ON is free over Z[Q8] ⇔ W (N/Q, χ) = +1.

This shows an unexpected connection between algebraic information (whether N has a normal integral basis) and anayltic information (root numbers of L-functions).

To ﬁt this into a general theory, we need to understand the signiﬁcance of tameness.

Back to quaternion extensions:

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 15 / 25 This shows an unexpected connection between algebraic information (whether N has a normal integral basis) and anayltic information (root numbers of L-functions).

To ﬁt this into a general theory, we need to understand the signiﬁcance of tameness.

Back to quaternion extensions: Theorem (Conjectured Serre 1971; proved Fr¨ohlich,1972)

Let N/Q be a tame Galois extension with Galois group Q8, and let χ be the irreducible symplectic character of Q8. Then

ON is free over Z[Q8] ⇔ W (N/Q, χ) = +1.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 15 / 25 To ﬁt this into a general theory, we need to understand the signiﬁcance of tameness.

Back to quaternion extensions: Theorem (Conjectured Serre 1971; proved Fr¨ohlich,1972)

Let N/Q be a tame Galois extension with Galois group Q8, and let χ be the irreducible symplectic character of Q8. Then

ON is free over Z[Q8] ⇔ W (N/Q, χ) = +1.

This shows an unexpected connection between algebraic information (whether N has a normal integral basis) and anayltic information (root numbers of L-functions).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 15 / 25 Back to quaternion extensions: Theorem (Conjectured Serre 1971; proved Fr¨ohlich,1972)

Let N/Q be a tame Galois extension with Galois group Q8, and let χ be the irreducible symplectic character of Q8. Then

ON is free over Z[Q8] ⇔ W (N/Q, χ) = +1.

This shows an unexpected connection between algebraic information (whether N has a normal integral basis) and anayltic information (root numbers of L-functions).

To ﬁt this into a general theory, we need to understand the signiﬁcance of tameness.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 15 / 25 Then Z(p) is a local ring, i.e. a ring with only one maximal ideal, namely pZ(p). The ﬁeld of fractions of Z(p) is Q.

The only prime number in Z which “survives” in Z(p) is p itself; all other primes become units in Z(p).

(We could also take completions and work with the p-adic integers Zp instead of Z(p). The ﬁeld of fractions of Zp is the the ﬁeld Qp of p-adic numbers.)

Locally free Z[G]-modules

For each prime number p, we deﬁne the localisation of Z at p: n a o = : a, b ∈ , p b . Z(p) b Z -

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 16 / 25 The only prime number in Z which “survives” in Z(p) is p itself; all other primes become units in Z(p).

(We could also take completions and work with the p-adic integers Zp instead of Z(p). The ﬁeld of fractions of Zp is the the ﬁeld Qp of p-adic numbers.)

Locally free Z[G]-modules

For each prime number p, we deﬁne the localisation of Z at p: n a o = : a, b ∈ , p b . Z(p) b Z -

Then Z(p) is a local ring, i.e. a ring with only one maximal ideal, namely pZ(p). The ﬁeld of fractions of Z(p) is Q.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 16 / 25 (We could also take completions and work with the p-adic integers Zp instead of Z(p). The ﬁeld of fractions of Zp is the the ﬁeld Qp of p-adic numbers.)

Locally free Z[G]-modules

For each prime number p, we deﬁne the localisation of Z at p: n a o = : a, b ∈ , p b . Z(p) b Z -

Then Z(p) is a local ring, i.e. a ring with only one maximal ideal, namely pZ(p). The ﬁeld of fractions of Z(p) is Q.

The only prime number in Z which “survives” in Z(p) is p itself; all other primes become units in Z(p).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 16 / 25 Locally free Z[G]-modules

For each prime number p, we deﬁne the localisation of Z at p: n a o = : a, b ∈ , p b . Z(p) b Z -

Then Z(p) is a local ring, i.e. a ring with only one maximal ideal, namely pZ(p). The ﬁeld of fractions of Z(p) is Q.

The only prime number in Z which “survives” in Z(p) is p itself; all other primes become units in Z(p).

(We could also take completions and work with the p-adic integers Zp instead of Z(p). The ﬁeld of fractions of Zp is the the ﬁeld Qp of p-adic numbers.)

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 16 / 25 Then

L/Q is tame ⇔ ON,(p) is a free Z(p)[G]-module for each prime p.

We then say ON is a locally free Z[G]-module. So N/Q has a normal integral basis if and only if the locally free Z[G]-module ON is free. Fröhlichconstructed a finite abelian group Cl(Z[G]) which classifies locally free Z[G]-modules (up to stable isomorphism).

We now consider tame Galois extensions N/K (with base ﬁeld K ⊇ Q), and consider ON as a locally free Z[G]-module of rank [K : Q].

For a Galois extension N/Q with Gal(N/Q) = G, we know that ON is a Z[G]-module. We can extend scalars from Z to Z(p), so that ON,(p) is a Z(p)[G]-module.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 17 / 25 So N/Q has a normal integral basis if and only if the locally free Z[G]-module ON is free. Fröhlichconstructed a finite abelian group Cl(Z[G]) which classifies locally free Z[G]-modules (up to stable isomorphism).

We now consider tame Galois extensions N/K (with base ﬁeld K ⊇ Q), and consider ON as a locally free Z[G]-module of rank [K : Q].

For a Galois extension N/Q with Gal(N/Q) = G, we know that ON is a Z[G]-module. We can extend scalars from Z to Z(p), so that ON,(p) is a Z(p)[G]-module. Then

L/Q is tame ⇔ ON,(p) is a free Z(p)[G]-module for each prime p.

We then say ON is a locally free Z[G]-module.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 17 / 25 Fröhlichconstructed a finite abelian group Cl(Z[G]) which classifies locally free Z[G]-modules (up to stable isomorphism).

We now consider tame Galois extensions N/K (with base ﬁeld K ⊇ Q), and consider ON as a locally free Z[G]-module of rank [K : Q].

For a Galois extension N/Q with Gal(N/Q) = G, we know that ON is a Z[G]-module. We can extend scalars from Z to Z(p), so that ON,(p) is a Z(p)[G]-module. Then

L/Q is tame ⇔ ON,(p) is a free Z(p)[G]-module for each prime p.

We then say ON is a locally free Z[G]-module. So N/Q has a normal integral basis if and only if the locally free Z[G]-module ON is free.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 17 / 25 We now consider tame Galois extensions N/K (with base ﬁeld K ⊇ Q), and consider ON as a locally free Z[G]-module of rank [K : Q].

For a Galois extension N/Q with Gal(N/Q) = G, we know that ON is a Z[G]-module. We can extend scalars from Z to Z(p), so that ON,(p) is a Z(p)[G]-module. Then

L/Q is tame ⇔ ON,(p) is a free Z(p)[G]-module for each prime p.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 17 / 25 For a Galois extension N/Q with Gal(N/Q) = G, we know that ON is a Z[G]-module. We can extend scalars from Z to Z(p), so that ON,(p) is a Z(p)[G]-module. Then

L/Q is tame ⇔ ON,(p) is a free Z(p)[G]-module for each prime p.

We now consider tame Galois extensions N/K (with base ﬁeld K ⊇ Q), and consider ON as a locally free Z[G]-module of rank [K : Q].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 17 / 25 The main theorem of tame Galois module structure

Theorem (Taylor, 1981)

For a tame Galois extension N/K of number ﬁelds, the class of ON in Cl(Z[G]) is determined by the W (N/K, χ) for the irreducible symplectic characters χ of G. In particular, if G has no such characters, then ON is a free Z[G]-module.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 18 / 25 It is locally free of rank 1, and determines a class in the locally free classgroup Cl(OK [G]). In general, this class is not determined by analytic invariants. Instead, we can ask which classes in Cl(OK [G]) are obtained as N varies over all tame G-extensions of K. This gives the set of realisable classes R(OK [G]).

McCulloh (1987) determined R(OK [G]) when G is abelian. There are similar results for some non-abelian groups G, e.g. for certain r metabelian groups G = (Cp ) o Cm (with ζp ∈ K); cf. Byott and Soda¨ıgui (2013).

It is expected that R(OK [G]) is always a subgroup of OK [G]. In general, we do not even know it is non-empty.

Relative tame Galois Module Structure:

What can we say about ON as an OK [G]-module (instead of Z[G]-module)?

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 In general, this class is not determined by analytic invariants. Instead, we can ask which classes in Cl(OK [G]) are obtained as N varies over all tame G-extensions of K. This gives the set of realisable classes R(OK [G]).

It is expected that R(OK [G]) is always a subgroup of OK [G]. In general, we do not even know it is non-empty.

Relative tame Galois Module Structure:

What can we say about ON as an OK [G]-module (instead of Z[G]-module)? It is locally free of rank 1, and determines a class in the locally free classgroup Cl(OK [G]).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 McCulloh (1987) determined R(OK [G]) when G is abelian. There are similar results for some non-abelian groups G, e.g. for certain r metabelian groups G = (Cp ) o Cm (with ζp ∈ K); cf. Byott and Soda¨ıgui (2013).

It is expected that R(OK [G]) is always a subgroup of OK [G]. In general, we do not even know it is non-empty.

Relative tame Galois Module Structure:

What can we say about ON as an OK [G]-module (instead of Z[G]-module)? It is locally free of rank 1, and determines a class in the locally free classgroup Cl(OK [G]). In general, this class is not determined by analytic invariants. Instead, we can ask which classes in Cl(OK [G]) are obtained as N varies over all tame G-extensions of K. This gives the set of realisable classes R(OK [G]).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 There are similar results for some non-abelian groups G, e.g. for certain r metabelian groups G = (Cp ) o Cm (with ζp ∈ K); cf. Byott and Soda¨ıgui (2013).

It is expected that R(OK [G]) is always a subgroup of OK [G]. In general, we do not even know it is non-empty.

Relative tame Galois Module Structure:

McCulloh (1987) determined R(OK [G]) when G is abelian.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 It is expected that R(OK [G]) is always a subgroup of OK [G]. In general, we do not even know it is non-empty.

Relative tame Galois Module Structure:

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 In general, we do not even know it is non-empty.

Relative tame Galois Module Structure:

It is expected that R(OK [G]) is always a subgroup of OK [G].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 Relative tame Galois Module Structure:

It is expected that R(OK [G]) is always a subgroup of OK [G]. In general, we do not even know it is non-empty.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 19 / 25 We can ask if ON is locally free over its associated order

AN/K = {λ ∈ K[G]: λ · ON ⊆ ON }.

This is a “local” question (i.e. it depends on one prime p at a time), so we can localise (and complete) at p and work in a p-adic setting.

Changing notation, let K be a ﬁnite extension of Qp, with ring of integers OK , and let N/K be a Galois extension with Galois group G. ∼ We just consider the simplest interesting case: G = Cp.

Wild extensions

If N/K is a wild Galois extension and G = Gal(N/K), then ON cannot be even locally free over OK [G].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 20 / 25 This is a “local” question (i.e. it depends on one prime p at a time), so we can localise (and complete) at p and work in a p-adic setting.

Changing notation, let K be a ﬁnite extension of Qp, with ring of integers OK , and let N/K be a Galois extension with Galois group G. ∼ We just consider the simplest interesting case: G = Cp.

Wild extensions

If N/K is a wild Galois extension and G = Gal(N/K), then ON cannot be even locally free over OK [G].

We can ask if ON is locally free over its associated order

AN/K = {λ ∈ K[G]: λ · ON ⊆ ON }.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 20 / 25 Changing notation, let K be a ﬁnite extension of Qp, with ring of integers OK , and let N/K be a Galois extension with Galois group G. ∼ We just consider the simplest interesting case: G = Cp.

Wild extensions

If N/K is a wild Galois extension and G = Gal(N/K), then ON cannot be even locally free over OK [G].

We can ask if ON is locally free over its associated order

AN/K = {λ ∈ K[G]: λ · ON ⊆ ON }.

This is a “local” question (i.e. it depends on one prime p at a time), so we can localise (and complete) at p and work in a p-adic setting.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 20 / 25 ∼ We just consider the simplest interesting case: G = Cp.

Wild extensions

If N/K is a wild Galois extension and G = Gal(N/K), then ON cannot be even locally free over OK [G].

We can ask if ON is locally free over its associated order

AN/K = {λ ∈ K[G]: λ · ON ⊆ ON }.

This is a “local” question (i.e. it depends on one prime p at a time), so we can localise (and complete) at p and work in a p-adic setting.

Changing notation, let K be a ﬁnite extension of Qp, with ring of integers OK , and let N/K be a Galois extension with Galois group G.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 20 / 25 Wild extensions

If N/K is a wild Galois extension and G = Gal(N/K), then ON cannot be even locally free over OK [G].

We can ask if ON is locally free over its associated order

AN/K = {λ ∈ K[G]: λ · ON ⊆ ON }.

This is a “local” question (i.e. it depends on one prime p at a time), so we can localise (and complete) at p and work in a p-adic setting.

Changing notation, let K be a ﬁnite extension of Qp, with ring of integers OK , and let N/K be a Galois extension with Galois group G. ∼ We just consider the simplest interesting case: G = Cp.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 20 / 25 Let πK be a generator for this ideal. (If K = Qp we can choose πK = p). m × If x ∈ K\{0}, we can write x = uπK for some unit u ∈ OK and some m ∈ Z.

We have x ∈ OK ⇔ m ≥ 0.

We deﬁne the valuation vK : K → Z ∪ {∞} by ( m if x 6= 0; vK (x) = ∞ if x = 0.

Similarly, ON has a unique maximal ideal πN ON , and we have a valuation vN : N Z ∪ {∞}.

OK is a Principal Ideal Domain with a unique maximal ideal.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 21 / 25 m × If x ∈ K\{0}, we can write x = uπK for some unit u ∈ OK and some m ∈ Z.

We have x ∈ OK ⇔ m ≥ 0.

We deﬁne the valuation vK : K → Z ∪ {∞} by ( m if x 6= 0; vK (x) = ∞ if x = 0.

Similarly, ON has a unique maximal ideal πN ON , and we have a valuation vN : N Z ∪ {∞}.

OK is a Principal Ideal Domain with a unique maximal ideal.

Let πK be a generator for this ideal. (If K = Qp we can choose πK = p).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 21 / 25 We have x ∈ OK ⇔ m ≥ 0.

We deﬁne the valuation vK : K → Z ∪ {∞} by ( m if x 6= 0; vK (x) = ∞ if x = 0.

Similarly, ON has a unique maximal ideal πN ON , and we have a valuation vN : N Z ∪ {∞}.

OK is a Principal Ideal Domain with a unique maximal ideal.

Let πK be a generator for this ideal. (If K = Qp we can choose πK = p). m × If x ∈ K\{0}, we can write x = uπK for some unit u ∈ OK and some m ∈ Z.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 21 / 25 We deﬁne the valuation vK : K → Z ∪ {∞} by ( m if x 6= 0; vK (x) = ∞ if x = 0.

Similarly, ON has a unique maximal ideal πN ON , and we have a valuation vN : N Z ∪ {∞}.

OK is a Principal Ideal Domain with a unique maximal ideal.

Let πK be a generator for this ideal. (If K = Qp we can choose πK = p). m × If x ∈ K\{0}, we can write x = uπK for some unit u ∈ OK and some m ∈ Z.

We have x ∈ OK ⇔ m ≥ 0.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 21 / 25 Similarly, ON has a unique maximal ideal πN ON , and we have a valuation vN : N Z ∪ {∞}.

OK is a Principal Ideal Domain with a unique maximal ideal.

Let πK be a generator for this ideal. (If K = Qp we can choose πK = p). m × If x ∈ K\{0}, we can write x = uπK for some unit u ∈ OK and some m ∈ Z.

We have x ∈ OK ⇔ m ≥ 0.

We deﬁne the valuation vK : K → Z ∪ {∞} by ( m if x 6= 0; vK (x) = ∞ if x = 0.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 21 / 25 OK is a Principal Ideal Domain with a unique maximal ideal.

Let πK be a generator for this ideal. (If K = Qp we can choose πK = p). m × If x ∈ K\{0}, we can write x = uπK for some unit u ∈ OK and some m ∈ Z.

We have x ∈ OK ⇔ m ≥ 0.

We deﬁne the valuation vK : K → Z ∪ {∞} by ( m if x 6= 0; vK (x) = ∞ if x = 0.

Similarly, ON has a unique maximal ideal πN ON , and we have a valuation vN : N Z ∪ {∞}.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 21 / 25 If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G]. ∼ Let G = hσi = Cp, and let

b = vN ((σ − 1) · πN ) − 1 ∈ Z>0. Then, for all x ∈ N\{0}, ( = vN (x) + b if p - vN (x), vN ((σ − 1) · x) > vN (x) + b if p | vN (x). b is called the ramiﬁcation break of N/K. It turns out that the possible values for b are: ep 1 ≤ b ≤ where e = v (π ), p − 1 Qp K p - b unless b = ep/(p − 1).

There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G]. ∼ Let G = hσi = Cp, and let

There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 ∼ Let G = hσi = Cp, and let

There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G].

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 Then, for all x ∈ N\{0}, ( = vN (x) + b if p - vN (x), vN ((σ − 1) · x) > vN (x) + b if p | vN (x). b is called the ramiﬁcation break of N/K. It turns out that the possible values for b are: ep 1 ≤ b ≤ where e = v (π ), p − 1 Qp K p - b unless b = ep/(p − 1).

There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G]. ∼ Let G = hσi = Cp, and let

b = vN ((σ − 1) · πN ) − 1 ∈ Z>0.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 b is called the ramiﬁcation break of N/K. It turns out that the possible values for b are: ep 1 ≤ b ≤ where e = v (π ), p − 1 Qp K p - b unless b = ep/(p − 1).

There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G]. ∼ Let G = hσi = Cp, and let

b = vN ((σ − 1) · πN ) − 1 ∈ Z>0. Then, for all x ∈ N\{0}, ( = vN (x) + b if p - vN (x), vN ((σ − 1) · x) > vN (x) + b if p | vN (x).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 It turns out that the possible values for b are: ep 1 ≤ b ≤ where e = v (π ), p − 1 Qp K p - b unless b = ep/(p − 1).

There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G]. ∼ Let G = hσi = Cp, and let

b = vN ((σ − 1) · πN ) − 1 ∈ Z>0. Then, for all x ∈ N\{0}, ( = vN (x) + b if p - vN (x), vN ((σ − 1) · x) > vN (x) + b if p | vN (x). b is called the ramiﬁcation break of N/K.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 There are two possibilities for our degree p extension N/K: either vN (πK ) = 1 or vN (πK ) = p.

If vN (πK ) = 1 then N/K is unramiﬁed. Then it is tame and ON is free over A(N/K) = OK [G].

If vN (πK ) = p then N/K is totally ramiﬁed and A(N/K) ) OK [G]. ∼ Let G = hσi = Cp, and let

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 22 / 25 ep b < − 1. p − 1

2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN . Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K). This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

To avoid special cases, assume that b is not too close to its upper bound

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN . Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K). This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

To avoid special cases, assume that b is not too close to its upper bound ep b < − 1. p − 1

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K). This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

To avoid special cases, assume that b is not too close to its upper bound ep b < − 1. p − 1

2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN .

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K). This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

To avoid special cases, assume that b is not too close to its upper bound ep b < − 1. p − 1

2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN . Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

To avoid special cases, assume that b is not too close to its upper bound ep b < − 1. p − 1

2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN . Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

To avoid special cases, assume that b is not too close to its upper bound ep b < − 1. p − 1

2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN . Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K). This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 To avoid special cases, assume that b is not too close to its upper bound ep b < − 1. p − 1

2 p−1 Then ON has OK -basis 1, πN , πN ,..., πN . Also, K[G] has a K-basis 1, σ − 1, (σ − 1)2,...,(σ − 1)p−1.

For 0 ≤ j ≤ p − 1, let r(j) ∈ Z be as large as possible with

−r(j) j πK (σ − 1) ∈ A(N/K). This means (σ − 1)j · πi ∈ O for 0 ≤ i ≤ p − 1, r(j) N N πK so that

−pr + bj + i ≥ 0 if i ≡ 0, b, 2b,..., (p − 1 − j)b (mod p).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 23 / 25 Using this we can show that A(N/K) has a basis of the form

−r(j) j πK (σ − 1) ∈ A(N/K) for 0 ≤ j ≤ p − 1, where the r(j) can be calculated, and we can check when ON is free over A(N/K). Theorem (Bertrandias and Ferton, 1972)

Let b = pb1 + b0 with 1 ≤ b0 ≤ p − 1. Then

ON is free over A(N/K) ⇔ b0 divides p − 1.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 24 / 25 We then have two ramiﬁcation breaks b2 ≥ b1 with

vN ((σ − 1) · x) = vN (x) + b1 if p - vN (x), vN ((τ − 1) · x) = vN (x) + b2 if p - vN (x),

(and with “=” replaced by “>” when p | vN (x)). This does not give us enough information to ﬁnd A(N/K). What we really need in order to generalise the calculations for the degree p case is two elements Ψ1,Ψ2 ∈ K[G] with the following property: 2 If vN (x) ≡ −a0b2 − pa1b1 (mod p ), with 0 ≤ a0, a1 ≤ p − 1, then

vN (Ψ1 · x) = vN (x) + b2 if a0 6= 0,

vN (Ψ2 · x) = vN (x) + pb1 if a1 6= 0.

Sometimes (but not always), it is possible to construct suitable Ψ1,Ψ2. This is the starting point for the theory of Galois Scaﬀolds.

What happens for degree p2 (or higher)? ∼ Now suppose Gal(N/K) = hσ, τi = CP × Cp with N/K totally ramiﬁed.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 25 / 25 This does not give us enough information to ﬁnd A(N/K). What we really need in order to generalise the calculations for the degree p case is two elements Ψ1,Ψ2 ∈ K[G] with the following property: 2 If vN (x) ≡ −a0b2 − pa1b1 (mod p ), with 0 ≤ a0, a1 ≤ p − 1, then

vN (Ψ1 · x) = vN (x) + b2 if a0 6= 0,

vN (Ψ2 · x) = vN (x) + pb1 if a1 6= 0.

Sometimes (but not always), it is possible to construct suitable Ψ1,Ψ2. This is the starting point for the theory of Galois Scaﬀolds.

What happens for degree p2 (or higher)? ∼ Now suppose Gal(N/K) = hσ, τi = CP × Cp with N/K totally ramiﬁed.

We then have two ramiﬁcation breaks b2 ≥ b1 with

vN ((σ − 1) · x) = vN (x) + b1 if p - vN (x), vN ((τ − 1) · x) = vN (x) + b2 if p - vN (x),

(and with “=” replaced by “>” when p | vN (x)).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 25 / 25 What we really need in order to generalise the calculations for the degree p case is two elements Ψ1,Ψ2 ∈ K[G] with the following property: 2 If vN (x) ≡ −a0b2 − pa1b1 (mod p ), with 0 ≤ a0, a1 ≤ p − 1, then

vN (Ψ1 · x) = vN (x) + b2 if a0 6= 0,

vN (Ψ2 · x) = vN (x) + pb1 if a1 6= 0.

Sometimes (but not always), it is possible to construct suitable Ψ1,Ψ2. This is the starting point for the theory of Galois Scaﬀolds.

What happens for degree p2 (or higher)? ∼ Now suppose Gal(N/K) = hσ, τi = CP × Cp with N/K totally ramiﬁed.

We then have two ramiﬁcation breaks b2 ≥ b1 with

vN ((σ − 1) · x) = vN (x) + b1 if p - vN (x), vN ((τ − 1) · x) = vN (x) + b2 if p - vN (x),

(and with “=” replaced by “>” when p | vN (x)). This does not give us enough information to ﬁnd A(N/K).

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 25 / 25 2 If vN (x) ≡ −a0b2 − pa1b1 (mod p ), with 0 ≤ a0, a1 ≤ p − 1, then

vN (Ψ1 · x) = vN (x) + b2 if a0 6= 0,

vN (Ψ2 · x) = vN (x) + pb1 if a1 6= 0.

Sometimes (but not always), it is possible to construct suitable Ψ1,Ψ2. This is the starting point for the theory of Galois Scaﬀolds.

What happens for degree p2 (or higher)? ∼ Now suppose Gal(N/K) = hσ, τi = CP × Cp with N/K totally ramiﬁed.

We then have two ramiﬁcation breaks b2 ≥ b1 with

vN ((σ − 1) · x) = vN (x) + b1 if p - vN (x), vN ((τ − 1) · x) = vN (x) + b2 if p - vN (x),

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 25 / 25 This is the starting point for the theory of Galois Scaﬀolds.

What happens for degree p2 (or higher)? ∼ Now suppose Gal(N/K) = hσ, τi = CP × Cp with N/K totally ramiﬁed.

We then have two ramiﬁcation breaks b2 ≥ b1 with

vN ((σ − 1) · x) = vN (x) + b1 if p - vN (x), vN ((τ − 1) · x) = vN (x) + b2 if p - vN (x),

vN (Ψ1 · x) = vN (x) + b2 if a0 6= 0,

vN (Ψ2 · x) = vN (x) + pb1 if a1 6= 0.

Sometimes (but not always), it is possible to construct suitable Ψ1,Ψ2.

Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 25 / 25 What happens for degree p2 (or higher)? ∼ Now suppose Gal(N/K) = hσ, τi = CP × Cp with N/K totally ramiﬁed.

We then have two ramiﬁcation breaks b2 ≥ b1 with

vN ((σ − 1) · x) = vN (x) + b1 if p - vN (x), vN ((τ − 1) · x) = vN (x) + b2 if p - vN (x),

vN (Ψ1 · x) = vN (x) + b2 if a0 6= 0,

vN (Ψ2 · x) = vN (x) + pb1 if a1 6= 0.

Sometimes (but not always), it is possible to construct suitable Ψ1,Ψ2. This is the starting point for the theory of Galois Scaﬀolds. Nigel Byott (University of Exeter) Galois Module Structure Omaha, May 2019 25 / 25