# Free $\mathbb{Q}$-groups are residually torsion-free nilpotent

Andrei Jaikin-Zapirain  Departamento de Matemáticas, Universidad Autónoma de
Madrid and Instituto de Ciencias Matemáticas, CSIC-UAM-UC3M-UCM
<EMAIL_ADDRESS>
(Date: November 2022)

###### Abstract.

We develop a method to show that some (abstract) groups can be embedded into a
free pro-$p$ group. In particular, we show that every finitely generated
subgroup of a free $\mathbb{Q}$-group can be embedded into a free pro-$p$
group for almost all primes $p$. This solves an old problem raised by G.
Baumslag: free $\mathbb{Q}$-groups are residually torsion-free nilpotent.

###### Key words and phrases:

Free $\mathbb{Q}$-groups, free pro-$p$ groups, mod-$p$ $L^{2}$-Betti numbers,
Lück approximation, universal division ring of fractions

###### 2010 Mathematics Subject Classification:

Primary: 20E06, Secondary: 16K40, 20C07, 20E18, 20E26

##  1\. Introduction

A group $G$ is called a $\mathbb{Q}$-group  if for any $n\in\mathbb{N}$ and
$g\in G$ there exists exactly one $h\in G$ satisfying $h^{n}=g$. These groups
were introduced by G. Baumslag in [4] under the name of
$\operatorname{\mathcal{D}}$-groups. He observed that $\mathbb{Q}$-groups may
be viewed as universal algebras, and as such they constitute a variety. Every
variety of algebras contains free algebras (in that variety). In the variety
of $\mathbb{Q}$-groups we call such free algebras free $\mathbb{Q}$-groups. G.
Baumslag dedicated several papers to the study of residual properties of free
$\mathbb{Q}$-groups [5, 7, 9]. For example, in [5] he showed that a free
$\mathbb{Q}$-group is residually periodic-by-soluble and locally residually
finite-by-soluble. He wrote in [5] “It is, of course, still possible that,
locally, free $\operatorname{\mathcal{D}}$-groups are, say, residually finite
$p$-groups” or in [7] “In particular it seems likely that free
$\operatorname{\mathcal{D}}$-groups are residually torsion-free nilpotent.
However the complicated nature of free $\operatorname{\mathcal{D}}$-groups
makes it difficult to substantiate such a remark.” This conjecture is part of
two main collections of problems in group theory ([10, Problem F12] and [38,
Problem 13.39 (a),(c)]), and in addition to mentioned works of Baumslag, it
was also studied in [15, 22]. In this paper we solve Baumslag’s conjecture.

######  Theorem 1.1.

A free $\mathbb{Q}$-group is residually torsion-free nilpotent.

The structure of a finitely generated subgroup of a free $\mathbb{Q}$-group
was studied already in [4] (see also [53, Section 8] and Proposition 5.3). It
was shown that it is the end result of repeatedly freely adjoining $n$th roots
to a finitely generated free group. The key point of our proof of Theorem 1.1
is to show that any finitely generated subgroup of a free $\mathbb{Q}$-group
can be embedded into a finitely generated free pro-$p$ group for some prime
$p$. We actually prove the following more precise result.

######  Theorem 1.2.

Let $p$ be a prime. Let $H_{0}$ be a finitely generated free group and let
$H_{0}\hookrightarrow\mathbf{F}$ be the canonical embedding of $H_{0}$ into
its pro-$p$ completion $\mathbf{F}$. Let $(H_{i})_{i\geq 0}$ be a sequence of
subgroups of $\mathbf{F}$ such that for $i\geq 0$,

  1. (1)

$H_{i+1}=\langle H_{i},B_{i}\rangle$, where $B_{i}$ is a finitely generated
abelian subgroup of $\mathbf{F}$ and

  2. (2)

$A_{i}=H_{i}\cap B_{i}$ is a maximal abelian subgroup of $H_{i}$.

Then for every $i\geq 0$, the canonical map

$H_{i}*_{A_{i}}B_{i}\to H_{i+1}$

is an isomorphism,

Theorem 1.2 is actually an application of the slightly more technical Theorem
5.1.

Let us make a few remarks about the groups $A_{i}$ and $B_{i}$. It is
relatively easy to describe abelian subgroups of amalgamated products. In
particular, the conclusion of the theorem implies that all abelian subgroups
of $H_{i}$ are finitely generated. Thus, an implicit hypothesis, which appears
in the theorem, that maximal abelian subgroups $A_{i}$ of $H_{i}$ are finitely
generated, is automatically fulfilled.

A maximal abelian subgroup of $\mathbf{F}$ is isomorphic to the additive group
of the ring of $p$-adic numbers $(\mathbb{Z}_{p},+)$. Therefore, for any
finitely generated (abstract) abelian subgroup $A$ of $\mathbf{F}$ and any
finitely generated torsion-free abelian group $B$ which contains $A$ and such
that $B/A$ has no $p$-torsion, it is posible to extend the embedding
$A\hookrightarrow\mathbf{F}$ to an embedding $B\hookrightarrow\mathbf{F}$.
This extension is unique if and only if $B/A$ is finite.

Given a commutative ring $A$, we will introduce in Section 5 the notions of
$A$-group and free $A$-group $F^{A}(X)$. For example, a free pro-$p$ group is
an example of a $\mathbb{Z}_{p}$-group. We have the following consequence of
Theorem 1.2.

######  Corollary 1.3.

Let $F(X)$ be the free group on a finite free generating set $X$, let
$\mathbf{F}$ be its pro-$p$ completion. Then the canonical homomorphism

$\phi:F^{\mathbb{Z}_{p}}(X)\to\mathbf{F}$

is injective.

Let $H$ be a group and $A$ the centralizer of a non-trivial element. Then the
group $G=H*_{A}(A\times\mathbb{Z}^{k})$ is said to be obtained from $H$ by
extension of a centralizer. A group is called an ICE group if it can be
obtained from a free group using iterated centralizer extensions. A group $G$
is a limit group if and only if it is a finitely generated subgroup of an ICE
group (see [37, 14]). All centralizers of non-trivial elements of an ICE group
are abelian. Thus, Theorem 1.1 provides explicit realizations of ICE groups
(and so limit groups) as subgroups of a non-abelian free pro-$p$-groups (for
this application we only need the case where all $B_{i}/A_{i}$ are torsion-
free). Non-explicit realizations of limit groups as subgroups of a non-abelian
free pro-$p$ group (in fact, as subgroups of every compact group containing a
non-abelian free group) was obtained in [2] (see also [12]).

In Section 5 we recall the definition of the $\mathbb{Q}$-completion of a
group $G$. For example, a free $\mathbb{Q}$-group is the
$\mathbb{Q}$-completion of a free group. Theorem 1.2 allows also to show that
the $\mathbb{Q}$-completion of a limit group is residually torsion-free
nilpotent.

######  Theorem 1.4.

The $\mathbb{Q}$-completion of a limit group is residually torsion-free
nilpotent.

A group $G$ is called parafree if it is residually nilpotent and for some free
group $F$, we have that for all $i$, $G/\gamma_{i}(G)\cong F/\gamma_{i}(F)$
where $\gamma_{i}(G)$ denotes the terms of the lower central series of $G$.
Baumslag introduced this family of groups and produced many examples of them
[6]. In [35] we apply the method of the proof of Theorem 1.2 in order to
construct new examples of finitely generated parafree groups.

Our proof of Theorem 1.2 is by induction on $i$. In the inductive step
argument we start with the following situation. We have a finitely generated
subgroup $H$ of $\mathbf{F}$, a maximal abelian subgroup $A$ of $H$ and an
abelian subgroup $B$ of $\mathbf{F}$ containing $A$. We want to show that the
canonical homomorphism $H*_{A}B\to\langle H,B\rangle$ is an isomorphism.
Unfortunately, we do not know how to show this statement in such a generality,
but we prove it in Theorem 5.1 under an additional assumption that the
embedding $H\hookrightarrow\mathbf{F}$ is strong (see Definition 3.8). Theorem
5.1 is the main result of the paper. Its proof uses in an essential way the
results of [32], where we proved a particular case of the Lück approximation
in positive characteristic.

The paper is organized as follows. In Section 2 we give basic preliminaries.
The proof of Theorem 5.1 uses the theory of mod-$p$ $L^{2}$-Betti numbers. In
Section 3 we explain how to define them for subgroups $G$ of a free pro-$p$
group. In Section 4 we introduce a technical notion of $\mathcal{D}$-torsion-
free modules and show that some relevant $\mathbb{F}_{p}[G]$-modules are
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free (see Proposition
4.10). In Section 5 we prove Theorem 5.1 and obtain all the results mentioned
in the introduction. In Section 6 we discuss the following two well-known
problems concerning linearity of free pro-$p$ groups and free
$\mathbb{Q}$-groups:

######  Question 1.5.

  1. (1)

(I. Kapovich) Is a free $\mathbb{Q}$-group linear?

  2. (2)

(A. Lubotzky) Is a free pro-$p$ group linear?

## Acknowledgments

I am very grateful to Pavel Zalesski who explained to me how a surface group
can be embedded into a free pro-$p$ group. This was a starting point of
Theorem 1.2. I would like to thank Warren Dicks for providing the references
needed in the proof of Proposition 2.13, Yago Antolin, Ashot Minasyan and a
referee for explaining to me the proof of Theorem 6.1, Emmanuel Breulliard for
pointing out the paper [2] and Ismael Morales and anonymous referees for
useful suggestions and comments.

This paper is partially supported by the grants MTM2017-82690-P and
PID2020-114032GB-I00 of the Ministry of Science and Innovation of Spain and by
the ICMAT Severo Ochoa project CEX2019-000904-S4.

##  2\. Preliminaries

###  2.1. $R$-rings

All rings in this paper are associative and have the identity element. All
ring homomorphisms send the identity to the identity. We denote the invertible
elements of a ring $R$ by $R^{*}$. An $R$-module means a left $R$-module. By
an $R$-ring we understand a ring homomorphism $\varphi:R\to S$. We will often
refer to $S$ as $R$-ring and omit the homomorphism $\varphi$ if $\varphi$ is
clear from the context. Two $R$-rings $\varphi_{1}:R\to S_{1}$ and
$\varphi_{2}:R\to S_{2}$ are said to be isomorphic if there exists a ring
isomorphism $\alpha:S_{1}\to S_{2}$ such that
$\alpha\circ\varphi_{1}=\varphi_{2}$.

If $Y=\\{y_{i}\colon i\in I\\}$, we denote by $A\langle\\!\langle
Y\rangle\\!\rangle$ the ring of of formal power series with coefficients in
$A$ on the non-commuting indeterminates $Y$.

###  2.2. Left ideals in group algebras

Let $G$ be a group and $k$ a commutative ring. We denote by $I_{G}$ the
augmentation ideal of $k[G]$. If $H$ is a subgroup of $G$ we denote by
$I_{H}^{G}$ the left ideal of $k[G]$ generated by $I_{H}$. The following lemma
gives an alternative description of the $k[G]$-module $I_{H}^{G}$.

######  Lemma 2.1.

Let $H\leq T$ be subgroups of $G$. Then the following holds.

  1. (a)

The canonical map

$k[G]\otimes_{k[H]}I_{H}\to I_{H}^{G}$

sending $a\otimes b$ to $ab$, is an isomorphism of $k[G]$-modules.

  2. (b)

The canonical map $k[G]\otimes_{k[T]}(I_{T}/I_{H}^{T})\to
I_{T}^{G}/I_{H}^{G}$, sending $a\otimes(b+I_{H}^{T})$ to $ab+I_{H}^{G}$, is an
isomorphism of $k[G]$-modules.

###### Proof.

(a) Consider an exact sequence

$0\to I_{H}\to k[H]\to k\to 0.$

The freeness of $k[G]$ as $k[H]$-module implies that the sequence

$0\to
k[G]\otimes_{k[H]}I_{H}\xrightarrow{\alpha}k[G]\xrightarrow{\beta}k[G]\otimes_{k[H]}k\to
0$

is also exact. Here $\alpha$ sends $a\otimes b$ to $ab$ and $\beta$ sends $a$
to $a\otimes 1$. Thus, $\alpha$ establishes an isomorphisms of $k[G]$-modules
between $k[G]\otimes_{k[H]}I_{H}$ and $\ker\beta=I_{H}^{G}$. This proves the
first claim of the lemma.

(b) Consider now the exact sequence

$0\to I_{H}^{T}\to I_{T}\to I_{T}/I_{H}^{T}\to 0.$

Applying $k[G]\otimes_{k[T]}$ and taking again into account that $k[G]$ is a
free $k[T]$-module, we obtain the exact sequence

$0\to I_{H}^{G}\to I_{T}^{G}\to k[G]\otimes_{k[T]}(I_{T}/I_{H}^{T})\to 0.$

This proves the second claim. ∎

###  2.3. Profinite modules over pro-$p$ groups

In this paper the letters $\mathbf{F}$, $\mathbf{G}$, $\mathbf{H}$, etc. will
denote pro-$p$ groups. When we speak about pro-$p$ groups, the finite
generation, the finite presentation, the freeness etc. will always be
considered in the category of pro-$p$ groups. For example, $d(\mathbf{G})$
denotes the minimal number of topological generators of $\mathbf{G}$.

Almost all pro-$p$ groups that we consider are free pro-$p$ groups. Recall
that a closed subgroup of a free pro-$p$ group is also free pro-$p$ ([56,
Corollary 7.7.5]). As a consequence we obtain the following result which we
will use often in this paper.

######  Lemma 2.2.

Every maximal abelian subgroup of a non-trivial free pro-$p$ group is
isomorphic to $(\mathbb{Z}_{p},+)$.

Let $\mathbf{G}$ be a pro-$p$ group. We denote by
$\mathbb{F}_{p}[[\mathbf{G}]]$ the inverse limit of
$\mathbb{F}_{p}[\mathbf{G}/\mathbf{U}]$, where the limit is taken over all
open normal subgroups $\mathbf{U}$ of $\mathbf{G}$.
$\mathbb{F}_{p}[[\mathbf{G}]]$ is called the completed group algebra of
$\mathbf{G}$ over $\mathbb{F}_{p}$. In the case where $\mathbf{G}$ is a free
pro-$p$ group we have the following useful description of
$\mathbb{F}_{p}[[\mathbf{G}]]$.

######  Proposition 2.3.

[39, Section II, Proposition 3.1.4] Let $\mathbf{F}$ be a finitely generated
free pro-$p$ group freely generated by $x_{1},\ldots,x_{d}$. Then the
continuous $\mathbb{F}_{p}$-algebra homomorphism

$\mathbb{F}_{p}\langle\\!\langle
y_{1},\ldots,y_{d}\rangle\\!\rangle\to\mathbb{F}_{p}[[\mathbf{F}]]$

that sends $y_{i}$ to $x_{1}-1$ (for $1\leq i\leq d$) is an isomorphism.

A discrete $\mathbb{F}_{p}[[\mathbf{G}]]$-module is an
$\mathbb{F}_{p}[[\mathbf{G}]]$-module $M$ such that for any $m\in M$,

$\operatorname{Ann}_{\mathbb{F}_{p}[[\mathbf{G}]]}(m):=\\{a\in\mathbb{F}_{p}[[\mathbf{G}]]\colon
am=0\\}$

is open in $\mathbb{F}_{p}[[\mathbf{G}]]$.

A profinite $\mathbb{F}_{p}[[\mathbf{G}]]$-module is an inverse limit of
finite discrete $\mathbb{F}_{p}[[\mathbf{G}]]$-modules.

######  Lemma 2.4.

Let $\mathbf{G}$ be a pro-$p$ group. Let $\alpha:M\to N$ be a homomorphism of
profinite $\mathbb{F}_{p}[[\mathbf{G}]]$-modules. If $M$ is finitely generated
as an $\mathbb{F}_{p}[[\mathbf{G}]]$-module, then $\alpha$ is continuous.

######  Remark 2.5.

This lemma resembles a well-known result of Nikolov and Segal [55] that says
that every homomorphism from a finitely generated profinite group to a
profinite group is always continuous. This is equivalent to that every
subgroup of finite index in a finitely generated profinite group is open. In
the case of $\mathbb{F}_{p}[[\mathbf{G}]]$-modules, the situation is
different: it is not always true that every left ideal of
$\mathbb{F}_{p}[[\mathbf{G}]]$ of finite index is open.

###### Proof.

Without loss of generality we may assume that $N$ is a finite discrete
$\mathbb{F}_{p}[[\mathbf{G}]]$-module and we have to show that $\ker\alpha$ is
open.

We put
$I=\operatorname{Ann}_{\mathbb{F}_{p}[[G]]}=\\{a\in\mathbb{F}_{p}[[\mathbf{G}]]\colon
aN=\\{0\\}\\}$. Then $I$ is an open, and so also closed, ideal of
$\mathbb{F}_{p}[[\mathbf{G}]]$. Let $IM$ be the submodule of $M$ generated by
$\\{a\cdot m\colon a\in I,m\in M\\}$. Since $M$ is finitely generated we can
write

$M=\sum_{i=1}^{s}\mathbb{F}_{p}[[\mathbf{G}]]m_{i}.$

Therefore, $IM=\sum_{i=1}^{s}Im_{i}$.

From the definition of a profinite $\mathbb{F}_{p}[[\mathbf{G}]]$-module it
follows that the multiplicative map $\mathbb{F}_{p}[[\bf G]]\times M\to M$ is
continuous. Hence, since $I$ is closed in $\mathbb{F}_{p}[[\mathbf{G}]]$,
$IM=\sum_{i=1}^{s}Im_{i}$ is a closed submodule of $M$. But $IM$ is also of
finite index. Hence, it is open in $M$. Since $IM\leq\ker\alpha$, $\ker\alpha$
is also open. ∎

If $M=\varprojlim_{i\in I}M_{i}$ and $N=\varprojlim_{j\in J}N_{i}$ are right
and left, respectively, profinite $\mathbb{F}_{p}[[\mathbf{G}]]$-modules
($M_{i}$ and $N_{j}$ are finite discrete
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules), then the profinite tensor product is
denoted by $\widehat{\otimes}$ and it is defined as the inverse limit of
$M_{i}\otimes_{\mathbb{F}_{p}[[\mathbf{G}]]}N_{j}$.

######  Lemma 2.6.

Let $\mathbf{H}$ be a closed subgroup of $\mathbf{G}$ and let $M$ be a
finitely presented $\mathbb{F}_{p}[[\mathbf{H}]]$-module. Then $M$ is a
profinite module and the canonical map

$\gamma:\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M\to\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}M$

is an isomorphism of $\mathbb{F}_{p}[[\mathbf{G}]]$-modules.

######  Remark 2.7.

It is claimed in [56, Proposition 5.5.3 (d)] that it is enough to assume that
$M$ is a finitely generated profinite $\mathbb{F}_{p}[[\mathbf{H}]]$-module.
However, we want to warn the reader that the proof is incorrect.

###### Proof.

Since $M$ is finitely presented, there exists an exact sequence of
$\mathbb{F}_{p}[[\mathbf{H}]]$-modules

(1)
$\mathbb{F}_{p}[[\mathbf{H}]]^{r}\xrightarrow{\alpha}\mathbb{F}_{p}[[\mathbf{H}]]^{d}\to
M\to 0.$

Thus, we can write $M\cong\mathbb{F}_{p}[[\mathbf{H}]]^{d}/I$, where
$I=\operatorname{Im}\alpha$. By Lemma 2.4, $\alpha$ is continuous. Since
$\mathbb{F}_{p}[[\mathbf{H}]]^{r}$ is compact and Hausdorff, $I$ is closed in
$\mathbb{F}_{p}[[\mathbf{H}]]^{d}$. Hence
$\mathbb{F}_{p}[[\mathbf{H}]]^{d}/I$, and so $M$, are profinite.

It is clear that

$\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}\mathbb{F}_{p}[[\mathbf{H}]]\cong\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\mathbb{F}_{p}[[\mathbf{H}]]\cong\mathbb{F}_{p}[[\mathbf{G}]].$

Thus, after applying
$\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}$ and
$\mathbb{F}_{p}[[\mathbf{G}]]{\widehat{\otimes}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$
to the sequence (1) we obtain a commutative diagram of
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules:

$\begin{array}[]{cccccc}\mathbb{F}_{p}[[\mathbf{G}]]^{r}&\to&\mathbb{F}_{p}[[\mathbf{G}]]^{d}&\to&\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M&\to
0\\\ ||&&||&&\downarrow^{\gamma}&\\\
\mathbb{F}_{p}[[\mathbf{G}]]^{r}&\to&\mathbb{F}_{p}[[\mathbf{G}]]^{d}&\to&\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}M&\to
0.\end{array}.$

Since $\otimes$ and $\widehat{\otimes}$ are right exact (see [57, Theorem
2.6.3] and [56, Proposition 5.5.3(a)]) the horizontal sequences are exact.
This clearly implies that $\gamma$ is an isomorphism. ∎

Recall that if a pro-$p$ group $\mathbf{G}$ is finitely generated, then the
augmentation ideal $I_{\mathbf{G}}$ is finitely generated as an
$\mathbb{F}_{p}[[\mathbf{G}]]$-module. In particular, the trivial
$\mathbb{F}_{p}[[\mathbf{G}]]$-module $\mathbb{F}_{p}$ is finitely presented.
We can extend this to all finite discrete
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules.

######  Lemma 2.8.

Let $\mathbf{G}$ be a finitely generated pro-$p$ group and let $M$ be a finite
discrete $\mathbb{F}_{p}[[\mathbf{G}]]$-module. Then $M$ is finitely presented
as an $\mathbb{F}_{p}[[G]]$-module.

###### Proof.

We have to show that every open submodule of a finitely generated profinite
$\mathbb{F}_{p}[[\mathbf{G}]]$-module $M$ is finitely generated. Arguing by
induction on the index of the open submodule, we easily obtain this statement
from the fact that $I_{\mathbf{G}}$ is finitely generated. ∎

###  2.4. Left ideals in completed group algebras

Now we apply the results of the previous subsection in a particular situation
that interests us.

Let $\mathbf{G}$ be a pro-$p$ group. The definition of free profinite
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules can be found in [56, Section 5.2]. We
will need the the following fact.

######  Lemma 2.9.

Let $\mathbf{G}$ be a pro-$p$ group and $\mathbf{H}$ a closed subgroup of
$\mathbf{G}$. Then $\mathbb{F}_{p}[[\mathbf{G}]]$ is a free profinite
$\mathbb{F}_{p}[[\mathbf{H}]]$-module. In particular the functor
$\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}-$
is exact.

###### Proof.

The first part follows from [56, Corollary 5.7.2] and the second one from [56,
Proposition 5.5.3 (e)]. ∎

We denote by $I_{\mathbf{G}}$ the augmentation ideal of
$\mathbb{F}_{p}[[\mathbf{G}]]$. The following lemma is well-known. We provide
a proof for the convenience of the reader.

######  Lemma 2.10.

Let $\mathbf{H}$ be a finitely presented pro-$p$ group. Then $I_{\mathbf{H}}$
is finitely presented as a $\mathbb{F}_{p}[[\mathbf{H}]]$-module.

###### Proof.

Since $\mathbb{F}_{p}[[\mathbf{H}]]$ is a local ring, a profinite
$\mathbb{F}_{p}[[\mathbf{H}]]$-module $M$ is finitely presented if and only if
$H_{0}(\mathbf{H};M)$ and $H_{1}(\mathbf{H},M)$ are finite. Since $\mathbf{H}$
is finitely presented as a pro-$p$ group, [56, Theorem 7.8.1 and Theorem
7.8.3] imply that $H_{1}(\mathbf{H};\mathbb{F}_{p})$ and
$H_{2}(\mathbf{H};\mathbb{F}_{p})$ are finite. However, we have that for
$i\geq 0$,
$H_{i}(\mathbf{H},I_{\mathbf{H}})=H_{i+1}(\mathbf{H};\mathbb{F}_{p})$. Thus,
$I_{\mathbf{H}}$ is finitely presented as a
$\mathbb{F}_{p}[[\mathbf{H}]]$-module. ∎

If $\mathbf{H}$ is a closed subgroup of $\mathbf{G}$, then
$I_{\mathbf{H}}^{\mathbf{G}}$ denotes the closed left ideal of
$\mathbb{F}_{p}[[\mathbf{G}]]$ generated by $I_{\mathbf{H}}$.

######  Lemma 2.11.

Let $\mathbf{G}$ be a pro-$p$ group and let $\mathbf{H}\leq\mathbf{T}$ be
closed subgroups of $\mathbf{G}$. Then the following holds.

  1. (a)

The continuous map

$\displaystyle\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}\to
I_{\mathbf{H}}^{\mathbf{G}}$

that sends $a\otimes b$ to $ab$, is an isomorphism of
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules.

  2. (b)

If $\mathbf{H}$ is finitely presented, the map

$\displaystyle\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}\to
I_{\mathbf{H}}^{\mathbf{G}}$

that sends $a\otimes b$ to $ab$, is an isomorphism of
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules.

  3. (c)

If $\mathbf{T}$ is finitely presented and $\mathbf{H}$ is finitely generated,
the map

$\displaystyle\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{T}]]}(I_{\mathbf{T}}/I_{\mathbf{H}}^{\mathbf{T}})\to
I_{\mathbf{T}}^{\mathbf{G}}/I_{\mathbf{H}}^{\mathbf{G}}$

sending $a\otimes(b+I_{\mathbf{H}}^{\mathbf{T}})$ to
$ab+I_{\mathbf{H}}^{\mathbf{G}}$, is an isomorphism of
$\mathbb{F}_{p}[[\mathbf{G}]]$-modules.

###### Proof.

(a) Consider an exact sequence

$0\to I_{\mathbf{H}}\to\mathbb{F}_{p}[[\mathbf{H}]]\to\mathbb{F}_{p}\to 0.$

By Lemma 2.9, the sequence

$0\to\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}\xrightarrow{\alpha}\mathbb{F}_{p}[[\mathbf{G}]]\xrightarrow{\beta}\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}\mathbb{F}_{p}\to
0$

is also exact. Thus,
$\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}$
is isomorphic to $\ker\beta=I_{\mathbf{H}}^{\mathbf{G}}$.

(b) Since $\mathbf{H}$ is finitely presented, by Lemma 2.10, $I_{\mathbf{H}}$
is finitely presented as $\mathbb{F}_{p}[[\mathbf{H}]]$-module, and so, by
Lemma 2.6,

$\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}\cong\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}.$

Thus we conclude that the natural map from
$\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}I_{\mathbf{H}}$
to $I_{\mathbf{H}}^{\mathbf{G}}$ is also an isomorphism by (a).

(c) Consider now the exact sequence

$0\to I_{\mathbf{H}}^{\mathbf{T}}\to I_{\mathbf{T}}\to
I_{\mathbf{T}}/I_{\mathbf{H}}^{\mathbf{T}}\to 0.$

Applying
$\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{T}]]}$
and taking again into account that $\mathbb{F}_{p}[[\mathbf{G}]]$ is a free
profinite $\mathbb{F}_{p}[[\mathbf{T}]]$-module (Lemma 2.10), we obtain the
exact sequence

$0\to I_{\mathbf{H}}^{\mathbf{G}}\to
I_{\mathbf{T}}^{\mathbf{G}}\to\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{T}]]}(I_{\mathbf{T}}/I_{\mathbf{H}}^{\mathbf{T}})\to
0.$

Since $\mathbf{T}$ is finitely presented and $\mathbf{H}$ is finitely
generated, by Lemma 2.10, $I_{\mathbf{T}}/I_{\mathbf{H}}^{\mathbf{T}}$ is
finitely presented as $\mathbb{F}_{p}[[\mathbf{T}]]$-module. Thus, by Lemma
2.6, the canonical map

$\mathbb{F}_{p}[[\mathbf{G}]]\otimes_{\mathbb{F}_{p}[[\mathbf{T}]]}(I_{\mathbf{T}}/I_{\mathbf{H}}^{\mathbf{T}})\to\mathbb{F}_{p}[[\mathbf{G}]]\widehat{\otimes}_{\mathbb{F}_{p}[[\mathbf{T}]]}(I_{\mathbf{T}}/I_{\mathbf{H}}^{\mathbf{T}})$

is an isomorphism. This proves the last claim. ∎

######  Corollary 2.12.

Let $\mathbf{H}$ be a finitely generated free subgroup of a pro-$p$ group
$\mathbf{G}$. Then $I_{\mathbf{H}}^{\mathbf{G}}$ is a free
$\mathbb{F}_{p}[[\mathbf{G}]]$-module of rank $d(\mathbf{H})$.

###### Proof.

By Proposition 2.3, $I_{\mathbf{H}}$ is a free
$\mathbb{F}_{p}[[\mathbf{H}]]$-module of rank $d(\mathbf{H})$. Therefore, by
Lemma 2.11(b), $I_{\mathbf{H}}^{\mathbf{G}}$ is a free
$\mathbb{F}_{p}[[\mathbf{G}]]$-module of rank $d(\mathbf{H})$. ∎

###  2.5. On amalgamated products of groups

Let $G$ be a group and $H_{1}$ and $H_{2}$ two subgroups that generate $G$ and
have intersection $A=H_{1}\cap H_{2}$. The following result gives an algebraic
condition for $G$ to be isomorphic to the amalgamated product of $H_{1}$ and
$H_{2}$ over $A$.

######  Proposition 2.13.

[40] Let $k$ be a non-trivial commutative ring. Then the canonical map
$H_{1}*_{A}H_{2}\to G$ is an isomorphism if and only if $I_{H_{1}}^{G}\cap
I_{H_{2}}^{G}=I_{A}^{G}$ in $k[G]$.

###### Proof.

This is [40, Theorem 1], which is proved for $k=\mathbb{Z}$ but the proof
works over an arbitrary nonzero commutative ring $k$. ∎

###  2.6. On convergence of Sylvester rank functions

Let $R$ be a ring. A Sylvester matrix rank function $\operatorname{rk}$ on $R$
is a function that assigns a non-negative real number to each matrix over $R$
and satisfies the following conditions.

  1. (SMat1)

$\operatorname{rk}(M)=0$ if $M$ is any zero matrix and
$\operatorname{rk}(1)=1$ (where 1 denotes the identity matrix of size one);

  2. (SMat2)

$\operatorname{rk}(M_{1}M_{2})\leq\min\\{\operatorname{rk}(M_{1}),\operatorname{rk}(M_{2})\\}$
for any matrices $M_{1}$ and $M_{2}$ which can be multiplied;

  3. (SMat3)

$\operatorname{rk}\left(\begin{array}[]{cc}M_{1}&0\\\
0&M_{2}\end{array}\right)=\operatorname{rk}(M_{1})+\operatorname{rk}(M_{2})$
for any matrices $M_{1}$ and $M_{2}$;

  4. (SMat4)

$\operatorname{rk}\left(\begin{array}[]{cc}M_{1}&M_{3}\\\
0&M_{2}\end{array}\right)\geq\operatorname{rk}(M_{1})+\operatorname{rk}(M_{2})$
for any matrices $M_{1}$, $M_{2}$ and $M_{3}$ of appropriate sizes.

We denote by $\mathbb{P}(R)$ the set of Sylvester matrix rank functions on
$R$, which is a compact convex subset of the space of functions on matrices
over $R$ considered with pointwise convergence.

Many problems can be reinterpreted in terms of convergence in $\mathbb{P}(R)$.
For example, if $G$ is group and $G\geq G_{1}\geq G_{2}\geq\ldots$ is a chain
of normal subgroups of $G$ of finite index with trivial intersection, then the
fact of the existence of the Lück approximation over $\mathbb{Q}$ and
$\mathbb{C}$ can be viewed as the convergence of $\operatorname{rk}_{G/G_{i}}$
to $\operatorname{rk}_{G}$ in $\mathbb{P}(\mathbb{Q}[G])$ and
$\mathbb{P}(\mathbb{C}[G])$ respectively (see [45, 30] for details).

An alternative way to introduce Sylvester rank functions is via Sylvester
module rank functions. A Sylvester module rank function $\dim$ on $R$ is a
function that assigns a non-negative real number to each finitely presented
$R$-module and satisfies the following conditions.

  1. (SMod1)

$\dim\\{0\\}=0$, $\dim R=1$;

  2. (SMod2)

$\dim(M_{1}\oplus M_{2})=\dim M_{1}+\dim M_{2}$;

  3. (SMod3)

if $M_{1}\to M_{2}\to M_{3}\to 0$ is exact then

$\dim M_{1}+\dim M_{3}\geq\dim M_{2}\geq\dim M_{3}.$

By [50, Theorem 4] (see also [43, Proposition 1.2.8]), there exists a natural
bijection between Sylvester matrix and module rank functions over a ring.
Given a Sylvester module rank function $\dim$ on $R$ and a finitely presented
$R$-module $M\cong R^{n}/R^{m}A$ ($A$ is a matrix over $R$), we define the
corresponding Sylvester matrix rank function $\operatorname{rk}$ by means of
$\operatorname{rk}(A)=n-\dim M$.

By a recent result of Li [41], any Sylvester module rank function $\dim$ on
$R$ can be extended to a unique function (satisfying some natural conditions)
on arbitrary modules over $R$. We will call this extension, the extended
Sylvester module rank function and denote it also by $\dim$. In this paper we
will mostly use this extension for finitely generated modules $M$. In this
case $\dim M$ is defined as

(2) $\dim M=\inf\\{\dim\widetilde{M}:\ \widetilde{M}\textrm{\ is finitely
presented and\ }M\textrm{\ is a quotient of\ }\widetilde{M}\\}.$

In the case of an arbitrary $R$-module $M$ the formula for $\dim M$ is more
complex. In this case we put

$\dim M=\sup_{M_{1}}\inf_{M_{2}}(\dim M_{2}-\dim(M_{2}/M_{1})),$

where $M_{1}\leq M_{2}$ are finitely generated $R$-submodules of $M$. Observe
that we allow $+\infty$ to be a value of $\dim M$.

######  Remark 2.14.

If $\operatorname{rk}$, $\operatorname{rk}_{i}\in\mathbb{P}(R)$
$(i\in\mathbb{N})$ are Sylvester matrix rank functions corresponding to
Sylvester module rank functions $\dim$, $\dim_{i}$, respectively, then
$\operatorname{rk}=\displaystyle\lim_{i\to\infty}\operatorname{rk}_{i}$ in the
space $\mathbb{P}(R)$ if and only if for any finitely presented $R$-module
$M$, $\dim M=\displaystyle\lim_{i\to\infty}\dim_{i}M$.

However, the existence of the limit
$\operatorname{rk}=\displaystyle\lim_{i\to\infty}\operatorname{rk}_{i}$ does
not imply that $\dim M=\displaystyle\lim_{i\to\infty}\dim_{i}M$ for any
finitely generated $R$-module $M$. This phenomenon is well-known. For example,
it explains why the Lück approximation of the first $L^{2}$-Betti numbers is
valid for finitely presented groups but not always valid for finitely
generated groups (see [46]).

If $M$ is a finitely generated $R$-module, we only always have that

(3) $\dim M\geq\limsup_{i\to\infty}\dim_{i}M.$

Indeed, let $\mathcal{F}$ be the set of all finitely presented $R$-modules
$\tilde{M}$ such that $M$ is a quotient of $\tilde{M}$. Then

$\dim
M=\inf_{\tilde{M}\in\mathcal{F}}\dim\tilde{M}=\inf_{\tilde{M}\in\mathcal{F}}\lim_{i\to\infty}\dim_{i}\tilde{M}\geq\\\
\limsup_{i\to\infty}\inf_{\tilde{M}\in\mathcal{F}}\dim_{i}\tilde{M}=\limsup_{i\to\infty}\dim_{i}M.$

In this subsection we will explain how to overcome this problem in some
situations. For two Sylvester rank functions $\operatorname{rk}_{1}$ and
$\operatorname{rk}_{2}\in\mathbb{P}(R)$ we write
$\operatorname{rk}_{1}\geq\operatorname{rk}_{2}$ if
$\operatorname{rk}_{1}(A)\geq\operatorname{rk}_{2}(A)$ for any matrix $A$ over
$R$. If $\dim_{1}$ and $\dim_{2}$ are the Sylvester module rank functions on
$R$ corresponding to $\operatorname{rk}_{1}$ and $\operatorname{rk}_{2}$, then
the condition $\operatorname{rk}_{1}\geq\operatorname{rk}_{2}$ is equivalent
to the condition $\dim_{1}\leq\dim_{2}$, meaning that $\dim_{1}M\leq\dim_{2}M$
for any finitely presented $R$-module $M$.

######  Proposition 2.15.

Let $R$ be a ring and let $\operatorname{rk}$,
$\operatorname{rk}_{i}\in\mathbb{P}(R)$ $(i\in\mathbb{N})$ be Sylvester matrix
rank functions on $R$ corresponding to (extended) Sylvester module rank
functions $\dim$, $\dim_{i}$ respectively. Assume that
$\operatorname{rk}=\displaystyle\lim_{i\to\infty}\operatorname{rk}_{i}$ and
for all $i$, $\operatorname{rk}\geq\operatorname{rk}_{i}$. Then, for any
finitely generated $R$-module $M$,

$\dim M=\displaystyle\lim_{i\to\infty}\dim_{i}M.$

###### Proof.

Fix $\varepsilon>0$. Let $k$ be such that

$\liminf_{i\to\infty}\dim_{i}M\geq\dim_{k}M-\varepsilon.$

There exists a finitely presented $R$-module $\tilde{M}$ such that $M$ is a
quotient of $\tilde{M}$ and $\dim_{k}M\geq\dim_{k}\tilde{M}-\varepsilon$.
Since $\operatorname{rk}\geq\operatorname{rk}_{k}$, we have that
$\dim_{k}\tilde{M}\geq\dim\tilde{M}$. Thus, we obtain

$\liminf_{i\to\infty}\dim_{i}M\geq\dim_{k}M-\varepsilon\geq\dim_{k}\tilde{M}-2\varepsilon\geq\dim\tilde{M}-2\varepsilon\geq\dim
M-2\varepsilon.$

Since $\varepsilon$ is arbitrary, we conclude that
$\displaystyle\liminf_{i\to\infty}\dim_{i}M\geq\dim M$. In view of (3), we are
done. ∎

###  2.7. Epic division $R$-rings

Let $R$ be a ring. An epic division $R$-ring is a $R$-ring
$\phi:R\to\operatorname{\mathcal{D}}$ where $\operatorname{\mathcal{D}}$ is a
division ring generated by $\phi(R)$. Moreover, we say that
$\operatorname{\mathcal{D}}$ is a division $R$-ring of fractions if $\phi$ is
injective. In this case we will normally omit $\phi$ and see $R$ as a subring
of $\operatorname{\mathcal{D}}$.

Each epic division $R$-ring $\operatorname{\mathcal{D}}$ induces a Sylvester
module rank function $\dim_{\operatorname{\mathcal{D}}}$ on $R$: for every a
finitely presented $R$-module $M$ we define
$\dim_{\operatorname{\mathcal{D}}}M$ to be equal to the dimension of
$\operatorname{\mathcal{D}}\otimes_{R}M$ as a
$\operatorname{\mathcal{D}}$-module.

The extended Sylvester rank function $\dim_{\operatorname{\mathcal{D}}}$ is
calculated in the same way: for an $R$-module $M$,
$\dim_{\operatorname{\mathcal{D}}}M$ is equal to the dimension of
$\operatorname{\mathcal{D}}\otimes_{R}M$ as a
$\operatorname{\mathcal{D}}$-module.

We will use $\dim_{\operatorname{\mathcal{D}}}$ for the (extended) Sylvester
module rank function on $R$ and for the $\operatorname{\mathcal{D}}$-dimension
of $\operatorname{\mathcal{D}}$-spaces. This is a coherent notation because,
since $\operatorname{\mathcal{D}}$ is epic,
$\operatorname{\mathcal{D}}\otimes_{R}\operatorname{\mathcal{D}}$ is
isomorphic to $\operatorname{\mathcal{D}}$ as $R$-bimodule (see [31, Section
4]).

The following result of P. M. Cohn will be used several times in the paper.

######  Proposition 2.16.

[18, Theorem 4.4.1] Two epic division $R$-rings
$\operatorname{\mathcal{D}}_{1}$ and $\operatorname{\mathcal{D}}_{2}$ are
isomorphic as $R$-rings if and only if
$\dim_{\operatorname{\mathcal{D}}_{1}}M=\dim_{\operatorname{\mathcal{D}}_{2}}M$
for every finitely presented $R$-module $M$.

###  2.8. Natural extensions of Sylvester rank functions

Let $G$ be a group with trivial element $e$. We say that a ring $R$ is
$G$-graded if $R$ is equal to the direct sum $\oplus_{g\in G}R_{g}$ and
$R_{g}R_{h}\subseteq R_{gh}$ for all $g$ and $h$ in $G$. If for each $g\in G$,
$R_{g}$ contains an invertible element $u_{g}$, then we say that $R$ is a
crossed product of $R_{e}$ and $G$ and we will write $R=S*G$ if $R_{e}=S$.

Let $R=S*G$ be a crossed product. Let $\operatorname{rk}$ be a Sylvester
matrix rank function on $S$ and $\dim$ its associated Sylvester module rank
function. We say that $\operatorname{rk}$ (and $\dim$) are $R$-compatible if
for any $g\in G$ and any matrix $A$ over $S$,
$\operatorname{rk}(A)=\operatorname{rk}(u_{g}Au_{g}^{-1})$. If $G$ is finite
and $M$ is a finitely presented $R$-module, then $M$ is also a finitely
presented $S$-module. Thus, if $\dim$ is $R$-compatible, we can define

(4) $\widetilde{\dim}\ M=\frac{\dim M}{|G|},$

where $M$ is a finitely presented $R$-module. This defines a Sylvester module
rank function on $R$, called the natural extension of $\dim$. This notion was
studied, for example, in [36]. We notice that the same formula (4) holds also
for extended Sylvester module rank functions (that is, when $M$ is an
arbitrary $R$-module). In this subsection we prove the following result.

######  Proposition 2.17.

Let $R=S*G$ be a crossed product with $G$ finite and let
$R\hookrightarrow\operatorname{\mathcal{D}}$ be a division $R$-ring of
fractions. Denote by $\operatorname{\mathcal{D}}_{e}$ the division closure of
$S$ in $\operatorname{\mathcal{D}}$. Then the following are equivalent.

  1. (a)

$\widetilde{\dim_{\operatorname{\mathcal{D}}_{e}}}=\dim_{\operatorname{\mathcal{D}}}$
as Sylvester functions on $R$.

  2. (b)

$\dim_{\operatorname{\mathcal{D}}_{e}}\operatorname{\mathcal{D}}=|G|$.

  3. (c)

$\operatorname{\mathcal{D}}$ is isomorphic to a crossed product
$\operatorname{\mathcal{D}}_{e}*G$.

  4. (d)

$\operatorname{\mathcal{D}}$ is isomorphic to
$\operatorname{\mathcal{D}}_{e}\otimes_{S}R$ as
$(\operatorname{\mathcal{D}}_{e},R)$-bimodule.

  5. (e)

$\operatorname{\mathcal{D}}$ is isomorphic to
$R\otimes_{S}\operatorname{\mathcal{D}}_{e}$ as
$(R,\operatorname{\mathcal{D}}_{e})$-bimodule.

###### Proof.

For any $h,g\in G$, define $\alpha(g,h)=u_{g}u_{h}(u_{gh})^{-1}\in R$. Since
the conjugation by $u_{g}$ fixes $S$, it also fixes
$\operatorname{\mathcal{D}}_{e}$. Therefore, we can define a ring structure on
$T=\oplus_{g\in G}\operatorname{\mathcal{D}}_{e}v_{g}$ defining the
multiplication on homogenous elements by means of

$(d_{1}v_{g})\cdot(d_{2}v_{h})=(d_{1}u_{g}d_{2}(u_{g})^{-1}\alpha(g,h))v_{gh},\
d_{1},d_{2}\in\operatorname{\mathcal{D}}_{e},\ g,h\in G.$

It is clear that $T=\operatorname{\mathcal{D}}_{e}*G$ is a crossed product, it
contains $R$ as a subring, and $T$ is isomorphic to
$\operatorname{\mathcal{D}}_{e}\otimes_{S}R$ as
$(\operatorname{\mathcal{D}}_{e},R)$-bimodule.

There exists a natural map $\gamma:T\to\operatorname{\mathcal{D}}$, sending
$\sum_{g\in G}d_{g}v_{g}$ ($d_{g}\in D_{e}$) to $\sum_{g\in G}d_{g}u_{g}$.
Observe that $\gamma(T)$ is a domain and of finite dimension over
$\operatorname{\mathcal{D}}_{e}$. Thus, $\gamma(T)$ is a division subring of
$\operatorname{\mathcal{D}}$. Since $T$ contains $R$,
$\operatorname{\mathcal{D}}=\gamma(T)$. This implies that (c) and (d) are
equivalent.

Since $\dim_{D_{e}}T=|G|$, (b) implies that $\gamma$ is an isomorphism, and
so, (b) implies (c).

Now, let us assume (d). Let $M$ be a finitely presented $R$-module. We have
the following.

$\widetilde{\dim_{\operatorname{\mathcal{D}}_{e}}}\
M=\frac{\dim_{\operatorname{\mathcal{D}}_{e}}(\operatorname{\mathcal{D}}_{e}\otimes_{S}M)}{|G|}=\frac{\dim_{\operatorname{\mathcal{D}}_{e}}(\operatorname{\mathcal{D}}_{e}\otimes_{S}(R\otimes_{R}M))}{|G|}=\\\
\frac{\dim_{\operatorname{\mathcal{D}}_{e}}((\operatorname{\mathcal{D}}_{e}\otimes_{S}R)\otimes_{R}M))}{|G|}\stackrel{{\scriptstyle\text{(d)}}}{{=}}\frac{\dim_{\operatorname{\mathcal{D}}_{e}}(\operatorname{\mathcal{D}}\otimes_{R}M)}{|G|}=\\\
\dim_{\operatorname{\mathcal{D}}}(\operatorname{\mathcal{D}}\otimes_{R}M)=\dim_{\operatorname{\mathcal{D}}}M.$

This proves (a).

Now, we assume that (a) holds. Since $\operatorname{\mathcal{D}}_{e}$ es an
epic $S$-ring
$\operatorname{\mathcal{D}}_{e}\otimes_{S}\operatorname{\mathcal{D}}_{e}$ is
isomorphic to $\operatorname{\mathcal{D}}_{e}$ as
$\operatorname{\mathcal{D}}_{e}$-bimodule and by the same reason,
$\operatorname{\mathcal{D}}\otimes_{R}\operatorname{\mathcal{D}}$ is
isomorphic to $\operatorname{\mathcal{D}}$ as
$\operatorname{\mathcal{D}}$-bimodule. Consider $M=\operatorname{\mathcal{D}}$
as an $R$-module and $N=\operatorname{\mathcal{D}}_{e}$ as a $S$-module. Then

$1=\dim_{\operatorname{\mathcal{D}}}(\operatorname{\mathcal{D}}\otimes_{R}M)=\dim_{\operatorname{\mathcal{D}}}M\stackrel{{\scriptstyle\text{(a)}}}{{=}}\widetilde{\dim_{\operatorname{\mathcal{D}}_{e}}}M=\\\
\frac{\dim_{\operatorname{\mathcal{D}}_{e}}(\operatorname{\mathcal{D}}_{e}\otimes_{S}M)}{|G|}=\frac{\dim_{\operatorname{\mathcal{D}}_{e}}(\operatorname{\mathcal{D}}_{e}\otimes_{S}(N^{\dim_{\operatorname{\mathcal{D}}_{e}}\operatorname{\mathcal{D}}}))}{|G|}=\frac{\dim_{\operatorname{\mathcal{D}}_{e}}\operatorname{\mathcal{D}}}{|G|}$

This implies (b).

Let $R^{op}$ denotes the opposite ring, that is the ring with the same
elements and addition operation, but with the multiplication performed in the
reverse order. Then $R^{op}\cong S^{op}*G$ and the condition (c) is equivalent
to

  1. (c’)

$\operatorname{\mathcal{D}}^{op}$ is isomorphic to a crossed product
$(\operatorname{\mathcal{D}}_{e})^{op}*G$.

Our previous proof gives that (c’) is equivalent to

  1. (d’)

$\operatorname{\mathcal{D}}^{op}$ is isomorphic to
$(\operatorname{\mathcal{D}}_{e})^{op}\otimes_{S^{op}}R^{op}$ as
$((\operatorname{\mathcal{D}}_{e})^{op},R^{op})$-bimodule.

Now, observe that (d’) is equivalent to (e). ∎

##  3\. On mod-$p$ $L^{2}$-Betti numbers of subgroups of a free pro-$p$ groups

###  3.1. Universal division ring of fractions

Given two epic division $R$-rings $\operatorname{\mathcal{D}}_{1}$ and
$\operatorname{\mathcal{D}}_{2}$ the condition
$\dim_{\operatorname{\mathcal{D}}_{1}}\leq\dim_{\operatorname{\mathcal{D}}_{2}}$
is equivalent to the existence of a specialization from
$\operatorname{\mathcal{D}}_{1}$ to $\operatorname{\mathcal{D}}_{2}$ in the
sense of P. Cohn ([18, Subsection 4.1]). We say that an epic division $R$-ring
$\operatorname{\mathcal{D}}$ is universal if for every division $R$-ring
$\mathcal{E}$, $\dim_{\operatorname{\mathcal{D}}}\leq\dim_{\mathcal{E}}$. If a
universal epic division $R$-ring exists, it is unique up to $R$-isomorphism.
We will denote it by $\operatorname{\mathcal{D}}_{R}$ and instead of
$\dim_{\operatorname{\mathcal{D}}_{R}}$ we will simply write $\dim_{R}$.

We say that a ring $R$ is a semifir if every finitely generated left ideal of
$R$ is free and free modules of distinct finite rank are non-isomorphic. For
example, if $K$ is a field, the ring $K\langle\\!\langle X\rangle\\!\rangle$
of non-commutative power series is a semifir ([17, Theorem 2.9.4]). By a
theorem of P. M. Cohn [16] a semifir $R$ has a universal division $R$-ring. P.
M. Cohn proved that in this case $\operatorname{\mathcal{D}}_{R}$ can be
obtained from $R$ by formally inverting all full matrices over $R$. In
particular, this implies the following result.

######  Proposition 3.1.

Let $R$ be a semifir. Then $\dim_{R}$ is the smallest Sylvester module rank
function among all the Sylvester module rank functions on $R$.

We will need the following result.

######  Proposition 3.2.

[32, Proposition 2.2] Let $R$ be a semifir. Then

$\operatorname{Tor}^{R}_{1}(\operatorname{\mathcal{D}}_{R},M)=0$

for any $R$-submodule of a $\operatorname{\mathcal{D}}_{R}$-module.

Let $G$ be a residually torsion-free nilpotent group (for example, $G$ is a
subgroup of a free pro-$p$ group). Let $K$ be a field. Then the universal
division ring of fractions $\operatorname{\mathcal{D}}_{K[G]}$ exists (see
[33]). It can be constructed in the following way. Since $G$ is residually
torsion free nilpotent, $G$ is bi-orderable. Fix a bi-invariant order
$\preceq$ on $G$. A. Malcev [49] and B. Neumann [54] (following an idea of H.
Hahn [24]) showed independently that the set $K((G,\preceq))$ of formal power
series over $G$ with coefficients in $K$ having well-ordered support has a
natural structure of a ring and, moreover, it is a division ring.
$\operatorname{\mathcal{D}}_{K[G]}$ can be defined as the division closure of
$K[G]$ in $K((G,\preceq))$. The universality of this division ring is shown in
[33, Theorem 1.1].

If $A$ is a torsion-free abelian group, then
$\operatorname{\mathcal{D}}_{K[A]}$ coincides with the classical ring of
fractions $\mathcal{Q}(K[A])$ of $K[A]$.

###  3.2. The division ring
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$

If $\mathbf{F}$ is a free pro-$p$ group freely generated by
$f_{1},\ldots,f_{d}$, then, by Proposition 2.3, the continuous homomorphism
$\mathbb{F}_{p}\langle\\!\langle
x_{1},\ldots,x_{d}\rangle\\!\rangle\to\mathbb{F}_{p}[[\mathbf{F}]]$ that sends
$x_{i}$ to $f_{1}-1$, is an isomorphism. Thus, there exists a universal
division ring of fraction
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$. Using results of
[32] we establish the following formula for
$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}$ which is one of main ingredients of the
proof of Theorem 1.2.

######  Proposition 3.3.

Let $\mathbf{F}=\mathbf{N}_{1}>\mathbf{N}_{2}>\mathbf{N}_{3}>\ldots$ be a
chain of open normal subgroups of a finitely generated free pro-$p$ group
$\mathbf{F}$ with trivial intersection. Let $M$ be a finitely generated
$\mathbb{F}_{p}[[\mathbf{F}]]$-module. Then

$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}M=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M)}{|\mathbf{F}:\mathbf{N}_{i}|}.$

###### Proof.

Let $\mathbf{N}$ be a normal open subgroup of $\mathbf{F}$ and let
$\dim_{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]}$ be a Sylvester module rank
function on $\mathbb{F}_{p}[[\mathbf{F}]]$ defined by

$\dim_{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]}L=\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}L)}{|\mathbf{F}:\mathbf{N}|},$

where $L$ is a finitely presented $\mathbb{F}_{p}[[\mathbf{F}]]$-module.

Let $M=\mathbb{F}_{p}[[\mathbf{F}]]^{n}/I$ be a finitely generated
$\mathbb{F}_{p}[[\mathbf{F}]]$-module. Since

${\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M\cong\mathbb{F}[[\mathbf{G}]]^{n}/(I+(I_{\mathbf{N}}^{\mathbf{G}})^{n})$

is finite, it is finitely presented by Lemma 2.10. Therefore, there exists a
finitely generated $\mathbb{F}_{p}[[\mathbf{G}]]$-submodule $J$ of $I$ such
that $I+(I_{N}^{G})^{n}=J+(I_{N}^{G})^{n}$. Put
$\widetilde{M}=\mathbb{F}_{p}[[\mathbf{F}]]^{n}/J$. Then $\widetilde{M}$ is a
finitely presented $\mathbb{F}_{p}[[\mathbf{F}]]$-module $\widetilde{M}$
satisfying the following conditions:

  1. (1)

$M$ is a quotient of $\widetilde{M}$ and

  2. (2)

${\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}\widetilde{M}\cong{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M.$

Therefore, from (2) we obtain that

$\dim_{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]}M=\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}\widetilde{M})}{|\mathbf{F}:\mathbf{N}|}=\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M)}{|\mathbf{F}:\mathbf{N}|}.$

In the case where $M$ is finitely presented, [32, Theorem 1.4] and Remark 2.14
impliy that

(5)
$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}M=\lim_{i\to\infty}\dim_{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]}M.$

By Proposition 3.1, we also have that
$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}M\leq\dim_{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]}M$
for all $i$. Therefore, Proposition 2.15 implies that (5) holds also when $M$
is finitely generated. Hence,

$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}M=\lim_{i\to\infty}\dim_{\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]}M=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M)}{|\mathbf{F}:\mathbf{N}_{i}|}.$

∎

In the following proposition we collect some basic properties of
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$.

######  Proposition 3.4.

Let $\mathbf{H}$ be a finitely generated closed subgroup of $\mathbf{F}$. The
following holds.

  1. (a)

Let $\operatorname{\mathcal{D}}_{\mathbf{H}}$ be the division closure of
$\mathbb{F}_{p}[[\mathbf{H}]]$ in
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$. Then
$\operatorname{\mathcal{D}}_{\mathbf{H}}$ is isomorphic to
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$ as an
$\mathbb{F}_{p}[[\mathbf{H}]]$-ring.

  2. (b)

If $M$ is a finitely generated $\mathbb{F}_{p}[[\mathbf{H}]]$-module, then

$\dim_{\mathbb{F}_{p}[[\mathbf{H}]]}(M)=\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M).$

  3. (c)

If $\mathbf{H}$ is open then,

$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\cong\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$

as $(\mathbb{F}_{p}[[\mathbf{F}]],\mathbb{F}_{p}[[\mathbf{H}]])$-bimodules.

###### Proof.

(a) Fix a normal chain
$\mathbf{F}=\mathbf{N}_{1}>\mathbf{N}_{2}>\mathbf{N}_{3}>\ldots$ of open
normal subgroups of $\mathbf{F}$, and let
$\mathbf{H}_{i}=\mathbf{N}_{i}\cap\mathbf{H}$. Let $M$ be a finitely generated
$\mathbb{F}_{p}[[\mathbf{H}]]$-module. Observe first, that by Proposition 3.3,

(6)
$\dim_{\mathbb{F}_{p}[[\mathbf{H}]]}M=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{H}/\mathbf{H}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)}{|\mathbf{H}:\mathbf{H}_{i}|}.$

Considering $\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]$ as a right
$\mathbb{F}_{p}[[\mathbf{H}]]$-module, we obtain that

$\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\cong\mathbb{F}_{p}[\mathbf{H}/\mathbf{H}_{i}]^{|\mathbf{F}:\mathbf{N}_{i}\mathbf{H}|}$

as right $\mathbb{F}_{p}[[\mathbf{H}]]$-modules. Thus,

(7)
$\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{H}/\mathbf{H}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)=\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)}{|\mathbf{F}:\mathbf{N}_{i}\mathbf{H}|}.$

Therefore, from (6), (7) and Proposition 3.3, we conclude that

(8)
$\dim_{\mathbb{F}_{p}[[\mathbf{H}]]}M=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)}{|\mathbf{F}:\mathbf{N}_{i}|}=\\\
\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M).$

On the other hand, we have

$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)=\\\
\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M))=\\\
\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M))=\dim_{\operatorname{\mathcal{D}}_{\mathbf{H}}}(\operatorname{\mathcal{D}}_{\mathbf{H}}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M).$

Thus, we conclude that

$\dim_{\mathbb{F}_{p}[[\mathbf{H}]]}M=\dim_{\operatorname{\mathcal{D}}_{\mathbf{H}}}(\operatorname{\mathcal{D}}_{\mathbf{H}}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M),$

and so, by Proposition 2.16, $\operatorname{\mathcal{D}}_{\mathbf{H}}$ is
isomorphic to $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$ as
an $\mathbb{F}_{p}[[\mathbf{H}]]$-ring.

(b) This is the equality (8).

(c) First assume that $\mathbf{H}$ is normal in $\mathbf{F}$. Observe that for
large $i$, $\mathbf{N}_{i}\leq\mathbf{H}$. Let $M$ be a finitely presented
$\mathbb{F}_{p}[[F]]$-module. Then we obtain that

$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}M=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[[\mathbf{N}_{i}]]}M)}{|\mathbf{F}:\mathbf{N}_{i}|}=\\\
\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[[\mathbf{N}_{i}]]}M)}{|\mathbf{F}:\mathbf{H}||\mathbf{H}:\mathbf{N}_{i}|}=\frac{\dim_{\mathbb{F}_{p}[[\mathbf{H}]]}M}{|\mathbf{F}:\mathbf{H}|}.$

Therefore,
$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}=\widetilde{\dim_{\mathbb{F}_{p}[[\mathbf{H}]]}}$.
Now, the result follows from Proposition 2.17.

Now we assume that $\mathbf{H}$ is arbitrary, We argue by induction on
$|\mathbf{F}:\mathbf{H}|$. If $\mathbf{H}$ has index $p$ in $\mathbf{F}$, then
it is normal. If $|\mathbf{F}:\mathbf{H}|>p$, we find $\mathbf{H}_{1}$ of
index $p$ in $\mathbf{F}$ containing $\mathbf{H}$. Then by induction,

$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\cong\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}_{1}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}_{1}]]}\cong\\\
\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}_{1}]]}(\mathbb{F}_{p}[[\mathbf{H}_{1}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]})\cong\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}.$

∎

In view of the previous proposition, we will identify
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$ and the division
closure of $\mathbb{F}_{p}[[\mathbf{H}]]$ in
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$, and see
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$ as a subring of
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$.

###  3.3. The division rings
$\operatorname{\mathcal{D}}(\mathbb{F}_{p}[G],\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]})$

Let $G$ be a subgroup of $\mathbf{F}$. As we have explained in Subsection 3.1
there exists the universal division $\mathbb{F}_{p}[G]$-ring of fractions
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$. Let
$\operatorname{\mathcal{D}}_{G}=\operatorname{\mathcal{D}}(\mathbb{F}_{p}[G],\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]})$
be the division closure of $\mathbb{F}_{p}[G]$ in
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$. In this
subsection we will show that $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$
and $\operatorname{\mathcal{D}}_{G}$ are isomorphic as
$\mathbb{F}_{p}[G]$-rings. In the case $G=F$ is a finitely generated free
group and $\mathbf{F}$ is the pro-$p$ completion of $F$, this result follows
from [17, Corollary 2.9.16].

######  Proposition 3.5.

Let $\mathbf{F}$ be a finitely generated free pro-$p$ group and let $G$ be a
finitely generated subgroup of $\mathbf{F}$. Let
$\mathbf{F}=\mathbf{N}_{1}>\mathbf{N}_{2}>\mathbf{N}_{3}>\ldots$ be a chain of
open normal subgroups of $\mathbf{F}$ with trivial intersection. We put
$G_{j}=G\cap\mathbf{N}_{j}$. Let
$\operatorname{\mathcal{D}}_{G}=\operatorname{\mathcal{D}}(\mathbb{F}_{p}[G],\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]})$
be the division closure of $\mathbb{F}_{p}[G]$ in
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$. Then for every
finitely generated $\mathbb{F}_{p}[G]$-module $M$,

$\dim_{\operatorname{\mathcal{D}}_{G}}(\operatorname{\mathcal{D}}_{G}\otimes_{\mathbb{F}_{p}[G]}M)=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[G_{i}]}M)}{|G:G_{i}|}=\\\
\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}\otimes_{\mathbb{F}_{p}[G]}M).$

In particular, the divison rings $\operatorname{\mathcal{D}}_{G}$ and
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$ are isomorphic as
$\mathbb{F}_{p}[G]$-rings.

###### Proof.

Without loss of generality we assume that $G$ is dense in $\mathbf{F}$. First
observe that

(9)
$\dim_{\operatorname{\mathcal{D}}_{G}}(\operatorname{\mathcal{D}}_{G}\otimes_{\mathbb{F}_{p}[G]}M)=\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[G]}M)=\\\
\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[G]}M))=\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[G]}M).$

Observe also that, since $G$ is dense in $\mathbf{F}$,
$|\mathbf{F}:\mathbf{N}_{i}|=|G:G_{i}|$ and

$\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[G_{i}]}M\cong\mathbb{F}_{p}[G/G_{i}]\otimes_{\mathbb{F}_{p}[G]}M\cong\\\
\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\otimes_{\mathbb{F}_{p}[G]}M\cong\mathbb{F}_{p}[\mathbf{F}/\mathbf{N}_{i}]\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[G]}M)).$

Thus, Proposition 3.3 implies that

(10)
$\dim_{\operatorname{\mathcal{D}}_{G}}(\operatorname{\mathcal{D}}_{G}\otimes_{\mathbb{F}_{p}[G]}M)=\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[G_{i}]}M)}{|G:G_{i}|}.$

Let $\mathbf{F}_{i}=\gamma_{i}(\mathbf{F})$ and we put
$H_{i}=G\cap\mathbf{F}_{i}$. The ring $\mathbb{F}_{p}[G/H_{i}]$ is a
Noetherian domain and its classical field of fractions
$\mathcal{Q}(\mathbb{F}_{p}[G/H_{i}])$ is universal. Moreover, by [33, Theorem
1.2], we have that

(11)
$\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}\otimes_{\mathbb{F}_{p}[G]}M)=\lim_{i\to\infty}\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G/H_{i}]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G/H_{i}]}\otimes_{\mathbb{F}_{p}[G]}M).$

Observe that $\mathbb{F}_{p}[[\mathbf{F}/\mathbf{F}_{i}]]$ is also a
Noetherian domain, and so the division closure of $\mathbb{F}_{p}[G/H_{i}]$ in
$\mathcal{Q}(\mathbb{F}_{p}[[\mathbf{F}/\mathbf{F}_{i}]])$ is isomorphic to
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G/H_{i}]}$ (as a
$\mathbb{F}_{p}[G]$-ring). Therefore,

(12)
$\dim_{\mathcal{Q}(\mathbb{F}_{p}[[\mathbf{F}/\mathbf{F}_{i}]])}(\mathcal{Q}(\mathbb{F}_{p}[[\mathbf{F}/\mathbf{F}_{i}]])\otimes_{\mathbb{F}_{p}[G]}M)=\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G/H_{i}]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G/H_{i}]}\otimes_{\mathbb{F}_{p}[G]}M).$

Using Proposition 3.3 and arguing as in the proof of [29, Theorem 2.3], we
obtain that

(13)
$\dim_{\mathbb{F}_{p}[[\mathbf{F}]]}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[G]}M)=\lim_{i\to\infty}\dim_{\mathcal{Q}(\mathbb{F}_{p}[[\mathbf{F}/F_{i}]])}(\mathcal{Q}(\mathbb{F}_{p}[[\mathbf{F}/\mathbf{F}_{i}]])\otimes_{\mathbb{F}_{p}[G]}M).$

Therefore, putting together (11), (12), (13), (9) and (10), we obtain that

$\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}\otimes_{\mathbb{F}_{p}[G]}M)=\dim_{\operatorname{\mathcal{D}}_{G}}(\operatorname{\mathcal{D}}_{G}\otimes_{\mathbb{F}_{p}[G]}M)=\\\
\lim_{i\to\infty}\frac{\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[G_{i}]}M)}{|G:G_{i}|}.$

Applying Proposition 2.16, we obtain that the divison rings
$\operatorname{\mathcal{D}}_{G}$ and
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$ are isomorphic as
$\mathbb{F}_{p}[G]$-rings. ∎

An alternative approach of proving that $\operatorname{\mathcal{D}}_{G}$ is
isomorphic to $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$ as
$\mathbb{F}_{p}[G]$-ring is taken in [51, Lemma 7.5.5], where the result is
proved by using a variation from [58, Theorem 6.3] of the uniqueness of
Hughes-free division rings [27] (see also [19]).

###  3.4. Mod-$p$ $L^{2}$-Betti numbers

$L^{2}$-Betti numbers play an important role in the solution of many problems
in group theory. In the last years there was an attempt to develop a theory of
mod-$p$ $L^{2}$-Betti numbers for different families of groups (see [31]). If
$G$ is torsion-free and satisfies the Atiyah conjecture, P. Linnell [42]
showed that $L^{2}$-Betti numbers of $G$ can be defined as

$b_{i}^{(2)}(G)=\dim_{\operatorname{\mathcal{D}}(G)}H_{i}(G;\operatorname{\mathcal{D}}(G)),$

where $\operatorname{\mathcal{D}}(G)$ is the division ring obtained as the
division closure of $\mathbb{Q}[G]$ in the ring of affilated operators
$\mathcal{U}(G)$. It turns out that if $G$ is residually torsion-free
nilpotent, $\operatorname{\mathcal{D}}(G)$ is isomorphic to the universal
division ring of fractions $\operatorname{\mathcal{D}}_{\mathbb{Q}[G]}$ (see,
for example, [34]). Therefore, we have

$b_{i}^{(2)}(G)=\dim_{\operatorname{\mathcal{D}}_{\mathbb{Q}[G]}}H_{i}(G;\operatorname{\mathcal{D}}_{\mathbb{Q}[G]}).$

Thus, by analogy, if $G$ is a residually torsion-free nilpotent group, we
define the $i$th mod-$p$ $L^{2}$-Betti number of $G$ as

$\beta^{\operatorname{mod}p}_{i}(G)=\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}}H_{i}(G;\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}).$

In the case, where $G$ is a subgroup of a free pro-$p$ group, we also obtain
the following formula, which can be seen as a mod-$p$ analogue of the Lück
approximation theorem [45].

######  Proposition 3.6.

Let $\mathbf{F}$ be a finitely generated free pro-$p$ group and let $G$ be a
subgroup of $\mathbf{F}$ of type $FP_{k}$ for some $k\geq 1$. Let
$\mathbf{F}=\mathbf{N}_{1}>\mathbf{N}_{2}>\mathbf{N}_{3}>\ldots$ be a chain of
open normal subgroups of $\mathbf{F}$ with trivial intersection. We put
$G_{j}=G\cap\mathbf{N}_{j}$. Then

$\beta_{k}^{\operatorname{mod}p}(G)=\displaystyle\lim_{j\to\infty}\frac{\dim_{\mathbb{F}_{p}}H_{k}(G_{j};\mathbb{F}_{p})}{|G:G_{j}|}.$

###### Proof.

There exists a resolution of the $\mathbb{F}_{p}[G]$-module $\mathbb{F}_{p}$

$0\to
R_{k}\to\mathbb{F}_{p}[G]^{n_{k}}\xrightarrow{\phi_{k}}\ldots\to\mathbb{F}_{p}[G]^{n_{1}}\xrightarrow{\phi_{1}}\mathbb{F}_{p}\xrightarrow{\phi_{0}}0$

with $R_{k}$ finitely generated. The relevant part of the sequence for
calculation of $H_{k}(G;*)$ is the following exact sequence

$0\to R_{k}\to\mathbb{F}_{p}[G]^{n_{k}}\to R_{k-1}\to 0,$

($R_{k-1}=\operatorname{Im}\phi_{k}=\ker\phi_{k-1}$), because for any right
$\mathbb{F}_{p}[G]$-module $M$ we have

$0\to H_{k}(G;M)\to M\otimes_{\mathbb{F}_{p}[G]}R_{k}\to M^{n_{k}}\to
M\otimes_{\mathbb{F}_{p}[G]}R_{k-1}\to 0$

Then we obtain that

$\beta_{k}^{\operatorname{mod}p}(G)=\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}}H_{k}(G;\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]})=\dim_{\mathbb{F}_{p}[G]}R_{k}-{n_{k}}+\dim_{\mathbb{F}_{p}[G]}R_{k-1}\textrm{\
and\ }$
$\dim_{\mathbb{F}_{p}}H_{k}(G_{j};\mathbb{F}_{p})=\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[G_{j}]}R_{k})-n_{k}|G:G_{j}|+\dim_{\mathbb{F}_{p}}(\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}[G_{j}]}R_{k-1}).$

Thus, Proposition 3.5 implies the proposition.∎

In this paper we will work only with $\beta_{1}^{\operatorname{mod}p}(G)$.
Observe that in this case, if $G$ is infinite, we have the formula

$\beta_{1}^{\operatorname{mod}p}(G)=\dim_{\mathbb{F}_{p}[G]}I_{G}-1.$

Also observe that if $A$ is a non-trivial torsion-free abelian group, then
since
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[A]}=\mathcal{Q}(\mathbb{F}_{p}[A])$
is the field of fractions of $\mathbb{F}_{p}[A]$,

(14)
$\beta_{1}^{\operatorname{mod}p}(A)=\dim_{\mathbb{F}_{p}[A]}I_{A}-1=\dim_{\mathcal{Q}(\mathbb{F}_{p}[A])}(\mathcal{Q}(\mathbb{F}_{p}[A])\otimes_{\mathbb{F}_{p}[A]}I_{A})-1=0$

###  3.5. Strong embeddings into free pro-$p$ groups

Assume that a finitely generated group $G$ is a subgroup of a free pro-$p$
group $\mathbf{F}$. Since a closed subgroup of a free pro-$p$ group is free
(see [56, Corollary 7.7.5]), changing $\mathbf{F}$ by the closure of $G$ in
$\mathbf{F}$, we may assume that $G$ is dense in $\mathbf{F}$. Let
$\mathbf{F}=\mathbf{N}_{1}>\mathbf{N}_{2}>\mathbf{N}_{3}>\ldots$ be a chain of
open normal subgroups of $\mathbf{F}$ with trivial intersection and put
$G_{j}=G\cap\mathbf{N}_{j}$. Observe that the closure of $G_{j}$ in
$\mathbf{F}$ is equal to $\mathbf{N}_{j}$, and so

(15)
$|G_{j}:G_{j}^{p}[G_{j},G_{j}]|\geq|\mathbf{N}_{j}:\mathbf{N}_{j}^{p}[\mathbf{N}_{j},\mathbf{N}_{j}]|.$

Denote by $d(\mathbf{F})$ the number of profinite generators of $\mathbf{F}$.
By [56, Theorem 7.8.1],

$d(\mathbf{F})=\log_{p}|\mathbf{F}:\mathbf{F}^{p}[\mathbf{F},\mathbf{F}]|$

and by the index Schreier formula

$d(\mathbf{N}_{j})=(d(\mathbf{F})-1)|\mathbf{F}:\mathbf{N}_{j}|+1.$

Therefore, we obtain

$\dim_{\mathbb{F}_{p}}H_{1}(G_{j};\mathbb{F}_{p})=\log_{p}|G_{j}:G_{j}^{p}[G_{j},G_{j}]|\geq\log_{p}|\mathbf{N}_{j}:\mathbf{N}_{j}^{p}[\mathbf{N}_{j},\mathbf{N}_{j}]|=\\\
d(\mathbf{N}_{j})=(d(\mathbf{F})-1)|\mathbf{F}:\mathbf{N}_{j}|+1=(d(\mathbf{F})-1)|G:G_{j}|+1.$

Thus, Proposition 3.6 implies the following corollary.

######  Corollary 3.7.

Let $G$ be a finitely generated dense subgroup of a free pro-$p$ group
$\mathbf{F}$. Then

(16) $\dim_{\mathbb{F}_{p}G}I_{G}=\beta_{1}^{\operatorname{mod}p}(G)+1\geq
d(\mathbf{F}).$

This result suggests the following definition.

######  Definition 3.8.

We say that an embedding $G\hookrightarrow\mathbf{F}$ of a finitely generated
group $G$ into a free pro-$p$ group $\mathbf{F}$ is strong if $G$ is dense in
$\mathbf{F}$ and
$\dim_{\mathbb{F}_{p}G}I_{G}=\beta_{1}^{\operatorname{mod}p}(G)+1=d(\mathbf{F})$.

A finitely generated group $G$ is called strongly embeddable in a free pro-$p$
group (SE($p$) for simplicity) if there are a free pro-$p$ group $\mathbf{F}$
and a strong embedding $G\hookrightarrow\mathbf{F}$.

Let $G$ be a parafree group. Observe that $G$ is residually-$p$ for every
prime $p$. Thus, if $G$ is finitely generated, its pro-$p$ completion
$G_{\widehat{p}}$ is a finitely generated free pro-$p$ group and $G$ is a
subgroup of $G_{\widehat{p}}$. In this case the inequality (15) is an
equality, and so in the same way as we obtained the inequality (16), we obtain
that $\beta_{1}^{\operatorname{mod}p}(G)=d(G_{\widehat{p}})-1$. Therefore, the
embedding $G\hookrightarrow G_{\widehat{p}}$ is strong. Thus, all finitely
generated parafree groups are SE($p$). On the other hand, not every subgroup
of a parafree group is SE($p$). For example, the fundamental group of an
oriented surface of genus greater than 1 is not SE($p$). However the
fundamental group of an oriented surface of genus greater than 2 can be
embedded in a parafree group (see [8, Section 4.1]).

By [13, Proposition 7.5], if $G$ is a finitely generated dense subgroup of a
finitely generated free pro-$p$ group $\mathbf{F}$, then $b_{1}^{(2)}(G)\geq
d(\mathbf{F})-1$. On the other hand, by [21, Theorem 1.6] and Proposition 3.6
, $b_{1}^{(2)}(G)\leq\beta_{1}^{\operatorname{mod}p}(G)$. Thus, if
$G\hookrightarrow\mathbf{F}$ is a strong embedding, we have
$b_{1}^{(2)}(G)=\beta_{1}^{\operatorname{mod}p}(G)=d(\mathbf{F})-1$.

##  4\. On $\mathcal{D}$-torsion-free modules.

###  4.1. General results

Let $R$ be a ring and let $R\hookrightarrow\mathcal{D}$ be an embedding of $R$
into a division ring $\mathcal{D}$. Let $M$ be a left $R$-module. We say that
$M$ is $\mathcal{D}$-torsion-free if the canonical map

$M\to\mathcal{D}\otimes_{R}M,\ m\mapsto 1\otimes m,$

is injective. The following lemma describes several equivalent definitions of
$\mathcal{D}$-torsion-free modules.

######  Lemma 4.1.

The following statements for a left $R$-module $M$ are equivalent.

  1. (a)

$M$ is $\mathcal{D}$-torsion-free.

  2. (b)

There are a $\operatorname{\mathcal{D}}$-module $N$ and an injective
homomorphism $\varphi:M\to N$ of $R$-modules.

  3. (c)

For any $0\neq m\in M$, there exists a homomorphism of $R$-modules
$\varphi:M\to\operatorname{\mathcal{D}}$, such that $\varphi(m)\neq 0$.

###### Proof.

The proof is straightforward and we leave it to the reader.∎

Let $M$ be a left $R$-module. Recall that we use
$\dim_{\operatorname{\mathcal{D}}}M$ to denote the dimension of
$\operatorname{\mathcal{D}}\otimes_{R}M$ as a
$\operatorname{\mathcal{D}}$-module. Observe that if
$\dim_{\operatorname{\mathcal{D}}}M$ is finite, it is also equal to the
dimension of $\operatorname{Hom}_{R}(M,\operatorname{\mathcal{D}})$ as a right
$\mathcal{D}$-module.

######  Lemma 4.2.

Let $0\to M_{1}\to M_{2}\to M_{3}\to 0$ be an exact sequence of $R$-modules.
Assume that

  1. (1)

$M_{1}$ and $M_{3}$ are $\operatorname{\mathcal{D}}$-torsion-free,

  2. (2)

$\dim_{\operatorname{\mathcal{D}}}M_{1}$ and
$\dim_{\operatorname{\mathcal{D}}}M_{3}$ are finite and

  3. (3)

$\dim_{\operatorname{\mathcal{D}}}M_{2}=\dim_{\operatorname{\mathcal{D}}}M_{1}+\dim_{\operatorname{\mathcal{D}}}M_{3}$.

Then $M_{2}$ is also $\operatorname{\mathcal{D}}$-torsion-free.

###### Proof.

We are going to use the third characterization of
$\operatorname{\mathcal{D}}$-torsion-free modules from Lemma 4.1. Consider the
following exact sequence of right $\mathcal{D}$-modules.

$0\to\operatorname{Hom}_{R}(M_{3},\operatorname{\mathcal{D}})\to\operatorname{Hom}_{R}(M_{2},\operatorname{\mathcal{D}})\to\operatorname{Hom}_{R}(M_{1},\operatorname{\mathcal{D}}).$

Since
$\dim_{\operatorname{\mathcal{D}}}M_{2}=\dim_{\operatorname{\mathcal{D}}}M_{1}+\dim_{\operatorname{\mathcal{D}}}M_{3}$,
the last map is surjective.

Let $m\in M_{2}$. If $m\not\in M_{1}$, then since $M_{3}$ is
$\mathcal{D}$-torsion-free, there exists
$\varphi\in\operatorname{Hom}_{R}(M_{2}/M_{1},\operatorname{\mathcal{D}})$
such that $\varphi(m+M_{1})\neq 0$. Hence there exists
$\widetilde{\varphi}\in\operatorname{Hom}_{R}(M_{2},\operatorname{\mathcal{D}})$,
such that $\widetilde{\varphi}(m)\neq 0$.

If $m\in M_{1}$, then since $M_{1}$ is $\mathcal{D}$-torsion-free, there
exists $\varphi\in\operatorname{Hom}_{R}(M_{1},\operatorname{\mathcal{D}})$
such that $\varphi(m)\neq 0$. Using that the restriction map

$\operatorname{Hom}_{R}(M_{2},\operatorname{\mathcal{D}})\to\operatorname{Hom}_{R}(M_{1},\operatorname{\mathcal{D}})$

is surjective, we obtain again that there exists
$\widetilde{\varphi}\in\operatorname{Hom}_{R}(M_{2},\operatorname{\mathcal{D}})$,
such that $\widetilde{\varphi}(m)\neq 0$.∎

In the calculations of $\dim_{\operatorname{\mathcal{D}}}$ the following
elementary lemma will be useful.

######  Lemma 4.3.

Let $\operatorname{\mathcal{D}}$ be a division $R$-ring and $M$ be a
$\operatorname{\mathcal{D}}$-torsion-free $R$-module of finite
$\operatorname{\mathcal{D}}$-dimension. Let $L$ be a non-trivial $R$-submodule
of $M$. Then
$\dim_{\operatorname{\mathcal{D}}}(M/L)<\dim_{\operatorname{\mathcal{D}}}M$.
Moreover, if $\dim_{\operatorname{\mathcal{D}}}L=1$, then
$\dim_{\operatorname{\mathcal{D}}}(M/L)=\dim_{\operatorname{\mathcal{D}}}M-1$.

###### Proof.

Since $M$ is $\operatorname{\mathcal{D}}$-torsion-free,
$\operatorname{\mathcal{D}}\otimes_{R}(M/L)$ is a proper quotient of
$\operatorname{\mathcal{D}}\otimes_{R}M$. Hence
$\dim_{\operatorname{\mathcal{D}}}M/L<\dim_{\operatorname{\mathcal{D}}}M$.

Now assume that $\dim_{\operatorname{\mathcal{D}}}L=1$. In this case
$\dim_{\operatorname{\mathcal{D}}}(M/L)\geq\dim_{\operatorname{\mathcal{D}}}M-1$.
Therefore,
$\dim_{\operatorname{\mathcal{D}}}(M/L)=\dim_{\operatorname{\mathcal{D}}}M-1$.∎

###  4.2. $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-
free modules

Let $\mathbf{F}$ be a finitely generated free pro-$p$ group. The main purpose
of this subsection is to prove the following result.

######  Proposition 4.4.

Assume that $1\neq z\in\mathbf{F}$ is not a proper $p$-power of an element of
$\mathbf{F}$. Denote by $\mathbf{Z}$ the closed subgroup of $\mathbf{F}$
generated by $z$. Then the module $I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$
is $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free.

Before proving the proposition we have to establish several preliminary
results.

######  Lemma 4.5.

Let $\mathbf{H}$ be an open subgroup of $\mathbf{F}$ Let $M$ be a
$\mathbb{F}_{p}[[\mathbf{H}]]$-module. Then
$\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free if
and only if $M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$-torsion-free.

###### Proof.

Assume that $M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$-torsion-free. We
have that the map
$M\to\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M$
is injective. Then, since $\mathbb{F}_{p}[[\mathbf{F}]]$ is a free right
$\mathbb{F}_{p}[[\mathbf{H}]]$-module, the map

$\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M\xrightarrow{\alpha}\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)$

is also injective.

Consider the canonical isomorphism between tensor products

$\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M)\xrightarrow{\beta}(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]})\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M.$

By Propositopn 3.4(c),

$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\cong\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$

as $(\mathbb{F}_{p}[[\mathbf{F}]],\mathbb{F}_{p}[[\mathbf{H}]])$-bimodules.
Thus, there exists an isomorphism of $\mathbb{F}_{p}[[\mathbf{F}]]$-modules

$(\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H})]]})\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M\xrightarrow{\gamma}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M.$

We put $\varphi=\gamma\circ\beta\circ\alpha$ and apply Lemma 4.1. Since
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M$
is a $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-module and
$\varphi$ is an injective $\mathbb{F}_{p}[[\mathbf{F}]]$-homomorphism,
$\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free.

Another direction of the proposition is clear because $M$ is a
$\mathbb{F}_{p}[[\mathbf{H}]]$-submodule of
$\mathbb{F}_{p}[[\mathbf{F}]]\otimes_{\mathbb{F}_{p}[[\mathbf{H}]]}M$.∎

######  Lemma 4.6.

Let $\mathbf{H}$ be an open normal subgroup of $\mathbf{F}$ and assume that
$1\neq z\in\mathbf{H}$. Let $\mathbf{Z}$ be the closed subgroup of
$\mathbf{H}$ generated by $z$. Then the following are equivalent.

  1. (a)

The $\mathbb{F}_{p}[[\mathbf{F}]]$-module
$I_{\mathbf{H}}^{\mathbf{F}}/I^{\mathbf{F}}_{\mathbf{Z}}$ is not
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free.

  2. (b)

The $\mathbb{F}_{p}[[\mathbf{H}]]$-module
$I_{\mathbf{H}}/I^{\mathbf{H}}_{\mathbf{Z}}$ is not
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$-torsion-free.

  3. (c)

There are $a\in I_{\mathbf{H}}^{\mathbf{F}}$ and $b\in I_{\mathbf{F}}$ such
that $ba=z-1$.

  4. (d)

There are $a\in I_{\mathbf{H}}^{\mathbf{F}}$ and $b\in I_{\mathbf{F}}$ such
that $ab=z-1$.

###### Proof.

(a)$\Longleftrightarrow$(b): This follows from Lemma 4.5 and Lemma 2.11.

(c)$\Longrightarrow$(a): Put
$N=\mathbb{F}_{p}[[\mathbf{F}]]a/I_{\mathbf{Z}}^{\mathbf{F}}$, Then

$N=\mathbb{F}_{p}[[\mathbf{F}]]a/\mathbb{F}_{p}[[\mathbf{F}]](z-1)=\mathbb{F}_{p}[[\mathbf{F}]]a/\mathbb{F}_{p}[[\mathbf{F}]]ba.$

Since $b$ is not invertible in $\mathbb{F}_{p}[[\mathbf{F}]]$, $N$ is a non-
trivial submodule of $I_{\mathbf{H}}^{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$
and since $\mathbb{F}_{p}[[\mathbf{F}]]$ does not have non-trivial zero-
divisors

$\mathbb{F}_{p}[[\mathbf{F}]]a/\mathbb{F}_{p}[[\mathbf{F}]]ba\cong\mathbb{F}_{p}[[\mathbf{F}]]/\mathbb{F}_{p}[[\mathbf{F}]]b.$

Clearly
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}N=0$,
and so (a) holds.

(a)$\Longrightarrow$(c): We put
$M=I_{\mathbf{H}}^{\mathbf{F}}/I^{\mathbf{F}}_{\mathbf{Z}}$ and let
$\phi:M\to\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M$
be the canonical map. Let $\overline{L}=L/I_{\mathbf{Z}}^{\mathbf{F}}$ be the
kernel of $\phi$ and $\overline{M}$ the image of $\phi$. Hence we have the
exact sequence

$0\to L\to I_{\mathbf{H}}^{\mathbf{F}}\to\overline{M}\to 0.$

After applying
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}$
we obtain the exact sequence

(17)
$0\to\operatorname{Tor}_{1}^{\mathbb{F}_{p}[[\mathbf{F}]]}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]},\overline{M})\to\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}L\to\\\
(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]})^{d(\mathbf{H})}\to\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}\overline{M}\to
0.$

By Corrollary 2.12, the $\mathbb{F}_{p}[[\mathbf{F}]]$-module
$I_{\mathbf{H}}^{\mathbf{F}}$ is free of rank $d(\mathbf{H})$ and the
$\mathbb{F}_{p}[[\mathbf{F}]]$-module $I_{\mathbf{Z}}^{\mathbf{F}}$ is cyclic.
Thus, by Lemma 4.3,

$\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M)=d(\mathbf{H})-1.$

Therefore,

$\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}\overline{M})=\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}M)=d(\mathbf{H})-1.$

Observe also that by Proposition 3.2,
$\operatorname{Tor}_{1}^{\mathbb{F}_{p}[[\mathbf{F}]]}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]},\overline{M})=0$.
Therefore, (17) implies that

$\dim_{\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}}(\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}\otimes_{\mathbb{F}_{p}[[\mathbf{F}]]}L)=1.$

By [29, Lemma 3.1] a profinite submodule of a free profinite
$\mathbb{F}_{p}[[\mathbf{F}]]$-module is again free. Hence $L$ is a free
profinite $\mathbb{F}_{p}[[\mathbf{F}]]$-module, and so $L$ should be a cyclic
$\mathbb{F}_{p}[[\mathbf{F}]]$-module. We write
$L=\mathbb{F}_{p}[[\mathbf{F}]]a$ for some $a\in I_{\mathbf{H}}^{\mathbf{F}}$.
Then there exists $b\in\mathbb{F}_{p}[[\mathbf{F}]]$ such that $ba=z-1$. By
our assumption $L\neq I_{\mathbf{Z}}^{\mathbf{F}}$. Thus, $b$ is not
invertible, and so $b\in I_{\mathbf{F}}$.

(c)$\Longrightarrow$(d): The map $g\mapsto g^{-1}$ on $\mathbf{F}$ can be
extended to a continuous anti-isomorphism
$\alpha:\mathbb{F}_{p}[[\mathbf{F}]]\to\mathbb{F}_{p}[[\mathbf{F}]]$. If
$z-1=ba$, then $z^{-1}-1=\alpha(z-1)=\alpha(a)\alpha(b)$ and so
$z-1=(-z\alpha(a))\alpha(b)$. Now note that $-z\alpha(a)\in
I_{\mathbf{H}}\mathbb{F}_{p}[[\mathbf{F}]]$ and $\alpha(b)\in I_{\mathbf{F}}$.
Since $\mathbf{H}$ is normal in $\mathbf{F}$,
$I_{\mathbf{H}}\mathbb{F}_{p}[[\mathbf{F}]]=I_{\mathbf{H}}^{\mathbf{F}}$, and
we obtain (d).

(d)$\Longrightarrow$(c): It is proved in the same way as
(c)$\Longrightarrow$(d).∎

Now we are ready to prove Proposition 4.4.

###### Proof of Proposition 4.4.

If $\mathbf{F}$ is cyclic, then $I_{\mathbf{F}}=I_{\mathbf{Z}}^{F}$ and so
$I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free.

Now we assume that $\mathbf{F}$ is not cyclic. There exists a normal open
subgroup $\mathbf{N}$ of $\mathbf{F}$ such that $z\mathbf{N}$ is not a
$p$-power in $\mathbf{F}/\mathbf{N}$. We will prove the proposition by
induction on $|\mathbf{F}/\mathbf{N}|$.

If $\mathbf{F}/\mathbf{N}$ is cyclic, then $z\not\in\Phi(\mathbf{F})$ and so
$z$ is a member of a free generating system of $\mathbf{F}$. If
$\\{z,x_{1},\ldots.x_{k}\\}$ is a free generating set of $F$, then

$I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}=(\mathbb{F}_{p}[[\mathbf{F}]](z-1)\oplus(\oplus_{i=1}^{k}\mathbb{F}_{p}[[\mathbf{F}]](x_{i}-1)))/\mathbb{F}_{p}[[\mathbf{F}]](z-1)\cong\mathbb{F}_{p}[[\mathbf{F}]]^{k}$

is a free $\mathbb{F}_{p}[[\mathbf{F}]]$-module and we are done.

Assume now that $\mathbf{F}/\mathbf{N}$ is not cyclic. Since
$\Phi(\mathbf{F}/\mathbf{N})=\mathbf{N}\Phi(\mathbf{F})/\mathbf{N}$, the
pro-$p$ group $\mathbf{F}/\mathbf{N}\Phi(\mathbf{F})$ is not cyclic as well.

Let $\mathbf{M}$ be the closed subgroup of $\mathbf{F}$ containing the
commutator subgroup $[\mathbf{F},\mathbf{F}]$ and the element $z$ and such
that $\mathbf{M}/([\mathbf{F},\mathbf{F}]\mathbf{Z})$ is the torsion part of
$\mathbf{F}/([\mathbf{F},\mathbf{F}]\mathbf{Z})$. Since $\mathbf{Z}$ is cyclic
and $\mathbf{F}/([\mathbf{F},\mathbf{F}])$ is torsion-free and abelian,
$\mathbf{M}\Phi(\mathbf{F})/\Phi(\mathbf{F})$ is non-trivial cyclic (if
$z\not\in\Phi(\mathbf{F})$) or trivial (if $z\in\Phi(\mathbf{F})$). Therefore,
since $\mathbf{F}/\mathbf{N}\Phi(\mathbf{F})$ is not cyclic,
$\mathbf{M}\mathbf{N}$ is a proper subgroup of $\mathbf{F}$.

By the construction of $\mathbf{M}$,
$\mathbf{F}/\mathbf{M}\cong\mathbb{Z}_{p}^{k}$ for some $k\geq 1$. Since
$\mathbf{M}\mathbf{N}$ is a proper subgroup of $\mathbf{F}$,
$\mathbf{M}\mathbf{N}/\mathbf{M}$ is a proper subgroup of
$\mathbf{F}/\mathbf{M}$. Therefore, there exists a surjective map
$\sigma:\mathbf{F}\to\mathbb{Z}_{p}$ such that $\mathbf{M}\leq\ker\sigma$ and
$\mathbf{N}\ker\sigma\neq\mathbf{F}$. We put $\mathbf{H}=\mathbf{N}\ker\sigma$
and extend $\sigma$ to the map
$\widetilde{\sigma}:\mathbb{F}_{p}[[\mathbf{F}]]\to\mathbb{F}_{p}[[\mathbb{Z}_{p}]]$.
Observe that $\ker\widetilde{\sigma}=I_{\ker\sigma}^{\mathbf{F}}$.

By way of contradiction, assume that
$I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$ is not
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free. Then
by Lemma 4.6, there are $a,b\in I_{\mathbf{F}}$ such that $ab=z-1$. Applying
$\widetilde{\sigma}$ we obtain that
$\widetilde{\sigma}(a)\widetilde{\sigma}(b)=0$. Since
$\mathbb{F}_{p}[[\mathbb{Z}_{p}]]$ is a domain, either $a$ or $b$ lie in
$\ker\widetilde{\sigma}=I_{\ker\sigma}^{\mathbf{F}}\subset
I_{\mathbf{H}}^{\mathbf{F}}$. Applying again Lemma 4.6, we conclude that
$I_{\mathbf{H}}/I_{\mathbf{Z}}^{\mathbf{H}}$ is not
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$-torsion-free.

However, observe that $\mathbf{N}$ is also a normal subgroup of $\mathbf{H}$,
$z\mathbf{N}$ is not a $p$-power in $\mathbf{H}/\mathbf{N}$ and
$|\mathbf{H}/\mathbf{N}|<|\mathbf{F}/\mathbf{N}|$. Thus, we can apply the
inductive assumption and conclude that
$I_{\mathbf{H}}/I_{\mathbf{Z}}^{\mathbf{H}}$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{H}]]}$-torsion-free. We
have arrived to a contradiction. Thus,
$I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-torsion-free. ∎

###  4.3. $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free
modules

In this subsection we assume that $\mathbf{F}$ is a finitely generated free
pro-$p$ group and $G$ is an (abstract) dense finitely generated subgroup of
$\mathbf{F}$. First we prove the following analogue of Lemma 4.5.

######  Lemma 4.7.

Let $H$ be a subgroup of $G$ and let $M$ be a
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[H]}$-torsion-free left
$\mathbb{F}_{p}[H]$-module. Then
$\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free.

###### Proof.

Let $\operatorname{\mathcal{D}}_{H}$ be the division closure of
$\mathbb{F}_{p}[H]$ in $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$.
Observe that $\operatorname{\mathcal{D}}_{H}$ and
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[H]}$ are isomorphic as
$\mathbb{F}_{p}[H]$-rings (it follows, for example, from Proposition 3.5).

We have that the map
$M\to\operatorname{\mathcal{D}}_{H}\otimes_{\mathbb{F}_{p}[H]}M$ is injective.
Then, since $\mathbb{F}_{p}[G]$ is a free right $\mathbb{F}_{p}[H]$-module,
the map

$\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}M\xrightarrow{\alpha}\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}(\operatorname{\mathcal{D}}_{H}\otimes_{\mathbb{F}_{p}[H]}M)$

is also injective.

Consider the canonical isomorphism between tensor products

$\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}(\operatorname{\mathcal{D}}_{H}\otimes_{\mathbb{F}_{p}[H]}M)\xrightarrow{\beta}(\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}\operatorname{\mathcal{D}}_{H})\otimes_{\mathbb{F}_{p}[H]}M.$

By [23], $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$ is strongly Hughes-
free. This means that the canonical map of
$(\mathbb{F}_{p}[G],\mathbb{F}_{p}[H])$-bimodules

$\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}\operatorname{\mathcal{D}}_{H}\to\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$

is injective. Moreover, the image of
$\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}\operatorname{\mathcal{D}}_{H}$
is a direct summand of $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$ as a
right $\operatorname{\mathcal{D}}_{H}$-submodule (and so, it is also a direct
summand as a right $\mathbb{F}_{p}[H]$-submodule). Thus, the following
canonical map of $\mathbb{F}_{p}[G]$-modules

$(\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}\operatorname{\mathcal{D}}_{H})\otimes_{\mathbb{F}_{p}[H]}M\xhookrightarrow{\gamma}\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}\otimes_{\mathbb{F}_{p}[H]}M$

is injective. We put $\varphi=\gamma\circ\beta\circ\alpha$ and apply Lemma
4.1. Since
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}\otimes_{\mathbb{F}_{p}[H]}M$
is a $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-module and $\varphi$ is
an injective $\mathbb{F}_{p}[G]$-homomorphism,
$\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free.

∎

Now we can present our first example of a
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free
$\mathbb{F}_{p}[G]$-module.

######  Proposition 4.8.

Let $H$ be a non-trivial subgroup of $G$ and $A$ a maximal abelian subgroup of
$H$. Then the $\mathbb{F}_{p}[G]$-module $I^{G}_{H}/I^{G}_{A}$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free.

###### Proof.

From Lemma 2.1 we know that

$I^{G}_{H}/I^{G}_{A}\cong\mathbb{F}_{p}[G]\otimes_{\mathbb{F}_{p}[H]}(I_{H}/I^{H}_{A}).$

Thus, in view of Lemma 4.7, it is enough to show that $I_{H}/I^{H}_{A}$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[H]}$-torsion-free.

Let $\mathbf{Z}=C_{\mathbf{F}}(A)$. Since $\mathbf{F}$ is a free pro-$p$
group, a centralizator of a non-trivial element is maximal cyclic pro-$p$
subgroup. Therefore, since $A$ is abelian and non-trivial, $\mathbf{Z}$ is a
maximal cyclic pro-$p$ subgroup of $\mathbf{F}$.

######  Claim 4.9.

The canonical map
$\mathbb{F}_{p}[H/A]\to\mathbb{F}_{p}[[\mathbf{F}/\mathbf{Z}]]$ is injective.

###### Proof.

Since $A$ is maximal abelian in $H$, we have that $A=\mathbf{Z}\cap H$. Hence
the obvious map $\mathbb{F}_{p}[H/A]\to\mathbb{F}_{p}[\mathbf{F}/\mathbf{Z}]$
in injective. The map
$\mathbb{F}_{p}[\mathbf{F}/\mathbf{Z}]\to\mathbb{F}_{p}[[\mathbf{F}/\mathbf{Z}]]$
is also injective, because $\mathbf{Z}$ is closed in $\mathbf{F}$. This
finishes the proof of the claim.∎

Observe that $\mathbb{F}_{p}[H/A]\cong\mathbb{F}_{p}[H]/I_{A}^{H}$ and
$\mathbb{F}_{p}[[\mathbf{F}]]/I_{\mathbf{Z}}^{\mathbf{F}}\cong\mathbb{F}_{p}[[\mathbf{F}/\mathbf{Z}]]$.
Therefore, by Claim 4.9, $I_{H}/I^{H}_{A}$ is a $\mathbb{F}_{p}[H]$-submodule
of $I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$. By Proposition 4.4, we can
embed $I_{\mathbf{F}}/I_{\mathbf{Z}}^{\mathbf{F}}$ in a
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-module. By
Proposition 3.5, every
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[[\mathbf{F}]]}$-module is also a
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[H]}$-module. Therefore, by Lemma
4.1, $I_{H}/I^{H}_{A}$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[H]}$-torsion-free. ∎

The following proposition shows another example of a
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free
$\mathbb{F}_{p}[G]$-module. This is the main result of this section.

######  Proposition 4.10.

Let $\mathbf{F}$ be a finitely generated free pro-$p$ group and let $G$ be an
(abstract) dense finitely generated subgroup of $\mathbf{F}$. Let $H$ be a
non-trivial subgroup of $G$ and $A$ a maximal abelian subgroup of $H$. Let $B$
be an abelian subgroup of $G$ containing $A$. We put

$J=\\{(x,-x)\in I^{G}_{H}\oplus I^{G}_{B}:\ x\in I^{G}_{A}\\}.$

Then $M=(I^{G}_{H}\oplus I^{G}_{B})/J$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free and
$\dim_{\mathbb{F}_{p}[G]}M=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}$.

###### Proof.

Let $L=(I^{G}_{A}\oplus I^{G}_{B})/J\leq M$. Then $L\cong I^{G}_{B}$ is a
submodule of $\mathbb{F}_{p}[G]$, and so it is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free. The
$\mathbb{F}_{p}[G]$-module $M/L$ is isomorphic to $I_{H}^{G}/I^{G}_{A}$, and
so it is $\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free by
Proposition 4.8.

By (14), $\dim_{\mathbb{F}_{p}[A]}I_{A}=1$. Therefore, by Proposition 3.4(b),

$\dim_{\mathbb{F}_{p}[G]}I^{G}_{A}=\dim_{\mathbb{F}_{p}[A]}I_{A}=1.$

In the same way we obtain that $\dim_{\mathbb{F}_{p}[G]}I^{G}_{B}=1$.

Since $\dim_{\mathbb{F}_{p}[G]}(I^{G}_{H}\oplus
I^{G}_{B})=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}+1,$ by Lemma 4.3,

$\dim_{\mathbb{F}_{p}[G]}M=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}+1-1=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}\textrm{\
and}\\\
\dim_{\mathbb{F}_{p}[G]}(I_{H}^{G}/I^{G}_{A})=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}-1=\dim_{\mathbb{F}_{p}[G]}M-1.$

Therefore,

$\dim_{\mathbb{F}_{p}[G]}(M/L)=\dim_{\mathbb{F}_{p}[G]}(I_{H}^{G}/I^{G}_{A})=\dim_{\mathbb{F}_{p}[G]}M-1.$

Thus, we have obtained that $M/L$ and $L$ are
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free and

$\dim_{\mathbb{F}_{p}[G]}M=\dim_{\mathbb{F}_{p}[G]}(M/L)+\dim_{\mathbb{F}_{p}[G]}L.$

Applying Lemma 4.2, we conclude that $M$ is also
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free. ∎

##  5\. Proof of main results

###  5.1. The inductive step in the proof of Theorem 1.2

The following theorem is the main result of the paper. Theorem 1.2 follows
from it directly.

######  Theorem 5.1.

Let $\mathbf{F}$ be a finitely generated free pro-$p$ group and let
$H\hookrightarrow\mathbf{F}$ be a strong embedding of a finitely generated
group $H$. Let $A$ be a maximal abelian subgroup of $H$ and let $B$ be an
abelian finitely generated subgroup of $\mathbf{F}$ containing $A$. Put
$G=\langle H,B\rangle$. Then the canonical homomorphism $H*_{A}B\to G$ is an
isomorphism, and the embedding $G\hookrightarrow\mathbf{F}$ is strong.

###### Proof.

In view of Proposition 2.13 we have to show that $I^{G}_{H}\cap
I_{B}^{G}=I_{A}^{G}$ in $\mathbb{F}_{p}[G]$. Let

$J=\\{(x,-x)\in I^{G}_{H}\oplus I^{G}_{B}:\ x\in I^{G}_{A}\\}\textrm{\ and\
}M=(I^{G}_{H}\oplus I^{G}_{B})/J.$

Then by Proposition 4.10,
$\dim_{\mathbb{F}_{p}[G]}M=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}$. Therefore,

$\dim_{\mathbb{F}_{p}[G]}M=\dim_{\mathbb{F}_{p}[G]}I^{G}_{H}=\dim_{\mathbb{F}_{p}[H]}I_{H}=\beta_{1}^{\operatorname{mod}p}(H)+1=d(\mathbf{F}).$

Since $I_{G}=I_{H}^{G}+I_{B}^{G}$, we have that the natural map $\alpha:M\to
I_{G}$ is surjective. In particular

$\beta^{\operatorname{mod}p}_{1}(G)=\dim_{\mathbb{F}_{p}[G]}I_{G}-1\leq\dim_{\mathbb{F}_{p}[G]}M-1=d(\mathbf{F})-1.$

Thus, using (16) we obtain that
$\beta^{\operatorname{mod}p}_{1}(G)=d(\mathbf{F})-1$ and
$\dim_{\mathbb{F}_{p}[G]}I_{G}=\dim_{\mathbb{F}_{p}[G]}M$. This shows that the
embedding $G\hookrightarrow\mathbf{F}$ is strong.

By Proposition 4.10, $M$ is
$\operatorname{\mathcal{D}}_{\mathbb{F}_{p}[G]}$-torsion-free. Therefore, by
Lemma 4.3, for any proper quotient $\overline{M}$ of $M$,
$\dim_{\mathbb{F}_{p}[G]}\overline{M}<\dim_{\mathbb{F}_{p}[G]}M$. This implies
that $\alpha$ is an isomorphism, and so $I^{G}_{H}\cap I_{B}^{G}=I_{A}^{G}$.
Hence, Proposition 2.13 implies that $G\cong H*_{A}B$.∎

Another direct consequence of Theorem 5.1 is the following corollary.

######  Corollary 5.2.

Let $\mathbf{F}$ be a finitely generated free pro-$p$ group and let
$H\hookrightarrow\mathbf{F}$ be a strong embedding of finitely generated group
$H$. Let $A$ be a maximal abelian subgroup of $H$. Assume that $A$ is finitely
generated. Let $B$ be a finitely generated torsion-free abelian group
containing $A$ and such that $B/A$ has no $p$-torsion. Then there exists an
embedding of $H*_{A}B$ into $\mathbf{F}$ that extends
$H\hookrightarrow\mathbf{F}$. In particular, $H*_{A}B$ is SE($p$) (see
Definition 3.8).

###### Proof.

Let $H\hookrightarrow\mathbf{F}$ be a strong embedding. Since $B/A$ has no
$p$-torsion and $B$ is torsion-free abelian, the embedding of $A$ into
$\mathbf{F}$ can be extended to an embedding of $B$ into $\mathbf{F}$. Now, we
can apply Theorem 5.1.∎

###  5.2. $A$-groups

Let $A$ be a commutative ring. An $A$-group is a group $G$ along with a map
$G\times A\to G$, called “action”, satisfying the following axioms:

  1. (a)

$g^{1_{A}}=g$, $g^{0_{A}}=1$, $1^{\alpha}=1$;

  2. (b)

$g^{\alpha}g^{\beta}=g^{\alpha+\beta}$,
$(g^{\alpha})^{\beta}=g^{\alpha\beta}$;

  3. (c)

$(h^{-1}gh)^{\alpha}=h^{-1}g^{\alpha}h$;

  4. (d)

If $gh=hg$, $(gh)^{\alpha}=g^{\alpha}h^{\alpha}$;

for all $g,h\in G$ and $\alpha,\beta\in A$. This definition generalizes the
definition of $\mathbb{Q}$-group which appears in the introduction.

Given a group $G$ and a commutative ring $A$, an $A$-completion of $G$ is an
$A$-group $G^{A}$ with a group homomorphism $\lambda:G\to G^{A}$ such that
$G^{A}$ is generated by $\lambda(G)$ as an $A$-group and for any $A$-group $H$
and any homomorphism $\phi:G\to H$ there exists a unique $A$-homomorphism
$\psi:G^{A}\to H$ such that $\phi=\psi\circ\lambda$. It is shown in [52,
Theorems 1 and 2] that an $A$-completion of $G$ exists and it is unique up to
an $A$-isomorphism.

An $A$-group $F^{A}(X)$ with the set of $A$-generators $X$ is said to be a
free $A$-group with base $X$, if for every $A$-group $G$ an arbitrary mapping
$\phi_{0}:X\to G$ can be extended to an $A$-homomorphism $\phi:F^{A}(X)\to G$.
Thus, $F^{A}(X)$ is the $A$-completion of the ordinary free group $F(X)$ with
free generating set $X$.

A CSA-group is a group in which the centralizer of every nontrivial element is
abelian and malnormal. In [53, Theorem 5] it is shown that every torsion-free
extension of centralizer of a torsion-free CSA group is again CSA. This is
used in [53, Theorem 8] to describe the group $F^{A}(X)$. In the following
proposition we extract the information that we will need later.

######  Proposition 5.3.

Let $X$ be a finite set and $A$ a commutative ring with a torsion-free
additive group. Then there are subgroups $\\{W_{n}\\}_{n\geq 0}$ of $F^{A}(X)$
such that

  1. (a)

$W_{0}=F(X)$;

  2. (b)

$W_{n+1}=\langle W_{n},z_{n}^{A}\rangle$, where $z_{n}\in W_{n}$ generates a
maximal abelian subgroup in $W_{n}$ and the canonical map $\displaystyle
W_{n}*_{z_{n}=1_{A}}A\to W_{n+1}$ is an isomorphism;

  3. (c)

$F(X)^{A}=\bigcup_{i=0}^{\infty}W_{n}$.

Now we are ready to prove Corollary 1.3.

###### Proof of Corollary 1.3.

If $\phi$ is not injective, then there exists a finitely generated subgroup
$G$ of $F^{\mathbb{Z}_{p}}(X)$ such that $G\cap\ker\phi$ is not trivial. By
Proposition 5.3, there exists a non-negative integer $k$ and a sequence
$G_{0}\leq G_{1}\leq\ldots\leq G_{k}$ of subgroups of $F^{A}(X)$, where

  1. (a)

$G_{0}=F(X_{0})$ is the group generated by a finite subset $X_{0}$ of $X$;

  2. (b)

for $0\leq i\leq k-1$ there exists an element $z_{i}\in G_{i}$, generating a
maximal abelian subgroup in $G_{i}$, and a finitely generated subgroup $T_{i}$
of $(A,+)$ containing $1_{A}$, such that $z_{i}^{T_{i}}\leq G_{i+1}$ and the
canonical map

$G_{i}*_{z_{i}=1_{A}}T_{i}\to G_{i+1}$

is an isomorphism;

  3. (c)

$G$ is a subgroup of $G_{k}$.

Put $H_{i}=\phi(G_{i})$ $A_{i}=\langle\phi(z_{i})\rangle$ and
$B_{i}=\phi(z_{i}^{T_{i}})$.

Let us show by induction on $i$ that $\phi_{|G_{i}}:G_{i}\to H_{i}$ is an
isomorphism. It is clear for $i=0$. Assume we have proved it for $i<k$.
Observe that, since $\phi(z_{i})\neq 1$,

$B_{i}=\phi(z_{i}^{T_{i}})=\phi(z_{i})^{T_{i}}\cong T_{i}\textrm{\ and \
}A_{i}=\phi(\langle z_{i}\rangle)=\phi(G_{i}\cap z_{i}^{T_{i}})=B_{i}\cap
H_{i}.$

Then by Theorem 1.2, $\phi_{|G_{i+1}}:G_{i+1}\to H_{i+1}$ is also an
isomorphism.

Thus, $\ker\phi\cap G_{k}=\\{1\\}$. This is a contradiction. ∎

We will also need the following consequence of Proposition 5.3.

######  Corollary 5.4.

[53, Corollary 5] Let $B$ be a ring with a torsion-free additive group and $A$
a subring of $B$. Then the canonical map $F^{A}(X)\to F^{B}(X)$ is injective.

###  5.3. Proof of Theorem 1.1

In this subsection we will prove that $F^{A}(X)$ is residually nilpotent for
every commutative ring $A$ with a torsion-free additive group. The proof is
based on the following general result communicated to us by a referee.

######  Proposition 5.5.

Let $G$ be a group. Assume that for each $n\geq 0$ there are subgroups $G_{n}$
and $B_{n}$ of $G$ such that

  1. (a)

$G_{n+1}=\langle G_{n},B_{n}\rangle$;

  2. (b)

$B_{n}$ is abelian and $G_{n}$ is residually torsion-free nilpotent;

  3. (c)

$A_{n}=G_{n}\cap B_{n}$ and the canonical map $G_{n}*_{A_{n}}B_{n}\to G_{n+1}$
is an isomorphism;

  4. (d)

$G=\bigcup_{i=0}^{\infty}G_{n}$.

Then $G$ is residually torsion-free nilpotent.

###### Proof.

Let $x\in G$. We want to show that $G$ has a torsion-free nilpotent quotient
such that the image of $x$ in this quotient is not trivial.

Let $x\in G_{m}$ for some $m\geq 0$. Since $G_{m}$ is residually torsion-free
nilpotent, there exists a torsion-free nilpotent group $N$ and the map
$\phi_{m}:G_{m}\to N$ such that $\phi_{m}(x)\neq 1$. We can embed $N$ in its
Malcev $\mathbb{Q}$-completion. Thus, without loss of generality, we may
assume that $N$ is also a $\mathbb{Q}$-group.

We will show that for each $n\geq m$ there exist $\phi_{n}:G_{n}\to N$ such
that the restriction of $\phi_{n}$ on $G_{n-1}$ is $\phi_{n-1}$. We construct
$\phi_{n}$ by induction. Assume that we constructed $\phi_{n}$ for $m\leq
n\leq k$. Since $N$ is a $\mathbb{Q}$-group, we can extend $\phi_{k}$ from
$A_{k}$ to $A_{k}^{\mathbb{Q}}$ and so to $B_{k}^{\mathbb{Q}}$ and $B_{k}$.
Thus we have a homomorphism $\tilde{\phi}_{k}:B_{k}\to N$ that extends
$(\phi_{k})_{|A_{k}}:A_{k}\to N$. Now, by the universal property of the
amalgamated product, there exists a homomorphism $\phi_{k+1}:G_{k+1}\to N$
whose restriction on $G_{k}$ is $\phi_{k}$ and on $B_{k}$ is
$\tilde{\phi}_{k}$.

Let $\phi:G\to N$ be the homomorphism satisfying $\phi(g)=\phi_{n}(g)$ if
$g\in G_{n}$. Then $\phi$ is well-defined and $\phi(x)\neq\\{1\\}$. ∎

######  Corollary 5.6.

Let $X$ be a set and $A$ a countable commutative ring with a torsion-free
additive group. Then $F^{A}(X)$ is residually torsion-free nilpotent.

###### Proof.

First assume that $X$ is finite. Since $A$ is countable, by Proposition 5.3
there are subgroups $\\{G_{n}\\}_{n\geq 0}$ and $\\{B_{n}\\}_{n\geq 0}$ of
$F^{A}(X)$ such that

  1. (a)

$G_{n+1}=\langle G_{n},B_{n}\rangle$;

  2. (b)

$B_{n}$ is abelian and finitely generated;

  3. (c)

$A_{n}=G_{n}\cap B_{n}$ is a maximal abelian subgroup of $G_{n}$ and the
canonical map $G_{n}*_{A_{n}}B_{n}\to G_{n+1}$ is an isomorphism;

  4. (d)

$F^{A}(X)=\bigcup_{i=0}^{\infty}G_{n}$.

By Theorem 1.2, $G_{n}$ are residually torsion-free nilpotent. Thus,
Proposition 5.5 implies that $F^{A}(X)$ is residually torsion-free nilpotent.

Now assume that $X$ is an arbitrary set. Given an element $1\neq g$ of
$F^{A}(X)$, it belongs to $F^{A}(X_{0})$ for some finite subset $X_{0}$ of
$X$, and $F^{A}(X_{0})$ is a retract of $F^{A}(X)$. As we have already proved,
there exists a homomorphism $\phi:F^{A}(X_{0})\to N$, where $N$ is torsion-
free nilpotent such that $\phi(g)\neq 1$. This finishes the proof. ∎

######  Remark 5.7.

The proof of the previous corollary can also be adapted to the case where $A$
is not countable. We prove it, using a different method, in Corollary 5.9.

Let $X=\\{x_{i}\colon i\in I\\}$ and $Y=\\{y_{i}\colon i\in I\\}$ be two sets
indexed by the elements of a set $I$. Let $A$ be a commutative ring. We say
that $A$ is a binomial domain if $A$ is a domain and $a\choose n$ belongs to
$A$ for every $a\in A$. For example $\mathbb{Z}_{p}$ and $\mathbb{Q}$-algebras
are binomial domains.

Assume that $A$ is a binomial domain. If $\Delta_{A}$ denotes the ideal of
$A\langle\\!\langle Y\rangle\\!\rangle$ generated by $Y$ then $1+\Delta_{A}$
is a subgroup of the group of units of $A\langle\\!\langle
Y\rangle\\!\rangle$. We can define an action of $A$ on $1+\Delta_{A}$ in the
following way:

$(1+f)^{a}=1+\sum_{n=1}^{\infty}{a\choose n}f^{n}\ (a\in A,f\in\Delta_{A}).$

Then, by [60], $1+\Delta_{A}$ provided with this action is an $A$-group.
Therefore, the map $x_{i}\mapsto 1+y_{i}$ can be uniquely extended to an
$A$-homomorphism

$\phi_{A}:F^{A}(X)\to 1+\Delta_{A}$

called the Magnus representation of $F^{A}(X)$. Magnus (see, for example,
[47]) proved that $\phi_{\mathbb{Z}}$ is injective. Now we prove the main
result of this subsection.

######  Theorem 5.8.

Let $X$ be an arbitrary set and $A$ be a binomial domain.Then the map
$\phi_{A}:F^{A}(X)\to 1+\Delta_{A}$ is injective.

###### Proof.

By Corollary 5.4, we can assume that $A$ is a $\mathbb{Q}$-algebra. If
$\ker\phi_{A}$ is not trivial, then there exists a finite subset $X_{0}$ of
$X$ and countable $\mathbb{Q}$-subalgebra of $A$ such that the kernel of the
map $F^{A_{0}}(X_{0})\to 1+\Delta_{A_{0}}$ is not trivial. Hence, we can also
assume that $X$ is finite and $A$ is countable.

Let $F_{n}=F(X)/\gamma_{n}(F(X))$. Denote by $K_{n}$ the kernel of the
canonical map $F^{A}(X)\to F_{n}^{A}$. Using [52, Property 2], we obtain that
$F_{n}^{A}$ is nilpotent of nilpotency class $n-1$. By Proposition 5.6,
$F^{A}(X)$ is residually torsion-free nilpotent. Hence
$\cap_{n}K_{n}=\\{1\\}$.

The map $F_{n}\to 1+\Delta_{A}/1+(\Delta_{A})^{n}$ induces the map
$\phi_{n}:F_{n}^{A}\to 1+\Delta_{A}/1+(\Delta_{A})^{n}$. It is known that
$\phi_{n}$ is injective (see, for example, [22, Theorem 23]). Thus $K_{n}$ is
the kernel of the canonical map $F^{A}(X)\to 1+\Delta_{A}/1+(\Delta_{A})^{n}$.
This implies that $\phi_{A}$ is injective. ∎

The following definition has been suggested to us by a referee. We say that an
$A$-group $G$ is $A$-torsion-free if for every $1\neq g\in G$, the map $A\to
G$ that sends $a$ to $g^{a}$ is injective.

######  Corollary 5.9.

Let $A$ be a commutative ring with a torsion-free additive group. Then the
group $F^{A}(X)$ is residually-($A$-torsion-free nilpotent).

###### Proof.

By Corollary 5.4, we can assume that $A$ is a $\mathbb{Q}$-algebra. Hence the
corollary follows from Theorem 5.8 because the $A$-groups
$1+\Delta_{A}/1+(\Delta_{A})^{n}$ are $A$-torsion-free and nilpotent. ∎

###  5.4. Proof of Theorem 1.4

The proof of Theorem 1.4 uses a particular case of [53, Theorem 8] that we
describe now.

######  Proposition 5.10.

Let $G$ be a countable torsion-free CSA-group. Then the
$\mathbb{Q}$-completion $G^{\mathbb{Q}}$ of $G$ is a direct union of subgroups
$W_{0}\leq W_{1}\leq\ldots$ such that

  1. (a)

$W_{0}=G$;

  2. (b)

for $i\geq 0$ there exists a maximal abelian subgroup $A_{i}$ of $W_{i}$ such
that $W_{i+1}$ is the image of the canonical map
$W_{i}*_{A_{i}}A_{i}^{\mathbb{Q}}\to G^{\mathbb{Q}}$.

Moreover, if $H$ is a subgroup $G^{\mathbb{Q}}$ and $A$ is a maximal abelian
subgroup of $H$, then the canonical map $H*_{A}A^{\mathbb{Q}}\to
G{{}^{\mathbb{Q}}}$ is injective.

###### Proof of Theorem 1.4.

Our definition of a limit group from the introduction and Proposition 5.3 show
that the limit groups are exactly finitely generated subgroups of
$F^{\mathbb{Z}[t]}(X)$. Since $F^{\mathbb{Z}[t]}(X)$ is a CSA-group, its
$\mathbb{Q}$-completion can be calculated using Proposition 5.10. In
particular, $G^{\mathbb{Q}}$ is a subgroup of the $\mathbb{Q}$-completion of
$F^{\mathbb{Z}[t]}(X)$ and so it is a subgroup of $F^{\mathbb{Q}[t]}(X)$.
Hence by Corollary 5.9, it is residually torsion-free nilpotent. ∎

##  6\. Linearity of free $\mathbb{Q}$-groups and free pro-$p$ groups

We finish this paper with a discussion on another two well-known problems
concerning linearity of free $\mathbb{Q}$-groups and free pro-$p$ groups.

The problem of whether a free $\mathbb{Q}$-group $F^{\mathbb{Q}}(X)$ is linear
appears in [10, Problem F13] and it is attributed to I. Kapovich (see also
[38, Problem 13.39(b)]). The problem of whether a free pro-$p$ group
$\mathbf{F}$ is linear is usually attributed to A. Lubotzky (for example, we
discussed this question in Jerusalem in November, 2001).

In the context of profinite groups, one can consider two kinds of linearity
(see, for example, [28]). On one hand, we say that a profinite group $G$ is
linear if it is linear as an abstract group, that is it has a faithful
representation by matrices of fixed degree over a field. On the other hand,
the concept of $t$-linear profinite group takes into account the topology of
$G$ and means that $G$ can be faithfully represented as a closed subgroup of
the group of invertible matrices of fixed degree over a profinite commutative
ring.

It is commonly believed that a non-abelian free pro-$p$ group is not
$t$-linear (see, [44, Conjecture 3.8], the discussion after [3, Theorem 1.1]
and [59, Section 5.3]). An equivalent reformulation of this statement is that
a $p$-adic analytic pro-$p$ group satisfies a non-trvial pro-$p$ identity. A.
Zubkov [64] proved that if $p>2$, then a non-abelian free pro-$p$ group cannot
be represented by 2-by-2 matrices over a profinite commutative ring. E.
Zelmanov announced that given a fixed $n$, a non-abelian free pro-$p$ group
cannot be represented by $n$-by-$n$ matrices over a profinite commutative ring
for every large enough prime $p>>n$ (see [62, 63]). Recently, D. El-Chai Ben-
Ezra, E. Zelmanov showed that a free pro-2 group cannot be represented by
2-by-2 matrices over a profinite commutative ring of characteristic 2 [20].

Recall that by a result of A. Malcev [48, Theorem IV], a group can be
represented by matrices of degree $n$ over a field if and only if every one of
its finitely generated subgroup has this property. Thus, in order to decide
whether $F^{\mathbb{Q}}(X)$ or $\mathbf{F}$ are linear, we have to analyze the
structure of their finitely generated (abstract) subgroups. In order to apply
the Malcev criterion one should find a uniform $n$ which does not depend on a
finitely generated subgroup. We may ask a weaker question of whether
$F^{\mathbb{Q}}(X)$ and $\mathbf{F}$ are locally linear. Using recent advances
in geometric group theory one can answer this positively in the case of
$F^{\mathbb{Q}}(X)$.

######  Theorem 6.1.

The groups $F^{\mathbb{Q}}(X)$ are locally linear over $\mathbb{Z}$.

###### Proof.

Let $G$ be a finitely generated subgroup of $F^{\mathbb{Q}}(X)$. Then by
Proposition 5.3, there exists $k\geq 0$ and a sequence $G_{0}\leq
G_{1}\leq\ldots\leq G_{k}$ of subgroups of $F^{\mathbb{Q}}(X)$, such that
$G_{0}$ is free, $G_{i+1}$ is obtained from $G_{i}$ by adjoining a root and
$G\leq G_{k}$. By [53, Theorem 5], the groups $G_{i}$ are CSA. Thus, from a
corollary on the page 100 of [11] we obtain that $G$ is hyperbolic. Therefore,
[26, Corollary C] ensures, by induction on $i$, that each $G_{i}$ acts
properly and cocompactly on a CAT(0) cube complex. Hence, by [1, Theorem 1.1],
$G_{k}$ has a finite index subgroup acting faithfully and specially on a
CAT(0) cube complex. Finally from [25, Theorem 1.1] it follows that $G_{k}$,
and so $G$, are linear over $\mathbb{Z}$.∎

I wouldn’t be surprised if the groups $F^{\mathbb{Q}[t]}(X)$ are also locally
linear. However, to decide whether $F^{\mathbb{Q}}(X)$ and
$F^{\mathbb{Q}[t]}(X)$ are linear will require a completely new approach.

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