2018
DOI: 10.1080/00927872.2018.1448845
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Automorphisms and isomorphisms of Jha-Johnson semifields obtained from skew polynomial rings

Abstract: We study the automorphisms of Jha-Johnson semifields obtained from a right invariant irreducible twisted polynomial f ∈ K[t; σ], where K = F q n is a finite field and σ an automorphism of K of order n, with a particular emphasis on inner automorphisms and the automorphisms of Sandler and Hughes-Kleinfeld semifields. We include the automorphisms of some Knuth semifields (which do not arise from skew polynomial rings).Isomorphism between Jha-Johnson semifields are considered as well.Every finite nonassociative P… Show more

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Cited by 5 publications
(9 citation statements)
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“…Then ker(N K/F ) is a cyclic group of order s = (q m − 1)/(q − 1) and any division algebra (K/F, σ, d) has exactly s inner automorphisms, all of them extending id K . The subgroup they generate is cyclic and isomorphic to ker(N K/F ) [10]. Hence if m divides s, which is the case if F contains a primitive mth root of unity, then there is a subgroup of automorphisms of order m extending id K and hence (K/F, σ, d) is a cyclic extension of K of degree m.…”
Section: 2mentioning
confidence: 99%
“…Then ker(N K/F ) is a cyclic group of order s = (q m − 1)/(q − 1) and any division algebra (K/F, σ, d) has exactly s inner automorphisms, all of them extending id K . The subgroup they generate is cyclic and isomorphic to ker(N K/F ) [10]. Hence if m divides s, which is the case if F contains a primitive mth root of unity, then there is a subgroup of automorphisms of order m extending id K and hence (K/F, σ, d) is a cyclic extension of K of degree m.…”
Section: 2mentioning
confidence: 99%
“…Theorem 8. All semifields of order q 4 with centre of order q and right nucleus of order q 2 are isotopic to a semifield of the form S n,s,1 (η, ρ), for some η ∈ F q n , ρ ∈ Aut(F q n ), [n, s] ∈ { [2,2], [4,1]}.…”
Section: New Constructionsmentioning
confidence: 99%
“…If n = 2, s = 2, k = 1, then we are in the case of Section 4.1. So assume n > 2 and [k, s] = [1,2]. Then g = g 0 + g 1 x + g 2n−1 (x 2n−1 + F 1 x n−1 ).…”
Section: Nuclei Of Codes In the Family S Nskmentioning
confidence: 99%
“…Let s = (q m − 1)/(q − 1). The following results are implied by the theorems on the automorphism groups for the related algebras S f [3,Theorems 19,20]:…”
Section: 2mentioning
confidence: 99%
“…K = F [x]/(x 2 − 2) = {0, 1, 2, x, 2x, x+1, x+2, 2x+1, 2x+2}. There are exactly two non-isomorphic semifields which are nonassociative quaternion algebras with nucleus K · 1, given by A 1 = (K/F, σ, x) with Aut F (A 1 ) ∼ = Z/4Z and by A 2 = (K/F, σ, x + 1) with Aut F (A 2 ) isomorphic to the group of quaternion units, the smallest dicyclic group Dic 2 (of order 8) [3,Example 32]. These are up to isomorphism the only two proper semifields of order 81 with center F · 1 and nucleus containing K · 1 [W, Theorem 1].…”
Section: 2mentioning
confidence: 99%