Hopf-Galois Structures on Galois Extensions of Fields

Hopf-Galois Structures on Galois Extensions of Fields Nigel Byott University of Exeter, UK 23 June 2017

Hopf Galois Extensions Let H be a finite dimensional cocommutative Hopf algebra over a field K, with comultiplication : H H K H, (h) = (h) h (1) h (2) = (h) h (2) h (1), counit ɛ : H K and antipode S : H H. An H-Galois extension of K is a K-algebra A with a K-linear action of H on A such that h (ab) = (h) (h (1) a)(h (2) b) for all h H and a, b A; h 1 A = ɛ(h)1 A for all h H; the K-linear map θ : A K H End K (A), given by θ(a h)(b) = a(h b), is bijective. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 2 / 11

Hopf Galois Extensions Let H be a finite dimensional cocommutative Hopf algebra over a field K, with comultiplication : H H K H, (h) = (h) h (1) h (2) = (h) h (2) h (1), counit ɛ : H K and antipode S : H H. An H-Galois extension of K is a K-algebra A with a K-linear action of H on A such that h (ab) = (h) (h (1) a)(h (2) b) for all h H and a, b A; h 1 A = ɛ(h)1 A for all h H; the K-linear map θ : A K H End K (A), given by θ(a h)(b) = a(h b), is bijective. Motivating Example: Let A = L be a finite Galois field extension of K with Galois group G = Gal(L/K). Then L is a K[G]-Galois extension of K. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 2 / 11

Why is this Interesting? Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 3 / 11

Why is this Interesting? 1 (Original motivation, Chase & Sweedler, 1969): Analogue of Galois theory for inseparable field extensions. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 3 / 11

Why is this Interesting? 1 (Original motivation, Chase & Sweedler, 1969): Analogue of Galois theory for inseparable field extensions. 2 (Algebraic Geometry): A is a commutative H-Galois extension of K Spec(A) is a G-torsor over Spec(K), where G is the affine group scheme Spec(H ) represented by the dual Hopf algebra H = Hom K (H, K) of H. 3 (Algebraic Number Theory): If L/K is a Galois extension of (say) number fields with Galois group G and rings of algebraic integers O L, O K, then arithmetic Galois module theory studies O L as an O K [G]-module. Other H-Galois extensions give interesting variants of this question. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 3 / 11

Why is this Interesting? 1 (Original motivation, Chase & Sweedler, 1969): Analogue of Galois theory for inseparable field extensions. 2 (Algebraic Geometry): A is a commutative H-Galois extension of K Spec(A) is a G-torsor over Spec(K), where G is the affine group scheme Spec(H ) represented by the dual Hopf algebra H = Hom K (H, K) of H. 3 (Algebraic Number Theory): If L/K is a Galois extension of (say) number fields with Galois group G and rings of algebraic integers O L, O K, then arithmetic Galois module theory studies O L as an O K [G]-module. Other H-Galois extensions give interesting variants of this question. 4 Hopf-Galois extensions are related to (skew) braces. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 3 / 11

The Greither-Pareigis Theorem Let L/K be a Galois extension of fields, G = Gal(L/K). Then L is H-Galois for H = K[G]. Question: Does it have any other Hopf-Galois structures? i.e. Are there other Hopf algebras H making L an H-Galois extension? Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 4 / 11

The Greither-Pareigis Theorem Let L/K be a Galois extension of fields, G = Gal(L/K). Then L is H-Galois for H = K[G]. Question: Does it have any other Hopf-Galois structures? i.e. Are there other Hopf algebras H making L an H-Galois extension? If so, the L-algebra L K L = Map(G, L) = L L L is an L K H-Galois extension. This can only happen if L K H is a group algebra L[N], where N acts regularly as permutations of G (i.e. given g, h G, there is a unique η N with η g = h). But not every regular subgroup N Perm(G) comes from a Hopf algebra over K. Theorem (Greither-Pareigis, 1987) Hopf-Galois structures on L/K correspond to regular subgroups N of Perm(G) normalized by left translations by G. The Hopf algebra corresponding to N is L[N] G. We call the isomorphism type of N the type of the Hopf-Galois structure. Note N = G, but in general N and G need not be isomorphic. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 4 / 11

Some Sample Results 1 (Kohl, 1998): If G = C p n for an odd prime p, then there are p n 1 Hopf-Galois structures, all of cyclic type. [For p = 2, dihedral and quaternion types also occur.] 2 (Childs, 2005): If G = (C p ) m with p > m 3, there are at least p m(m 1) 1 (p 1) Hopf-Galois structures of type (C p ) m and also some of nonabelian type. 3 (NB, 2004) For primes p, q with p 1 (mod q). There are 2 groups of order pq, viz. C pq and the metabelian group M pq = C p C q. If G = Cpq there is one cyclic Hopf-Galois structure and p metabelian ones. If G = M pq there are 2 + 2p(q 2) metabelian Hopf-Galois structures and p cyclic ones. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 5 / 11

Some Sample Results 1 (Kohl, 1998): If G = C p n for an odd prime p, then there are p n 1 Hopf-Galois structures, all of cyclic type. [For p = 2, dihedral and quaternion types also occur.] 2 (Childs, 2005): If G = (C p ) m with p > m 3, there are at least p m(m 1) 1 (p 1) Hopf-Galois structures of type (C p ) m and also some of nonabelian type. 3 (NB, 2004) For primes p, q with p 1 (mod q). There are 2 groups of order pq, viz. C pq and the metabelian group M pq = C p C q. If G = Cpq there is one cyclic Hopf-Galois structure and p metabelian ones. If G = M pq there are 2 + 2p(q 2) metabelian Hopf-Galois structures and p cyclic ones. 4 (NB, 2004) If G is nonabelian simple, there are just 2 Hopf-Galois structures, both of type G. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 5 / 11

Some Sample Results 1 (Kohl, 1998): If G = C p n for an odd prime p, then there are p n 1 Hopf-Galois structures, all of cyclic type. [For p = 2, dihedral and quaternion types also occur.] 2 (Childs, 2005): If G = (C p ) m with p > m 3, there are at least p m(m 1) 1 (p 1) Hopf-Galois structures of type (C p ) m and also some of nonabelian type. 3 (NB, 2004) For primes p, q with p 1 (mod q). There are 2 groups of order pq, viz. C pq and the metabelian group M pq = C p C q. If G = Cpq there is one cyclic Hopf-Galois structure and p metabelian ones. If G = M pq there are 2 + 2p(q 2) metabelian Hopf-Galois structures and p cyclic ones. 4 (NB, 2004) If G is nonabelian simple, there are just 2 Hopf-Galois structures, both of type G. 5 (NB, 2016) If G is abelian, all Hopf-Galois structures are of soluble type. See also talks of Ali Alabdali and Kayvan Nejabati Zenouz this afternoon. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 5 / 11

The Holomorph Formulation: We want to find the number e(g, N) of Hopf-Galois structures of type N on a Galois extension L/K with Galois group G. By Greither-Pareigis, these correspond to regular subgroups N Perm(G) normalised by the group λ(g) of left translations by G. Given such a subgroup, we can identify the sets N and G by η η 1 G. Starting with abstract groups N, G this yields a bijection in which {Regular embeddings α : N Perm(G )} {Regular embeddings β : G Perm(N )} α(n ) is normalised by λ(g ) β(g ) Hol(N ) where Hol(N ) = λ(n ) Aut(N ) Perm(N ). Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 6 / 11

The Holomorph Formulation: Composing a regular embedding α : N Perm(G) with θ Aut(N) gives a different embedding but the same subgroup of Perm(G). So # (regular embeddings α with α(n) nomalised by λ(g) ) = Aut(N) #( regular subgroups α(n) normalised by λ(g)) Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 7 / 11

The Holomorph Formulation: Composing a regular embedding α : N Perm(G) with θ Aut(N) gives a different embedding but the same subgroup of Perm(G). So # (regular embeddings α with α(n) nomalised by λ(g) ) = Aut(N) #( regular subgroups α(n) normalised by λ(g)) = Aut(N) e(g, N) Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 7 / 11

The Holomorph Formulation: Composing a regular embedding α : N Perm(G) with θ Aut(N) gives a different embedding but the same subgroup of Perm(G). So Similarly # (regular embeddings α with α(n) nomalised by λ(g) ) = Aut(N) #( regular subgroups α(n) normalised by λ(g)) = Aut(N) e(g, N) # (regular embeddings β : G Hol(N) ) = Aut(G) #( regular subgroups β(g) Hol(N)) Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 7 / 11

The Holomorph Formulation: Composing a regular embedding α : N Perm(G) with θ Aut(N) gives a different embedding but the same subgroup of Perm(G). So Similarly # (regular embeddings α with α(n) nomalised by λ(g) ) = Aut(N) #( regular subgroups α(n) normalised by λ(g)) = Aut(N) e(g, N) # (regular embeddings β : G Hol(N) ) So we get the formula = Aut(G) #( regular subgroups β(g) Hol(N)) e(g, N) = Aut(G) Aut(N) # (regular subgroups = G in Hol(N)). Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 7 / 11

The Holomorph Formulation: Giving a regular embedding β : G Hol(N) = N Aut(N) amounts to giving a homomorphism G Aut(N), i.e. an action of G on the group N; a bijection π : G N satisfying the 1-cocycle relation for this action: π(gh) = π(g)(g π(h)). Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 8 / 11

The Connection with Braces A (left) brace is a set B with binary operations +, such that (B, +) is an abelian group; (B, ) is a group; a (b + c) + a = a b + a c a, b, c B. We will call (B, +) the additive group, and (B, ) the multiplicative group, of B. We then have group homomorphism (B, ) Aut(B, +) with a λ a where λ a (b) = a b a. Then id : (B, ) (B, +) is a bijective cocycle for the corresponding action. Thus, given finite groups G, N with G = N and N abelian, finding a brace with additive group N and multiplicative group G amounts to finding a regular subgroup in Hol(N) isomorphic to G. Two such subgroups give isomorphic braces iff they are conjugate under Aut(N). Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 9 / 11

The Connection with Braces So counting braces and counting Hopf-Galois structures are closely related. Both involve finding the regular subgroups in Hol(N) isomorphic to G. But these two problems are not the same: to count braces, find the number of orbits of these subgroup under Aut(N); to count Hopf-Galois structures, find the number of these subgroups times Aut(G) / Aut(N). Example: G = N = C p n = σ for an odd prime p, so Aut(N) = {θ a : a (Z/p n Z) } where θ a (σ) = σ a. Any regular subgroup in Hol(N) is cyclic and has a unique generator of the form (σ, θ a ) N Aut(N) where a 1 (mod p). Hence there are p n 1 regular subgroups, giving p n 1 Hopf-Galois structures. But the groups (σ, θ a ) and (σ, θ b ) are conjugate under Aut(N) iff θ a, θ b have the same order, so there are n braces, corresponding to a = 1, 1 + p n 1, 1 + p n 2,..., 1 + p. Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 10 / 11

Thank you for your attention! Nigel Byott (University of Exeter, UK ) Hopf-Galois Structures 23 June 2017 11 / 11