# Non-Abelian extensions of degree $p^{3}$ and $p^{4}$ in characteristic $p>2$

Grant Moles

###### Abstract.

This paper describes in terms of Artin-Schreier equations field extensions
whose Galois group is isomorphic to any of the four non-cyclic groups of order
$p^{3}$ or the ten non-Abelian groups of order $p^{4}$, $p$ an odd prime, over
a field of characteristic $p$.

##  1\. Introduction

Let $K$ be a field of characteristic $p$, a prime, and $G$ be a $p$-group. The
main theorem of [Wit36] tells us under what circumstances there will exist a
field extension $L$ of $K$ such that $L/K$ is Galois with Galois group $G$.
Furthermore, in [Sal78], it was shown that such a field extension can be
described using polynomials of a particular form, as we will describe shortly
(in fact, Saltman showed this more generally for ring extensions). The purpose
of this paper is to serve as a natural extension of these important results.
For field extensions of this form of degree $p^{3}$ or $p^{4}$, we will
construct the polynomials based on the presentation of the Galois group $G$.

Let $K_{4}/K_{0}$ be a non-Abelian Galois extension of degree $p^{4}$, where
$K_{0}$ is a field of characteristic $p$, an odd prime. According to Burnside
[Bur55], there are ten non-Abelian groups of order $p^{4}$. Nine of these
groups can be expressed as:

$\displaystyle G_{\vec{a},\vec{b}}=\langle\sigma_{1}$
$\displaystyle,\sigma_{2},\sigma_{3},\sigma_{4}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p}=\sigma_{3}^{a_{1}}\sigma_{4}^{b_{1}},\sigma_{2}^{p}=\sigma_{4}^{b_{2}},\sigma_{3}^{p}=\sigma_{4}^{p}=1,\sigma_{4}\in
Z(G_{\vec{a},\vec{b}}),$
$\displaystyle\sigma_{3}\sigma_{2}=\sigma_{2}\sigma_{3},\sigma_{3}\sigma_{1}=\sigma_{1}\sigma_{3}\sigma_{4}^{b_{3}},\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\sigma_{4}^{b_{0}}\rangle,$

where $a_{i}\in\\{0,1\\}$, $b_{j}\in\\{0,1\\}$ for $j\neq 2$, and
$b_{2}\in\\{0,1,\alpha\\}$, where $\alpha$ is a quadratic non-residue mod $p$.
The tenth group is expressed as

$H=\langle\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{3}=\sigma_{1}^{p},\sigma_{4}=\sigma_{1}^{p^{2}},\sigma_{1}^{p^{3}}=\sigma_{2}^{p}=1,\sigma_{2}\sigma_{1}=\sigma_{1}^{1+p^{2}}\sigma_{2}\rangle.$

In any of these groups, we can construct the composition series

$\\{1\\}\trianglelefteq\langle\sigma_{4}\rangle\trianglelefteq\langle\sigma_{3},\sigma_{4}\rangle\trianglelefteq\langle\sigma_{2},\sigma_{3},\sigma_{4}\rangle\trianglelefteq\langle\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\rangle,$

with $\langle\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\rangle$ being the
entire group. The reader will note that this is a central series in any of the
given groups, so we can define intermediate fields $K_{3},K_{2},K_{1}$ such
that $K_{3}$ is the fixed field of $\langle\sigma_{4}\rangle$, $K_{2}$ is the
fixed field of $\langle\sigma_{3},\sigma_{4}\rangle$, and $K_{1}$ is the fixed
field of $\langle\sigma_{2},\sigma_{3},\sigma_{4}\rangle$. In general, we have
$\mathrm{Gal}(K_{i}/K_{i-1})=\langle\sigma_{i}|_{K_{i}}\rangle$, the subgroup
of automorphisms of $K_{i}$ generated by $\sigma_{i}$ restricted to $K_{i}$.
When the Galois group of $K_{4}/K_{0}$ is $G_{\vec{a},\vec{b}}$ as above, we
have that $K_{3}/K_{0}$ is a non-cyclic Galois extension of degree $p^{3}$
with Galois group

$G_{\vec{a}}=\langle\sigma_{1},\sigma_{2},\sigma_{3}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p}=\sigma_{3}^{a_{1}},\sigma_{2}^{p}=\sigma_{3}^{p}=1,\sigma_{3}\in
Z(G_{\vec{a}}),\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\rangle,$

with $\vec{a}$ the same vector as in the representation of
$G_{\vec{a},\vec{b}}$. In fact, this form also holds when the Galois group is
$H$, with $a_{0}=0$, $a_{1}=1$. Now since each extension $K_{i}/K_{i-1}$ is
cyclic of degree $p$ with Galois group $\langle\sigma_{i}|_{K_{i}}\rangle$,
there is some $x_{i}\in K_{i}$ such that $K_{i}=K_{i-1}(x_{i})$ which
satisfies an Artin-Schreier polynomial with constant coefficient in $K_{i-1}$
such that $(\sigma_{i}-1)x_{i}=1$. Note that since $\sigma_{i}$ fixes $K_{j}$
for $0\leq j<i\leq 4$, $(\sigma_{i}-1)x_{j}=0$ for $1\leq j<i\leq 4$.
Furthermore, we have that
$\mathrm{Gal}(K_{2}/K_{0})=\langle\sigma_{1}|_{K_{2}},\sigma_{2}|_{K_{2}}\rangle\cong
C_{p}^{2}$, so we may choose $x_{2}\in K_{2}$ to be such that
$(\sigma_{1}-1)x_{2}=0$. Then we arrive at the main theorem of this paper:

######  Theorem 1.1.

Let $K_{4}/K_{0}$ be a non-Abelian Galois extension of degree $p^{4}$ with
$\sigma_{i}$, $K_{i}$, and $x_{i}$ for $1\leq i\leq 4$ as defined above. Then
$x_{1}$ and $x_{2}$ can be chosen so that $\wp(x_{1})\in K_{0}$ and
$\wp(x_{2})\in K_{0}$, where $\wp$ is the Weierstrass $\wp$ function, defined
by $\wp(x)=x^{p}-x$. Denote $\wp(x_{1})=\beta_{1}$ and $\wp(x_{2})=\beta_{2}$.
If $\mathrm{Gal}(K_{4}/K_{0})\cong G_{\vec{a},\vec{b}}$ as above, then we can
select $x_{3}$ and $x_{4}$ such that

$\displaystyle\wp(x_{3})$
$\displaystyle=a_{0}\beta_{2}x_{1}+a_{1}D_{1}(x_{1})+\beta_{3},$
$\displaystyle\wp(x_{4})$
$\displaystyle=b_{0}\beta_{2}x_{1}+b_{1}D_{1}(x_{1})+b_{2}D_{1}(x_{2})+b_{3}\left(a_{0}\beta_{2}\binom{x_{1}}{2}+\beta_{3}x_{1}\right)+\beta_{4},$

for some $\beta_{i}\in K_{0}$, where $D_{1}\in\mathbb{F}_{p}[x]$ the Witt
polynomial

$D_{1}(x)=\frac{x^{p}+(x^{p}-x)^{p}-(x+(x^{p}-x))^{p}}{p}.$

For each group, the specific values of $a_{i}$ and $b_{i}$ are shown in Table
1. Otherwise, $\mathrm{Gal}(K_{4}/K_{0})\cong H$ as above, and we can select
$x_{3}$ and $x_{4}$ such that

$\displaystyle\wp(x_{3})$ $\displaystyle=D_{1}(x_{1})+\beta_{3},$
$\displaystyle\wp(x_{4})$
$\displaystyle=\beta_{2}x_{1}+D_{2}(x_{1},x_{3})+\beta_{4}.$

Again, $\beta_{i}\in K_{0}$, $D_{1}$ is as above, and
$D_{2}(x_{1},x_{3})\in\mathbb{F}_{p}[x_{1},x_{3}]$ is the Witt polynomial

$D_{2}(x_{1},x_{3})=\frac{x_{1}^{p^{2}}+\beta_{1}^{p^{2}}-(x_{1}+\beta_{1})^{p^{2}}+p\left(x_{3}^{p}+\beta_{3}^{p}-(x_{3}+D_{1}(x_{1})+\beta_{3})^{p}\right)}{p^{2}}$

Moreover, the converse holds; that is, any algebraic extension of such a form
(the particular form described in more detail later) is a Galois extension
with Galois group as presented here.

Table 1: Translation to Burnside and James with $\vec{a}$ and $\vec{b}$ Values
Burnside James $\sigma_{1}$ $\sigma_{2}$ $\sigma_{3}$ $\sigma_{4}$ $a_{0}$
$a_{1}$ $b_{0}$ $b_{1}$ $b_{2}$ $b_{3}$ (xiv) $\phi_{2}(1^{4})$ $S$ $R$ $Q$
$P$ $0$ $0$ $1$ $0$ $0$ $0$ (ix) $\phi_{2}(211)a$ $P$ $R^{-1}$ $Q$ $P^{p}$ $0$
$0$ 1 1 0 0 (vii) $\phi_{2}(211)b$ $R$ $P$ $Q$ $P^{p}$ 0 0 0 0 1 1 (xv),$p=3$
$\phi_{3}(1^{4})$ $R$ $P^{-1}$ $Q^{-1}$ $P^{p}$ 1 0 0 0 $\alpha$ 1 (xv),$p>3$
$\phi_{3}(1^{4})$ $S$ $R$ $Q$ $P$ 1 0 0 0 0 1 (x) $\phi_{2}(211)$ $P$ $R$
$Q^{-1}$ $P^{p}$ 1 0 0 1 0 0 (xi) $\phi_{3}(211)b_{1},p=3$
$\phi_{3}(211)a,p>3$ $P$ $R$ $Q^{-1}$ $P^{p}$ 1 0 0 1 0 1 (xii)
$\phi_{3}(211)b_{\nu},p=3$ $\phi_{3}(211)b_{1},p>3$ $P$ $R$ $Q^{-1}$ $P^{p}$ 1
0 0 1 1 1 (xiii) $\phi_{3}(211)a,p=3$ $\phi_{3}(211)b_{\nu},p>3$ $P$ $R$
$Q^{-1}$ $P^{p}$ 1 0 0 1 $\alpha$ 1 (viii) $\phi_{2}(22)$ $P$ $Q^{-1}$ $P^{p}$
$Q^{-p}$ 1 1 0 0 1 0 (vi) $\phi_{2}(31)$ $P$ $Q^{-1}$ $P^{p}$ $P^{p^{2}}$ - -
- - - -

In Table 1, $\alpha$ refers to a quadratic non-residue mod $p$ (recall that
when $p=3$, the only quadratic non-residue is $\alpha=-1$). The value for
$\wp(x_{3})$ is determined below in Proposition 2.2, and the value for
$\wp(x_{4})$ in Propositions 3.1 and 3.3. It is worth noting that not every
combination of vectors $\vec{a}$ and $\vec{b}$ is considered here, though
every non-Abelian group is represented. Notably, some other combinations of
$\vec{a}$ and $\vec{b}$ will yield Abelian groups, and some will yield
equivalent representations of the non-Abelian groups. For example, when both
$\vec{a}$ and $\vec{b}$ are zero vectors, $G_{\vec{a},\vec{b}}\cong
C_{p}^{4}$. Since the combinations presented cover all non-Abelian groups,
only these combinations will be considered in this paper. Also shown in the
table are the notations for the groups used by Burnside [Bur55] and James
[Jam80], as well as the translation from the group elements $P,Q,R,S$ used by
Burnside to the $\sigma_{i}$ defined above. The order has been changed to
place groups with similar corresponding Artin-Schreier equations near each
other. For more information on the groups presented here, as well as other
groups of order $p^{n}$ for $n\leq 6$, see [Jam80].

At this point, it should be noted that although the work presented here
follows naturally from Saltman’s work, the notation used will be that
presented above, rather than that used in [Sal78]. In particular, the elements
$x_{i}$ and $\beta_{i}$ in this paper are referred to as $\theta_{i}$ and
$r_{i}$, respectively, by Saltman. In his notation, we are constructing the
particular polynomials $s_{i}$ corresponding to each choice of Galois group
$G$. It it worth noting that, in Theorem 1.1 above, we have
$\beta_{i}=x_{i}^{p}-x_{i}$ for $i=1,2$, and so $\beta_{3}$ is actually equal
to $\wp(x_{3})$ plus a polynomial in $\mathbb{F}_{p}[x_{1},x_{2}]$ (and
similarly for $\beta_{4}$), as we would expect.

The work presented here has potential applications in group theory, namely the
study of $p$-groups and related structures, as well as Galois theory. Perhaps
most obviously, the methods used to determine group actions and Artin-Schreier
equations throughout this paper could be extended to extensions defined by
other Galois groups. Most immediately, this could be applied to Galois
extensions of degree $p^{n}$ for $n\geq 5$. However, this could also extend to
other subgroups of the Nottingham group, especially those which are non-
Abelian (see [Cam97], [BCT22]). The results in this paper could also be used
to approach the local lifting problem for the groups outlined above. For more
information on this problem, as well as a characterization of extensions with
Galois group $D_{4}$ in characteristic 2, see [Wea18]. Finally, results such
as those found here are of particular interest and utility when considering
ramification indices of local field extensions with these Galois groups (see
[EK23a], [Eld23], [EK23b]).

##  2\. Extensions of degree $p^{3}$

Before presenting the arguments for extensions of degree $p^{4}$, it will help
to consider those of degree $p^{3}$. As discussed, we will only consider non-
cyclic groups of order $p^{3}$, of which there are four: $C_{p}^{3}$,
$C_{p^{2}}\times C_{p}$, the Heisenberg group $H(p^{3})$, and the Metacyclic
group $M(p^{3})$. These can all be expressed as

(1)
$G_{\vec{a}}=\langle\sigma_{1},\sigma_{2},\sigma_{3}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p}=\sigma_{3}^{a_{1}},\sigma_{2}^{p}=\sigma_{3}^{p}=1,\sigma_{3}\in
Z(G_{\vec{a}}),\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\rangle,$

where $\vec{a}=(a_{0},a_{1})$ takes on the values shown in Table 2.

Table 2: Values of $\vec{a}$
Group $a_{0}$ $a_{1}$ $C_{p}^{3}$ 0 0 $C_{p^{2}}\times C_{p}$ 0 1 $H(p^{3})$ 1
0 $M(p^{3})$ 1 1

Although in these groups, $a_{0}$ and $a_{1}$ only take on values of 0 or 1,
it will later be convenient to consider more general $a_{i}\in\mathbb{F}_{p}$
(any such choice of $a_{1}$ and $a_{2}$ will lead to a different
representation of one of these same groups). As such, we will allow $a_{1}$
and $a_{2}$ to take on values other than 0 or 1 in our results. This
representation will lead us to the desired result on the polynomials defining
the extension $K_{0}/K_{0}$, but first we need a lemma.

######  Lemma 2.1.

Let $K$ be a field of characteristic $p$ and $L/K$ a Galois extension of
degree $p$ with Galois group $G=\langle\sigma_{1}\rangle$. Let $x_{1}\in L$ be
such that $L=K(x_{1})$, $(\sigma_{1}-1)x_{1}=1$ and $\wp(x_{1})\in K$ (call
this element $\wp(x_{1})=\beta_{1}$). Let $C_{1}$ be the polynomial

$C_{1}(x_{1})=\frac{x_{1}^{p}+1-(x_{1}+1)^{p}}{p}=-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x_{1}^{i}\in\mathbb{F}_{p}[x_{1}].$

Then
$\mathrm{Tr}_{L/K}(C_{1}(x_{1}))=(1+\sigma_{1}+\dots+\sigma_{1}^{p-1})C_{1}(x_{1})=1$
and $\wp(C_{1}(x_{1}))=(\sigma_{1}-1)D_{1}(x_{1})$, where

$D_{1}(x_{1})=\frac{x_{1}^{p}+\beta_{1}^{p}-(x_{1}+\beta_{1})^{p}}{p}=-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x_{1}^{i}\beta_{1}^{p-i}\in\mathbb{F}_{p}[x_{1}].$

###### Proof.

First, note that $(\sigma_{1}-1)x_{1}=1$, i.e. $\sigma_{1}(x_{1})=x_{1}+1$,
implies that $\sigma_{1}^{k}(x_{1})=x_{1}+k$ for each $k\in\mathbb{N}$. Then
using the fact that $\sigma_{1}$ is an automorphism of $L$ which fixes $K$
(and thus $\mathbb{F}_{p}$), we have

$\mathrm{Tr}_{L/K}(C_{1}(x_{1}))=(1+\sigma_{1}+\dots+\sigma_{1}^{p-1})C_{1}(x_{1})=\sum_{j=0}^{p-1}\left(-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x_{1}+j)^{i}\right).$

To get the desired result, it will now be easiest to show a polynomial
identity in the ring $\mathbb{Q}[x]$.

$\displaystyle\sum_{j=0}^{p-1}\left(-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x+j)^{i}\right)$
$\displaystyle=\sum_{j=0}^{p-1}\frac{(x+j)^{p}+1-(x+j+1)^{p}}{p}$
$\displaystyle=\frac{x^{p}+1-(x+1)^{p}}{p}+\dots+\frac{(x+p-1)^{p}+1-(x+p)^{p}}{p}$
$\displaystyle=\frac{x^{p}-(x+p)^{p}+p}{p}$
$\displaystyle=1-\sum_{i=0}^{p-1}\frac{\binom{p}{i}}{p}x^{i}p^{p-i}.$

Note that since the two ends of this string of identities lie in
$\mathbb{Z}[x]$, this identity will still hold in this context. Furthermore,
we can reduce modulo $p$ and plug in $x=x_{1}$ to return to the context of
$\mathbb{F}[x_{1}]$:

$\mathrm{Tr}_{L/K}(C_{1}(x_{1}))=\sum_{j=0}^{p-1}\left(-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x_{1}+j)^{i}\right)=1.$

This shows the desired result for the trace; we now turn our attention to
$\wp(C_{1}(x_{1})).$ First, recall that $\wp$ is an $\mathbb{F}_{p}$-linear
map; then

$\displaystyle\wp(C_{1}(x_{1}))$ $\displaystyle=C_{1}(x_{1}^{p})-C_{1}(x_{1})$
$\displaystyle=C_{1}(x_{1}+\beta_{1})-C_{1}(x_{1})$
$\displaystyle=-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x_{1}+\beta_{1})^{i}+\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x_{1}^{i}.$

As before, simplifying this expression will be easier to do after some
intermediate steps carried out in $\mathbb{Q}[x,y]$:

$\displaystyle-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x+y)^{i}+\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x^{i}$
$\displaystyle=\frac{(x+y)^{p}+1-(x+y+1)^{p}}{p}-\frac{x^{p}+1-(x+1)^{p}}{p}$
$\displaystyle=\frac{(x+y)^{p}-(x+y+1)^{p}-x^{p}+(x+1)^{p}}{p}$
$\displaystyle=\frac{(x+1)^{p}+y^{p}-(x+y+1)^{p}}{p}-\frac{x^{p}+y^{p}-(x+y)^{p}}{p}$
$\displaystyle=-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x+1)^{i}y^{p-i}+\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x^{i}y^{p-i}.$

Again, note that both ends of this string of identities lie in
$\mathbb{Z}[x]$, so the identity will still hold true after reducing modulo
$p$ and substituting $x=x_{1}$ and $y=\beta_{1}$. Then

$\displaystyle\wp(C_{1}(x_{1}))$
$\displaystyle=-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x_{1}+\beta_{1})^{i}+\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x_{1}^{i}$
$\displaystyle=-\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}(x_{1}+1)^{i}\beta_{1}^{p-i}+\sum_{i=1}^{p-1}\frac{\binom{p}{i}}{p}x_{1}^{i}\beta_{1}^{p-i}$
$\displaystyle=(\sigma_{1}-1)D_{1}(x_{1}).$

∎

######  Proposition 2.2.

Let $K_{3}/K_{0}$ be a Galois extension of degree $p^{3}$ with Galois group
$\mathrm{Gal}(K_{3}/K_{0})\cong G_{\vec{a}}$ as in (1), with
$a_{0},a_{1}\in\mathbb{F}_{p}$ and $K_{2}/K_{0}$ as defined in the
introduction. That is, $K_{2}=K_{0}(x_{1},x_{2})$ is the fixed field of
$\langle\sigma_{3}\rangle\leq G_{\vec{a}}$, $K_{1}=K_{0}(x_{1})$ is the fixed
field of $\langle\sigma_{2},\sigma_{3}\rangle$, and $(\sigma_{i}-1)x_{i}=1$
for $1\leq i\leq 3$, with $\wp(x_{1})=\beta_{1}$ and $\wp(x_{2})=\beta_{2}$
for some $\beta_{1},\beta_{2}\in K_{0}$. Then we can select $x_{3}$ such that
$K_{3}=K_{0}(x_{1},x_{2},x_{3})$, with
$x_{3}^{p}-x_{3}=a_{0}\beta_{2}x_{1}+a_{1}D_{1}(x_{1})+\beta_{3}$ for some
$\beta_{3}\in K_{0}$, where $D_{1}$ is the Witt polynomial from Theorem 1.1.

Moreover, the converse holds; that is, if $K_{3}/K_{0}$ is an algebraic
extension of degree $p^{3}$ with $K_{3}=K_{0}(x_{1},x_{2},x_{3})$ and
$\wp(x_{i})$ as above for $i=1,2,3$, then $K_{3}/K_{0}$ is a Galois extension
with Galois group $G_{\vec{a}}$.

###### Proof.

First, recall that since $\mathrm{Gal}(K_{2}/K_{0})$ is a $C_{p}^{2}$ group,
we may choose $x_{2}\in K_{2}$ such that $(\sigma_{1}-1)x_{2}=0$. Also note
that $\mathrm{Gal}(K_{3}/K_{1})=\langle\sigma_{2},\sigma_{3}\rangle\cong
C_{p}^{2}$. Then when considering the extension $K_{3}/K_{1}$, we can choose
$x_{3}\in K_{3}$ such that $(\sigma_{2}-1)x_{3}=0$. Then the only unknown part
of the group action is $\sigma_{1}$ on $x_{3}$. Let $A=(\sigma_{1}-1)x_{3}\in
K_{3}$. Since $\sigma_{3}\in Z(G_{\vec{a}})$,
$(\sigma_{3}-1)A=(\sigma_{1}-1)(\sigma_{3}-1)x_{3}=(\sigma_{1}-1)1=0$. Thus,
$A\in K_{2}$. Now using a known identity from the group definition, we can
derive the following identity:

$\displaystyle x_{3}+\sigma_{2}(A)$ $\displaystyle=\sigma_{2}(x_{3}+A)$
$\displaystyle=\sigma_{2}\sigma_{1}(x_{3})$
$\displaystyle=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}(x_{3})$
$\displaystyle=\sigma_{1}\sigma_{2}(x_{3}+a_{0})$
$\displaystyle=\sigma_{1}(x_{3}+a_{0})$ $\displaystyle=x_{3}+A+a_{0}.$

Then $(\sigma_{2}-1)A=a_{0}$. Therefore, $(\sigma_{2}-1)(A-a_{0}x_{2})=0$, so
$A-a_{0}x_{2}\in K_{1}$.

Now
$\mathrm{Tr}_{K_{1}/K_{0}}(A-a_{0}x_{2})=(\sigma_{1}^{p}-1)x_{3}=(\sigma_{3}^{a_{1}}-1)x_{3}=a_{1}$.
Then by the lemma,
$\mathrm{Tr}_{K_{1}/K_{0}}(A-a_{0}x_{2}-a_{1}C_{1}(x_{1}))=0$. Using the
additive version of Hilbert’s Theorem 90, we then know that there is some
$k_{1}\in K_{1}$ such that
$(\sigma_{1}-1)k_{1}=A-a_{0}x_{2}-a_{1}C_{1}(x_{1})$. Then
$(\sigma_{1}-1)(x_{3}-k_{1})=a_{0}x_{2}+a_{1}C_{1}(x_{1})$. We can then
replace $x_{3}$ with $x_{3}-k_{1}$, which maintains all prior relationships
and establishes

(2) $(\sigma_{1}-1)x_{3}=a_{0}x_{2}+a_{1}C_{1}(x_{1}).$

It will also be helpful to use (2) to establish another identity. For this,
note that in characteristic $p$,
$1+\sigma_{1}+\dots+\sigma_{1}^{p-1}=(\sigma_{1}-1)^{p-1}$.

(3) $\displaystyle(1+\sigma_{1}+\dots+\sigma_{1}^{p-1})x_{3}$
$\displaystyle=(\sigma_{1}-1)^{p-2}(a_{0}x_{2}+a_{1}C_{1}(x_{1}))$
$\displaystyle=a_{1}(\sigma_{1}-1)^{p-2}C_{1}(x_{1}).$

Now consider $B=\wp(x_{3})\in K_{2}$. Recall that $\wp$ is an
$\mathbb{F}_{p}$-linear map and an additive homomorphism which commutes with
the elements of $G_{\vec{a}}$; then
$(\sigma_{2}-1)B=\wp((\sigma_{2}-1)x_{3})=0$, so $B\in K_{1}$. Now note that:

$\displaystyle(\sigma_{1}-1)B$ $\displaystyle=\wp((\sigma_{1}-1)x_{3})$
$\displaystyle=\wp(a_{0}x_{2}+a_{1}C_{1}(x_{1}))$
$\displaystyle=a_{0}\beta_{2}+a_{1}(\sigma_{1}-1)D_{1}(x_{1})$
$\displaystyle=(\sigma_{1}-1)(a_{0}\beta_{2}x_{1}+a_{1}D_{1}(x_{1})),$

so $(\sigma_{1}-1)(B-a_{0}\beta_{2}x_{1}-a_{1}D_{1}(x_{1}))=0$. For ease of
notation, define $A_{1}\mathrel{\mathop{\mathchar 58\relax}}=\beta_{2}x_{1}$.
Then $B-a_{0}A_{1}-a_{1}D_{1}(x_{1})=\beta_{3}$ for some $\beta_{3}\in K_{0}$.
Therefore,

$B=\wp(x_{3})=a_{0}A_{1}+a_{1}D_{1}(x_{1})+\beta_{3}.$

Now for the converse, let $K_{3}/K_{0}$ be an algebraic extension of degree
$p^{3}$ and $K_{3}=K_{0}(x_{1},x_{2},x_{3})$, with
$\wp(x_{3})=a_{0}A_{1}+a_{1}D_{1}(x_{1})+\beta_{3}$, $\wp(x_{2})=\beta_{2}$,
and $\wp(x_{1})=\beta_{1}$, where $\beta_{i}\in K_{0}$,
$A_{1}=\beta_{2}x_{1}$, and $a_{i}\in\\{0,1\\}$. Define
$G=\mathrm{Gal}(K_{3}/K_{0})$ (the group of automorphisms of $K_{3}$ which fix
$K_{0}$ pointwise), $K_{1}=K_{0}(x_{1})$ and $K_{2}=K_{1}(x_{2})$ as used
previously, so that $K_{3}=K_{2}(x_{3})$. Then each $K_{i}/K_{i-1}$ is an
extension of degree $p$ for $1\leq i\leq 3$, and the Artin-Schreier equations
given are exactly the minimal polynomials for each $x_{i}$ over $K_{i-1}$.
Immediately, we know that $K_{2}/K_{0}$ is a Galois extension of order $p^{2}$
(and in fact, is a $C_{p}^{2}$ extension), since $x_{1}$ and $x_{2}$ each
satisfy an Artin-Schreier polynomial over $K_{0}$. We also have that $K_{3}$
is trivially Galois over both $K_{2}$ and $K_{1}(x_{3})$. Then we need to show
that $K_{3}$ is Galois over $K_{0}$.

Let $\sigma_{3}\in\mathrm{Gal}(K_{3}/K_{2})\subseteq G$ be such that
$\sigma_{3}(x_{3})=x_{3}+1$. Then $\sigma_{3}(x_{1})=x_{1}$, and
$\sigma_{3}(x_{2})=x_{2}$. Also let
$\sigma_{2}\in\mathrm{Gal}(K_{3}/K_{1}(x_{3}))\subseteq G$ such that
$\sigma_{2}(x_{2})=x_{2}+1$. Then $\sigma_{2}(x_{3})=x_{3}$, and
$\sigma_{2}(x_{1})=x_{1}$. Now let $\sigma_{1}$ be an embedding of $K_{3}$
into its algebraic closure which fixes $K_{0}$ pointwise and for which
$\sigma_{1}(x_{1})\neq x_{1}$. Then

$\sigma_{1}(x_{1})^{p}-\sigma_{1}(x_{1})-\beta_{1}=\sigma_{1}(x_{1}^{p}-x_{1}-\beta_{1})=0,$

so $\sigma_{1}(x_{1})$ must be another root of $x^{p}-x-\beta_{1}$; then
$\sigma_{1}(x_{1})=x_{1}+n$ for some $n\in\mathbb{F}_{p}$. Without loss of
generality, assume $\sigma_{1}(x_{1})=x_{1}+1$ (if not, some power of
$\sigma_{1}$ works). Furthermore, we know that $\sigma_{1}(x_{2})=x_{2}+m$ for
some $m\in\mathbb{F}_{p}$. Without loss of generality, we can assume that
$m=0$ (if not, multiply $\sigma_{1}$ by a power of $\sigma_{2}$). Finally,
note that

$\displaystyle\wp(\sigma_{1}(x_{3}))$ $\displaystyle=\sigma_{1}(\wp(x_{3}))$
$\displaystyle=\sigma_{1}(a_{0}A_{1}+a_{1}D_{1}(x_{1})+\beta_{3})$
$\displaystyle=a_{0}A_{1}+a_{0}\beta_{2}+a_{1}D_{1}(x_{1}+1)+\beta_{3}.$

That is, $\sigma_{1}(x_{3})$ is a root of the polynomial
$x^{p}-x-(a_{0}\beta_{2}(x_{1}+1)+a_{1}D_{1}(x_{1}+1)+\beta_{3})$, a degree
$p$ polynomial over $K_{1}$. The $p$ roots of this polynomial are of the form
$x_{3}+a_{0}x_{2}+a_{1}C_{1}(x_{1})+n$, where $n\in\mathbb{F}_{p}$ (using
identities shown above). Then $\sigma_{1}(x_{3})$ must be of this form, and
therefore lies in $K_{3}$. Without loss of generality, we can assume that
$n=0$ (if not, multiply $\sigma_{1}$ by the appropriate power of
$\sigma_{3}$). Now note that $\sigma_{1}(x_{1})=x_{1}+1$,
$\sigma_{1}(x_{2})=x_{2}$, and
$\sigma_{1}(x_{3})=x_{3}+a_{0}x_{2}+a_{1}C_{1}(x_{1})$ (and $\sigma_{1}$ fixes
$K_{0}$ and respects addition and multiplication), so $\sigma_{1}$ is an
automorphism of $K_{3}$ which fixes $K_{0}$, i.e. $\sigma_{1}\in G$. However,
since $\sigma_{1}$ does not fix $x_{1}$,
$\sigma_{1}\notin\langle\sigma_{2},\sigma_{3}\rangle$. Then since
$\mathinner{\\!\left\lvert\langle\sigma_{2},\sigma_{3}\rangle\right\rvert}=p^{2}$
and $\mathinner{\\!\left\lvert G\right\rvert}$ must divide $p^{3}$, it follows
that $\mathinner{\\!\left\lvert G\right\rvert}=p^{3}$ and
$G=\langle\sigma_{1},\sigma_{2},\sigma_{3}\rangle$; in particular, this means
that $K_{3}$ is Galois over $K_{0}$. Furthermore, we have

$\displaystyle\sigma_{1}^{p}(x_{1})$ $\displaystyle=x_{1}+p$
$\displaystyle=x_{1},$ $\displaystyle\sigma_{1}^{p}(x_{2})$
$\displaystyle=x_{2},$ $\displaystyle\sigma_{1}^{p}(x_{3})$
$\displaystyle=x_{3}+a_{0}(px_{2})+a_{1}(C_{1}(x_{1})+C_{1}(x_{1}+1)+\dots+C_{1}(x_{1}+p-1))$
$\displaystyle=x_{3}+a_{1}.$

Therefore, $\sigma_{1}^{p}=\sigma_{3}^{a_{1}}$. Finally, note that
$\sigma_{1}$ commutes with $\sigma_{3}$ and
$\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}$. Therefore, we
have

$G=G_{\vec{a}}=\langle\sigma_{1},\sigma_{2},\sigma_{3}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p}=\sigma_{3}^{a_{1}},\sigma_{2}^{p}=\sigma_{3}^{p}=1,\sigma_{3}\in
Z(G_{\vec{a}}),\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\rangle.$

This completes the proof of the converse.

∎

##  3\. Extensions of degree $p^{4}$

We can now discuss the Artin-Schreier equations for the non-Abelian degree
$p^{4}$ extensions discussed in the introduction. Note that since
$K_{3}/K_{0}$ is a degree $p^{3}$ extension, we already know $\wp(x_{3})$ (and
more trivially, $\wp(x_{1})$ and $\wp(x_{2})$). We will first focus on the
first nine cases. As discussed in the introduction, these Galois groups are
all of the form

(4) $\displaystyle G_{\vec{a},\vec{b}}=\langle\sigma_{1}$
$\displaystyle,\sigma_{2},\sigma_{3},\sigma_{4}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p}=\sigma_{3}^{a_{1}}\sigma_{4}^{b_{1}},\sigma_{2}^{p}=\sigma_{4}^{b_{2}},\sigma_{3}^{p}=\sigma_{4}^{p}=1,\sigma_{4}\in
Z(G_{\vec{a},\vec{b}}),$
$\displaystyle\sigma_{3}\sigma_{2}=\sigma_{2}\sigma_{3},\sigma_{3}\sigma_{1}=\sigma_{1}\sigma_{3}\sigma_{4}^{b_{3}},\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\sigma_{4}^{b_{0}}\rangle,$

where $\vec{a}=(a_{0},a_{1})$ and $\vec{b}=(b_{0},b_{1},b_{2},b_{3})$ take on
the values shown in Table 1 above. Recall that $\alpha$ refers to any
quadratic non-residue (mod p), and in the case of $p=3$, the only non-residue
is $\alpha=2\equiv-1$. This representation leads us to our main proposition:

######  Theorem 3.1.

Let $K_{4}/K_{0}$ be a Galois extension of degree $p^{4}$ with Galois group
$G_{\vec{a},\vec{b}}$ as shown in equation (4) above, where
$a_{0},a_{1},b_{0},b_{1},b_{3}\in\\{0,1\\}$, $b_{2}\in\\{0,1,\alpha\\}$,
$\alpha$ is a quadratic non-residue modulo $p$, and all notation is as above.
Assume that $K_{3}/K_{0}$, $x_{1}$, $x_{2}$, and $x_{3}$ are as in Proposition
2.2. Then we can select $x_{4}$ such that $K_{4}=K_{3}(x_{4})$ and

$x_{4}^{p}-x_{4}=b_{0}\beta_{2}x_{1}+b_{1}D_{1}(x_{1})+b_{2}D_{1}(x_{2})+b_{3}\left(a_{0}\beta_{2}\binom{x_{1}}{2}+\beta_{3}x_{1}\right)+\beta_{4},$

where $\beta_{4}\in K_{0}$ and $D_{1}$ is the Witt polynomial from Theorem
1.1.

Moreover, the converse holds; that is, if $K_{4}/K_{0}$ is an algebraic
extension of degree $p^{4}$ with $K_{4}=K_{0}(x_{1},x_{2},x_{3},x_{4})$ and
$\wp(x_{i})$ as above for $i=1,2,3,4$, then $K_{4}/K_{0}$ is a Galois
extension with Galois group $G_{\vec{a},\vec{b}}$.

###### Proof.

First, since $\mathrm{Gal}(K_{3}/K_{0})$ is as in Proposition 2.2, we have
$\wp(x_{3})=a_{0}A_{1}+a_{1}D_{1}(x_{1})+\beta_{3}$, with
$(\sigma_{2}-1)x_{3}=0$ and
$(\sigma_{1}-1)x_{3}=a_{0}x_{2}+a_{1}C_{1}(x_{1})$. Note that $K_{4}/K_{1}$ is
another extension of degree $p^{3}$. From the expression of
$G_{\vec{a},\vec{b}}$, we can see that the Galois group for $K_{4}/K_{1}$ is
as shown in equation (1) in Section 2, with $a_{0}=0$ and $a_{1}=b_{2}$ (i.e.
the Galois group is either $C_{p}^{3}$ or $C_{p^{2}}\times C_{p}$, both of
which are Abelian). Then by Proposition 2.2, we can choose $x_{4}\in K_{4}$
such that $(\sigma_{3}-1)x_{4}=0$ and $(\sigma_{2}-1)x_{4}=b_{2}C_{1}(x_{2})$.

Now all that remains is to determine the action of $\sigma_{1}$ on $x_{4}$. To
that end, let $A=(\sigma_{1}-1)x_{4}\in K_{4}$. Since $\sigma_{4}\in
Z(G_{\vec{a},\vec{b}})$,
$(\sigma_{4}-1)A=(\sigma_{1}-1)(\sigma_{4}-1)x_{4}=(\sigma_{1}-1)1=0$. Then
$A\in K_{3}$. Now using a known identity from the group definition, we can
derive the following identity:

$\displaystyle x_{4}+\sigma_{3}(A)$ $\displaystyle=\sigma_{3}(x_{4}+A)$
$\displaystyle=\sigma_{3}\sigma_{1}(x_{4})$
$\displaystyle=\sigma_{1}\sigma_{3}\sigma_{4}^{b_{3}}(x_{4})$
$\displaystyle=\sigma_{1}\sigma_{3}(x_{4}+b_{3})$
$\displaystyle=\sigma_{1}(x_{4}+b_{3})$ $\displaystyle=x_{4}+A+b_{3}$

Then $(\sigma_{3}-1)A=b_{3}$. Therefore, $(\sigma_{3}-1)(A-b_{3}x_{3})=0$, so
$A-b_{3}x_{3}\in K_{2}$. Now using another identity from the group definition:

$\displaystyle x_{4}+b_{2}C_{1}(x_{2})+\sigma_{2}(A)$
$\displaystyle=\sigma_{2}(x_{4}+A)$
$\displaystyle=\sigma_{2}\sigma_{1}(x_{4})$
$\displaystyle=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\sigma_{4}^{b_{0}}(x_{4})$
$\displaystyle=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}(x_{4}+b_{0})$
$\displaystyle=\sigma_{1}\sigma_{2}(x_{4}+b_{0})$
$\displaystyle=\sigma_{1}(x_{4}+b_{2}C_{1}(x_{2})+b_{0})$
$\displaystyle=x_{4}+A+b_{2}C_{1}(x_{2})+b_{0}$

Then $(\sigma_{2}-1)A=b_{0}$. Since $x_{3}$ is fixed by $\sigma_{2}$,
$(\sigma_{2}-1)(A-b_{0}x_{2}-b_{3}x_{3})=0$, so $A-b_{0}x_{2}-b_{3}x_{3}\in
K_{1}$.

Now using the action of $\sigma_{1}$ on each element as well as equation (3)
in Proposition 2.2,
$\mathrm{Tr}_{K_{1}/K_{0}}(A-b_{0}x_{2}-b_{3}x_{3})=(\sigma_{1}^{p}-1)x_{4}-p(b_{0}x_{2})-a_{1}b_{3}(\sigma_{1}-1)^{p-2}C_{1}(x_{1})=(\sigma_{3}^{a_{1}}\sigma_{4}^{b_{1}}-1)x_{4}=b_{1}$.
It should be noted that this holds because either $a_{1}$ or $b_{3}$ must be
0, as seen in Table 3 (the fact that one of these constants must be 0 can be
shown more generally from the group definition, though this is left to the
reader). Then using the identity $\mathrm{Tr}_{K_{1}/K_{0}}(C_{1}(x_{1}))=1$,
shown in Lemma 2.1,
$\mathrm{Tr}_{K_{1}/K_{0}}(A-b_{0}x_{2}-b_{1}C_{1}(x_{1})-b_{3}x_{3})=0$.
Using the additive version of Hilbert’s Theorem 90, we then know that there is
some $k_{1}\in K_{1}$ such that
$(\sigma_{1}-1)k_{1}=A-b_{0}x_{2}-b_{1}C_{1}(x_{1})-b_{3}x_{3}$. Then
$(\sigma_{1}-1)(x_{4}-k_{1})=b_{0}x_{2}+b_{1}C_{1}(x_{1})+b_{3}x_{3}$. We can
now replace $x_{4}$ with $x_{4}-k_{1}$, which maintains all prior
relationships and establishes the identity

(5) $(\sigma_{1}-1)x_{4}=b_{0}x_{2}+b_{1}C_{1}(x_{1})+b_{3}x_{3}.$

Now consider $B=\wp(x_{4})\in K_{3}$. Recall that $\wp$ is an
$\mathbb{F}_{p}$-linear map and an additive homomorphism which commutes with
the elements of $G_{\vec{a},\vec{b}}$. Then
$(\sigma_{3}-1)B=\wp((\sigma_{3}-1)x_{4})=0$. Then $B\in K_{2}$. Now note that

$\displaystyle(\sigma_{2}-1)B$ $\displaystyle=\wp((\sigma_{2}-1)x_{4})$
$\displaystyle=\wp(b_{2}C_{1}(x_{2}))$
$\displaystyle=(\sigma_{2}-1)(b_{2}D_{1}(x_{2})),$

so $(\sigma_{2}-1)(B-b_{2}D_{1}(x_{2}))=0$. Then $B-b_{2}D_{1}(x_{2})\in
K_{1}$. Now note the following, recalling that $a_{0}b_{3}=0$:

$\displaystyle(\sigma_{1}-1)(B-b_{2}D_{1}(x_{2}))$
$\displaystyle=\wp((\sigma_{1}-1)x_{4})$
$\displaystyle=\wp(b_{0}x_{2}+b_{1}C_{1}(x_{1})+b_{3}x_{3})$
$\displaystyle=b_{0}\beta_{2}+b_{1}(\sigma_{1}-1)D_{1}(x_{1})+b_{3}(a_{0}\beta_{2}x_{1}+a_{1}D_{1}(x_{1})+\beta_{3})$
$\displaystyle=(\sigma_{1}-1)\left(b_{0}\beta_{2}x_{1}+b_{1}D_{1}(x_{1})+b_{3}\left(a_{0}\beta_{2}\binom{x_{1}}{2}+\beta_{3}x_{1}\right)\right).$

For ease of notation, define $A_{1}=\beta_{2}x_{1}$,
$A_{2}=\beta_{2}\binom{x_{1}}{2}$, and $A_{3}=\beta_{3}x_{1}$. Then
$(\sigma_{1}-1)(B-b_{0}A_{1}-b_{1}D_{1}(x_{1})-b_{2}D_{1}(x_{2})-b_{3}(a_{0}A_{2}+A_{3}))=0$.
Therefore,
$B-b_{0}A_{1}-b_{1}D_{1}(x_{1})-b_{2}D_{1}(x_{2})-b_{3}(a_{0}A_{2}+A_{3})=\beta_{4}$
for some $\beta_{4}\in K_{0}$. Then

$B=b_{0}A_{1}+b_{1}D_{1}(x_{1})+b_{2}D_{1}(x_{2})+b_{3}(a_{0}A_{2}+A_{3})+\beta_{4}.$

Now for the converse, let $K_{4}/K_{0}$ be an algebraic extension of degree
$p^{4}$ and $K_{4}=K_{0}(x_{1},x_{2},x_{3},x_{4})$, with
$\wp(x_{4})=b_{0}A_{1}+b_{1}D_{1}(x_{1})+b_{2}D_{1}(x_{2})+b_{3}(a_{0}A_{2}+A_{3})+\beta_{4}$,
$\wp(x_{3})=a_{0}A_{1}+a_{1}D_{1}(x_{1})+\beta_{3}$, $\wp(x_{2})=\beta_{2}$,
$\wp(x_{1})=\beta_{1}$, where $\beta_{i}\in K_{0}$, $A_{1}=\beta_{2}x_{1}$,
$A_{2}=\beta_{2}\binom{x_{1}}{2}$, $A_{3}=\beta_{3}x_{1}$,
$a_{i}\in\\{0,1\\}$, $b_{i}\in\\{0,1\\}$ for $i\neq 2$, and
$b_{2}\in\\{0,1,\alpha\\}$, where $\alpha$ is a quadratic non-residue modulo
$p$. Define $G=\mathrm{Gal}(K_{4}/K_{0})$ (the group of automorphisms of
$K_{4}$ which fix $K_{0}$ pointwise), $K_{1}=K_{0}(x_{1})$,
$K_{2}=K_{1}(x_{2})$, and $K_{3}=K_{2}(x_{3})$ as used previously, so that
$K_{4}=K_{3}(x_{4})$. Then each $K_{i}/K_{i-1}$ is an extension of degree $p$
for $1\leq i\leq 4$, and the Artin-Schreier equations given are exactly the
minimal polynomials for each $x_{i}$ over $K_{i-1}$. Immediately from the
proof of Proposition 2.2, we know that $K_{3}/K_{0}$ and $K_{4}/K_{1}$ are
Galois extensions of order $p^{3}$. We also have that $K_{4}$ is trivially
Galois over both $K_{3}$ and $K_{2}(x_{4})$. Then we need to show that $K_{4}$
is Galois over $K_{0}$.

Let $\sigma_{4}\in\mathrm{Gal}(K_{4}/K_{3})\subset G$ such that
$\sigma_{4}(x_{4})=x_{4}+1$. Then $\sigma_{4}$ fixes $x_{1}$, $x_{2}$, and
$x_{3}$. Also let $\sigma_{3}\in\mathrm{Gal}(K_{4}/K_{2}(x_{4}))$ such that
$\sigma_{3}(x_{3})=x_{3}+1$. Then $\sigma_{3}$ fixes $x_{1}$, $x_{2}$, and
$x_{4}$. Using the proof of Proposition 2.2, we also have
$\sigma_{2}\in\mathrm{Gal}(K_{4}/K_{1})$ such that
$\sigma_{2}(x_{2})=x_{2}+1$, $\sigma_{2}(x_{3})=x_{3}$, and
$\sigma_{2}(x_{4})=x_{3}+b_{2}C_{1}(x_{2})$. Now let $\sigma_{1}$ be an
embedding of $K_{4}$ into its algebraic closure which fixes $K_{0}$ pointwise
and for which $\sigma_{1}(x_{1})\neq x_{1}$. Then $\sigma_{1}|_{K_{3}}$ is an
element of $\mathrm{Gal}(K_{3}/K_{0})$ which does not fix $x_{1}$. Then we can
again use the proof of Proposition 2.2 to assume that
$\sigma_{1}(x_{1})=x_{1}+1$, $\sigma_{1}(x_{2})=x_{2}$, and
$\sigma_{1}(x_{3})=x_{3}+a_{0}x_{2}+a_{1}C_{1}(x_{1})$. Now note the
following, recalling that $a_{0}b_{3}=0$:

$\displaystyle\wp(\sigma_{1}(x_{4}))=$ $\displaystyle\sigma_{1}(\wp(x_{4}))$
$\displaystyle=$
$\displaystyle\sigma_{1}(b_{0}A_{1}+b_{1}D_{1}(x_{1})+b_{2}D_{1}(x_{2})+b_{3}(a_{0}A_{2}+A_{3})+\beta_{4})$
$\displaystyle=$ $\displaystyle
b_{0}(A_{1}+\beta_{2})+b_{1}D_{1}(x_{1}+1)+b_{2}D_{1}(x_{2})$
$\displaystyle+b_{3}\left(a_{0}\beta_{2}\binom{x_{1}+1}{2}+A_{3}+\beta_{3}\right)+\beta_{4}$

That is, $\sigma_{1}(x_{4})$ is a root of the polynomial
$x^{p}-x-\wp(\sigma_{1}(x_{4}))$, where $\wp(\sigma_{1}(x_{4}))$ is the
expression derived above. This is a degree $p$ polynomial over $K_{2}$, and
the $p$ distinct elements $x_{4}+b_{0}x_{2}+b_{1}C_{1}(x_{1})+b_{3}x_{3}+n$,
with $n\in\mathbb{F}_{p}$, are roots. Then these are the only roots, and so
$\sigma_{1}(x_{4})$ is of this form for some $n\in\mathbb{F}_{p}$ (and thus
lies in $K_{4}$). Without loss of generality, we will assume that $n=0$ (if
not, multiply $\sigma_{1}$ by the appropriate power of $\sigma_{4}$). Now note
that $\sigma_{1}(x_{1})=x_{1}+1$, $\sigma_{1}(x_{2})=x_{2}$,
$\sigma_{1}(x_{3})=x_{3}+a_{0}x_{2}+a_{1}C_{1}(x_{1})$, and
$\sigma_{1}(x_{4})=x_{4}+b_{0}x_{2}+b_{1}C_{1}(x_{1})+b_{3}x_{3}$ (and
$\sigma_{1}$ fixes $K_{0}$ and respects addition and multiplication), so
$\sigma_{1}$ is an automorphism of $K_{4}$ which fixes $K_{0}$, i.e.
$\sigma_{1}\in G$. However, since $\sigma_{1}$ does not fix $x_{1}$,
$\sigma_{1}\notin\langle\sigma_{2},\sigma_{3},\sigma_{4}\rangle=\mathrm{Gal}(K_{4}/K_{1})$.
Then since $\mathinner{\\!\left\lvert
G\right\rvert}>\mathinner{\\!\left\lvert\mathrm{Gal}(K_{4}/K_{1})\right\rvert}=p^{3}$
and $\mathinner{\\!\left\lvert G\right\rvert}$ must divide $p^{4}$, it follows
that $\mathinner{\\!\left\lvert G\right\rvert}=p^{4}$ and
$G=\langle\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\rangle$; in particular,
this means that $K_{4}$ is Galois over $K_{0}$. Furthermore, we have

$\displaystyle\sigma_{1}^{p}(x_{1})=$ $\displaystyle x_{1},$
$\displaystyle\sigma_{1}^{p}(x_{2})=$ $\displaystyle x_{2},$
$\displaystyle\sigma_{1}^{p}(x_{3})=$ $\displaystyle x_{3}+a_{1},$
$\displaystyle\sigma_{1}^{p}(x_{4})=$ $\displaystyle
x_{4}+pb_{0}x_{2}+b_{1}(C_{1}(x_{1})+\dots+C_{1}(x_{1}+p-1))$
$\displaystyle+a_{1}b_{3}\left((p-1)C_{1}(x_{1})+(p-2)C_{1}(x_{1}+1)+\dots+C_{1}(x_{1}+p-2)\right)$
$\displaystyle=$ $\displaystyle x_{4}+b_{1}.$

Note that the first three equalities here come directly from the proof of
Proposition 2.2. Then $\sigma_{1}^{p}=\sigma_{3}^{a_{1}}\sigma_{4}^{b_{1}}$.
Furthermore, it can be quickly verified that
$\sigma_{4}\sigma_{1}=\sigma_{1}\sigma_{4}$,
$\sigma_{3}\sigma_{1}=\sigma_{1}\sigma_{3}\sigma_{4}^{b_{3}}$, and
$\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\sigma_{4}^{b_{0}}$.
Then using these facts and those we know about
$\mathrm{Gal}(K_{4}/K_{1})\subset G$ from Proposition 2.2, we have

$\displaystyle G=G_{\vec{a},\vec{b}}=\langle\sigma_{1}$
$\displaystyle,\sigma_{2},\sigma_{3},\sigma_{4}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p}=\sigma_{3}^{a_{1}}\sigma_{4}^{b_{1}},\sigma_{2}^{p}=\sigma_{4}^{b_{2}},\sigma_{3}^{p}=\sigma_{4}^{p}=1,\sigma_{4}\in
Z(G_{\vec{a},\vec{b}}),$
$\displaystyle\sigma_{3}\sigma_{2}=\sigma_{2}\sigma_{3},\sigma_{3}\sigma_{1}=\sigma_{1}\sigma_{3}\sigma_{4}^{b_{3}},\sigma_{2}\sigma_{1}=\sigma_{1}\sigma_{2}\sigma_{3}^{a_{0}}\sigma_{4}^{b_{0}}\rangle,$

exactly as presented in equation (4) above. This completes the proof of the
converse.

∎

We will now consider degree $p^{4}$ extensions with the final Galois group. To
do so, we will first need a lemma similar to Lemma 2.1.

######  Lemma 3.2.

Let $K$ be a field of characteristic $p$ and $L/K$ a Galois extension of
degree $p^{2}$ with Galois group $G=\langle\sigma_{1}\rangle$. Let
$x_{1},x_{3}\in L$ be such that $L=K(x_{1},x_{3})$, $(\sigma_{1}-1)x_{1}=1$,
$(\sigma_{1}-1)x_{3}=C_{1}(x_{1})$, $\wp(x_{1})=\beta_{1}\in K$, and
$\wp(x_{3})=D_{1}(x_{1})+\beta_{3}$ for some $\beta_{3}\in K$, with
$C_{1},D_{1}\in\mathbb{F}_{p}[x_{1}]$ as defined in Lemma 2.1. Let $C_{2}$ be
the polynomial

$C_{2}(x_{1},x_{3})=\frac{x_{1}^{p^{2}}+1-(x_{1}+1)^{p^{2}}+p(x_{3}^{p}-(x_{3}+C_{1}(x_{1}))^{p})}{p^{2}}\in\mathbb{F}[x_{1},x_{3}].$

Then
$\mathrm{Tr}_{L/K}(C_{2}(x_{1},x_{3}))=(1+\sigma_{1}+\dots+\sigma_{1}^{p^{2}-1})C_{2}(x_{1},x_{3})=1$
and $\wp(C_{2}(x_{1},x_{3}))=(\sigma_{1}-1)D_{2}(x_{1},x_{3})$, where

$D_{2}(x_{1},x_{3})=\frac{x_{1}^{p^{2}}+\beta_{1}^{p^{2}}-(x_{1}+\beta_{1})^{p^{2}}+p\left(x_{3}^{p}+\beta_{3}^{p}-(x_{3}+D_{1}(x_{1})+\beta_{3})^{p}\right)}{p^{2}}\in\mathbb{F}_{p}[x_{1},x_{3}].$

###### Proof.

This proof will follow much like that of Lemma 2.1. As before, some steps will
actually require using polynomial identities in polynomial rings over
$\mathbb{Q}$. For details on how such methods work, see Lemma 2.1; here, a
simplified version that slightly abuses notation will be presented to avoid
overly complicated expressions.

First, note that as before, $\sigma_{1}^{k}(x_{1})=x_{1}+k$ for every
$k\in\mathbb{N}$. Furthermore, since $\sigma_{1}$ is an automorphism of $L$
which fixes $K$ and $\sigma_{1}(x_{3})=x_{3}+C_{1}(x_{1})$, it is
straightforward to see that
$\sigma_{1}^{k}(x_{3})=x_{3}+\sum_{i=0}^{k-1}C_{1}(x_{1}+i)$ for
$k\in\mathbb{N}$. Then

$\displaystyle\mathrm{Tr}_{L/K}(C_{2}(x_{1},x_{3}))$
$\displaystyle=\sum_{i=0}^{p^{2}-1}\frac{(x_{1}+i)^{p^{2}}+1-(x_{1}+i+1)^{p^{2}}}{p^{2}}$
$\displaystyle\quad+\sum_{i=0}^{p^{2}-1}\frac{p((x_{3}+\sum_{j=0}^{i-1}C_{1}(x_{1}+j))^{p}-(x_{3}+\sum_{j=0}^{i}C_{1}(x_{1}+j))^{p})}{p^{2}}$
$\displaystyle=1+\frac{x_{1}^{p^{2}}-(x_{1}+p^{2})^{p^{2}}}{p^{2}}+\frac{x_{3}^{p}-(x_{3}+\sum_{j=0}^{p^{2}-1}C_{1}(x_{1}+j))^{p}}{p}$

Note that
$x_{1}^{p^{2}}-(x_{1}+p^{2})^{p^{2}-1}=-\sum_{i=0}^{p^{2}}\binom{p^{2}-1}{i}x_{1}^{i}p^{2(p^{2}-i)}$.
Note that aside from when $i=p^{2}-1$, each of these summands has at least a
factor of $p^{3}$. Since $\binom{p^{2}}{p^{2}-1}=p^{2}$, then in fact every
summand has a factor of at least $p^{3}$. Then since $L$ is a field of
characteristic $p$, the first fraction in the above expression is equal to 0.
For the second fraction, it will help to use the definition of $C_{1}$:

$\displaystyle\sum_{j=0}^{p^{2}-1}C_{1}(x_{1}+j)$
$\displaystyle=\sum_{j=0}^{p^{2}-1}\frac{(x_{1}+j)^{p}+1-(x_{1}+j+1)^{p}}{p}$
$\displaystyle=\frac{x_{1}^{p}+p^{2}-(x_{1}+p^{2})^{p}}{p}$
$\displaystyle=p-\sum_{i=0}^{p-1}\frac{\binom{p}{i}}{p}x_{1}^{i}p^{2(p-i)}$
$\displaystyle=pz.$

Here, we simply note that the expression is divisible by $p$, then denote by
$z$ the result after factoring out $p$. Then the second fraction in the above
expression for $\mathrm{Tr}_{L/K}(x_{1},x_{3})$ becomes

$\frac{x_{3}^{p}-(x_{3}+pz)^{p}}{p}=\frac{x_{3}^{p}-x_{3}^{p}-\sum_{i=0}^{p-1}\binom{p}{i}x_{3}^{i}(pz)^{p-i}}{p}\equiv
0\textup{ (mod }p).$

Then $\mathrm{Tr}_{L/K}(C_{1}(x_{1},x_{3}))=1$.

We must now show that
$\wp(C_{2}(x_{1},x_{3}))=(\sigma_{1}-1)D_{2}(x_{1},x_{3})$. To do so, it will
help to keep in mind that $\wp$ is an $\mathbb{F}_{p}$ linear map. Since
$C_{2}(x_{1},x_{3})\in\mathbb{F}_{p}[x_{1},x_{3}]$, we know that
$\wp(C_{2}(x_{1},x_{3}))=C_{2}(x_{1}^{p},x_{3}^{p})-C_{2}(x_{1},x_{3})$. Then:

$\displaystyle\wp(C_{2}(x_{1},x_{3}))$
$\displaystyle=-C_{2}(x_{1},x_{3})+\frac{(x_{1}+\beta_{1})^{p^{2}}+1-(x_{1}+\beta_{1}+1)^{p^{2}}}{p^{2}}$
$\displaystyle\quad+\frac{p((x_{3}+D_{1}(x_{1})+\beta_{3})^{p}-(x_{3}+C_{1}(x_{1})+D_{1}(x_{1}+1)+\beta_{3})^{p})}{p^{2}}$
$\displaystyle=\frac{(x_{1}+\beta_{1})^{p^{2}}-x_{1}^{p^{2}}-(x_{1}+\beta_{1}+1)^{p^{2}}+(x_{1}+1)^{p^{2}}}{p^{2}}$
$\displaystyle\quad+\frac{p((x_{3}+D_{1}(x_{1})+\beta_{3})^{p}-x_{3}^{p})}{p^{2}}$
$\displaystyle\quad+\frac{p(-(x_{3}+C_{1}(x_{1})+D_{1}(x_{1}+1)+\beta_{3})^{p}+(x_{3}+C_{1}(x_{1}))^{p})}{p^{2}}$

Finally, one will note that
$(\sigma_{1}-1)D(x_{1},x_{3})=D(x_{1}+1,x_{3}+C_{1}(x_{1}))-D(x_{1},x_{3})$.
Inspection will show that this is exactly the expression obtained here for
$\wp(C_{2}(x_{1},x_{3}))$. ∎

######  Theorem 3.3.

Let $K_{4}/K_{0}$ be a Galois extension of degree $p^{4}$ with Galois group of
the form

$H=\langle\sigma_{1},\sigma_{2}\mathrel{\mathop{\mathchar
58\relax}}\sigma_{1}^{p^{3}}=\sigma_{2}^{p}=1,\sigma_{2}\sigma_{1}=\sigma_{1}^{1+p^{2}}\sigma_{2}\rangle,$

and all notation as above. Assume that $K_{3}/K_{0}$, $x_{1}$, $x_{2}$, and
$x_{4}$ are as in Proposition 2.2. Then we can select $x_{4}$ such that
$K_{4}=K_{3}(x_{4})$, and

$x_{4}^{p}-x_{4}=\beta_{2}x_{1}+D_{2}(x_{1},x_{3})+\beta_{4},$

with $\beta_{4}\in K_{0}$ and $D_{2}$ is the Witt polynomial from Theorem 1.1.

Moreover, the converse holds; that is, if $K_{4}/K_{0}$ is an algebraic
extension of degree $p^{4}$ with $K_{4}=K_{0}(x_{1},x_{2},x_{3},x_{4})$ and
$\wp(x_{i})$ as above for $i=1,2,3,4$, then $K_{4}/K_{0}$ is a Galois
extension with Galois group $H$.

###### Proof.

As before, define $\sigma_{3}=\sigma_{1}^{p}$ and
$\sigma_{4}=\sigma_{1}^{p^{2}}$. Then note that
$H/\langle\sigma_{4}\rangle\cong C_{p^{2}}\times{C_{p}}$. Then by Proposition
2.2, we can choose $x_{3}\in K_{3}$ such that
$\wp(x_{3})=D_{1}(x_{1})+\beta_{3}$, with $(\sigma_{2}-1)x_{3}=0$ and
$(\sigma_{1}-1)x_{3}=C_{1}(x_{1})$ (where $C_{1}$ and $D_{1}$ are as defined
in Lemma 2.1). Define $J_{2}=K_{1}(x_{3})$ and note that $J_{2}$ is the fixed
field of $\langle\sigma_{1}^{p^{2}},\sigma_{2}\rangle$. Then since
$\langle\sigma_{1}^{p^{2}},\sigma_{2}\rangle\cong C_{p}^{2}$, we can choose
$x_{4}\in K_{4}$ such that $(\sigma_{2}-1)x_{4}=0$.

Now all that remains is to determine the action of $\sigma_{1}$ on $x_{4}$. To
that end, let $A=(\sigma_{1}-1)x_{4}\in K_{4}$. Since $\sigma_{4}\in Z(H)$,
$(\sigma_{4}-1)A=(\sigma_{1}-1)(\sigma_{4}-1)x_{4}=(\sigma_{1}-1)1=0$. Then
$A\in K_{3}$. Now using a known identity from the group definition, we derive:

$\displaystyle x_{4}+\sigma_{2}(A)$ $\displaystyle=\sigma_{2}(x_{4}+A)$
$\displaystyle=\sigma_{2}\sigma_{1}(x_{4})$
$\displaystyle=\sigma_{1}^{1+p^{2}}\sigma_{2}(x_{4})$
$\displaystyle=\sigma_{1}^{1+p^{2}}(x_{4})$
$\displaystyle=\sigma_{1}(x_{4}+1)$ $\displaystyle=x_{4}+A+1$

Then $(\sigma_{2}-1)A=1$. Therefore, $(\sigma_{2}-1)(A-x_{2})=0$, so
$A-x_{2}\in J_{2}$.

Now
$\mathrm{Tr}_{J_{2}/K_{0}}(A-x_{2})=(\sigma_{1}^{p^{2}}-1)x_{4}=(\sigma_{4}-1)x_{4}=1$.
Then by the lemma, $\mathrm{Tr}_{J_{2}/K_{0}}(A-x_{2}-C_{2}(x_{1},x_{3})))=0$.
Using the additive version of Hilbert’s Theorem 90, we then know that there is
some $j_{2}\in J_{2}$ such that
$(\sigma_{1}-1)j_{2}=A-x_{2}-C_{2}(x_{1},x_{3})$. Then
$(\sigma_{1}-1)(x_{4}-j_{2})=x_{2}+C_{2}(x_{1},x_{3})$. Without loss of
generality, we can now replace $x_{4}$ with $x_{4}-j_{2}$, which maintains all
prior relationships and establishes the identity
$(\sigma_{1}-1)x_{4}=x_{2}+C_{2}(x_{1},x_{3})$. This also gives us that
$(\sigma_{3}-1)x_{4}=(\sigma_{1}^{p}-1)x_{4}=\mathrm{Tr}_{K_{1}/K_{0}}A=C_{1}(x_{3})$.

Now consider $B=\wp(x_{4})\in K_{3}$. Since $\wp$ commutes with the elements
of $H$, $(\sigma_{2}-1)B=\wp((\sigma_{2}-1)x_{4})=0$. Then $B\in J_{2}$. Now
note that

$\displaystyle(\sigma_{1}-1)B$ $\displaystyle=\wp(x_{2}+C_{2}(x_{1},x_{3}))$
$\displaystyle=\beta_{2}+(\sigma_{1}-1)D_{2}(x_{1},x_{3})$
$\displaystyle=(\sigma_{1}-1)(\beta_{2}x_{1}+D_{2}(x_{1},x_{3})),$

so $(\sigma_{1}-1)(B-A_{1}-D_{2}(x_{1},x_{3}))=0$, with $A_{1}=\beta_{2}x_{1}$
as before. Therefore, $B-A_{1}-D_{2}(x_{1},x_{3})=\beta_{4}$ for some
$\beta_{4}\in K_{0}$. Then $B=A_{1}+D_{2}(x_{1},x_{3})+\beta_{4}$.

The converse holds as well, though this part will not be shown in detail.
However, it follows in much the same way as in the proof of Theorem 3.1.

∎

As a final note, one will note that the result stated in Theorem 1.1 simply
follows by combining Proposition 2.2 and Theorems 3.1 and 3.3.

######  Acknowledgments.

This work was partially supported by a NASA Nebraska Space Grant Fellowship. I
would also like to thank my undergraduate advisor, Dr. Griff Elder, for his
mentorship, support, and guidance in this project.

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