# Induced Hopf Galois Structures and their Local Hopf Galois Modules

Daniel Gil-Muñoz, Anna Rio

###### Abstract

The regular subgroup determining an induced Hopf Galois structure for a Galois
extension $L/K$ is obtained as the direct product of the corresponding regular
groups of the inducing subextensions. We describe here the associated Hopf
algebra and Hopf action of an induced structure and we prove that they are
obtained by tensoring the corresponding inducing objects. In order to deal
with their associated orders we develop a general method to compute bases and
free generators in terms of matrices coming from representation theory of Hopf
modules. In the case of an induced Hopf Galois structure it allows us to
decompose the associated order, assuming that inducing subextensions are
arithmetically disjoint.

##  1 Introduction

A finite extension of fields $L/K$ is said to be Hopf Galois if there is a
$K$-Hopf algebra $H$ and a $K$-linear action of $H$ in $L$ which endows $L$
with an $H$-module structure $H\to\operatorname{End}_{K}(L)$, such that the
induced map $j\colon L\otimes_{K}H\longrightarrow\mathrm{End}_{K}(L)$ is
bijective. In that case, the pair formed by the Hopf algebra and the Hopf
action is said to be a Hopf Galois structure of $L/K$. For simplicity, we may
say that $L/K$ is $H$-Galois. This notion was introduced by Chase and Sweedler
in their book [5] and it generalizes the notion of Galois extension of fields
since the group algebra of the Galois group together with the Galois action on
$L$ provides a Hopf Galois structure.

Although Hopf Galois structures are difficult to compute in general, the ones
of separable extensions can be labeled by objects of group theory thanks to a
theorem of Greither and Pareigis. This result was introduced in their paper
[11]. If $L/K$ is a separable extension, let $\widetilde{L}$ be its normal
closure, $G=\mathrm{Gal}(\widetilde{L}/K)$,
$G^{\prime}=\mathrm{Gal}(\widetilde{L}/L)$ and $X=G/G^{\prime}$. Let
$\lambda\colon G\longrightarrow\mathrm{Perm}(X)$ be the left translation map,
namely the embedding given by the group action of $G$ on $X$ by left
translation.

######  Theorem 1.1 (Greither-Pareigis).

Hopf Galois structures of $L/K$ are in one-to-one correspondence with regular
(i.e, simply transitive) subgroups of $\mathrm{Perm}(X)$ normalized by
$\lambda(G)$. Moreover, if $N$ is some such subgroup, the corresponding Hopf
Galois structure is given by the $K$-Hopf algebra $\widetilde{L}[N]^{G}$ and
its action on $L$ defined by

$\left(\sum_{i=1}^{r}c_{i}n_{i}\right)\cdot
x=\sum_{i=1}^{r}c_{i}n_{i}^{-1}(\,\overline{1}_{G}\,)(x)$

See [6, Theorem 6.8].

If a Hopf Galois structure corresponds to a group $N$, its type is defined as
the isomorphism class of $N$.

Hopf Galois theory can be used to generalize Galois module theory, giving rise
to a Hopf Galois module theory. In the approach introduced by Leopoldt, Galois
module theory studies the structure of the valuation ring $\mathcal{O}_{L}$ of
a Galois extension $L/K$ of local fields over its associated order
$\mathfrak{A}_{K[G]}$, defined as the maximal $\mathcal{O}_{K}$-order in
$K[G]$ that acts on $\mathcal{O}_{L}$. If $L/K$ is $H$-Galois, the associated
order of $\mathcal{O}_{L}$ in $H$ is defined as

$\mathfrak{A}_{H}=\\{\alpha\in H\,|\,\alpha\cdot x\in\mathcal{O}_{L}\hbox{ for
every }x\in\mathcal{O}_{L}\\}.$

In the context of the existence of a normal integral basis, it is interesting
to determine the freeness of $\mathcal{O}_{L}$ as $\mathfrak{A}_{H}$-module as
$H$ runs through the different Hopf Galois structures of $L/K$.

In this paper we consider Hopf Galois structures of a semidirect extension
$L/K$ of fields, that is, a Galois extension whose Galois group $G$ is a
semidirect product. We will write $G=J\rtimes G^{\prime}$ with $J$ a normal
subgroup of $G$ and $E=L^{G^{\prime}}$, $F=L^{J}$ and we will focus on induced
Hopf Galois structures. These are structures built from a Hopf Galois
structure on $E/K$ and a Hopf Galois structure on $L/E$. This notion was
introduced by Crespo, Rio and Vela in their paper [10]. The type of an induced
Hopf Galois structure is the direct product of the types of the corresponding
Hopf Galois structures of $L/E$ and $E/K$. As a consequence, the regular
subgroup $N$ which determines such a Hopf Galois structure is a direct product
$N=N_{1}\times N_{2}$.

Our first results show that certain properties are directly related to the
type, since we have a complete analogy with the situation of classical Galois
extensions whose Galois group is a direct product of subgroups. For such
extensions we can use standard Galois theory to prove the following:

######  Proposition 1.2.

Let $L/K$ be a finite Galois extension with Galois group $G$ and assume that
$G$ can be written as a direct product $G=J\times G^{\prime}$. Let
$E=L^{G^{\prime}}$ and $F=L^{J}$. Then:

  * 1.

$E/K$ and $F/K$ are Galois extensions.

  * 2.

$L=EF$ and $E\cap F=K$.

  * 3.

$E/K$ and $F/K$ are linearly disjoint, namely the canonical map
$E\otimes_{K}F\longrightarrow EF$ is a $K$-isomorphism.

  * 4.

$K[G]=K[J]\otimes_{K}K[G^{\prime}]$.

  * 5.

The Galois action of $K[G]$ on $L$ is the Kronecker product of the Galois
actions of $K[J]$ on $E$ and $K[G^{\prime}]$ on $F$.

In the setting of Hopf Galois theory, Proposition 1.2 gives information about
the classical Galois structure of $L/K$. We shall check that induced Hopf
Galois structures of Galois extensions $L/K$ with Galois group of the form
$G=J\rtimes G^{\prime}$ present a similar behaviour.

We note that the condition that $G^{\prime}$ has a normal complement $J$ is
equivalent to the condition (from [11]) that $E/K$ is an almost classical Hopf
Galois extension. On the other hand, $L/E$ is Galois and therefore Hopf
Galois. By [10, Theorem 3], every choice of Hopf Galois structures of $E/K$
and $L/E$ determines a Hopf Galois structure of $L/K$. On the other hand,
since $J$ is normal $F/K$ is Galois and hence a Hopf Galois extension. We
shall see in 5.1 the relationship between structures of $L/E$ and $F/K$.

######  Theorem 1.3.

Let $L/K$ be a finite Galois extension with Galois group $G=J\rtimes
G^{\prime}$. Let $E=L^{G^{\prime}}$ and $F=L^{J}$. Then:

  * 1.

$E/K$ and $F/K$ are Hopf Galois extensions.

  * 2.

$L=EF$ and $E\cap F=K$.

  * 3.

$E/K$ and $F/K$ are linearly disjoint.

Let $E/K$ be $H_{1}$-Galois and let $L/E$ be $H_{2}$-Galois. We consider the
corresponding induced Hopf Galois structure of $L/K$ . Let $H$ be its
associated Hopf algebra. Then:

  * 4.

$H=H_{1}\otimes_{K}\overline{H}$, where $\overline{H}$ is the Hopf algebra of
the Hopf Galois structure of $F/K$ such that $\overline{H}\otimes_{K}E=H_{2}$
(see Proposition 5.3).

$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
L\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H_{2}}$$\scriptstyle{H}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H_{1}}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
F}$$\scriptstyle{\overline{H}}$$\textstyle{K}$

  * 5.

The Hopf action of $H$ on $L$ is the Kronecker product of the Hopf actions of
$H_{1}$ on $E$ and $\overline{H}$ on $F$.

The second and third items are again straightforward by using basic Galois
theory. We shall prove the remaining assertions in Section 5.

In Sections 3 and 4 we deal with the Hopf Galois module structure at the
integer level. $K$ will be the quotient field of a principal ideal domain
$\mathcal{O}_{K}$, $L/K$ will be a finite separable Hopf Galois extension and
$\mathcal{O}_{L}$ will be the integral closure of $\mathcal{O}_{K}$ in $L$. As
in linear representation theory, we study Hopf actions through matrices and we
show how to use matrices to obtain a basis for the associated order and to
check for the existence of a free generator of $\mathcal{O}_{L}$ as module
over the associated order.

When we restrict ourselves to induced Hopf Galois structures, we can ask
whether the induced Hopf Galois module structure of $\mathcal{O}_{L}$ can be
described from the Hopf Galois module structures of $\mathcal{O}_{E}$ and
$\mathcal{O}_{F}$. With regard to the classical Galois module structure of
$\mathcal{O}_{L}$ when $L/K$ is a direct extension (i.e, with Galois group
isomorphic to a direct product), Byott and Lettl included in [2] the
following:

######  Proposition 1.4.

Let $K$ be the quotient field of a Dedekind domain $\mathcal{O}_{K}$ and let
$E/K$, $F/K$ be finite Galois extensions. Put $L=EF$ and suppose that $E/K$
and $F/K$ are arithmetically disjoint. Then:

  * 1.

$\mathfrak{A}_{L/F}=\mathfrak{A}_{E/K}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$
and
$\mathfrak{A}_{L/K}=\mathfrak{A}_{E/K}\otimes_{\mathcal{O}_{K}}\mathfrak{A}_{F/K}$.

  * 2.

If there exists some $\gamma\in\mathcal{O}_{E}$ with
$\mathcal{O}_{E}=\mathfrak{A}_{E/K}\cdot\gamma$, then
$\mathcal{O}_{L}=\mathfrak{A}_{L/F}\cdot(\gamma\otimes 1)$.
If there also exists $\delta\in\mathcal{O}_{F}$ with
$\mathcal{O}_{F}=\mathfrak{A}_{F/K}\cdot\delta$, then
$\mathcal{O}_{L}=\mathfrak{A}_{L/K}\cdot(\gamma\otimes\delta)$.

See [2, Lemma 5].

Recall that $E/K$ and $F/K$ are said to be arithmetically disjoint over $K$ if
$\mathcal{O}_{EF}=\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$, or
equivalently, if their discriminants are coprime and $E$ is linearly disjoint
from $F$ over $K$. The decomposition turns out to be very useful in order to
study the Galois module structure of $\mathcal{O}_{EF}$ in terms of
$\mathcal{O}_{E}$ and $\mathcal{O}_{F}$.

Going back to the semidirect case, if we keep the condition that $E/K$ and
$F/K$ are arithmetically disjoint, we are able to obtain analogous results for
induced Hopf Galois structures. We will describe the induced Hopf Galois
module structure of $\mathcal{O}_{L}$ by considering the corresponding module
structures of $\mathcal{O}_{E}$ and $\mathcal{O}_{F}$ as follows.

######  Theorem 1.5.

Let $K$ be the quotient field of a principal ideal domain $\mathcal{O}_{K}$,
$L/K$ a finite separable Hopf Galois extension and $\mathcal{O}_{L}$ the
integral closure of $\mathcal{O}_{K}$ in $L$. Assume that the structure is an
induced one and its Hopf algebra is $H=H_{1}\otimes_{K}\overline{H}$. If $E/K$
and $F/K$ are arithmetically disjoint then the following statements hold:

  * 1.

$\mathfrak{A}_{H}=\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathfrak{A}_{\overline{H}}$.

  * 2.

If $\mathcal{O}_{E}$ is $\mathfrak{A}_{H_{1}}$-free and $\mathcal{O}_{F}$ is
$\mathfrak{A}_{\overline{H}}$-free, then $\mathcal{O}_{L}$ is
$\mathfrak{A}_{H}$-free. Moreover, an $\mathfrak{A}_{H}$-generator of
$\mathcal{O}_{L}$ is the product of an $\mathfrak{A}_{H_{1}}$-generator of
$\mathcal{O}_{E}$ and an $\mathfrak{A}_{\overline{H}}$-generator of
$\mathcal{O}_{F}$.

Since $E/K$ and $F/K$ are linearly disjoint, it is well known that for every
Hopf Galois structure $H_{1}$ of $E/K$, $H_{1}\otimes_{K}F$ is a Hopf Galois
structure of $L/F$. We will also describe the module structure of
$\mathcal{O}_{L}$ in this Hopf Galois structures in the following way:

######  Theorem 1.6.

Under the same hypothesis,

  * 1.

$\mathfrak{A}_{H_{1}\otimes_{K}F}=\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$.

  * 2.

If $\mathcal{O}_{E}$ is $\mathfrak{A}_{H_{1}}$-free, then $\mathcal{O}_{L}$ is
$\mathfrak{A}_{H_{1}\otimes_{K}F}$-free.

##  2 Hopf Galois linear representations

Let $A$ be an $n$-dimensional $K$-algebra and let $V$ be a $K$-vector space. A
linear representation of $A$ in $V$ is a $K$-algebra homomorphism
$\rho:A\textrightarrow\mathrm{End}_{K}(V)$ or, by choosing a $K$-basis of V, a
$K$-algebra homomorphism $\rho:A\textrightarrow\mathrm{Mat}_{n}(K)$. This is
the notion of $A$-module that we have mentioned in the introduction. The
tensor product or Kronecker product of two representations
$\rho^{1}:A_{1}\textrightarrow\mathrm{End}_{K}(V_{1})$ and
$\rho^{2}:A_{2}\textrightarrow\mathrm{End}_{K}(V_{2})$ is

$\rho^{1}\otimes\rho^{2}:A_{1}\otimes_{K}A_{2}\textrightarrow\mathrm{End}_{K}(V_{1}\otimes
V_{2})$

defined by

$\rho^{1}\otimes\rho^{2}(a_{1}\otimes a_{2})(v_{1}\otimes
v_{2})=\rho^{1}(a_{1})(v_{1})\otimes\rho^{2}(a_{2})(v_{2}).$

If we have a finite field extension $L/K$, then $L$ is a $K$\- algebra and we
put $1:L\to\mathrm{End}_{K}(L)$ for the representation given by multiplication
in $L$.

In this section we consider Hopf Galois structures from the point of view of
representations, having in mind that when we consider induced Hopf Galois
structures we will deal with the Kronecker product of actions and also, since
the Hopf action appears in the definition of the associated order, that
matrices become useful for explicit computations.

Let $L/K$ be a Galois extension of degree $n$ with Galois group $G$. Since $G$
acts in $L$ we have a Galois representation $\rho:G\to\mathrm{Aut}_{K}(L)$.
Observe that a Galois representation gives a $K[G]$-module structure, that is,
a linear representations of the Hopf algebra $K[G]$, since it can be extended
to a $K$-linear representation $\rho\colon
K[G]\longrightarrow\operatorname{End}_{K}(L).$ This representation provides
full information about the action of $K[G]$ on $L$. In fact, $L/K$ being
Galois is equivalent to
$(1,\rho):L\otimes_{K}K[G]\to\operatorname{End}_{K}(L).$ being an isomorphism.

In a complete analogy, a Hopf-Galois structure for a field extension $L/K$ of
degree $n$ corresponds to the existence of a $K$-Hopf algebra $H$ and a linear
representation $\rho_{H}:H\textrightarrow\mathrm{End}_{K}(L)$ such that
$j=(1,\rho_{H}):L\otimes_{K}H\to\mathrm{End}_{K}(L)$ is an isomorphism. Recall
that in the separable case, Greither and Pareigis show that
$L\otimes_{K}H=L[N]$ ($H$ is an $L$-form of $K[N]$) with $N$ a group of order
$n$.

The Hopf action of $H$ in $L$ gives rise to a $K$-bilinear map

$\begin{array}[]{ccc}H\times L&\to&L\\\ (h,x)&\mapsto&h\cdot x\end{array}$

with values in $L$. Therefore, given a $K$-basis $\\{w_{i}\\}_{i=1}^{n}$ of
$H$ and a $K$-basis $\\{\gamma_{j}\\}_{j=1}^{n}$ of $L$, the action can be
described through the $n\times n$ matrix
$\left(w_{i}\cdot\gamma_{j}\right)_{ij}$ with entries in $L$. We will write
these elements in the chosen basis of $L$ and we will usually represent this
matrix in a table in order to make clear the bases under consideration. For
$1\leq i\leq n$, the $i$-th row of the table corresponds to the matrix
$\rho_{H}(w_{i})$ representing $w_{i}$.

######  Example 2.1.

Let $L/K$ be a finite degree $n$ Galois extension and let $G$ be its Galois
group. Let us consider the classical Galois representation $\rho\colon
K[G]\longrightarrow\operatorname{End}_{K}(L).$

Let us write $G=\\{\sigma_{1},...,\sigma_{n}\\}$ and take these elements as
$K$-basis for $K[G]$. On the other hand, we fix a normal basis
$\\{\alpha_{j}=\sigma_{j}(\alpha)\\}_{j=1}^{n}$ for $L$. We have
$\sigma_{i}(\alpha_{j})=\sigma_{i}\sigma_{j}(\alpha)$ and via the regular
representation of $G$ we can identify $\sigma_{i}$ with a permutation of the
set $\\{1,\dots,n\\}$, namely $\sigma_{i}(j)=k$ if
$\sigma_{i}\sigma_{j}=\sigma_{k}$. With this identification, the action of
$K[G]$ on $L$ is the regular representation of the group $G$. Hence, for every
$1\leq i\leq n$, $\rho(\sigma_{i})$ is an $n\times n$ permutation matrix.

######  Example 2.2.

In the previous example, assume that $G\cong C_{n}$. Let $\sigma$ be a
generator of $G$ and take $\\{\mathrm{Id},\sigma,...,\sigma^{n-1}\\}$ as
$K$-basis of $K[G]$. Then, the action of $K[G]$ on $L$ is given by:

|  | $\alpha$
---
| $\sigma(\alpha)$
---
| $...$
---
| $\sigma^{n-1}(\alpha)$
---
$\mathrm{Id}$ | $\alpha$ | $\sigma(\alpha)$ | $...$ | $\sigma^{n-1}(\alpha)$
$\sigma$ | $\sigma(\alpha)$ | $\sigma^{2}(\alpha)$ | $...$ | $\alpha$
$...$ | $...$ | $...$ | $...$ | $...$
$\sigma^{n-1}$ | $\sigma^{n-1}(\alpha)$ | $\alpha$ | $...$ | $\sigma^{n-2}(\alpha)$

and for every $i\in\mathbb{Z}/n\mathbb{Z}$, $\rho(\sigma^{i})$ is the
circulant matrix whose first column is the the $(i+1)$-th vector of the
canonical basis of $K^{n}$.

######  Example 2.3.

Let $E=\mathbb{Q}_{3}(\alpha)$, where $\alpha$ is a root of the polynomial
$f(x)=x^{3}+3\in\mathbb{Q}_{3}[x]$. Then $E/\mathbb{Q}_{3}$ is a non-Galois
extension with normal closure $L=E(z)$, where $z=\sqrt{-3}$.

The roots of $f$ are $\alpha,\xi\alpha,\xi^{2}\alpha$, where
$\xi=\frac{-1+z}{2}$. Let $r$ be the $\mathbb{Q}_{3}$-automorphism of $L$
given by $r(\alpha)=\xi\alpha$ and $r(z)=z$, and let $s$ be the one given by
$s(z)=-z$ and $s(\alpha)=\alpha$. As permutations of the roots of $f$,
$r=(\alpha,\xi\alpha,\xi^{2}\alpha)$ and $s=(\xi\alpha,\xi^{2}\alpha)$. These
two elements generate $\mathrm{Gal}(L/\mathbb{Q}_{3})$, which is dihedral of
order $6$. Additionally, $s$ generates $G^{\prime}=\mathrm{Gal}(L/E)$.

Let $X=G/G^{\prime}$ and consider $\lambda\colon
G\longrightarrow\mathrm{Perm}(X)$ the morphism given by the action of $G$ on
$X$ by left translation. By Greither-Pareigis theorem, since $Perm(X)\cong
S_{3}$ has a unique subgroup of order $3$ and it is normalized by
$\lambda(G)$, we know that there is a unique Hopf Galois structure of
$E/\mathbb{Q}_{3}$. The corresponding $\mathbb{Q}_{3}$-Hopf algebra $H_{1}$
has $\mathbb{Q}_{3}$-basis

$w_{1}=\mathrm{Id},\,w_{2}=(\lambda(r)-\lambda(r)^{-1})z,\,w_{3}=\lambda(r)+\lambda(r)^{-1}.$

The action of $H_{1}$ on $E$ is given by

$\begin{tabular}[]{c|ccc}&\begin{tabular}[c]{@{}l@{}}$1$\end{tabular}&\begin{tabular}[c]{@{}l@{}}$\alpha$\end{tabular}&\begin{tabular}[c]{@{}l@{}}$\alpha^{2}$\end{tabular}\\\
\hline\cr$w_{1}$&$1$&$\alpha$&$\alpha^{2}$\\\
$w_{2}$&$0$&$3\alpha$&$-3\alpha^{2}$\\\ $w_{3}$&$2$&$-\alpha$&$-\alpha^{2}$\\\
\end{tabular},$ (1)

Then, the matrices representing $w_{1},w_{2}$ and $w_{3}$ are

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\\\
\end{pmatrix}\qquad\begin{pmatrix}0&0&0\\\ 0&3&0\\\ 0&0&-3\\\
\end{pmatrix}\qquad\begin{pmatrix}2&0&0\\\ 0&-1&0\\\ 0&0&-1\end{pmatrix}.$

On the other hand, the matrices representing multiplication by $1$, $\alpha$
and $\alpha^{2}$ are

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\\\
\end{pmatrix}\qquad\begin{pmatrix}0&0&-3\\\ 1&0&0\\\
0&1&0\end{pmatrix}\qquad\begin{pmatrix}0&-3&0\\\ 0&0&-3\\\
1&0&0\end{pmatrix}.$

By multiplying these two families we can see that we obtain nine linearly
independent matrices and explicitly check that
$j:E\otimes_{\mathbb{Q}_{3}}H_{1}\to\operatorname{End}_{\mathbb{Q}_{3}}(E)$ is
an isomorphism.

When we consider the classical Galois action, since elements of $G$ act as
automorphisms in $L$, we have a canonical basis of the Hopf algebra whose
action is represented with nonsingular matrices. As we have seen in the last
example, for a general Hopf Galois action we can have elements in a basis of
the Hopf algebra which act through non invertible matrices.

##  3 Associated orders of Hopf Galois actions

Let $\mathcal{O}_{K}$ be a principal ideal domain with field of fractions $K$.
Let $L/K$ be a separable Hopf Galois extension of degree $n$ and let
$\mathcal{O}_{L}$ be the integral closure of $\mathcal{O}_{K}$ in $L$. If $H$
is the Hopf algebra of a Hopf Galois structure of $L/K$, then the associated
$\mathcal{O}_{K}$-order to $\mathcal{O}_{L}$ in $H$ is

$\mathfrak{A}_{H}=\\{\alpha\in H\,|\,\alpha\cdot x\in\mathcal{O}_{L}\,\hbox{
for all }x\in\mathcal{O}_{L}\\}.$

In this section we establish a general method to compute an
$\mathcal{O}_{K}$-basis of $\mathfrak{A}_{H}$.

###  3.1 Motivating example

In order to compute an $\mathcal{O}_{K}$-basis of $\mathfrak{A}_{H}$, we will
use the space of solutions of a linear system with coefficient matrix that
will be called $M(H,L)$. To motivate the construction let us go back to
Example 2.3. There, since the irreducible polynomial of $\alpha$ is
$3$-Eisenstein, we have $\mathcal{O}_{E}=\mathbb{Z}_{3}[\alpha]$. An element
$h_{1}w_{1}+h_{2}w_{2}+h_{3}w_{3}\in H_{1}$ belongs to the associated order
if, and only if,
$(h_{1}w_{1}+h_{2}w_{2}+h_{3}w_{3})(x_{1}+x_{2}\alpha+x_{3}\alpha^{2})\in\mathbb{Z}_{3}[\alpha]$
for all $x_{1},x_{2},x_{3}\in\mathbb{Z}_{3}$. This condition becomes

$\left(h_{1}\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\\\
\end{pmatrix}+h_{2}\begin{pmatrix}0&0&0\\\ 0&3&0\\\ 0&0&-3\\\
\end{pmatrix}+h_{3}\begin{pmatrix}2&0&0\\\ 0&-1&0\\\
0&0&-1\end{pmatrix}\right)\begin{pmatrix}x_{1}\\\ x_{2}\\\
x_{3}\end{pmatrix}\in\mathbb{Z}_{3}^{3}$

for all $x_{1},x_{2},x_{3}\in\mathbb{Z}_{3}$. We could write this condition as

$(h_{1},h_{2},h_{3})\begin{pmatrix}w_{1}\\\ w_{2}\\\
w_{3}\end{pmatrix}\in\mathrm{Mat}_{3}(\mathbb{Z}_{3})$

using matrix representation of the basis elements $w_{i}$ and considering the
second matrix as a block matrix. The condition we obtain is

$\begin{pmatrix}h_{1}+2h_{3}&0&0\\\ 0&h_{1}+3h_{2}-h_{3}&0\\\
0&0&h_{1}-3h_{2}-h_{3}\end{pmatrix}\in\mathrm{Mat}_{3}(\mathbb{Z}_{3})$

and instead of the previous action on block matrices we are going to consider
matrix actions of standard linear algebra:

$\begin{pmatrix}1&0&2\\\ 0&0&0\\\ 0&0&0\\\ 0&0&0\\\ 1&3&-1\\\ 0&0&0\\\
0&0&0\\\ 0&0&0\\\ 1&-3&-1\end{pmatrix}\begin{pmatrix}h_{1}\\\ h_{2}\\\
h_{3}\end{pmatrix}\in\mathbb{Z}_{3}^{9}$

Therefore, in the general situation, once we have fixed a $K$-basis of $L$ and
a $K$-basis of $H$, we are going to use the $n^{2}\times n$ matrix whose $j$th
column is formed by concatenating the columns of the matrix for $w_{j}$.

Let us observe that in this example, after deleting zero rows, this last
condition becomes

$\begin{pmatrix}1&0&2\\\ 1&3&-1\\\ 1&-3&-1\\\
\end{pmatrix}\begin{pmatrix}h_{1}\\\ h_{2}\\\
h_{3}\end{pmatrix}\in\mathbb{Z}_{3}^{3}$

Now, this matrix is invertible with inverse

$\frac{1}{6}\begin{pmatrix}2&2&2\\\ 0&1&-1\\\ 2&-1&-1\end{pmatrix}.$

Then, we have that $h=\sum_{i=1}^{3}h_{i}w_{i}\in\mathfrak{A}_{H}$ if and only
if there are some $c_{1},c_{2},c_{3}\in\mathbb{Z}_{3}$ such that

$\begin{pmatrix}h_{1}\\\ h_{2}\\\
h_{3}\end{pmatrix}=\frac{1}{6}\begin{pmatrix}2&2&2\\\ 0&1&-1\\\
2&-1&-1\end{pmatrix}\begin{pmatrix}c_{1}\\\ c_{2}\\\ c_{3}\end{pmatrix},$

that is, if and only if

$\begin{split}h&=\frac{1}{6}(2c_{1}+2c_{2}+2c_{3})w_{1}+\frac{1}{6}(c_{2}-c_{3})w_{2}+\frac{1}{6}(2c_{1}-c_{2}-c_{3})w_{3}\\\
&=\frac{w_{1}+w_{3}}{3}c_{1}+\frac{2w_{1}+w_{2}-w_{3}}{6}c_{2}+\frac{2w_{1}-w_{2}-w_{3}}{6}c_{3}\end{split}$

for some $c_{1},c_{2},c_{3}\in\mathbb{Z}_{3}$. We find a basis for the
associated order

$\mathfrak{A}_{H_{1}}=\mathbb{Z}_{3}\left[\frac{w_{1}+w_{3}}{3},\frac{2w_{1}+w_{2}-w_{3}}{6},\frac{2w_{1}-w_{2}-w_{3}}{6}\right].$

The action of this basis on the basis of $\mathcal{O}_{E}$ is

|  | $1$
---
| $\alpha$
---
| $\alpha^{2}$
---
$b_{1}=\dfrac{w_{1}+w_{3}}{3}$ | $1$ | $0$ | $0$
$b_{2}=\dfrac{2w_{1}+w_{2}-w_{3}}{6}$ | $0$ | $\alpha$ | $0$
$b_{3}=\dfrac{2w_{1}-w_{2}-w_{3}}{6}$ | $0$ | $0$ | $\alpha^{2}$
(2)

This example is related to the non-classical Hopf Galois structure of Chase
and Sweedler ([5, section 4, pp. 35-39]). The ring $\mathbb{Z}_{3}[\alpha]$ is
a $C_{3}$-graded $\mathbb{Z}_{3}$-algebra and the associated order
$\mathfrak{A}_{H_{1}}$ is $\mathbb{Z}_{3}[C_{3}]^{*}$ (the dual of the group
ring); $\mathbb{Z}_{3}[C_{3}]^{*}$ acts on $\mathcal{O}_{E}$ as
$b_{i}\cdot\alpha^{j}=\delta_{ij}\alpha^{j}$.

###  3.2 $n^{2}\times n$ matrix attached to a Hopf Galois representation

For $1\leq i,j\leq n$, let $E_{ij}\in\mathrm{Mat}_{n}(K)$ be the matrix having
$1$ in position $(i,j)$ and $0$ elsewhere. We consider this canonical basis of
$\mathrm{Mat}_{n}(K)$ in the following order: $E_{11},E_{21},\dots
E_{n1},E_{12},\dots E_{nn}$. Therefore, we have an isomorphism
$\varphi:\mathrm{Mat}_{n}(K)\to K^{n^{2}}$ where the image of a matrix becomes
the $n^{2}$ column vector formed by its ordered columns. We have
$\varphi(E_{ij})=e_{n(j-1)+i}$, vector of the canonical basis of $K^{n^{2}}$.

######  Definition 3.1.

Let $L/K$ be a degree $n$ $H$-Galois field extension. Given a $K$-basis
$\\{\gamma_{j}\\}_{j=1}^{n}$ of $L$ and a $K$-basis $\\{w_{i}\\}_{i=1}^{n}$ of
$H$, as in Section 2 we denote by $w_{i}\in\mathrm{Mat}_{n}(K)$ the matrix of
$\rho_{H}(w_{i})$ in base $\\{\gamma_{1},\dots,\gamma_{n}\\}$. We define the
matrix of the action of $H$ on $L$ as

$M(H,L)=\begin{pmatrix}|&|&\dots&|\\\
\varphi(w_{1})&\varphi(w_{2})&\dots&\varphi(w_{n})\\\ |&|&\dots&|\\\
\end{pmatrix}\in\mathrm{Mat}_{n^{2}\times n}(K),$

namely, the columns of $M(H,L)$ are obtained transforming into column vectors
the matrices of a basis of $H$ acting on a given basis of $L$.

######  Proposition 3.2.

Let $L/K$ be a degree $n$ Hopf Galois extension with Hopf algebra $H$. Then,
the matrix $M(H,L)$ has rank $n$.

###### Proof.

Since $w_{1},\dots,w_{n}$ is a $K$-basis of $H$ and
$j:L\otimes_{K}H\to\operatorname{End}_{K}(L)$ is an isomorphism, the
corresponding matrices $w_{i}$ are $K$-linearly independent and the same
happens to the vectors $\varphi(w_{i})$, which are the columns of $M(H,L)$. ∎

This is a convenient way to store all the information of the Hopf action in a
single object. If we think of it as a matrix defined by blocks

$M(H,L)=\begin{pmatrix}\\\\[-4.30554pt] M_{1}(H,L)\\\\[4.30554pt]
\hline\cr\\\\[-4.30554pt] \cdots\\\\[4.30554pt] \hline\cr\\\\[-4.30554pt]
M_{n}(H,L)\\\ \\\\[-4.30554pt] \end{pmatrix}$

then the block $M_{j}(H,L)$ provides the action of the Hopf algebra on the
element $\gamma_{j}$ of the basis, namely it is the matrix of the $K$-linear
map

$\begin{array}[]{ccc}H&\to&L\\\ h&\mapsto&h\cdot\gamma_{j}\end{array}$

in the chosen bases of $H$ and $L$.

###  3.3 Computing a basis of ${\mathfrak{A}}_{H}$

Let us return to the problem of basis computation for the associated order
$\mathfrak{A}_{H}$ attached to a Hopf Galois structure of an extension $L/K$
with the hypothesis and notations stated at the beginning of this section,
which ensure the existence of an integral basis.

We fix again a $K$-basis $\\{w_{i}\\}_{i=1}^{n}$ of $H$, but now we take an
$\mathcal{O}_{K}$-basis $\\{\gamma_{j}\\}_{j=1}^{n}$ of $\mathcal{O}_{L}$.
This is, in particular, a $K$-basis of $L$.

######  Theorem 3.3.

Let $h=\sum_{i=1}^{n}h_{i}w_{i}\in H$, $h_{i}\in K$. Then,

$h\in\mathfrak{A}_{H}\iff M(H,L)\begin{pmatrix}h_{1}\\\ \vdots\\\
h_{n}\end{pmatrix}\in\mathcal{O}_{K}^{n^{2}}.$

###### Proof.

By definition, $h\in\mathfrak{A}_{H}$ if and only if $h\cdot
x\in\mathcal{O}_{L}$ for all $x\in\mathcal{O}_{L}$. Fix $x\in\mathcal{O}_{L}$.
Since $\\{\gamma_{j}\\}_{j=1}^{n}$ is an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$, we can write $x=\sum_{j=1}^{n}x_{j}\gamma_{j}$, with
$x_{j}\in\mathcal{O}_{K}$, $1\leq j\leq n$. We compute

$h\cdot
x=\left(\sum_{i=1}^{n}h_{i}w_{i}\right)\cdot\left(\sum_{j=1}^{n}x_{j}\gamma_{j}\right)=\sum_{i=1}^{n}\sum_{j=1}^{n}h_{i}x_{j}w_{i}\cdot\gamma_{j}=\sum_{j=1}^{n}x_{j}M_{j}(H,L)\begin{pmatrix}h_{1}\\\
\vdots\\\ h_{n}\end{pmatrix}.$

Therefore, $h\in\mathfrak{A}_{H}$ if and only if

$M_{j}(H,L)\begin{pmatrix}h_{1}\\\ \vdots\\\
h_{n}\end{pmatrix}\in\mathcal{O}_{K}^{n}$

for all $1\leq j\leq n$, which is equivalent to the condition of the
statement. ∎

In order to characterize elements
$h=\sum_{i=1}^{n}h_{i}w_{i}\in\mathfrak{A}_{H}$, we look for an expression for
the coordinate vector $(h_{1},...,h_{n})$. The previous result says that it
becomes a vector of integers when multiplied by $M(H,L)$. If we could replace
$M(H,L)$ with an invertible matrix we would be able to express such vector as
the inverse matrix applied to a vector of integers. To this end we can take
for example the Hermite normal form of the integral matrix obtained from
$M(H,L)$ removing common denominator.

######  Theorem 3.4 (Hermite normal form).

Let $A$ be a PID and $M\in M_{m\times n}(A)$ a matrix of rank $n$. Then, there
exists an $m\times m$ unimodular matrix $U$, that is
$U\in\operatorname{GL}_{m}(A)$, such that $UM$ is a matrix in Hermite normal
form (in particular in row echelon form).

###### Proof.

See [1, Chapter 5, Theorem 3.1]. ∎

Since $K$ is the field of fractions of $\mathcal{O}_{K}$, we can write
$M(H,L)=d_{M}\,M$ with $d_{M}\in K$ and $M\in M_{m\times n}(\mathcal{O}_{K})$
such that coefficients in $M$ are coprime. Let
$U\in\operatorname{GL}_{n^{2}}(\mathcal{O}_{K})$ an unimodular matrix such
that $UM$ is in Hermite normal form, and let
$D\in\mathrm{Mat}_{n}(\mathcal{O}_{K})$ the matrix obtained deleting the zero
rows of $UM$. This matrix belongs to $\operatorname{GL}_{n}(K)$. Then,

$\begin{split}h\in\mathfrak{A}_{H}&\iff M(H,L)\begin{pmatrix}h_{1}\\\
\vdots\\\ h_{n}\end{pmatrix}\in\mathcal{O}_{K}^{n^{2}}\iff
d_{M}\,M\begin{pmatrix}h_{1}\\\ \vdots\\\
h_{n}\end{pmatrix}\in\mathcal{O}_{K}^{n^{2}}\\\ &\iff
d_{M}\,UM\begin{pmatrix}h_{1}\\\ \vdots\\\
h_{n}\end{pmatrix}\in\mathcal{O}_{K}^{n^{2}}\iff d_{M}D\begin{pmatrix}h_{1}\\\
\vdots\\\ h_{n}\end{pmatrix}\in\mathcal{O}_{K}^{n}\\\
&\iff\begin{pmatrix}h_{1}\\\ \vdots\\\
h_{n}\end{pmatrix}=\dfrac{1}{d_{M}}D^{-1}\begin{pmatrix}c_{1}\\\ \vdots\\\
c_{n}\end{pmatrix}\end{split}$

for some $c_{1},\dots,c_{n}$ in $\mathcal{O}_{K}$.

######  Theorem 3.5.

If $\dfrac{1}{d_{M}}D^{-1}=(d_{ij})_{i,j=1}^{n}$, then the elements

$v_{i}=\sum_{l=1}^{n}d_{li}w_{l},\quad 0\leq i\leq n-1$

form an $\mathcal{O}_{K}$-basis of $\mathfrak{A}_{H}$. Namely, the columns of
$\dfrac{1}{d_{M}}D^{-1}$ are coordinates of basis elements for
$\mathfrak{A}_{H}$ in the fixed basis of $H$. The action of
$v_{i}\in\mathfrak{A}_{H}$ on $\mathcal{O}_{L}$ is given by

$v_{i}\cdot\gamma_{j}=M_{j}(H,L)\begin{pmatrix}d_{1i}\\\ \vdots\\\
d_{ni}\end{pmatrix}$

where resulting coordinates refer to the chosen $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$.

###### Proof.

Let $h=\sum_{l=1}^{n}h_{l}w_{l}\in H$. By Theorem 3.4,

$h\in\mathfrak{A}_{H}\iff\begin{pmatrix}h_{1}\\\ \vdots\\\
h_{n}\end{pmatrix}=\dfrac{1}{d_{M}}D^{-1}\begin{pmatrix}c_{1}\\\ \vdots\\\
c_{n}\end{pmatrix}\iff h=\sum_{i=1}^{n}c_{i}\begin{pmatrix}d_{1i}\\\ \vdots\\\
d_{ni}\end{pmatrix}\iff h\in\langle v_{1},...,v_{n}\rangle_{\mathcal{O}_{K}}.$

Then $\\{v_{1},...,v_{n}\\}$ is a system of generators of $\mathfrak{A}_{H}$.
Now, it is $K$-linearly independent because
$\dfrac{1}{d_{M}}D^{-1}\in\operatorname{GL}_{n}(K)$ and
$\\{w_{i}\\}_{i=1}^{n}$ is a $K$-basis, so it is also
$\mathcal{O}_{K}$-linearly independent and hence an $\mathcal{O}_{K}$-basis
for $\mathfrak{A}_{H}$.

As we said before, $M_{j}(H,L)$ is the matrix of the $K$-linear map

$\begin{array}[]{ccc}H&\to&L\\\ h&\mapsto&h\cdot\gamma_{j}\end{array}$

in the chosen basis. Therefore, in order to obtain $v_{i}\cdot\gamma_{j}$ we
just need the coordinates of $v_{i}$ in basis $w_{1},\dots,w_{n}$ and these
are precisely the elements of the $i$-th column of $\dfrac{1}{d_{M}}D^{-1}$. ∎

######  Remark 3.6.

We haven’t made use of the upper triangular nature of $\dfrac{1}{d_{M}}D^{-1}$
because the result is independent from that. For any
$D_{U}\in\mathrm{Mat}_{n}(K)$ such that

$U\,M(H,L)=\begin{pmatrix}D_{U}\\\ \hline\cr\\\\[-8.61108pt] O\end{pmatrix}$

with $U\in GL_{n^{2}}(\mathcal{O}_{K})$, we will have
$D_{U}\in\operatorname{GL}_{n}(K)$ and the columns of $D_{U}^{-1}$ will
provide an $\mathcal{O}_{K}$-basis for the associated order.

######  Example 3.7.

Let $L=K(\sqrt{a}\,)$ be a quadratic Galois extension. Then, its unique Hopf
Galois structure is the classical one, which has $H=K[1,\sigma]$ with
$\sigma(\sqrt{a}\,)=-\sqrt{a}$. Consider the basis $\\{1,\sigma\\}$ of $H$ and
assume that $\\{1,\sqrt{a}\\}$ is an integral basis, then the matrix of the
action is

$M(H,L)=\begin{pmatrix}1&1\\\ 0&0\\\ 0&0\\\ 1&-1\\\ \end{pmatrix}.$

With a permutation of rows, we get

$D_{U}=\begin{pmatrix}1&1\\\ 1&-1\end{pmatrix}.$

and the columns of the inverse matrix provide the basis of the associated
order (which is the maximal order) of orthogonal idempotents

$\dfrac{1+\sigma}{2},\dfrac{1-\sigma}{2}.$

On the other hand, computation of the Hermite normal form gives

$D=\begin{pmatrix}1&1\\\ 0&2\end{pmatrix}.$

which provides the basis $1,\dfrac{-1+\sigma}{2}$. If we transform $D$ into

$\begin{pmatrix}1&-1\\\ 0&2\end{pmatrix}.$

we obtain basis $1,\dfrac{1+\sigma}{2}$.

######  Example 3.8.

In the above motivating example, we took a matrix $U$ which was just a
permutation matrix acting on the rows of $M(H,L)$ and compute a basis using
the corresponding $D_{U}$. We have $d_{M}=1$ and when we compute the Hermite
normal form of $M(H,L)$ we obtain

$D=\begin{pmatrix}1&0&2\\\ 0&3&3\\\ 0&0&6\end{pmatrix}\,,\qquad
D^{-1}=\begin{pmatrix}1&0&-1/3\\\ 0&1/3&-1/6\\\ 0&0&1/6\end{pmatrix}.$

Therefore, another basis for the associated order would be

$w_{1},\ \dfrac{1}{3}w_{2},\ -\dfrac{1}{6}(2w_{1}+w_{2}-w_{3}).$

If we perform further reduction steps on matrix $D$

$\begin{pmatrix}1&0&2\\\ 0&3&3\\\ 0&0&6\end{pmatrix}\to\begin{pmatrix}1&0&2\\\
0&3&3\\\ 0&0&3\end{pmatrix}\to\begin{pmatrix}1&0&-1\\\ 0&3&0\\\
0&0&3\end{pmatrix}$

we obtain yet another basis:

$w_{1},\ \dfrac{w_{2}}{3},\ \dfrac{w_{1}+w_{3}}{3}.$

######  Example 3.9.

Now we consider $E=\mathbb{Q}_{3}(\alpha)$ with $\alpha$ a root of
$f(x)=x^{3}+3x^{2}+3\in\mathbb{Q}_{3}[x]$. As in Example 2.3, the normal
closure $L/\mathbb{Q}_{3}$ is dihedral of degree $6$, and the cubic extension
$E/\mathbb{Q}_{3}$ has an unique Hopf Galois structure with Hopf algebra
$H_{1}$ generated by elements $w_{1}$, $w_{2}$ and $w_{3}$ defined in a
similar way; namely, we take a $3$-cycle $r$ of
$\mathrm{Gal}(L/\mathbb{Q}_{3})$ and $z=\sqrt{-1}$. With respect to this
$\mathbb{Q}_{3}$-basis of $H_{1}$ and the $\mathbb{Q}_{3}$-basis of $L$
corresponding to the powers of $\alpha$, the action of $H_{1}$ on $E$ is given
by

$\begin{tabular}[]{c|ccc}&\begin{tabular}[c]{@{}l@{}}$1$\end{tabular}&\begin{tabular}[c]{@{}l@{}}$\alpha$\end{tabular}&\begin{tabular}[c]{@{}l@{}}$\alpha^{2}$\end{tabular}\\\
\hline\cr$w_{1}$&$1$&$\alpha$&$\alpha^{2}$\\\
$w_{2}$&$0$&$3+9\alpha+2\alpha^{2}$&$-3-30\alpha-9\alpha^{2}$\\\
$w_{3}$&$2$&$-3-\alpha$&$9-\alpha^{2}$\\\ \end{tabular},$ (3)

Since $\alpha$ is a root of a $3$-Eisenstein polynomial, powers of $\alpha$
give a $\mathbb{Z}_{3}$-basis of $\mathcal{O}_{E}$. We can apply the previous
method to compute the associated order $\mathfrak{A}_{H_{1}}$. We have

$\begin{pmatrix}1&0&0&0&0&0&0&0&0\\\ 0&0&0&-1&0&2&0&0&0\\\
0&0&0&-2&0&3&0&0&0\\\ 0&1&0&0&0&0&0&0&0\\\ 0&0&1&0&0&0&0&0&0\\\
-1&0&0&-1&1&-3&0&0&0\\\ 0&0&0&3&0&-3&1&0&0\\\ 0&0&0&0&0&15&0&1&0\\\
-1&0&0&-1&0&6&0&0&1\end{pmatrix}\begin{pmatrix}1&0&2\\\ 0&0&0\\\ 0&0&0\\\
0&3&-3\\\ 1&9&-1\\\ 0&2&0\\\ 0&-3&9\\\ 0&-30&0\\\
1&-9&-1\end{pmatrix}=\begin{pmatrix}1&0&2\\\ 0&1&3\\\ 0&0&6\\\ 0&0&0\\\
0&0&0\\\ 0&0&0\\\ 0&0&0\\\ 0&0&0\\\ 0&0&0\\\ \end{pmatrix}.$

Hence,

$\mathfrak{A}_{H_{1}}=\mathbb{Z}_{3}\left[w_{1},w_{2},\frac{-2w_{1}-3w_{2}+w_{3}}{6}\right].$

If we perform further reduction steps on matrix $D$

$\begin{pmatrix}1&0&2\\\ 0&1&3\\\ 0&0&6\end{pmatrix}\to\begin{pmatrix}1&0&2\\\
0&1&3\\\ 0&0&3\end{pmatrix}\to\begin{pmatrix}1&0&-1\\\ 0&1&0\\\
0&0&3\end{pmatrix}$

we obtain

$\mathfrak{A}_{H_{1}}=\mathbb{Z}_{3}\left[w_{1},w_{2},\dfrac{w_{1}+w_{3}}{3}\right].$

##  4 Freeness of the ring of integers over the associated order

We keep the hypothesis of Section 3. In this part we deal with the problem of
the structure of $\mathcal{O}_{L}$ as $\mathfrak{A}_{H}$-module. Namely, we
would like to find conditions for the freeness of such module.

###  4.1 Characterization of freeness

Let $H$ be the Hopf algebra acting on $L$ and let $\\{w_{i}\\}_{i=1}^{n}$ be a
$K$-basis of $H$ and $\\{\gamma_{j}\\}_{j=1}^{n}$ an $\mathcal{O}_{K}$-basis
of $\mathcal{O}_{L}$. We proved in Theorem 3.5 that the elements

$v_{i}=\sum_{l=1}^{n}d_{li}w_{l},\,1\leq i\leq n$

form an $\mathcal{O}_{K}$-basis of $\mathfrak{A}_{H}$. Standard linear algebra
yields that an element $\beta\in\mathcal{O}_{L}$ is a free generator of
$\mathcal{O}_{L}$ as $\mathfrak{A}_{H}$-module if and only if the elements
$v_{i}\cdot\beta\in\mathcal{O}_{L}$ form an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$, that is, such set is both $\mathcal{O}_{K}$-linearly
independent and an $\mathcal{O}_{K}$-system of generators of
$\mathcal{O}_{L}$.

For a given element $\beta\in\mathcal{O}_{L}$, we denote by
$\mathcal{D}_{\beta}(H,L)$ the matrix whose $i$-th column are the coordinates
of $v_{i}\cdot\beta$ with respect to the fixed $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$. That is, if $\beta=\sum_{j=1}^{n}\beta_{j}\gamma_{j}$ and
$D^{-1}$ is the matrix whose columns provide the basis of the associated
order, then

$\mathcal{D}_{\beta}(H,L)=\sum_{j=1}^{n}\beta_{j}M_{j}(H,L)D^{-1}.$

It is trivial that $\det{\mathcal{D}_{\beta}(H,L)}\neq 0$ is a necessary
condition for the elements $v_{i}\cdot\beta$ being an $\mathfrak{A}_{H}$-basis
of $\mathcal{O}_{L}$. However, it does not imply that
$\\{v_{i}\cdot\beta\\}_{i=1}^{n}$ is a system of generators. What is true is
that it is equivalent to the unimodularity of $\mathcal{D}_{\beta}(H,L)$. In
order to prove this, we assume that $\mathcal{D}_{\beta}(H,L)$ is unimodular
and we recall the following.

######  Proposition 4.1.

Let $R$ be a noetherian integral domain with field of fractions $K$ and let
$A\subset B$ be $R$-algebras which are $R$-orders in a finite separable
$K$-algebra $L$. Then $A=B$ if and only if
$\mathrm{disc}_{R}(A)=\mathrm{disc}_{R}(B)$.

###### Proof.

See [6, (22.4)]. ∎

In our case, $\mathfrak{A}_{H}\cdot\beta\subset\mathcal{O}_{L}$, and whenever
the $v_{i}\cdot\beta$ are $\mathcal{O}_{K}$-linearly independent
$\mathfrak{A}_{H}\cdot\beta$ is an $\mathcal{O}_{K}$-order of $L$. Under the
hypothesis $\mathcal{D}_{\beta}(H,L)\in\operatorname{GL}_{n}(\mathcal{O}_{K})$
this is the case. By [6, (22.2)],

$\mathrm{disc}_{\mathcal{O}_{K}}(\mathfrak{A}_{H}\cdot\beta)=[\mathcal{O}_{L}:\mathfrak{A}_{H}\cdot\beta]^{2}\,\mathrm{disc}_{\mathcal{O}_{K}}(\mathcal{O}_{L}),$

where $[\mathcal{O}_{L}:\mathfrak{A}_{H}\cdot\beta]$ is the generalized module
index of the free $\mathcal{O}_{K}$-modules $\mathcal{O}_{L}$ and
$\mathfrak{A}_{H}\cdot\beta$ as defined in [4, Page 10]. Then, the two
discriminants coincide if and only if the index
$[\mathcal{O}_{L}:\mathfrak{A}_{H}\cdot\beta]$ is in $\mathcal{O}_{K}^{*}$. We
compute the index by taking the $\mathcal{O}_{K}$-bases
$\\{v_{i}\cdot\beta\\}_{i=1}^{n}$ of $\mathfrak{A}_{H}\cdot\beta$ and
$\\{\gamma_{j}\\}_{j=1}^{n}$ of $\mathcal{O}_{L}$, and we find that
$[\mathcal{O}_{L}:\mathfrak{A}_{H}\cdot\beta]=\mathrm{det}(\mathcal{D}_{\beta}(H,L))$,
which belongs to $\mathcal{O}_{K}^{*}$. Conversely, if
$\\{v_{i}\cdot\beta\\}_{i=0}^{n-1}$ is an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$, then $\mathcal{D}_{\beta}(H,L)$ is a change of basis
matrix, so it is unimodular. Hence, we have the following criterion.

######  Proposition 4.2.

An element $\beta\in\mathcal{O}_{L}$ is an $\mathfrak{A}_{H}$-free generator
of $\mathcal{O}_{L}$ if and only if the associated matrix
$\mathcal{D}_{\beta}(H,L)$ is unimodular.

######  Example 4.3.

Let $L/K$ be a quadratic extension of $p$-adic fields, which is known to have
the classical Galois structure $K[G]$ as its unique Hopf Galois structure.
When $p\geq 3$, $L/K$ is tamely ramified and then $\mathcal{O}_{L}$ is
$\mathfrak{A}_{K[G]}$-free. Let $z\in L$ such that $z\notin K$, $z^{2}\in K$
and $\mathcal{O}_{L}=\mathcal{O}_{K}[z]$.

By Example 3.7, a basis of the associated order is $\eta_{1}=1$,
$\eta_{2}^{\prime}=\dfrac{-1+\sigma}{2}$. The action of this basis over $L$ is

$\mathfrak{A}_{K[G]}$ | $1$ | $z$
---|---|---
$\eta_{1}$ | $1$ | $z$
$\eta_{2}^{\prime}$ | $0$ | $-z$

Thus,

$\mathcal{D}_{1+z}(K[G],L)=\begin{pmatrix}1&0\\\ 1&-1\end{pmatrix}$

which is in $\operatorname{GL}_{n}(\mathcal{O}_{K})$. Therefore $1+z$
generates $\mathcal{O}_{L}$ as $\mathfrak{A}_{K[G]}$-module.

##  5 Induced Hopf Galois structures

Let $L/K$ be a finite Galois extension of fields with Galois group $G$. The
Greither-Pareigis theorem applied to this situation gives that Hopf Galois
structures of $L/K$ have Hopf algebra $H=L[N]^{G},$ where $N$ is a regular
subgroup of $\mathrm{Perm}(G)$ normalized by $\lambda(G)$. Here,
$\lambda:G\to\operatorname{Perm}(G)$ is the left regular representation of
$G$.

Assume that $G$ decomposes as a semidirect product $G=J\rtimes G^{\prime},$
where $J$ is a normal subgroup of $G$, normal complement for the subgroup
$G^{\prime}$. Let us denote $E=L^{G^{\prime}}$, $F=L^{J}$, $r=[E:K]$,
$u=[F:K]$. We have a lattice of intermediate fields

$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u}$$\scriptstyle{G^{\prime}}$$\scriptstyle{J}$$\scriptstyle{r}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
F}$$\scriptstyle{u}$$\textstyle{K}$

with $E\cap F=K$ and $L=EF$. Since $L/K$ is Galois, $L/E$ is Galois with
Galois group $G^{\prime}$. By the Greither-Pareigis theorem, Hopf Galois
structures of $L/E$ are in one-to-one correspondence with regular subgroups of
$\mathrm{Perm}(G^{\prime})$ normalized by $\lambda^{\prime}(G^{\prime})$.
Here, $\lambda^{\prime}:G^{\prime}\to\mathrm{Perm}(G^{\prime})$ is the left
regular representation of $G^{\prime}$.

On the other hand, if $\widetilde{E}/K$ is the normal closure of $E/K$ in
$L/K$, then Hopf Galois structures of $E/K$ are in one-to-one correspondence
with regular subgroups of $\mathrm{Perm}(\widetilde{X})$ normalized by
$\widetilde{\lambda}(G)$, where
$\widetilde{X}=\mathrm{Gal}(\widetilde{E}/K)/\mathrm{Gal}(\widetilde{E}/E)$
and
$\tilde{\lambda}:\mathrm{Gal}(\widetilde{E}/K)\to\mathrm{Perm}(\widetilde{X})$
is given by the action of $\mathrm{Gal}(\widetilde{E}/K)$ by left translation
on left cosets. But the Fundamental Theorem of Galois Theory gives us the
isomorphisms

$\mathrm{Gal}(\widetilde{E}/K)\cong\mathrm{Gal}(L/K)/\mathrm{Gal}(L/\widetilde{E}),$
$\mathrm{Gal}(\widetilde{E}/E)\cong\mathrm{Gal}(L/E)/\mathrm{Gal}(L/\widetilde{E}).$

Hence, there is a bijection
$\widetilde{X}\cong\mathrm{Gal}(L/K)/\mathrm{Gal}(L/E)=G/G^{\prime}=X\cong J$.
Taking the actions into account, the action of $\mathrm{Gal}(\widetilde{E}/K)$
by left translation on left cosets corresponds to the action of $G$ on left
cosets. As for the second isomorphism, if we let $\sigma_{1},\dots,\sigma_{r}$
of $J$ be a left transversal of $G^{\prime}$, then the action of an element
$g=\sigma\tau\in G$, with $\sigma\in J$ and $\tau\in G^{\prime}$, is the
following:

$\sigma\tau\cdot\sigma_{i}G^{\prime}=\sigma\tau\sigma_{i}\tau^{-1}\tau
G^{\prime}=\sigma(\tau\sigma_{i}\tau^{-1})G^{\prime}$

This gives an action $\lambda_{c}:G\to\mathrm{Perm}(J)$.

We build a Hopf Galois structure of $L/K$ from Hopf Galois structures of $L/E$
and $E/K$ as follows:

######  Theorem 5.1 (Induction).

Assume that $N_{1}\subseteq\mathrm{Perm}(X)=\mathrm{Perm}(J)$ is regular and
normalized by $\lambda_{c}(G)$, namely it gives $E/K$ a Hopf Galois structure,
and $N_{2}\subseteq\mathrm{Perm}(G^{\prime})$ is regular and normalized by
$\lambda^{\prime}(G^{\prime})$, namely it gives $L/E$ a Hopf Galois structure.
Then, $L/K$ has a Hopf Galois structure of type $N_{1}\times N_{2}$.

###### Proof.

See [10, Theorem 3]. ∎

Hopf Galois structures of $L/K$ are given by the regular subgroups of
$\mathrm{Perm}(G)$ normalized by $\lambda(G)$. The definition of such a group
relies on the following factorization of $\lambda$

$\begin{array}[]{ccccc}G&\to&\mathrm{Perm}(J)\times\mathrm{Perm}(G^{\prime})&\hookrightarrow&\mathrm{Perm}(G)\\\
\sigma\tau&\mapsto&\big{(}\lambda_{c}(\sigma\tau),\lambda^{\prime}(\tau)\big{)}&&\\\
&&(\varphi,\psi)&\mapsto&\iota(\varphi,\psi)\end{array}$

where $\iota(\varphi,\psi)(\sigma\tau)=\varphi(\sigma)\psi(\tau).$ The group
$N=\iota(N_{1}\times N_{2})$ provides a Hopf Galois structure for $L/K$.

######  Remark 5.2.

Let us remark that this notion of induction includes the case when $G$ is a
direct product. By iteration we could deal for example with families like
nilpotent groups, which are direct product of their Sylow subgroups. In this
sense, Theorem 5 of [3] would say that for nilpotent Galois extensions all
Hopf Galois structures of nilpotent type are induced.

Although a regular subgroup of $\mathrm{Perm}(G)$ normalized by $\lambda(G)$
is enough to completely determine a Hopf Galois structure of $L/K$, our goal
is to give a more precise description of the Hopf algebra and the Hopf action
of an induced Hopf Galois structure for a better understanding of how
induction works and how we can apply the methods of previous section to study
associated orders and freeness of the ring of integers.

###  5.1 Induced Hopf algebras

Our first aim is to describe the Hopf algebra of an induced Hopf Galois
structure of $L/K$ in terms of the Hopf algebras of the Hopf Galois structures
from which it is built. To this end, first of all we prove that in fact
induction can also be defined in terms of Hopf Galois structures of the
subextensions with base field $K$.

######  Proposition 5.3.

Let $L/K$ be a Galois extension with Galois group $G=J\rtimes G^{\prime}$ and
let $E=L^{G^{\prime}}$, $F=L^{J}$. Then, there is an one-to-one correspondence
between Hopf Galois structures of $F/K$ and Hopf Galois structures of $L/E$.

###### Proof.

Since $J$ is a normal subgroup of $G$, the extension $F/K$ is Galois with
Galois group $G/J$. Its Hopf Galois structures are in one to one
correspondence with regular subgroups of $\operatorname{Perm}(G/J)$ normalized
by $\lambda^{G/J}(G/J)$, where $\lambda^{G/J}:G/J\to\operatorname{Perm}(G/J)$
is the left regular representation of this quotient group. Via the isomorphism

$\begin{array}[]{c}G/J\longleftrightarrow G^{\prime}\\\
\overline{\tau}\longleftrightarrow\tau\end{array}$

we can identify $\operatorname{Perm}(G/J)$ with
$\operatorname{Perm}(G^{\prime})$ and under this identification
$\lambda^{G/J}(G/J)$ identifies with $\lambda^{\prime}(G^{\prime})$. Therefore
regular subgroups of $\operatorname{Perm}(G^{\prime})$ normalized by
$\lambda^{\prime}(G^{\prime})$ are in one-to-one correspondence with regular
subgroups of $\operatorname{Perm}(G/J)$ normalized by $\lambda^{G/J}(G/J)$. ∎

If we use the same name $N_{2}$ for such a regular subgroup the corresponding
Hopf algebras are $\overline{H}=F[N_{2}]^{G/J}$ and
$H_{2}=L[N_{2}]^{G^{\prime}}$, so that $H_{2}=E\otimes_{K}\overline{H}$ and
$\overline{H}=H_{2}^{J}$ (see [12, Section 3] for a more general framework).

######  Remark 5.4.

The actions of the Hopf algebras involved work as follows:

  * •

$E\otimes_{K}\overline{H}$ acts on $L=E\otimes_{K}F$ through the product on
$E$ in the first term and the Hopf action in the second one.

  * •

The action of the Hopf Galois structure $H_{2}=L[N_{2}]^{G^{\prime}}$ of $L/E$
on $L$ is $J$-equivariant, namely $\sigma(h\cdot x)=\sigma(h)\cdot\sigma(x)$
for $\sigma\in J$, $h\in H_{2}$ and $x\in L$. Indeed, $J$ acts on $L$ by the
classical Galois action and on $N_{2}$ by conjugation, but this last action
turns out to be trivial (see the proof of [12, Theorem 3.1]). Consequently,
the restricted action of $H_{2}^{J}$ on $F=L^{J}$ makes sense, and this is the
Hopf Galois action of $\overline{H}$ on $F$.

$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
L\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H_{2}}$$\scriptstyle{H}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H_{1}}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
F}$$\scriptstyle{\overline{H}}$$\textstyle{K}$

According to [12, Theorem 3.1], the Hopf algebra of an induced Hopf Galois
structure fits in a short exact sequence with the Hopf Galois structures
$H_{1}$ and $\overline{H}$ from which it is induced. But in our case there is
a deeper relation: $H$ is actually the tensor product
$H_{1}\otimes_{K}\overline{H}$.

######  Proposition 5.5.

Let $N_{1}\subseteq\mathrm{Perm}(J)$ be regular and normalized by
$\lambda_{c}(G)$ and let $N_{2}\subseteq\mathrm{Perm}(G/J)$ be regular and
normalized by $\lambda^{G/J}(G/J)$. Then $N=\iota(N_{1}\times
N_{2})\subseteq\mathrm{Perm}(G)$ gives the induced Hopf Galois structure of
$L/K$. Therefore, the corresponding Hopf algebras are $H=L[N]^{G}$,
$H_{1}=L[N_{1}]^{G}$ and $\overline{H}=F[N_{2}]^{G/J}$. Then,

$H=H_{1}\otimes_{K}\overline{H}.$

###### Proof.

$\begin{array}[]{rcl}H&=&L[N]^{G}=L[\iota(N_{1}\times
N_{2})]^{G}=\bigl{(}L[\iota(N_{1}\times 1)\times\iota(1\times
N_{2})]\bigr{)}^{G}\\\\[8.61108pt] &=&\bigl{(}L[\iota(N_{1}\times
1)]\otimes_{K}L[\iota(1\times
N_{2})]\bigr{)}^{G}=\bigl{(}L[N_{1}]\otimes_{K}L[N_{2}]\bigr{)}^{G}\end{array}$

Since the action of conjugation by $\lambda(G)$ on $\operatorname{Perm}(G)$
factors through $\operatorname{Perm}(J)\times\operatorname{Perm}(G/J)$ as
conjugation by $\lambda_{c}(G)$ on the first component and conjugation by
$\lambda^{G/J}(G/J)$ on the second one, we have

$\begin{array}[]{rcl}H&=&L[N_{1}]^{G}\otimes_{K}F[N_{2}]^{G/J}=H_{1}\otimes\overline{H}\end{array}$

∎

######  Example 5.6.

Let us assume that $G\cong D_{2p}$, the dihedral group of order $2p$ with $p$
an odd prime, and let us fix a presentation $G=\langle r,s\mid
r^{p}=s^{2}=1,\,sr=r^{-1}s\rangle.$ Then $G=J\rtimes G^{\prime}$ with
$J=\langle r\rangle$ the unique order $p$ subgroup of $G$ and $G^{\prime}$ any
of the $p$ different order $2$ subgroups $G^{\prime}_{d}=\langle
r^{d}s\rangle$ with $0\leq d\leq p-1$. Therefore, $F=L^{J}/K$ is the unique
degree $2$ subextension of $L/K$, while there are $p$ possible degree $p$
subextensions $E_{d}/K$ of $L/K$, $E_{d}=L^{\langle r^{d}s\rangle}$. Hence,
$L/K$ has $p$ induced Hopf Galois structures, which are of cyclic type
$C_{2p}$ and have respective Hopf algebras

$H_{d}=L[\langle\lambda(r)\rangle]^{G}\otimes_{K}F[\langle\lambda^{\prime}(r^{d}s)\rangle]^{G/J},\
0\leq d\leq p-1.$

The action on the second factor is obtained taking the elements of
$G^{\prime}_{d}$ as a system of representatives for $G/J$ and make them act by
conjugation. Therefore, the action is trivial and we have the group algebra
$K[G/J]$. In fact, the classical Galois structure is the unique Hopf Galois
structure of a quadratic extension.

######  Example 5.7.

Let $F$ be an imaginary quadratic field, and let $\mathcal{O}$ be the order of
the conductor $f$ in $F$. Let $L$ be the ring class field associated to
$\mathcal{O}$. Then $L/F$ is an abelian extension and $L/\mathbb{Q}$ is
Galois. By [8, Lemma 9.3],
$\mathrm{Gal}(L/\mathbb{Q})=\mathrm{Gal}(L/F)\rtimes G^{\prime},$ where
$G^{\prime}$ is a group of order $2$. Let $E=L^{G^{\prime}}$. Then,
$J=\mathrm{Gal}(L/K)$ is a normal complement of $G^{\prime}$, so the Hopf
algebras of the induced Hopf Galois structures of $L/\mathbb{Q}$ are
$H=H_{1}\otimes_{K}\overline{H},$ where $H_{1}$ gives the Hopf Galois
structure of $E/\mathbb{Q}$ and $\overline{H}$ is the group algebra of
$\mathrm{Gal}(F/\mathbb{Q})$, since the classical one is again the unique Hopf
Galois structure of this extension.

###  5.2 Induced Hopf actions and their representations

In this section we study the Hopf action of an induced Hopf Galois structure
$H=L[N]^{G}$ of $L/K$ in terms of the action of $H_{1}=L[N_{1}]^{G}$ on $E$
and the action of $\overline{H}=F[N_{2}]^{G/J}$ on $F$. We prove item $5$ of
Theorem 1.3 in the following result. The terminology for representations
follows [13, section 3.2] which covers the case of $G$ being a direct product
and considering classical Galois actions.

######  Proposition 5.8.

With the previous notations for induced Hopf Galois structures, let
$\rho_{H}:H\to\operatorname{End}_{K}(L)$,
$\rho_{H_{1}}:H_{1}\to\operatorname{End}_{K}(E)$ and
$\rho_{\overline{H}}:\overline{H}\to\operatorname{End}_{K}(F)$ be the
representations obtained from the respective Hopf actions. We have
$L=E\otimes_{K}F$, $H=H_{1}\otimes\overline{H}$ and

$\rho_{H}=\rho_{H_{1}}\otimes\rho_{\overline{H}}$

That is, for $w\in H_{1}$, $\eta\in\overline{H}$, $\alpha\in E$ and $z\in F$,
$(w\otimes\eta)\cdot(\alpha\otimes z)=(w\cdot\alpha)\otimes(\eta\cdot z).$

###### Proof.

As $w\in L[N_{1}]^{G}$ and $\eta\in F[N_{2}]^{G/J}$, let us write

$w=\sum_{i=1}^{r}c_{i}n_{i}^{(1)},\,c_{i}\in
L,\qquad\eta=\sum_{j=1}^{u}d_{j}n_{j}^{(2)},\,d_{j}\in F,$

where $N_{1}=\\{n_{i}^{(1)}\\}_{i=1}^{r}$ and
$N_{2}=\\{n_{j}^{(2)}\\}_{j=1}^{u}$. Recall that
$\iota(n_{i}^{(1)},n_{j}^{(2)})(Id_{G})=n_{i}^{(1)}(\mathrm{Id}_{J})n_{j}^{(2)}(\mathrm{Id}_{G^{\prime}})$.
Then:

$\begin{split}(w\otimes\eta)\cdot(\alpha\otimes
z)&=\left(\sum_{i=1}^{r}\sum_{j=1}^{u}c_{i}d_{j}\iota\bigl{(}(n_{i}^{(1)},n_{j}^{(2)})\bigr{)}\right)\cdot(\alpha\otimes
z)\\\
&=\sum_{i=1}^{r}\sum_{j=1}^{u}c_{i}d_{j}\iota\bigl{(}(n_{i}^{(1)},n_{j}^{(2)})\bigr{)}^{-1}(\mathrm{Id_{G}})(\alpha
z)\\\
&=\sum_{i=1}^{r}\sum_{j=1}^{u}c_{i}d_{j}\iota\bigl{(}((n_{i}^{(1)})^{-1},(n_{j}^{(2)})^{-1})\bigr{)}(\mathrm{Id_{G}})(\alpha
z)\\\
&=\sum_{i=1}^{r}\sum_{j=1}^{u}c_{i}d_{j}(n_{i}^{(1)})^{-1}(\mathrm{Id_{J}})(\alpha)(n_{j}^{(2)})^{-1}(\mathrm{Id_{G^{\prime}}})(z)\\\
&=\left(\sum_{i=1}^{r}c_{i}(n_{i}^{(1)})^{-1}(\mathrm{Id_{J}})(\alpha)\right)\left(\sum_{j=1}^{u}d_{j}(n_{j}^{(2)})^{-1}(\mathrm{Id_{G^{\prime}}})(z)\right)\\\
&=\left(\sum_{i=1}^{r}c_{i}n_{i}^{(1)}\right)\cdot\alpha\left(\sum_{j=1}^{u}d_{j}n_{j}^{(2)}\right)\cdot
z=(w\cdot\alpha)\otimes(\eta\cdot z).\end{split}$

∎

Therefore, by suitable choices of basis, the matrices of the action of $H$ are
Kronecker products of the matrices of the actions of $H_{1}$ and
$\overline{H}$.

######  Example 5.9.

We consider again Example 2.3. We have

$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
L=\mathbb{Q}_{3}(\alpha,z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H}$$\textstyle{E=\mathbb{Q}_{3}(\alpha)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H_{1}}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
F=\mathbb{Q}_{3}(z)}$$\scriptstyle{\overline{H}}$$\textstyle{\mathbb{Q}_{3}}$

and the action of $H_{1}$ on $E$ and the action of $\overline{H}$ on $F$ are
given by

$\begin{tabular}[]{c|ccc}&\begin{tabular}[c]{@{}l@{}}$1$\end{tabular}&\begin{tabular}[c]{@{}l@{}}$\alpha$\end{tabular}&\begin{tabular}[c]{@{}l@{}}$\alpha^{2}$\end{tabular}\\\
\hline\cr$w_{1}$&$1$&$\alpha$&$\alpha^{2}$\\\
$w_{2}$&$0$&$3\alpha$&$-3\alpha^{2}$\\\ $w_{3}$&$2$&$-\alpha$&$-\alpha^{2}$\\\
\end{tabular}\qquad\quad\begin{tabular}[]{ c |c c }&1&$z$\\\
\hline\cr$\eta_{1}$&1&$z$\\\ $\eta_{2}$&1&$-z$\\\ \end{tabular},$ (4)

respectively. The Hopf action of the induced structure is immediately computed
from these two actions. One can also check that the structure is given by
$N=\langle g\rangle\subseteq\operatorname{Perm}(G)$, where
$g=(1,rs,r^{2},s,r,r^{2}s)$, and the Hopf action goes as follows:

| $1$ | $z$ | $\alpha$ | $\alpha z$ | $\alpha^{2}$ | $\alpha^{2}z$
---|---|---|---|---|---|---
$id=w_{1}\otimes\eta_{1}$ | $1$ | $z$ | $\alpha$ | $\alpha z$ | $\alpha^{2}$ | $\alpha^{2}z$
$g^{3}=w_{1}\otimes\eta_{2}$ | $1$ | $-z$ | $\alpha$ | $-\alpha z$ | $\alpha^{2}$ | $-\alpha^{2}z$
$z(g^{-2}-g^{2})=w_{2}\otimes\eta_{1}$ | $0$ | $0$ | $3\alpha$ | $3\alpha z$ | $-3\alpha^{2}$ | $-3\alpha^{2}z$
$z(g-g^{-1})=w_{2}\otimes\eta_{2}$ | $0$ | $0$ | $3\alpha$ | $-3\alpha z$ | $-3\alpha^{2}$ | $3\alpha^{2}z$
$g^{2}+g^{-2}=w_{3}\otimes\eta_{1}$ | $2$ | $2z$ | $-\alpha$ | $-\alpha z$ | $-\alpha^{2}$ | $-\alpha^{2}z$
$g+g^{-1}=w_{3}\otimes\eta_{2}$ | $2$ | $-2z$ | $-\alpha$ | $\alpha z$ | $-\alpha^{2}$ | $\alpha^{2}z$

By fixing the basis $\\{1,z,\alpha,\alpha z,\alpha^{2},\alpha^{2}z\\}$ of $L$,
we can compute the matrix representing the element $w_{i}\otimes\eta_{j}$ from
the previous table, which coincides with the Kronecker product of the ones
representing $w_{i}$ and $\eta_{j}$. For example, the matrix representing
$w_{2}\otimes\eta_{2}$ is

$\left(\begin{array}[]{cccccc}0&0&0&0&0&0\\\ 0&0&0&0&0&0\\\ 0&0&3&0&0&0\\\
0&0&0&-3&0&0\\\ 0&0&0&0&-3&0\\\ 0&0&0&0&0&3\\\
\end{array}\right)=\begin{pmatrix}0&0&0\\\ 0&3&0\\\ 0&0&-3\\\
\end{pmatrix}\otimes\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

###  5.3 Associated order of an induced Hopf Galois action

Let us assume that $L/K$ is any field extension with an induced Hopf Galois
structure. We recall the setting of Section 5.1: $G=J\rtimes G^{\prime}$
$E=L^{G^{\prime}}$ and $F=L^{J}$, $E/K$ is $H_{1}$-Galois and $F/K$ is
$\overline{H}$-Galois and we have $H=H_{1}\otimes_{K}\overline{H}$.

We assume that the Hopf Galois structure is induced and has
$H=H_{1}\otimes_{K}\overline{H}$. First of all we analyze the behaviour of the
matrix of the action with respect to the tensor product. Let us fix a
$K$-basis $\\{w_{i}\\}_{i=1}^{r}$ of $H_{1}$ and a $K$-basis
$\\{\eta_{j}\\}_{j=1}^{u}$ of $\overline{H}$. We also fix $K$-bases
$\\{\alpha_{k}\\}_{k=1}^{r}$ of $E$ and $\\{z_{l}\\}_{l=1}^{u}$ of $F$

######  Theorem 5.10.

We consider product bases in $H$ and $L$. Then, there exists a permutation
matrix $P$ of size $n^{2}$ such that

$PM(H,L)=M(H_{1},E)\otimes M(\overline{H},F).$

###### Proof.

Since $H=H_{1}\otimes_{K}\overline{H}$, $\\{w_{i}\eta_{j}\,|\,1\leq i\leq
r,\,1\leq j\leq u\\}$ is a $K$-basis of $H$. As we have seen in 5.8
$\rho_{H}=\rho_{H_{1}}\otimes\rho_{\overline{H}}$. Considering the fixed bases
of $E$ and $F$ we can identify $w_{i}$ and $\eta_{j}$ with matrices in
$\mathrm{Mat}_{r}(K)$ and $\mathrm{Mat}_{u}(K)$ respectively. Then, fixing the
$K$-basis $\\{\alpha_{k}z_{l}\\}$ of $L$, the matrix of $w_{i}\eta_{j}\in H$
is

$w_{i}\eta_{j}=w_{i}\otimes\eta_{j}$

For the block matrix of the action of $H$ we have

$\begin{pmatrix}w_{1}\otimes\eta_{1}\\\ \dots\\\ w_{1}\otimes\eta_{u}\\\
w_{2}\otimes\eta_{1}\\\ \dots\\\
w_{r}\otimes\eta_{u}\end{pmatrix}=\begin{pmatrix}w_{1}\\\ w_{2}\\\ \dots\\\
w_{r}\end{pmatrix}\circ\begin{pmatrix}\eta_{1}\\\ \eta_{2}\\\ \dots\\\
\eta_{u}\end{pmatrix},$

which is the Tracy–Singh product of partitioned matrices. To obtain the matrix
of the action, we have to take coordinates with respect to the canonical basis
of the respective matrix spaces: $M(H,L)$ has columns
$\varphi(w_{i}\eta_{j})=\varphi_{n}(w_{i}\otimes\eta_{j})\in K^{n^{2}}$ while
$M(H_{1},E)$ has columns $\varphi_{r}(w_{i})\in K^{r^{2}}$ and
$M(\overline{H},F)$ has columns $\varphi_{u}(\eta_{j})\in K^{u^{2}}$.

Working with canonical bases of matrices and vectors we find the permutation.

$\varphi(E^{r}_{ab}\otimes
E^{u}_{cd})=\varphi(E^{n}_{u(a-1)+c,u(b-1)+d})=e_{nu(b-1)+n(d-1)+u(a-1)+c}\in
K^{n^{2}}$

Here, we use $e_{m}$ for vectors of the canonical basis of $K^{n^{2}}$ and
$E^{r}_{a,b}$ denotes the $r\times r$ matrix having a $1$ in position $(a,b)$
and coefficients $0$ elsewhere. Notations for the other matrices is analogous.
On the other hand, if we indicate with superindexes vectors of the canonical
basis of $K^{r^{2}}$ and $K^{u^{2}}$ respectively,

$\varphi_{r}(E^{r}_{ab})\otimes\varphi_{u}(E^{u}_{cd})=e^{r}_{r(b-1)+a}\otimes
e^{u}_{u(d-1)+c}=e_{nu(b-1)+u^{2}(a-1)+u(d-1)+c}$

Therefore, $P$ is the permutation matrix which places row
$nu(b-1)+n(d-1)+u(a-1)+c$ in position $nu(b-1)+u^{2}(a-1)+u(d-1)+c$ for $1\leq
a,b\leq r$ and $1\leq c,d\leq u$. ∎

For example, if $r=u=2$, then the matrix $P$ has size $16$ and corresponds to
the permutation $(3,5)(4,6)(11,13)(12,14)\in\mathfrak{S}_{16}$. If $r=3,u=2$,
then the matrix $P$ has size $36$ and corresponds to the permutation $(3\ 5\
9\ 7)(4\ 6\ 10\ 8)(15\ 17\ 21\ 19)(16\ 18\ 22\ 20)(27\ 29\ 33\ 31)(28\ 30\ 34\
32)\in\mathfrak{S}_{36}.$

Now we assume the hypothesis of section 3 and we assume that we deal with and
induced Hopf Galois structure with $H=H_{1}\otimes_{K}\overline{H}$. In this
context we will think of $P$ as an unimodular matrix.

We consider the three Hopf actions involved and the corresponding associated
orders $\mathfrak{A}_{H}$, $\mathfrak{A}_{H_{1}}$ and
$\mathfrak{A}_{\overline{H}}$ of $\mathcal{O}_{L}$, $\mathcal{O}_{E}$ and
$\mathcal{O}_{F}$, respectively. In what follows we study the relationship
between these objects. The main goal is to prove the following:

######  Theorem 5.11.

If $E/K$ and $F/K$ are arithmetically disjoint, then

$\mathfrak{A}_{H}=\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathfrak{A}_{\overline{H}}.$

The idea is to use the previous method to compute bases of the associated
orders involved and prove that the product of suitable bases of
$\mathfrak{A}_{H_{1}}$ and $\mathfrak{A}_{\overline{H}}$ gives a basis for
$\mathfrak{A}_{H}$.

We fix an $\mathcal{O}_{K}$-basis $\\{\alpha_{k}\\}_{k=1}^{r}$ of
$\mathcal{O}_{E}$ and an $\mathcal{O}_{K}$-basis $\\{z_{l}\\}_{l=1}^{u}$ of
$\mathcal{O}_{F}$. Since we assume that $E/K$ and $F/K$ are arithmetically
disjoint, we have
$\mathcal{O}_{L}=\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$, so
the products $\\{\alpha_{k}z_{l}\\}_{k,l}$ form an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$. These integral bases are also $K$-bases for the respective
fields and Theorem 3.3 applies.

######  Proposition 5.12.

Let us write $M(H,L)=d_{M}M$ with $d_{M}\in K$ and
$M\in\mathcal{M}_{n}(\mathcal{O}_{K})$ with coprime coefficients. The integral
matrices $\dfrac{1}{d_{M}}\,M(H,L)$ and $\dfrac{1}{d_{M}}(M(H_{1},E)\otimes
M(\overline{H},F))$ have the same Hermite normal form.

###### Proof.

We apply Theorem 5.10 and observe that since product by $P$ is just a
permutation of the rows, the content $d_{M}$ does not change.

We have an unimodular matrix $U\in GL_{n}(\mathcal{O}_{K})$ such that

$U\,M(H,L)={d_{M}}\begin{pmatrix}D\\\ \hline\cr O\\\ \end{pmatrix}$

where $D$ has coefficients in $\mathcal{O}_{K}$ and is in Hermite normal form.
Then,

$UP^{-1}\,PM(H,L)={d_{M}}\begin{pmatrix}D\\\ \hline\cr O\\\ \end{pmatrix}$

Since $UP^{-1}$ is unimodular and the Hermite normal form is unique, we have
obtained the Hermite normal form of $\dfrac{1}{d_{M}}(M(H_{1},E)\otimes
M(\overline{H},F))$. ∎

######  Proposition 5.13.

If $\dfrac{1}{d_{1}}M(H_{1},E)$ and $\dfrac{1}{\overline{d}}M(\overline{H},F)$
have integral coprime coefficients, then $d_{M}=d_{1}\overline{d}$ up to
multiplication by an unit and if

$U_{1}\,M(H_{1},E)=d_{1}\begin{pmatrix}D_{1}\\\ \hline\cr O\\\
\end{pmatrix}\qquad\overline{U}\,M(\overline{H},F)={\overline{d}}\begin{pmatrix}\
\overline{D}\ \\\ \hline\cr O\\\ \end{pmatrix}$

with $U_{1}\in GL_{r^{2}}(\mathcal{O}_{K})$, $\overline{U}\in
GL_{u^{2}}(\mathcal{O}_{K})$ and $D_{1},\overline{D}$ in Hermite normal form,
then

$D=D_{1}\otimes\overline{D}.$

###### Proof.

The relation between $d_{M}$, $d_{1}$ and $\overline{d}$ comes from the fact
that the Kronecker product of matrices with coprime coefficients is also a
matrix with coprime coefficients. Since $d_{M}$, $d_{1}$ and $\overline{d}$
are determined theirselves up to multiplication by an unit, the equality also
holds up to multiplication by an unit. We have

$(U_{1}M(H_{1},E))\otimes(\overline{U}M(\overline{H},F))=(U_{1}\otimes\overline{U})\left(M(H_{1},E)\otimes
M(\overline{H},F)\right)$

by the mixed product property of the Kronecker product. Therefore,

${d_{1}}\begin{pmatrix}D_{1}\\\ \hline\cr O\\\
\end{pmatrix}\otimes{\overline{d}}\begin{pmatrix}\ \overline{D}\ \\\ \hline\cr
O\\\ \end{pmatrix}=(U_{1}\otimes\overline{U})PM(H,L)$
${d_{M}}\begin{pmatrix}D_{1}\otimes\overline{D}\\\ \hline\cr O\\\
\end{pmatrix}=(U_{1}\otimes\overline{U})PM(H,L)$

Since
$(U_{1}\otimes\overline{U})P\in\operatorname{GL}_{n^{2}}(\mathcal{O}_{K})$ and
$D_{1}\otimes\overline{D}$ is in Hermite normal form, unicity gives
$D=D_{1}\otimes\overline{D}.$ ∎

######  Corollary 5.14.

If for a matrix $U_{1}\in GL_{r^{2}}(\mathcal{O}_{K})$ we have

$U_{1}\,M(H_{1},E)=\begin{pmatrix}D_{1}\\\ \hline\cr O\\\ \end{pmatrix}$

with $D_{1}\in\operatorname{GL}_{r}(K)$, and for a matrix $\overline{U}\in
GL_{u^{2}}(\mathcal{O}_{K})$ we have

$\overline{U}\,M(\overline{H},F)=\begin{pmatrix}\ \overline{D}\ \\\ \hline\cr
O\\\ \end{pmatrix}$

with $\overline{D}\in\operatorname{GL}_{u}(K)$, then

$(U_{1}\otimes\overline{U})PM(H,L)=\begin{pmatrix}D_{1}\otimes\overline{D}\\\
\hline\cr O\\\ \end{pmatrix}$

with
$(U_{1}\otimes\overline{U})P\in\operatorname{GL}_{n^{2}}(\mathcal{O}_{K})$ and
$D=D_{1}\otimes\overline{D}\in\operatorname{GL}_{n}(K)$.

Either in Hermite normal form or in general, since

$(D_{1}\otimes\overline{D})^{-1}=D_{1}^{-1}\otimes\overline{D}^{\,-1}$

and columns of the corresponding matrices provide bases for
$\mathfrak{A}_{H}$, $\mathfrak{A}_{H_{1}}$ and $\mathfrak{A}_{\overline{H}}$
respectively, this finishes the proof of Theorem 5.11.

######  Example 5.15.

In example 3.9, the normal closure of $E$ is $L=\mathbb{Q}_{3}(\alpha,z)$,
where $z=\sqrt{-1}$. In analogy with Example 5.9, if we call
$F=\mathbb{Q}_{3}(z)$, then $L=EF$ and we can induce a Hopf Galois structure
$H$ of $L/K$ from the Hopf Galois structure $H_{1}$ of $E/\mathbb{Q}_{3}$ and
$\overline{H}$ of $F/\mathbb{Q}_{3}$.

Since $F/\mathbb{Q}_{3}$ is unramified, the extensions $E/\mathbb{Q}_{3}$ and
$F/\mathbb{Q}_{3}$ are arithmetically disjoint and we can obtain the
associated order to the induced Hopf Galois structure of $L/\mathbb{Q}_{3}$
using the above method. We take

$D_{1}=\begin{pmatrix}1&0&-1\\\ 0&1&0\\\ 0&0&3\end{pmatrix},\qquad\overline{D}=\begin{pmatrix}1&-1\\\ 0&2\end{pmatrix}.$ $\mathfrak{A}_{H_{1}}$ |  | $1$
---
| $\alpha$
---
| $\alpha^{2}$
---
$w_{1}$ | $1$ | $\alpha$ | $\alpha^{2}$
$w_{2}$ | $0$ | $3+9\alpha+2\alpha^{2}$ | $-3-30\alpha-9\alpha^{2}$
$w_{3}^{\prime}=\dfrac{w_{1}+w_{3}}{3}$ | $1$ | $-1$ | $3$
$\mathfrak{A}_{\overline{H}}$ | $1$ | $z$
---|---|---
$\eta_{1}$ | $1$ | $z$
$\eta_{2}^{\prime}=\dfrac{\eta_{1}+\eta_{2}}{2}$ | $1$ | $0$

Then,

$D=D_{1}\otimes\overline{D}=\begin{pmatrix}1&-1&0&0&-1&1\\\ 0&2&0&0&0&-2\\\
0&0&1&-1&0&0\\\ 0&0&0&2&0&0\\\ 0&0&0&0&3&-3\\\ 0&0&0&0&0&6\end{pmatrix}\quad
D^{-1}=\begin{pmatrix}1&1/2&0&0&1/3&1/6\\\ 0&1/2&0&0&0&1/6\\\ 0&0&1&1/2&0&0\\\
0&0&0&1/2&0&0\\\ 0&0&0&0&1/3&1/6\\\ 0&0&0&0&0&1/6\end{pmatrix}$

gives

$\mathfrak{A}_{H}=\mathbb{Z}_{3}\left[\,w_{1}\eta_{1},\dfrac{w_{1}\eta_{1}+w_{1}\eta_{2}}{2},w_{2}\eta_{1},\dfrac{w_{2}\eta_{1}+w_{2}\eta_{2}}{2},\dfrac{w_{1}\eta_{1}+w_{3}\eta_{1}}{3},\dfrac{w_{1}\eta_{1}+w_{1}\eta_{2}+w_{3}\eta_{1}+w_{3}\eta_{2}}{6}\,\right]$
$\mathfrak{A}_{H}=\mathbb{Z}_{3}\left[\,w_{1}\eta_{1},w_{1}\eta_{2}^{\prime},w_{2}\eta_{1},w_{2}\eta_{2}^{\prime},w_{3}^{\prime}\eta_{1},w_{3}^{\prime}\eta_{2}^{\prime}\,\right].$

Taking the basis $1,z,\alpha,\alpha z,\alpha_{2},\alpha^{2}z$ for
$\mathcal{O}_{L}$ we can compute the action of $\mathfrak{A}_{H}$ using the
tables above.

###  5.4 Induced Hopf Galois module structure

Our aim to relate the freeness of $\mathcal{O}_{L}$ as
$\mathfrak{A}_{H}$-module with the freeness of $\mathcal{O}_{E}$ as
$\mathfrak{A}_{H_{1}}$-module and the freeness of $\mathcal{O}_{F}$ as
$\mathfrak{A}_{\overline{H}}$-module.

######  Theorem 5.16.

Let us assume that $E/K$ and $F/K$ are arithmetically disjoint. If
$\mathcal{O}_{E}$ is $\mathfrak{A}_{H_{1}}$-free and $\mathcal{O}_{F}$ is
$\mathfrak{A}_{\overline{H}}$-free, then $\mathcal{O}_{L}$ is
$\mathfrak{A}_{H}$-free.

Moreover, if $\gamma$ is a $\mathfrak{A}_{H_{1}}$-free generator of
$\mathcal{O}_{E}$ and $\delta$ is a $\mathfrak{A}_{\overline{H}}$-free
generator of $\mathcal{O}_{F}$, then $\gamma\delta$ is a
$\mathfrak{A}_{H}$-free generator of $\mathcal{O}_{L}$.

###### Proof.

Let $\\{v_{i}\\}_{i=1}^{r}$ be an $\mathcal{O}_{K}$-basis of
$\mathfrak{A}_{H_{1}}$ and let $\\{\mu_{j}\\}_{j=1}^{u}$ be an
$\mathcal{O}_{K}$-basis of $\mathfrak{A}_{\overline{H}}$. Then,
$\\{v_{i}\cdot\gamma\\}_{i=1}^{r}$ is an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{E}$ and $\\{\mu_{j}\cdot\delta\\}_{j=1}^{u}$ is an
$\mathcal{O}_{K}$-basis of $\mathcal{O}_{F}$. Since
$\mathcal{O}_{L}=\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$, the
product of these bases is an $\mathcal{O}_{K}$-basis of $\mathcal{O}_{L}$. But
that basis is formed by the elements

$(v_{i}\cdot\gamma)(\mu_{j}\cdot\delta)=(v_{i}\mu_{j})\cdot(\gamma\delta),\,1\leq
i\leq r,\,0\leq j\leq u.$

Since
$\mathfrak{A}_{H}=\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathfrak{A}_{\overline{H}}$,
this amounts to say that $\gamma\delta$ is a $\mathfrak{A}_{H}$-free generator
of $\mathcal{O}_{L}$. ∎

######  Example 5.17.

Let us consider once again the situation of Example 3.9. We know by Example
5.15 that the action of the associated order over $\mathcal{O}_{E}$ is given
by

$\mathfrak{A}_{H_{1}}$ |  | $1$
---
| $\alpha$
---
| $\alpha^{2}$
---
$w_{1}$ | $1$ | $\alpha$ | $\alpha^{2}$
$w_{2}$ | $0$ | $3+9\alpha+2\alpha^{2}$ | $-3-30\alpha-9\alpha^{2}$
$w_{3}^{\prime}$ | $0$ | $-1$ | $3$
(5)

Hence, for an element
$\beta=\beta_{0}+\beta_{1}\alpha+\beta_{2}\alpha^{2}\in\mathcal{O}_{E}$, we
have:

$w_{1}\cdot\beta=\beta_{0}+\beta_{1}\alpha+\beta_{2}\alpha^{2},$
$w_{2}\cdot\beta=3(\beta_{1}-\beta_{2})+(9\beta_{1}-30\beta_{2})\alpha+(2\beta_{1}-9\beta_{2})\alpha^{2}$
$w_{3}^{\prime}\cdot\beta=\beta_{0}-\beta_{1}+3\beta_{2}.$

If we take $\beta=\alpha$, the associated matrix is

$\mathcal{D}_{\alpha}(H_{1},E)=\begin{pmatrix}0&3&-1\\\ 1&9&0\\\
0&2&0\end{pmatrix},$

which has determinant $-2$. Then, $\alpha$ is a generator of $\mathcal{O}_{E}$
as $\mathfrak{A}_{H_{1}}$-module and since $F/\mathbb{Q}_{3}$ is quadratic, by
Example 4.3, $1+z$ is an $\mathfrak{A}_{\overline{H}}$-generator of
$\mathcal{O}_{F}$. We know by Example 5.15 that $E/\mathbb{Q}_{3}$ and
$F/\mathbb{Q}_{3}$ are arithmetically disjoint. Then, we can apply Theorem
5.16 to obtain that $\mathcal{O}_{L}$ is $\mathfrak{A}_{H}$-free with
generator $\alpha(1+z)$.

###  5.5 Freeness after tensoring by $\mathcal{O}_{F}$

We finally compute the associated order of $\mathcal{O}_{L}$ in
$H_{1}\otimes_{K}F$ and discuss the freeness of $\mathcal{O}_{L}$ in
$\mathfrak{A}_{H_{1}\otimes_{K}F}$. Note that this is actually a Hopf Galois
structure of $L/F$ because $H_{1}$ is a Hopf Galois structure of $E/K$ and $F$
is $K$-flat. Moreover, the action of $H_{1}\otimes_{K}F$ on $L$ is obtained by
extending $F$-linearly the one of $H_{1}$ on $E$.

We study the relation between $\mathfrak{A}_{H_{1}}$ and
$\mathfrak{A}_{H_{1}\otimes_{K}F}$, as well as the
$\mathfrak{A}_{H_{1}}$-freeness of $\mathcal{O}_{E}$ and the
$\mathfrak{A}_{H_{1}\otimes_{K}F}$-freeness of $\mathcal{O}_{L}$. In order to
do this, we need a suitable description of elements of $\mathcal{O}_{L}$. For
this reason, we make again the hypothesis that $E/K$ and $F/K$ are
arithmetically disjoint, which implies that
$\mathcal{O}_{L}=\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$.

Let $\\{\alpha_{i}\\}_{i=1}^{r}$ be an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{E}$ and let $\\{z_{j}\\}_{j=1}^{u}$ be an
$\mathcal{O}_{K}$-basis of $\mathcal{O}_{F}$. Since
$\mathcal{O}_{L}=\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$,
$\\{\alpha_{i}z_{j}\\}_{i,j}$ is an $\mathcal{O}_{K}$-basis of
$\mathcal{O}_{L}$.

######  Proposition 5.18.

If $E/K$ and $F/K$ are arithmetically disjoint, then
$\mathfrak{A}_{H_{1}\otimes_{K}F}=\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$.

###### Proof.

Let $H^{(1)}=H_{1}\otimes_{K}F$. First, we prove that
$\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}\subset\mathfrak{A}_{H^{(1)}}$.
It is clearly contained in
$H_{1}\otimes_{\mathcal{O}_{K}}F=H_{1}\otimes_{K}F=H^{(1)}$. On the other
hand, it acts $\mathcal{O}_{K}$-linearly on $\mathcal{O}_{L}$ componentwise
since
$\mathcal{O}_{L}=\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$.
This proves the claim.

For the reverse inclusion, let $h\in\mathfrak{A}_{H^{(1)}}$. Trivially, $h\in
H^{(1)}=H_{1}\otimes_{K}F$. Since $\\{z_{j}\\}_{j=1}^{u}$ is a $K$-basis of
$F$ and $H_{1}$ is $K$-flat, it is also an $H_{1}$-basis of $H^{(1)}$. Then,

$h=\sum_{j=1}^{u}h^{(j)}z_{j},\,h^{(j)}\in H_{1}.$

The result will follow from the fact that $h^{(j)}\in\mathfrak{A}_{H_{1}}$ for
all $1\leq j\leq u$. In order to prove this, we may check that
$h^{(j)}\cdot\gamma\in\mathcal{O}_{E}$ for all $\gamma\in\mathcal{O}_{E}$.
Take any such $\gamma\in\mathcal{O}_{E}$. In particular
$\gamma\in\mathcal{O}_{L}$, and since $h\in\mathfrak{A}_{H^{(1)}}$, we have
that $h\cdot_{L}\gamma\in\mathcal{O}_{L}$. But

$h\cdot_{L}\gamma=\left(\sum_{j=1}^{u}h^{(j)}z_{j}\right)\cdot_{L}\gamma=\sum_{j=1}^{u}(h^{(j)}\cdot_{E}\gamma)z_{j}\in\mathcal{O}_{L}.$

Now, $\\{z_{j}\\}_{j=1}^{u}$ is an $\mathcal{O}_{E}$-basis of
$\mathcal{O}_{L}$ because $\mathcal{O}_{E}$ is $\mathcal{O}_{K}$-flat. Hence,
the previous expression yields that $h^{(j)}\cdot_{E}\gamma\in\mathcal{O}_{E}$
for all $1\leq j\leq u$. ∎

With this result, we have determined how the associated order changes when we
tensor with an arithmetically disjoint extension. Now, we move to the question
of the freeness of $\mathcal{O}_{L}$ as
$\mathfrak{A}_{H_{1}\otimes_{K}F}$-module. We will prove $2$ of Theorem 1.6.

######  Corollary 5.19.

Assume that $E/K$ and $F/K$ are arithmetically disjoint. If $\mathcal{O}_{E}$
is $\mathfrak{A}_{H_{1}}$-free, then $\mathcal{O}_{L}$ is
$\mathfrak{A}_{H_{1}\otimes_{K}F}$-free.

###### Proof.

Since $\mathcal{O}_{F}$ is $\mathcal{O}_{K}$-flat,
$\mathcal{O}_{E}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}=\mathcal{O}_{L}$ is
$\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}$-free. By the
previous result,
$\mathfrak{A}_{H_{1}}\otimes_{\mathcal{O}_{K}}\mathcal{O}_{F}=\mathfrak{A}_{H_{1}\otimes_{K}F}$
and the claim follows. ∎

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