Nonassociative cyclic extensions of fields and central simple algebras

Abstract

We define nonassociative cyclic extensions of degree m of both fields and central simple algebras over fields. If a suitable field contains a primitive mth (resp., qth) root of unity, we show that suitable nonassociative generalized cyclic division algebras yield nonassociative cyclic extensions of degree m (resp., qs). Some of Amitsur's classical results on non-commutative associative cyclic extensions of both fields and central simple algebras are obtained as special cases.2010 Mathematics Subject Classification. Primary: 17A35; Secondary: 17A60, 16S36.

The automorphisms of generalized cyclic Azumaya algebras

“…The proof is similar to the one of [7,Theorem 6], but works also when the algebra is non-associative, as it does not rely on the right nucleus being D, whereas the proof of [7,Theorem 6] Thus…”

“…Note that associative generalized cyclic algebras over fields were investigated by Amitsur in [3], and nonassociative generalized cyclic algebras over fields in [7]. Another generalization of associative generalized cyclic algebras over fields called associative cyclic extensions of simple rings was considered by Kishimoto [14].…”

The automorphisms of generalized cyclic Azumaya algebras

“…The proof is similar to the one of [7,Theorem 6], but works also when the algebra is associative, as it does not rely on the right nucleus being D, whereas the proof of [7,Theorem 6] Thus…”

How a Nonassociative Algebra Reflects the Properties of a Skew Polynomial

“…When choosing f and R in the right way, it can be also seen as a generalization of certain crossed product algebras and some Azumaya algebra constructions. First results on the structure of the algebras S f which initially were defined by Petit in [29] have appeared in [29,30,4,5,32,33,31]. First applications to coding theory have appeared for instance in [34,35,36].…”

“…The condition that f is bounded is necessary here, as is shown in [15, Example 3] where f ∈ Q(x)[t; d/dt] is reducible in the differential operator ring Q(x)[t; d/dt], but Nuc r (S f ) is a division algebra. For instance, if D is a finite field and δ = 0, all polynomials are bounded and hence f is irreducible if and only if E( f ) is a finite field [13, Theorem 3.3].The argument leading up to[30, Section 2,(6)] implies that S f has no zero divisors if and only if f is irreducible, which is in turn equivalent to S f being a right division ring (i.e., right multiplication R a in S f is bijective for all 0 = a ∈ S f ):THEOREM 12. ([30, (6)], but without a full proof).…”