polynomial ring $R=D[t;\sigma]$, and their nuclei. As a Corollary to Theorem
58, we obtain:

######  Theorem 65.

Let $f(t)=t^{m}-\sum\limits_{i=0}^{m-1}a_{i}t^{i}\in R$ and $c\in D$. Then
$c\in L$ if and only if $\sigma^{m}(c)a_{k}=a_{k}\sigma^{k}(c)$ for all
$k\in\\{0,1,\dots,m-1\\}$.

###### Proof.

For $\delta=0$, $\Delta_{j,i}=0$ for all non-negative integers $i,j$ with
$i\neq j$, and $\Delta_{j,j}=\sigma^{j}$. We apply this to Theorem 58. ∎

From now on we denote the indices of the nonzero coefficients $a_{i}$ in
$f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in D[t;\sigma]$ by
$\lambda_{1},\dots,\lambda_{r}$, $1\leq r\leq m$. We denote the set of these
indices by

$\Lambda_{f}=\\{\lambda_{1},\lambda_{2},\dots,\lambda_{r}\\}\subset\\{0,1,\dots,m-1\\}.$

If it is clear from the context which $f$ is used, we simply write $\Lambda$.

######  Corollary 66.

Let $f(t)=t^{m}-\sum\limits_{i=0}^{m-1}a_{i}t^{i}\in R$ and assume that
$a_{i}\in C$ for all $i\in\\{0,1,\dots,m-1\\}$. Then

$L=\bigcap\limits_{\lambda_{j}\in\Lambda}{\rm Fix}(\sigma^{m-\lambda_{j}}).$

In the case that $D$ is commutative (i.e. $D$ is a field and $D=C$) Petit
states that $t$ lies in the right nucleus if and only if the coefficients of
$f$ are all fixed by $\sigma$ [Pet66, (16)]. We extend this result to the case
that $D$ is a (noncommutative) unital division ring:

######  Theorem 67.

[Pet66, (16) for $D$ commutative] Let
$f(t)=t^{m}-\sum\limits_{i=0}^{m-1}a_{i}t^{i}\in R.$ Then $a_{i}\in{\rm
Fix}(\sigma)$ for all $i\in\\{0,1,\dots,m-1\\}$ if and only if $t\in{\rm
Nuc}_{r}(S_{f})$.

###### Proof.

If $a_{i}\in{\rm Fix}(\sigma)$ for all $i$, then $t\in{\rm Nuc}_{r}(S_{f})$ by
Theorem 60 with $\delta=0$. Conversely, suppose that $t\in{\rm
Nuc}_{r}(S_{f})$, then $[t,t^{m-1},t]=0$ which implies that

$(\sum_{i=0}^{m-1}a_{i}t^{i})t=t(\sum_{i=0}^{m-1}a_{i}t^{i}),$

i.e.,

$(a_{m-1}t^{m-1}+\sum_{i=0}^{m-2}a_{i}t^{i})t=t(a_{m-1}t^{m-1}+\sum_{i=0}^{m-2}a_{i}t^{i}).$

This yields

$a_{m-1}t^{m}+\sum_{i=0}^{m-2}a_{i}t^{i+1}=\sigma(a_{m-1})t^{m}+\sum_{i=0}^{m-2}\sigma(a_{i})t^{i+1}.$

We note that

$t^{m}=(\sum\limits_{i=0}^{m-1}a_{i}t^{i})\,{\rm mod}_{r}f.$

Suppose that $k$ is the smallest element of $\\{0,1,\dots,m-1\\}$ such that
$a_{k}\neq 0$, i.e. $a_{i}=0$ for all $i<k$. Then

$\sum_{i=k}^{m-1}a_{m-1}a_{i}t^{i}+\sum_{i=k}^{m-2}a_{i}t^{i+1}=\sum_{i=k}^{m-1}\sigma(a_{m-1})a_{i}t^{i}+\sum_{i=k}^{m-2}\sigma(a_{i})t^{i+1}.$

Equating the $t^{k}$ terms gives $a_{m-1}a_{k}=\sigma(a_{m-1})a_{k}\Rightarrow
a_{m-1}\in{\rm Fix}(\sigma).$ Therefore

$\displaystyle a_{m-1}\sum_{i=k}^{m-1}a_{i}t^{i}+\sum_{i=k}^{m-2}a_{i}t^{i+1}$
$\displaystyle=a_{m-1}\sum_{i=k}^{m-1}a_{i}t^{i}+\sum_{i=k}^{m-2}\sigma(a_{i})t^{i+1}$
$\displaystyle\Rightarrow\sum_{i=k}^{m-2}a_{i}t^{i+1}$
$\displaystyle=\sum_{i=k}^{m-2}\sigma(a_{i})t^{i+1}$

and we have $a_{i}\in{\rm Fix}(\sigma)$ for all $i\in\\{k,k+1,\dots,m-2\\}$.
Since $a_{i}=0$ for $i<k$, and $a_{m-1}\in{\rm Fix}(\sigma)$, we have
$a_{i}\in{\rm Fix}(\sigma)$ for all $i\in\\{0,1,\dots,m-1\\}.$ ∎

The following result of Brown and Pumplün [BP18, Proposition 2] is an
immediate Corollary to Theorem 67:

######  Corollary 68.

[BP18, Proposition 2] Let $f(t)\in F[t]\subset R$. Then $F[t]/(f(t))$ is a
commutative subring of ${\rm Nuc}_{r}(S_{f}),$ and a subfield of degree $m$ if
$f$ is irreducible in $F[t]$.

###### Proof.

$S_{f}$ contains the commutative subring $F[t]/(f)$ which is isomorphic to the
ring consisting of the elements $\sum_{i=0}^{m-1}a_{i}t^{i}$ with $a_{i}\in
F$. In particular, the powers of $t$ are associative. This implies that
$t\in{\rm Nuc}_{r}(S_{f})$ [Pet66, (5)]. Clearly $F\subset{\rm
Nuc}_{r}(S_{f})$, so $F\oplus Ft\oplus\dots\oplus Ft^{m-1}\subset{\rm
Nuc}_{r}(S_{f})$, hence we obtain the assertion. ∎

######  Corollary 69.

Suppose that $f(t)\in F[t]\subset R$ is not right-invariant such that
$f(t)\not=t$, and let $h=\hat{h}(t^{n})$ be its minimal central left multiple.
If $f$ is irreducible and $[L:F]=n/k$, where $k$ is the number of irreducible
factors of the minimal central left multiple $h$ of $f$, then ${\rm
Nuc}_{r}(S_{f})=L_{f}.$ In particular, then $f$ is irreducible in
$K[t;\sigma]$ if and only if $f$ is irreducible in $L[t;\sigma]$.

###### Proof.

We know that $L_{f}=L[t;\sigma]/L[t;\sigma]f(t)$ is a subalgebra of ${\rm
Nuc}_{r}(S_{f})$ (Theorem 64). If $f$ is irreducible then ${\rm
Nuc}_{r}(S_{f})$ has degree $ms$ over $F$ by Theorem 19, therefore comparing
the degrees of the field extensions we obtain the assertion. ∎

Below are two examples showing how we may compute the right nucleus of $S_{f}$
using the preceding theory as well as the Wedderburn theory of Chapter 2.

Example 1 Let $D=(-1,-1)_{\mathbb{Q}}$, $\sigma:D\longrightarrow D,$
$\sigma(d)=udu^{-1}$ with $u=i+1$ and $R=D[t;\sigma]$. Thus $\sigma$ has inner
order one, ${\rm Fix}(\sigma)=\mathbb{Q}(i)$, and $C(R)=\mathbb{Q}[x]$ with
$x=u^{-1}t$. Also since $\sigma^{4}=id$ and ${\rm Fix}(\sigma^{2})={\rm
Fix}(\sigma^{3})={\rm Fix}(\sigma)=\mathbb{Q}(i)$, every $f\in R$ is bounded.
Let $f\in R$ be monic and irreducible of degree $m>1$ such that $(f,t)_{r}=1$.
Let $h(t)=\hat{h}(u^{-1}t)$ be its minimal central left multiple, such that
$\hat{h}(x)\not=x$. Then one of the following holds:

  * •

${\rm Nuc}_{r}(S_{f})\cong E_{\hat{h}}$ is a field extension of degree $2m$,
$k=2$, and ${\rm deg}(h)=2m$, or

  * •

${\rm Nuc}_{r}(S_{f})$ is a central division algebra over $E_{\hat{h}}$ of
degree $2$, $h$ is irreducible in $R$, ${\rm deg}(h)=m={\rm deg}(\hat{h})$ and
$[{\rm Nuc}_{r}(S_{f}):F]=4m$ (Corollary 22), in which case $f$ must be right-
invariant checking the dimensions.

So assume that $f$ is not right-invariant, monic and irreducible, such that
$(f,t)_{r}=1$. Then ${\rm Nuc}_{r}(S_{f})\cong E_{\hat{h}}$ is a field
extension of $\mathbb{Q}$ of degree $2m$. In particular, let
$f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in\mathbb{Q}(i)[t]\subset D[t;\sigma]$
then $\mathbb{Q}(i)\subset L^{(\sigma,f)}$ (Proposition 55) and
$t,t^{2},\dots,t^{m-1}\in{\rm Nuc}_{r}(S_{f})$. Thus we obtain that

$\mathbb{Q}(i)[t]/\mathbb{Q}(i)[t]f(t)\subset{\rm Nuc}_{r}(S_{f})$

and that

${\rm Nuc}_{r}(S_{f})=\mathbb{Q}(i)[t]/\mathbb{Q}(i)[t]f(t)$

for irreducible $f\in\mathbb{Q}(i)[t]$, by comparing dimensions of the vector
spaces. Let $f(t)=it^{2}+(k+1)t+j+k$, then

$f^{*}(t)=-4t^{4}+(4+4i)t^{3}-8it^{2}+(8-8i)t+8$

[GTLN13, Example 2.13]. This means that

$\hat{h}(x)=16x^{4}-16x^{3}+16x^{2}+16x+8\in\mathbb{Q}[x]$

and

$h(t)=t^{4}-(1+i)t^{3}+2it^{2}-(2-2i)t-2.$

Scaling $f$ to make it monic does not change its minimal central left
multiple, so w.l.o.g. assume $f(t)=t^{2}+i(k+1)t+k+ik$. If $f$ is irreducible
then the above result tells us that

${\rm Nuc}_{r}(S_{f})\cong\mathbb{Q}[x]/(x^{4}-x^{3}+x^{2}+x+1/2).$

Example 2 Let $D$ be a quaternion algebra over a field $F$ of characteristic
not 2 and $\sigma\in{\rm Aut}_{F}(D)$ of order two. Then $K={\rm Fix}(\sigma)$
is a quadratic field extension in $D$. We know that $\sigma(d)=udu^{-1}$ and
$u^{2}\in F$. W.l.o.g. choose $u\in K^{\times}$. Then $C(R)=F[x]$ with
$x=u^{-1}t$. Let $f\in R$ be monic and irreducible of degree $m>1$, such that
$(f,t)_{r}=1$. Let $h(t)=\hat{h}(u^{-1}t)$ be its minimal central left
multiple. Then either

  * •

${\rm Nuc}_{r}(S_{f})\cong E_{\hat{h}}$ is a field extension of degree $2m$,
$k=2$, and ${\rm deg}(h)=2m$, or

  * •

${\rm Nuc}_{r}(S_{f})$ is a central division algebra over $E_{\hat{h}}$ of
degree $2$, $h$ is irreducible in $R$, ${\rm deg}(h)=m={\rm deg}(\hat{h})$ and
$[{\rm Nuc}_{r}(S_{f}):F]=4m$ (Corollary 22), in which case $f$ must be
invariant checking the dimensions.

So assume that $f$ is not right-invariant, monic and irreducible. Then ${\rm
Nuc}_{r}(S_{f})\cong E_{\hat{h}}$ is a field extension of degree $2m$. In
particular, let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in{\rm
Fix}(\sigma)[t]\subset D[t;\sigma]$ then ${\rm Fix}(\sigma)\subset
L^{(\sigma,f)}$ (Proposition 55) and $t,t^{2},\dots,t^{m-1}\in{\rm
Nuc}_{r}(S_{f})$. Thus we obtain that

${\rm Fix}(\sigma)[t]/(f(t))\subset{\rm Nuc}_{r}(S_{f}),$

with equality if $f$ is irreducible and not right-invariant.

####  5.2.1 The Right Nucleus of $t^{m}-a\in D[t;\sigma]$

In the particular case $f(t)=t^{m}-a\in R=D[t;\sigma]$ we obtain stronger
results on the right nucleus than in the general case for any $f\in R$. Later
we will see how this can be used to obtain irreducibility criteria for the
polynomials of the form $t^{m}-a$ where $a\in F^{\times}$ where $F$ denotes
the field $C\cap{\rm Fix}(\sigma)$.

For $f(t)=t^{m}-a\in F[t]\subset D[t;\sigma]$ we get the reverse inclusion of
Theorem 64, i.e. we can show that the right nucleus of $S_{f}$ is equal to the
subalgebra

$L_{f}=L[t;\sigma]/L[t;\sigma]f(t):$

######  Theorem 70.

Let $f(t)=t^{m}-a\in F[t]\subset R$. Then

${\rm Nuc}_{r}(S_{f})=L[t;\sigma]/L[t;\sigma]f(t).$

###### Proof.

If $f(t)\in F[t]\subset D[t;\sigma]$, then

$L[t;\sigma]/L[t;\sigma]f(t)$

is a subalgebra of ${\rm Nuc}_{r}(S_{f})$, by Theorem 64 with $\delta=0$.
Now, we suppose that $g(t)=\sum\limits_{i=0}^{m-1}g_{i}t^{i}\in{\rm
Nuc}_{r}(S_{f})$. Then $fg\in Rf$, i.e. there exists a polynomial
$g^{\prime}\in D[t;\sigma]$ such that $fg=g^{\prime}f$. It is easy to see that
${\rm deg}(g)={\rm deg}(g^{\prime})$, so we let
$g^{\prime}(t)=\sum\limits_{j=0}^{m-1}g^{\prime}_{j}t^{j}$ for some
$g^{\prime}_{j}\in D$. Then we have

$\displaystyle(t^{m}-a)\sum\limits_{i=0}^{m-1}g_{i}t^{i}$
$\displaystyle=(\sum\limits_{j=0}^{m-1}g^{\prime}_{j}t^{j})(t^{m}-a)$
$\displaystyle\Rightarrow\sum\limits_{i=0}^{m-1}\sigma^{m}(g_{i})t^{m+i}-\sum\limits_{i=0}^{m-1}ag_{i}t^{i}$
$\displaystyle=\sum\limits_{j=0}^{m-1}g^{\prime}_{j}t^{m+j}-\sum\limits_{j=0}^{m-1}g^{\prime}_{j}at^{j}.$

Comparing coefficients of $t^{m+j}$ for $j\in\\{0,1,...,m-1\\}$ yields
$g^{\prime}_{j}=\sigma^{m}(g_{j})$, and comparing coefficients of $t^{j}$ for
$j\in\\{0,1,\dots,m-1\\}$ yields $g^{\prime}_{j}a=ag_{j}$. Combining these
gives

$\sigma^{m}(g_{j})a=ag_{j}$

for all $j\in\\{0,1,\dots,m-1\\}$, that is $g_{j}\in L$ for all $j$. Hence
$g(t)\in L[t;\sigma]/L[t;\sigma]f(t)$. The result follows immediately. ∎

####  5.2.2 $L^{(\sigma,f)}$ for $f\in K[t;\sigma]$ with $K$ a Cyclic Field
Extension of $F$

For the remainder of the section (Section 5.2) suppose that $K/F$ is a Galois
extension of $F$ of degree $n$ with cyclic Galois group ${\rm
Gal}(K/F)=\langle\sigma\rangle$, let $R=K[t;\sigma]$ and let

$f(t)=t^{m}-\sum\limits_{i=0}^{m-1}a_{i}t^{i}\in R.$

######  Proposition 71.

$L=\\{u\in K:u\in{\rm Fix}(\sigma^{m-\lambda_{j}})\text{ for all
}\lambda_{j}\in\Lambda\\}=\bigcap\limits_{\lambda_{j}\in\Lambda}{\rm
Fix}(\sigma^{m-\lambda_{j}}).$

###### Proof.

Let

$u\in L=\\{u\in K:\sigma^{m}(u)a_{i}=a_{i}\sigma^{i}(u)\text{ for all
}i\in\\{0,1,\dots,m-1\\}\\}.$

Then

$\displaystyle\,\sigma^{m}(u)a_{i}=a_{i}\sigma^{i}(u)\text{ for each }i$
$\displaystyle\Leftrightarrow$
$\displaystyle\,\sigma^{m}(u)a_{i}=\sigma^{i}(u)a_{i}\text{ for each }i$
$\displaystyle\Leftrightarrow$ $\displaystyle\,\sigma^{m-i}(u)=u\text{ for
each }i\text{ such that }a_{i}\neq 0$ $\displaystyle\Leftrightarrow$
$\displaystyle\,\sigma^{m-\lambda_{j}}(u)=u\text{ for each
}\lambda_{j}\in\Lambda.$

This yields the assertion. ∎

######  Corollary 72.

If $a_{m-1}\neq 0$, then $L=F$.

###### Proof.

Let $u\in L$, then $\sigma^{m}(u)a_{m-1}=a_{m-1}\sigma^{i}(u)$ yields
$\sigma(u)=u$, hence $u\in F$. This implies immediately that $L=F.$ ∎

######  Lemma 73.

(i) Let $E={\rm Fix}(\sigma^{u_{1}})\cap{\rm Fix}(\sigma^{u_{2}})$ for some
$u_{1},u_{2}\in\mathbb{Z}$ such that $1\leq u_{1},u_{2}<n$. If ${\alpha}={\rm
gcd}(u_{1},u_{2},n)$, then ${\rm Gal}(K/E)=\langle\sigma^{\alpha}\rangle$ and
$E={\rm Fix}(\sigma^{\alpha})$.
(ii) Let $E^{\prime}={\rm Fix}(\sigma^{u_{1}})\cap{\rm
Fix}(\sigma^{u_{2}})\cap\dots\cap{\rm Fix}(\sigma^{u_{k}})$ for some
$u_{1},u_{2},\dots,u_{k}\in\mathbb{Z}$ such that $1\leq
u_{1},u_{2},\dots,u_{k}<n$. If $\alpha={\rm gcd}(u_{1},u_{2},\dots,u_{k},n)$,
then ${\rm Gal}(K/E^{\prime})=\langle\sigma^{\alpha}\rangle$, and
$E^{\prime}={\rm Fix}(\sigma^{\alpha})$.

###### Proof.

(i) Let $v_{1}={\rm gcd}(u_{1},n)$ and $v_{2}={\rm gcd}(u_{2},n)$. Now

${\rm ord}(\sigma^{u_{1}})=\frac{n}{{\rm gcd}(u_{1},n)}=\frac{n}{v_{1}}$

and

${\rm ord}(\sigma^{v_{1}})=\frac{n}{{\rm gcd}(v_{1},n)}=\frac{n}{v_{1}}.$

Therefore $\langle\sigma^{u_{1}}\rangle=\langle\sigma^{v_{1}}\rangle$, and
similarly $\langle\sigma^{u_{2}}\rangle=\langle\sigma^{v_{2}}\rangle$ (note
that also $E={\rm Fix}(\sigma^{v_{1}})\cap{\rm Fix}(\sigma^{v_{2}})$). We have
that ${\rm Gal}(K/E)$ is a cyclic subgroup of $\langle\sigma\rangle$. Thus
there exists a smallest integer $\beta$, $1\leq\beta<n$, such that ${\rm
Gal}(K/E)=\langle\sigma^{\beta}\rangle$.

$E$ is a subfield of ${\rm Fix}(\sigma^{v_{1}})$, which means that
$\langle\sigma^{v_{1}}\rangle$ is a subgroup of
$\langle\sigma^{\beta}\rangle$, and $\beta$ divides $v_{1}$. Similarly, we get
that $\beta$ also divides $v_{2}$. Therefore $\beta$ is a common divisor of
$v_{1}$ and $v_{2}$. By [Lan04, Chapter VI, Section 1, Corollary 1.4] we know
that ${\rm Gal}(K/E)=\langle\sigma^{v_{1}},\sigma^{v_{2}}\rangle$, i.e. if
$\tau\in{\rm Gal}(K/E)$, then

$\tau=\sigma^{\tau_{1}v_{1}}\sigma^{\tau_{2}v_{2}}=\sigma^{\tau_{1}v_{1}+\tau_{2}v_{2}},$

for some $\tau_{1},\tau_{2}\in\mathbb{Z}$.

Now $\langle\sigma^{\beta}\rangle={\rm
Gal}(K/E)=\langle\sigma^{v_{1}},\sigma^{v_{2}}\rangle$, and so
$\sigma^{\beta}\in\langle\sigma^{v_{1}},\sigma^{v_{2}}\rangle$. Hence
$\sigma^{\beta}=\sigma^{xv_{1}+yv_{2}}$ for some $x,y\in\mathbb{Z}$, and so we
take $\beta=xv_{1}+yv_{2}+zn$ for some $z\in\mathbb{Z}$. Let $\alpha={\rm
gcd}(v_{1},v_{2},n)$. Then $\alpha$ divides $v_{1}$, $v_{2}$ and $n$, by
definition. Therefore $\alpha$ divides $x_{0}v_{1}+y_{0}v_{2}+z_{0}n$ for any
$x_{0},y_{0},z_{0}\in\mathbb{Z}$; in particular $\alpha$ divides
$xv_{1}+yv_{2}+zn=\beta$. In summary, we have shown that $\beta$ is a common
divisor of $v_{1}$, $v_{2}$, and $n$, and $\alpha={\rm gcd}(v_{1},v_{2},n)$
divides $\beta$; so we must have that $\alpha=\beta$. Moreover,

$\alpha={\rm gcd}(v_{1},v_{2},n)={\rm gcd}({\rm gcd}(u_{1},n),{\rm
gcd}(u_{2},n),n)={\rm gcd}(u_{1},u_{2},n).$

Hence we obtain that ${\rm Gal}(K/E)=\langle\sigma^{\alpha}\rangle,$ and
$E={\rm Fix}(\sigma^{\alpha})$ with $\alpha={\rm gcd}(u_{1},u_{2},n)$ as
claimed.
(ii) This follows by induction from (i). ∎

######  Theorem 74.

If $\alpha={\rm gcd}(m-\lambda_{1},m-\lambda_{2},\dots,m-\lambda_{r},n)$, then

$L={\rm Fix}(\sigma^{\alpha})$

and $[L:F]=\alpha$. In particular, $L=F$ if and only if $\alpha=1$.

###### Proof.

By Proposition 71, we have

$L=\bigcap\limits_{\lambda_{j}\in\Lambda}{\rm
Fix}(\sigma^{m-\lambda_{j}})={\rm Fix}(\sigma^{m-\lambda_{1}})\cap{\rm
Fix}(\sigma^{m-\lambda_{2}})\cap\dots\cap{\rm Fix}(\sigma^{m-\lambda_{r}}).$

Therefore it follows that $L={\rm Fix}(\sigma^{\alpha})$ where $\alpha={\rm
gcd}(m-\lambda_{1},m-\lambda_{2},\dots,m-\lambda_{r},n)$ by Lemma 73.
Obviously $L=F$ if and only if ${\rm Fix}(\sigma^{\alpha})=F$, if and only if
$\langle\sigma^{\alpha}\rangle=\langle\sigma\rangle$, which is true if and
only if $\sigma^{\alpha}$ has order $n$. Now ${\rm
ord}(\sigma^{\alpha})=\frac{n}{{\rm gcd}(n,\alpha)}=n$ if and only if ${\rm
gcd}(n,\alpha)=1$. That is, $L=F$ if and only if ${\rm gcd}(n,\alpha)=1$, if
and only if $\alpha=1$. ∎

######  Theorem 75.

(i) $L=K$ if and only if $m-\lambda_{j}$ is a multiple of $n$ for all
$\lambda_{j}\in\Lambda$.
(ii) Suppose that $n$ is prime. Then $L=F$ if and only if there exists
$\lambda_{j}\in\Lambda$ such that $m-\lambda_{j}$ is not divisible by $n$.

###### Proof.

We have $L={\rm Fix}(\sigma^{\alpha})$ where $\alpha={\rm
gcd}(m-\lambda_{1},m-\lambda_{2},\dots,m-\lambda_{r},n)$ (Lemma 73).
(i) $\alpha={\rm gcd}(m-\lambda_{1},m-\lambda_{2},\dots,m-\lambda_{r},n)=n$ if
and only if $m-\lambda_{j}$ is a multiple of $n$ for all
$\lambda_{j}\in\Lambda$. That is $L={\rm Fix}(\sigma^{n})=K$ if and only if
$m-\lambda_{j}$ is a multiple of $n$ for all $\lambda_{j}\in\Lambda$.
(ii) If $n$ is prime, then $\alpha|n$ $\Rightarrow$ $\alpha\in\\{1,n\\}$. The
result now follows immediately from (i). ∎

Recall that $C(R)=F[x]$ with the identification $x=t^{n}$. We obtain the
following well-known result (cf. [Jac09]) rewritten in terms of the Petit
algebra $S_{f}$:

######  Corollary 76.

Suppose that $f$ has the form $f(t)=ag(t^{n})t^{l}$ for some scalar $a\in K$,
some polynomial $g(t)=\hat{g}(t^{n})$ such that $\hat{g}\in F[x]$ and some
integer $l\geq 0$. Then $S_{f}$ is associative and hence $f$ is right-
invariant.

###### Proof.

By Theorem 64, we know that $L[t;\sigma]/L[t;\sigma]f(t)$ is a subalgebra of
the right nucleus. Since $m-\lambda_{j}$ is a multiple of $n$ for all
$\lambda_{j}\in\Lambda$, it follows that $L=K$ by Theorem 75 (i). Thus
$K[t;\sigma]/K[t;\sigma]f(t)$ is a subalgebra of ${\rm Nuc}_{r}(S_{f})$, which
implies that $S_{f}={\rm Nuc}_{r}(S_{f})$. Hence $S_{f}$ is an associative
algebra which is true if and only if $f$ in right-invariant in $R$. ∎

####  5.2.3 The Case $n<m$.

Let $K/F$ be a cyclic Galois extension of degree $n$ with Galois group ${\rm
Gal}(K/F)=\langle\sigma\rangle$ and

${\rm ord}(\sigma)=n<m={\rm deg}(f).$

Then there exist integers $q,r$ such that $q\neq 0$, and $m=qn+r$ where $0\leq
r<n$. Moreover, $K={\rm Fix}(\sigma^{n})$ hence the coefficients of $f\in R$
always lie in ${\rm Fix}(\sigma^{n})$. Applying this to Theorem 61 yields the
following:

(i) If $m=qn$, then

$L\oplus Lt^{n}\oplus Lt^{2n}\oplus\dots\oplus Lt^{(q-1)n}$

is an $F$-sub vector space of ${\rm Nuc}_{r}(S_{f})$ of dimension $q[L:F]$ and

$t^{qn}=t^{m}=\sum_{i=0}^{m-1}a_{i}t^{i}\in{\rm Nuc}_{r}(S_{f}).$

(ii) If $m=qn+r$ for some positive integers $q,r$ with $0<r<n$, then

$L\oplus Lt^{n}\oplus Lt^{2n}\oplus\dots\oplus Lt^{qn}$

is an $F$-sub vector space of ${\rm Nuc}_{r}(S_{f})$ of dimension
$(q+1)[L:F]$. If we assume that $n$ is either prime or ${\rm gcd}(m,n)=1$, $f$
is not right-invariant and $(f,t)_{r}=1$, as well as $[L:F]=n$, then $f$ is
reducible.

######  Theorem 77.

Suppose that $[L:F]=\rho$ where $n=b\rho$ for some $b\in\mathbb{N}$. Then
$m=q\rho+r$ for some integers $q,r$ with $0\leq r<\rho$, and
$f(t)=g(t^{\rho})t^{r}$, where $g$ is a polynomial of degree $q$ in
$K[t^{\rho};\sigma^{\rho}]$.

###### Proof.

By Theorem 74, we have that $L={\rm Fix}(\sigma^{\alpha})$, where $\alpha={\rm
gcd}(m-\lambda_{1},m-\lambda_{2},\dots,m-\lambda_{r},n)$. Now $\alpha=\rho$ if
and only if $m-\lambda_{j}$ is a multiple of $\rho$ for all
$\lambda_{j}\in\Lambda$. But $m-\lambda_{j}$ is equal to a multiple of $\rho$
if and only if $\lambda_{j}=r+\rho l$ for some integer $l$ such that $0\leq
l<q$ (since $m=q\rho+r$). Therefore we obtain
$\Lambda\subset\\{r,r+\rho,r+2\rho,\dots,r+(q-1)\rho\\}.$ Thus

$\displaystyle f(t)$
$\displaystyle=t^{q\rho+r}-a_{(q-1)\rho+r}t^{(q-1)\rho+r}-\dots-
a_{r+\rho}t^{r+\rho}-a_{r}t^{r}$
$\displaystyle=[(t^{\rho})^{q}-a_{(q-1)\rho+r}(t^{\rho})^{(q-1)}-\dots-
a_{r+\rho}t^{\rho}-a_{r}]t^{r}$ $\displaystyle=g(t^{\rho})t^{r}$

where $g$ has degree $q$ in $K[t^{\rho};\sigma^{\rho}]$. ∎

###  5.3 The Special Case $R=D[t;\delta]$

Throughout this section we consider the Petit algebras associated with the
polynomial ring $R=D[t;\delta]$, and their nuclei. As a Corollary to Theorem
58 we obtain:

######  Theorem 78.

Let $f(t)=t^{m}-\sum\limits_{i=0}^{m-1}a_{i}t^{i}\in R$ and $c\in D$. Then
$c\in L$ if and only if

$ca_{k}=\sum\limits_{i=k}^{m-1}a_{i}\binom{i}{k}\delta^{i-k}(c)-\binom{m}{k}\delta^{m-k}(c)$

for all $k\in\\{0,1,\dots,m-1\\}$.

###### Proof.

Since here $\sigma={\rm id}$, the map $\Delta_{j,i}$ is equal to
$\binom{j}{i}\delta^{j-i}$ for any $i,j\in\mathbb{Z}$ such that $0\leq i\leq
j$. The result follows immediately by Theorem 58. ∎

######  Corollary 79.

Suppose that $D$ has prime characteristic $p$. Let
$f(t)=t^{p}-\sum\limits_{i=0}^{p-1}a_{i}t^{i}\in R$, and let $c\in D$. Then
$c\in L$ if and only if

$ca_{k}=\sum\limits_{i=k}^{p-1}a_{i}\binom{i}{k}\delta^{i-k}(c)$

for all $k\in\\{1,2\dots,m-1\\}$, and

$ca_{0}=\sum\limits_{i=0}^{p}a_{i}\delta^{i}(c)$

where $a_{p}=-1$.

###### Proof.

We note that $\binom{p}{k}$ is divisible by $p$ for all
$k\in\\{1,2,\dots,p-1\\}$, and $\binom{p}{0}=1$. The result follows
immediately. ∎

We can also prove the converse to Theorem 60, for the special case
$\sigma={\rm id}$:

######  Theorem 80.

Let $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in R=D[t;\delta].$ Then
$a_{i}\in{\rm Const}(\delta)$ for all $i\in\\{0,1,\dots,m-1\\}$ if and only if
$t\in{\rm Nuc}_{r}(S_{f})$.

###### Proof.

By Theorem 60 with $\sigma={\rm id}$, if $a_{i}\in{\rm Const}(\delta)$ for all
$i\in\\{0,1,\dots,m-1\\}$, then $t\in{\rm Nuc}_{r}(S_{f})$. On the other hand,
if $t\in{\rm Nuc}_{r}(S_{f})$, then $[t,t^{m-1},t]=0$ which gives

$(\sum_{i=0}^{m-1}a_{i}t^{i})t=t(\sum_{i=0}^{m-1}a_{i}t^{i}),$

that is

$a_{m-1}t^{m}+\sum\limits_{i=0}^{m-2}a_{i}t^{i+1}=a_{m-1}t^{m}+\sum\limits_{i=0}^{m-2}a_{i}t^{i}+\sum\limits_{i=0}^{m-1}\delta(a_{i})t^{i}.$
(5.8)

We note that $t^{m}=(\sum\limits_{i=0}^{m-1}a_{i}t^{i})\,{\rm mod}_{r}f$,
however, we need not explicitly substitute this into Equation 5.8, since only
$a_{m-1}t^{m}$ appears on each side of 5.8, hence all terms in $t^{m}$ sum to
$0$. We are left with

$\sum\limits_{i=0}^{m-1}\delta(a_{i})t^{i}=0.$ (5.9)

Since Equation 5.9 is an equality in $S_{f}$, we have that
$\delta(f(t)):=\sum\limits_{i=0}^{m-1}\delta(a_{i})t^{i}\in Rf$, i.e that
$\delta(f)=gf$ for some $g(t)\in R$. Then

${\rm deg}(gf)={\rm deg}(g)+{\rm deg}(f)={\rm deg}(g)+m={\rm
deg}(\delta(f))\leq m-1,$

which forces $g=0$, and $\delta(f)=gf=0$. Therefore Equation 5.9 is also true
in $R$, and so $\delta(a_{i})=0$ for each $i\in\\{0,1,\dots,m-1\\}$. The
result follows immediately. ∎

####  5.3.1 The Right Nucleus of $t^{p}-a\in D[t;\delta]$ when ${\rm
Char}(D)=p>0$

For the remainder of the chapter we suppose that $D$ has prime characteristic
$p$, and we consider the polynomial $f(t)=t^{p}-a\in D[t;\delta]$.

In a similar fashion to the results of Section 5.2.1, we obtain stronger
results on the right nucleus of $S_{f}$ for the particular polynomial
$f(t)=t^{p}-a$ under certain conditions on the constant coefficient $a\in
D^{\times}$. In particular Theorem 82 is analogous to Theorem 70. Again we
will make use of the following results in the coming chapters to determine
irreducibility criteria for the polynomial $f=t^{p}-a\in D[t;\delta]$. Recall
that the bracket defined by $[b,c]=bc-cb$ for $b,c\in D$ is the commutator on
$D$.

######  Lemma 81.

$L=\\{c\in D:[a,c]=\delta^{p}(c)\\}.$

In particular, if $a\in C$, then

$L=\\{c\in D:\delta^{p}(c)=0\\}={\rm Const}(\delta^{p}).$

###### Proof.

Let $c\in L$. By definition, this is true if and only if $fc=c^{\prime}f$ for
some $c^{\prime}\in D$, i.e. $(t^{p}-a)c=c^{\prime}(t^{p}-a)$. Computing both
sides gives

$\sum\limits_{i=0}^{p}\binom{p}{i}\delta^{p-i}(c)t^{i}-ac=c^{\prime}t^{p}-c^{\prime}a.$

Now, since $\binom{p}{i}$ is divisible by $p$ for all $i\neq 0,p$, this
simplifies to

$ct^{p}+\delta^{p}(c)-ac=c^{\prime}t^{p}-c^{\prime}a.$

This yields $c=c^{\prime}$ and $\delta^{p}(c)-ac=-ca$, i.e.
$\delta^{p}(c)=[a,c]$. Therefore $L\subset\\{c\in D:[a,c]=\delta^{p}(c)\\}$.
Conversely, let $c\in\\{c\in D:[a,c]=\delta^{p}(c)\\}$. Then

$fc=\sum\limits_{i=0}^{p}\binom{p}{i}\delta^{p-i}(c)t^{i}-ac=ct^{p}+\delta^{p}(c)-ac=ct^{p}-ca=cf.$

Therefore $c\in L$, and $\\{c\in D:[a,c]=\delta^{p}(c)\\}\subset L$. The
second part follows easily, since $[a,c]=0$ for all $c\in D$ when $a$ is in
the center $C$ of $D$. ∎

######  Theorem 82.

If $a\in{\rm Const}(\delta)$, then

${\rm Nuc}_{r}(S_{f})=L[t;\delta|_{L}]/L[t;\delta|_{L}]f(t).$

###### Proof.

First we note that $a\in L$, since $[a,a]=\delta^{p}(a)=0$ (Lemma 73), i.e.
$f(t)\in L[t;\delta|_{L}]$. Then

$L_{f}=L[t;\delta|_{L}]/L[t;\delta|_{L}]f(t)$

is a well-defined Petit algebra. Now since $a\in{\rm Const}(\delta)$, the
algebra $L_{f}$ is an associative subalgebra of ${\rm Nuc}_{r}(S_{f})$. To
complete the proof we show that the right nucleus ${\rm Nuc}_{r}(S_{f})$ is
contained in the set

$L\oplus Lt\oplus\cdots\oplus Lt^{p-1}.$

Let $g(t)=\sum\limits_{i=0}^{p-1}g_{i}t^{i}\in{\rm Nuc}_{r}(S_{f})$. Then
$fg=g^{\prime}f$ for some $g^{\prime}\in D[t;\delta]$. Assume w.l.o.g. that
$g^{\prime}(t)=\sum\limits_{i=0}^{p-1}g_{i}^{\prime}t^{i}$, then we have

$\displaystyle(t^{p}-a)\sum\limits_{i=0}^{p-1}g_{i}t^{i}=\sum\limits_{i=0}^{p-1}g_{i}^{\prime}t^{i}(t^{p}-a)$
$\displaystyle\Rightarrow$
$\displaystyle\sum\limits_{i=0}^{p-1}\sum\limits_{j=0}^{p}\binom{p}{j}\delta^{p-j}(g_{i})t^{i+j}-\sum\limits_{i=0}^{p-1}ag_{i}t^{i}=\sum\limits_{i=0}^{p-1}g_{i}^{\prime}t^{p+i}-\sum\limits_{i=0}^{p-1}g_{i}^{\prime}\sum\limits_{j=0}^{i}\binom{i}{j}\delta^{i-j}(a)t^{j}$

Since $\binom{p}{j}$ is divisible by $p$ for all $j\neq 0,p$, and $D$ has
characteristic $p$, and $\delta(a)=0$ we reduce to the following:

$\displaystyle\sum\limits_{i=0}^{p-1}g_{i}t^{p+i}+\sum\limits_{i=0}^{p-1}\delta^{p}(g_{i})t^{i}-\sum\limits_{i=0}^{p-1}ag_{i}t^{i}=\sum\limits_{i=0}^{p-1}g_{i}^{\prime}t^{p+i}-\sum\limits_{i=0}^{p-1}g_{i}^{\prime}at^{i}.$

This yields $g_{i}=g_{i}^{\prime}$ and $[a,g_{i}]=\delta^{p}(g_{i})$ for all
$i=0,1,\dots,p-1$, that is $g_{i}\in L$ for each $i=0,1,\dots,p-1$. Hence
$g(t)\in L\oplus Lt\oplus\cdots\oplus Lt^{p-1}$, and the result follows. ∎

##  Chapter 6 The Right Nucleus of $S_{f}$ for Low Degree Polynomials in
$K[t;\sigma]$

In this chapter we assume that $K$ is a cyclic Galois field extension of
finite degree $n$ over $F$ with ${\rm Gal}(K/F)=\langle\sigma\rangle$. We
repeatedly use that $[{\rm Fix}(\sigma^{\rho}):F]={\rm gcd}(n,\rho).$ We now
use the previous results to explore the structure of ${\rm Nuc}_{r}(S_{f})$
for some polynomials of low degree in $K[t;\sigma]$. The same arguments then
apply for those of higher degree as well but of course quickly become tedious.
The results of this chapter also appear in [OP19].

####  6.1.1 ${\rm deg}(f)=2$

Let $f(t)=t^{2}-a_{1}t-a_{0}\in K[t;\sigma]$. Then
$L^{(\sigma,f)}=\bigcap\limits_{\lambda_{j}\in\Lambda}{\rm
Fix}(\sigma^{2-\lambda_{j}})$ by Proposition 71.

  1. (1)

If $f(t)=t^{2}-a_{0}$ with $a_{0}\in K^{\times}$, then $L^{(\sigma,f)}={\rm
Fix}(\sigma^{2})$.

  2. (2)

If $f(t)=t^{2}-a_{1}t-a_{0}$ with $a_{1}\in K^{\times}$, then
$L^{(\sigma,f)}=F$.

Note that if $n=[K:F]$ is even, then $\sigma^{2}$ has order $\frac{n}{2}$ in
${\rm Gal}(K/F)$, which means that $F\neq{\rm Fix}(\sigma^{2})$. If $n$ is
odd, then ${\rm gcd}(n,2)=1$, therefore ${\rm Fix}(\sigma^{2})=F$.

If we additionally assume that $f(t)\in F[t]$, then we obtain:

######  Proposition 83.

Let $f(t)=t^{2}-a_{0}\in F[t]\subset K[t;\sigma]$, $a_{0}\not=0$ then

${\rm Fix}(\sigma^{2})[t;\sigma]/{\rm Fix}(\sigma^{2})[t;\sigma]f(t)$

is a subalgebra of ${\rm Nuc}_{r}(S_{f})$ of dimension $2[{\rm
Fix}(\sigma^{2}):F]$ over $F$. In particular, if $n$ is odd, $f$ not right-
invariant such that $(f,t)_{r}=1$, and $\hat{h}(x)\not=x$, then $f$ is
reducible.

###### Proof.

By Theorem 64, $L^{(\sigma,f)}[t;\sigma]/L^{(\sigma,f)}[t;\sigma]f(t)$ is a
subalgebra of ${\rm Nuc}_{r}(S_{f})$ and $L^{(\sigma,f)}={\rm
Fix}(\sigma^{2})$ by (1), which yields the first assertion. The second
assertion follows from the fact that the right nucleus has dimension 2 over
$F$ for irreducible right-invariant $f$ under our assumptions. ∎

####  6.1.2 ${\rm deg}(f)=3$

Let $f(t)\in K[t;\sigma]$ be monic of degree $3$. Then
$L^{(\sigma,f)}=\bigcap\limits_{\lambda_{j}\in\Lambda}{\rm
Fix}(\sigma^{3-\lambda_{j}})$ by Proposition 71.

  1. (1)

If $f(t)=t^{3}-a_{0}\in K[t;\sigma]$, where $a_{0}\in K^{\times}$, then
$L^{(\sigma,f)}={\rm Fix}(\sigma^{3})$.

  2. (2)

If $f(t)=t^{3}-a_{1}t\in K[t;\sigma]$, where $a_{1}\in K^{\times}$, then
$L^{(\sigma,f)}={\rm Fix}(\sigma^{2})$.

  3. (3)

In all other cases, we obtain $L^{(\sigma,f)}=F$.

######  Proposition 84.

(i) If $f(t)=t^{3}-a_{0}$ with $0\not=a_{0}\in F$, then

${\rm Fix}(\sigma^{3})[t;\sigma]/{\rm Fix}(\sigma^{3})[t;\sigma]f(t)$

is a subalgebra of ${\rm Nuc}_{r}(S_{f})$ of dimension $3[{\rm
Fix}(\sigma^{3}):F]$ over $F$. In particular, if $f$ is not right-invariant,
$n$ is prime or not divisible by 3, such that $(f,t)_{r}=1$, and
$\hat{h}(x)\not=x$ then $f$ is reducible.
(ii) If $f(t)=t^{3}-a_{1}t$ with $0\not=a_{1}\in F$, then

${\rm Fix}(\sigma^{2})[t;\sigma]/{\rm Fix}(\sigma^{2})[t;\sigma]f(t)$

is a subalgebra of ${\rm Nuc}_{r}(S_{f})$ of dimension $3[{\rm
Fix}(\sigma^{2}):F]$ over $F$.

###### Proof.

By Theorem 64,
$L^{(\sigma,f)}[t;\sigma]/L^{(\sigma,f)}[t;\sigma]f(t)\subset{\rm
Nuc}_{r}(S_{f})$.
(i) If $f(t)=t^{3}-a_{0}\in F[t]$ with $a_{0}\neq 0$, then
$L^{(\sigma,f)}={\rm Fix}(\sigma^{3})$ which proves the assertion looking at
the dimensions.
(ii) If $f(t)=t^{3}-a_{1}t\in F[t]$ with $a_{1}\neq 0$, then
$L^{(\sigma,f)}={\rm Fix}(\sigma^{2})$. ∎

####  6.1.3 ${\rm deg}(f)=4$

Let $f(t)\in K[t;\sigma]$ be monic of degree $m=4$. Then
$L^{(\sigma,f)}=\bigcap\limits_{\lambda_{j}\in\Lambda}{\rm
Fix}(\sigma^{4-\lambda_{j}})$ by Proposition 71.

  1. (1)

If $f(t)=t^{4}-a_{0}$ with $a_{0}\in K^{\times}$ then $L^{(\sigma,f)}={\rm
Fix}(\sigma^{4})$.

  2. (2)

If $f(t)=t^{4}-a_{1}t$ with $a_{1}\in K^{\times}$ then $L^{(\sigma,f)}={\rm
Fix}(\sigma^{3})$.

  3. (3)

If $f(t)=t^{4}-a_{2}t^{2}$ with $a_{2}\in K^{\times}$ then
$L^{(\sigma,f)}={\rm Fix}(\sigma^{2})$.

  4. (4)

If $f(t)=t^{4}-a_{2}t^{2}-a_{0}$ with $a_{0},a_{2}\in K^{\times}$, then
$L^{(\sigma,f)}={\rm Fix}(\sigma^{4})\cap{\rm Fix}(\sigma^{2})={\rm
Fix}(\sigma^{2})$.

  5. (5)

In all other cases, $L^{(\sigma,f)}={\rm Fix}(\sigma)=F$.

Observe that:

  * •

If $n$ is odd then ${\rm Fix}(\sigma^{4})={\rm Fix}(\sigma^{2})=F$.

  * •

If $n\equiv 0(mod\text{ }4)$, then $[{\rm Fix}(\sigma^{4}):F]=4$.

  * •

If $n\equiv 2(mod\text{ }4)$, then $[{\rm Fix}(\sigma^{4}):F]=2$.

  * •

If $n\equiv 0(mod\text{ }3)$, then $[{\rm Fix}(\sigma^{3}):F]=3$.

  * •

If $n\equiv 1\text{ or }2(mod\text{ }3)$, then ${\rm Fix}(\sigma^{3})=F$.

  * •

If $n\equiv 0(mod\text{ }2)$ then $[{\rm Fix}(\sigma^{2}):F]=2$.

The above cases for $n$ are not mutually exclusive.

######  Proposition 85.

(i) If $f(t)=t^{4}-a_{0}\in F[t]$ with $0\not=a_{0}$, then the subalgebra

${\rm Fix}(\sigma^{4})[t;\sigma]/{\rm Fix}(\sigma^{4})[t;\sigma]f(t)$

of the right nucleus of $S_{f}$ has dimension $4[{\rm
Fix}(\sigma^{4}):F]=4{\rm gcd}(n,4)$ over $F$. In particular:
(a) If $f$ is irreducible, not right-invariant and either $n\not=2$ is prime
or ${\rm gcd}(n,4)=1$, and $f(t)\not=t$, then

${\rm Nuc}_{r}(S_{f})\cong{\rm Fix}(\sigma^{4})[t;\sigma]/{\rm
Fix}(\sigma^{4})[t;\sigma]f(t).$

(b) If $f$ is not right-invariant and $n=2$, then $f$ is reducible.
(ii) If $f(t)=t^{4}-a_{1}t\in F[t;\sigma]$ with $0\not=a_{1}$, then the
subalgebra

${\rm Fix}(\sigma^{3})[t;\sigma]/{\rm Fix}(\sigma^{3})[t;\sigma]f(t)$

of the right nucleus of $S_{f}$ has dimension $4[{\rm
Fix}(\sigma^{3}):F]=4{\rm gcd}(n,3)$ over $F$.
(iii) If $f(t)=t^{4}-a_{2}t-a_{0}\in F[t;\sigma]$ with $0\not=a_{2}$, then the
subalgebra

${\rm Fix}(\sigma^{2})[t;\sigma]/{\rm Fix}(\sigma^{2})[t;\sigma]f(t)$

of the right nucleus of $S_{f}$ has dimension $4[{\rm
Fix}(\sigma^{2}):F]=4{\rm gcd}(n,2)$ over $F$. In particular:
(a) If $f$ is irreducible, not right-invariant and either $n\not=2$ is prime
or ${\rm gcd}(n,4)=1$, and $f(t)\not=t$, then

${\rm Nuc}_{r}(S_{f})\cong{\rm Fix}(\sigma^{2})[t;\sigma]/{\rm
Fix}(\sigma^{2})[t;\sigma]f(t).$

(b) If $f$ is not right-invariant and $n=2$, $f(t)\not=t$, then $f$ is
reducible.

###### Proof.

$L^{(\sigma,f)}[t;\sigma]/L^{(\sigma,f)}[t;\sigma]f(t)\subset{\rm
Nuc}_{r}(S_{f})$ by Theorem 64.
(i) Here $L^{(\sigma,f)}={\rm Fix}(\sigma^{4})$ by (1), and thus

${\rm Fix}(\sigma^{4})[t;\sigma]/{\rm
Fix}(\sigma^{4})[t;\sigma]f(t)\subset{\rm Nuc}_{r}(S_{f}).$

(ii) We know $L^{(\sigma,f)}={\rm Fix}(\sigma^{3})$ by (2), and hence

${\rm Fix}(\sigma^{3})[t;\sigma]/{\rm
Fix}(\sigma^{3})[t;\sigma]f(t)\subset{\rm Nuc}_{r}(S_{f}).$

(iii) We have $L^{(\sigma,f)}={\rm Fix}(\sigma^{2})$ by (3), and so

${\rm Fix}(\sigma^{2})[t;\sigma]/{\rm
Fix}(\sigma^{2})[t;\sigma]f(t)\subset{\rm Nuc}_{r}(S_{f}).$

∎

###  6.2 Conclusion

Let $K/F$ be a cyclic Galois extension of degree $n$ with Galois group ${\rm
Gal}(K/F)=\langle\sigma\rangle$. We assume that $n$ is either prime or that
${\rm gcd}(m,n)=1$ for $m$ the degree of the polynomial $f$ we look at. For
certain types of skew polynomials $f(t)=t^{m}-\sum_{i=0}^{m-1}a_{i}t^{i}\in
K[t;\sigma]$ such that $\hat{h}(x)\not=x$ and $(f,t)_{r}=1$ which are not
right-invariant, we can decide if they are reducible based on the following
“algorithm” with output TRUE if $f$ is reducible and STOP if we cannot decide:

  1. (1)

If $f\in F[t]$ then: if $f$ is not right-invariant and $f$ is reducible in
$F[t]$, then $f$ is reducible in $K[t;\sigma]$ TRUE, else STOP. If $f\not\in
F[t]$ then go to (2).

  2. (2)

Compute $L=L^{(\sigma,f)}={\rm Fix}(\sigma^{d})$, where $d={\rm
gcd}(m-\lambda_{1},m-\lambda_{2},\dots,m-\lambda_{r},n)$.
If $[L:F]>m$, then $f$ is reducible TRUE.
If $[L:F]\leq m$ then go to (3).

  3. (3)

Find the smallest integer $\rho$, such that $a_{i}\in{\rm Fix}(\sigma^{\rho})$
for all $i$, and such that ${\rm Fix}(\sigma^{\rho})$ is a proper subfield of
$K$.
If ${\rm Fix}(\sigma^{\rho})=L$ then $f$ is reducible TRUE.
If $m=q\rho$ and $[L:F]>\rho$, then $f$ is reducible TRUE.
If $m=q\rho+r$ with $0<r<\rho$, and $[L:F]\geq\rho$ then $f$ is reducible
TRUE.
In all other cases, go to (4).

  4. (4)

If all $a_{i}$ are not contained in a proper subfield of $K$, then we _cannot
decide_ if $f$ is reducible STOP.

Furthermore, if $f(t)\in F[t]$ then we can use the fact that

$L^{(\sigma,f)}[t;\sigma]/L^{(\sigma,f)}[t;\sigma]f(t)$

is a subalgebra of ${\rm Nuc}_{r}(S_{f})$ to look for zero divisors in ${\rm
Nuc}_{r}(S_{f})$ in order to factor $f$.

##  Chapter 7 When is the Eigenring of $f$ a Central Simple Algebra over $F$?

Let $K$ be a field of characteristic $0$ and $R=K[t;\delta]$ be the ring a
differential polynomials with coefficients in $K$. In order to derive results
on the structure of the left $R$-modules $R/Rf$, Amitsur studied spaces of
linear differential operators via differential transformations [Ami53, Ami54,
Ami55]. He observed that every central simple algebra $B$ over a field $F$ of
characteristic $0$ that is split by an algebraically closed field extension
$K$ of $F$, is isomorphic to the eigenring 111Amitsur refers to the eigenring
of $f$ as the invariant ring of $f$ in [Ami54] of some polynomial $f\in
K[t;\delta]$, for a suitable derivation $\delta$ of $K$. This identification
of a central simple algebra $B$ with a suitable differential polynomial $f\in
K[t;\delta]$ he called A-polynomial also holds when $K$ has prime
characteristic $p$ [Ami53, Section 10], [Pum18]. Let $D$ be a central division
algebra of degree $d$ over its center $C$, $\sigma$ an endomorphism of $D$ and
$\delta$ a left $\sigma$-derivation of $D$. Our aim is to provide a partial
answer to the following generalisation of Amitsur’s investigation:

“For which polynomials $f$ in a skew polynomial ring $D[t;\sigma,\delta]$ is
the eigenring $\mathcal{E}(f)$ a central simple algebra over its subfield
$F=C\cap{\rm Fix}(\sigma)\cap{\rm Const}(\delta)$?”

After a brief review of the Amitsur’s theory in Section 7.1, we investigate
three different setups, always assuming that $f$ has degree $m\geq 1$ and that
the minimal central left multiple of $f$ is square-free. We look at
generalised A-polynomials in $D[t;\sigma]$ in Section 7.2, where $\sigma$ is
an automorphism of $D$ of finite inner order $n$ with $\sigma^{n}=\iota_{u}$
for some $u\in D^{\times}$. In Section 7.3, we study generalised A-polynomials
in $D[t;\delta]$ where $C$ has characteristic $0$ and $\delta$ is the inner
derivation $\delta_{c}$ in $D$ for some $c\in D$. Finally, in Section 7.4, we
investigate A-polynomials in $D[t;\delta]$ where $C$ has prime characteristic
$p$ and $\delta$ is an algebraic derivation of $D$ with minimum polynomial
$g\in F[t]$ of degree $p^{e}$ such that $g(\delta)=\delta_{c}$ for some $c\in
D$. The contents of this chapter will also appear in [OP21].

###  7.1 Differential Transformations and Amitsur’s A-polynomials

All of the definitions and results in this section can be found in [Ami54]
unless stated otherwise, although some are attributed to Jacobson, and have
appeared earlier in [Jac37].

Let $K$ be a field of any characteristic, $\sigma$ be an automorphism of $K$,
$\delta$ be a left $\sigma$-derivation of $K$, and let $F={\rm
Fix}(\sigma)\cap{\rm Const}(\delta)$. Let $V$ be a vector space over $K$ of
dimension $m$. A pseudo-linear transformation (p.l.t.) of $V$ is an additive
map $T:V\longrightarrow V$ such that

$T(av)=\sigma(a)T(v)+\delta(a)v.$

The ring of transformations in $V$ generated by $T$ and the elements of $K$ is
equal to the polynomial ring $K[T]$. For the rest of this section we focus
only on the case $\sigma={\rm id}_{K}$, in which case $(K,\delta)$ is a
differential field, and we call the p.l.t. $T$ a differential transformation
(d.t.). The ring $K[T]$ of polynomials in $T$ is then a ring of left
differential operators on $V$, and we write $V=(V,T)$ to make clear this
association.

Given a basis $(b_{1},\dots,b_{m})$ of $V$ we can represent the d.t. $T$ by a
matrix $B(T)=(a_{ij})\in M_{m}(K)$, where the entries $a_{ij}$ are determined
by the images of the basis vectors $b_{i}$ under $T$, i.e.

$T(b_{i})=\sum\limits_{j=0}^{m}a_{ij}b_{j}.$

If we write $B=B(T)$, then

$(V,T)\cong\frac{K[t;\delta]}{K[t;\delta]g_{1}(t)}\times\frac{K[t;\delta]}{K[t;\delta]g_{2}(t)}\times\cdots\times\frac{K[t;\delta]}{K[t;\delta]g_{m}(t)},$

where $g_{1},\dots,g_{m}\in K[t;\delta]$ are the invariant factors of the
matrix $B-tI_{m}$.

Additionally, we assume from now on that $K$ has characteristic zero for the
remainder of the section. In this case the matrix $(B-tI_{m})$ has only one
proper invariant factor [Jac37, p. 499] and

$(V,T)\cong\frac{K[t;\delta]}{K[t;\delta]g(t)}$

for some $g\in K[t;\delta]$ of degree $m$. We call the polynomial $g(t)$ the
characteristic polynomial of both the matrix $A$ and the differential
transformation $T$, and we note that the characteristic polynomial of any
differential transformation is unique up to similarity. On the other hand, for
any differential polynomial $f\in R$ of degree $m$, write
$V_{f}=K[t;\delta]/K[t;\delta]f(t).$ Then $V_{f}$ is a $K$-vector space of
dimension $m$, and $f(t)$ is the characteristic polynomial of the differential
transformation $T$ on $V_{f}$ defined by $p(t)\mapsto tp(t)$. Now let
$V^{\prime}$ be a vector space over $K$ of dimension $m^{\prime}$ equipped
with a differential transformation $T^{\prime}$. We define a differential
transformation $T\times T^{\prime}$ on the tensor product space
$V\otimes_{K}V^{\prime}$ by

$(T\times T^{\prime})(u)=(T\times T^{\prime})(\sum\limits_{i}v_{i}\otimes
v_{i}^{\prime})=\sum\limits_{i}T(v_{i})\otimes
v_{i}^{\prime}+\sum\limits_{j}v_{j}\otimes T^{\prime}(v_{j}^{\prime}),$

for all $u=\sum\limits_{i}v_{i}\otimes v_{i}^{\prime}\in
V\otimes_{K}V^{\prime}$. Related to this ‘product’ differential
transformation, Amitsur also introduced the related notion for characteristic
polynomials. The so-called resultant is defined in the following way:

######  Definition 86.

Let $f,g\in K[t;\delta]$ be the characteristic polynomials of the differential
transformations $T$ and $T^{\prime}$ respectively. Then the resultant of $f$
and $g$, denoted by $f\times g$, is defined as the characteristic polynomial
of the differential transformation $T\times T^{\prime}$ over
$V\otimes_{K}V^{\prime}$.

######  Remark.

In the notation of Definition 86, with ${\rm deg}(f)=m$ and ${\rm
deg}(g)=m^{\prime}$, the resultant $f\times g$ is a differential polynomial of
degree $mm^{\prime}$ uniquely determined up to similarity.

For $r\in\mathbb{N}$, we denote by $e_{r}(t)$ a fixed characteristic
polynomial of the $r\times r$ zero matrix.

######  Definition 87.

Let $f\in R$ have degree $m$. Then $f$ is called an A-polynomial if there
exists some polynomial $g\in R$, such that $f\times g\sim e_{mn}(t)$, where
$n={\rm deg}(g)$.

######  Theorem 88.

  1. (1)

If $f\in K[t;\delta]$ is an A-polynomial, and $f\sim g$, then $g$ is also an
A-polynomial.

  2. (2)

If $f,g\in K[t;\delta]$ are A-polynomials, then the resultant $f\times g$ is
an A-polynomial.

  3. (3)

If $f,g\in K[t;\delta]$ are A-polynomials such that $f\sim g$, and $f\times
p\sim g\times q$ for some $p,q\in K[t;\delta]$, then $p\sim q$.

  4. (4)

If $f\in K[t;\delta]$ is an A-polynomial, then $f\sim g\times e_{r}(t)$ for
some $r\in\mathbb{N}$ and some irreducible A-polynomial $g\in R$ uniquely
determined up to similarity.

  5. (5)

If $f\in K[t;\delta]$ is an A-polynomial and $f\sim g\times h$ or $f=gh$ for
some $g,h\in K[t;\delta]$, then $g$ and $h$ are also A-polynomials.

In fact, the set of all A-polynomials in $K[t;\delta]$, denoted by
$\mathcal{A}(K,\delta)$, forms a semigroup under the product $\times$ (i.e. if
$f,g\in\mathcal{A}(K,\delta)$, we take $f\times g$ to be their product in
$\mathcal{A}(K,\delta)$) and the relation $\sim$ (similarity of polynomials in
$K[t;\delta]$), with the polynomial $t$ acting as the identity in
$\mathcal{A}(K,\delta)$. In [Ami53] it is shown that the A-polynomials are
precisely the polynomials in $K[t;\delta]$ such that $\mathcal{E}(f)$ is a
central simple algebra over $F$ of degree $m={\rm deg}(f)$, which is split by
$K$. Below are the main results relating such A-polynomials to central simple
algebras:

######  Theorem 89.

  1. (1)

If $f,g\in K[t;\delta]$ are A-polynomials, then $h=f\times g$ is an
A-polynomial, and

$\mathcal{E}(h)\cong\mathcal{E}(f)\otimes_{F}\mathcal{E}(g).$

  2. (2)

If $f\in K[t;\delta]$ is an A-polynomial of degree $m$, then $\mathcal{E}(f)$
is a central simple algebra over $F$ of degree $m$, which is split by $K$.

  3. (3)

Every central simple algebra over $F$ of degree $m$ which is split by $K$, is
isomorphic to $\mathcal{E}(f)$ for some A-polynomial $f\in K[t;\delta]$ of
degree $m$.

Suppose instead that we take $F$ to be a field of characteristic $p\neq 0$,
with $K$ a finite purely inseparable field extension of exponent 1, i.e.
$K^{p}\subseteq F\subset K$. Then the theory of differential transformations
also holds in the setting. The related theory of A-polynomials and central
simple algebras holds when the A-polynomial $f\in K[t;\delta]$ involved has
degree $m<[K:F]=p^{e}$ for some $e\in\mathbb{N}\cup\\{\infty\\}$ (with slight
abuse of notation writing $e=\infty$ for $[K:F]=\infty$). In particular, we
note that the results of Theorem 89 hold in this setting, for A-polynomials
$f$ of degree $m<[K:F]$, and for central simple algebras over $F$ of degree
$m$ which are split by $K$, satisfying $m<[K:F]$ [Ami54, Section 10]. This is
the most important case that we can say more on.

###  7.2 Generalised A-polynomials

For $f\in D[t;\sigma,\delta]$ we are interested in answering the question:

“When is $\mathcal{E}(f)$ a central simple algebra over the field $F=C\cap{\rm
Fix}(\sigma)\cap{\rm Const}(\delta)$?”

To this end we define a generalised A-polynomial as follows:

######  Definition 90.

Let $f\in D[t;\sigma,\delta]$. We call $f$ a generalised A-polynomial if
$\mathcal{E}(f)$ is a central simple algebra over $F$.

If $D$ is commutative (i.e. a field), then $d=1$ and without loss of
generality we can take $\sigma={\rm id}_{D}$ or $\delta=0$. Then a polynomial
$f$ in $D[t;\sigma]$ (resp. $D[t;\delta]$) is a generalised A-polynomial if
$\mathcal{E}(f)$ is a central simple algebra over $F={\rm Fix}(\sigma)$ (resp.
$F={\rm Const}(\delta)$).

For each $v\in D^{\times}$, we define a map $\Omega_{v}:D\longrightarrow D$ by

$\Omega_{v}(\alpha)=\sigma(v)\alpha v^{-1}+\delta(v)v^{-1}.$

If $D$ is commutative, we note that $\delta=0$ yields
$\Omega_{v}(\alpha)=\sigma(v)\alpha v^{-1}$ and $\sigma={\rm id}$ yields
$\Omega_{v}(\alpha)=\alpha+\delta(v)v^{-1}$ for all $v\in D^{\times}$.

The following is an easy consequence of the definition of similarity of skew
polynomials in $R$, and will prove to be very useful:

######  Lemma 91.

[Ami54, Lemma 2 for $\sigma={\rm id}$] Let $\alpha,\beta\in D$. Then
$(t-\alpha)\sim(t-\beta)$ in $D[t;\sigma,\delta]$ if and only if
$\Omega_{v}(\alpha)=\beta$ for some $v\in D^{\times}$.

###### Proof.

$(t-\alpha)\sim(t-\beta)$ is equivalent to the existence of $v,w\in
D^{\times}$ such that $w(t-\alpha)=(t-\beta)v$ [Jac43, pg. 33], i.e. there
exists $v,w\in D^{\times}$ such that

$w(t-\alpha)=\sigma(v)t+\delta(v)-\beta v.$

This is the case if and only if $w=\sigma(v)$ and
$w\alpha=\sigma(v)\alpha=\beta v-\delta(v)$. The result follows immediately. ∎

###  7.3 Generalised A-polynomials in $D[t;\sigma]$

Let $R$ denote the twisted polynomial ring $D[t;\sigma]$ with $D$ a central
division algebra over $C$ of degree $d$ and $\sigma$ an automorphism of $D$.
We assume that $\sigma$ has finite inner order $n$, with
$\sigma^{n}=\iota_{u}$ for some $u\in D^{\times}$.

Recall that $R$ has center $F[u^{-1}t^{n}]\cong F[x]$. For the remainder of
this section we suppose that $f\in R$ is a monic polynomial of degree $m\geq
1$ such that $(f,t)_{r}=1$. Then $f$ has a unique minimal central left
multiple $h(t)=\hat{h}(u^{-1}t^{n})$ for some $\hat{h}\in F[x]$ of degree at
most $dm$. Since $(f,t)_{r}=1$, the polynomial $h(t)$ is a bound of $f$.

We aim to provide some necessary, and some sufficient conditions for $f\in R$
to be a generalised A-polynomial.

First suppose that

$\hat{h}(x)=\hat{\pi}_{1}(x)^{e_{1}}\hat{\pi}_{2}(x)^{e_{2}}\cdots\hat{\pi}_{z}(x)^{e_{z}}$

for some irreducible polynomials
$\hat{\pi}_{1},\hat{\pi}_{2},\dots\hat{\pi}_{z}\in F[x]$ such that
$\hat{\pi}_{i}\neq\hat{\pi}_{j}$ for $i\neq j$, and some exponents
$e_{1},e_{2},\dots,e_{z}\geq 1$, and let
$\pi_{i}(t)=\hat{\pi}_{i}(u^{-1}t^{n})$ for each $i=1,2,\dots,z$.

We assume that the decomposition of $\hat{h}$ is square-free, i.e. that
$e_{1}=e_{2}=\cdots=e_{z}=1$ and

$\hat{h}(x)=\hat{\pi}_{1}(x)\hat{\pi}_{2}(x)\cdots\hat{\pi}_{z}(x).$

By Theorem 53, $\mathcal{E}(f)$ has center $E_{\hat{h}}=F[x]/(\hat{h}(x))$,
and by the Chinese Remainder Theorem for commutative rings (see for example
[Coh63, §5])

$E_{\hat{h}}\cong E_{\hat{\pi}_{1}}\oplus E_{\hat{\pi}_{2}}\oplus\cdots\oplus
E_{\hat{\pi}_{z}},$

where $E_{\hat{\pi}_{i}}=F[x]/(\hat{\pi}_{i}(x))$ for each $i$. Hence
$\mathcal{E}(f)$ has center $F$ if and only if $E_{\hat{\pi}_{1}}\oplus
E_{\hat{\pi}_{2}}\oplus\cdots\oplus E_{\hat{\pi}_{z}}=F$, which is true if and
only if $z=1$ and $E_{\hat{\pi}_{1}}=F$. Therefore we can assume $\hat{h}$ is
irreducible in $F[x]$ without loss of generality, and apply the results of
Chapter 2.

In particular by Theorem 20, for $f\in R$ with $\hat{h}\in F[x]$ irreducible,
we have that $f=f_{1}f_{2}\cdots f_{l}$ for $f_{1},f_{2},\dots,f_{l}\in R$
irreducible polynomials, which are all similar to each other (i.e. $f_{i}\sim
f_{j}$ for all $i,j$). Moreover

$\mathcal{E}(f)\cong M_{l}(\mathcal{E}(f_{i})),$

and $\mathcal{E}(f_{i})$ is a central division algebra over $E_{\hat{h}}$ of
degree $s^{\prime}=\frac{dn}{k}$, ${\rm deg}(\hat{h})=\frac{dm}{ls^{\prime}}$,
and $[\mathcal{E}(f_{i}):F]=\frac{dms^{\prime}}{l}$ where $k$ is the number of
irreducible factors of $h$ in any complete factorisation in $R$. Hence,
$\mathcal{E}(f)$ is a central simple algebra over $E_{\hat{h}}$ of degree
$s=ls^{\prime}$ and $[\mathcal{E}(f):F]=dms$.

After considering the above we are left almost immediately with the following
result:

######  Theorem 92.

Suppose that $\hat{h}(x)$ is irreducible in $F[x]$. Then $f$ is a generalised
A-polynomial in $R$ if and only if $\hat{h}(x)=x-a$ for some $a\in F$ if and
only if $f$ right divides $u^{-1}t^{n}-a$ for some $a\in F$. In particular, if
$f$ is a generalised A-polynomial, then $m\leq n$.

###### Proof.

Suppose that $f$ is a generalised A-polynomial in $R$. By the discussion
preceding this result, for $f$ to be a generalised A-polynomial it is
necessary that $\hat{h}(x)=x-a$ for some $a\in F$. Conversely if
$\hat{h}(x)=x-a\in F[x]$, then $E_{\hat{h}}=F[x]/(x-a)=F$. Hence
$\mathcal{E}(f)$ is a central simple algebra over $F$ by Theorem 20, i.e. $f$
is a generalised A-polynomial. It is easy to see that $\hat{h}(x)=x-a$ is
equivalent to $f$ being a right divisor of $u^{-1}t^{n}-a$ by definition of
the minimal central left multiple. Moreover, if $f$ right divides
$u^{-1}t^{n}-a$, then ${\rm deg}(f)\leq n$. ∎

For $n$ prime we are able to provide a more concrete description of $f$:

######  Theorem 93.

Suppose that $\hat{h}$ is irreducible in $F[x]$. Suppose that $n$ is prime and
not equal to $d$. Then $f$ is a generalised A-polynomial in $R$ if and only if
one of the following holds:

  1. (1)

There exists some $a\in F^{\times}$ such that
$ua\neq\prod\limits_{j=1}^{n}\sigma^{n-j}(b)$ for every $b\in D$, and

$f(t)=t^{n}-ua.$

In this case $f$ is an irreducible polynomial in $R$.

  2. (2)

$m\leq n$ and there exist some constants $c_{1},c_{2},\dots,c_{m-1},c_{m},b\in
D^{\times}$, with $c_{m}=1$, such that the product
$u^{-1}\prod\limits_{j=0}^{n-1}\sigma^{n-j}(b)$ lies in $F^{\times}$, and

$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(b)).$

In this case $f$ is a reducible polynomial unless $m=1$.

###### Proof.

By Theorem 92 $f$ is a generalised A-polynomial in $R$ if and only if $f$
right divides $u^{-1}t^{n}-a$ for some $a\in F^{\times}$. So suppose that $f$
is a generalised A-polynomial in $R$, then there exists some $a\in F^{\times}$
and some nonzero $g\in R$ such that

$u^{-1}t^{n}-a=gf$ (7.1)

In the notation of Theorem 20, $ldn=ks$ and since $f$ is a generalised
A-polynomial ${\rm deg}(\hat{h})=\frac{dm}{s}=1$, i.e. $dm=s$. Combining these
yields $\frac{n}{k}=\frac{m}{l}\in\mathbb{N}$. That is $k$ must divide $n$,
and so we must have that $k=1$ or $k=n$ as $n$ is prime. We analyse the cases
$k=1$ and $k=n$ separately.

First suppose that $k=1$, then $h(t)$ is irreducible. Therefore Equation (7.1)
becomes

$u^{-1}t^{n}-a=gf(t)$ (7.2)

for some $a\in F^{\times}$ and some $g\in D^{\times}$. This yields $g=u^{-1}$
and $f(t)=t^{n}-ua$ for some $a\in F^{\times}$. Suppose that $f$ were
reducible, then $f$ would be the product of $n$ linear factors as $n$ is
prime, hence $f$ is irreducible if and only if
$ua\neq\prod\limits_{j=1}^{n}\sigma^{n-j}(b)$ for any $b\in D$, by [Bro18,
Corollary 3.4].

On the other hand, if $k=n$, then $h(t)$ is equal to a product of $n$ linear
factors in $R$, all of which are mutually similar to one another. Also, since
$\frac{n}{k}=\frac{m}{l}$ and $n=k$, we have $m=l\leq n$. Hence $f$ is the
product of $m\leq n$ linear factors in $R$, all of which are similar.

So there exist constants $b_{1},b_{2},\dots,b_{m}\in D^{\times}$ such that
$(t-b_{i})\sim(t-b_{j})$ for all $i,j\in\\{1,2,\dots,m\\}$, and

$f(t)=\prod\limits_{i=1}^{m}(t-b_{i}).$

In particular $(t-b_{i})\sim(t-b_{m})$ for all $i\neq m$, which is true if and
only if there exist constants $c_{1},c_{2},\dots,c_{m-1},c_{m}\in D^{\times}$
such that $b_{i}=\Omega_{c_{i}}(b_{m})$ for all $i$ by Lemma 91. Hence setting
$b=b_{m}$ and $c_{m}=1$ yields

$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(b)).$

Finally, we note that $(t-b)|_{r}(t^{n}-ua)$ for some $a\in F^{\times}$ if and
only if $u^{-1}\prod\limits_{j=0}^{n-1}\sigma^{n-j}(b)=a\in F^{\times}$, by
[Bro18, Corollary 3.4]. ∎

So there are no generalised A-polynomials in $R$ of degree greater than $n$
under the assumption that $\hat{h}$ is irreducible in $F[x]$.
If $e_{i}>1$ for at least one $i$, then it is not clear to the author when
$\mathcal{E}(f)$ is a central simple algebra over the field $F$.

####  7.3.1 Generalised A-polynomials in $K[t;\sigma]$ and
$\mathbb{F}_{q^{n}}[t;\sigma]$

Throughout this section we suppose that $R=K[t;\sigma]$ with $K$ a field, and
that $\sigma$ an automorphism of $K$ of finite order $n$ with fixed field $F$.
Now the center of $R$ is $F[t^{n}]\cong F[x]$. Let $f\in R$ be of degree
$m\geq 1$ and satisfy $(f,t)_{r}=1$, and suppose that $f$ has minimal central
left multiple $h(t)=\hat{h}(t^{n})$ for some irreducible polynomial
$\hat{h}\in F[x]$.

######  Theorem 94.

$f$ is a generalised A-polynomial in $R$ if and only if $\hat{h}(x)=x-a$ for
some $a\in F[x]$ if and only if $f$ right divides $t^{n}-a$ in $R$. In
particular, if $f$ is a generalised A-polynomial in $R$, then $m\leq n$.

This follows from Theorem 93. In the case that $n$ is prime, we obtain the
following as an immediate corollary to both Theorem 93 and Theorem 94:

######  Theorem 95.

Let $n$ be prime. Then $f$ is a generalised A-polynomial in $R$ if and only if
one of the following holds:

  1. (1)

There exists some $a\in F^{\times}$ such that $a\neq N_{K/F}(b)$ for any $b\in
K$, and

$f(t)=t^{n}-a.$

In this case $f$ is an irreducible polynomial in $R$.

  2. (2)

$m\leq n$ and there exist some constants $c_{1},c_{2},\dots,c_{m-1},c_{m},b\in
K^{\times}$ with $c_{m}=1$, such that

$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(b)).$

###### Proof.

The proof is identical to the proof of Theorem 93 with $d=u=1$. We also
dropped the condition that $\prod\limits_{j=0}^{n-1}\sigma^{j}(b)$ lie in
$F^{\times}$ since this product is equal to the field norm $N_{K/F}(b)$ which
is always in $F^{\times}$ for $b\in K^{\times}$. ∎

In particular, let $K=\mathbb{F}_{q^{n}}$, where $q=p^{e}$ for some prime $p$
and exponent $e\geq 1$, and where $\sigma:K\longrightarrow K$, $a\mapsto
a^{q}$, is the Frobenius automorphism of order $n$, with fixed field
$F=\mathbb{F}_{q}$. In this case the only central division algebra over the
field $\mathbb{F}_{q}$ is $\mathbb{F}_{q}$ itself. The following result is
therefore an easy consequence of Theorem 94:

######  Theorem 96.

Suppose that $f\in\mathbb{F}_{q^{n}}[t;\sigma]$ satisfies $(f,t)_{r}=1$, and
has minimal central left multiple $h(t)=\hat{h}(t^{n})$ for some irreducible
polynomial $\hat{h}\in\mathbb{F}_{q}[x]$. Then $f$ is an A-polynomial if and
only if $m\leq n$ and there exist some constants
$c_{1},c_{2},\dots,c_{m-1},c_{m},b\in K^{\times}$ with $c_{m}=1$, such that

$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(b)).$

In particular, $f$ is a reducible polynomial in $\mathbb{F}_{q^{n}}[t;\sigma]$
unless $m=1$.

###### Proof.

First we note that $E_{\hat{h}}=\frac{\mathbb{F}_{q}[x]}{(\hat{h}(x))}$ is a
field extension of $\mathbb{F}_{q}$ of finite degree, hence it is also a
finite field. In the notation of Theorem 32, $\mathcal{E}(f)\cong
M_{l}(E_{\hat{h}})$ since the only central division algebra over the finite
field $E_{\hat{h}}$ is $E_{\hat{h}}$ itself. Hence $\mathcal{E}(f)$ is a
central simple algebra over $E_{\hat{h}}$ of degree $s=l$. By Theorem 32,
${\rm deg}(h)=\frac{mn}{s}=\frac{mn}{l}$. On the other hand, we know that
${\rm deg}(h)=k{\rm deg}(f_{i})$ where $f_{i}$ is any irreducible divisor of
$f$ in $R$, and ${\rm deg}(f_{i})=\frac{m}{l}$. Combining the expressions for
${\rm deg}(h)$ yields $n=k$.
The rest of the proof can be taken verbatim to be the proof of Theorem 93 from
the point that we examine the $n=k$ case. We note that the part of the proof
of Theorem 93 that we invoke here does not rely on $n$ being a prime. ∎

###  7.4 Generalised A-polynomials in $D[t;\delta]$

This section is dedicated to finding necessary and sufficient conditions for
certain polynomials $f\in D[t;\delta]$. To do this, we focus on the cases
${\rm Char}(D)=0$ and ${\rm Char}(D)=p\neq 0$, separately. Again, it can be
seen that all of the results in this section follow from a similar argument to
those use throughout the previous sections of this chapter.

####  7.4.1 ${\rm Char}(D)=0$

In this section we impose that $D$ has characteristic $0$, and that $\delta$
is the inner derivation defined by some element $c\in D^{\times}$, that is
$\delta(a)=[c,a]=ca-ac$ for all $a\in D$.

Recall that $R$ has center $F[t-c]\cong F[x]$ where $F=C\cap{\rm
Const}(\delta)$, and that all polynomials $f\in R=D[t;\delta]$ have a unique
minimal central left multiple $h(t)=\hat{h}(t-c)$ for some $\hat{h}\in F[x]$,
which is equal to a bound of $f$.

######  Remark.

As in the search for generalised A-polynomials in a twisted polynomial ring
$D[t;\sigma]$, we restrict ourselves to the case $\hat{h}$ is square-free in
$F[x]$, otherwise the author was unable to determine when the eigenring of
$f\in R$ is indeed a central simple algebra over $F$. Then by a similar
argument to Section 3 (pg. 74), we lose no generality in the search for
A-polynomials in $R$, if we take $\hat{h}$ to be an irreducible polynomial in
$F[x]$.

By Theorem 38, for any $f\in R$ of degree $m\geq 1$ with $\hat{h}\in F[x]$
irreducible, we have that $f=f_{1}f_{2}\cdots f_{l}$ for $f_{1},\dots,f_{l}$
irreducible polynomials in $R$, which are all mutually similar to each other,

$\mathcal{E}(f)\cong M_{l}(\mathcal{E}(f_{i})),$

and $\mathcal{E}(f_{i})$ is a central division algebra over $E_{\hat{h}}$ of
degree $s^{\prime}=\frac{d}{k}$, where $k$ is the number of irreducible
factors of $h$ in $R$. Moreover, ${\rm deg}(\hat{h})=\frac{dm}{ls^{\prime}}$,
and $[\mathcal{E}(f_{i}):F]=\frac{dms^{\prime}}{l}$. Hence, $\mathcal{E}(f)$
is a central simple algebra over $E_{\hat{h}}$ of degree $s=ls^{\prime}$ and
$[\mathcal{E}(f):F]=dms$.

This yields the following result almost immediately:

######  Theorem 97.

Suppose that $f\in R$ has degree $m$, and that $f$ has minimal central left
multiple $h(t)=\hat{h}(t-c)$ for some irreducible polynomial $\hat{h}\in
F[x]$. Then $f$ is a generalised A-polynomial in $R$ if and only if
$f(t)=t-(c+a)$ for some $a\in F$.

###### Proof.

Suppose that $f$ is a generalised A-polynomial in $R$, that is
$\mathcal{E}(f)$ is a central simple algebra over $F$ of degree $dm$. This is
true if and only if $E_{\hat{h}}=F$, i.e. if and only if ${\rm
deg}(\hat{h})=\frac{md}{s}=1$. Having $\hat{h}$ of degree one is equivalent to
having $h(t)=t-(c+a)$ for some $a\in F$, and by definition $h(t)=g(t)f(t)$ for
some $g\in R$. ∎

######  Remark.

The above result shows that there are in fact no generalised A-polynomials in
$R$ with a square-free minimal central left multiple, other than the central
element $t-c$ up to a possible shift by some element of $F$ and a possible
scalar multiplication by some element of $D^{\times}$.

####  7.4.2 ${\rm Char}(D)=p>0$

From now on let $R=D[t;\delta]$ where $D$ is a central division algebra of
degree $d$ over $C$. Assume that $C$ has prime characteristic $p$, and that
$\delta$ is an algebraic derivation of $D$ with minimum polynomial
$g(t)=t^{p^{e}}+\gamma_{1}t^{p^{e-1}}+\cdots+\gamma_{e}t\in F[t]$, such that
$g(\delta)(a)=[c,a]=ca-ac$ for some nonzero $c\in D$ and for all $a\in D$.
Here, $F=C\cap{\rm Const}(\delta)$ ($D=K$ is a field is included here as
special case). Then $R$ has center $F[g(t)-c]\cong F[x]$. For every $f\in R$,
the minimal central left multiple of $f$ in $R$ is the unique polynomial of
minimal degree $h\in C(R)=F[x]$ such that $h=gf$ for some $g\in R$, and such
that $h(t)=\hat{h}(g(t)-c)$ for some monic $\hat{h}\in F[x]$. All $f\in
R=D[t;\delta]$ have a unique minimal central left multiple, which is a bound
of $f$. Again we can restrict our investigation to the case $\hat{h}$ is
square-free in $F[x]$, and note that it is necessary that $\hat{h}$ be
irreducible in $F[x]$ for $f$ to be a generalised A-polynomial in $R$.

By Theorem 43, for any $f\in R$ of degree $m\geq 1$ with $\hat{h}\in F[x]$
irreducible, we have that $f=f_{1}f_{2}\cdots f_{l}$ for $f_{1},\dots,f_{l}$
irreducible polynomials in $R$, which are all mutually similar to each other,

$\mathcal{E}(f)\cong M_{l}(\mathcal{E}(f_{i})),$

and $\mathcal{E}(f_{i})$ is a central division algebra over $E_{\hat{h}}$ of
degree $s^{\prime}=\frac{dp^{e}}{k}$, where $k$ is the number of irreducible
factors of $h$ in $R$. Moreover, ${\rm deg}(\hat{h})=\frac{dm}{ls^{\prime}}$,
and $[\mathcal{E}(f_{i}):F]=\frac{dms^{\prime}}{l}$. Hence, $\mathcal{E}(f)$
is a central simple algebra over $E_{\hat{h}}$ of degree $s=ls^{\prime}$ and
$[\mathcal{E}(f):F]=dms$. We obtain the following:

######  Theorem 98.

$f$ is a generalised A-polynomial in $R$ if and only if $f$ right divides
$g(t)-(b+c)$ for some $b\in F$. In particular, ${\rm deg}(f)\leq p^{e}$.

###### Proof.

Suppose that $f$ is a generalised A-polynomial in $R$, that is
$\mathcal{E}(f)$ is a central simple algebra over $F$ of degree $dm$. This is
true if and only if $E_{\hat{h}}=F$, i.e. if and only if ${\rm
deg}(\hat{h})=\frac{md}{s}=1$. Having $\hat{h}$ of degree one is equivalent to
having $h(t)=g(t)-(b+c)$ for some $b\in F$, and by definition $h(t)=r(t)f(t)$
for some $r\in R$. ∎

In $D[t;\delta]$

$(t-\alpha)^{p}=t^{p}-V_{p}(\alpha),\;V_{p}(\alpha)=\alpha^{p}+\delta^{p-1}(\alpha)+\nabla_{\alpha}$
(7.3)

for all $\alpha\in D$, where $\nabla_{\alpha}$ is a sum of commutators of
$\alpha,\delta(\alpha),\delta^{2}(\alpha),\dots,\delta^{p-2}(\alpha)$ [Jac09,
pg. 17-18]. In particular, if $D$ is commutative, then $\nabla_{\alpha}=0$ and

$V_{p}(\alpha)=\alpha^{p}+\delta^{p-1}(\alpha)$ (7.4)

for all $\alpha\in D$. Using the identities
$t^{p}=(t-\alpha)^{p}+V_{p}(\alpha)$ and $t=(t-\alpha)+\alpha$ for all
$\alpha\in D$, we arrive at:

######  Lemma 99.

[Jac09, Proposition 1.3.25 ($e=1$)] Let $f(t)=t^{p}-\gamma_{1}t-\gamma\in
D[t;\delta]$ and $\alpha\in D$. Then $(t-\alpha)|_{r}f(t)$ if and only if
$V_{p}(\alpha)-\gamma_{1}\alpha-\gamma=0.$

The identity (7.3) can be iterated to obtain

$(t-\alpha)^{p^{i}}=t^{p^{i}}-V_{p^{i}}(\alpha)$ (7.5)

for all $\alpha\in D$ and all integer exponents $i\geq 1$, where

$V_{p^{i}}(\alpha)=V_{p}^{i}(\alpha)=\underbrace{(V_{p}\circ
V_{p}\circ\cdots\circ V_{p})}_{\text{i terms}}(\alpha).$

In a similar fashion to Lemma 99 the identities
$t^{p^{i}}=(t-\alpha)^{p^{i}}+V_{p^{i}}(\alpha)$ and $t=(t-\alpha)+\alpha$ may
be used to obtain:

######  Lemma 100.

[Jac09, Proposition 1.3.25] Let
$f(t)=t^{p^{e}}+\gamma_{1}t^{p^{e-1}}+\cdots+\gamma_{e}t-\gamma\in
D[t;\delta]$ and $\alpha\in D$. Then $(t-\alpha)|_{r}f(t)$ if and only if

$V_{p^{e}}(\alpha)+\gamma_{1}V_{p^{e-1}}+\cdots+\gamma_{e}\alpha-\gamma=0.$

If $e=1$ (i.e. $\delta$ is an algebraic derivation of $D$ of degree $p$), we
can determine necessary and sufficient conditions for $f$ to be an
A-polynomial in $R$:

######  Theorem 101.

Suppose that $\delta$ has minimum polynomial $g(t)=t^{p}-\gamma_{1}t$ and that
$\hat{h}$ is irreducible in $F[x]$. Then $f$ is a generalised A-polynomial in
$R$ if and only if one of the following holds:

  1. (1)

$f(t)=h(t)=t^{p}-\gamma_{1}t-(\gamma+c)$ for some $\gamma\in F$, and

$V_{p}(\alpha)-\gamma_{1}\alpha-(\gamma+c)\neq 0$

for all $\alpha\in D$. In this case $f$ is irreducible in $R$.

  2. (2)

$h(t)=t^{p}-\gamma_{1}t-(\gamma+c)$ for some $\gamma\in F$, $m\leq p$ and

$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(\alpha))$

for some $c_{1},c_{2},\dots,c_{m},\alpha\in D^{\times}$ with $c_{m}=1$
(w.l.o.g.) such that

$V_{p}(\alpha)-\gamma_{1}\alpha-(\gamma+c)=0.$

In particular, $f$ is a reducible polynomial in $R$ unless $m=1$.

###### Proof.

By Theorem 98, $f$ is a generalised A-polynomial in $R$ if and only if $f$
right divides $t^{p}-\gamma_{1}t-(\gamma+c)$ for some $\gamma\in F$. So
suppose that $f$ is a generalised A-polynomial in $R$, then there exists some
$\gamma\in F$ and some nonzero $f^{\prime}\in R$ such that
$t^{p}-\gamma_{1}t-(\gamma+c)=f^{\prime}f.$ In the notation of Theorem 43,
$ldp=ks$ and since $f$ is a generalised A-polynomial, ${\rm
deg}(\hat{h})=\frac{dm}{s}=1$, i.e. $dm=s$. Combining these yields
$\frac{p}{k}=\frac{m}{l}\in\mathbb{N}$. That is $k$ must divide $p$, and so we
must have that $k=1$ or $k=p$ as $p$ is prime.
First suppose that $k=1$, then $h(t)$ is irreducible in $R$ and
$t^{p}-\gamma_{1}t-(\gamma+c)=f^{\prime}f$ yields $f^{\prime}=1$, i.e.
$f(t)=t^{p}-\gamma_{1}t-(\gamma+c)$. Suppose that $f$ were reducible, then $f$
would be the product of $p$ linear factors as $p$ is prime, hence $f$ is
irreducible if and only if $V_{p}(\alpha)-\gamma_{1}\alpha-(\gamma+c)\neq 0$
for any $\alpha\in D$, by Lemma 99.
On the other hand, if $k=p$, then $h(t)$ is equal to a product of $p$ linear
factors in $R$, all of which are similar to one another. Also, since
$\frac{p}{k}=\frac{m}{l}$ and $p=k$, we have $m=l\leq p$. Hence $f$ is the
product of $m\leq p$ linear factors in R, all of which are mutually similar to
each other. So there exist constants
$\alpha_{1},\alpha_{2},\dots,\alpha_{m}\in D^{\times}$ such that
$f(t)=\prod\limits_{i=1}^{m}(t-\alpha_{i})$, and
$(t-\alpha_{i})\sim(t-\alpha_{j})$ for all $i,j\in\\{1,2,\dots,m\\}$. In
particular $(t-\alpha_{i})\sim(t\alpha_{m})$ for all $i\neq m$, which is true
if and only if there exist constants $c_{1},c_{2},\dots,c_{m-1},c_{m}\in
D^{\times}$ such that $\alpha_{i}=\Omega_{c_{i}}(\alpha_{m})$ for all $i$ by
Lemma 91. Hence setting $\alpha=\alpha_{m}$ and $c_{m}=1$ yields
$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(\alpha))$. Finally, we note that
$(t-\alpha)$ right divides $t^{p}-\gamma_{1}t-(\gamma+c)$ if and only if
$V_{p}(\alpha)-\gamma_{1}\alpha-(\gamma+c)=0$ by Lemma 99. ∎

In the following, $\mathbb{K}$ denotes a finite field of characteristic $p$
and $\delta$ is an algebraic derivation of $\mathbb{K}$ with fixed field
$\mathbb{F}$ and minimum polynomial
$g(t)=t^{p^{e}}+\gamma_{1}t^{p^{e-1}}+\cdots+\gamma_{e}t\in\mathbb{F}[t]$ such
that $g(\delta)=0$. Then $\mathbb{K}[t;\delta]$ has centre
$\mathbb{F}[g(t)]\cong\mathbb{F}[x]$. In this particular case we obtain a more
detailed description on the generalised A-polynomials in
$\mathbb{K}[t;\delta]$.

######  Theorem 102.

Suppose that $f\in\mathbb{K}[t;\delta]$ has degree $m\geq 1$ and minimal
central left multiple $h(t)=\hat{h}(g(t))$ for some irreducible polynomial
$\hat{h}\in\mathbb{F}[x]$. Then $f$ is a generalised A-polynomial if and only
if $h(t)=g(t)-\gamma$ for some $\gamma\in\mathbb{F}$, $m\leq p^{e}$, and

$f(t)=\prod\limits_{i=1}^{m}(t-\Omega_{c_{i}}(\alpha)),$

for some $c_{1},c_{2},\dots,c_{m-1},c_{m},\alpha\in\mathbb{K}^{\times}$ with
$c_{m}=1$ (w.l.o.g.), such that

$V_{p^{e}}(\alpha)+\gamma_{1}V_{p^{e-1}}(\alpha)+\cdots+\gamma_{e}\alpha-\gamma=0.$

In particular, $f$ is a reducible polynomial in $\mathbb{K}[t;\delta]$ unless
$m=1$.

###### Proof.

First we note that $E_{\hat{h}}=\frac{\mathbb{F}[x]}{(\hat{h}(x))}$ is a field
extension of $\mathbb{F}$ of finite degree, hence it is also a finite field.
In the notation of Theorem 43, $\mathcal{E}(f)\cong M_{l}(E_{\hat{h}})$ since
the only central division algebra over the finite field $E_{\hat{h}}$ is
$E_{\hat{h}}$ itself. Hence $\mathcal{E}(f)$ is a central simple algebra over
$E_{\hat{h}}$ of degree $s=l$. By Theorem 43, ${\rm
deg}(h)=\frac{mp^{e}}{s}=\frac{mp^{e}}{l}$. On the other hand, we know that
${\rm deg}(h)=k{\rm deg}(f_{i})$ where $f_{i}$ is any irreducible divisor of
$f$ in $R$, and ${\rm deg}(f_{i})=\frac{m}{l}$. Combining the expressions for
${\rm deg}(h)$ yields $p^{e}=k$.
Now by Theorem 98, $f$ is a generalised A-polynomial in $R$ if and only if
$f|_{r}(g(t)-\gamma)$ for some $\gamma\in\mathbb{F}$ which is true if and only
if $h(t)=g(t)-\gamma$ for some $\gamma\in\mathbb{F}$, i.e. ${\rm
deg}(\hat{h})=1$. Since ${\rm deg}(\hat{h})=\frac{m}{s}=\frac{m}{l}$ we have
that $m=l$, that is $f$ is the product of $m$ irreducible polynomials in $R$
that are all similar to each other. Hence there exist elements
$\alpha_{1},\alpha_{2},\dots,\alpha_{m}\in\mathbb{K}$ such that

$f(t)=\prod\limits_{i=1}^{m}(t-\alpha_{i})$

and elements $c_{1},c_{2},\dots,c_{m}\in\mathbb{K}^{\times}$ such that
$\Omega_{c_{i}}(\alpha_{m})=\alpha_{i}$ for each $i$ by Lemma 91. We use the
notation $\alpha=\alpha_{m}$ and without loss of generality we may take
$c_{m}=1$. Finally we note that $(t-\alpha)$ right divides $h(t)=g(t)-\gamma$
if and only if
$V_{p^{e}}(\alpha)+\gamma_{1}V_{p^{e-1}}(\alpha)+\cdots+\gamma_{e}\alpha-\gamma=0$
by Lemma 100. ∎

.

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