The automorphisms of Petit's algebras

TL;DR

This paper investigates the automorphisms of a class of nonassociative algebras generalizing certain quotient algebras, providing explicit automorphism group descriptions under specific conditions and exploring isomorphism criteria.

Contribution

It computes automorphism groups of Petit algebras when the automorphism commutes with all automorphisms of the base field and extends results to finite Galois extensions.

Findings

Automorphism groups are fully characterized when $\sigma$ commutes with all automorphisms.

Partial results are obtained for cases where the order of $\sigma$ is less than the degree of $f$.

In finite Galois extensions, detailed structure of automorphism groups is described.

Abstract

Let $\sigma$ be an automorphism of a field $K$ with fixed field $F$. We study the automorphisms of nonassociative unital algebras which are canonical generalizations of the associative quotient algebras $K[t;\sigma]/fK[t;\sigma]$ obtained when the twisted polynomial $f\in K[t;\sigma]$ is invariant, and were first defined by Petit. We compute all their automorphisms if $\sigma$ commutes with all automorphisms in ${\rm Aut}_F(K)$ and $n\geq m-1$, where $n$ is the order of $\sigma$ and $m$ the degree of $f$,and obtain partial results for $n<m-1$. In the case where $K/F$ is a finite Galois field extension, we obtain more detailed information on the structure of the automorphism groups of these nonassociative unital algebras over $F$. We also briefly investigate when two such algebras are isomorphic.

Equations

$$H_{\tau ,k}(\sum _{i=0}^{m-1}{x}_i{t}^i)=\tau (x_0)+\tau (x_1)kt+\tau (x_2)k\sigma (k)t^2+\cdots +\tau (x_{m-1})k\sigma (k)\cdots {\sigma }^{m-2}(k)t^{m-1},$$

$$\tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}a_{i}$$

$$H(tz)=H(t)H(z)=(\sum _{i=0}^{m-1}{k}_i{t}^i)\tau (z)=\sum _{i=0}^{m-1}{k}_i{\sigma }^i(\tau (z))t^i$$

$$H(tz)=H(\sigma (z)t)=\sum _{i=0}^{m-1}\tau (\sigma (z))k_i{t}^i$$

$$k_i{\sigma }^i(\tau (z))=k_i\tau ({\sigma }^i(z))=\tau (\sigma (z))k_i=k_i\tau (\sigma (z))$$

$$k_i=0\text{ or }{\sigma }^i(z)=\sigma (z)$$

$$H_{\tau ,k}(\sum _{i=0}^{m-1}{x}_i{t}^i)=\tau (x_0)+\sum _{i=1}^{m-1}\tau (x_i)\prod _{l=0}^{i-1}{\sigma }^l(k)t^i,$$

$$H(t^{m})=H\Big{(}\sum_{i=0}^{m-1}a_{i}t^{i}\Big{)}=\sum_{i=0}^{m-1}H(a_{i})H(t)^{i}=\tau(a_{0})+\sum_{i=1}^{m-1}\tau(a_{i})\big{(}\prod_{l=0}^{i-1}\sigma^{l}(k)\big{)}t^{i}$$

$$H(t{)}^m=H(t)H(t{)}^{m-1}=k\sigma (k)\cdots {\sigma }^{m-1}(k)t^m=k\sigma (k)\cdots {\sigma }^{m-1}(k)\sum _{i=0}^{m-1}{a}_i{t}^i.$$

$$H(\sum _{i=0}^{m-1}{x}_i{t}^i)=\sum _{i=0}^{m-1}\tau (x_i)g(t{)}^i.$$

$$(1)\quad\quad\quad\tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}a_{i}$$

$$a_{i}=\Big{(}\prod_{q=i}^{m-1}\sigma^{q}(k)\Big{)}a_{i}$$

$$1=\prod_{q=l}^{m-1}\sigma^{q}(k)=\prod_{q=l}^{m-1}\sigma^{q}\big{(}\frac{\sigma(\alpha)}{\alpha}\big{)}=\frac{\prod_{q=l+1}^{m}\sigma^{q}(\alpha)}{\prod_{q=l}^{m-1}\sigma^{q}(\alpha)}=\frac{\alpha}{\sigma^{l}(\alpha)}.$$

$${\mathrm{Dic}}_l=⟨x,y|y^{2l}=1,x^2=y^l,x^{-1}yx=y^{-1}⟩$$

$$G_{c^l}(x_0+x_{1}t)$$

$$=x_0+x_1{\mu }_1^{-1}{b}^{-1}\sqrt{b}(-{\mu }_1\sqrt{b})t$$

$$=x_0-x_{1}t=H_{id,-1}(x_0+x_{1}t)$$

$$\displaystyle H_{\sigma,-k}\big{(}G_{c}\big{(}H_{\sigma,k}\big{(}x_{0}+x_{1}t\big{)}\big{)}\big{)}$$

$$\displaystyle=H_{\sigma,-k}\big{(}\sigma(x_{0})+\sigma(x_{1})kc^{-1}\sigma(c)t\big{)}$$

$$=x_0-x_1\sigma (k)\sigma (c^{-1})ckt$$

$$\displaystyle=x_{0}+x_{1}\sigma(c^{-1})ct=G_{\sigma(c)}\big{(}x_{0}+x_{1}t\big{)}$$

$$\tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}b_{i}$$

$$G_{\tau ,k}(\sum _{i=0}^{m-1}{x}_i{t}^i)=\tau (x_0)+\sum _{i=1}^{m-1}\tau (x_i)\prod _{l=0}^{i-1}{\sigma }^l(k)t^i.$$

$$G_{\tau ,k}(\sum _{i=0}^{m-1}{x}_i{t}^i)=\tau (x_0)+\sum _{i=1}^{m-1}\tau (x_i)\prod _{l=0}^{i-1}{\sigma }^l(k)t^i$$

Full text

Introduction

Let $D$ be a division algebra, $\sigma$ an injective endomorphism of $D$, $\delta$ a left $\sigma$-derivation and $R=D[t;\sigma,\delta]$ a skew polynomial ring (for instance, c.f. [16, § 3.4]). For an invariant skew polynomial $f\in R$, i.e. when the ideal $Rf$ is a two-sided principal ideal, the quotient algebra $R/Rf$ appears in classical constructions of associative central simple algebras, usually employing an irreducible $f\in R$ to get examples of division algebras, e.g. see [15].

In 1967, Petit [22, 23] introduced a class of unital nonassociative algebras $S_{f}$, which canonically generalize the quotient algebras $R/Rf$ obtained when factoring out an invariant $f\in R$ of degree $m$. The algebra $S_{f}=D[t;\sigma,\delta]/D[t;\sigma,\delta]f$ is defined on the additive subgroup $\{h\in R\,|\,{\rm deg}(h)<m\}$ of $R$ by using right division by $f$ to define the algebra multiplication $g\circ h=gh\,\,{\rm mod}_{r}f$. The properties of the algebras $S_{f}$ were studied in detail in [22, 23], and for $D$ a finite base field (hence w.l.o.g. $\delta=0$) in [20].

Even earlier, the algebra $S_{f}$ with $f(t)=t^{2}-i\in\mathbb{C}[t;\overline{\phantom{x}}]$, $\overline{\phantom{x}}$ the complex conjugation, appeared in [8] as the first example of a nonassociative division algebra.

Although the algebras themselves have received little attention so far, the right nucleus of $S_{f}$ (the eigenspace of $f\in R$) already appeared implicitly in classical constructions by Amitsur [2, 3, 4], but also in results on computational aspects of operator algebras; they are for instance used in algorithms factoring skew polynomials over $\mathbb{F}_{q}(t)$ or finite fields, cf. [11, 12, 13, 14]. The role of classical algebraic constructions in coding theory is well known (cf. [16, Chapter 9], [17, 1, 7]).

Moreover, recently space-time block codes, coset codes and wire-tap codes were obtained employing the algebras $S_{f}$ over number fields, cf. [9, 10, 21, 24, 27, 28, 29], and they also appear useful for linear cyclic codes [25, 26].

If $K$ is a finite field, $F$ the fixed field of $\sigma$, $K/F$ a finite Galois field extension and $f\in K[t;\sigma]=K[t;\sigma,0]$ irreducible and invariant, the $S_{f}$ are Jha-Johnson semifields (also called cyclic semifields) [20, Theorem 15], and were studied for instance by Wene [35] and more recently by Lavrauw and Sheekey [20]. The main motivation for our paper comes from the question how the automorphism groups of Jha-Johnson semifields look like. The results presented here are applied to some Jha-Johnson semifields in [6].

The structure of this paper is as follows: In Section 1, we introduce the terminology and define the algebras $S_{f}$. We limit our observations to the algebras which are not associative. Given a field extension $K$, $\sigma\in{\rm Aut}(K)$ of order $n$ with fixed field $F$, such that $\sigma$ commutes with all $\tau\in{\rm Aut}_{F}(K)$, and $f\in K[t;\sigma]$ of degree $m$ not invariant, we compute the automorphisms of $S_{f}$ in Section 2. We obtain all automorphisms for $n\geq m-1$ and some partial results for $n<m-1$ (Theorems 4 and 5). For $n\geq m-1$, the automorphisms in ${\rm Aut}_{F}(S_{f})$ are canonically induced by the $F$-automorphism $G$ of $R=K[t;\sigma]$ which satisfy $G(f(t))=af(t)$ for some $a\in K^{\times}$, and on $K$ restrict to an automorphism that commutes with $\tau$.

The automorphisms groups of $S_{f}$ where $f(t)=t^{m}-a\in K[t;\sigma]$, $a\in K\setminus F$, play a special role, as for all nonassociative $S_{g}$ with $g(t)=t^{m}-\sum_{i=0}^{m-1}b_{i}t^{i}\in K[t;\sigma]$ and $b_{0}=a$, ${\rm Aut}_{F}(S_{g})$ is a subgroup of ${\rm Aut}_{F}(S_{f})$ when $n\geq m-1$.

We then focus on the situation that $K/F$ is a finite Galois field extension such that $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$. In many cases, either ${\rm Aut}_{F}(S_{f})\cong{\rm Gal}(K/F)$ or is trivial (Theorem 10). Necessary conditions for extending Galois automorphisms $\tau\in{\rm Gal}(K/F)$ to $S_{f}$ are studied in Sections 3 and 4. The existence of cyclic subgroups of ${\rm Aut}_{F}(S_{f})$ is investigated in Section 5.

For $f(t)=t^{m}-a\in K[t;\sigma]$ and $K/F$ a cyclic field extension of degree $m$, the algebra $S_{f}$ is also called a nonassociative cyclic algebra and denoted by $(K/F,\sigma,a)$. These algebras are canonical generalizations of associative cyclic algebras, but also generalizations of the algebras in [3, 15]. The automorphisms of nonassociative cyclic algebras are investigated in Section 6. All the automorphisms of $A=(K/F,\sigma,a)$ extending $id_{K}$ are inner and form a cyclic subgroup of ${\rm Aut}_{F}(A)$ isomorphic to ${\rm ker}(N_{K/F})$. In some cases, this is the whole automorphism group, e.g. if $F$ has no $m$th root of unity. In these cases, every automorphism of $A$ leaves $K$ fixed and is inner. We explain when the automorphism group of a nonassociative quaternion algebra $A$ (where $m=2$) contains a dicyclic group and when it contains a subgroup isomorphic to the semidirect product of two cyclic groups.

In Section 7 we briefly investigate isomorphisms between two algebras $S_{f}$ and $S_{g}$.

This work is part of the first author’s PhD thesis [5] written under the supervision of the second author. For results on the automorphisms of the more general algebras defined using $f\in D[t;\sigma]$, or a more detailed study and the (less relevant) cases left out in this paper the reader is referred to [5]. For examples of applications of the associated classical constructions the readers are referred to [1, 7, 16, 17].

1. Preliminaries

1.1. Nonassociative algebras

Let $F$ be a field and let $A$ be an $F$-vector space. $A$ is an algebra over $F$ if there exists an $F$-bilinear map $A\times A\to A$, $(x,y)\mapsto x\cdot y$, denoted simply by juxtaposition $xy$, the multiplication of $A$. An algebra $A$ is called unital if there is an element in $A$, denoted by 1, such that $1x=x1=x$ for all $x\in A$. We will only consider unital algebras from now on without explicitly saying so.

Associativity in $A$ is measured by the associator $[x,y,z]=(xy)z-x(yz)$. The left nucleus of $A$ is defined as ${\rm Nuc}_{l}(A)=\{x\in A\,|\,[x,A,A]=0\}$, the middle nucleus of $A$ is ${\rm Nuc}_{m}(A)=\{x\in A\,|\,[A,x,A]=0\}$ and the right nucleus of $A$ is defined as ${\rm Nuc}_{r}(A)=\{x\in A\,|\,[A,A,x]=0\}$. ${\rm Nuc}_{l}(A)$, ${\rm Nuc}_{m}(A)$, and ${\rm Nuc}_{r}(A)$ are associative subalgebras of $A$. Their intersection ${\rm Nuc}(A)=\{x\in A\,|\,[x,A,A]=[A,x,A]=[A,A,x]=0\}$ is the nucleus of $A$. ${\rm Nuc}(A)$ is an associative subalgebra of $A$ containing $F1$ and $x(yz)=(xy)z$ whenever one of the elements $x,y,z$ lies in ${\rm Nuc}(A)$. The center of $A$ is ${\rm C}(A)=\{x\in A\,|\,x\in\text{Nuc}(A)\text{ and }xy=yx\text{ for all }y\in A\}$.

An $F$-algebra $A\not=0$ is called a division algebra if for any $a\in A$, $a\not=0$, the left multiplication with $a$, $L_{a}(x)=ax$, and the right multiplication with $a$, $R_{a}(x)=xa$, are bijective. If $A$ has finite dimension over $F$, then $A$ is a division algebra if and only if $A$ has no zero divisors [31, pp. 15, 16]. An element $0\not=a\in A$ has a left inverse $a_{l}\in A$, if $R_{a}(a_{l})=a_{l}a=1$, and a right inverse $a_{r}\in A$, if $L_{a}(a_{r})=aa_{r}=1$.

An automorphism $G\in{\rm Aut}_{F}(A)$ is an inner automorphism if there is an element $m\in A$ with left inverse $m_{l}$ such that $G(x)=(m_{l}x)m$ for all $x\in A$. Given an inner automorphism $G_{m}\in{\rm Aut}_{F}(A)$ and some $H\in{\rm Aut}_{F}(A)$, then clearly $H^{-1}\circ G_{m}\circ H\in{\rm Aut}_{F}(A)$ is an inner automorphism. [34, Lemma 2, Theorem 3, 4] generalize to any nonassociative algebra:

Proposition 1**.**

*Let $A$ be an algebra over $F$.

(i) For all invertible $n\in{\rm Nuc}(A)$, $G_{n}(x)=(n^{-1}x)n$ is an inner automorphism of $A$.

(ii) If $G_{m}$ is an inner automorphism of $A$, then so is $G_{nm}(x)=((m_{l}n^{-1})x)(nm)$ for all invertible $n\in{\rm Nuc}(A)$.

(iii) If $G_{m}$ is an inner automorphism of $A$, and $a,b\in{\rm Nuc}(A)$ are invertible, then $G_{am}=G_{bm}$ if and only if $ab^{-1}\in C(A).$

(iv) For invertible $n,m\in{\rm Nuc}(A)$, $G_{m}=G_{n}$ if and only if $n^{-1}m\in C(A)$.*

The set $\{G_{m}\,|\,m\in{\rm Nuc}(A)\text{ invertible}\}$ is a subgroup of ${\rm Aut}_{F}(A)$. For each invertible $m\in{\rm Nuc}(A)\setminus C(A)$, $G_{m}$ generates a cyclic subgroup which has finite order $s$ if $m^{s}\in C(A)$, so in particular if $m$ has order $s$.

Note that if the nucleus is commutative, then for all invertible $n\in{\rm Nuc}(A)$, $G_{n}(x)=(n^{-1}x)n$ is an inner automorphism of $A$ such that ${G_{n}}|_{{\rm Nuc}(A)}=id_{{\rm Nuc}(A)}$.

1.2. Twisted polynomial rings

Let $K$ be a field and $\sigma$ an automorphism of $K$. The twisted polynomial ring $K[t;\sigma]$ is the set of polynomials $a_{0}+a_{1}t+\dots+a_{n}t^{n}$ with $a_{i}\in K$, where addition is defined term-wise and multiplication by $ta=\sigma(a)t$ for all $a\in K$. For $f=a_{0}+a_{1}t+\dots+a_{n}t^{n}$ with $a_{n}\not=0$ define ${\rm deg}(f)=n$ and put ${\rm deg}(0)=-\infty$. Then ${\rm deg}(fg)={\rm deg}(f)+{\rm deg}(g).$ An element $f\in R$ is irreducible in $R$ if it is not a unit and it has no proper factors, i.e. if there do not exist $g,h\in R$ with ${\rm deg}(g),{\rm deg}(h)<{\rm deg}(f)$ such that $f=gh$.

$R=K[t;\sigma]$ is a left and right principal ideal domain and there is a right division algorithm in $R$: for all $g,f\in R$, $g\not=0$, there exist unique $r,q\in R$ with ${\rm deg}(r)<{\rm deg}(f)$, such that $g=qf+r.$ There is also a left division algorithm in $R$ [15, p. 3 and Prop. 1.1.14]. (Our terminology is the one used by Petit [22] and Lavrauw and Sheekey [20]; it is different from Jacobson’s, who calls what we call right a left division algorithm and vice versa.) Define $F={\rm Fix}(\sigma)$.

1.3. Nonassociative algebras obtained from skew polynomial rings

Let $K$ be a field, $\sigma$ an automorphism of $K$ with $F={\rm Fix}(\sigma)$, and $f\in R=K[t;\sigma]$ of degree $m$. Let ${\rm mod}_{r}f$ denote the remainder of right division by $f$. Then then additive abelian group $R_{m}=\{g\in K[t;\sigma]\,|\,{\rm deg}(g)<m\}$ together with the multiplication $g\circ h=gh\,\,{\rm mod}_{r}f$ is a unital nonassociative algebra $S_{f}=(R_{m},\circ)$ over $F_{0}=\{a\in K\,|\,ah=ha\text{ for all }h\in S_{f}\}.$ $F_{0}$ is a subfield of $K$ [22, (7)] and it is straightforward to see that if $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}$ and $a_{0}\not=0$ then $F_{0}=F$ [25, Remark 9]. The algebra $S_{f}$ is also denoted by $R/Rf$ [22, 23] if we want to make clear which ring $R$ is involved in the construction. In the following, we call the algebras $S_{f}$ Petit algebras and denote their multiplication simply by juxtaposition.

Using left division by $f$ and the remainder ${\rm mod}_{l}f$ of left division by $f$ instead, we can analogously define the multiplication for another unital nonassociative algebra on $R_{m}$ over $F_{0}$, called $\,{}_{f}S$. We will only consider the Petit algebras $S_{f}$, since every algebra $\,{}_{f}S$ is the opposite algebra of some Petit algebra [22, (1)].

Theorem 2**.**

*(cf. [22, (2), (5), (9)]) Let $f(t)\in R=K[t;\sigma]$.

(i) If $S_{f}$ is not associative then ${\rm Nuc}_{l}(S_{f})={\rm Nuc}_{m}(S_{f})=K$ and ${\rm Nuc}_{r}(S_{f})=\{g\in R\,|\,fg\in Rf\}.$

(ii) The powers of $t$ are associative if and only if $t^{m}t=tt^{m}$ if and only if $t\in{\rm Nuc}_{r}(S_{f})$ if and only if $ft\in Rf.$

(iii) Let $f\in R$ be irreducible and $S_{f}$ a finite-dimensional $F$-vector space or free of finite rank as a right ${\rm Nuc}_{r}(S_{f})$-module. Then $S_{f}$ is a division algebra.

Conversely, if $S_{f}$ is a division algebra then $f$ is irreducible.

(iv) $S_{f}$ is associative if and only if $f$ is invariant. In that case, $S_{f}$ is the usual quotient algebra.

(v) Let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in R=K[t;\sigma]$. Then $f$ is invariant if and only if $\sigma^{m}(z)a_{i}=a_{i}\sigma^{i}(z)$ for all $z\in K$, $i\in\{0,\dots,m-1\}$ and $a_{i}\in F$ for all $i\in\{0,\dots,m-1\}$.*

Note that if $f$ is not invariant, then the nucleus of any $S_{f}=K[t;\sigma]/K[t;\sigma]f$ is a subfield of $K={\rm Nuc}_{l}(S_{f})$. If ${\rm Nuc}(S_{f})$ is larger than $F$, then $\{G_{m}\,|\,0\not=m\in{\rm Nuc}(A)\}$ is a non-trivial subgroup of ${\rm Aut}_{F}(S_{f})$ and each inner automorphism $G_{m}$ in this subgroup extends $id_{{\rm Nuc}(A)}$ by Proposition 1.

Proposition 3**.**

*Let $f(t)\in F[t]=F[t;\sigma]\subset K[t;\sigma]$.

(i) $F[t]/(f(t))$ is a commutative subring of $S_{f}$ and $F[t]/(f(t))\cong F\oplus Ft\oplus\dots\oplus Ft^{m-1}\subset{\rm Nuc}_{r}(S_{f}).$ In particular, then $ft\in Rf$ which is equivalent to the powers of $t$ being associative, which again is equivalent to $t^{m}t=tt^{m}$.

(ii) If $f(t)$ is irreducible in $F[t]$, $F[t]/(f(t))$ is an algebraic subfield of degree $m$ contained in the right nucleus.*

Proof.

$S_{f}$ contains the commutative subring $F[t]/(f(t)).$ If $f(t)$ is irreducible in $F[t]$, this is an algebraic field extension of $F$. This subring is isomorphic to the ring consisting of the elements $\sum_{i=0}^{m-1}a_{i}t^{i}$ with $a_{i}\in F$.

Clearly $F\subset{\rm Nuc}_{r}(S_{f})$. For all $a,b,c\in K$, $i,j\in\{0,\dots,m-1\}$ we have $[at^{i},bt^{j},t]=(a\sigma^{i}(b)t^{i+j})t-(at^{i})(bt^{j+1})=a\sigma^{i}(b)t^{i+j+1}-a\sigma^{i}(bc)t^{i+j}=0.$ Thus $t\in{\rm Nuc}_{r}(S_{f})$ which implies that $F\oplus Ft\oplus\dots\oplus Ft^{m-1}\subset{\rm Nuc}_{r}(S_{f})$, hence the assertion. The rest is obvious. ∎

We will assume throughout the paper that $\text{deg}(f)=m\geq 2$ (since if $f$ is constant then $S_{f}\cong K$) and that $\sigma\not=id$. Without loss of generality, we only consider monic polynomials $f$, since $S_{f}=S_{af}$ for all non-zero $a\in K$.

2. Automorphisms of $S_{f}$

2.1.

Let $K$ be a field, $\sigma$ an automorphism of $K$ of order $n$ (which may be infinite), $F={\rm Fix}(\sigma)$, and $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ a twisted polynomial which is not invariant.

Theorem 4**.**

Suppose $\sigma$ commutes with all $\tau\in{\rm Aut}_{F}(K)$. Let $n\geq m-1$. Then $H$ is an automorphism of $S_{f}$ if and only if $H=H_{\tau,k}$ with

[TABLE]

where $\tau\in{\rm Aut}_{F}(K)$ and $k\in K^{\times}$ is such that

[TABLE]

for all $i\in\{0,\ldots,m-1\}$.

Proof.

Let $H:S_{f}\rightarrow S_{f}$ be an automorphism. Since $S_{f}$ is not associative, $\text{Nuc}_{l}(S_{f})=K$ by Theorem 2 (i). Since any automorphism preserves the left nucleus, $H(K)=K$ and so $H|_{K}=\tau$ for some $\tau\in\text{Aut}_{F}(K)$. Suppose $H(t)=\sum_{i=0}^{m-1}k_{i}t^{i}$ for some $k_{i}\in K$. Then we have

[TABLE]

and

[TABLE]

for all $z\in K$. Comparing the coefficients of $t^{i}$ in (2) and (3) we obtain

[TABLE]

for all $i\in\{0,\ldots,m-1\}$ and all $z\in K$. This implies k_{i}(\tau\big{(}\sigma^{i}(z)-\sigma(z)\big{)})=0 for all $i\in\{0,\ldots,m-1\}$ and all $z\in K$ since $\sigma$ and $\tau$ commute, i.e.

[TABLE]

for all $i\in\{0,\ldots,m-1\}$ and all $z\in K$.

Since $\sigma$ has order $n\geq m-1$, which means $\sigma^{i}\not=\sigma$ for all $i\in\{0,\ldots,m-1\}$, $i\not=1$, (5) implies $k_{i}=0$ for all $i\in\{0,\ldots,m-1\}$, $i\not=1$. Therefore $H(t)=kt$ for some $k\in K^{\times}$. Furthermore, we have H(zt^{i})=H(z)H(t)^{i}=\tau(z)(kt)^{i}=\tau(z)\Big{(}\prod_{l=0}^{i-1}\sigma^{l}(k)\Big{)}t^{i} for all $i\in\{1,\ldots,m-1\}$ and $z\in K$. Thus $H$ has the form

[TABLE]

for some $k\in K^{\times}$. Moreover, with $t^{m}=tt^{m-1}$, also

[TABLE]

and $H(tt^{m-1})=H(t)H(t^{m-1})=H(t)H(t)^{m-1}$, i.e.

[TABLE]

Comparing (7) and (8) gives \tau(a_{i})=\Big{(}\prod_{q=i}^{m-1}\sigma^{q}(k)\Big{)}a_{i} for all $i\in\{0,\ldots,m-1\}$. Thus $H$ is as in (6) where $k\in K^{\times}$ is such that (1) holds for all $i\in\{0,\ldots,m-1\}$.

The $H_{\tau,k}$ are indeed automorphisms of $S_{f}$: Let $G$ be an automorphism of $R=K[t;\sigma]$. Then for $h(t)=\sum_{i=0}^{r}b_{i}t^{i}\in K[t;\sigma]$ we have $G(h(t))=\tau(b_{0})+\sum_{i=i}^{m-1}\tau(b_{i})\prod_{l=0}^{i-1}\sigma^{l}(k)t^{i}$ for some $\tau\in{\rm Aut}(K)$ such that $\sigma\circ\tau=\tau\circ\sigma$ and some $k\in K^{\times}$ (the proof of [20, Lemma 1] works for any $R=K[t;\sigma]$, or cf. [18, p. 75]). It is straightforward to see that $S_{f}\cong S_{G(f)}$ (cf. [20, Theorem 7], the proof works for any $R=K[t;\sigma]$). In particular, this means that if $k\in K^{\times}$ satisfies (1) then G(f(t))=\big{(}\prod_{l=0}^{m-1}\sigma^{l}(k)\big{)}f(t)=af(t) with $a\in K^{\times}$ being the product of the $\sigma^{l}(k)$, and so $G$ induces an isomorphism of $S_{f}$ with $S_{af}=S_{f}$, i.e. an automorphism of $S_{f}$. The automorphisms of ${\rm Aut}_{F}(S_{f})$ are therefore all canonically induced by the $F$-automorphisms $G$ of $R=K[t;\sigma]$ which satisfy (1). ∎

The assumption that $n\geq m-1$ is needed in (4) to conclude that $k_{i}=0$ for $i=0,2,3,\ldots,m-1$ and so $H(t)=kt$. If $n<m-1$ we still obtain:

Theorem 5**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Aut}_{F}(K)$. Let $n<m-1$.

(i) For all $k\in K^{\times}$ satisfying (1) for all $i\in\{0,\ldots,m-1\}$, the maps $H_{\tau,k}$ from Theorem 4 are automorphisms of $S_{f}$ and form a subgroup of ${\rm Aut}_{F}(S_{f})$.

(ii) Let $H\in{\rm Aut}_{F}(S_{f})$ and $N={\rm Nuc}_{r}(S_{f})$. Then $H|_{K}=\tau$ for some $\tau\in{\rm Aut}_{F}(K)$, $H|_{N}\in{\rm Aut}_{F}(N)$ and $H(t)=g(t)$ with $g(t)=k_{1}t+k_{1+n}t^{1+n}+k_{1+2n}t^{1+2n}+\ldots+k_{1+sn}t^{1+sn}$ for some $k_{1+ln}\in K$, $0\leq l\leq s$. Moreover, $g(t)^{i}$ is well defined for all $i\leq m-1$, i.e., all powers of $g(t)$ are associative for all $i\leq m-1$, and $g(t)g(t)^{m-1}=\sum_{i=0}^{m-1}\tau(a_{i})g(t)^{i}.$ Thus*

[TABLE]

Proof.

(i) is straightforward, using the relevant parts of the proof of Theorem 4. Note that the inverse of $H_{\tau,k}$ is $H_{\tau^{-1},\tau^{-1}(k^{-1})}$ and $H_{\tau,k}\circ H_{\rho,b}=H_{\tau\rho,\tau(b)k}.$

(ii) Let $H:S_{f}\rightarrow S_{f}$ be an automorphism. As in Theorem 4, $H|_{K}=\tau$ for some $\tau\in{\rm Aut}_{F}(K)$, and $H|_{N}\in{\rm Aut}_{F}(N)$. Suppose $H(t)=\sum_{i=0}^{m-1}k_{i}t^{i}$ for some $k_{i}\in K$. Comparing the coefficients of $t$ in $H(tz)=H(t)H(z)=H(\sigma(z)t)$ we obtain (5) for all $i\in\{0,\ldots,m-1\}$ and all $z\in K$. Since $\sigma$ has order $n<m-1$, here, (5) only implies $k_{i}=0$ for $i\in\{0,\ldots,n\}$, $i\not=1$. Therefore $H(t)=k_{1}t+\sum_{i=n+1}^{m-1}k_{i}t^{i}$ for some $k_{i}\in K$. However, $\sigma^{i}(z)=\sigma(z)$ for all $z\in K$ if and only if $i=nl+1$ for some $l\in\mathbb{Z}$ since $\sigma$ has order $n$. Therefore (5) implies $k_{i}=0$ for every $i\neq 1+nl$, $l\in\mathbb{N}_{0}$, $i\in\left\{0,\ldots,m-1\right\}$. Thus $H(t)=k_{1}t+k_{1+n}t^{n+1}+\ldots+k_{1+sn}t^{1+sn}$ for some $s$, $sn<m-1$. Furthermore, $H(t^{m})=H(\sum_{i=0}^{m-1}a_{i}t^{i})=\sum_{i=0}^{m-1}\tau(a_{i})(k_{1}t+k_{1+n}t^{1+n}+\ldots+k_{1+sn}t^{1+sn})^{i}$ and $H(t^{m})=(k_{1}t+k_{1+n}t^{1+n}+\ldots+k_{1+sn}t^{1+sn})^{m}.$ Similarly, $H(t)^{i}=(k_{1}t+k_{1+n}t^{1+n}+\ldots+k_{1+sn}t^{1+sn})^{i}.$ Together these imply the assertion. ∎

A closer look at the proof of Theorems 4 and 5 reveals that in fact the following holds without requiring $\sigma$ to commute with all $\tau\in{\rm Aut}_{F}(K)$:

Proposition 6**.**

*(i) For every $k\in K^{\times}$ satisfying (1) for all $i\in\{0,\ldots,m-1\}$ for $\tau=id$, $H_{id,k}$ is an automorphism of $S_{f}$ and generates a subgroup of ${\rm Aut}_{F}(S_{f})$.

(ii) If any $H\in{\rm Aut}_{F}(S_{f})$ restricts to some $\tau\in{\rm Aut}_{F}(K)$ such that $\tau\circ\sigma=\sigma\circ\tau$ then $H=H_{\tau,k}$ with $k\in K^{\times}$ as in Theorem 4. Moreover, $\{H_{\tau,k}\,|\,\tau\in{\rm Aut}_{F}(S_{f}),\tau\circ\sigma=\sigma\circ\tau,k\in K^{\times}\text{ with }\tau(a_{i})=(\prod_{l=i}^{m-1}\sigma^{l}(k))a_{i}\text{ for all }i\in\{0,\ldots,m-1\}\}$ is a subgroup of ${\rm Aut}_{F}(S_{f})$.

(iii) If $m=2$, $H\in{\rm Aut}(S_{f})$ if and only if $H=H_{\tau,k}$ with $\tau\circ\sigma=\sigma\circ\tau$, $\tau(a_{0})=k\sigma(k)a_{0},$ and $\tau(a_{1})=\sigma(k)a_{1}.$*

2.2.

The automorphisms groups of $S_{f}$ for $f(t)=t^{m}-a\in K[t;\sigma]$, $a\in K\setminus F$, are crucial in the understanding of the automorphism groups of all the algebras $S_{g}$, as for all nonassociative $S_{g}$ with $g(t)=t^{m}-\sum_{i=0}^{m-1}b_{i}t^{i}\in K[t;\sigma]$ such that $b_{0}=a$, ${\rm Aut}_{F}(S_{g})$ is a subgroup of ${\rm Aut}_{F}(S_{f})$:

Theorem 7**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$. Let $n\geq m-1$ and $g(t)=t^{m}-\sum_{i=0}^{m-1}b_{i}t^{i}\in K[t;\sigma]$ not be invariant.

(i) If $f(t)=t^{m}-b_{0}\in K[t;\sigma]$, $b_{0}\in K\setminus F$, then ${\rm Aut}_{F}(S_{g})\subset{\rm Aut}_{F}(S_{f})$ is a subgroup.

(ii) Let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ not be invariant and assume $b_{i}\in\left\{0,a_{i}\right\}$ for all $i\in\{0,\ldots,m-1\}$. Then ${\rm Aut}_{F}(S_{g})\subset{\rm Aut}_{F}(S_{f})$ is a subgroup.*

Proof.

(i) Let $H\in{\rm Aut}_{F}(S_{g})$. By Theorem 4, $H$ has the form $H(\sum_{i=0}^{m-1}x_{i}t^{i})=\tau(x_{0})+\sum_{i=1}^{m-1}\tau(x_{i})\prod_{l=0}^{i-1}\sigma^{l}(k)t^{i},$ where $\tau\in{\rm Aut}_{F}(K)$ and $k\in K^{\times}$ satisfy \tau(b_{i})=\Big{(}\prod_{j=i}^{m-1}\sigma^{j}(k)\Big{)}b_{i} for all $i=0,\ldots,m-1$. In particular, $\tau(b_{0})=k\sigma(k)\cdots\sigma^{m-1}(k)b_{0}$ and so $H$ is also an automorphism of $S_{f}$, again by Theorem 4.

(ii) The proof is analogous to (i). ∎

Similarily, for $n<m-1$ employing Theorem 5 we obtain:

Theorem 8**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Aut}_{F}(K)$. Let $n<m-1$ and $g(t)=t^{m}-\sum_{i=0}^{m-1}b_{i}t^{i}\in K[t;\sigma]$ not be invariant.

(i) If $f(t)=t^{m}-b_{0}\in K[t;\sigma]$, $b_{0}\in K\setminus F$, then $\{H\in{\rm Aut}_{F}(S_{g})\,|\,H=H_{\tau,k}\}$ is a subgroup of $\{H\in{\rm Aut}_{F}(S_{f})\,|\,H=H_{\tau,k}\}.$

(ii) Let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ not be invariant such that $b_{i}\in\left\{0,a_{i}\right\}$ for all $i\in\{0,\ldots,m-1\}$ Then $\{H\in{\rm Aut}_{F}(S_{g})\,|\,H=H_{\tau,k}\}$ is a subgroup of $\{H\in{\rm Aut}_{F}(S_{f})\,|\,H=H_{\tau,k}\}.$*

The automorphism groups of $S_{f}$ with $f(t)=t^{m}-a\in K[t;\sigma]$ are therefore particularly relevant.

3. Necessary conditions for extending Galois automorphisms to $S_{f}$

From now on we restrict ourselves to the situation that $R=K[t;\sigma]$ and $F={\rm Fix}(\sigma)$, where $K/F$ is a finite Galois field extension and $\sigma$ of order $n$.

We take a closer look at Equality (1), which gives necessary conditions on how to choose the elements $k\in K^{\times}$ used to extend $\tau\in{\rm Gal}(K/F)$ to ${\rm Aut}_{F}(S_{f})$. These become more restrictive for the choice of the elements $k$, the more coefficients in $f(t)$ are non-zero. Let $N_{K/F}:K\rightarrow F$ be the norm of $K/F$. All monic polynomials $f$ considered in the following are assumed to not be invariant and of degree $m$.

Proposition 9**.**

Suppose that $\sigma$ and $\tau$ commute. Let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ and $k\in K^{\times}$ such that

[TABLE]

*for all $i\in\{0,\ldots,m-1\}$. Then:

(i) For all $i\in\{0,\ldots,m-1\}$ with $a_{i}\not=0$, $N_{K/F}(k)$ is an $(m-i)$th root of unity.

In particular, if $a_{0}\not=0$ (e.g., if $f(t)$ is irreducible) then $N_{K/F}(k)$ is an $m$th root of unity, and if $a_{m-1}\not=0$ then $N_{K/F}(k)=1$.

If $a_{m-1}\in{\rm Fix}(\tau)^{\times}$ then $k=1$.

(ii) If $\tau\neq id$ and there is some $i$ such that $a_{i}$ is not contained in ${\rm Fix}(\tau)$, then $k\not=1$.

(iii) Suppose that there is some $a_{i}\not=0$ and $F$ does not contain any non-trivial $(m-i)$th roots of unity. Then $N_{K/F}(k)=1$.

(iv) If there is an $i\in\{0,\ldots,m-1\}$ such that $a_{i}\in{\rm Fix}(\tau)^{\times}$, then $1=\prod_{l=i}^{m-1}\sigma^{l}(k).$ In particular, if $n=m$, $\sigma$ generates ${\rm Gal}(K/F)$, and $a_{0}\in{\rm Fix}(\tau)^{\times}$ then $k\in{\rm ker}(N_{K/F})$.

(v) Suppose $\tau=id_{K}$. Then for all $i\in\{0,\ldots,m-1\}$ with $a_{i}\not=0$, $1=\prod_{l=i}^{m-1}\sigma^{l}(k).$ In particular, if $n=m$, $\sigma$ generates ${\rm Gal}(K/F)$ and $a_{0}\not=0$ then $k\in{\rm ker}(N_{K/F})$. In this case, the automorphisms extending $id_{K}$ are in one-one correspondence with those $k\in{\rm ker}(N_{K/F})$ satisfying (1).*

Proof.

(i) Equality (1) states that \tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}a_{i} for all $i\in\{0,\ldots,m-1\}$. Thus $N_{K/F}(a_{i})=\prod_{l=i}^{m-1}N_{K/F}(\sigma^{l}(k))N_{K/F}(a_{i})$ (apply $N_{K/F}$ to both sides of (1)), and therefore $N_{K/F}(a_{i})=N_{K/F}(k)^{m-i}N_{K/F}(a_{i})$ for all $i\in\{0,\ldots,m-1\}$ is a necessary condition on $k$. For all $a_{i}\not=0$, this yields $1=N_{K/F}(k)^{m-i}$ therefore $N_{K/F}(k)\in F^{\times}$ must be an $(m-i)$th root of unity, for all $i\in\{0,\ldots,m-1\}$, with $a_{i}\not=0$. Hence if $a_{m-1}\not=0$ then $\tau(a_{m-1})=\sigma^{m-1}(k)a_{m-1}$, thus $N_{K/F}(a_{m-1})=N_{K/F}(k)N_{K/F}(a_{m-1})$, i.e. $N_{K/F}(k)=1$. If even $a_{m-1}\in{\rm Fix}(\tau)^{\times}$ then $a_{m-1}=\sigma^{m-1}(k)a_{m-1}$ means $\sigma^{m-1}(k)=1$, i.e. $k=1$.

(ii) $k=1$ implies $\tau(a_{i})=a_{i}$, i.e. $a_{i}\in{\rm Fix}(\tau)$ for all $i\in\{0,\ldots,m-1\}$.

(iii) By (i), $N_{K/F}(k)\in F^{\times}$ is an $(m-i)$th root of unity, for all $i\in\{0,\ldots,m-1\}$ with $a_{i}\not=0$. If $F$ does not contain any non-trivial $(m-i)$th roots of unity, then $N_{K/F}(k)=1$.

(iv) If there is an $i\in\{0,\ldots,m-1\}$ such that $a_{i}\in{\rm Fix}(\tau)^{\times}$, then (1) becomes $1=\prod_{l=i}^{m-1}\sigma^{l}(k)$. In particular, if $a_{0}\in{\rm Fix}(\tau)^{\times}$, $m=n$ and $\sigma$ generates ${\rm Gal}(K/F)$, then $N_{K/F}(k)=1$ is a necessary condition on $k$.

(v) Here, (1) becomes $1=\prod_{l=i}^{m-1}\sigma^{l}(k)$ for all $i\in\{0,\ldots,m-1\}$ with $a_{i}\not=0$. In particular, if $n=m$, $\sigma$ generates ${\rm Gal}(K/F)$ and $a_{0}\not=0$ (which happens if $f(t)$ is irreducible) then $N_{K/F}(k)=1$ is a necessary condition on $k$. ∎

For instance, Proposition 9 (i) yields that $k=1$ if $a_{m-1}\in{\rm Fix}(\tau)^{\times}$ and so Theorems 4 and 5 become:

Theorem 10**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$ and $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ is not invariant with $a_{m-1}\in F^{\times}$.

(i) Let $n\geq m-1$. If $a_{i}\not\in{\rm Fix}(\tau)$ for all $\tau\not=id$ and all non-zero $a_{i}$, $i\not=m-1$, then ${\rm Aut}_{F}(S_{f})=\{id\}.$

If $f(t)\in F[t;\sigma]$, any automorphism $H$ of $S_{f}$ has the form $H_{\tau,1}$ where $\tau\in{\rm Gal}(K/F)$, and ${\rm Aut}_{F}(S_{f})\cong{\rm Gal}(K/F).$

(ii) Let $n<m-1$. If $f(t)\in F[t;\sigma]$ is not invariant, the maps $H_{\tau,1}$ are automorphisms of $S_{f}$ for all $\tau\in{\rm Gal}(K/F)$ and ${\rm Gal}(K/F)$ is isomorphic to a subgroup of ${\rm Aut}_{F}(S_{f})$.*

Proof.

(i) $H$ is an automorphism of $S_{f}$ if and only if $H$ has the form $H_{\tau,k}$, where $\tau\in{\rm Gal}(K/F)$ and $k\in K^{\times}$ is such that \tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}a_{i} for all $i\in\{0,\ldots,m-1\}$. Since $a_{m-1}\in F^{\times}$ we have $a_{m-1}\in{\rm Fix}(\tau)^{\times}$ for all $\tau$ which forces $k=1$ as the only possibility for any $\tau\in{\rm Gal}(K/F)$ by Proposition 9 (i). This in turn means that any extension $H_{\tau,k}$ has the form $H_{\tau,1}$. In particular, the existence of an extension $H_{\tau,k}$, $\tau\not=id$, implies $\tau(a_{i})=a_{i}$ for all non-zero $a_{i}$, $i\not=m-1$, that is $a_{i}\in{\rm Fix}(\tau)$ for all non-zero $a_{i}$.

Thus if $a_{i}\not\in{\rm Fix}(\tau)$ for all $\tau\not=id$ and all $i\in\{0,\ldots,m-2\}$ then there is no non-trivial $\tau$ that extends to an automorphism of $S_{f}$ and ${\rm Aut}_{F}(S_{f})=\{H_{id,1}\}=\{id\}$.

If $f(t)\in F[t;\sigma]$ then ${\rm Aut}_{F}(S_{f})=\{H_{\tau,1}\}\cong{\rm Gal}(K/F)$.

(ii) follows from (i) and Theorem 8. ∎

Note that indeed Condition (1) heavily restricts the choice of available $k$ to $k=1$ in most cases.

Corollary 11**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$. Let $n\geq m-1$ and $f(t)=t^{m}-a_{0}\in K[t;\sigma]$, $a_{0}\in K\setminus F$.

(i) $H\in{\rm Aut}_{F}(S_{f})$ if and only if $H=H_{\tau,k}$ where $k\in K^{\times}$ is such that \tau(a_{0})=\Big{(}\prod_{l=0}^{m-1}\sigma^{l}(k)\Big{)}a_{0}. In particular, here $N_{K/F}(k)$ is an $m$th root of unity.

(ii) For all $g(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ with $a_{0}\in K\setminus F$, ${\rm Aut}_{F}(S_{g})$ is a subgroup of ${\rm Aut}_{F}(S_{f})$.*

Proof.

(i) follows from Theorem 4 and Proposition 9.

(ii) follows from Theorem 7. ∎

For $f(t)=t^{m}-a_{0}\in K[t;\sigma]$, $a_{0}\in K\setminus F$, the automorphisms $H_{id,k}$ extending $id_{K}$ thus are in one-to-one correspondence with those $k$ satisfying $\prod_{l=0}^{m-1}\sigma^{l}(k)=1$ (in particular, we have $N_{K/F}(k)^{m}=1$). Analogously, we still obtain for $n<m-1$ employing Theorem 5 and Theorem 8:

Corollary 12**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$. Let $n<m-1$ and $f(t)=t^{m}-a_{0}\in K[t;\sigma]$, $a_{0}\in K\setminus F$.

(i) For all $k\in K^{\times}$ with $N_{K/F}(k)$ an $m$th root of unity and \tau(a_{0})=\Big{(}\prod_{l=0}^{m-1}\sigma^{l}(k)\Big{)}a_{0}, the maps $H_{\tau,k}$ are automorphisms of $S_{f}$.

(ii) For all $g(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ with $a_{0}\in K\setminus F$, $\{H\in{\rm Aut}_{F}(S_{g})\,|\,H=H_{\tau,k}\}$ is a subgroup of $\{H\in{\rm Aut}_{F}(S_{f})\,|\,H=H_{\tau,k}\}.$*

For $m=n$ and $K/F$ a cyclic field extension, the algebras considered in Corollary 11 are called nonassociative cyclic algebras of degree $m$, as they can be seen as canonical generalizations of associative cyclic algebras. These algebras are treated in Section 6.

4. Automorphisms extending $id_{K}$ when $K/F$ is a cyclic field extension

Let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ not be invariant. In general, we know that if $S_{f}$ has nucleus $K$ then every inner automorphism $G_{c}$ with $c\in K^{\times}$, extends $id_{K}$. Conversely, an extension $H_{id,k}$ of $id_{K}$ is inner for the right choice of $k$:

Lemma 13**.**

Let $k=c^{-1}\sigma(c)$ with $c\in K^{\times}$, then $H_{id,k}\in{\rm Aut}_{F}(S_{f})$ is an inner automorphism.

Proof.

A simple calculation shows that G_{c}\Big{(}\sum_{i=0}^{m-1}x_{i}t^{i}\Big{)}=\Big{(}c^{-1}\sum_{i=0}^{m-1}x_{i}t^{i}\Big{)}c=x_{0}+\\ \sum_{i=1}^{m-1}x_{i}c^{-1}\sigma^{i}(c)t^{i}=H_{id,k}\Big{(}\sum_{i=0}^{m-1}x_{i}t^{i}\Big{)}. ∎

Let now $K/F$ be a cyclic Galois field extension of degree $n$ with ${\rm Gal}(K/F)=\langle\sigma\rangle$ and norm $N_{K/F}:K\rightarrow F,\quad N_{K/F}(k)=k\sigma(k)\sigma^{2}(k)\cdots\sigma^{n-1}(k).$ By Hilbert’s Theorem 90, ${\rm ker}(N_{K/F})=\Delta^{\sigma}(1),$ where $\Delta^{\sigma}(l)=\{\sigma(c)lc^{-1}\,|\,c\in K^{\times}\}$ is the $\sigma$-conjugacy class of $l\in K^{\times}$ [19].

Theorem 14**.**

*(i) Every automorphism $H_{id,k}\in{\rm Aut}_{F}(S_{f})$ such that $N_{K/F}(k)=1$ is an inner automorphism.

(ii) If $n\geq m-1$ and $a_{m-1}\not=0$, or if $n=m$, $a_{i}=0$ for all $i\not=0$ and $a_{0}\in K\setminus F$, then these are all the automorphisms extending $id_{K}$.*

Proof.

(i) Suppose there is $H_{id,k}\in{\rm Aut}_{F}(S_{f})$ with $N_{K/F}(k)=1$, then by Hilbert 90, there exists $c\in K^{\times}$ such that $k=c^{-1}\sigma(c)$. Thus $H_{id,k}=H_{id,c^{-1}\sigma(c)}$ for $c\in K^{\times}$ and so $G_{c}=H_{id,k}$ by Lemma 13.

(ii) By Theorem 4 and Proposition 9 (i), these are all the automorphisms extending $id_{K}$ when $n\geq m-1$ if $a_{m-1}\not=0$. The remaining assertion is proved analogoulsy. ∎

5. Cyclic subgroups of ${\rm Aut}_{F}(S_{f})$

For any Galois field extension $K/F$ and $\sigma\in{\rm Gal}(K/F)$ of order $n$, we now give some conditions for ${\rm Aut}_{F}(S_{f})$ to have cyclic subgroups.

Theorem 15**.**

Suppose $F$ contains an $s$th root of unity $\omega$. Suppose that either $f(t)=t^{s}-a\in K[t;\sigma]$ where $a\in K\setminus F$, or $f(t)=t^{sl}-\sum_{i=0}^{l-1}a_{is}t^{is}\in K[t;\sigma]$ such that $S_{f}$ is not associative. Then $\langle H_{id,\omega}\rangle$ is a cyclic subgroup of ${\rm Aut}_{F}(S_{f})$ of order at most $s$ and of order $s$, if $\omega$ is a primitive root of unity.

Proof.

(i) Let $f(t)=t^{s}-a$. Then $\omega^{j}\sigma(\omega^{j})\cdots\sigma^{s-1}(\omega^{j})=\omega^{js}=1$ and so $H_{id,\omega^{j}}\in{\rm Aut}_{F}(S_{f})$ for all $0\leq j\leq s-1$ by Proposition 6.

(ii) Let $f(t)=t^{sl}-\sum_{i=0}^{l-1}a_{is}t^{is}$. Then we have $\prod_{q=is}^{ls-1}\sigma^{q}(\omega^{j})=\omega^{j(ls-is)}=1$ for all $i=0,\ldots,l-1$. Hence a_{is}=\Big{(}\prod_{q=is}^{ls-1}\sigma^{q}(\omega^{j})\Big{)}a_{is} for all $i=0,\ldots,l-1$ and so $H_{id,\omega^{j}}\in{\rm Aut}_{F}(S_{f})$ for all $0\leq j\leq s-1$ by Proposition 6.

In both (i) and (ii), $\langle H_{id,\omega}\rangle$ is a cyclic subgroup of ${\rm Aut}(S_{f})$ of order less or equal to $s$, since $H_{id,\omega^{j}}\circ H_{id,\omega^{r}}=H_{id,\omega^{j+r}}$ for all $0\leq j,r\leq sl-1$. ∎

Lemma 16**.**

Let $F$ have characteristic not two, $m$ be even and $f(t)=t^{m}-\sum_{i=0}^{(m-2)/2}a_{2i}t^{2i}\in K[t;\sigma]$ not invariant. Then $\left\{H_{id,1},H_{id,-1}\right\}$ is a subgroup of $S_{f}$ of order $2$.

Proof.

The maps $H_{id,1}$ and $H_{id,-1}$ are automorphisms of $S_{f}$ by Proposition 6, and $H_{id,-1}\circ H_{id,-1}=H_{id,1}$. ∎

If $f\in F[t]\subset K[t;\sigma]$, we obtain:

Theorem 17**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$, and $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in F[t;\sigma]\subset K[t;\sigma]$ is not invariant.

(i) $\langle H_{\sigma,1}\rangle\cong\mathbb{Z}/n\mathbb{Z}$ is a cyclic subgroup of ${\rm Aut}_{F}(S_{f})$.

(ii) Suppose ${\rm Gal}(K/F)=\langle\sigma\rangle$, $n=m$ is prime, $a_{0}\neq 0$ and not all of $a_{1},\ldots,a_{m-1}$ are zero. Then ${\rm Aut}_{F}(S_{f})=\langle H_{\sigma,1}\rangle\cong\mathbb{Z}/m\mathbb{Z}.$*

Proof.

Let $j\in\left\{0,\ldots,n-1\right\}$. Since $\tau(a_{i})=a_{i}$ for all $i$, here (1) becomes

[TABLE]

for all $i\in\left\{0,\ldots,m-1\right\}$.

(i) Clearly, (9) is satisfied for $k=1$ and all $i\in\left\{0,\ldots,m-1\right\}$, therefore the maps $H_{\tau,1}$ are automorphisms of $S_{f}$ for all $\tau\in{\rm Gal}(K/F)$ by Theorems 4 and 5. We have $H_{\sigma^{j},1}\circ H_{\sigma^{l},1}=H_{\sigma^{j+l},1}$ and $H_{\sigma^{n},1}=H_{id,1}$. Hence $\langle H_{\sigma,1}\rangle=\left\{H_{id,1},H_{\sigma,1},\ldots,H_{\sigma^{m-1},1}\right\}$ is a cyclic subgroup of order $n$.

(ii) By Theorem 4, the automorphisms of $S_{f}$ are exactly the maps $H_{\sigma^{j},k}$ where $j\in\left\{0,\ldots,n-1\right\}$ and $k\in K^{\times}$ satisfies (9) for all $i\in\left\{0,\ldots,m-1\right\}$. The maps $H_{\sigma^{j},1}$ are therefore automorphisms of $S_{f}$ for all $j\in\left\{0,\ldots,n-1\right\}$. We prove that these are the only automorphisms of $S_{f}$: $a_{0}\neq 0$ and so $N_{K/F}(k)=1$ by (9). Therefore, by Hilbert 90, there exists $\alpha\in K$ such that $k=\sigma(\alpha)/\alpha$. Let $l\in\left\{1,\ldots,m-1\right\}$ be such that $a_{l}\neq 0$. Then by (9),

[TABLE]

Thus $\alpha\in\text{Fix}(\sigma^{j})=F$ since $m$ is prime. Therefore $k=\sigma(\alpha)/\alpha=\alpha/\alpha=1$ as required. ∎

This complements our results from Theorem 10, which in case ${\rm Gal}(K/F)$ is cyclic of degree $n$ mean the following:

Corollary 18**.**

*Suppose ${\rm Gal}(K/F)$ is cyclic of degree $n$ and $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in F[t;\sigma]$ not invariant with $a_{m-1}\in F^{\times}$.

(i) Let $n\geq m-1$ then for all $\tau\in{\rm Gal}(K/F)$ the maps $H_{\tau,1}$ are exactly the automorphisms of $S_{f}$ and ${\rm Aut}_{F}(S_{f})\cong{\rm Gal}(K/F)\cong\mathbb{Z}/n\mathbb{Z}.$

(ii) Let $n<m-1$ then for all $\tau\in{\rm Gal}(K/F)$ the maps $H_{\tau,1}$ are automorphisms of $S_{f}$ and ${\rm Gal}(K/F)\cong\mathbb{Z}/n\mathbb{Z}$ is isomorphic to a subgroup of ${\rm Aut}_{F}(S_{f})$.*

6. Nonassociative cyclic algebras

6.1.

Let $K/F$ be a cyclic Galois extension of degree $m$ with ${\rm Gal}(K/F)=\langle\sigma\rangle$ and $f(t)=t^{m}-a\in K[t;\sigma]$. Then $(K/F,\sigma,a)=K[t;\sigma]/K[t;\sigma](t^{m}-a)$ is called a nonassociative cyclic algebra of degree $m$ over $F$. It is not associative for all $a\in K\setminus F$ and a cyclic associative central simple algebra over $F$ for $a\in F^{\times}$. We will only consider the case that $a\in K\setminus F$. If $1,a,a^{2},\ldots,a^{m-1}$ are linearly independent over $F$ then $(K/F,\sigma,a)$ is a division algebra (cf. [32], [30] for finite $F$). In particular, if $K/F$ is of prime degree then $(K/F,\sigma,a)$ is a division algebra for every $a\in K\setminus F$.

Theorem 19**.**

*Let $A=(K/F,\sigma,a)$ be a nonassociative cyclic algebra of degree $m$.

(i) All the automorphisms of $A$ which extend $id_{K}$ are inner automorphisms and of the form $H_{id,l}$ for all $l\in K^{\times}$ such that $N_{K/F}(l)=1$.

The subgroup they generate in ${\rm Aut}_{F}(A)$ is isomorphic to ${\rm ker}(N_{K/F})$.

(ii) An automorphism $\sigma^{j}\not=id$ can be extended to $H\in{\rm Aut}_{F}(A)$, if and only if there is some $l\in K$ such that $\sigma^{j}(a)=N_{K/F}(l)a.$ In that case, $H=H_{\sigma^{j},l}$ and if $m$ is prime then $N_{K/F}(l)=\omega$ for an $m$th root of unity $1\not=\omega\in F$.

(iii) Let $c\in K\setminus F$ and suppose there exists $r\in\mathbb{N}$ such that $c^{r}\in F^{\times}$. Let $r$ be minimal. Then $\langle G_{c}\rangle$ is a cyclic subgroup of ${\rm Aut}_{F}(S_{f})$ of order $r$.*

Proof.

Theorem 4, Theorem 14 (ii), and Proposition 9 imply (i) and (ii).

(iii) Let $c\in K\setminus F$. Then $G_{c}$ is an automorphism, because $K$ is the nucleus of $A$. Since $G_{c}\circ G_{c}=G_{c^{2}}$, $G_{c}\circ G_{c}\circ G_{c}=G_{c^{3}}$ and so on, we have $G_{c^{r}}=id$ if and only if $c^{r}\in F$. If $r\in\mathbb{N}$ is smallest possible then $\langle G_{c}\rangle$ is a cyclic subgroup of ${\rm Aut}_{F}(S_{f})$ of order $r$. ∎

Note that different roots of unity yield different $l$ in Theorem 19 (ii). This yields:

Theorem 20**.**

*Let $A=(K/F,\sigma,a)$ be a nonassociative cyclic algebra of degree $m$. Suppose $F$ contains a non-trivial $m$th root of unity $\omega$.

(i) $\langle H_{id,\omega}\rangle$ is a cyclic subgroup of ${\rm Aut}_{F}(A)$ of order at most $m$. If $\omega$ is a primitive $m$th root of unity, then $\langle H_{id,\omega}\rangle$ has order $m$.

(ii) If there is an element $l\in K$, such that $N_{K/F}(l)=\omega$ for $\omega$ a primitive $m$th root of unity and $\sigma(d)=\omega d$, then the subgroup generated by $H_{\sigma,l}$ has order $m^{2}$.*

Proof.

(i) follows from Theorem 15.

(ii) Suppose $\sigma$ can be extended to an $F$-automorphism $H$ of $A$. Then by Theorem 19, there is an element $l\in K$, such that $N_{K/F}(l)=\omega$, $\omega\not=1$ and $\sigma(d)=\omega d$, and $H=H_{\sigma,l}$. (If $1=N_{K/F}(l)$, then $\sigma(d)=d$, contradiction.)

The subgroup generated by $H=H_{\sigma,l}$ has order greater than $m$, since $H_{\sigma,l}\circ\cdots\circ H_{\sigma,l}$ ($m$-times) becomes $H_{\sigma^{m},b}=H_{id,\omega}$ with $\omega=N_{K/F}(l)$. $H_{id,\omega}$ has order $m$, so the subgroup generated by $H=H_{\sigma,l}$ has order $m^{2}$. ∎

6.2. The case that $m$ is prime

Let us now assume that the cyclic field extension $K/F$ has prime degree $m={\rm deg}(f)$. Suppose that $F$ contains a primitive $m$th root of unity, where $m$ is prime to the characteristic of $F$. Then $K=F(d)$, where $d$ is a root of an irreducible polynomial $t^{m}-c\in F[t]$.

Lemma 21**.**

(cf. [33, Lemma 6.2.7]) The eigenvalues of $\sigma^{j}\in{\rm Gal}(K/F)$ are precisely the $m$th roots of unity. Moreover, the only possible eigenvectors are of the form $ed^{i}$ for some $i$, $0\leq i\leq m-1$ and some $e\in F$.

Let $f(t)=t^{m}-a\in K[t;\sigma],\quad a\not\in F.$ Then we get the following strong restriction for automorphisms of $S_{f}$:

Theorem 22**.**

$H$* is an automorphism of $S_{f}$ extending $\sigma^{j}\not=id$ if and only if $H=H_{\sigma^{j},k}$ for some $k\in K^{\times}$, where $N_{K/F}(k)$ is an $m$th root of unity and $a=ed^{s}$ for some $e\in F^{\times}$ and some $d^{s}$.*

Proof.

$H$ is an automorphism of $S_{f}$ if and only if $H=H_{\sigma^{j},k}$ where $j\in\left\{0,\ldots,m-1\right\}$ and $k\in K^{\times}$ is such that \sigma^{j}(a)=\Big{(}\prod_{l=0}^{m-1}\sigma^{l}(k)\Big{)}a=N_{K/F}(k)a. For all $\sigma^{j}\not=id$, by Lemma 21 this condition is equivalent to $N_{K/F}(k)$ being an $m$th root of unity and $a=ed^{s}$ for some $d^{s}$ and $e\in F^{\times}$, for all $k\in K^{\times}$. ∎

Applying Theorem 7, our results for the automorphisms of a nonassociative cyclic algebra $A=(K/F,\sigma,a)$ of degree $m$ yield the following observations for more general algebras $S_{g}$:

Corollary 23**.**

Suppose ${\rm Gal}(K/F)=\langle\sigma\rangle$ is cyclic of degree $m$ and $g(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in K[t;\sigma]$ is not invariant with $a_{0}\in K\setminus F$. Suppose one of the following holds:

• •

$F$* has no *$m$th root of unity.

• •

$m$* is prime and $F$ contains a primitive *$m$th root of unity, where $m$ is prime to the characteristic of $F$. Let $K=F(d)$ as in Section 6.2 and $a_{0}\not=ed^{i}$, $e\in F^{\times}$.

Then every $F$-automorphism of $S_{g}$ leaves $K$ fixed, is inner and ${\rm Aut}_{F}(S_{g})$ is a subgroup of ${\rm ker}(N_{K/F})$, thus cyclic. In particular, if ${\rm ker}(N_{K/F})$ has prime order, then either ${\rm Aut}_{F}(S_{g})$ is trivial or ${\rm Aut}_{F}(S_{g})\cong{\rm ker}(N_{K/F})$.

6.3. The automorphism groups of nonassociative quaternion algebras

Recall the dicyclic group

[TABLE]

of order $4l$. The semidirect product $\mathbb{Z}/s\mathbb{Z}\rtimes_{l}\mathbb{Z}/n\mathbb{Z}$ between the cyclic groups $\mathbb{Z}/s\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$ corresponds to a choice of an integer $l$ such that $l^{n}\equiv 1\ {\rm mod}\ s$. It can be described by the presentation $\mathbb{Z}/s\mathbb{Z}\rtimes_{l}\mathbb{Z}/n\mathbb{Z}=\langle x,y\ |\ x^{s}=1,\ y^{n}=1,\ yxy^{-1}=x^{l}\rangle.$

We obtain the following result for the automorphism groups of nonassociative quaternion algebras (where $m=2$):

Theorem 24**.**

Suppose $K=F(\sqrt{b})$ is a quadratic field extension of $F$, $char(F)\neq 2$, and consider the nonassociative quaternion algebra $A=(K/F,\sigma,\lambda\sqrt{b})$ for some $\lambda\in F^{\times}$. Suppose there exists $k\in K^{\times}$ such that $k\sigma(k)=-1$.

*For every $c\in K\setminus F$ for which there is a positive integer $j$ such that $c^{j}\in F^{\times}$, pick the smallest such $j$.

(i) If $j$ is even then ${\rm Aut}_{F}(S_{f})$ contains the dicyclic group of order $2j$.

(ii) If $j$ is odd then ${\rm Aut}_{F}(S_{f})$ contains a subgroup isomorphic to the semidirect product $\mathbb{Z}/j\mathbb{Z}\rtimes_{j-1}\mathbb{Z}/4\mathbb{Z}.$ In particular, ${\rm Aut}_{F}(A)$ always contains a subgroup isomorphic to $\mathbb{Z}/4\mathbb{Z}$.*

Proof.

Since $\sigma(\sqrt{b})=-\sqrt{b}$ and $k\sigma(k)=-1$, $H_{\sigma,k}\in{\rm Aut}_{F}(S_{f})$ by Theorem 19. A simple calculation shows that $\langle H_{\sigma,k}\rangle=\{H_{\sigma,k},H_{id,-1},H_{\sigma,-k},H_{id,1}\}.$ $\langle G_{c}\rangle$ is a cyclic subgroup of ${\rm Aut}_{F}(S_{f})$ of order $j$ by Theorem 19 (iii).

(i) Suppose $j$ is even and write $j=2l$. We prove first that $G_{c^{l}}=H_{id,-1}$. Write $c^{l}=\mu_{0}+\mu_{1}\sqrt{b}$ for some $\mu_{0},\mu_{1}\in F$. Then $c^{j}=c^{2l}=\mu_{0}^{2}+\mu_{1}^{2}b+2\mu_{0}\mu_{1}\sqrt{b}\in F$ which implies $2\mu_{0}\mu_{1}=0$. Hence $\mu_{0}=0$ or $\mu_{1}=0$. Since $j$ is minimal, $c^{l}\notin F$ so $\mu_{0}=0$ and $c^{l}=\mu_{1}\sqrt{b}$. We obtain

[TABLE]

which implies $G_{c^{l}}=H_{id,-1}$. Next we prove $(H_{\sigma,k})^{-1}G_{c}H_{\sigma,k}=G_{c}^{-1}$. Simple calculations show $(H_{\sigma,k})^{-1}=H_{\sigma,-k}$ and $G_{c}^{-1}=G_{\sigma(c)}$. We have

[TABLE]

and so $(H_{\sigma,k})^{-1}G_{c}H_{\sigma,k}=G_{c}^{-1}$.

Thus $H_{\sigma,k}^{2}=H_{id,-1}=G_{c^{l}}=G_{c}^{l},\ G_{c}^{2l}=id$ and $(H_{\sigma,k})^{-1}G_{c}H_{\sigma,k}=G_{c}^{-1}$. Hence $\langle H_{\sigma,k},G_{c}\rangle$ has the presentation (10) as required.

(ii) Suppose $j$ is odd. Then $\langle G_{c}\rangle$ does not contain $H_{id,-1}$ as $H_{id,-1}$ has order $2$ which implies $\langle H_{\sigma,k}\rangle\cap\langle G_{c}\rangle=\{id\}$. Furthermore $(H_{\sigma,k})^{-1}G_{c}H_{\sigma,k}=G_{c}^{-1}=G_{c}^{j-1}=G_{c^{j-1}}$ can be shown similarly as in (i). Note that $(j-1)^{4}=j^{4}-4j^{3}+6j^{2}-4j+1\equiv 1\text{ mod }(j).$ Thus ${\rm Aut}_{F}(S_{f})$ contains the subgroup $\langle G_{c}\rangle\rtimes_{j-1}\langle H_{\sigma,k}\rangle\cong\mathbb{Z}/j\mathbb{Z}\rtimes_{j-1}\mathbb{Z}/4\mathbb{Z}$ as required.

In particular, choose $c=\sqrt{b}$ in (i), so that $j=2$. This implies ${\rm Aut}_{F}(A)$ contains the dicyclic group of order $4$, which is the cyclic group of order $4$. ∎

Example 25**.**

(i) Let $F=\mathbb{Q}(i)$, $K=F(\sqrt{-3})$, $\sigma(\sqrt{-3})=-\sqrt{-3}$ and $A=(K/F,\sigma,\lambda\sqrt{-3})$ be a nonassociative quaternion algebra with some $\lambda\in F^{\times}$. Note that for $k=i$ we have $i\sigma(i)=-1$. Let $c=1+\sqrt{-3}$. Then $c^{2}=-2+2\sqrt{-3}\text{ and }c^{3}=-8$ which implies $j=3$ here. Therefore ${\rm Aut}_{F}(S_{f})$ contains a subgroup isomorphic to the semidirect product $\mathbb{Z}/3\mathbb{Z}\rtimes_{2}\mathbb{Z}/4\mathbb{Z}$ by Theorem 24.

(ii) Let $F=\mathbb{Q}(i)$, $K=F(\sqrt{-1/12})$, $\sigma(\sqrt{-1/12})=-\sqrt{-1/12}$ and $A=(K/F,\sigma,\lambda\sqrt{-1/12})$ be a nonassociative quaternion algebra for some $\lambda\in F^{\times}$. Again for $k=i$ we have $i\sigma(i)=-1$. Let $c=1+2\sqrt{-1/12}$. Then $c^{2}=2/3+2i/\sqrt{3},$ $c^{3}=8i/3\sqrt{3},$ $c^{4}=-8/9+8i/3\sqrt{3},$ $c^{5}=-16/9+16i/9\sqrt{3}$ and $c^{6}=-64/27.$ Hence $c,c^{2},c^{3},c^{4},c^{5}\in K\setminus F$ and $c^{6}\in F$. Therefore ${\rm Aut}_{F}(A)$ contains the dicyclic group of order $12$ by Theorem 24.

7. Isomorphisms between $S_{f}$ and $S_{g}$

The proofs of the previous sections can be adapted to check when two Petit algebras are isomorphic and when not. This is not the main focus of this paper so we just point out how some of the results can be transferred.

If $K$ and $L$ are fields, and $S_{f}=K[t;\sigma]/K[t;\sigma]f(t)\cong L[t;\sigma^{\prime}]/L[t;\sigma^{\prime}]g(t)=S_{g},$ then $K\cong L,$ $Nuc_{r}(S_{f})\cong Nuc_{r}(S_{g}),$ ${\rm deg}(f)={\rm deg}(g),$ and ${\rm Fix}(\sigma)\cong{\rm Fix}(\sigma^{\prime}),$ since isomorphic algebras have the same dimensions, and isomorphic nuclei and center.

If $G$ is an automorphism of $R=K[t;\sigma]$ which restricts to an automorphism $\tau$ on $K$ which commutes with $\sigma$, $f\in R$ is irreducible and $g(t)=G(f(t))$, then $G$ induces an isomorphism $S_{f}\cong S_{G(g)}$ [20, Theorem 7] (the proof works for any base field).

From now on let $F$ be the fixed field of $\sigma$, $\sigma$ have order $n$, and both $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}$ and $g(t)=t^{m}-\sum_{i=0}^{m-1}b_{i}t^{i}\in K[t;\sigma]$ be not invariant. Then the following is proved analogously to Theorem 4, Theorem 5 and Proposition 6:

Theorem 26**.**

Suppose $\sigma$ commutes with all $\tau\in{\rm Aut}_{F}(K)$ and $n\geq m-1$. Then $S_{f}\cong S_{g}$ if and only if there exists $\tau\in{\rm Aut}_{F}(K)$ and $k\in K^{\times}$ such that

[TABLE]

for all $i\in\{0,\ldots,m-1\}$. Every such $\tau$ and $k$ yield a unique isomorphism $G_{\tau,k}:S_{f}\rightarrow S_{g}$,

[TABLE]

If $n<m-1$ we still get a partial result:

Theorem 27**.**

Suppose there exists $\tau\in{\rm Aut}_{F}(K)$ and $k\in K^{\times}$ such that $\tau\circ\sigma=\sigma\circ\tau$ and such that (11) holds for all $i\in\{0,\ldots,m-1\}$. Then $S_{f}\cong S_{g}$ with an isomorphism given by

[TABLE]

as in Theorem 26.

Corollary 28**.**

For every $k\in K^{\times}$ such that a_{i}=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}b_{i} for all $i\in\{0,\ldots,m-1\}$, $G_{id,k}:S_{f}\rightarrow S_{g}$ is an isomorphism.

As a direct consequence of Theorem 26 we obtain:

Theorem 29**.**

Suppose $\sigma$ commutes with all $\tau\in{\rm Aut}_{F}(K)$ and $n\geq m-1$. If $S_{f}\cong S_{g}$, then $a_{i}=0$ if and only if $b_{i}=0$, for all $i\in\{0,\ldots,m-1\}$.

Proof.

If $S_{f}\cong S_{g}$ then by Theorem 26, there exists $j\in\left\{0,\ldots,n-1\right\}$ and $k\in K^{\times}$ such that \tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}b_{i} for all $i\in\{0,\ldots,m-1\}$. This implies $a_{i}=0$ if and only if $b_{i}=0$, for all $i\in\{0,\ldots,m-1\}$. ∎

From now on we restrict ourselves to the situation that $R=K[t;\sigma]$ and $F={\rm Fix}(\sigma)$, where $K/F$ is a finite Galois field extension and $\sigma$ of order $n$. We take a closer look at the consequences of Equality (11):

Proposition 30**.**

*Let $k\in K^{\times}$ such that \tau(a_{i})=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}b_{i} for all $i\in\{0,\ldots,m-1\}$. Then $a_{i}=0$ if and only if $b_{i}=0$ and:

(i) For all $i\in\{0,\ldots,m-1\}$ with $a_{i}\not=0$, $N_{K/F}(a_{i})=N_{K/F}(k)^{m-i}N_{K/F}(b_{i}).$

(ii) If there is some $i\in\{0,\ldots,m-1\}$ such that $a_{i}\in{\rm Fix}(\tau)^{\times}$, then $a_{i}/b_{i}=\prod_{l=i}^{m-1}\sigma^{l}(k).$ In particular, if $a_{m-1}\in F^{\times}$ and $b_{m-1}\in F^{\times}$, then $k\in F^{\times}$ and $a_{i}=k^{m-i}b_{i}$ for all $i\in\{0,\ldots,m-1\}$.

(iii) If $a_{0}\in{\rm Fix}(\tau)^{\times}$, $m=n$ and ${\rm Gal}(K/F)=\langle\sigma\rangle$ then $a_{0}=N_{K/F}(k)b_{0}.$*

Proof.

(i) Equality (11) implies that $N_{K/F}(a_{i})=\prod_{l=i}^{m-1}N_{K/F}(\sigma^{l}(k))N_{K/F}(b_{i})$ (simply apply $N_{K/F}$ to both sides of (11)), therefore $N_{K/F}(a_{i})=N_{K/F}(k)^{m-i}N_{K/F}(b_{i})$ for all $i\in\{0,\ldots,m-1\}$ is a necessary condition on $k$.

(ii) If there is an $i\in\{0,\ldots,m-1\}$ such that $a_{i}\in{\rm Fix}(\tau)^{\times}$, then (11) implies that a_{i}=\Big{(}\prod_{l=i}^{m-1}\sigma^{l}(k)\Big{)}b_{i}, so that we obtain $a_{i}/b_{i}=\prod_{l=i}^{m-1}\sigma^{l}(k).$

Alternatively, if $a_{m-1}\in F^{\times}$ and $b_{m-1}\in F^{\times}$, then $a_{m-1}=\sigma^{m-1}(k)b_{m-1}$ imply $k\in F^{\times}$, hence $a_{i}=k^{m-1-i}b_{i}$ for all $i\in\{0,\ldots,m-1\}$.

(iii) In particular, if $a_{0}\in{\rm Fix}(\tau)^{\times}$, $m=n$ and $\sigma$ generates ${\rm Gal}(K/F)$, then $a_{0}/b_{0}=\prod_{l=0}^{m-1}\sigma^{l}(k)=N_{K/F}(k)$ is a necessary condition on $k$. ∎

Corollary 31**.**

*Suppose $\sigma$ commutes with all $\tau\in{\rm Gal}(K/F)$ and $n\geq m-1$. Assume that one of the following holds:

(i) There exists $i\in\{0,\ldots,m-1\}$ such that $b_{i}\not=0$ and $N_{K/F}(a_{i}b_{i}^{-1})\not\in N_{K/F}(K^{\times})^{m-i};$

(ii) $m=n$, $a_{0}\in F^{\times}$ and $b_{0}\in K\setminus F$.

Then $S_{f}\ncong S_{g}$.*

Corollary 32**.**

*Suppose ${\rm Gal}(K/F)=\langle\sigma\rangle$ and $n=m$. Let $f(t)=t^{m}-a,$ $g(t)=t^{m}-b\in K[t;\sigma]$ where $a,b\in K\setminus F$.

(i) $S_{f}\cong S_{g}$ if and only if there exists $\tau\in{\rm Gal}(K/F)$ and $k\in K^{\times}$ such that $\tau(a)=N_{K/F}(k)b.$ (ii) If $\sigma^{j}(a)\neq N_{K/F}(k)b$ for all $k\in K^{\times}$, $j=0,\ldots,m-1$, then $S_{f}\not\cong S_{g}$.*

These follow from Proposition 30. Note that Corollary 32 canonically generalizes well-known criteria for associative cyclic algebras.