# On finite embedding problems with abelian kernels

François Legrand<EMAIL_ADDRESS>Normandie Univ., UNICAEN, CNRS,
Laboratoire de Mathématiques Nicolas Oresme, 14000 Caen, France

###### Abstract.

Given a Hilbertian field $k$ and a finite set $\mathcal{S}$ of Krull
valuations of $k$, we show that every finite split embedding problem
$G\rightarrow{\rm{Gal}}(L/k)$ over $k$ with abelian kernel has a solution
${\rm{Gal}}(F/k)\rightarrow G$ such that every $v\in\mathcal{S}$ is totally
split in $F/L$. Two applications are then given. Firstly, we solve a non-
constant variant of the Beckmann–Black problem for solvable groups: given a
field $k$ and a non-trivial finite solvable group $G$, every Galois field
extension $F/k$ of group $G$ is shown to occur as the specialization at some
$t_{0}\in k$ of some Galois field extension $E/k(T)$ of group $G$ with
$E\not\subseteq\overline{k}(T)$. Secondly, we contribute to inverse Galois
theory over division rings, by showing that, for every division ring $H$ and
every automorphism $\sigma$ of $H$ of finite order, all finite semiabelian
groups occur as Galois groups over the skew field of fractions $H(T,\sigma)$
of the twisted polynomial ring $H[T,\sigma]$.

##  1\. Introduction

###  1.1. Main result

The inverse Galois problem over a field $k$, a question going back to Hilbert
and Noether, asks whether every finite group occurs as a Galois group over
$k$, i.e., as the Galois group of a Galois field extension of $k$. A stronger
version asks for solutions to finite embedding problems over $k$. As in, e.g.,
[FJ08, §16.4], say that a finite embedding problem over $k$ is an epimorphism
$\alpha:G\rightarrow{\rm{Gal}}(L/k)$, where $G$ is a finite group and $L/k$ a
Galois field extension, and that $\alpha$ splits if there is an embedding
$\alpha^{\prime}:{\rm{Gal}}(L/k)\rightarrow G$ such that
$\alpha\circ\alpha^{\prime}={\rm{id}}_{{\rm{Gal}}(L/k)}$. A solution to
$\alpha$ is an isomorphism $\beta:{\rm{Gal}}(F/k)\rightarrow G$, where $F$ is
a Galois field extension of $k$ containing $L$, such that $\alpha\circ\beta$
is the restriction map ${\rm{Gal}}(F/k)\rightarrow{\rm{Gal}}(L/k)$.

The following conjecture is a special case of the Dèbes–Deschamps conjecture
(see [DD97]):

######  Conjecture 1.1.

Each finite split embedding problem over any Hilbertian field has a solution.

Recall that a field $k$ is Hilbertian if Hilbert’s irreducibility theorem
holds over $k$. For example, all global fields are Hilbertian. We refer to,
e.g., [FJ08] for more on Hilbertian fields.

The main interest of Conjecture 1.1 is that it generalizes and unifies several
conjectures in classical inverse Galois theory. For example, it allows to
solve both the inverse Galois problem over every Hilbertian field (in
particular, over $\mathbb{Q}$) and the Shafarevich conjecture, the latter
asserting that the absolute Galois group of the maximal cyclotomic extension
of $\mathbb{Q}$ is profinite free and being intensively studied (see, e.g.,
[Pop96, HS05, Par09, Des15]).

So far, Conjecture 1.1 has been proved if the base field is a rational
function field of one variable over an ample field111Recall that a field $k$
is ample (or large) if every smooth geometrically irreducible $k$-curve has
zero or infinitely many $k$-rational points. Ample fields include
algebraically closed fields, the complete valued fields $\mathbb{Q}_{p}$,
$\mathbb{R}$, $\kappa((Y))$, the field $\mathbb{Q}^{\rm{tr}}$ of totally real
numbers, etc. See, e.g., [Jar11, BSF13, Pop14] for more details. or if the
base field is ample and Hilbertian (see [Pop96, HJ98]). On the other hand, no
counter-example to Conjecture 1.1 is known.

There is, however, a weaker result, originating in a work of Ikeda (see
[Ike60]), which holds over every Hilbertian field (see, e.g., [FJ08,
Proposition 16.4.5]):

($*$) Given a Hilbertian field $k$, every finite split embedding problem
$\alpha:G\rightarrow{\rm{Gal}}(L/k)$ over $k$ with abelian kernel has a
solution $\beta:{\rm{Gal}}(F/k)\rightarrow G$.

Our main aim is to show that there are solutions $\beta$ to $\alpha$ which are
“totally split” at any finitely many given Krull valuations of $k$. In the
next result, which improves on Statement ($*$), we let $k_{v}^{h}$ denote the
Henselization of a field $k$ at a Krull valuation $v$ of $k$ (see §2.3).

######  Theorem 1.2.

Let $k$ be a Hilbertian field, $\mathcal{S}$ a finite set of Krull valuations
of $k$, and $\alpha:G\rightarrow{\rm{Gal}}(L/k)$ a finite embedding problem
over $k$. Assume $\alpha$ splits and ${\rm{ker}}(\alpha)$ is abelian. Then
$\alpha$ has a solution ${\rm{Gal}}(F/k)\rightarrow G$ such that $F\subseteq
L\cdot k_{v}^{h}$ for every $v\in\mathcal{S}$.

We point out that there are other results on solving finite embedding problems
over Hilbertian fields with local conditions (see, e.g., [KM04, JR18, JR19,
JR20, FL20, CF21]), possibly with a wider range of local conditions and/or
kernels. However, those results hold only over some specific classes of
Hilbertian fields and consider more restrictive classes of valuations.

The proof of Theorem 1.2 (see §3) follows the main lines of that of ($*$).
Recall that the latter mainly consists in viewing $G$ as a quotient of the
wreath product ${\rm{ker}}(\alpha)\wr{\rm{Gal}}(L/k)$, in realizing the latter
as the Galois group of a Galois field extension of $k(T_{1},\dots,T_{n})$
containing $L$, where $n=[L:k]$, and in using the Hilbertianity of $k$ to
specialize $T_{1},\dots,T_{n}$ suitably. To get our extra local conclusion, we
use various tools, such as ramification in higher dimension, several lemmas of
general interest on specializations of function field extensions in several
variables that we establish, and the fact that, given a field $k$, every
finite abelian group is the Galois group of a Galois extension $E$ of the
rational function field $k(T)$ with $E\subseteq k((T))$.

###  1.2. Applications

We then give two applications of Theorem 1.2, to the Beckmann–Black problem
(see §1.2.1) and inverse Galois theory over division rings (see §1.2.2).

####  1.2.1. On the Beckmann–Black problem

The geometric approach to realize a finite group $G$ as a Galois group over a
number field $k$ consists in realizing $G$ as a regular Galois group over $k$,
i.e., as the Galois group of a Galois field extension $E/k(T)$ with
$E\cap\overline{k}=k$. For such an extension $E/k(T)$, Hilbert’s
irreducibility theorem provides infinitely many $t_{0}\in k$ such that the
specialization $E_{t_{0}}/k$ of $E/k(T)$ at $t_{0}$ has Galois group $G$ (see
§2.1 for basic terminology). Several finite groups, including non-abelian
simple ones, have been realized by this method. See [Ser92, Völ96, FJ08, MM18]
for more details and references within.

The Beckmann–Black problem asks whether this approach is optimal: given a
field $k$ and a finite group $G$, is every Galois field extension $F/k$ of
group $G$ the specialization at some $t_{0}\in k$ of some Galois extension
$E/k(T)$ of group $G$ with $E\cap\overline{k}=k$? If $k$ is ample, the answer
is Yes for every finite group $G$ (see [CT00, MB01]). If $k$ is a number
field, the answer is known to be Yes for only a few groups $G$, including
abelian groups (see [Bec94, Dèb99]), symmetric groups, and alternating groups
(see [Mes90, KM01]), and no counter-example is known. See also théorème 2.2 of
the survey paper [Dèb01].

Given a number field $k$ and a finite solvable group $G$, it is widely unknown
whether $G$ is a regular Galois group over $k$. As $G$ is a Galois group over
$k$ by Shafarevich’s theorem (see [NSW08, Theorem 9.6.1]), a positive answer
to the Beckmann–Black problem for $G$ over $k$ is even harder to reach. The
next theorem (see Corollary 4.5), which was not available in the literature,
establishes a necessary condition for such an affirmative answer:

######  Theorem 1.3.

Let $G$ be a non-trivial finite solvable group, $k$ an arbitrary field, and
$F/k$ a Galois field extension of group $G$. There exist $t_{0}\in k$ and a
Galois field extension $E/k(T)$ of group $G$ with
$E\not\subseteq\overline{k}(T)$ such that the specialization $E_{t_{0}}/k$ of
$E/k(T)$ at $t_{0}$ equals $F/k$.

In fact, we have the following stronger result. Given a finite embedding
problem $\alpha:G\rightarrow{\rm{Gal}}(L/k)$ over a field $k$, denote the
restriction map ${\rm{Gal}}(L(T)/k(T))\rightarrow{\rm{Gal}}(L/k)$ by
${\rm{res}}$, and say that a solution to the finite embedding problem
${\rm{res}}^{-1}\circ\alpha$ over $k(T)$ is a geometric solution to $\alpha$.
We derive from Theorem 1.2 that, if ${\rm{ker}}(\alpha)$ is abelian, then
every solution ${\rm{Gal}}(F/k)\rightarrow G$ to $\alpha$ occurs as the
specialization at some $t_{0}\in k$ of some geometric solution
${\rm{Gal}}(E/k(T))\rightarrow G$ to $\alpha$ with $E\cap\overline{k}=L$ (see
§2.2 for more details on specializations at the level of finite embedding
problems and Theorem 4.1 for our precise result).

That is, our result solves the Beckmann–Black problem for finite embedding
problems with abelian kernels over arbitrary fields and, in particular, widely
extends both Dèbes’ theorem (see [Dèb99]) solving the Beckmann–Black problem
for abelian groups over arbitrary fields (the case $L=k$ of our result) and
the affirmative answer to the Beckmann–Black problem for Brauer embedding
problems (see, e.g., [MM18, Chapter IV, §7.1] for more details). To our
knowledge, only one other result about the Beckmann–Black problem for finite
embedding problems was available in the literature. Namely, the answer is
affirmative if the base field is PAC222Recall that a field $k$ is Pseudo
Algebraically Closed (PAC) if every non-empty geometrically irreducible
$k$-variety has a Zariski-dense set of $k$-rational points. See, e.g., [FJ08]
for more on PAC fields., as a combination of a result of Bary-Soroker (see
[BS09, Corollary 3.4]) and the result asserting that every finite embedding
problem $G\rightarrow{\rm{Gal}}(L/k)$ over a PAC field $k$ has a geometric
solution ${\rm{Gal}}(E/k(T))\rightarrow G$ with $E\cap\overline{k}=L$ (see
[Pop96]).

####  1.2.2. Inverse Galois theory over division rings

According to Artin, if $H\subseteq L$ are division rings333A division ring is
a non-zero ring $H$ such that every non-zero element of $H$ is invertible in
$H$. Of course, commutative division rings are nothing but fields., $L/H$ is
Galois if every element of $L$ which is fixed under all automorphisms of $L$
fixing $H$ pointwise is in $H$ (see [Jac64, Coh95] for more on Galois theory
of division rings). With the latter definition, inverse Galois theory can be
studied over arbitrary division rings. Yet, it is striking that not much is
known in the non-commutative case (see [DL20, ALP20, Beh21, BDL20, Des21,
FL20] for recent works on that topic).

Given an automorphism $\sigma$ of a division ring $H$, let $H[T,\sigma]$ be
the ring of polynomials $a_{0}+a_{1}T+\cdots+a_{n}T^{n}$ with $n\geq 0$ and
$a_{0},\dots,a_{n}\in H$, whose addition is defined componentwise and
multiplication fulfills $Ta=\sigma(a)T$ for $a\in H$. By $H(T,\sigma)$, we
mean the unique division ring containing $H[T,\sigma]$ and each element of
which can be written as $ab^{-1}$ with $a\in H[T,\sigma]$, $b\in
H[T,\sigma]\setminus\\{0\\}$. If $H$ is a field and $\sigma={\rm{id}}_{H}$, we
retrieve the usual commutative polynomial ring $H[T]$ and the rational
function field $H(T)$, respectively. See §5.1.1 for more details.

In [Beh21, théorème A], Behajaina shows that, for $\sigma$ of finite order,
every finite group is a Galois group over $H(T,\sigma)$, if the fixed field of
$\sigma$ in the center $h$ of $H$ contains an ample field, thus yielding a
non-commutative analogue of Pop’s result solving the regular inverse Galois
problem over ample fields. Under weaker assumptions on $h$ (e.g., $h$ of
characteristic zero), some “usual” finite groups in inverse Galois theory,
like abelian, symmetric, and alternating groups, are shown to occur as Galois
groups over $H(T,\sigma)$ (see [Beh21, remarque du §2.1]).

We combine Theorem 1.2 and Behajaina’s method to realize another family of
usual finite groups as Galois groups over $H(T,\sigma)$, without making any
assumption on the center (see §5.2 for the proof of the next theorem). Recall
that a finite group is semiabelian if it belongs to the smallest non-empty
class $\mathcal{C}$ of finite groups closed under quotients and such that, if
$G\in\mathcal{C}$ and $A$ is finite abelian, then every semidirect product
$A\rtimes G$ is in $\mathcal{C}$.

######  Theorem 1.4.

Let $H$ be a division ring and $\sigma$ an automorphism of $H$ of finite
order. Every finite semiabelian group is a Galois group over $H(T,\sigma)$.

We also contribute to solving finite embedding problems over division rings (a
topic introduced in [BDL20]), by showing that every finite split embedding
problem with abelian kernel over any given division ring $H$ of finite
dimension over its center acquires a solution over the division ring
$H(T)=H(T,{\rm{id}}_{H})$. See §5.1.2 for more on finite embedding problems
over division rings and Theorem 5.6 for our precise result.

Acknowledgements. Part of this work was done during a research visit at the
Université Caen Normandie in March 2020. We would like to thank the
Laboratoire de Mathématiques Nicolas Oresme for kind hospitality and financial
support. Another part of this work fits into Project TIGANOCO, which is funded
by the European Union within the framework of the Operational Programme
ERDF/ESF 2014-2020. We also wish to thank Angelot Behajaina and Arno Fehm for
helpful discussions about this work.

##  2\. Preliminaries

In this section, we collect the material on specializations of function field
extensions, finite embedding problems over fields, and Krull valuations which
will be used in the sequel.

###  2.1. Specializations of function field extensions

Let $k$ be a field, ${\bf{T}}=(T_{1},\dots,T_{n})$ a tuple of indeterminates,
$E$ a Galois field extension of the rational function field $k({\bf{T}})$ of
degree $d$, and ${B}$ the integral closure of $k[{\bf{T}}]$ in ${E}$. For
${\bf{t}}=(t_{1},\dots,t_{n})\in k^{n}$, the residue field of ${B}$ at a
maximal ideal $\mathfrak{P}$ lying over $\langle
T_{1}-t_{1},\dots,T_{n}-t_{n}\rangle$ is denoted by ${E}_{\bf{t}}$, and the
extension ${E}_{\bf{t}}/k$ is called the specialization of ${E}/k({\bf{T}})$
at ${\bf{t}}$. Since ${E}/k({\bf{T}})$ is Galois, the field ${E}_{\bf{t}}$
does not depend on the choice of the maximal ideal $\mathfrak{P}$ lying over
$\langle T_{1}-t_{1},\dots,T_{n}-t_{n}\rangle$. Moreover, the field extension
${E}_{\bf{t}}/k$ is finite and normal, and its automorphism group is
isomorphic to $D_{\mathfrak{P}}/I_{\mathfrak{P}}$, where $D_{\mathfrak{P}}$
(resp., $I_{\mathfrak{P}}$) denotes the decomposition group (resp., the
inertia group) of $E/k({\bf{T}})$ at $\mathfrak{P}$. More precisely,

$\varphi_{{\bf{t}}}:\left\\{\begin{array}[]{ccc}D_{\mathfrak{P}}&\rightarrow&{\rm{Aut}}({E}_{{\bf{t}}}/k)\\\
\sigma&\mapsto&\overline{\sigma}\\\ \end{array}\right.,$ (2.1)

where
$\overline{\sigma}(x\,{\rm{mod}}\,\mathfrak{P})=\sigma(x)\,{\rm{mod}}\,\mathfrak{P}$
for every $x\in{B}$, is an epimorphism whose kernel is $I_{\mathfrak{P}}$. If
${E}_{\bf{t}}/k$ is Galois of degree $d$, then $\varphi_{\bf{t}}$ is an
isomorphism
${\rm{Gal}}({E}/k({\bf{T}}))\rightarrow{\rm{Gal}}({E}_{{\bf{t}}}/k)$. For
${\bf{t}}$ outside a Zariski-closed proper subset (depending only on
${E}/k({\bf{T}})$), ${E}_{\bf{t}}/k$ is Galois.

###  2.2. Finite embedding problems over fields

Given finite Galois field extensions $L/k$ and $E/F$ with $k\subseteq F$ and
$L\subseteq E$, we let ${\rm{res}}^{E/F}_{L/k}$ denote the natural restriction
map ${\rm{Gal}}(E/F)\rightarrow{\rm{Gal}}(L/k)$ (i.e.,
${\rm{res}}^{E/F}_{L/k}(\sigma)(x)=\sigma(x)$ for all
$\sigma\in{\rm{Gal}}(E/F)$ and $x\in L$).

A finite embedding problem over a field $k$ is an epimorphism
$\alpha:G\rightarrow{\rm{Gal}}(L/k)$, where $G$ is a finite group and $L/k$ a
Galois field extension. Say that $\alpha$ splits if there is an embedding
$\alpha^{\prime}:{\rm{Gal}}(L/k)\rightarrow G$ with
$\alpha\circ\alpha^{\prime}={\rm{id}}_{{\rm{Gal}}(L/k)}$. A solution to
$\alpha$ is an isomorphism $\beta:{\rm{Gal}}(F/k)\rightarrow G$, where $F$ is
a Galois field extension of $k$ containing $L$, such that
$\alpha\circ\beta={\rm{res}}^{F/k}_{L/k}$. Given a finite tuple ${\bf{T}}$ of
indeterminates, the composed map

$\alpha_{k({\bf{T}})}=({\rm{res}}^{L({\bf{T}})/k({\bf{T}})}_{L/k})^{-1}\circ\alpha:G\rightarrow{\rm{Gal}}(L({\bf{T}})/k({\bf{T}}))$

is a finite embedding problem over $k({\bf{T}})$. A geometric solution to
$\alpha$ is a solution to $\alpha_{k(T)}$.

We now recall the notion of specializations at the level of finite embedding
problems. To that end, we need the following lemma:

######  Lemma 2.1.

Let ${\bf{T}}$ be an $n$-tuple of indeterminates,
$\alpha:G\rightarrow{\rm{Gal}}(L/k)$ a finite embedding problem over a field
$k$, and $\beta:{\rm{Gal}}(E/k({\bf{T}}))$ $\rightarrow G$ a solution to
$\alpha_{k({\bf{T}})}$. For ${\bf{t}}\in k^{n}$ such that the specialization
$E_{\bf{t}}/k$ of $E/k({\bf{T}})$ at ${\bf{t}}$ is Galois of degree
$[E:k({\bf{T}})]$, the map
$\beta\circ\varphi^{-1}_{{\bf{t}}}:{\rm{Gal}}(E_{{\bf{t}}}/k)\rightarrow G$ is
a solution to $\alpha$, where
$\varphi_{{\bf{t}}}:{\rm{Gal}}(E/k({\bf{T}}))\rightarrow{\rm{Gal}}(E_{{\bf{t}}}/k)$
is defined in (2.1).

###### Proof.

As $\beta$ is a solution to $\alpha_{k({\bf{T}})}$, we have

$({\rm{res}}^{L({\bf{T}})/k({\bf{T}})}_{L/k})^{-1}\circ\alpha\circ\beta={\rm{res}}^{E/k({\bf{T}})}_{L({\bf{T}})/k({\bf{T}})},$

thus yielding

$\alpha\circ\beta\circ\varphi_{{\bf{t}}}^{-1}={\rm{res}}_{L/k}^{L({\bf{T}})/k({\bf{T}})}\circ{\rm{res}}^{E/k({\bf{T}})}_{L({\bf{T}})/k({\bf{T}})}\circ\varphi_{{\bf{t}}}^{-1}={\rm{res}}^{E/k({\bf{T}})}_{L/k}\circ\varphi_{{\bf{t}}}^{-1}.$

It then suffices to show
${\rm{res}}^{E/k({\bf{T}})}_{L/k}={\rm{res}}^{E_{{\bf{t}}}/k}_{L/k}\circ\varphi_{{\bf{t}}}$.
But the latter equality holds as $\varphi_{{\bf{t}}}(\sigma)(x)=\sigma(x)$ for
every $\sigma\in{\rm{Gal}}(E/k({\bf{T}}))$ and every $x\in L$, by the
definition of $\varphi_{{\bf{t}}}$. ∎

The previous lemma motivates the following definition:

######  Definition 2.2.

Let ${\bf{T}}$ be an $n$-tuple of indeterminates,
$\alpha:G\rightarrow{\rm{Gal}}(L/k)$ a finite embedding problem over a field
$k$, and $\beta:{\rm{Gal}}(E/k({\bf{T}}))$ $\rightarrow G$ a solution to
$\alpha_{k({\bf{T}})}$. For ${\bf{t}}\in k^{n}$ such that the specialization
$E_{\bf{t}}/k$ of $E/k({\bf{T}})$ at ${\bf{t}}$ is Galois of degree
$[E:k({\bf{T}})]$, the solution
$\beta\circ\varphi^{-1}_{{\bf{t}}}:{\rm{Gal}}(E_{{\bf{t}}}/k)\rightarrow G$ to
$\alpha$ from Lemma 2.1 is the specialization of $\beta$ at ${\bf{t}}$,
denoted $\beta_{\bf{t}}$.

Finally, we observe that, to realize solutions as specialized solutions, it
suffices to work at the level of fields:

######  Lemma 2.3.

Let $\alpha:G\rightarrow{\rm{Gal}}(L/k)$ be a finite embedding problem over a
field $k$ and $\gamma:{\rm{Gal}}(F/k)\rightarrow G$ a solution to $\alpha$.
Given an $n$-tuple ${\bf{T}}$ of indeterminates, let $E$ be a Galois field
extension of $k({\bf{T}})$ containing $L$ with $[E:k({\bf{T}})]=|G|$, and let
${\bf{t}}\in k^{n}$ be such that the specialization $E_{\bf{t}}/k$ of
$E/k({\bf{T}})$ at ${\bf{t}}$ is $F/k$. Then there is an isomorphism
$\beta:{\rm{Gal}}(E/k({\bf{T}}))\rightarrow G$ which is a solution to
$\alpha_{k({\bf{T}})}$ and such that the specialization $\beta_{\bf{t}}$ is
$\gamma$.

###### Proof.

As $E_{\bf{t}}=F$ and $[E:k({\bf{T}})]=|G|$, we have
$[E_{\bf{t}}:k]=[E:k({\bf{T}})]$ and $E_{\bf{t}}/k$ is Galois. Hence, the map
$\varphi_{\bf{t}}$ from (2.1) is an isomorphism
${\rm{Gal}}({E}/k({\bf{T}}))\rightarrow{\rm{Gal}}({E}_{{\bf{t}}}/k)$. Using
again $E_{\bf{t}}=F$, we get that
$\gamma\circ\varphi_{{\bf{t}}}:{\rm{Gal}}(E/k({\bf{T}}))\rightarrow G$ is a
well-defined isomorphism. Then $\beta=\gamma\circ\varphi_{{\bf{t}}}$ is a
solution to $\alpha_{k({\bf{T}})}$ as
$\alpha\circ\beta={\rm{res}}^{F/k}_{L/k}\circ\varphi_{{\bf{t}}}={\rm{res}}^{E_{\bf{t}}/k}_{L/k}\circ\varphi_{{\bf{t}}}={\rm{res}}^{E/k({\bf{T}})}_{L/k}$,
the latter equality being observed in the proof of Lemma 2.1. Finally, from
our choice of $\beta$, the claimed equality $\beta_{\bf{t}}=\gamma$ is clear.
∎

###  2.3. Valuations

See, e.g., [Jar91] for more on the following. An ordered group is an abelian
(additive) group together with a total ordering $<$ such that $\alpha<\beta$
implies $\alpha+\gamma<\beta+\gamma$ for all $\alpha,\beta,\gamma\in\Gamma$. A
(Krull) valuation of a field $k$ is a surjective map
$v:k^{*}\rightarrow\Gamma$, with $\Gamma$ a non-trivial ordered group (the
value group of $v$), such that $v(ab)=v(a)+v(b)$ for all $a,b\in k^{*}$ and
$v(a+b)\geq{\rm{min}}\\{v(a),v(b)\\}$ for all $a,b\in k^{*}$ with $a+b\not=0$.
Add the symbol $\infty$ to $\Gamma$ with the conventions
$\infty+\infty=\alpha+\infty=\infty+\alpha=\infty$ and $\alpha<\infty$ for
every $\alpha\in\Gamma$. Extend $v$ to $k$ by defining $v(0)=\infty$. Then $v$
fulfills $v(ab)=v(a)+v(b)$ and $v(a+b)\geq{\rm{min}}\\{v(\alpha),v(\beta)\\}$
for all $a,b\in k$. The pair $(k,v)$ is a valued field. Two valuations
$v_{1}:k^{*}\rightarrow\Gamma_{1}$ and $v_{2}:k^{*}\rightarrow\Gamma_{2}$ of a
field $k$ are equivalent if there is an order preserving isomorphism
$f:\Gamma_{1}\rightarrow\Gamma_{2}$ with $f\circ v_{1}=v_{2}$.

Let $v:k^{*}\rightarrow\Gamma$ be a valuation of a field $k$. A valuation
$w:L^{*}\rightarrow\Delta$ of a field extension $L$ of $k$ extends $v$ if
there is an order preserving monomorphism $f:\Gamma\rightarrow\Delta$ such
that $w(a)=f\circ v(a)$ for $a\in k^{*}$; say that $(L,w)$ is an extension of
$(k,v)$. The valued field $(k,v)$ is Henselian if $v$ extends uniquely (up to
equivalence) to each algebraic field extension of $k$. The valued field
$(k,v)$ has a separable algebraic extension $(k_{v}^{h},v^{h})$ which is
Henselian and such that, if $(L,w)$ is an extension of $(k,v)$ which is
Henselian, then $(L,w)$ extends $(k_{v}^{h},v^{h})$. The valued field
$(k_{v}^{h},v^{h})$ is unique up to $k$-isomorphism; it is the Henselization
of $(k,v)$.

##  3\. Proof of Theorem 1.2

The aim of this section is to prove Theorem 1.2 from the introduction (see
§3.3). As already alluded to in §1.1, the proof requires several lemmas on
ramification in higher dimension and specializations of function field
extensions in several variables that we establish in §3.1 and §3.2,
respectively.

###  3.1. Ramification in higher dimension

Let $F/k$ be a finite Galois field extension. Suppose $k$ is the fraction
field of an integrally closed Noetherian domain $A$. If there is no possible
confusion, we will say that a $k$-basis of $F$ whose elements are integral
over $A$ is an integral basis. A maximal ideal $\mathfrak{p}$ of $A$ is
unramified in $F/k$ if there exists an integral $k$-basis $y_{1},\dots,y_{d}$
of $F$ whose discriminant $\Delta_{F/k}(y_{1},\dots,y_{d})$ fulfills
$\Delta_{F/k}(y_{1},\dots,y_{d})\in A\setminus\mathfrak{p}$.

######  Lemma 3.1.

Let $k$ be the fraction field of an integrally closed Noetherian domain $A$,
let $\mathfrak{p}$ be a maximal ideal of $A$, and let $E/k$ be a finite Galois
field extension. Let $F$ be a field extension of $k$ which is linearly
disjoint from $E$ over $k$.

(1) Assume that $F/k$ is finite Galois, and that $\mathfrak{p}$ is unramified
in both $E/k$ and $F/k$. Then $\mathfrak{p}$ is unramified in $EF/k$.

(2) Assume that $\mathfrak{p}$ is unramified in $E/k$, and let $\mathfrak{P}$
be a maximal ideal of the integral closure of $A$ in $F$ lying over
$\mathfrak{p}$. Then $\mathfrak{P}$ is unramified in $EF/F$.

We need the following classical lemma about the discriminant of a composite
basis.

######  Lemma 3.2.

Let $F/k$ and $L/k$ be finite Galois field extensions with $L\subseteq F$. Let
$y_{1},\dots,y_{d}$ (resp., $z_{1},\dots,z_{e}$) be a $k$-basis (resp., an
$L$-basis) of $L$ (resp., of $F$). If $N_{L/k}$ denotes the norm in $L/k$,
then the discriminant of the $k$-basis
$y_{1}z_{1},\dots,y_{1}z_{e},\dots,y_{d}z_{1},\dots,y_{d}z_{e}$ of $F$ equals
$\Delta_{L/k}(y_{1},\dots,y_{d})^{e}\cdot
N_{L/k}(\Delta_{F/L}(z_{1},\dots,z_{e})).$

###### Proof of Lemma 3.1.

We show (1) and (2) simultaneously. As $\mathfrak{p}$ is unramified in $E/k$
(in both statements), there is an integral $k$-basis $y_{1},\dots,y_{e}$ of
$E$ such that $\Delta_{E/k}(y_{1},\dots,y_{e})\in A\setminus\mathfrak{p}$.
Moreover, $E$ and $F$ are linearly disjoint over $k$. Then $y_{1},\dots,y_{e}$
yield an integral $F$-basis of $EF$, and the discriminant of the latter basis
cannot be in $\mathfrak{P}$. Hence, $\mathfrak{P}$ is unramified in $EF/F$,
thus establishing (2). Now, in the situation of (1), there is an integral
$k$-basis $z_{1},\dots,z_{f}$ of $F$ such that
$\Delta_{F/k}(z_{1},\dots,z_{f})\in A\setminus\mathfrak{p}$. From our
assumption that $E$ and $F$ are linearly disjoint over $k$, the elements
$y_{1}z_{1},\dots,y_{1}z_{f},\dots,y_{e}z_{1},\dots,y_{e}z_{f}$ provide an
integral $k$-basis of $EF$. Lemma 3.2 then yields that the discriminant
$\Delta$ of the latter basis equals

$\begin{array}[]{ccl}\Delta_{F/k}(z_{1},\dots,z_{f})^{e}\cdot
N_{F/k}(\Delta_{EF/F}(y_{1},\dots,y_{e}))&=&\Delta_{F/k}(z_{1},\dots,z_{f})^{e}\cdot
N_{F/k}(\Delta_{E/k}(y_{1},\dots,y_{e}))\\\
&=&\Delta_{F/k}(z_{1},\dots,z_{f})^{e}\cdot\Delta_{E/k}(y_{1},\dots,y_{e})^{f}.\end{array}$

Since neither $\Delta_{E/k}(y_{1},\dots,y_{e})$ nor
$\Delta_{F/k}(z_{1},\dots,z_{f})$ is in $\mathfrak{p}$, we get
$\Delta\not\in\mathfrak{p}$, as needed. ∎

###  3.2. Specialization lemmas

We start with specializations of composite field extensions:

######  Lemma 3.3.

Let $E/k({\bf{T}})=E/k(T_{1},\dots,T_{n})$ be a finite Galois field extension
and let ${\bf{t}}=(t_{1},\dots,t_{n})\in k^{n}$ be such that $\langle
T_{1}-t_{1},\dots,T_{n}-t_{n}\rangle$ is unramified in $E/k({\bf{T}})$.

(1) Let $F/k({\bf{T}})$ be a finite Galois field extension such that $E$ and
$F$ are linearly disjoint over $k({\bf{T}})$. Then the specialization
$(EF)_{{\bf{t}}}/k$ of $EF/k({\bf{T}})$ at ${\bf{t}}$ equals
$E_{\bf{t}}F_{\bf{t}}/k$, where $E_{\bf{t}}/k$ (resp., $F_{\bf{t}}/k$) is the
specialization of $E/k({\bf{T}})$ (resp., of $F/k({\bf{T}})$) at ${\bf{t}}$.

(2) Let $U$ be a new indeterminate. Then the specialization
$(E(U))_{(u,{\bf{t}})}/k$ of $E(U)/k(U,{\bf{T}})$ at $(u,{\bf{t}})$ equals the
specialization $E_{\bf{t}}/k$ of $E/k({\bf{T}})$ at ${\bf{t}}$ for every $u\in
k$.

###### Proof.

(1) Let $\mathfrak{P}$ be a maximal ideal of the integral closure of
$k[{\bf{T}}]$ in $E$ containing $T_{1}-t_{1},\dots,T_{n}-t_{n}$. As $\langle
T_{1}-t_{1},\dots,T_{n}-t_{n}\rangle$ is unramified in $E/k({\bf{T}})$, there
is an integral $k({\bf{T}})$-basis $y_{1},\dots,y_{e}$ of $E$ such that
$\Delta({\bf{t}})\not=0$, where $\Delta({\bf{T}})\in k[{\bf{T}}]$ is the
discriminant of $y_{1},\dots,y_{e}$. Let $B$ denote the integral closure of
$k[{\bf{T}}]$ in $EF$. As $E$ and $F$ are linearly disjoint over
$k({\bf{T}})$, the elements $y_{1},\dots,y_{e}$ yield an integral $F$-basis of
$EF$. Now, given $x\in B$, there are $\lambda_{1},\dots,\lambda_{e}\in F$ with
$x=\lambda_{1}y_{1}+\cdots+\lambda_{e}y_{e}.$ For $i\in\\{1,\dots,e\\}$, we
then have

${\rm{tr}}_{EF/F}(xy_{i})=\lambda_{1}{\rm{tr}}_{EF/F}(y_{1}y_{i})+\cdots+\lambda_{e}{\rm{tr}}_{EF/F}(y_{e}y_{i}),$

with ${\rm{tr}}_{EF/F}$ the trace in $EF/F$. As $x,y_{1},\dots,y_{e}\in B$, we
get that $\Delta({\bf{T}})\lambda_{i}$ is in the integral closure of
$k[{\bf{T}}]$ in $F$ for every $i\in\\{1,\dots,e\\}$, by applying Kramer’s
law. Hence, if $\mathfrak{Q}$ denotes a maximal ideal of $B$ lying over
$\mathfrak{P}$, then the reduction modulo $\mathfrak{Q}$ of

$x=\lambda_{1}y_{1}+\cdots+\lambda_{e}y_{e}=\frac{\Delta({\bf{T}})\lambda_{1}}{\Delta({\bf{T}})}y_{1}+\cdots+\frac{\Delta({\bf{T}})\lambda_{e}}{\Delta({\bf{T}})}y_{e}$

lies in the compositum of $E_{\bf{t}}$ and $F_{\bf{t}}$, thus showing
$(EF)_{\bf{t}}\subseteq E_{\bf{t}}F_{\bf{t}}$. As to the converse, it suffices
to notice that $E\subseteq EF$ and $F\subseteq EF$ yield
$E_{\bf{t}}\subseteq(EF)_{\bf{t}}$ and $F_{\bf{t}}\subseteq(EF)_{\bf{t}}$.

(2) As above, there is an integral $k({\bf{T}})$-basis $y_{1},\dots,y_{e}$ of
$E$ with $\Delta({\bf{t}})\not=0$, where $\Delta({\bf{T}})\in k[{\bf{T}}]$ is
the discriminant of $y_{1},\dots,y_{e}$. As $E$ and $k({\bf{T}},U)$ are
linearly disjoint over $k({\bf{T}})$, the elements $y_{1},\dots,y_{e}$ provide
a $k({\bf{T}},U)$-basis of $E(U)$, which is contained in the integral closure
$B$ of $k[U,{\bf{T}}]$ in $E(U)$. Moreover, we have
$\Delta({\bf{T}})B\subseteq k[U,{\bf{T}}]{y_{1}}\oplus\cdots\oplus
k[U,{\bf{T}}]{y_{e}}.$ Let $u\in k$. As $\Delta({\bf{T}})$ does not depend on
$U$ and $\Delta({\bf{t}})\not=0$, the above inclusion yields that
$(E(U))_{(u,{\bf{t}})}$ is the field generated over $k$ by the reductions of
$y_{1},\dots,y_{e}$ modulo a maximal ideal of $B$ lying over $\langle
U-u,T_{1}-t_{1},\dots,T_{n}-t_{n}\rangle$, i.e., equals $E_{\bf{t}}$. ∎

The next lemma deals with successive specializations:

######  Lemma 3.4.

Let $E/k({\bf{T}})=E/k(T_{1},\dots,T_{n})$ be a finite Galois field extension,
$t_{1}\in k$, and ${\bf{t^{\prime}}}=(t_{2}(T_{1}),\dots,t_{n}(T_{1}))\in
k[T_{1}]^{n-1}$ such that the specialization $E_{\bf{t^{\prime}}}/k(T_{1})$ of
$E/k(T_{1})(T_{2},\dots,T_{n})$ at ${\bf{t^{\prime}}}$ is Galois. Then the
specialization $E_{\bf{t}}/k$ of $E/k({\bf{T}})$ at
${\bf{t}}=(t_{1},t_{2}(t_{1}),\dots,t_{n}(t_{1}))$ is a subextension of the
specialization $(E_{\bf{t^{\prime}}})_{t_{1}}/k$ of
$E_{\bf{t^{\prime}}}/k(T_{1})$ at $t_{1}$.

###### Proof.

Denote the integral closure of $k(T_{1})[T_{2},\dots,T_{n}]$ (resp., of
$k[{\bf{T}}]$) in $E$ by $B_{2}$ (resp., by $B$). Clearly, $B\subseteq B_{2}$.
For a maximal ideal $\mathfrak{P}_{2}$ of $B_{2}$ lying over $\langle
T_{2}-t_{2}(T_{1}),\dots,T_{n}-t_{n}(T_{1})\rangle$ and $x\in B_{2}$, we
denote the reduction of $x$ modulo $\mathfrak{P}_{2}$ by $\overline{x}$. Then
the reduction $\overline{x}$ of any element $x$ of $B$ is integral over
$k[T_{1}]$, i.e., is an element of the integral closure $B_{1}$ of $k[T_{1}]$
in $E_{\bf{t^{\prime}}}=B_{2}/\mathfrak{P}_{2}$. Let $\mathfrak{P}_{1}$ be a
maximal ideal of $B_{1}$ containing $T_{1}-t_{1}$. Composing the inclusion
$B/(B\cap\mathfrak{P}_{2})\subseteq B_{1}$ with the canonical projection
$B_{1}\rightarrow B_{1}/\mathfrak{P}_{1}$ yields a well-defined homomorphism
$\psi:B\rightarrow B_{1}/\mathfrak{P}_{1}$ satisfying $\langle
T_{1}-t_{1},T_{2}-t_{2}(t_{1}),\dots,T_{n}-t_{n}(t_{1})\rangle\subseteq{\rm{ker}}(\psi)\cap
k[{\bf{T}}].$ Hence, $\langle
T_{1}-t_{1},T_{2}-t_{2}(t_{1}),\dots,T_{n}-t_{n}(t_{1})\rangle={\rm{ker}}(\psi)\cap
k[{\bf{T}}],$ as the ideal in the left-hand side is maximal and
$k[{\bf{T}}]\not\subseteq{\rm{ker}}(\psi)$. Consequently, the ideal
${\rm{ker}}(\psi)$ of $B$ lies over $\langle
T_{1}-t_{1},T_{2}-t_{2}(t_{1}),\dots,T_{n}-t_{n}(t_{1})\rangle$ and it is
maximal. Hence, $E_{\bf{t}}=B/{\rm{ker}}(\psi)\subseteq
B_{1}/\mathfrak{P}_{1}=(E_{\bf{t^{\prime}}})_{t_{1}}$. ∎

Our last lemma gives a situation where the inclusion of Lemma 3.4 is actually
an equality:

######  Lemma 3.5.

Let $E/k({\bf{T}})=E/k(T_{1},\dots,T_{n})$ be a finite Galois field extension,
$t_{1}\in k$, and ${\bf{t^{\prime}}}=(t_{2}(T_{1}),\dots,t_{n}(T_{1}))\in
k[T_{1}]^{n-1}$. Suppose the following three conditions hold:

$\bullet$ $\langle T_{2}-t_{2}(T_{1}),\dots,T_{n}-t_{n}(T_{1})\rangle$ is
unramified in $E/k(T_{1})(T_{2},\dots,T_{n})$,

$\bullet$ the specialization $E_{\bf{t^{\prime}}}/k(T_{1})$ of
$E/k(T_{1})(T_{2},\dots,T_{n})$ at ${\bf{t^{\prime}}}$ is Galois,

$\bullet$ $t_{1}$ is outside some finite set $\mathcal{S}$ depending on
$E/k({\bf{T}})$ and ${\bf{t^{\prime}}}$.

Then the specialization $E_{\bf{t}}/k$ of $E/k({\bf{T}})$ at
${\bf{t}}=(t_{1},t_{2}(t_{1}),\dots,t_{n}(t_{1}))$ equals the specialization
$(E_{\bf{t^{\prime}}})_{t_{1}}/k$ of $E_{\bf{t^{\prime}}}/k(T_{1})$ at
$t_{1}$.

###### Proof.

Denote the integral closure of $k(T_{1})[T_{2},\dots,T_{n}]$ (resp., of
$k[{\bf{T}}]$) in $E$ by $B_{2}$ (resp., by $B$). For a maximal ideal
$\mathfrak{P}_{2}$ of $B_{2}$ lying over $\langle
T_{2}-t_{2}(T_{1}),\dots,T_{n}-t_{n}(T_{1})\rangle$ and $x\in B_{2}$, we
denote the reduction of $x$ modulo $\mathfrak{P}_{2}$ by $\overline{x}$. As
$\langle T_{2}-t_{2}(T_{1}),\dots,T_{n}-t_{n}(T_{1})\rangle$ is unramified in
$E/k(T_{1})(T_{2},\dots,T_{n})$, there exist $y_{1},\dots,y_{e}\in B_{2}$ such
that $E=k({\bf{T}})y_{1}\oplus\cdots\oplus k({\bf{T}})y_{e},$

$\Delta(y_{1},\dots,y_{e})B_{2}\subseteq
k(T_{1})[T_{2},\dots,T_{n}]{y_{1}}\oplus\cdots\oplus
k(T_{1})[T_{2},\dots,T_{n}]{y_{e}},$

and such that the discriminant $\Delta(y_{1},\dots,y_{e})\in
k(T_{1})[T_{2},\dots,T_{n}]$ of $y_{1},\dots,y_{e}$ is not in the ideal
$\langle T_{2}-t_{2}(T_{1}),\dots,T_{n}-t_{n}(T_{1})\rangle$. In particular,
we have

$E_{\bf{t^{\prime}}}=B_{2}/\mathfrak{P}_{2}=k(T_{1})\frac{\overline{y_{1}}}{\overline{\Delta(y_{1},\dots,y_{e})}}+\cdots+k(T_{1})\frac{\overline{y_{e}}}{\overline{\Delta(y_{1},\dots,y_{e})}}.$
(3.1)

Pick $b(T_{1})\in k[T_{1}]\setminus\\{0\\}$ such that $z_{i}=b(T_{1})y_{i}$ is
in $B$ for every $i\in\\{1,\dots,e\\}$. Then, from (3.1), we have
$E_{\bf{t^{\prime}}}=k(T_{1})\overline{z_{1}}+\cdots+k(T_{1})\overline{z_{e}}.$
Up to reordering, there exists $c\leq e$ such that

$E_{\bf{t^{\prime}}}=k(T_{1})\overline{z_{1}}\oplus\cdots\oplus
k(T_{1})\overline{z_{c}}.$ (3.2)

Let $\Delta(T_{1})\in k[T_{1}]\setminus\\{0\\}$ be the discriminant of
$\overline{z_{1}},\dots,\overline{z_{c}}$ and let $\mathcal{S}$ be the set of
all roots of $\Delta(T_{1})$. For $t_{1}\in k\setminus\mathcal{S}$ and a
maximal ideal $\mathfrak{P}_{1}$ of the integral closure $B_{1}$ of $k[T_{1}]$
in $E_{\bf{t^{\prime}}}$ containing $T_{1}-t_{1}$, (3.2) and
$t_{1}\not\in\mathcal{S}$ yield that
$(E_{\bf{t^{\prime}}})_{t_{1}}=B_{1}/\mathfrak{P}_{1}$ is the field generated
over $k$ by the reductions of $\overline{z_{1}},\dots,\overline{z_{c}}$ modulo
$\mathfrak{P}_{1}$. Since these reductions are necessarily in $E_{\bf{t}}$, we
get $(E_{\bf{t^{\prime}}})_{t_{1}}\subseteq E_{\bf{t}}$. Lemma 3.4 then yields
$(E_{\bf{t^{\prime}}})_{t_{1}}=E_{\bf{t}}$, thus concluding the proof. ∎

######  Remark 3.6.

Under the extra assumptions that
$[E_{{\bf{t^{\prime}}}}:k(T_{1})]=[E:k({\bf{T}})]$ and $y_{1},\dots,y_{e}$ are
integral over $k[{\bf{T}}]$, the set $\mathcal{S}$ is reduced to the set of
all roots of $\overline{\Delta(y_{1},\dots,y_{e})}\in k[T_{1}]$.

###  3.3. Proof of Theorem 1.2

Since $\alpha$ splits, there exists an embedding
$\alpha^{\prime}:{\rm{Gal}}(L/k)\rightarrow G$ satisfying

$\alpha\circ\alpha^{\prime}={\rm{id}}_{{\rm{Gal}}(L/k)}.$ (3.3)

Set $G^{\prime}=\alpha^{\prime}({\rm{Gal}}(L/k))\subseteq G$, and
$A={\rm{ker}}(\alpha)\subseteq G$. Then

$\psi:\left\\{\begin{array}[]{ccc}A\rtimes G^{\prime}&\longrightarrow&G\\\
(a,g^{\prime})&\longmapsto&a\cdot g^{\prime}\end{array}\right.,$ (3.4)

where $G^{\prime}$ acts by conjugation on $A$, is an isomorphism. Let
$c_{1},\dots,c_{n}$ be a $k$-basis of $L$ and set
${\rm{Gal}}(L/k)=\\{\sigma_{1},\dots,\sigma_{n}\\}$.

Let $T$ be an extra indeterminate, and let $F/k(T)$ be a Galois extension of
group $A$ in which $\langle T\rangle$ is unramified and whose specialization
$F_{0}/k$ at 0 equals $k/k$; such an extension exists by the twisting lemma
(see [Dèb99]). From the unramifiedness condition, there is a $k(T)$-basis
$z_{1},\dots,z_{|A|}$ of $F$, which is integral over $k[T]$ and such that
$\Delta(0)\in k^{*}$, where $\Delta(T)\in k[T]$ denotes the discriminant of
$z_{1},\dots,z_{|A|}$. Then consider the Galois field extension
$E/L(T)=FL/L(T)$. As $F\cap\overline{k}=k$, we have ${\rm{Gal}}(E/L(T))=A$.
Let $P(T,Y)\in L[T][Y]$ be the minimal polynomial of a primitive element of
$E/L(T)$, assumed to be integral over $L[T]$. Given $i\in\\{1,\dots,n\\}$, let
$U_{i}$ be a new indeterminate and $E^{(i)}$ the field generated over
$L(U_{i})$ by one root, say $y_{i}$, of $\sigma_{i}(P)(U_{i},Y)\in
L[U_{i}][Y]$, where $\sigma_{i}(P)(U_{i},Y)$ is obtained from $P(U_{i},Y)$ by
applying $\sigma_{i}$ to the coefficients. We then let $M^{(i)}$ denote the
field $E^{(i)}(U_{1},\dots,U_{i-1},U_{i+1},\dots U_{n})$ and set
$M=M^{(1)}\cdots M^{(n)}$. The extension
$M/L({\bf{U}})=M/L(U_{1},\dots,U_{n})$ is Galois of group $A^{n}$.

######  Lemma 3.7.

(1) There is an $L({\bf{U}})$-basis of $M$ which is integral over
$L[{\bf{U}}]$ and whose discriminant equals $(\Delta(U_{1})$
$\cdots\Delta(U_{n}))^{|A|^{n-1}}$ for some $\Delta(T)\in k[T]$ with
$\Delta(0)\in k^{*}$. In particular, $\langle U_{1},\dots,U_{n}\rangle$ is
unramified in $M/L({\bf{U}})$.

(2) The specialization $M_{(0,\dots,0)}/L$ of $M/L({\bf{U}})$ at $(0,\dots,0)$
equals $L/L$.

###### Proof.

We work with the $k(T)$-basis $z_{1},\dots,z_{|A|}$ of $F$ from the above.
Since $F\cap\overline{k}=k$, the elements $z_{1},\dots,z_{|A|}$ yield an
$L(T)$-basis of $E$ which is integral over $L[T]$ and whose discriminant
equals $\Delta(T)$. In particular, $\langle T\rangle$ is unramified in
$E/L(T)$. By Lemma 3.3(1), the specialization $E_{0}/L$ of $E/L(T)$ at 0 is
$L/L$. Next, let $i=1,\dots,n$. The elements
$\sigma_{i}(z_{1}),\dots,\sigma_{i}(z_{|A|})$ of $E^{(i)}$ are an
$L(U_{i})$-basis of $E^{(i)}$ which is integral over $L[U_{i}]$. Its
discriminant $\Delta_{i}(U_{i})$ is obtained from $\Delta(T)$ by replacing $T$
by $U_{i}$ and by applying $\sigma_{i}$ to the coefficients. But, since
$\Delta(T)\in k[T]$, we have $\Delta_{i}(U_{i})=\Delta(U_{i})$. In particular,
$\Delta_{i}(0)\not=0$ and $\langle U_{i}\rangle$ is unramified in
$E^{(i)}/L(U_{i})$. Moreover, the specialization $(E^{(i)})_{0}/L$ of
$E^{(i)}/L(U_{i})$ at 0 equals $L/L$. Lemma 3.1(2) then shows that $\langle
U_{1},\dots,U_{n}\rangle$ is unramified in $M^{(i)}/L({\bf{U}})$. Moreover, by
Lemma 3.3(2), the specialization $(M^{(i)})_{(0,\dots,0)}/L$ of
$M^{(i)}/L({\bf{U}})$ at $(0,\dots,0)$ equals $L/L$. Finally, by Lemma 3.1(1),
$\langle U_{1},\dots,U_{n}\rangle$ is unramified in $M/L({\bf{U}})$. Then the
$L({\bf{U}})$-basis
$\\{\sigma_{1}(z_{i_{1}})\cdots\sigma_{n}(z_{i_{n}})\\}_{1\leq
i_{1},\dots,i_{n}\leq|A|}$ of $M$ is integral over $L[{\bf{U}}]$. By the
construction and Lemma 3.2, its discriminant equals
$(\Delta(U_{1})\cdots\Delta(U_{n}))^{|A|^{n-1}}$, thus proving (1). As to (2),
as $\langle U_{1},\dots,U_{n}\rangle$ is unramified in $M^{(i)}/L({\bf{U}})$
for each $i\in\\{1,\dots,n\\}$, Lemma 3.3(1) yields
$M_{(0,\dots,0)}=(M^{(1)})_{(0,\dots,0)}\cdots(M^{(n)})_{(0,\dots,0)}$, i.e.,
$M_{(0,\dots,0)}=L$. This completes the proof. ∎

Now, there is a unique tuple ${\bf{T}}=(T_{1},\dots,T_{n})$ of indeterminates
fulfilling:

$\left\\{\begin{array}[]{ccc}\sigma_{1}(c_{1})T_{1}+\sigma_{1}(c_{2})T_{2}+\cdots+\sigma_{1}(c_{n})T_{n}&=&U_{1}\\\
\dots&=&\dots\\\
\sigma_{n}(c_{1})T_{1}+\sigma_{n}(c_{2})T_{2}+\cdots+\sigma_{n}(c_{n})T_{n}&=&U_{n}\end{array}\right..$
(3.5)

######  Lemma 3.8.

(1) The extension $M/k({\bf{T}})$ is Galois.

(2) The specialization of $M/k({\bf{T}})$ at $(0,\dots,0)$ equals $L/k$.

(3) The maximal ideal $\langle T_{1},\dots,T_{n}\rangle$ of $k[{\bf{T}}]$ is
unramified in $M/k({\bf{T}})$.

###### Proof.

(1) The extensions $M/L({\bf{T}})$ and $L({\bf{T}})/k({\bf{T}})$ are Galois.
Given $i\in\\{1,\dots,n\\}$, it then suffices to find
$\tau_{i}\in{\rm{Aut}}(M/k({\bf{T}}))$ such that the restriction of $\tau_{i}$
to $L$ is $\sigma_{i}$. To that end, note that the multiplication on the left
by $\sigma_{i}$ yields a pemutation $\epsilon_{i}$ of $\\{1,\dots,n\\}$. Then
consider the automorphism $\tau_{i}:M\rightarrow M$ defined by
$\tau_{i}(U_{j})=U_{\epsilon_{i}(j)}$ and
$\tau_{i}(y_{j})=y_{\epsilon_{i}(j)}$ for $j\in\\{1,\dots,n\\}$, and
$\tau_{i}|{{}_{L}}=\sigma_{i}$. The automorphism $\tau_{i}$ permutes the
equations of (3.5), i.e., $(\tau_{i}(T_{1}),\dots,\tau_{i}(T_{n}))$ is a
solution to (3.5). As the solution to (3.5) is unique, we have
$\tau_{i}(T_{i})=T_{i}$ for every $i\in\\{1,\dots,n\\}$. Hence, $\tau_{i}$ is
in ${\rm{Aut}}(M/k({\bf{T}}))$ and extends $\sigma_{i}$.

(2) By the definition of ${\bf{T}}$, the integral closure of $L[{\bf{U}}]$ in
$M$ equals the integral closure of $L[{\bf{T}}]$ in $M$, and a maximal ideal
of the latter integral closure contains $T_{1},\dots,T_{n}$ if and only if it
contains $U_{1},\dots,U_{n}$. The claim then follows from Lemma 3.7(2).

(3) By Lemma 3.7(1), there is an integral $L({\bf{U}})$-basis
$b_{1},\dots,b_{|A|^{n}}$ of $M$ whose discriminant is
$(\Delta(U_{1})\cdots\Delta(U_{n}))^{|A|^{n-1}}$ for some $\Delta(T)\in k[T]$
with $\Delta(0)\in k^{*}$. Clearly, $b_{1},\dots,b_{|A|^{n}}$ yield an
integral $L({\bf{T}})$-basis of $M$ whose discriminant
$\widetilde{\Delta}({\bf{T}})$, as an element of $L[{\bf{T}}]$, equals

$\widetilde{\Delta}({\bf{T}})=\prod_{i=1}^{n}(\Delta(\sigma_{i}(c_{1})T_{1}+\cdots+\sigma_{i}(c_{n})T_{n}))^{|A|^{n-1}}.$

As applying any element of ${\rm{Gal}}(L({\bf{T}})/k({\bf{T}}))$ to
$\widetilde{\Delta}({\bf{T}})$ permutes the factors, we have
$\widetilde{\Delta}({\bf{T}})\in k[{\bf{T}}]$, and
$\widetilde{\Delta}({\bf{T}})$ does not vanish at $(0,\dots,0)$. Moreover,
$c_{1},\dots,c_{n}$ yield a $k({\bf{T}})$-basis of $L({\bf{T}})$, which is
integral over $k[{\bf{T}}]$ and whose discriminant, which is in $k^{*}$,
cannot be in $\langle T_{1},\dots,T_{n}\rangle$. Lemma 3.2 then shows that the
discriminant of the integral $k({\bf{T}})$-basis $\\{b_{i}c_{j}\\}_{1\leq
i\leq{|A|^{n}},1\leq j\leq n}$ of $M$ is not in $\langle
T_{1},\dots,T_{n}\rangle$. Hence, $\langle T_{1},\dots,T_{n}\rangle$ is
unramified in $M/k({\bf{T}})$. ∎

Given groups $G_{1}$ and $G_{2}$, recall that ${G_{1}}^{G_{2}}$ denotes the
group of all functions $f:G_{2}\rightarrow G_{1}$, and that $G_{2}$ acts on
${G_{1}}^{G_{2}}$ by $f^{\tau}(\sigma)=f(\tau\sigma)$ ($\tau\in
G_{2},\sigma\in G_{2}$). The corresponding semidirect product will be denoted
by ${G_{1}}^{G_{2}}\rtimes G_{2}$, and the projection map
${G_{1}}^{G_{2}}\rtimes G_{2}\rightarrow G_{2}$ by ${\rm{pr}}$.

The first part of the following lemma is a special case of [FJ08, Remarks
13.7.6 and 13.7.7]. As to the second part, it is [FJ08, Lemma 16.4.3].

######  Lemma 3.9.

(1) There exists an isomorphism $\phi:{\rm{Gal}}(M/k({\bf{T}}))\rightarrow
A^{G^{\prime}}\rtimes G^{\prime}$ satisfying

${\rm{pr}}\circ\phi=\alpha^{\prime}\circ{\rm{res}}^{M/k({\bf{T}})}_{L/k}.$
(3.6)

(2) The following map is an epimorphism:

$\varphi:\left\\{\begin{array}[]{ccc}A^{G^{\prime}}\rtimes
G^{\prime}&\longrightarrow&A\rtimes G^{\prime}\\\
(f,\alpha^{\prime}(\sigma))&\longmapsto&(f(\alpha^{\prime}(\sigma_{1}))^{\alpha^{\prime}(\sigma_{1})^{-1}}\cdots
f(\alpha^{\prime}(\sigma_{n}))^{\alpha^{\prime}(\sigma_{n})^{-1}},\alpha^{\prime}(\sigma))\end{array}.\right.$
(3.7)

Let us next consider $N=M^{{\rm{ker}}(\psi\circ\varphi\circ\phi)},$ where
$\psi$, $\phi$, and $\varphi$ are defined in (3.4), (3.6), and (3.7),
respectively. By Lemma 3.9(1),
${\rm{Gal}}(M/L({\bf{T}}))=\phi^{-1}(A^{G^{\prime}}\rtimes\\{1\\})$. Hence,
$L\subseteq N$ by the definition of $\varphi$. Moreover, by Lemma 3.8(2), the
specialization $N_{(0,\dots,0)}/k$ of $N/k({\bf{T}})$ at $(0,\dots,0)$ equals
$L/k$. Finally, there is an isomorphism
$\beta:{\rm{Gal}}(N/k({\bf{T}}))\rightarrow G$ satisfying

$\beta\circ{\rm{res}}^{M/k({\bf{T}})}_{N/k({\bf{T}})}=\psi\circ\varphi\circ\phi.$
(3.8)

We then have $\alpha\circ\beta={\rm{res}}^{N/k({\bf{T}})}_{L/k}$. Indeed,
given $\sigma\in{\rm{Gal}}(N/k({\bf{T}}))$, let
$\tau\in{\rm{Gal}}(M/k({\bf{T}}))$ be such that
${\rm{res}}^{M/k({\bf{T}})}_{N/k({\bf{T}})}(\tau)=\sigma$. Since (3.7) gives
$\psi\circ\varphi(A^{G^{\prime}}\times\\{1\\})\subseteq A$
($={\rm{ker}}(\alpha)$), we have:

$\alpha\circ\beta(\sigma)\stackrel{{\scriptstyle(\ref{eqbetafin})}}{{=}}\alpha\circ\psi\circ\varphi\circ\phi(\tau)\stackrel{{\scriptstyle(\ref{eqlemma2})}}{{=}}\alpha\circ\psi(1,{\rm{pr}}\circ\phi(\tau))\stackrel{{\scriptstyle(\ref{eqlemma})}}{{=}}\alpha\circ\psi(1,\alpha^{\prime}\circ{\rm{res}}^{M/k({\bf{T}})}_{L/k}(\tau))$
$\hskip
241.84842pt\stackrel{{\scriptstyle(\ref{eqdefpsi})}}{{=}}\alpha\circ\alpha^{\prime}\circ{\rm{res}}^{M/k({\bf{T}})}_{L/k}(\tau)$
$\hskip
202.01474pt\stackrel{{\scriptstyle(\ref{eqid})}}{{=}}{\rm{res}}^{M/k({\bf{T}})}_{L/k}(\tau)$
$\hskip 202.01474pt={\rm{res}}^{N/k({\bf{T}})}_{L/k}(\sigma).$

######  Lemma 3.10.

Let $k$ be a Hilbertian field, $\mathcal{S}$ a finite set of valuations of
$k$, and $\alpha:G\rightarrow{\rm{Gal}}(L/k)$ a finite embedding problem over
$k$. Assume there are a tuple ${\bf{T}}=(T_{1},\dots,T_{n})$ of indeterminates
and a solution $\beta:{\rm{Gal}}(N/k({\bf{T}}))\rightarrow G$ to
$\alpha_{k({\bf{T}})}$ such that the ideal $\langle T_{1},\dots,T_{n}\rangle$
is unramified in $N/k({\bf{T}})$. Then the specialization $N_{(0,\dots,0)}/k$
of $N/k({\bf{T}})$ at $(0,\dots,0)$ is Galois and there is a solution
${\rm{Gal}}(F/k)\rightarrow G$ to $\alpha$ such that $F\cdot
k_{v}^{h}=N_{(0,\dots,0)}\cdot k_{v}^{h}$ for every $v\in\mathcal{S}$.

Addendum 3.10. The field $F$ can be chosen as follows. Given a new
indeterminate $T$, consider the Galois field extension $N(T)$ of
$k(T,{\bf{T}})$. Then there exist ${\bf{a}}\in k[T]^{n}$ and $t\in k$ such
that

$\bullet$ the specialization $(N(T))_{\bf{a}}/k(T)$ of $N(T)/k(T)({\bf{T}})$
at ${\bf{a}}$ is Galois of group $G$,

$\bullet$ the specialization $((N(T))_{\bf{a}})_{t}/k$ of
$(N(T))_{\bf{a}}/k(T)$ at $t$ is Galois of group $G$,

$\bullet$ $F=((N(T))_{\bf{a}})_{t}$.

###### Proof.

We break the proof into five parts.

(a) Let $P({\bf{T}},Y)\in k[{\bf{T}}][Y]$ be the minimal polynomial of a
primitive element of $N/k({\bf{T}})$, assumed to be integral over
$k[{\bf{T}}]$. As $k$ is assumed to be Hilbertian, there is
${\bf{t}}=(t_{1},\dots,t_{n})\in k^{n}$ such that $P({\bf{t}},Y)\in k[Y]$ is
irreducible over $k$ and separable. Hence, the specialization $N_{\bf{t}}/k$
of $N/k({\bf{T}})$ at ${\bf{t}}$ is Galois of degree $[N:k({\bf{T}})]$ and
$N_{\bf{t}}$ is the splitting field over $k$ of $P({\bf{t}},Y)$. Given a new
indeterminate $T$, for $i\in\\{1,\dots,n\\}$, fix $a_{i}(T)\in k[T]$ with
$a_{i}(0)=0$ and $a_{i}(1)=t_{i}$. Consider the Galois extension
$N(T)/k(T)({\bf{T}})$ of group $G$ and its specialization
$(N(T))_{\bf{a}}/k(T)$ at ${\bf{a}}=(a_{1}(T),\dots,a_{n}(T))$. As
$P({\bf{t}},Y)$ is separable and $a_{i}(1)=t_{i}$ for every
$i\in\\{1,\dots,n\\}$, the polynomial $P({\bf{a}},Y)$ is separable. Hence,
$(N(T))_{\bf{a}}$ is the splitting field over $k(T)$ of $P({\bf{a}},Y)$ and
$(N(T))_{\bf{a}}/k(T)$ is separable. Using again that $P({\bf{t}},Y)$ is
separable and $a_{i}(1)=t_{i}$ for every $i\in\\{1,\dots,n\\}$, we get that
$((N(T))_{\bf{a}})_{1}$ is the splitting field over $k$ of $P({\bf{t}},Y)$,
i.e., equals $N_{\bf{t}}$. As $[N_{\bf{t}}:k]=[N(T):k(T,{\bf{T}})]$, we get
the equality $[(N(T))_{\bf{a}}:k(T)]=[N(T):k(T)({\bf{T}})]$.

(b) The maximal ideal $\langle T\rangle$ is unramified in
$(N(T))_{\bf{a}}/k(T)$. Indeed, as $\langle T_{1},\dots,T_{n}\rangle$ is
unramified in $N/k({\bf{T}})$, there is a $k({\bf{T}})$-basis
$y_{1},\dots,y_{d}$ of $N$ which is integral over $k[{\bf{T}}]$ and whose
discriminant $\Delta({\bf{T}})\in k[{\bf{T}}]$ fulfills $\Delta(0,\dots,0)\in
k^{*}$. As $k({\bf{T}},T)$ and $N$ are linearly disjoint over $k({\bf{T}})$,
the elements $y_{1},\dots,y_{d}$ yield a $k({\bf{T}},T)$-basis of $N(T)$ which
is contained in the integral closure $B$ of $k(T)[{\bf{T}}]$ in $N(T)$ and
with $\Delta({\bf{T}})B\subseteq k(T)[{\bf{T}}]y_{1}\oplus\cdots\oplus
k(T)[{\bf{T}}]y_{d}.$ As
$\Delta(a_{1}(0),\dots,a_{n}(0))=\Delta(0,\dots,0)\not=0$, we have
$\Delta({\bf{a}})\not=0$ (in particular, $\langle
T_{1}-a_{1}(T),\dots,T_{n}-a_{n}(T)\rangle$ is unramified in
$N(T)/k(T)({\bf{T}})$). Hence, if $\mathfrak{P}$ is a maximal ideal of $B$
lying over $\langle T_{1}-a_{1}(T),\dots,T_{n}-a_{n}(T)\rangle$, we have
$(N(T))_{\bf{a}}=B/\mathfrak{P}=k(T)\overline{y_{1}}\oplus\cdots\oplus
k(T)\overline{y_{d}},$ where $\overline{y_{i}}$ is the reduction modulo
$\mathfrak{P}$ of $y_{i}$ ($1\leq i\leq d$). As the discriminant
$\Delta(a_{1}(T),\dots,a_{n}(T))$ of $\overline{y_{1}},\dots,\overline{y_{d}}$
does not vanish at 0, the claim holds.

(c) The specialization $((N(T))_{\bf{a}})_{0}/k$ of $(N(T))_{\bf{a}}/k(T)$ at
0 equals $N_{(0,\dots,0)}/k$ (in particular, $N_{(0,\dots,0)}/k$ is Galois by
(b)). Indeed, as $\langle T_{1},\dots,T_{n}\rangle$ is unramified in
$N/k({\bf{T}})$, Lemma 3.3(2) yield that the specialization
$N_{(0,0,\dots,0)}/k$ of $N(T)/k(T,{\bf{T}})$ at $(0,0,\dots,0)$ equals
$N_{(0,\dots,0)}/k$. Moreover, $\langle
T_{1}-a_{1}(T),\dots,T_{n}-a_{n}(T)\rangle$ is unramified in
$N(T)/k(T)({\bf{T}})$ (see (b)). The claim then follows from Lemma 3.5, if 0
is not in the exceptional set from that lemma, when applied to
$N(T)/k(T,T_{1},\dots,T_{n})$. But, in the present situation, the latter set
is reduced to the set of roots of $\Delta(a_{1}(T),\dots,a_{n}(T))$ (see
Remark 3.6), which does not contain 0 (see (b)).

(d) We now specialize $T$ suitably. As $\langle T\rangle$ is unramified in
$(N(T))_{\bf{a}}/k(T)$ (see (b)) and $k$ is infinite, there is a primitive
element of $(N(T))_{\bf{a}}$ over $k(T)$, which is integral over $k[T]$ and
whose minimal polynomial $Q(T,Y)\in k[T][Y]$ is such that $Q(0,Y)\in k[Y]$ is
separable (see, e.g., [Dèb09, corollaire 1.5.16]). Krasner’s lemma and the
“continuity of roots” (see, e.g., [Jar91, Proposition 12.3]), applied to the
separable polynomial $Q(0,Y)$ over the Henselian valued field
$(k_{v}^{h},v^{h})$, provide an element $\alpha_{v}$ of the value group of $v$
such that, if $t\in k$ fulfills $v(t)\geq\alpha_{v}$, then the splitting
fields of $Q(0,Y)$ and $Q(t,Y)$ over $k_{v}^{h}$ coincide.

Next, as $k$ is Hilbertian, [Jar91, Proposition 19.6] yields $t\in k$ such
that $v(t)\geq\alpha_{v}$ for every $v\in\mathcal{S}$, and such that
$Q(t,Y)\in k[Y]$ is irreducible over $k$ and separable. Let $F$ be the field
generated over $k$ by one root of $Q(t,Y)$. As the latter is irreducible over
$k$, we have $[F:k]=[(N(T))_{\bf{a}}:k(T)]$ and $F/k$ is the specialization
$((N(T))_{\bf{a}})_{t}/k$ of $(N(T))_{\bf{a}}/k(T)$ at $t$. In particular,
$((N(T))_{\bf{a}})_{t}/k$ is Galois of degree $[((N(T))_{\bf{a}}:k(T)]$.

Finally, as $Q(0,Y)$ is separable, the splitting field of $Q(0,Y)$ over $k$
equals $((N(T))_{\bf{a}})_{0}$, i.e., equals $N_{(0,\dots,0)}$ (see (c)).
Hence, for $v\in\mathcal{S}$, we have $N_{(0,\dots,0)}\cdot k_{v}^{h}=F\cdot
k_{v}^{h}$.

(e) Finally, as $\beta:{\rm{Gal}}(N/k({\bf{T}}))\rightarrow G$ is a solution
to $\alpha_{k({\bf{T}})}$, the map
$\beta\circ{\rm{res}}^{N(T)/k(T,{\bf{T}})}_{N/k({\bf{T}})}:{\rm{Gal}}(N(T)/k(T,{\bf{T}}))\rightarrow
G$ is a solution to $\alpha_{k(T,{\bf{T}})}=(\alpha_{k(T)})_{k(T)({\bf{T}})}$.
Since $(N(T))_{\bf{a}}/k(T)$ is a Galois field extension of degree
$[N(T):k(T)({\bf{T}})]$, Lemma 2.1 shows that
$(\beta\circ{\rm{res}}^{N(T)/k(T,{\bf{T}})}_{N/k({\bf{T}})})_{\bf{a}}:{\rm{Gal}}((N(T))_{\bf{a}}/k(T))\rightarrow
G$ is a solution to $\alpha_{k(T)}$. Moreover, $F=((N(T))_{\bf{a}})_{t}$ is a
Galois field extension of $k$ of degree $[(N(T))_{\bf{a}}:k(T)]$. Applying
Lemma 2.1 once more then shows that
$((\beta\circ{\rm{res}}^{N(T)/k(T,{\bf{T}})}_{N/k({\bf{T}})})_{\bf{a}})_{t}:{\rm{Gal}}(F/k)\rightarrow
G$ is a solution to $\alpha$. This concludes the proof. ∎

By Lemma 3.10 and the above, to conclude the proof of Theorem 1.2, it would
suffice to show that the ideal $\langle T_{1},\dots,T_{n}\rangle$ is
unramified in $N/k({\bf{T}})$. As proved above, $\langle
T_{1},\dots,T_{n}\rangle$ is unramified in $M/k({\bf{T}})$ and $N\subseteq M$.
However, with our definition of “unramified”, which is designed for computing
residue fields, it is not clear that unramifiedness is closed under
subextensions. To conclude the proof of Theorem 1.2, we proceed as follows.

With $\Gamma={\rm{Gal}}(M/k({\bf{T}}))$, consider the finite embedding problem
${\rm{res}}^{M/k({\bf{T}})}_{L/k}:\Gamma\rightarrow{\rm{Gal}}(L/k)$ over $k$.
Clearly, ${\rm{id}}:{\rm{Gal}}(M/k({\bf{T}}))\rightarrow\Gamma$ is a solution
to $({\rm{res}}^{M/k({\bf{T}})}_{L/k})_{k({\bf{T}})}$. By Lemma 3.8, we may
apply Lemma 3.10 and its addendum at the level of $M$. Namely, given a new
indeterminate $T$, there exist ${\bf{a}}\in k[T]^{n}$, $t\in k$, and a
solution ${\rm{Gal}}(F/k)\rightarrow\Gamma$ to
${\rm{res}}^{M/k({\bf{T}})}_{L/k}$ such that the following conditions hold:

$\bullet$ the specialization $(M(T))_{\bf{a}}/k(T)$ of $M(T)/k(T)({\bf{T}})$
at ${\bf{a}}$ is Galois of group $\Gamma$,

$\bullet$ the specialization $((M(T))_{\bf{a}})_{t}/k$ of
$(M(T))_{\bf{a}}/k(T)$ at $t$ is Galois of group $\Gamma$,

$\bullet$ $F=((M(T))_{\bf{a}})_{t}$ and $F\subseteq L\cdot k_{v}^{h}$ for
every $v\in\mathcal{S}$.

Finally, we proceed as in the last part of the proof of Lemma 3.10. Consider
the solution $\beta:{\rm{Gal}}(N/k({\bf{T}}))\rightarrow G$ to
$\alpha_{k({\bf{T}})}$. Then

$\beta\circ{\rm{res}}^{N(T)/k(T,{\bf{T}})}_{N/k({\bf{T}})}:{\rm{Gal}}(N(T)/k(T,{\bf{T}}))\rightarrow
G$

is a solution to $(\alpha_{k(T)})_{k(T)({\bf{T}})}$. As $(M(T))_{\bf{a}}/k(T)$
is Galois of degree $[M(T):k(T)({\bf{T}})]$, the specialization
$(N(T))_{\bf{a}}/k(T)$ of $N(T)/k(T)({\bf{T}})$ at ${\bf{a}}$, which is a
subextension of $(M(T))_{\bf{a}}/k(T)$, is Galois of degree
$[N(T):k(T)({\bf{T}})]$. Lemma 2.1 then yields a solution
${\rm{Gal}}((N(T))_{\bf{a}}/k(T))\rightarrow G$ to $\alpha_{k(T)}$. Again,
$((M(T))_{\bf{a}})_{t}/k$ is Galois of degree $[(M(T))_{\bf{a}}:k(T)]$. Hence,
$((N(T))_{\bf{a}})_{t}/k$ is Galois of degree $[(N(T))_{\bf{a}}:k(T)]$. Lemma
2.1 then yields a solution ${\rm{Gal}}(((N(T))_{\bf{a}})_{t}/k)\rightarrow G$
to $\alpha$ such that
$((N(T))_{\bf{a}})_{t}\subseteq((M(T))_{\bf{a}})_{t}\subseteq L\cdot
k_{v}^{h}$ for every $v\in\mathcal{S}$, thus concluding the proof.

##  4\. On the Beckmann–Black problem

The following theorem, which solves the Beckmann–Black problem for finite
embedding problems with abelian kernels over arbitrary fields, is the aim of
the present section:

######  Theorem 4.1.

Let $\alpha:G\rightarrow{\rm{Gal}}(L/k)$ be a finite embedding problem with
abelian kernel over any field $k$ and $\gamma:{\rm{Gal}}(F/k)\rightarrow G$ a
solution to $\alpha$. There are $t_{0}\in k$ and a geometric solution
$\beta:{\rm{Gal}}(E/k(T))\rightarrow G$ to $\alpha$ with $E\cap\overline{k}=L$
such that $E_{t_{0}}/k=F/k$ and $\beta_{t_{0}}=\gamma$.

The present section is organized as follows. In §4.1, we prove Theorem 4.1
under the extra assumption that $\alpha$ splits, and then show in §4.2 that
the latter assumption is redundant. Finally, in §4.3, we derive a more precise
version of Theorem 1.3.

###  4.1. Proof of Theorem 4.1 under an extra assumption

######  Proposition 4.2.

Let $\alpha:G\rightarrow{\rm{Gal}}(L/k)$ be a finite split embedding problem
with abelian kernel over an arbitrary field $k$ and
$\gamma:{\rm{Gal}}(F/k)\rightarrow G$ a solution to $\alpha$. There are a
geometric solution $\beta:{\rm{Gal}}(E/k(T))\rightarrow G$ to $\alpha$ with
$E\cap\overline{k}=L$, such that the specialization $E_{0}/k$ of $E/k(T)$ at
$0$ equals $F/k$, and such that the specialization $\beta_{0}$ of $\beta$ at
$0$ equals $\gamma$.

###### Proof.

First, as $\alpha$ splits, there is an embedding
$\alpha^{\prime}:{\rm{Gal}}(L/k)\rightarrow G$ with
$\alpha\circ\alpha^{\prime}={\rm{id}}_{{\rm{Gal}}(L/k)}$. Set
$G^{\prime}=\alpha({\rm{Gal}}(L/k))$. Then $G^{\prime}$ acts on
${\rm{ker}}(\alpha)$ by conjugation and

$\phi:\left\\{\begin{array}[]{ccc}{\rm{ker}}(\alpha)\rtimes
G^{\prime}&\longrightarrow&G\\\ (a,g^{\prime})&\longmapsto&a\cdot
g^{\prime}\end{array}\right.$

is an isomorphism. Moreover, since $\alpha$ splits and has abelian kernel, we
may apply Theorem 1.2 to get the existence of a geometric solution
$\beta:{\rm{Gal}}(E/k(T))\rightarrow G$ to $\alpha$ such that the inclusion
$E\subseteq L((T))$ holds.

Now, we identify ${\rm{Gal}}(EF/k(T))$ and a subgroup of
$({\rm{ker}}(\alpha)\rtimes G^{\prime})\times({\rm{ker}}(\alpha)\rtimes
G^{\prime})$. Since $E\cap\overline{k}=L$, the fields $E$ and $F(T)$ are
linearly disjoint over $L(T)$. Hence, with

$\widetilde{G}=\\{(\sigma,\tau)\in{\rm{Gal}}(E/k(T))\times{\rm{Gal}}(F(T)/k(T))\,:\,{\rm{res}}^{E/k(T)}_{L(T)/k(T)}(\sigma)={\rm{res}}^{F(T)/k(T)}_{L(T)/k(T)}(\tau)\\},$
$\psi:\left\\{\begin{array}[]{ccc}{\rm{Gal}}(EF/k(T))&\longrightarrow&\widetilde{G}\\\
\sigma&\longmapsto&({\rm{res}}^{EF/k(T)}_{E/k(T)}(\sigma),{\rm{res}}^{EF/k(T)}_{F(T)/k(T)}(\sigma))\end{array}\right.$

is an isomorphism. Next,
$\phi^{-1}\circ\beta:{\rm{Gal}}(E/k(T))\rightarrow{\rm{ker}}(\alpha)\rtimes
G^{\prime}$ and
$\phi^{-1}\circ\gamma\circ{\rm{res}}^{F(T)/k(T)}_{F/k}:{\rm{Gal}}(F(T)/k(T))\rightarrow{\rm{ker}}(\alpha)\rtimes
G^{\prime}$ are isomorphisms. Then consider the monomorphism

$\theta:\left\\{\begin{array}[]{ccc}\widetilde{G}&\longrightarrow&({\rm{ker}}(\alpha)\rtimes
G^{\prime})\times({\rm{ker}}(\alpha)\rtimes G^{\prime})\\\
(\sigma,\tau)&\longmapsto&(\phi^{-1}\circ\beta(\sigma),\phi^{-1}\circ\gamma\circ{\rm{res}}^{F(T)/k(T)}_{F/k}(\tau))\end{array}\right..$

We claim that the image of $\theta$ equals
$\mathcal{G}=\\{((a,g^{\prime}),(b,g^{\prime}))\,:\,(a,b)\in{\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha),\,g^{\prime}\in
G^{\prime}\\}$. Hence, from the above, the composed map
$\theta\circ\psi:{\rm{Gal}}(EF/k(T))\rightarrow\mathcal{G}$ is an isomorphism.

To show the claim, for
$(\sigma,\tau)\in{\rm{Gal}}(E/k(T))\times{\rm{Gal}}(F(T)/k(T))$, set
$\phi^{-1}\circ\beta(\sigma)=(a,g^{\prime}_{1})\in{\rm{ker}}(\alpha)\rtimes
G^{\prime}$ and
$\phi^{-1}\circ\gamma\circ{\rm{res}}^{F(T)/k(T)}_{F/k}(\tau)=(b,g^{\prime}_{2})\in{\rm{ker}}(\alpha)\rtimes
G^{\prime}$. Set $g^{\prime}_{i}=\alpha^{\prime}(g_{i})$ with $g_{i}\in G$ for
$i=1,2$. We then have

$g_{1}=\alpha\circ\alpha^{\prime}(g_{1})=\alpha(g^{\prime}_{1})=\alpha(ag^{\prime}_{1})=\alpha\circ\beta(\sigma)={\rm{res}}^{E/k(T)}_{L/k}(\sigma)={\rm{res}}^{L(T)/k(T)}_{L/k}({\rm{res}}^{E/k(T)}_{L(T)/k(T)}(\sigma))$

and, similarly,
$g_{2}={\rm{res}}^{L(T)/k(T)}_{L/k}({\rm{res}}^{F(T)/k(T)}_{L(T)/k(T)}(\tau)).$
Hence, the image of $\theta$ equals $\mathcal{G}$, as claimed.

Finally, consider the subgroup
$\mathcal{H}=\\{((a,1),(a,1))\,:\,a\in{\rm{ker}}(\alpha)\\}$ of $\mathcal{G}$.
As ${\rm{ker}}(\alpha)$ is abelian, $\mathcal{H}$ is normal in $\mathcal{G}$.
Hence, with $M=(EF)^{(\theta\circ\psi)^{-1}(\mathcal{H})}$, the extension
$M/k(T)$ is Galois of degree $[EF:k(T)]/|{\rm{ker}}(\alpha)|=|G|$. Moreover,
as
$\theta\circ\psi({\rm{Gal}}(EF/L(T)))=\\{((a,1),(b,1))\,:\,(a,b)\in{\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha)\\}$,
we have $L(T)\subseteq M$. Furthermore, as $M$ and $E$ are linearly disjoint
over $L(T)$, and as $[M:L(T)]=|{\rm{ker}}(\alpha)|$, we get $EF=EM$. Hence,
the specialization $(EF)_{0}/k$ of $EF/k(T)$ at 0 is the specialization
$(EM)_{0}/k$ of $EM/k(T)$ at 0. But, as $E\subseteq L((T))$, the ideal
$\langle T\rangle$ is unramified in $E/k(T)$. By, e.g., [FJ08, Lemma 2.4.8]
(or Lemma 3.3(1)), we then have $(EF)_{0}=E_{0}(F(T))_{0}=E_{0}F$ and
$(EM)_{0}=E_{0}M_{0}$. Using again $E\subseteq L((T))$, we get $E_{0}=L$ and
so $F=M_{0}$. It then remains to apply Lemma 2.3 to conclude the proof of the
proposition, up to showing $M\cap\overline{k}=L$. But $M\overline{k}\supseteq
MF=EF$, i.e., $E\overline{k}\subseteq M\overline{k}$. As
$[E\overline{k}:\overline{k}(T)]=[E:L(T)]=[M:L(T)]$, we get
$[M\overline{k}:\overline{k}(T)]=[M:L(T)]$, thus showing the desired equality
$M\cap\overline{k}=L$. ∎

###  4.2. Proof of Theorem 4.1

Denote the canonical projection $G\rightarrow G/{\rm{ker}}(\alpha)$ by
$\varphi$. Set $G_{\varphi}^{2}=\\{(g_{1},g_{2}))\in
G^{2}\,:\,\varphi(g_{1})=\varphi(g_{2})\\}$ and, for $i\in\\{1,2\\}$, consider

$\pi_{i}:\left\\{\begin{array}[]{ccc}G_{\varphi}^{2}&\longrightarrow&G\\\
(g_{1},g_{2})&\longmapsto&g_{i}\end{array}\right..$

We also denote the kernel of $\pi_{i}$ by $N_{i}$ for $i\in\\{1,2\\}$.

Since we are given the solution $\gamma$ to $\alpha$, whose kernel is abelian,
we may combine the weak$\rightarrow$split reduction (see [Pop96, §1 B) 2)] and
[DD97, §2.1.2]) and, e.g., [FJ08, §16.4] to get that $\alpha$ has a geometric
solution $\delta:{\rm{Gal}}(F_{2}/k(T_{1}))\rightarrow G$ fulfilling
$F_{2}\cap\overline{k}=L$.

######  Lemma 4.3.

(1) The following map is a well-defined isomorphism:

$\epsilon:\left\\{\begin{array}[]{ccc}{\rm{Gal}}(FF_{2}/k(T_{1}))&\longrightarrow&G_{\varphi}^{2}\\\
\sigma&\longmapsto&(\gamma\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F/k}(\sigma),\delta\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F_{2}/k(T_{1})}(\sigma))\end{array}\right..$

(2) We have
$\epsilon({\rm{Gal}}(FF_{2}/L(T_{1})))={\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha)$.

###### Proof.

Let ${\theta}:{\rm{Gal}}(L/k)\rightarrow G/{\rm{ker}}(\alpha)$ be the unique
isomorphism satisfying ${\theta}\circ\alpha=\varphi$. As

$\varphi\circ\gamma\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F/k}=\theta\circ\alpha\circ\gamma\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F/k}=\theta\circ{\rm{res}}^{F/k}_{L/k}\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F/k}=\theta\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{L/k},$
$\varphi\circ\delta\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F_{2}/k(T_{1})}=\theta\circ\alpha\circ\delta\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F_{2}/k(T_{1})}=\theta\circ{\rm{res}}_{L/k}^{F_{2}/k(T_{1})}\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F_{2}/k(T_{1})}=\theta\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{L/k},$

$\epsilon$ is well-defined. Moreover, $\epsilon$ is clearly injective and,
since $F_{2}\cap\overline{k}=L$, the fields $F(T_{1})$ and $F_{2}$ are
linearly disjoint over $L(T_{1})$, thus showing that $\epsilon$ is also
surjective. Hence, (1) holds. As for (2), for
$\sigma\in{\rm{Gal}}(FF_{2}/L(T_{1}))$, we have
$\alpha\circ\gamma\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F/k}(\sigma)={\rm{res}}^{F/k}_{L/k}\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F/k}(\sigma)={\rm{id}}_{{\rm{Gal}}(L/k)}$
and, similarly,
$\alpha\circ\delta\circ{\rm{res}}^{FF_{2}/k(T_{1})}_{F_{2}/k(T_{1})}(\sigma)={\rm{id}}_{{\rm{Gal}}(L/k)}$,
thus showing
$\epsilon({\rm{Gal}}(FF_{2}/L(T_{1})))\subseteq{\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha)$.
Since both groups have the same order, we are done. ∎

Now, consider the finite embedding problem
$\delta^{-1}\circ\pi_{2}:G_{\varphi}^{2}\rightarrow{\rm{Gal}}(F_{2}/k(T_{1}))$
over $k(T_{1})$. Then $\delta^{-1}\circ\pi_{2}$ splits and its kernel $N_{2}$
is abelian. Moreover, the isomorphism $\epsilon$ from Lemma 4.3(1) is a
solution to $\delta^{-1}\circ\pi_{2}$. Consequently, we may use Proposition
4.2 to lift the solution $\epsilon$ to a geometric solution. Namely, there
exists a geometric solution $\beta:{\rm{Gal}}(E/k(T_{1})(T_{2}))\rightarrow
G_{\varphi}^{2}$ to $\delta^{-1}\circ\pi_{2}$ with
$E\cap\overline{k(T_{1})}=F_{2}$, such that the specialization
$E_{0}/k(T_{1})$ of $E/k(T_{1})(T_{2})$ at 0 equals $FF_{2}/k(T_{1})$, and
such that the specialization of $\beta$ at $0$ is $\epsilon$.

######  Lemma 4.4.

(1) We have $E^{\beta^{-1}(N_{1})}\cap\overline{k(T_{1})}=L(T_{1})$.

(2) The specialization $(E^{\beta^{-1}(N_{1})})_{0}/k(T_{1})$ of
$E^{\beta^{-1}(N_{1})}/k(T_{1})(T_{2})$ at $0$ equals $F(T_{1})/k(T_{1})$.

###### Proof.

Let
$\varphi_{0}:{\rm{Gal}}(E/k(T_{1})(T_{2}))\rightarrow{\rm{Gal}}(E_{0}/k(T_{1}))$
be the isomorphism defined in (2.1).

(1) By the definition of $\varphi_{0}$, we have

$\varphi_{0}({\rm{Gal}}(E/L(T_{1})(T_{2})))={\rm{Gal}}(E_{0}/L(T_{1}))={\rm{Gal}}(FF_{2}/L(T_{1}))=\epsilon^{-1}({\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha)),$

the last equality being Lemma 4.3(2). As
$\beta\circ\varphi_{0}^{-1}=\epsilon$, we get
${\rm{Gal}}(E/L(T_{1},T_{2}))=\beta^{-1}({\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha))$
and, as $N_{1}\subseteq{\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha)$, we get
$L(T_{1},T_{2})\subseteq E^{\beta^{-1}(N_{1})}$. In particular,
$L(T_{1})\subseteq E^{\beta^{-1}(N_{1})}\cap\overline{k(T_{1})}$.

For the converse, the equality $E\cap\overline{k(T_{1})}=F_{2}$ yields
$E^{\beta^{-1}(N_{1})}\cap\overline{k(T_{1})}=E^{\beta^{-1}(N_{1})}\cap
F_{2}.$ Hence, $(E^{\beta^{-1}(N_{1})}\cap\overline{k(T_{1})})(T_{2})$ is a
subfield of

$E^{\beta^{-1}(N_{1})}\cap F_{2}(T_{2})=E^{\beta^{-1}(N_{1})}\cap
E^{\beta^{-1}(N_{2})}{=}E^{\langle\beta^{-1}(N_{1}),\beta^{-1}(N_{2})\rangle}=E^{\beta^{-1}({\rm{ker}}(\alpha)\times{\rm{ker}}(\alpha))},$

i.e., a subfield of $L(T_{1},T_{2})$. We then have
$E^{\beta^{-1}(N_{1})}\cap\overline{k(T_{1})}\subseteq L(T_{1})$.

(2) By the definition of $\epsilon$, we have
$F(T_{1})=(FF_{2})^{\epsilon^{-1}(N_{1})}=(E_{0})^{\epsilon^{-1}(N_{1})}=(E_{0})^{\varphi_{0}\circ\beta^{-1}(N_{1})}.$
Hence, $(E^{\beta^{-1}(N_{1})})_{0}=F(T_{1})$ if and only if
$(E^{\beta^{-1}(N_{1})})_{0}=(E_{0})^{\varphi_{0}\circ\beta^{-1}(N_{1})},$
i.e., if and only if
${\rm{Gal}}(E_{0}/(E^{\beta^{-1}(N_{1})})_{0})={\varphi_{0}\circ\beta^{-1}(N_{1})}$.
As the latter equality holds by the definition of $\varphi_{0}$, we have
$(E^{\beta^{-1}(N_{1})})_{0}=F(T_{1})$, as needed. ∎

Next, as in the fourth part of the proof of Lemma 3.10, we specialize $T_{2}$
suitably. By Lemma 4.4(2), the ideal $\langle T_{2}\rangle$ is unramified in
$E^{\beta^{-1}(N_{1})}/L(T_{1})(T_{2})$. Then, as $L(T_{1})$ is infinite,
there exists a primitive element of $E^{\beta^{-1}(N_{1})}$ over
$L(T_{1})(T_{2})$ which is integral over $L(T_{1})[T_{2}]$ and whose minimal
polynomial $P(T_{2},Y)\in L(T_{1})[T_{2}][Y]$ is such that $P(0,Y)\in
L(T_{1})[Y]$ is separable. Krasner’s lemma and the “continuity of roots” (see,
e.g., [Jar91, Proposition 12.3]), applied to the separable polynomial $P(0,Y)$
over the complete valued field $L((T_{1}))$, yield a positive integer $N$ such
that, for every $t_{2}(T_{1})\in L(T_{1})$ of valuation at least $N$ with
respect to $\langle T_{1}\rangle$, the splitting fields of $P(0,Y)$ and
$P(t_{2}(T_{1}),Y)$ over $L((T_{1}))$ coincide.

Finally, by Lemma 4.4(1), we have $E^{\beta^{-1}(N_{1})}\cap\overline{k}=L$.
Hence, $P(T_{2},Y)$ is irreducible over $\overline{k}(T_{1})(T_{2})$. In
particular, $P(T_{1}^{N}T_{2},Y)$ and $P(T_{1}^{N}T_{2}^{-1},Y)$ are also
irreducible over $\overline{k}(T_{1})(T_{2})$. Then apply either [FJ08,
Proposition 13.2.1] if $k$ is infinite or [FJ08, Theorem 13.4.2 and
Proposition 16.11.1] if $k$ is finite to get the existence of $t_{2}(T_{1})\in
k(T_{1})$ such that $P(T_{1}^{N}t_{2}(T_{1}),Y)$ $\in L(T_{1})[Y]$ and
$P(T_{1}^{N}t_{2}(T_{1})^{-1},Y)\in L(T_{1})[Y]$ are irreducible over
$\overline{k}(T_{1})$ and separable. Without loss, we may assume that
$t_{2}(T_{1})$ is of non-negative valuation with respect to $\langle
T_{1}\rangle$. Let $M$ be the field generated over $L(T_{1})$ by one root of
$P(T_{1}^{N}t_{2}(T_{1}),Y)$. As the latter is irreducible over $L(T_{1})$,
one has $[M:k(T_{1})]=[E^{\beta^{-1}(N_{1})}:k(T_{1})(T_{2})]$ and,
consequently, the fields $M$ and $(E^{\beta^{-1}(N_{1})})_{t_{2}(T_{1})}$
coincide. It then remains to use the irreducibility of
$P(T_{1}^{N}t_{2}(T_{1}),Y)$ over $\overline{k}(T_{1})$ to get
$(E^{\beta^{-1}(N_{1})})_{t_{2}(T_{1})}\cap\overline{k}=L$. Moreover, since
$P(T_{1}^{N}t_{2}(T_{1}),Y)$ is separable, its splitting field over $L(T_{1})$
equals $(E^{\beta^{-1}(N_{1})})_{t_{2}(T_{1})}$. Similarly, as $P(0,Y)$ is
separable, its splitting field over $L(T_{1})$ equals
$(E^{\beta^{-1}(N_{1})})_{0}$, i.e., equals $F(T_{1})$ (see Lemma 4.4(2)).
Hence, from the previous paragraph, the completion of
$(E^{\beta^{-1}(N_{1})})_{t_{2}(T_{1})}$ with respect to any prime extending
$\langle T_{1}\rangle$ equals $F((T_{1}))$. In particular, the specialization
$((E^{\beta^{-1}(N_{1})})_{t_{2}(T_{1})})_{0}/k$ of
$(E^{\beta^{-1}(N_{1})})_{t_{2}(T_{1})}/k(T_{1})$ at $0$ equals $F/k$. It then
remains to apply Lemma 2.3 to conclude the proof of Theorem 4.1.

###  4.3. Proof of Theorem 1.3

We actually have the following more precise consequence:

######  Corollary 4.5.

Let $k$ be an arbitrary field, $G$ a finite group, and $F/k$ a Galois
extension of group $G$. Assume $G$ has a non-trivial solvable normal subgroup.
Then there exist $t_{0}\in k$ and a Galois field extension $E/k(T)$ of group
$G$, with $E\not\subset\overline{k}(T)$, and such that $F/k=E_{t_{0}}/k$.

###### Proof.

By the assumption, $G$ has a non-trivial solvable normal subgroup $H$. The
smallest non-trivial term $H^{\prime}$ of the derived series of $H$ then is an
abelian characteristic subgroup of $H$, and so is a non-trivial abelian normal
subgroup of $G$. Let $\theta:G\rightarrow{\rm{Gal}}(F/k)$ be an isomorphism.
Consider the finite embedding problem

$\alpha={\rm{res}}^{F/k}_{F^{\theta(H^{\prime})}/k}\circ\theta:G\rightarrow{\rm{Gal}}(F^{\theta(H^{\prime})}/k)$

over $k$; it has kernel $H^{\prime}$, which is abelian, and
$\theta^{-1}:{\rm{Gal}}(F/k)\rightarrow G$ is a solution. Theorem 4.1 then
yields $t_{0}\in k$ and a geometric solution ${\rm{Gal}}(E/k(T))\rightarrow G$
to $\alpha$ with $E_{t_{0}}/k=F/k$ and
$E\cap\overline{k}=F^{\theta(H^{\prime})}$. From the latter, we have
$[E\cap\overline{k}:k]=|G/H^{\prime}|$ and, as $H^{\prime}$ is not trivial,
this implies $E\not\subset\overline{k}(T)$, thus concluding the proof. ∎

######  Remark 4.6.

(1) For $n\geq 1$, let $D_{2^{n}}$ be the dihedral group of cardinality
$2^{n}$. The Beckmann–Black problem for $D_{2^{n}}$ over $\mathbb{Q}$ is known
to have a positive answer only for $n\leq 5$ (see [Dèb01, théorème 2.2] for
references), and no counter-example is known. Our method gives that, for every
field $k$, every $n\geq 1$, and every Galois extension $F/k$ of group
$D_{2^{n}}$, there are $t_{0}\in k$ and a Galois extension $E/k(T)$ of group
$D_{2^{n}}$, with $[E\cap\overline{k}:k]=2$, and such that $F/k=E_{t_{0}}/k$.

(2) Say that a finite group $G$ is a non-constant Galois group over a field
$k$ if there exists a Galois field extension $E/k(T)$ of group $G$ with
$E\not\subset\overline{k}(T)$. Corollary 4.5 allows to reobtain that, if a
finite group $G$ is the Galois group of a Galois field extension of $k$ and if
$G$ has a non-trivial solvable normal subgroup, then $G$ is a non-constant
Galois group over $k$.

##  5\. Inverse Galois theory over division rings

The present section is organized as follows. In §5.1, we collect some material
from previous papers about division rings and finite embedding problems in
this context. We then prove Theorem 1.4 in §5.2. Finally, in §5.3, we present
our result about finite embedding problems with abelian kernels over division
rings alluded to at the end of §1.2.2.

###  5.1. Preliminaries

We refer to [DL20, Beh21, BDL20] for more on the following.

####  5.1.1. Division rings

Let $H\subseteq L$ be division rings. The group of all automorphisms of $L$
fixing $H$ pointwise is the automorphism group ${\rm{Aut}}(L/H)$ of $L/H$.
Following Artin, we say that $L/H$ is Galois if each element of $L$ which is
fixed under all elements of ${\rm{Aut}}(L/H)$ is in $H$. If $L/H$ is Galois,
${\rm{Aut}}(L/H)$ is the Galois group ${\rm{Gal}}(L/H)$ of $L/H$.

The following theorem, which is [DL20, corollaire 2], will be used implicitly
in the sequel:

######  Theorem 5.1.

Let $L/H$ be a Galois extension of division rings with finite Galois group.
Assume $H$ is of finite dimension over its center $h$ and let $\ell$ denote
the center of $L$. Then $\ell$ is a Galois field extension of $h$ and
${\rm{Gal}}(\ell/h)\cong{\rm{Gal}}(L/H)$.

A ring $R\not=\\{0\\}$ with $ab\not=0$ for $a\not=0$ and $b\not=0$ is a right
Ore domain if, for all $x\not=0$ and $y\not=0$, there are $r,s\in R$ with
$xr=ys\not=0$. For a right Ore domain $R$, there is a division ring
$H\supseteq R$ each element of which can be written as $ab^{-1}$ with $a\in
R$, $b\in R\setminus\\{0\\}$ (see, e.g., [GW04, Theorem 6.8]), and $H$ is
unique up to isomorphism (see [Coh95, Proposition 1.3.4]).

Let $H$ be a division ring and $\sigma$ an automorphism of $H$. We let
$H[T,\sigma]$ be the ring of all polynomials $a_{0}+a_{1}T+\cdots+a_{n}T^{n}$
with $n\geq 0$ and $a_{0},\dots,a_{n}\in H$, whose addition is defined
componentwise and multiplication fulfills $Ta=\sigma(a)T$ for every $a\in H$.
Note that $H[T,\sigma]$ is commutative if and only if $H$ is a field and
$\sigma={\rm{id}}_{H}$. In the sense of Ore (see [Ore33]), $H[T,\sigma]$ is
the polynomial ring $H[T,\sigma,\delta]$ in the variable $T$, where the
derivation $\delta$ is 0. The ring $H[T,\sigma]$ fulfills $ab\not=0$ for $a\in
H[T,\sigma]\setminus\\{0\\}$ and $b\in H[T,\sigma]\setminus\\{0\\}$, as the
degree is additive on products. Moreover, $H[T,\sigma]$ is a right Ore domain
(see, e.g., [GW04, Theorem 2.6 and Corollary 6.7]). By $H(T,\sigma)$, we then
mean the unique division ring containing $H[T,\sigma]$ and every element of
which can be written as $ab^{-1}$ with $a\in H[T,\sigma]$ and $b\in
H[T,\sigma]\setminus\\{0\\}$. If $\sigma={\rm{id}}_{H}$, we write $H[T]$ and
$H(T)$ instead of $H[T,{\rm{id}}_{H}]$ and $H(T,{\rm{id}}_{H})$, respectively.

####  5.1.2. Finite embedding problems

First, let $L/H$ and $F/M$ be Galois extensions of division rings with finite
Galois groups, and such that the two inclusions $L\subseteq F$ and $H\subseteq
M$ hold. We let ${\rm{res}}^{F/M}_{L/H}$ denote the restriction map
${\rm{Gal}}(F/M)\rightarrow{\rm{Gal}}(L/H)$ (i.e.,
${\rm{res}}^{F/M}_{L/H}(\sigma)(x)=\sigma(x)$ for $\sigma\in{\rm{Gal}}(F/M)$
and $x\in L$), if it is well-defined.

Unlike the field case, ${\rm{res}}^{F/M}_{L/H}$ is not always well-defined.
The next proposition, which relies on the special cases III) and IV) of
[BDL20, §3.1], gives two situations where it is well-defined:

######  Proposition 5.2.

Let $H$ be a division ring of finite dimension over its center $h$.

(1) Let $L/H$ and $F/H$ be two Galois extensions of division rings with finite
Galois groups and such that $L\subseteq F$. Then ${\rm{res}}^{F/H}_{L/H}$ is
well-defined.

(2) Let $\sigma$ be an automorphism of $H$ of finite order $m$, let $L/H$ be a
Galois extension of division rings with finite Galois group, and let $\tau$ be
an automorphism of $L$ of order $m$ extending $\sigma$. Denote the restriction
of $\tau$ to the center $\ell$ of $L$ by $\widetilde{\tau}$, and assume

$\langle\widetilde{\tau},{\rm{Gal}}(\ell/h)\rangle\cong\langle\widetilde{\tau}\rangle\times{\rm{Gal}}(\ell/h)\,\,.$

Then $L(T,\tau)/H(T,\sigma)$ is Galois with finite Galois group and
${\rm{res}}^{L(T,\tau)/H(T,\sigma)}_{L/H}$ is a well-defined isomorphism.

Now, a finite embedding problem over a division ring $H$ of finite dimension
over its center is an epimorphism $\alpha:G\rightarrow{\rm{Gal}}(L/H)$, with
$G$ a finite group and $L/H$ a Galois extension. Say that $\alpha$ splits if
there is an embedding $\alpha^{\prime}:{\rm{Gal}}(L/H)\rightarrow G$ with
$\alpha\circ\alpha^{\prime}={\rm{id}}_{{\rm{Gal}}(L/H)}$. A weak solution to
$\alpha$ is a monomorphism $\beta:{\rm{Gal}}(F/H)\rightarrow G$, where $F/H$
is a Galois extension with $L\subseteq F$, such that $\alpha\circ\beta$ is the
map ${\rm{res}}^{F/H}_{L/H}$ (which is well-defined by Proposition 5.2(1)). If
$\beta$ is an isomorphism, we say solution instead of weak solution.

######  Remark 5.3.

Let $L/H$ be a Galois extension of division rings with ${\rm{Gal}}(L/H)$
finite. Then $L$ is a field if and only if $H$ is a field (see [BDL20, lemme
2.1]). In particular, the above terminology of finite embedding problems
generalizes that from the commutative case recalled in §2.2, and there is no
possible confusion with it.

Next, let $H$ be a division ring of finite dimension over its center $h$ and
$\sigma$ an automorphism of $H$ of finite order $m$. Let $\widetilde{\sigma}$
be the restriction of $\sigma$ to $h$. Let
$\alpha:G\rightarrow{\rm{Gal}}(L/H)$ be a finite embedding problem over $H$
and $\tau$ an automorphism of $L$ of order $m$ extending $\sigma$. Let
$\widetilde{\tau}$ denote the restriction of $\tau$ to the center $\ell$ of
$L$ and assume
$\langle\widetilde{\tau},{\rm{Gal}}(\ell/h)\rangle\cong\langle\widetilde{\tau}\rangle\times{\rm{Gal}}(\ell/h)$.
Then, by Proposition 5.2(2), $L(T,\tau)/H(T,\sigma)$ is Galois with finite
Galois group and ${\rm{res}}^{L(T,\tau)/H(T,\sigma)}_{L/H}$ is a well-defined
isomorphism. Hence,

$\alpha_{\sigma,\tau}=({\rm{res}}^{L(T,\tau)/H(T,\sigma)}_{L/H})^{-1}\circ\alpha:G\rightarrow{\rm{Gal}}(L(T,\tau)/H(T,\sigma))$

is a finite embedding problem over $H(T,\sigma)$. A $(\sigma,\tau)$-geometric
solution to $\alpha$ is a solution ${\rm{Gal}}(E/H(T,\sigma))\rightarrow G$ to
$\alpha_{\sigma,\tau}$. If $\tau={\rm{id}}_{L}$ (and so
$\sigma={\rm{id}}_{H}$), we say geometric solution instead of
$({\rm{id}}_{H},{\rm{id}}_{L})$-geometric solution. As in Remark 5.3,
$H(T,\sigma)$ is a field if and only if $E$ is.

Moreover, we let $h^{\langle\widetilde{\sigma}\rangle}$ (resp.,
$\ell^{\langle\widetilde{\tau}\rangle}$) denote the field which consists of
all elements of $h$ (resp., of $\ell$) which are fixed under
$\widetilde{\sigma}$ (resp., under $\widetilde{\tau}$). By [BDL20, lemmes 2.4
& 3.5], $\ell^{\langle\widetilde{\tau}\rangle}$ is a Galois field extension of
$h^{\langle\widetilde{\sigma}\rangle}$ and the (usual) restriction map
${\rm{res}}^{\ell/h}_{\ell^{\langle\widetilde{\tau}\rangle}/h^{\langle\widetilde{\sigma}\rangle}}$
is an isomorphism. Furthermore, the special case II) of [BDL20, §3.1] shows
that the restriction map ${\rm{res}}^{L/H}_{\ell/h}$ is a well-defined
isomorphism. We may then consider this finite embedding problem over
$h^{\langle\widetilde{\sigma}\rangle}$:

$\overline{\alpha}_{\sigma,\tau}={\rm{res}}^{\ell/h}_{\ell^{\langle\widetilde{\tau}\rangle}/h^{\langle\widetilde{\sigma}\rangle}}\circ{\rm{res}}^{L/H}_{\ell/h}\circ\alpha:G\rightarrow{\rm{Gal}}(\ell^{\langle\widetilde{\tau}\rangle}/h^{\langle\widetilde{\sigma}\rangle}).$
(5.1)

The next lemma, which is [BDL20, lemme 4.2], shows why
$\overline{\alpha}_{\sigma,\tau}$ is of interest:

######  Lemma 5.4.

Let $H$ be a division ring of finite dimension over its center $h$ and
$\sigma$ an automorphism of $H$ of finite order $m$. Denote the restriction of
$\sigma$ to $h$ by $\widetilde{\sigma}$. Let
$\alpha:G\rightarrow{\rm{Gal}}(L/H)$ be a finite embedding problem over $H$
and $\tau$ an automorphism of $L$ of order $m$ extending $\sigma$. Denote the
restriction of $\tau$ to the center $\ell$ of $L$ by $\widetilde{\tau}$.
Assume
$\langle\widetilde{\tau},{\rm{Gal}}(\ell/h)\rangle\cong\langle\widetilde{\tau}\rangle\times{\rm{Gal}}(\ell/h).$
Then $\alpha$ has a $(\sigma,\tau)$-geometric solution, if
$\overline{\alpha}_{\sigma,\tau}$ has a geometric solution
${\rm{Gal}}(E/h^{\langle\widetilde{\sigma}\rangle}(T^{m}))\rightarrow G$ such
that $E\subseteq\ell^{\langle\widetilde{\tau}\rangle}((T^{m}))$.

In the trivial case $L=H$ (and so $\tau=\sigma$), Lemma 5.4 reduces to this
statement: if $H$ is of finite dimension over its center $h$, then $G$ is a
Galois group over $H(T,\sigma)$ if there is a Galois field extension
$E/h^{\langle\widetilde{\sigma}\rangle}(T^{m})$ of group $G$ with $E\subseteq
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$. The latter statement actually
holds without the assumption that $H$ is of finite dimension over $h$, as
shown in [Beh21, §2.1]:

######  Lemma 5.5.

Let $H$ be a division ring of center $h$, let $\sigma$ be an automorphism of
$H$ of finite order $m$, and let $\widetilde{\sigma}$ be the restriction of
$\sigma$ to $h$. Given a finite group $G$, there is a Galois extension of
$H(T,\sigma)$ of group $G$, if there is a Galois field extension
$E/h^{\langle\widetilde{\sigma}\rangle}(T^{m})$ of group $G$ with $E\subseteq
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$.

###  5.2. Proof of Theorem 1.4

Denote the order of $\sigma$ by $m$ and the restriction of $\sigma$ to the
center $h$ of $H$ by $\widetilde{\sigma}$. Let $\mathcal{C}$ be the class of
finite groups $G$ which occur as the Galois group of a Galois field extension
$E/h^{\langle\widetilde{\sigma}\rangle}(T^{m})$ with $E\subseteq
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$. Then $\mathcal{C}$ is non-
empty and it is clear that, if $G\in\mathcal{C}$ and $H\trianglelefteq G$,
then $G/H\in\mathcal{C}$. Moreover, let $G\in\mathcal{C}$ and consider any
semidirect product $A\rtimes G$ with $A$ finite abelian. As $G\in\mathcal{C}$,
there is a Galois field extension
$F/h^{\langle\widetilde{\sigma}\rangle}(T^{m})$ of group $G$ with $F\subseteq
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$. Theorem 1.2 then yields a
Galois field extension $E/h^{\langle\widetilde{\sigma}\rangle}(T^{m})$ of
group $A\rtimes G$ with $E\subseteq F\cdot
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$. Since $F\subseteq
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$, we get $E\subseteq
h^{\langle\widetilde{\sigma}\rangle}((T^{m}))$, which implies that $A\rtimes
G$ is in $\mathcal{C}$. Consequently, every finite semiabelian group is in
$\mathcal{C}$. As every element of $\mathcal{C}$ is a Galois group over
$H(T,\sigma)$ by Lemma 5.5, we are done.

###  5.3. Finite embedding problems with abelian kernels over division rings

Let $H$ be a division ring of finite dimension over its center $h$. As shown
in [BDL20, corollaire 5.4], any given finite embedding problem $\alpha$ over
$H$ with a weak solution has a geometric solution, if $h$ is ample. Taking
$\sigma={\rm{id}}_{H}$ (and so
$(\tau,\tau^{\prime})=({\rm{id}}_{L},{\rm{id}}_{L^{\prime}})$) in the next
theorem, which relies on Theorem 1.2 and the material from §5.1, yields that
the same holds if we replace the assumption “$h$ is ample” by the one
“${\rm{ker}}(\alpha)$ is abelian”, as alluded to at the end of §1.2.2.

######  Theorem 5.6.

Let $H$ be a division ring of finite dimension over its center $h$ and
$\sigma$ an automorphism of $H$ of finite order $m$. Let
$\alpha:G\rightarrow{\rm{Gal}}(L/H)$ be a finite embedding problem over $H$,
and let $\tau$ be an automorphism of $L$ of order $m$ extending $\sigma$ and
such that
$\langle{\widetilde{\tau}},{\rm{Gal}}(\ell/h)\cong\langle\widetilde{\tau}\rangle\times{\rm{Gal}}(\ell/h),$
where $\widetilde{\tau}$ denotes the restriction of $\tau$ to the center
$\ell$ of $L$. Assume the following two conditions hold:

(1) ${\rm{ker}}(\alpha)$ is abelian,

(2) there exist a weak solution ${\rm{Gal}}(L^{\prime}/H)\rightarrow G$ to
$\alpha$, and an automorphism $\tau^{\prime}$ of $L^{\prime}$ of order $m$
extending $\tau$ and such that, if $\widetilde{\tau^{\prime}}$ is the
restriction of $\tau^{\prime}$ to the center $\ell^{\prime}$ of $L^{\prime}$,
then

$\langle\widetilde{\tau^{\prime}},{\rm{Gal}}(\ell^{\prime}/h)\rangle\cong\langle\widetilde{\tau^{\prime}}\rangle\times{\rm{Gal}}(\ell^{\prime}/h).$
(5.2)

Then $\alpha$ has a $(\sigma,\tau)$-geometric solution.

###### Proof.

As (2) holds, we may apply the weak$\rightarrow$split reduction for finite
embedding problems over division rings (see [BDL20, proposition 5.3]) to get
the existence of a finite split embedding problem
$\alpha^{\prime}:G^{\prime}\rightarrow{\rm{Gal}}(L^{\prime}/H)$ over $H$ such
that ${\rm{ker}}(\alpha)\cong{\rm{ker}}(\alpha^{\prime})$ and such that every
$(\sigma,\tau^{\prime})$-geometric solution to $\alpha^{\prime}$ yields a
$(\sigma,\tau)$-geometric solution to $\alpha$. By the former conclusion and
(1), ${\rm{ker}}(\alpha^{\prime})$ is abelian. Now, as (5.2) holds, we may
apply Lemma 5.4: $\alpha^{\prime}$ has a $(\sigma,\tau^{\prime})$-geometric
solution, if the finite embedding problem
${\overline{\alpha^{\prime}}}_{\sigma,\tau^{\prime}}:G^{\prime}\rightarrow{\rm{Gal}}(\ell^{\prime\langle{\widetilde{\tau^{\prime}}}\rangle}/h^{\langle\widetilde{\sigma}\rangle})$
over $h^{\langle\widetilde{\sigma}\rangle}$ (see (5.1)) has a geometric
solution ${\rm{Gal}}(E/h^{\langle\widetilde{\sigma}\rangle}(T^{m}))\rightarrow
G^{\prime}$ with
$E\subseteq\ell^{\prime\langle\widetilde{\tau^{\prime}}\rangle}((T^{m}))$.
Note that ${\overline{\alpha^{\prime}}}_{\sigma,\tau^{\prime}}$ splits (as
this holds for $\alpha^{\prime}$) and
${\rm{ker}}({\overline{\alpha^{\prime}}}_{\sigma,\tau^{\prime}})$ is abelian
(as ${\rm{ker}}(\alpha^{\prime})$ is). Hence, such a geometric solution to
${\overline{\alpha^{\prime}}}_{\sigma,\tau^{\prime}}$ exists by Theorem 1.2.
This concludes the proof. ∎

##  6\. Concluding remarks

Given a finite split embedding problem $\alpha:G\rightarrow{\rm{Gal}}(L/k)$
over a field $k$, producing a geometric solution
$\beta:{\rm{Gal}}(E/k(T))\rightarrow G$ with $E\subseteq L((T))$, as Theorem
1.2 does if ${\rm{ker}}(\alpha)$ is abelian, has several interests, as we have
already seen in §4 and §5. Here are two others.

(1) Recall that the level of a field $k$ is equal to either $\infty$ if -1
cannot be written as a sum of finitely many squares in $k$ or the smallest
positive integer $n$ such that there exist $x_{1},\dots,x_{n}\in k$ with
$-1=x_{1}^{2}+\cdots+x_{n}^{2}$ otherwise. By a well-known result of Pfister
(see, e.g., [Lam05, Chapter XI, Theorem 2.2]), the level of a field is either
$\infty$ or a power of 2.

($*$) Let $k$ be a number field of level at least 4 and $G$ a finite group
which occurs as the Galois group of a Galois field extension $E/k(T)$ with
$E\subseteq k((T))$. Then $G$ is the Galois group of a Galois extension of the
division ring $H_{k}$ of quaternions with coefficients in $k$ (i.e.,
$H_{k}=k\oplus k\mathbf{i}\oplus k\mathbf{j}\oplus k\mathbf{k}$ with
$\mathbf{i}^{2}=\mathbf{j}^{2}=\mathbf{k}^{2}=\mathbf{i}\mathbf{j}\mathbf{k}=-1$).

Indeed, as $k$ has level $\geq$ 4, the Hasse–Minkowski theorem (see, e.g.,
[Lam05, p. 170]) yields an equivalence class $v$ of absolute values on $k$
such that the completion $k_{v}$ of $k$ at $v$ has level $\geq 4$. As
$E_{0}/k=k/k$, we have $E_{0}\subseteq k_{v}$. Then pick $t_{0}\in k$ such
that the specialization $E_{t_{0}}/k$ of $E/k(T)$ at $t_{0}$ is of group $G$
and such that $E_{t_{0}}\subseteq k_{v}$. Hence, $E_{t_{0}}$ has level $\geq
4$ and it remains to use [DL20, théorème 7] to conclude the proof of ($*$).

(2) Let $h:\overline{\mathbb{Q}}\rightarrow\mathbb{R}_{\geq 0}$ be the
(absolute logarithmic) Weil height. Recall that a field
$L\subseteq\overline{\mathbb{Q}}$ has the Northcott property (Property (N)) if
$\\{x\in L\,\,:\,\,h(x)<T\\}$ is finite for every $T>0$. By Northcott’s
theorem (see [Nor49, Theorem 1]), every number field has Property (N). But
examples of infinite algebraic extensions of $\mathbb{Q}$ with Property (N)
are more sparse.

In [CF21], Checcoli and Fehm give many such examples by proving that, given a
sequence $(G_{n})_{n\geq 1}$ of finite solvable groups, there is an infinite
Galois field extension $L$ of $\mathbb{Q}$ such that
${\rm{Gal}}(L/\mathbb{Q})=\prod_{n=1}^{\infty}G_{n}$, the completion of
$L/\mathbb{Q}$ at $p$ is a finite extension of $\mathbb{Q}_{p}$ for every
prime number $p$, and $L$ has Property (N). From the proof of [CF21, Theorem
1.3] (see also [CF21, Remark 2.5]), the assumption that $G_{n}$ is solvable is
used only to guarantee the existence of “many” Galois extensions of
$\mathbb{Q}$ of group $G_{n}$ which are totally split at any finitely many
given prime numbers. Hence, the result of Checcoli and Fehm remains true if
each $G_{n}$ ($n\geq 1$) is either solvable or the Galois group of a Galois
extension $E/\mathbb{Q}(T)$ with $E\subseteq\mathbb{Q}((T))$.

## References

  * [ALP20] Gil Alon, François Legrand, and Elad Paran.  Galois groups over rational function fields over skew fields.  C. R. Math. Acad. Sci. Paris, 358(7):785–790, 2020.
  * [BDL20] Angelot Behajaina, Bruno Deschamps, and François Legrand.  Problèmes de plongement finis sur les corps non commutatifs. (French).  2020\.  To appear in Israel J. Math., arXiv:2008.08333v2.
  * [Bec94] Sybilla Beckmann.  Is every extension of $\mathbb{Q}$ the specialization of a branched covering?  J. Algebra, 164(2):430–451, 1994.
  * [Beh21] Angelot Behajaina.  Théorie inverse de Galois sur les corps de fractions rationnelles tordus. (French).  J. Pure Appl. Algebra, 225(4), 2021.  Article 106549, 10 pp.
  * [BS09] Lior Bary-Soroker.  On pseudo algebraically closed extensions of fields.  J. Algebra, 322(6):2082–2105, 2009.
  * [BSF13] Lior Bary-Soroker and Arno Fehm.  Open problems in the theory of ample fields.  In Geometric and differential Galois theories, volume 27 of Sémin. Congr., pages 1–11. Soc. Math. France, Paris, 2013.
  * [CF21] Sara Checcoli and Arno Fehm.  On the Northcott property and local degrees.  Proc. Amer. Math. Soc., 149(6):2403–2414, 2021.
  * [Coh95] Paul Moritz Cohn.  Skew fields. Theory of general division rings.  Encyclopedia of Mathematics and its Applications, 57. Cambridge University Press, Cambridge, 1995.  xvi+500 pp.
  * [CT00] Jean-Louis Colliot-Thélène.  Rational connectedness and Galois covers of the projective line.  Ann. of Math. (2), 151(1):359–373, 2000.
  * [DD97] Pierre Dèbes and Bruno Deschamps.  The regular inverse Galois problem over large fields.  In Geometric Galois actions, 2, volume 243 of London Math. Soc. Lecture Note Ser., pages 119–138. Cambridge Univ. Press, Cambridge, 1997.
  * [Dèb99] Pierre Dèbes.  Galois covers with prescribed fibers: the Beckmann–Black problem.  Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 28(2):273–286, 1999.
  * [Dèb01] Pierre Dèbes.  Théorie de Galois et géométrie : une introduction. (French).  In Arithmétique de revêtements algébriques (Saint-Étienne, 2000), volume 5 of Sémin. Congr., pages 1–26. Soc. Math. France, Paris, 2001.
  * [Dèb09] Pierre Dèbes.  Arithmétique des revêtements de la droite.  Lecture notes, 2009.  At https://math.univ-lille1.fr/~pde/rev_www.pdf.
  * [Des15] Bruno Deschamps.  Minimalité et abyssalité des extensions abéliennes et projectives de $\mathbb{Q}$. (French).  J. Algebra, 441:1–20, 2015.
  * [Des21] Bruno Deschamps.  La méthode Behajaina appliquée aux corps de fractions tordus par une dérivation. (French).  Res. Number Theory, 7(2), 2021.  Paper No. 39, 11 pp.
  * [DL20] Bruno Deschamps and François Legrand.  Le problème inverse de Galois sur les corps des fractions tordus à indéterminée centrale. (French).  J. Pure Appl. Algebra, 224(5), 2020.  Article 106240, 13 pp.
  * [FJ08] Michael D. Fried and Moshe Jarden.  Field arithmetic.  Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics], 11. Springer-Verlag, Berlin, third edition, 2008.  Revised by Jarden. xxiv+792 pp.
  * [FL20] Arno Fehm and François Legrand.  A note on finite embedding problems with nilpotent kernel.  2020\.  To appear in J. Théor. Nombres Bordeaux, arXiv:2011.07536v3.
  * [GW04] Kenneth R. Goodearl and Robert Breckenridge Warfield, Jr.  An Introduction to noncommutative Noetherian rings.  London Mathematical Society Student Texts, 61. Cambridge University Press, Cambridge, 2004.  Second edition. xxiv+344 pp.
  * [HJ98] Dan Haran and Moshe Jarden.  Regular split embeddings problems over function fields of one variable over ample fields.  J. Algebra, 208(1):147–164, 1998.
  * [HS05] David Harbater and Katherine F. Stevenson.  Local Galois theory in dimension two.  Adv. Math., 198(2):623–653, 2005.
  * [Ike60] Masatoshi Ikeda.  Zur Existenz eigentlicher galoisscher Körper beim Einbettungsproblem für galoissche Algebren. (German).  Abh. Math. Sem. Univ. Hamburg, 24:126–131, 1960.
  * [Jac64] Nathan Jacobson.  Structure of rings, volume 37 of American Mathematical Society Colloquium Publications.  American Mathematical Society, Providence, R.I., 1964.  Revised edition. ix+299 pp.
  * [Jar91] Moshe Jarden.  Intersections of local algebraic extensions of a Hilbertian field.  In Generators and relations in groups and geometries, volume 333 of NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., pages 343–405. Kluwer Acad. Publ., Dordrecht, 1991.
  * [Jar11] Moshe Jarden.  Algebraic patching.  Springer Monographs in Mathematics. Springer, Heidelberg, 2011.  xxiv+290 pp.
  * [JR18] Moshe Jarden and Cynthia Nantsoina Ramiharimanana.  Solving embedding problems with bounded ramification.  Proc. Lond. Math. Soc. (3), 117(1):149–191, 2018.
  * [JR19] Moshe Jarden and Cynthia Nantsoina Ramiharimanana.  Embedding problems with bounded ramification over global fields of positive characteristic.  J. Lond. Math. Soc. (2), 100(1):323–340, 2019.
  * [JR20] Moshe Jarden and Cynthia Nantsoina Ramiharimanana.  Embedding problems with bounded ramification over function fields of positive characteristic.  New York J. Math., 26:1422–1443, 2020.
  * [KM01] Jürgen Klüners and Gunter Malle.  A database for field extensions of the rationals.  LMS J. Comput. Math., 4:182–196, 2001.
  * [KM04] Jürgen Klüners and Gunter Malle.  Counting nilpotent Galois extensions.  J. Reine Angew. Math., 572:1–26, 2004.
  * [Lam05] Tsit Yuen Lam.  Introduction to quadratic forms over fields.  Graduate Studies in Mathematics, 67. American Mathematical Society, Providence, RI, 2005.  xxii+550 pp.
  * [MB01] Laurent Moret-Bailly.  Construction de revêtements de courbes pointées. (French).  J. Algebra, 240(2):505–534, 2001.
  * [Mes90] Jean-François Mestre.  Extensions régulières de $\mathbb{Q}({T})$ de groupe de Galois $\tilde{A}_{n}$. (French).  J. Algebra, 131(2):483–495, 1990.
  * [MM18] Gunter Malle and B. Heinrich Matzat.  Inverse Galois theory.  Springer Monographs in Mathematics. Springer, Berlin, 2018.  Second edition. xvii+532 pp.
  * [Nor49] Douglas Geoffrey Northcott.  An inequality in the theory of arithmetic on algebraic varieties.  Proc. Cambridge Phil. Soc., 45:502–509, 1949.
  * [NSW08] Jürgen Neukirch, Alexander Schmidt, and Kay Wingberg.  Cohomology of number fields, volume 323 of Grundlehren der mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].  Springer-Verlag, Berlin, second edition, 2008.  xvi+825 pp. Electronic version 2.3 (as of May 2020) available at https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/NSW2.3.pdf.
  * [Ore33] Oystein Ore.  Theory of non-commutative polynomials.  Ann. of Math. (2), 34(3):480–508, 1933.
  * [Par09] Elad Paran.  Split embedding problems over complete domains.  Ann. of Math. (2), 170(2):899–914, 2009.
  * [Pop96] Florian Pop.  Embedding problems over large fields.  Ann. of Math. (2), 144(1):1–34, 1996.
  * [Pop14] Florian Pop.  Little survey on large fields - old $\&$ new.  In Valuation theory in interaction, EMS Ser. Congr. Rep., pages 432–463. Eur. Math. Soc., Zürich, 2014.
  * [Ser92] Jean-Pierre Serre.  Topics in Galois Theory, volume 1 of Research Notes in Mathematics.  Jones and Bartlett Publishers, Boston, MA, 1992.  Lecture notes prepared by Henri Darmon [Henri Darmon]. With a foreword by Darmon and the author. xvi+117 pp.
  * [Völ96] Helmut Völklein.  Groups as Galois groups. An introduction, volume 53 of Cambridge Studies in Advanced Mathematics.  Cambridge University Press, Cambridge, 1996.  xviii+248 pp.


a_{i}^{c}(\mathbf{g}_{i})\sum_{j}g_{ij}a_{j}^{c}(\mathbf{g}_{j})\right]$
$\displaystyle W^{e}(\mathbf{g})$ $\displaystyle=\sum_{i\in
N}\left[a^{e}(\mathbf{g}_{i})-\frac{1}{2}a_{i}^{e}(\mathbf{g}_{i})^{2}+\lambda
a_{i}^{e}(\mathbf{g}_{i})\sum_{j}g_{ij}a_{j}^{e}(\mathbf{g}_{j})\right]$

where $a_{i}^{*}(\mathbf{g}_{i})$ is the BNE action of agent $i$ who is of
type $\mathbf{g}_{i}$, $a_{i}^{c}(\mathbf{g}_{i})$ is complete information
Nash equilibrium action of agent $i$ whose row is also $\mathbf{g}_{i}$, and
$a_{i}^{e}(\mathbf{g}_{i})$ the complete information efficient action level of
agent $i$ whose row is also $\mathbf{g}_{i}$.

For any $n>7$, $n_{co}\leq\frac{n}{2}$ and $\lambda\leq\frac{1}{2(n-1)}$,
compared to complete infomration, welfare under incomplete information is
closer to the efficient one. While this claim in non-trivial to show
analytically, it can be numerically verified that it holds over a large set of
parameter values via the following algorithm.

Figure 6: n=50, $\lambda=0.005$

Figure 7: n=50, $\lambda=0.0101$

Figures 6 and 7 plot complete, incomplete and efficient welfare levels against
different core sizes when $n=50$. We see that all three welfare levels are
monotonically increasing with the core size. For small, as well as large
$\lambda$ efficient welfare dominates both the incomplete and complete
information welfare. Moreover, when the core size is less than half the
population size, incomplete information welfare is larger than complete
information. Under smaller values of the complementarity parameter $\lambda$,
network externalities are smaller which in turn results in smaller differences
between the three welfare levels. On the other hand, when $\lambda$ and hence
network externalities are high, we see a significant difference between the
three welfare levels.