# Groups elementarily equivalent to the classical matrix groups

Alexei Myasnikov111Address: Stevens Institute of Technology, Department of
Mathematical Sciences, Hoboken, NJ 07087, USA. Email<EMAIL_ADDRESS>,
Mahmood Sohrabi222Address: Stevens Institute of Technology, Department of
Mathematical Sciences, Hoboken, NJ 07087, USA. Email<EMAIL_ADDRESS>
###### Abstract

In this paper we describe all groups that are first-order (elementarily)
equivalent to the classical matrix groups such as $GL_{n}(F),SL_{n}(F)$ and
$T_{n}(F)$ over a field $F$ provided $n\geq 3$.

2010 MSC: 03C60, 20F16

Keywords: First-order equivalence, Elementary equivalence,
Bi-interpretability, Special linear group, General linear group, Abelian
deformation.

To Gerhard Rosenberger as a token of our friendship

##  1 Introduction

In this paper, we solve the first-order classification problem for the
classical matrix groups $GL_{n}(F),SL_{n}(F)$ and $T_{n}(F)$ over fields $F$
provided $n\geq 3$, i.e., we describe the algebraic structure of all groups
$H$ such that $H\equiv GL_{n}(F)$ or $H\equiv SL_{n}(F)$, or $H\equiv
T_{n}(F)$. Note that this gives a far reaching generalization of the famous
Malcev’s theorems stating that $GL_{n}(F)\equiv GL_{m}(L)$ or $SL_{n}(F)\equiv
SL_{m}(L)$ imply that $n=m$ and $F\equiv L$ for all fields $F,L$ and numbers
$n,m\geq 3$ [28]. To prove this, Malcev showed that for
$G\in\\{GL_{n}(F),SL_{n}(F),T_{n}(F)\\}$ the parameter $n$ and the field $F$
can be described by finitely many sentences that hold in $G$. More precisely,
in the modern terminology, he showed that $n$ and $F$ are interpreted in $G$
in some nice “uniform” way (regularly interpreted). Here, we go further and
show that for such groups $G$ not only $n$ and $F$, but also the algebraic
scheme itself $GL_{n},SL_{n}$ or $T_{n}$, perhaps up to a cohomological
abelian deformation, is regularly interpretable in $G$. It follows that any
group $H$ with $H\equiv G$ is an abelian deformation of the same algebraic
group as $G$ over a field $L$ with $L\equiv F$. In fact, in the case
$G=SL_{n}(F)$ no abelian deformations are required, while for $G=GL_{n}(F)$ or
$G=T_{n}(F)$ they are unavoidable. These groups obtained by abelian
deformations from $GL_{n}(L)$ or $T_{n}(L)$ are interesting in their own right
and deserve a separate study.

Our choice of $GL_{n},SL_{n}$ and $T_{n}$ was mostly influenced by the desire
to show various types of algebraic descriptions that may occur when one
describes all models of complete theories of such groups using regular
interpretations. In [31] the authors showed that these methods work for
classical matrix groups over rings of algebraic integers. Recently, Bunina and
Gvozdevsky used this approach successfully for a very large class of Chevalley
groups $G$ on rings, thus solving the first-order classification problem for
such $G$ [12]. These strengthened the earlier results [37] on bi-
interpretability with parameters for Chevalley groups due to Segal and Tent.
Avni, Lubotzky, and Meiri studied the model theory of high-rank arithmetic
groups applying methods of interpretation [2, 3]. On the other hand, some time
ago, the first-order classification problem was solved, using essentially the
same technique, for algebraic nilpotent $k$-groups (unipotent $k$-groups) over
arbitrary fields $k$ of characteristic zero [29]. Later, these results were
extended to arbitrary finitely generated nilpotent groups [32]. The model
theory of unitriangular groups was studied in detail in [4, 5]. All the
results mentioned above indicate that there might be a general approach to the
first-order classification problem of algebraic groups, based on the methods
of interpretations. Furthermore, it seems that these methods and results
provide an outline of the emerging unified model theory of algebraic groups,
which organically includes the theory of abstract isomorphisms of such groups
[6, 7, 8, 9, 10, 11, 13].

###  1.1 General approach

Here, we give a general description of our approach. Throughout this paper, we
assume that $n\geq 3$.

This approach is based on recent advances in the theory of interpretability
(we refer to a survey [25] for details), especially on the techniques of bi-
interpretability [30, 24, 35, 1, 37, 26, 27, 34, 16]. The notion of regular
interpretability is somewhat in the middle between absolute interpretability
and interpretability with parameters. The former one fits extremely well for
questions related to the first-order equivalence, but occurs rather rarely.
The interpretability with parameters is a more common one, but the parameters
are usually not in the language of $G$, which presents a problem when
describing models of $Th(G)$. The regular interpretability (or regular bi-
interpretability) combines the useful properties of both: it can be used in
the first-order classification problem in a way similar to the absolute one,
and occurs almost as often as interpretability with parameters. To go further
we need the following notation (see Section 4): if a structure $\mathcal{A}$
is interpreted in a structure $\mathcal{B}$ by a set of formulas $\Gamma$ and
a tuple of parameters $\bar{p}\in\mathcal{B}^{n}$ we write
$\mathcal{A}\simeq\Gamma(\mathcal{B},\bar{p})$; if there is a formula
$\phi(\bar{x})$ in the language of $\mathcal{B}$ such that
$\mathcal{A}\simeq\Gamma(\mathcal{B},\bar{p})$ for every $\bar{p}$ satisfying
$\phi(\bar{x})\in\mathcal{B}$, then we say that $\mathcal{A}$ is _regularly
interpretable_ in $\mathcal{B}$ and write
$\mathcal{A}\simeq\Gamma(\mathcal{B},\phi)$. However, showing regular
interpretability (regular bi-interpretability) requires one to go deeper into
model theoretic properties of the structures. A bi-interpretability (regular
bi-interpretability) of $\mathcal{A}$ and $\mathcal{B}$ is a very strong
version of a mutual interpretability (regular interpretability) of
$\mathcal{A}$ and $\mathcal{B}$ in each other (see Section 4). The crucial
point in using interpretability in first-order classification problems can be
seen as follows: if $\mathcal{A}\simeq\Gamma(\mathcal{B},\phi)$ and
$\mathcal{B}_{1}\equiv\mathcal{B}$ then for
$\mathcal{A}_{1}\simeq\Gamma(\mathcal{B}_{1},\phi)$ one has
$\mathcal{A}_{1}\equiv\mathcal{A}$. Thus, regular interpretability
$\mathcal{A}\simeq\Gamma(\mathcal{B},\phi)$ gives some models
$\mathcal{A}_{1}$ of the theory $Th(\mathcal{A})$, but, in general, not all
the models. To get all models of $Th(\mathcal{A})$ in the form
$\mathcal{A}_{1}\simeq\Gamma(\mathcal{B}_{1},\phi)$, where
$\mathcal{B}_{1}\equiv\mathcal{B}$ one needs regular bi-interpretability (or
at least regular invertible interpretability) of $\mathcal{A}$ and
$\mathcal{B}$ in each other. By construction, interpretability
$\mathcal{A}\simeq\Gamma(\mathcal{B},\phi)$ gives an algebraic description of
the structure $\mathcal{A}$ through the structure $\mathcal{B}$. The efficacy
of the description depends on the interpretation $\Gamma$ and how well one
knows the models $\mathcal{B}_{1}$ of $Th(\mathcal{B})$.

Let $G\in\\{SL_{n}(F),GL_{n}(F),T_{n}(F)\\}$ or just an algebraic group of
$F$-points over a field $F$. With $G$ we associate an algebraic structure
$C(G)$, _the core of $G$_ which satisfy the following conditions:

  1. 1)

$C(G)$ determines the essential algebraic properties of $G$;

  2. 2)

$C(G)$ is regularly interpretable in $G$, so $H\equiv G$ implies $C(H)\equiv
C(G)$;

  3. 3)

$C(G)$ is bi-interpretable (or invertible interpretable) with $F$, i.e.,
$C(G)\simeq\Gamma(F,\phi)$. Hence

$C(H)\equiv C(G)\Longleftrightarrow C(H)\simeq\Gamma(L,\phi)\ for\ some\
L\equiv F.$

Now if $H\equiv G$ then by 2) $C(H)\equiv C(G)$, hence by 3) $C(H)$ has the
same structure over some field $L$, as $C(G)$ over the field $F$, where
$L\equiv F$. This gives the algebraic structure of the core $C(H)$ over $L$.
Knowing the core $C(H)$ one deduces the algebraic structure of $H$. This
description of $H$ depends on how well the core $C(G)$ ($C(H)$) reflects the
structure of $G$ ($H$). Our main results below explain how this approach works
in particular cases when $G\in\\{SL_{n}(F),GL_{n}(F),T_{n}(F)\\}$.

The structure of the paper is as follows. In Section 2 we collect all the
basic notation and results on matrix groups. In Section 3, we recall some
results from [33] which show that many important subgroups of the classical
matrix groups $GL_{n}(R)$ and $PGL_{n}(R)$ are Diophantine, that is, definable
by Diophantine formulas. These are the basis for our bi-interpretability
results. In passing, we review the principal results on the decidability of
Diophantine problems in the classical matrix groups $G$, which generally state
that Diophantine problems in $G$ and $F$ are P-time equivalent. Notice that
recently similar results were obtained for Chevalley groups over arbitrary
rings [14], namely, if $\Phi$ is an irreducible root system of rank $>1$, $R$
is an arbitrary commutative ring with $1$, then the Diophantine problem in any
Chevalley group $G_{\pi}(\Phi,R)$ is Karp equivalent to the Diophantine
problem in $R$. Surprisingly, this type of reduction of Diophantine problems
in groups to Diophantine problems in rings occurs not only in algebraic
groups, where it is of course very natural, but also in various classes of
discrete groups. Thus, in [20] it was shown that the Diophantine problem in
wide classes of finitely generated solvable groups $G$ is closely related to
the Diophantine problem in the corresponding rings of algebraic integers.
Furthermore, a similar approach works for various non-commutative $R$-algebras
$A$, in which case the Diophantine problem in $A$ is equivalent to the
Diophantine problem in a ring $R$ or in a ring of polynomials $R[X]$, or in
some other commutative ring associated with $A$ (see, for example, [29, 21,
22, 26, 27]).

The rest of the paper is devoted to proving the main results of this paper,
which we describe without technical details in the following three
subsections.

###  1.2 Groups first-order equivalent to $SL_{n}(F)$

Let $G=SL_{n}(F)$. In this case, the core of $SL_{n}(F)$ is $SL_{n}(F)$
itself, so conditions 1) and 2) above on the core $C(G)$ obviously hold, while
condition 3) follows from the result below.

Theorem A1 Let $F$ be a field and $n\geq 3$. Then the group $SL_{n}(F)$ is
regularly bi-interpretable with $F$.

Based on this, we describe all groups first-order equivalent to $SL_{n}(F)$.

Theorem A2 Let $F$ be a field and $n\geq 3$. Then for a group $H$ the
following conditions are equivalent:

  * 1)

$H\equiv SL_{n}(F)$;

  * 2)

$H\simeq SL_{n}(L)$ for some field $L$ such that $F\equiv L$.

###  1.3 Groups first-order equivalent to $T_{n}(F)$

Let $G=T_{n}(F)$. In this case, we define the core $C(T_{n}(F))$ as the
quotient $P\textrm{T}_{n}(F)=\textrm{T}_{n}(F)/Z(\textrm{T}_{n}(F))$. Clearly,
$P\textrm{T}_{n}(F)$ is absolutely interpretable in $T_{n}(F)$.

The following is condition 3) on the core $C(T_{n}(F))$.

Lemma 5.3. The quotient group
$P\textrm{T}_{n}(F)=\textrm{T}_{n}(F)/Z(\textrm{T}_{n}(F))$ is regularly bi-
interpretable with $F$.

As a corollary we get the following crucial technical result.

Corollary 5.4. If $H$ is a group and $H\equiv\textrm{T}_{n}(F)$, then

$H/Z(H)\cong PT_{n}(L)$

for some field $L\equiv F$.

The following is the main structural theorem for groups of type $T_{n}$.

Theorem A3. Assume that $F$ is an infinite field of characteristic $\neq 2$.
If $H$ is any group $H\equiv\textrm{T}_{n}(F)$, then $H\cong T_{n}(L,f,Z)$ for
some field $L\equiv F$, symmetric 2-cocycle $f\in S^{2}(B_{n}(L),Z)$ and an
abelian group $Z=Z(H)\equiv Z(\textrm{T}_{n}(F))\cong F^{\times}$.

In the statement, $F^{\times}$ denotes the multiplicative group of units in
the field $F$, $B_{n}(L)\cong(L^{\times})^{n-1}$, and $Z(H)$ is the center of
the group $H$. The group $T_{n}(L,f,Z)$ is the so-called abelian deformation
of $T_{n}(L)$. The maximal torus $D_{n}(L)$ consisting of all diagonal
matrices in $T_{n}(L)\cong(L^{\times})^{n}$ has been deformed into an abelian
extension of $Z=Z(H)$ by $B_{n}(L)$, while the action of the deformed torus on
the unipotent radical $UT_{n}(L)$ has remained “the same”.

The proof of this theorem is given in Section 5. In that section, we provide
simpler characterizations of the groups $H$ of Theorem A3 in the cases where
$F$ is a real closed field or an algebraically closed field (see Theorem
5.13). If $F$ is a number field or a similar type of field and there are some
restrictions on the type of 2-cocycles, then we prove that the description in
Theorem A3 is also sufficient (see Theorem 5.12). Finally, we prove the
existence of a group $H$ elementarily equivalent to $T_{n}(\mathbb{Q})$, where
$\mathbb{Q}$ is the field of rationals, but the group $H$ is not of type
$T_{n}$ (see Theorem 5.15).

###  1.4 Groups first-order equivalent to $GL_{n}(F)$

Let $G=GL_{n}(F)$. In this case, we put $C(G)=SL_{n}(F)=G^{\prime}$. Note that
there is a natural number $w=w(n)$ (the width of the commutator of
$G^{\prime}$) that depends only on $n$ and such that every element of
$G^{\prime}=SL_{n}(F)$ is a product of at most $w$ transvections (see Sections
2.1 and 2.2), so it is a product of $w$ commutators in $G$. It follows that
$SL_{n}(F)$ is absolutely and uniformly (i.e., the defining formula does not
depend on $F$) definable in $G$. This addresses conditions 2) and 3) of the
definition of the core.

Observe that

$G=\textrm{GL}_{n}(F)\cong\textrm{SL}_{n}(F)\rtimes d_{1}(F^{\times}),$

where $F^{\times}$ is the multiplicative group of $F$, and
$d_{1}(F^{\times})=\\{diag(\alpha,1,\ldots,1)\mid\alpha\in F^{\times}\\}$.
Moreover,

$G/(G^{\prime}\cdot Z(G))\cong F^{\times}/(F^{\times})^{n}.$

This allows us to describe the algebraic structure of groups $H$ such that
$H\equiv\textrm{GL}_{n}(F)$. We are not going to provide the main technical
statements here in the introduction. However, a simpler, yet crucial, result
is as follows.

Theorem A4. If $H$ is any group such that $H\equiv\textrm{GL}_{n}(F)$, then
$H$ fits into the short exact sequence:

$1\rightarrow H^{\prime}\cdot Z(H)\rightarrow
H\rightarrow\frac{L^{\times}}{(L^{\times})^{n}}\rightarrow 1$

where $H^{\prime}\cong\textrm{SL}_{n}(L)$, $L\equiv F$, as fields and
$Z(H)\equiv F^{\times}$.

The proof of Theorem A4 and the rest of the main results on $GL_{n}$ are given
in Section 6. Similarly to the case of $T_{n}$, there are much simpler
descriptions of the group $H$ from Theorem A4 when $F$ is an algebraically
closed field or a real closed field (see Theorems 6.1 and 6.3). The most
technical case belongs to number fields and similar fields, called NFT-fields
(see Theorem 6.8). All these more precise results are improvements of Theorem
A4 for particular fields. Finally, in Section 6.3.2 we prove the existence of
a group that is first-order equivalent to $GL_{n}(\mathbb{Q})$ but not of the
type $GL_{n}$.

##  2 Preliminaries on classical matrix groups

In this section, we fix some notation and recall technical results that are
used throughout the paper.

For a group $G$ and $x,y\in G$ we denote by $x^{y}$ the conjugate $y^{-1}xy$
of $x$ by $y$, and by $[x,y]$ the commutator $x^{-1}y^{-1}xy$. For a subset
$A\subseteq G$ by $C_{G}(A)$ we denote the centralizer $\\{x\in G\mid\forall
a\in A~{}([x,a]=1)\\}$, in particular, $Z(G)=\\{x\in G\mid\forall y\in
G~{}([x,y]=1)\\}$ is the center of $G$. For the subsets $X,Y\subseteq G$ by
$[X,Y]$, we denote the subgroup of $G$ generated by all commutators $[x,y]$,
where $x\in X,y\in Y$. Then $[G,G]$ is the derived subgroup $G^{\prime}$ of
$G$ (the commutant of $G$). The lower central series of $G$ is defined as
$G=\gamma_{1}(G)\geq\gamma_{2}(G)\geq\ldots$, where
$\gamma_{i+1}(G)=[G,\gamma_{i}(G)]$.

In the rest of the paper by $R$ we denote an arbitrary associative ring with
identity 1. By $R^{\times}$ we denote the multiplicative group of invertible
(unit) elements of $R$ and by $R^{+}$ the additive group of $R$.

Now we define the groups we study in this paper and list some of their
properties that we use often.

###  2.1 $GL_{n}(R)$

Fix $n\in\mathbb{N}$. By $GL_{n}(R)$ we denote the group of all invertible
$n\times n$ matrices over an associative unitary ring $R$.

Let $e_{ij}$ be an $n\times n$ matrix where the $(i,j)$ entry is $1$ and all
other entries are $0$. For $1\leq i\neq j\leq n$, the matrix
$t_{ij}(\alpha)=I_{n}+\alpha e_{ij}$, where $\alpha\in R$ and $I_{n}$ is the
$n\times n$ identity matrix, is called a _transvection_. Sometimes we denote
transvection $t_{ij}(1)$ simply by $t_{ij}$. Put
$\mathcal{T}_{n}=\\{t_{ij}\mid 1\leq i\neq j\leq n\\}$.

Transvections $t_{ij}(\alpha),t_{kl}(\beta)$, for $\alpha,\beta\in R$, satisfy
the following well-known (Steinberg) relations:

  1. 1)

$t_{ij}(\alpha)t_{ij}(\beta)=t_{ij}(\alpha+\beta).$

  2. 2)

$[t_{ik}(\alpha),t_{kl}(\beta)]=t_{il}(\alpha\beta)$, for $i\neq l$.

  3. 3)

$[t_{ik}(\alpha),t_{jl}(\beta)]=1$ for $i\neq l,j\neq k$.

Observe that 2) implies that
$[t_{ij}(\alpha),t_{ki}(\beta)]=t_{kj}(-\alpha\beta)$, for $j\neq k$.

Let $diag(\alpha_{1},\ldots,\alpha_{n})$ be the $n\times n$ diagonal matrix
with $(i,i)$-entry $\alpha_{i}\in R^{\times}$. Then
$diag(\alpha_{1},\ldots,\alpha_{n})\in GL_{n}(R)$ and the set of all such
matrices forms a subgroup $D_{n}(R)$ of $GL_{n}(R)$. Note that for any
$\alpha_{1},\ldots,\alpha_{n}\in R^{\times}$, $\beta\in R$ the following
holds:

$diag(\alpha_{1},\ldots,\alpha_{n})^{-1}t_{ij}(\beta)diag(\alpha_{1},\ldots,\alpha_{n})=t_{ij}(\alpha_{i}^{-1}\beta\alpha_{j}).$
(2.1)

In particular,

$[t_{ij}(\beta),diag(\alpha_{1},\ldots,\alpha_{n})]=t_{ij}(\alpha_{i}^{-1}\beta\alpha_{j}-\beta).$
(2.2)

By $d(\alpha)$ we denote the scalar matrix $diag(\alpha,\ldots,\alpha)=\alpha
I_{n}$, where $\alpha\in R^{\times}$. The set of all scalar matrices forms a
subgroup $R^{\times}I_{n}\leq D_{n}(R)$ which is isomorphic to $R^{\times}$.
It follows from (2.2) that the subgroup $Z_{n}(R)$ of $R^{\times}I_{n}$, which
consists of all scalar matrices $d(\alpha)$, where $\alpha$ is in the center
of the group $R^{\times}$, forms the center of the groups $R^{\times}I_{n}$,
$D_{n}(R)$, as well as the group $GL_{n}(R)$.

Now consider the following diagonal matrices for $\alpha\in R^{\times}$:

$d_{i}(\alpha)\stackrel{{\scriptstyle\text{def}}}{{=}}diag(1,\ldots,\underbrace{\alpha}_{i^{\prime}\text{th}},\ldots,1).$

It is known (see, for example [23]) that if $R$ is a field then there are
natural numbers $r$ and $s$, which depend only on $n$, such that every element
$g\in GL_{n}(R)$ can be presented as a product of the type

$g=x_{1}\ldots x_{r}d_{n}(\beta)y_{1}\dots y_{s}$ (2.3)

where $x_{i},y_{j}$ are transvections and $\beta\in R^{\times}$.

###  2.2 $SL_{n}(R)$ and $E_{n}(R)$

Denote by $E_{n}(R)$ the subgroup of $GL_{n}(R)$ generated by all
transvections, i.e., $E_{n}(R)=\langle t_{ij}(\alpha)\mid\alpha\in R,1\leq
i\neq j\leq n\rangle$.

######  Definition 2.1.

We say that a subgroup $G\leq GL_{n}(R)$ is _large_ if $G$ contains
$E_{n}(R)$.

Throughout the paper, we assume $n\geq 3$.

If the ring $R$ is commutative, then, as usual, the group $SL_{n}(R)$ consists
of all matrices of $GL_{n}(R)$ with determinant 1. One can define $SL_{n}(R)$
for arbitrary division rings (using the Dieudonne determinant), but we do not
consider such groups here. In every case, when we mention a group $SL_{n}(R)$
we assume that $R$ is commutative. Clearly, $E_{n}(R)$ is a subgroup of
$SL_{n}(R)$, so $SL_{n}(R)$ is a large subgroup of $GL_{n}(R)$. If $R$ is a
field or Euclidean domain, then $SL_{n}(R)=E_{n}(R)$, but in general, this is
not the case. For the fields $R$ we have

$[GL_{n}(R),GL_{n}(R)]=SL_{n}(R).$

Now (2.3) implies that

$GL_{n}(R)\simeq SL_{n}(R)\rtimes d_{n}(R^{\times})\simeq
GL_{n}(R)^{\prime}\rtimes R^{\times}.$

Note also that in this case

$[SL_{n}(R),SL_{n}(R)]=SL_{n}(R).$

Following [15] we say that $SL_{n}(R)$ has _bounded elementary generation_ if
there is a natural number $w$ such that every element of $SL_{n}(R)$ is a
product of at most $w$ transvections. Order all pairs of indices
$(i,j),i,j=1,\ldots,n$ into a sequence $\sigma$ in some arbitrary but fixed
way, say $\sigma=(1,1),\ldots,(n,n)$. Repeat this sequence consequently $w$
times, obtaining a new sequence
$\sigma^{*}=\sigma,\sigma,\ldots,\sigma=(i_{1},j_{1}),\ldots,(i_{m},j_{m})$,
where $m=wn^{2}$. Then every element $g\in SL_{n}(R)$ can be decomposed into a
product

$g=t_{i_{1}j_{1}}(\alpha_{1})\ldots t_{i_{m}j_{m}}(\alpha_{m})$

for some $\alpha_{1},\ldots,\alpha_{m}\in R$ (note that we allow here elements
$\alpha_{i}=0$), which is uniform in order of transvections.

For $1\leq i\neq j\leq n$ denote by $T_{ij}$ the one-parametric subgroup
$\\{t_{ij}(\alpha)\mid\alpha\in R\\}$. Then the bounded elementary generation
of $SL_{n}(R)$ is equivalent to the statement that there is sequence of pairs
$(i_{1},j_{1}),\ldots,(i_{m},j_{m})$, where $1\leq i_{k}\neq j_{k}\leq n$ such
that

$SL_{n}(R)=T_{i_{1}j_{1}}\ldots T_{i_{m}j_{m}}.$

If $R$ is a field, then formula (2.3) implies that $SL_{n}(R)$ has bounded
elementary generation. A much more difficult argument shows that
$SL_{n}(\mathcal{O})$ over a ring of algebraic integers $\mathcal{O}$ has
bounded elementary generation [15]. However, this is not the case for even
arbitrary domains. In fact, it was shown in [38, 18] that if $F$ is a field of
infinite transcendence degree over its prime subfield (for example:
$F=\mathbb{C}$) then for every number $c$ there is a matrix in the group
$SL_{n}(F[x])$ that cannot be written as a product of $c$ commutators.

###  2.3 Unitriangular groups $UT_{n}(R)$

Transvections $t_{ij}(\alpha)$, where $1\leq i<j\leq n$, $\alpha\in R$,
generate the subgroup $UT_{n}(R)$ of all (upper) unitriangular matrices in
$GL_{n}(R)$.

For $m=1,\ldots,n$ denote by $UT^{m}(n,R)$ the subgroup of $UT_{n}(R)$
consisting of all matrices with $m-1$ zero diagonals above the main one. Then

$UT_{n}(R)=UT_{n}^{1}(R)>UT_{n}^{2}(R)>\ldots>UT_{n}^{n}(R)=1.$ (2.4)

Furthermore, for any positive $r,s\in\mathbb{N}$ one has

$[UT_{n}^{r}(R),UT_{n}^{s}(R)]=UT_{n}^{r+s}(R).$

This implies that the series (2.4) is the lower central series of $UT_{n}(R)$,
in particular, $\gamma_{k}(UT_{n}(R))=UT_{n}^{k}(R)$. Direct computations show
that any element $g\in UT_{n}^{m}(R)$ can be uniquely written as a product of
the following type:

$g=t_{n-m,n}(\alpha_{n-m})t_{n-m-1,n-1}(\alpha_{n-m-1})\ldots
t_{1,1+m}(\alpha_{1})h,$

where $\alpha_{i}\in R$ and $h\in UT_{n}^{m+1}(R)$. Therefore,

$UT_{n}^{m}(R)=T_{n-m,n}T_{n-m-1,n-1}\ldots T_{1,1+m}UT_{n}^{m+1}(R)$ (2.5)

This implies that $UT_{n}(R)$ is a finite product of one parametric subgroups
$T_{ij}$.

###  2.4 Triangular groups $T_{n}(R)$

Recall that $T_{n}(R)$ consists of all upper triangular matrices $x=(x_{ij})$
over $R$ with units on the main diagonal, that is, $x_{ij}=0$ for $i>j$,
$x_{ij}\in R$ for $i<j$, and $x_{ii}\in R^{\times}$ for $1\leq i\leq n$.
Clearly, $UT_{n}(R)\leq T_{n}(R)$.

Note that any matrix $x\in T_{n}(R)$ can be represented as a product
$x=d_{x}u_{x}$, where $d_{x}=diag(x_{11},\ldots,x_{nn})$, and $u_{x}\in
UT_{n}(R)$, where $u_{x}=(y_{ij})$, with $y_{ij}=x_{ii}^{-1}x_{ij}$ for $i<j$.
Therefore, $T_{n}(R)=D_{n}(R)UT_{n}(R)$ and $D_{n}(R)\cap UT_{n}(R)=1$.
Furthermore, $UT_{n}(R)$ is a normal subgroup in $T_{n}(R)$, see (2.1), hence

$T_{n}(R)\simeq UT_{n}(R)\rtimes D_{n}(R).$

######  Definition 2.2.

We call $G\leq T_{n}(R)$ a large subgroup of $T_{n}(R)$ if $G$ contains
$UT_{n}(R)$.

Obviously, $UT_{n}(R)$ is a large subgroup of $T_{n}(R)$, but it is not a
large subgroup of $GL_{n}(R)$.

###  2.5 Groups $PGL_{n}(R)$ and $PSL_{n}(R)$

The general and special projective linear groups $PGL_{n}(R)$ and $PSL_{n}(R)$
are defined as quotients $PGL_{n}(R)=GL_{n}(R)/Z(GL_{n}(R))$ and
$PSL_{n}(R)=SL_{n}(R)/Z(SL_{n}(R))$ of the groups by their centers. Note that
$Z(GL_{n}(R))$ is the subgroup $Z_{n}(R)$ of all scalar matrices $d(\alpha)$,
where $\alpha$ belongs to the center of the group $R^{\times}$. Respectively,
$Z(SL_{n}(R))=Z_{n}(R)\cap SL_{n}(R)$.

######  Definition 2.3.

Let $\phi:GL_{n}(R)\to PGL_{n}(R)$ be the canonical homomorphism. We call a
subgroup $G$ of $PGL_{n}(R)$ _large_ if it contains $\phi(E_{n}(R))$.

##  3 Diophantine structure in large subgroups of classical matrix groups

In this section, we recall some results from [33] which show that many
important subgroups of the classical matrix groups $GL_{n}(R)$ and
$PGL_{n}(R)$ are Diophantine. We freely use the notation from Preliminaries.

We start with the following key technical result of [33].

######  Theorem 3.1.

[33] Let $G$ be a large subgroup of $GL_{n}(R)$, $n\geq 3$. Then for any
$1\leq k\neq m\leq n$ the subgroup $T_{km}$ is Diophantine in $G$ (defined
with parameters $\mathcal{T}_{n}$).

######  Corollary 3.2.

[33] Let $G=GL_{n}(R)$, $n\geq 3$. Then for any $1\leq k\neq m\leq n$ the
subgroup $T_{km}$ is Diophantine in $G$ (defined with parameters
$\mathcal{T}_{n}$).

######  Corollary 3.3.

[33] Let $R$ be a commutative ring and $G=SL_{n}(R)$, $n\geq 3$. Then for any
$1\leq k\neq m\leq n$ the subgroup $T_{km}$ is Diophantine in $G$ (defined
with parameters $\mathcal{T}_{n}$).

######  Proposition 3.4 ([33]).

Let $G$ be a large subgroup of $GL_{n}(R)$, $n\geq 3$. Then for every $1\leq
m\leq n-1$ the subgroup $UT_{n}^{m}(R)$ is Diophantine in $G$. In particular,
$UT_{n}(R)$ is Diophantine in $G$.

######  Lemma 3.5 ([33]).

Let $G$ be a large subgroup of $GL_{n}(R)$, $n\geq 3$. Then the set $R_{G}$ of
all scalar matrices from $G$ is Diophantine in $G$.

######  Proposition 3.6 ([33]).

Let $G$ be a large subgroup of $GL_{n}(R)$, $n\geq 3$. If $R^{+}$ does not
have elements of order 2, then the following hold:

  * 1)

$G\cap D_{n}(R)$ is Diophantine in $G$.

  * 2)

$G\cap T_{n}(R)$ is Diophantine in $G$.

###  3.1 Mutual Diophantine interpretability of the classical matrix groups
and $R$

####  3.1.1 The Diophantine interpretability of $R$ in classical linear groups

######  Theorem 3.7.

Let $G$ be a large subgroup of $GL_{n}(R)$, $PGL_{n}(R)$ (assuming that $R$
has no zero divisors in this case), or $T_{n}(R)$, $n\geq 3$. Then the ring
$R$ is e-interpretable in $G$.

###### Proof.

There are three cases to consider.

Case 1. $G$ is a large subgroup of $GL_{n}(R)$. By Theorem 3.1 the subgroups
$T_{12}$, $T_{2n}$, and $T_{1n}$ are Diophantine in $G$.

We e-interpret $R$ on $T_{1n}$, turning it into a ring $\langle
T_{1n};\oplus,\otimes\rangle$ as follows.

For $x,y\in T_{1n}$ define

$x\oplus y=x\cdot y$ (3.1)

Note that if $x=t_{1n}(\alpha),y=t_{1n}(\beta)$, then $x\cdot
y=t_{1n}(\alpha+\beta)$, which corresponds to the addition in $R$.

To define $x\otimes y$ for given $x,y\in T_{1n}$ we need some notation. Let
$x_{1},y_{1}\in G$ be such that

$x_{1}\in T_{12}\text{ and }[x_{1},t_{2n}]=x,\ \ y_{1}\in T_{2n}\text{ and
}[t_{12},y_{1}]=y.$ (3.2)

Note that such $x_{1},y_{1}$ always exist and unique, for if
$x=t_{1n}(\alpha),y=t_{1n}(\beta)$ then
$x_{1}=t_{12}(\alpha),y_{1}=t_{2n}(\beta)$. Now, define

$x\otimes y=[x_{1},y_{1}].$

Observe, that in this case

$[x_{1},y_{1}]=[t_{12}(\alpha),t_{2n}(\beta)]=t_{1n}(\alpha\beta),$ (3.3)

so $\otimes$ corresponds to the multiplication in $R$. To finish the proof, we
need two claims.

Claim 1. The map $\alpha\to t_{1n}(\alpha)$ gives rise to a ring isomorphism
$R\to\langle T_{1n};\oplus,\otimes\rangle$.

This is clear from the argument above.

Claim 2. The ring $\langle T_{1n};\oplus,\otimes\rangle$ is e-interpretable in
$G$.

To see this, observe first that, as mentioned above, $T_{1n}$ is Diophantine
in $G$. The addition $\oplus$ defined in (3.1) is clearly Diophantine in $G$.
Since the subgroups $T_{12}$ and $T_{2n}$ are Diophantine in $G$ the
conditions (3.2) are Diophantine in $G$, as well as the condition (3.3). This
shows that the multiplication $\otimes$ is also Diophantine in $G$. This
proves case 1.

Case 2. Let $G$ be a large subgroup of $PGL_{n}(R)$. The proof is similar to
the one in Case 1. Only instead of Theorem 3.1, one uses a similar result for
$PGL_{n}$ (see [33]).

Case 3. Let $G$ be a large subgroup of $T_{n}(R)$. To prove that $R$ is
e-interpretable in $G$ we adjust the above argument by making a few changes.
That is, we replace the subgroup $T_{12}$ by the Diophantine subgroup
$T_{12}^{\prime}=R_{G}T_{12}T_{1n}$ if $n>3$. If $n=3$, then we also replace
$T_{2n}=T_{23}$ with $T_{23}^{\prime}=R_{G}T_{23}T_{13}$. In both cases, the
argument works word by word except that in these cases the elements $x_{1}\in
T_{12}^{\prime}$ and $y_{1}\in T_{23}^{\prime}$ if $n=3$, are not unique.
However, this does not matter, since any such $x_{1}$ and $y_{1}$ give the
same commutator $[x_{1},y_{1}]$.

This proves the theorem. ∎

######  Corollary 3.8.

The ring $R$ is e-interpretable in the groups $GL_{n}(R)$, $SL_{n}(R)$
(assuming that in this case $R$ is commutative), $E_{n}(R)$, $T_{n}(R)$ and
$UT_{n}(R)$, where $n\geq 3$. If in addition $R$ has no zero divisors, then
$R$ is e-interpretable in $PGL_{n}(R)$ and $PSL_{n}(R)$ (as before, $R$ is
also commutative in this case).

######  Remark 3.9.

If $n\geq 3$, the ring $R$ is e-interpretable in the groups $GL_{n}(R)$,
$PGL_{n}(R)$, $E_{n}(R)$ on each of the one-parameter subgroups $T_{ij}$. The
same holds for $T_{n}(R)$ and $UT_{n}(R)$, if $j-i\geq 2$. If in addition $R$
is commutative, then it can be e-interpretable in $SL_{n}(R)$ and $PSL_{n}(R)$
on each of the one-parameter subgroups $T_{ij}$.

####  3.1.2 The Diophantine absolute interpretability of classical linear
groups in $R$

Now we prove the converse of Theorem 3.7 (with the exception of $E_{n}(R)$).
We believe that the result is known in folklore.

######  Proposition 3.10.

The groups $GL_{n}(R)$, $T_{n}(R)$, and $UT_{n}(R)$, are all e-interpretable
in $R$. If $R$ is commutative, then the groups $PGL_{n}(R)$, $SL_{n}(R)$, and
$PSL_{n}(R)$ are all e-interpretable in $R$.

###### Proof.

We represent an $n\times n$ matrix $x=(x_{ij})$ with entries in $R$ by an
$n^{2}$-tuple $\bar{x}$ over $R$, where

$\bar{x}=(x_{11},\ldots,x_{1n},x_{21},\ldots,x_{n1},\ldots,x_{nn}).$

The matrix multiplication $\odot$ on tuples from $R^{n^{2}}$ is defined by

$\bar{x}\odot\bar{y}=\bar{z}\Longleftrightarrow\bigwedge_{i,j=1}^{n}z_{ij}=P_{ij}(\bar{x},\bar{y}),$

where $P_{ij}(\bar{x},\bar{y})$ is an integer polynomial
$\Sigma_{s=1}^{n}x_{is}y_{sj}$. The multiplication $\odot$ is clearly
Diophantine. To finish the description of the interpretations of the groups
$GL_{n}(R)$, $SL_{n}(R)$, $T_{n}(R)$ and $UT_{n}(R)$ in $R$, it suffices to
define the corresponding subsets of $R^{n^{2}}$ using Diophantine formulas. We
do it case by case.

The case of $GL_{n}(R)$. A matrix $x=(x_{ij})$ belongs to $GL_{n}(R)$ if it is
invertible, that is, there exists a matrix $y=(y_{ij})$ such that $xy=I_{n}$.
In the language of the tuples $\bar{x}$ and $\bar{y}$, this is expressed as

$\exists\bar{y}(\bar{x}\odot\bar{y})=\bar{I}_{n}.$

This condition is Diophantine, so $GL_{n}(R)$ is e-interpretable in $R$.

The case of $SL_{n}(R)$. In this case $R$ is a commutative ring. Recall that
the determinant of a $n\times n$ matrix $(x_{ij})$ with entries in a
commutative ring $R$ can be calculated as the value of some fixed polynomial
$Det_{n}(\bar{x})$ in the tuple $\bar{x}$. Therefore, the matrix $x=(x_{ij})$
belongs to $SL_{n}(R)$ if and only if $Det_{n}(\bar{x})=1$. This is a
Diophantine equation in $R$.

The case of $PGL_{n}(R)$ and $PSL_{n}(R)$. Observe that if the ring $R$ is
commutative, then the subgroup of all scalar matrices in $GL_{n}(R)$ and
$SL_{n}(R)$ is Diophantine as the centralizer of the finite set of all
transvections $t_{ij}$. Therefore, the factor groups $PGL_{n}(R)$ and
$PSL_{n}(R)$ are e-interpretable, correspondingly, in the groups $GL_{n}(R)$
and $SL_{n}(R)$. Now, the result follows from the previous two cases by the
transitivity of e-interpretability.

The case of $T_{n}(R)$. By definition $x=(x_{ij})\in T_{n}(R)$ if and only if
$x_{ij}=0$ for all $i<j$ and there exists $y$ in $R$ such that $x_{11}\ldots
x_{nn}y=1$. All these conditions are Diophantine, so $T_{n}(R)$ is
e-interpretable in $R$.

The case of $UT_{n}(R)$ is similar to the one above. This proves the
proposition.

∎

######  Theorem 3.11.

Let $n\geq 3$. The Diophantine problem in each of the groups $GL_{n}(R)$,
$SL_{n}(R)$ (assuming that in this case $R$ is commutative), $T_{n}(R)$, and
$UT_{n}(R)$, as well as in the groups $PGL_{n}(R)$ and $PSL_{n}(R)$ (in this
case we additionally assume that the ring $R$ has no zero divisors) is Karp
equivalent to the Diophantine problem in $R$. In particular, the Diophantine
problem in all these groups is decidable if and only if it is decidable in
$R$.

###### Proof.

The result follows from Corollary 3.8, Proposition 3.10, and the properties of
e-interpretability (see the complete details in [33]). ∎

We cannot show that $E_{n}(R)$ is e-interpretable in $R$ for any associative
ring $R$. However, the following holds.

######  Theorem 3.12.

If $E_{n}(R)$ has bounded elementary generation then $E_{n}(R)$ is
e-interpretable in $R$. In this case, the Diophantine problem in $E_{n}(R)$ is
Karp equivalent to the Diophantine problem in $R$.

######  Corollary 3.13.

If $n\geq 3$, and $E_{n}(R)$ has bounded elementary generation, then the
Diophantine problem in $E_{n}(R)$ is Karp equivalent to the Diophantine
problem in $R$.

For an arbitrary associative ring $R$, we have the following consequence of
Corollary 3.8.

######  Proposition 3.14.

If the Diophantine problem is undecidable in a ring $R$, then it is also
undecidable in the group $E_{n}(R)$ for any $n\geq 3$.

##  4 Groups first-order equivalent to $SL_{n}(F)$

###  4.1 Bi-interpretability of $R$ and $SL_{n}(R)$

Let $\mathcal{T}_{n}$ be a tuple of all transvections $t_{ij}(1),1\leq i\neq
j\leq n$ ordered in some arbitrary but fixed way.

######  Lemma 4.1.

Let $R$ be an associative commutative unitary ring. Then $R$ is interpretable
in every large subgroup of $SL_{n}(R)$, $n\geq 3$ using Diophantine formulas
with constants from $\mathcal{T}_{n}$.

###### Proof.

Let $G$ be a large subgroup of $SL_{n}(R)$. By Theorem 3.1, the subgroups
$T_{12}$, $T_{2n}$, and $T_{1n}$ are Diophantine in $G$ with parameters
$\mathcal{T}_{n}$.

Similarly to the proof of Case 1 in Theorem 3.7 we interpret $R$ on $T_{1n}$
turning it into a ring $\langle T_{1n};\oplus,\otimes\rangle$. This proves the
lemma, but we would like to note that similar to the ring $\langle
T_{1n};\oplus,\otimes\rangle$ one can interpret the ring $R$ in $G$ as
$\langle T_{ik};\oplus,\otimes\rangle$ for any $1\leq i\neq k\leq n$. In fact,
in this case one may use the Diophantine subgroups $T_{ij}$, $T_{jk}$ and
$T_{ik}$, instead of $T_{12},T_{2n}$ and $T_{1n}$ where $j$ is an arbitrary
index not equal to $i$ and $k$. For further references we will use notation

$R_{ik}=\langle T_{ik};\oplus,\otimes\rangle.$ (4.1)

∎

######  Lemma 4.2.

Let $R$ be an associative commutative unitary ring, $n\geq 3$, and $1\leq
i\neq j\leq n$, $1\leq k\neq m\leq n$. Then the following hold:

  * 1)

The map $f_{i,j,k,m}:T_{ij}\to T_{km}$ defined by

$f_{i,j,k,m}:t_{ij}(\alpha)\to t_{km}(\alpha),\ \alpha\in R,$

is Diophantine in $G$ with parameters $\mathcal{T}_{n}$;

  * 2)

The map $f_{i,j,k,m}$ defined above gives an isomorphism of interpretations
$R_{ij}\to R_{km}$ of the ring $R$ in $G$.

###### Proof.

Fix arbitrary $1\leq i\neq j\leq n$ and $1\leq k\neq m\leq n$.

Case 1. Let $i=k$. Note that in this case $j\neq m$. Take $x=t_{ij}(\alpha)\in
T_{ij}$. Then $f_{ijkm}(x)=y$ where
$y=t_{km}(\alpha)=[t_{ij}(\alpha),t_{jm}(1)]=[x,t_{jm}(1)]$.

Case 2. Suppose $j=m$. Note that in this case $i\neq k$. Take
$x=t_{ij}(\alpha)\in T_{ij}$. Then $f_{ijkm}(x)=y$ where
$y=t_{km}(\alpha)=[t_{ki}(1),t_{im}(\alpha)]=[t_{ki}(1),x]$.

Observe that both maps $f_{ijim}$ and $f_{imkm}$ are Diophantine in $G$ with
parameters $\mathcal{T}_{n}$.

Case 3. Assume $i\neq k$ and $j\neq m$. In this case, the map $f_{ijkm}$ is
the composition of the two maps $f_{ijkm}=f_{imkm}\circ f_{ijim}$ each of
which is Diophantine in $G$. Hence, $f_{ijkm}$ is also Diophantine in $G$ as
the composition of Diophantine maps.

∎

######  Lemma 4.3.

Let $R$ be an associative commutative unitary ring and $n\geq 3$. Then the
group $SL_{n}(R)$ is absolutely interpretable by equations in $R$.

###### Proof.

We represent an $n\times n$ matrix $x=(x_{ij})$ with entries in $R$ by an
$n^{2}$-tuple $\bar{x}$ over $R$, where

$\bar{x}=(x_{11},\ldots,x_{1n},x_{21},\ldots,x_{n1},\ldots,x_{nn}).$

The matrix multiplication $\odot$ on tuples from $R^{n^{2}}$ is defined by

$\bar{x}\odot\bar{y}=\bar{z}\Longleftrightarrow\bigwedge_{i,j=1}^{n}z_{ij}=P_{ij}(\bar{x},\bar{y}),$

where $P_{ij}(\bar{x},\bar{y})$ is integer polynomial
$\Sigma_{s=1}^{n}x_{is}y_{sj}$. Thus, the multiplication $\odot$ is
interpretable by the equations in $R$. To finish the description of the
interpretation, it suffices to define the subset of $R^{n^{2}}$ which
corresponds to $SL_{n}(R)$ by equations. We do it case by case.

Recall that the determinant of an $n\times n$-matrix $(x_{ij})$ with entries
in a commutative ring $R$ can be calculated as the value of some fixed
polynomial $Det_{n}(\bar{x})$ in the tuple $\bar{x}$. Therefore, the matrix
$x=(x_{ij})$ belongs to $SL_{n}(R)$ if and only if $Det_{n}(\bar{x})=1$. This
gives an equation that defines $SL_{n}(R)$ in $R$.

∎

######  Remark 4.4.

We refer to the interpretation of $SL_{n}(R)$ in $R$ from Lemma 4.3 as the
_standard interpretation_.

Theorem A1. Let $F$ be a field, and $n\geq 3$. Then the group $SL_{n}(F)$ is
regularly bi-interpretable with $F$.

###### Proof.

Let $G=SL_{n}(F)$. Any field, hence $F$, has the elementary bounded generation
property. Therefore, there is a number $m$ such that every element $g\in G$ is
a product of at most $k$ transvections. It follows that there is fixed
sequence $\sigma$ of pairs of indexes $(i_{1},j_{1}),\ldots,(i_{m},j_{m})$
such that for every element $g\in G$ there exists a sequence of elements
$\alpha_{1},\ldots,\alpha_{m}\in F$ such that

$g=t_{i_{1}j_{1}}(\alpha_{1})\ldots t_{i_{m}j_{m}}(\alpha_{m}).$ (4.2)

In this case, we will write $g=t_{\sigma}(\bar{\alpha})$

Claim 1. There is a Diophantine formula $\Phi(x,\bar{y},\mathcal{T}_{n})$ with
parameters $\mathcal{T}_{n}$ (here $y=(y_{1},\ldots,y_{m})$) such that for any
elements $g,h_{1},\ldots,h_{m}\in G$ the following equivalence holds:

$G\models\Phi(g,\bar{h},\mathcal{T}_{n})\Longleftrightarrow\bigwedge_{i=1}^{m}h_{i}=t_{1n}(\alpha_{i})\wedge
g=t_{\sigma}(\bar{\alpha}).$

Indeed, the subgroup $T_{1n}$ is defined by the equations in $G$. For each
element $t_{1n}(\alpha_{k})\in T_{1n}$ by Lemma 4.2 the definable map (by
Diophantine formulas with parameters in $\mathcal{T}$)
$f_{1ni_{k}j_{k}}:T_{1n}\to T_{i_{k}j_{k}}$ give the transvection
$t_{i_{k}j_{k}}(\alpha_{k})$ as the image
$f_{1ni_{k}j_{k}}(t_{1n}(\alpha_{k}))$. Now to finish the proof of the claim,
it suffices to note that

$g=t_{\sigma}(\bar{\alpha})\Longleftrightarrow
g=\Pi_{k=1}^{m}f_{1ni_{k}j_{k}}(t_{1n}(\alpha_{k})).$

Now consider the interpretation $F_{1n}=\Gamma(G,\mathcal{T}_{n})$ of $F$ in
$G$ given in Lemma 4.1 and also the standard matrix interpretation
$G^{*}=\Delta(F_{1n})$ of $SL_{n}(F)$ in $F_{1n}$ from Lemma 4.3.

Claim 2. There is an isomorphism $\lambda:G\to G^{*}$ definable in $G$ by
Diophantine formulas with parameters $\mathcal{T}_{n}$.

Let $M(t_{\sigma}(\bar{\alpha}))$ be the matrix given by the product of
transvections
$t_{\sigma}(\bar{\alpha})=\Pi_{k=1}^{m}f_{1ni_{k}j_{k}}(t_{1n}(\alpha_{k})$.
Since the sequence $\sigma$ is fixed then for every $(i,j)$ with $1\leq i\neq
j\leq n$, there exists a polynomial $P_{ij}$ with integer coefficients such
that for any $\bar{\alpha}\in F^{m}$ the entry $a_{ij}$ of the matrix
$M(t_{\sigma}(\bar{\alpha}))$ is equal to $P_{ij}(\bar{\alpha})$. Now we have
an equivalence

$g=t_{\sigma}(\bar{\alpha})\Longleftrightarrow\lambda(g)=(a_{ij})\bigwedge_{i,j=1}^{n}a_{ij}=P_{ij}(\bar{\alpha}).$

which gives an isomorphism $\lambda$. The equality
$g=t_{\sigma}(\bar{\alpha})$ is described by the formula
$\Phi(x,\bar{y},\mathcal{T}_{n})$ of claim 1. Hence, the following formula
defines the isomorphism $\lambda$ (here by $\bar{t}_{\sigma}(\bar{\alpha})$ we
denote the sequence
$(t_{i_{1}j_{1}}(\alpha_{1}),\ldots,t_{i_{m}j_{m}}(\alpha_{m}))$:

$\exists\bar{t}_{\sigma}(\bar{\alpha})[\Phi(g,\bar{t}_{\sigma}(\bar{\alpha}),\mathcal{T}_{n})\wedge\lambda(g)=(a_{ij})\bigwedge_{i,j=1}^{n}a_{ij}=P_{ij}(\bar{\alpha})].$

Note that the defining formula is Diophantine with parameters
$\mathcal{T}_{n}$.

Consider now the standard interpretations $G^{*}=\Delta(F)$ and the
interpretation $F^{*}=\Gamma(G^{*},\mathcal{T}_{n}^{*})$ from Lemma 4.1, where
$\mathcal{T}_{n}^{*}$ is the set of transvections in the group $G^{*}\simeq
SL_{n}(F)$.

Claim 3. There is an isomorphism $\mu:F\to F^{*}$ definable in $F$ by
Diophantine formulas with parameters $\mathcal{T}_{n}^{*}$.

We define the isomorphism $\mu:F\to F^{*}$ by $\mu(\alpha)=t_{1n}(\alpha)$
where $t_{1n}(\alpha)$ is the corresponding matrix in the group $G^{*}$. Hence
the condition that defines $\mu(\alpha)$ in $F$ is the following:

$\mu(\alpha)=(a_{ij})\wedge
a_{1n}=\alpha\bigwedge_{j=1}^{n}a_{ii}=1\bigwedge_{1\leq i\neq j\leq
n,i\&(i,j)\neq(1,n)}a_{ij}=0.$

The claims 1 and 2 show that $F$ and $G=SL_{n}(F)$ are bi-interpretable with
parameters $\mathcal{T}_{n}$. Now we need to show that one can take any tuple
of parameters from some definable subset of tuples in $G$ to bi-interpret $F$
and $G=SL_{n}(F)$.

Now we need to show that the bi-interpretation of $G$ and $F$ described above
is regular. Note first that the interpretations $G^{*}=\Delta(F)$ and
$F^{*}=\Gamma(\Delta(F),\mathcal{T}_{n}^{*})$ are absolute (the second is
absolute because the tuple $\mathcal{T}_{n}^{*}$ is absolutely definable in
$F$. Therefore, it suffices to show that instead of $\mathcal{T}_{n}$ one may
use any tuple from a definable subset of tuples in $G$.

Consider the following conditions:

  * •

the interpretation $\Delta(F)$ gives a group;

  * •

the interpretation $F\simeq\Gamma(G,\mathcal{T}_{n})$ gives a field;

  * •

the formula that defines $\lambda$ gives an isomorphism of groups;

  * •

the formula that defines $\mu$ gives an isomorphism of fields.

As we have seen above, all the conditions above can be described by a formula
with parameters $\mathcal{T}_{n}$. Let $\Psi(\bar{z})$ be the conjunction of
formulas that describe the conditions above and where the parameters
$\mathcal{T}_{n}$ are replaced by a tuple of new variables $\bar{z}$. Denote
by $D$ the definable set $\\{S\mid G\models\Psi(S)\\}$. Observe that
$G\models\Psi(\mathcal{T}_{n})$, $D\neq\emptyset$.

Let $S\in D$. Then $F_{S}=\Gamma(G,S)$ is a field, $G_{S}=\Delta(F_{S})$ is
the group isomorphic to $SL_{n}(F_{S})$ (by construction of $\Delta(F_{A})$).
Furthermore, the formula that defines the isomorphism $\lambda:G\to G^{*}$ now
defines some isomorphism $\lambda_{S}:G\to G_{S}$, so $SL_{n}(F)\simeq
SL_{n}(F_{S})$. By the abstract isomorphism theorem (see the book[13]) one has
$F\simeq F_{S}$. This shows that the interpretation $F\simeq\Gamma(G,D)$ of
$F$ in $G$ is regular with respect to the definable set $D$. As we have
mentioned above, the interpretation $G\simeq\Delta(F)$ is absolute and hence
also regular. The formulas that define the isomorphisms $\lambda_{S}$ and
$\mu_{S}$ are also uniform in $S\in D$. This shows that $G$ and $F$ are
regularly bi-interpretable.

This proves the theorem.

∎

######  Remark 4.5.

In the proof of Theorem A1. above one can replace the field $F$ with any
associative commutative unitary ring $R$ which satisfy the following
conditions:

  * •

$R$ admits bounded elementary generation (in fact, this could be an even
weaker condition);

  * •

$R$ satisfies abstract isomorphism theorem for $SL_{n}$, $n\geq 3$ (there are
many such rings, see [13]).

We call such rings $SL$-capable.

###  4.2 The characterization theorem

######  Theorem 4.6.

Let $F$ be a field, and $n\geq 3$. Then for a group $H$ the following
conditions are equivalent:

  * 1)

$H\equiv SL_{n}(F)$;

  * 2)

$H\simeq SL_{n}(L)$ for some field $L$ such that $F\equiv L$

###### Proof.

We use notation from Theorem A1. 1) $\to$ 2). Suppose that $H$ is a group such
that $H\equiv SL_{n}(F)$. Since $F\simeq\Gamma(G,S)$ for any tuple $S$ over
$G$ such that $G\models\Psi(S)$ it follows that for any tuple $\tilde{S}$ in
$H$ such that $H\models\Psi(\tilde{S})$ the same formulas that interpret $F$
in $G$ with parameters $\mathcal{T}$ interpret a field $L_{\tilde{S}}$ in $H$
with parameters $\tilde{S}$, that is,
$L_{\tilde{S}}\simeq\Gamma(H,\tilde{S})$. Since this is true for any
$\tilde{S}$ that satisfies $\Psi$ in $H$ it follows that $L_{\tilde{S}}\equiv
F$ (see [25]). Since the isomorphism $\lambda_{S}:G\to G_{S}$ is definable by
some formula $\theta(x,\bar{y},S)$ with parameters $S$ for any $S\in D$ one
finds that for any such $\tilde{S}$ the formula $\theta(x,\bar{y},\tilde{S})$
defines some isomorphism $\lambda^{*}_{\tilde{S}}:H\to H^{*}$, where
$H^{*}=\Delta(\Gamma(H,\tilde{S}))$. By construction of $\Delta$ $H^{*}\simeq
SL_{n}(\Gamma(H,\tilde{S})\simeq SL_{n}(L_{\tilde{S}})$, which proves that 1)
implies 2).

2) $\to$ 1). In fact, if $L\equiv F$ then $SL_{n}(F)\simeq SL_{n}(L)$ since
$SL_{n}(F)$ and $SL_{n}(L)$ are interpretable, respectively, in $F$ and $L$ by
the same formulas without parameters.

∎

##  5 Groups first-order equivalent to $\textrm{T}_{n}(F)$

In this section $F$ is an infinite field of characteristic $\neq 2$, unless
otherwise stated.

To keep the notation simple in this section by $\mathcal{T}$ we denote the set
of all upper triangular transvections, $t_{ij}$, $i>j$. Let $\mathcal{T}^{*}$
denote $\mathcal{T}\cup\\{d_{i}:i=1,\ldots,n\\}$.

######  Lemma 5.1.

Let $H$ be the subgroup of $\textrm{T}_{n}(F)$ generated by
$\textrm{UT}_{n}(F)$ and all the $d_{i}=d_{i}(-1)$, $i=1,\ldots,n$, where $F$
is a characteristic zero field. Then, the field $F$ is regularly bi-
interpretable with $H$.

###### Proof.

By Theorem 3.7, $F$ is e-interpretable in $H$ and by (an easy generalization)
of the proof of Proposition 3.10, $H$ is e-interpretable in $F$. All the
$T_{ij}$ are in Diophantine in $H$, with the possible exception, when $j=i+1$,
where $T_{i,i+1}^{\prime}=R_{H}T_{i,i+1}T_{1n}$ is so. Since $n\geq 3$, either
$i\neq 1$ or $i+1\neq n$. Assume without loss of generality that $i\neq 1$.
Then $T_{i,i+1}=[d_{i},T^{\prime}_{i,i+1}]$. Hence, $T_{i,i+1}$ is also
Diophantine in $H$.

Note that there is a unique sequence $(i_{m},j_{m})$, $1\leq i_{m}<j_{m}\leq
n$, $m=1,\ldots,n(n-1)/2$, such that for any $g$ in $H$, there is a unique
sequence $x$ in $\langle d_{1},\ldots,d_{n}\rangle$, and a unique sequence of
$\alpha_{k}\in F$, such that

$g=xt_{i1,j1}(\alpha_{1})\cdots t_{im,jm}(\alpha_{m})$

The rest of the proof mimics the proof of Theorem A1. ∎

######  Lemma 5.2.

The subgroup $D_{n}(F)$ is Diophantine in $\textrm{T}_{n}(F)$.

###### Proof.

Clear, by noting that

$x\in D_{n}(F)\Leftrightarrow\bigwedge_{i=1}^{n}[x,d_{i}]=1$

∎

######  Lemma 5.3.

The quotient group $P\textrm{T}_{n}(F)=\textrm{T}_{n}(F)/Z(\textrm{T}_{n}(F))$
is regularly bi-interpretable with $F$ with respect to the constants
$\bar{\mathcal{T}^{*}}$, the images of the constants $\mathcal{T}^{*}$ in
$P\textrm{T}_{n}(F)$ under the canonical epimorphism.

###### Proof.

Recall that $Z(G)$, where $G=\textrm{T}_{n}(F)$, is a direct factor of
$D_{n}(R)$, that is, there is a subgroup, say $B_{n}$ of $D_{n}$ such that
$D_{n}=B_{n}\times Z(G)$. For example, we might set $\displaystyle
B_{n}=\prod_{i=1}^{n-1}d_{i}(F^{\times}).$ Let us fix the subgroup $B_{n}$ as
defined above. To emphasize the dependency on the field $F$, we also refer to
this subgroup as $B_{n}(F)$. Since

$G=\textrm{T}_{n}(F)=\textrm{UT}_{n}(F)\rtimes
D_{n}=\textrm{UT}_{n}(F)\rtimes(B_{n}\times Z(G))=(\textrm{UT}_{n}(F)\rtimes
B_{n})\times Z(G)$

clearly, $PT_{n}(F)\cong\textrm{UT}_{n}(F)\rtimes B_{n}$. We prove the
statement for

$K_{n}(F)=\textrm{UT}_{n}(F)\rtimes B_{n}$

where the constants are from $\mathcal{T}^{*}\setminus\\{d_{n}\\}$. We note
that

$d_{1}\cdots d_{n-1}Z(G)=d_{n}^{-1}Z(G)=d_{n}Z(G)$

Hence, the inclusion of all the constants $d_{i}Z(G)$ in the statement is not
necessary.

We proceed by proving a few claims.

_Claim 1._ The subgroup $B_{n}$ and all the subgroups $T_{ij}$ are Diophantine
in $K_{n}$.

_Proof of the claim._ For $B_{n}$ simply observe that $x\in
B_{n}\Leftrightarrow\bigwedge_{i=1}^{n-1}[x,d_{i}]=1$. For $T_{ij}$, note that
one simply does not need the constant $d_{n}$ in any of the formulas in the
proof of Lemma 5.1. In fact, any $n-1$ number of distinct $d_{i}$’s suffices.
∎

_Claim 2._ Each subgroup $d_{k}(F^{\times})$, $k=1,\ldots,n-1$ is Diophantine
in $K_{n}$.

_Proof of the claim._ Note that:

$\displaystyle x\in d_{k}(F^{\times})\Leftrightarrow(x\in
B_{n}\wedge\bigwedge_{i\neq k,j\neq k}[x,t_{ij}]=1)$

In addition, we observe that $Z(K_{n})=\textbf{1}$. ∎

_Claim 3._ For each $1\leq i<j\leq n$, the map

$f_{ij}:d_{i}(F^{\times})\to T_{ij}^{\times},\quad d_{i}(\alpha)\mapsto
t_{ij}(\alpha),\alpha\in F^{\times}$

when $T_{ij}$ is regarded as a ring, is Diophantine.

_Proof of the claim._ Take $x=d_{i}(\alpha)$ for some $\alpha\in F^{\times}$,
then

$y=f_{ij}(x)\Leftrightarrow y=xt_{ij}x^{-1}$

∎

We note that $K_{n}$ is still a group of matrices and given the claims above
one can basically repeat the proof of Theorem A1. here. In the step to prove
regularity, we note that abstract isomorphisms of groups of type $K_{n}$ will
map the derived subgroup $(K_{n}(F))^{\prime}=UT_{n}(F)$ onto the derived
subgroup of $(K_{n}(F_{s}))^{\prime}=UT_{n}(F_{s})$ and then $F\cong F_{S}$
follows. ∎

######  Corollary 5.4.

If $H$ is a group and $H\equiv\textrm{T}_{n}(F)$, then

$\frac{H}{Z(H)}\cong PT_{n}(L)$

for some field $L\equiv F$.

###### Proof.

This follows from Lemma 5.3 and an argument similar to that of proof of
Theorem 4.6. ∎

###  5.1 Non-standard tori and abelian deformations of $\textrm{T}_{n}(R)$

For preliminaries on extension theory and the notation used here we refer the
reader to [31] for further details

Consider $\textrm{T}_{n}(R)$ and the torus $D_{n}(R)$. The subgroup $D_{n}(R)$
is a direct product $(R^{\times})^{n}$ of $n$ copies of the multiplicative
group of units $R^{\times}$ of $R$. The center $Z(G)$ of $G$ consists of
diagonal scalar matrices $Z(G)=\\{\alpha\cdot\textbf{1}:\alpha\in
R^{\times}\\}\cong R^{\times}$, where 1 is the identity matrix. Recall that
$Z(G)$ is a direct factor of $D_{n}(R)$, i.e. there is a subgroup, say $B_{n}$
of $D_{n}$ such that $D_{n}=B_{n}\times Z(G)$. For example, one might set

$B_{n}=\prod_{i=1}^{n-1}d_{i}(R^{\times})$

In the sequel we fix the subgroup $B_{n}$ as defined above. To emphasize the
dependency on the ring $R$, we also refer to this subgroup as
$B_{n}(R^{\times})$.

First, we define a new abelian group $D_{n}(R,f,z)$, where $f\in
S^{2}(B_{n}(R),Z)$ as the abelian extension:

$1\rightarrow Z\rightarrow D_{n}(R,f,Z)\rightarrow B_{n}(R)\rightarrow 1$

In our applications, typically $Z\cong Z(G)$ is elementarily equivalent to
$R^{\times}$ but not necessarily isomorphic to it, the group $G$ to be
constructed below. If $Z\cong R^{\times}$ we simply denote $D_{n}(R,f,Z)$ by
$D_{n}(R,f)$.

We are now ready to define the abelian deformations $T_{n}(R,f,Z)$.

Recall that $G=\textrm{T}_{n}(R)=\textrm{UT}_{n}(R)\rtimes_{\psi}D_{n}$, where
$\psi:D_{n}\to Aut(\textrm{UT}_{n}(R))$, describes the action of $D_{n}$ on
$\textrm{UT}_{n}(R)$ by conjugation and consider the monomorphism
$\phi:D_{n}(R)/Z(G)\to Aut(\textrm{UT}_{n}(R))$ induced by $\psi$. Since
$D_{n}(R)/Z(G)\cong D_{n}(R,f,Z)/Z$, we can assume that
$\phi:D_{n}(R,f,Z)/Z\to Aut(\textrm{UT}_{n}(R))$. Clearly, elements of
$D_{n}(R,f,Z)$ can be written as a product $g=(z,b)$, where $z\in A$ and $b$
is any representative of $gZ$ in $D_{n}(R,Z,f)$. Now define:

$T_{n}(R,f,Z)\stackrel{{\scriptstyle\text{def}}}{{=}}\textrm{UT}_{n}(R)\rtimes
D_{n}(R,f,Z)$

where the product is defined by:

$(u_{1},(z_{1},b_{1}))(u_{2},(z_{2},b_{2}))=(u_{1}u_{2}^{\phi(b_{1})},(z_{1}z_{2}f(b_{1},b_{2}),b_{1}b_{2}))$

It is not very hard to show that this is indeed a well-defined group product
and $Z=Z(T_{n}(R,f,Z))$.

######  Remark 5.5.

Let $G=T_{n}(R,f,Z)$, where $Z=Z(G)$. Note that since

$\displaystyle
Ext((R^{\times})^{n-1},Z)\cong\prod_{i=1}^{n-1}Ext(R^{\times},Z)$

$f$ is cohomologous to $\displaystyle\prod_{i=1}^{n-1}f_{i}$ for some
$f_{i}\in S^{2}(R^{\times},Z)$. So, each 2-cocycle $f_{i}$, $i=1,\ldots,n-1$,
defines a subgroup

$\Delta_{i}(R^{\times})=d_{i}(R^{\times})\cdot Z$

as an abelian extension of $Z$ by $\Delta_{i}(R^{\times})/Z\cong R^{\times}$.
For $\alpha,\beta\in R^{\times}$,

$f(\alpha,\beta)=f_{1}(\alpha,\beta)\cdots f_{n-1}(\alpha,\beta)$

and assume that $f_{n}\in S^{2}(R^{\times},Z)$ defines the subgroup
$\Delta_{n}(R^{\times})$ as an extension of $Z$ by $R^{\times}$. Since
$d_{1}(\alpha)\cdots d_{n}(\alpha)\in Z(G)$, for $\alpha\in R^{\times}$ by
abuse of notation we write it as $\textrm{diag}(\alpha)$. Hence, there exists
a function $\textrm{diag}:R^{\times}\to Z$ such that

$\displaystyle diag(\alpha)diag(\beta)$ $\displaystyle=d_{1}(\alpha)\cdots
d_{n}(\alpha)d_{1}(\beta)\cdots d_{n}(\beta)$
$\displaystyle=d_{1}(\alpha)d_{1}(\beta)\cdots d_{n}(\alpha)d_{n}(\beta)$
$\displaystyle=d_{1}(\alpha\beta)f_{1}(\alpha,\beta)\cdots
d_{n-1}(\alpha\beta)f_{n-1}(\alpha,\beta)d_{n}(\alpha\beta)f_{n}(\alpha,\beta)$
$\displaystyle=diag(\alpha\beta))f(\alpha,\beta)f_{n}(\alpha,\beta)$

Hence $f_{n}$ and $f^{-1}$ are cohomologous, or simply we write $f_{n}\equiv
f^{-1}$, $``\equiv"$ not to be mistaken with elementary equivalence here.

###  5.2 Characterization Theorem

Theorem A3. Assume that $F$ is an infinite field of characteristic $\neq 2$.
If $H$ is any group $H\equiv\textrm{T}_{n}(F)$, then $H\cong T_{n}(L,f,Z)$ for
some field $L\equiv F$, 2-cocycle $f\in S^{2}(B_{n}(L),Z)$ and an abelian
group $Z=Z(H)\equiv Z(\textrm{T}_{n}(F))\cong F^{\times}$ .

###### Proof.

Let $G=\textrm{T}_{n}(F)$. The quotient group $G/Z(G)$ is absolutely
interpretable in $G$ and the same formulas that interpret $G/Z(G)$ in $G$,
interpret $H/Z(H)$ in $H$. Hence, by Corollary 5.4,

$H/Z(H)\cong PT_{n}(L)$

for some $L\equiv F$. Clearly, $Z(H)\equiv F^{\times}\equiv L^{\times}$.

The fact that $D_{n}\cap G^{\prime}=\textbf{1}$ is uniformly expressible by
$\mathcal{L}_{\text{groups}}$-formulas. Now moving to $H$, the same formulas
define a subgroup $E_{n}=E_{n}(\bar{e})$ of $H$, so that
$E_{n}(\bar{e})/Z(H)\cong(L^{\times})^{(n-1)}$ and $Z(H)\equiv L^{\times}$,
$H^{\prime}\cap E_{n}=\textbf{1}$ and the action of $E_{n}$ on
$H^{\prime}\cong UT_{n}(L)$ is the action of the action of $D_{n}$ on the
$t_{ij}\in\textrm{UT}_{n}(L)$ extended to all of $\textrm{UT}_{n}(L)$. This
shows that $H\cong\textrm{UT}_{n}(L)\rtimes E_{n}$, where $E_{n}\cong
D_{n}(L,f,Z)$, where $Z=Z(H)$, and some $f\in S^{2}(B_{n}(L),Z)$.

∎

###  5.3 Sufficiency of the characterization

In this section, we prove a sufficiency statement for the class of fields
where $F^{\times}$ has a finite maximal torsion subgroup $T$. As $T$ is a
product of finite cyclic groups, it is known that $F^{\times}$ splits over
$T$.

We will need to state a few well-known definitions and results.

Let $B$ be an abelian group and $A$ a subgroup of $B$. Then $A$ is called a
_pure subgroup of $B$_ if $\forall n\in\mathbb{N}$, $nA=nB\cap A$.

######  Lemma 5.6.

Let $A\leq B$ be abelian groups such that the quotient group $B/A$ is torsion-
free. Then $A$ is a pure subgroup of $B$.

An abelian group $A$ is called _pure-injective_ if $A$ is a direct summand in
any abelian group $B$ that contains $A$ as a pure subgroup.

The following theorem expresses a connection between pure-injective groups and
uncountably saturated abelian groups.

######  Theorem 5.7 ([17], Theorem 1.11).

Let $\kappa$ be any uncountable cardinal. Then any $\kappa$-saturated abelian
group is pure-injective.

######  Definition 5.8 (NFT-field).

A field $F$ is called a field of _number field type_ , if the torsion subgroup
$T$ of $F^{\times}$ is finite. We refer to these fields as NFT-fields.

Note that if $F$ is an NFT-field then $F^{\times}$ splits over the torsion
subgroup $T$.

######  Definition 5.9.

Let $F$ be an NFT-field with $F^{\times}=T(F^{\times})\times B$ and let $Z$ be
an abelian group with a similar decomposition $Z=T(Z)\times B(Z)$ where the
maximal torsion subgroup is denoted by $T(Z)$ and the complement $B(Z)$ is
torsion-free. We know that $Ext(F^{\times},Z)\cong Ext(T(F^{\times}),A)\times
Ext(B(F^{\times}),Z)$, so any symmetric 2-cocycle $f\in S^{2}(F^{\times},Z)$
can be written as $f\equiv f_{1}\cdot f_{2}$, where $f_{1}\in
S^{2}(T(F^{\times}),Z)$, $f_{2}\in S^{2}(B(F^{\times}),Z)$. The symmetric
2-cocycle $f$ is said to be a coboundary on torsion or CoT if $f_{1}$ is a
2-coboundary.

######  Remark 5.10.

Assume for abelian groups $Z_{i}$, $i=1,2$ we have $Z_{i}=T_{i}\times B_{i}$
where $T_{i}$ and $B_{i}$ are some subgroups of the $Z_{i}$. Consider a
symmetric 2-cocycle $f:Z_{1}\to Z_{2}$. By abuse of notation, we consider $f$
as $f:T_{1}\cdot B_{1}\to T_{2}\times B_{2}$. Then $f$ is cohomologous to
$(g_{1}g_{2},h_{1}h_{2})$ where $g_{1}\in S^{2}(T_{1},T_{2}))$, $g_{2}\in
S^{2}(T_{1},B_{2})$, $h_{1}\in S^{2}(B_{1},T_{2})$ and finally $h_{2}\in
S^{2}(B_{1},B_{2})$. We will use this notation in the following.

######  Lemma 5.11.

Let $F$ be an NFT-field, and $F^{\times}=T(F^{\times})\times B(F^{\times})$
and $Z\equiv F^{\times}$. Then $Z=T(Z)\times B(Z)$, where $f\in
S^{2}(F^{\times},Z)$ is CoT and $(I,D)$ is an ultrafilter so that the
ultrapowers $(F^{\times})^{*}$ and $Z^{*}$ over $D$ are
$\aleph_{1}$-saturated. Then the 2-cocycle $f^{*}\in
S^{2}((F^{\times})^{*},(Z)^{*})$ induced by $f$ is a 2-coboundary.

###### Proof.

Since $T(F^{\times})$ is finite, $T(Z)$ is also finite and $Z$ splits over it.
Note that $(F^{\times})^{*}=(F^{*})^{\times}$,
$T((F^{\times})^{*})=(T(F^{\times}))^{*}\cong T(F^{\times})$. Then
$(F^{\times})^{*}\cong T\times B^{*}$. A similar decomposition occurs in
$Z^{*}$. The assumption that $f$ is CoT implies that the induced cocycle
$f^{*}\in S^{2}((F^{\times})^{*},Z^{*})$ is CoT. Then, using the notation of
Remark 5.10, we see that $g_{1}^{*}\in S^{2}(T^{*}(F^{\times}),T^{*}(Z))$ and
$g^{*}_{2}\in S^{2}(T^{*}(F^{\times}),B^{*})$ are both coboundaries. The
2-cocycle $h_{1}^{*}\in S^{2}(B(F^{\times})^{*},T(Z)^{*})$ is a coboundary
since $B(F^{\times})^{*}$ is torsion-free abelian and $T(Z)^{*}$ is finite
abelian. The 2-cocycle $h_{2}^{*}\in S^{2}(B(F^{\times})^{*},B(Z)^{*})$ is
also a coboundary since $A^{*}$ is pure-injective by Theorem 5.7 and also
$A^{*}$ is a pure subgroup in an extension represented by $h_{2}^{*}$ because
of Lemma 5.6. Consequently, $f^{*}$ is a coboundary. ∎

######  Theorem 5.12.

Let $H=T_{n}(F,f,Z)$, where $F$ is an NFT-field and $f$ is CoT. Then
$H\equiv\textrm{T}_{n}(F)$.

###### Proof.

Let $G=\textrm{T}_{n}(F)$, and Let $B_{n}(F^{\times})$ be the subgroup
introduced at the beginning of Subsection 5.1.

Let $(I,D)$ be an $\aleph_{1}$-incomplete ultrafilter, and let $M^{*}$ denote
the ultraproduct $M^{I}/D$. Then
$B_{n}((F^{*})^{\times})=B_{n}((F^{\times})^{*})=B_{n}^{*}(F^{\times})$ is
$\aleph_{1}$-saturated. Recall that $f$ is cohomologous with $f_{1}\cdots
f_{n-1}$, $f_{i}\in S^{2}(F^{\times},Z)$. The hypothesis implies that each
$f_{i}$ is CoT. If $f^{*}_{i}\in S^{2}((F^{\times})^{*},A^{*})$ denotes the
2-cocycle induced by $f_{i}$, then for each $i=1,\ldots,n-1$, $f^{*}_{i}$ is a
2-coboundary by the Lemma 5.11. The fact that

$T^{*}_{n}(F,f,Z)\cong T_{n}(F^{*},f^{*},Z^{*})\cong PT_{n}(F^{*})\times
Z^{*}$

requires only some routine checking.

Therefore,

$T^{*}_{n}(F,f,Z)\cong PT_{n}(F^{*})\times Z^{*}\equiv
PT_{n}(F^{*})\times(F^{\times})^{*}\cong T_{n}^{*}(F)$

Hence, $T_{n}(F,f,Z)\equiv T_{n}(F)$ ∎

###  5.4 Real closed fields and algebraically closed fields

######  Theorem 5.13.

Let $F$ be a real closed field or an algebraically closed field of
characteristic $\neq 2$. Let $G=T_{n}(F)$ and $H$ be a group. Then the
following are equivalent

  1. 1.

$G\equiv H$

  2. 2.

$H\cong PT_{n}(L)\times Z(H)$ for a filed $L\equiv F$ and $Z(H)\equiv
F^{\times}$.

###### Proof.

We shall only prove $(1.)\Rightarrow(2.)$, as the other direction is clear.

So assume that $F$ is an algebraically closed field of characteristic $\neq
2$. Then $Z(H)\equiv F^{\times}\cong Z(G)$ is a radicable (divisible) abelian
group and as $H^{\prime}\cong PT_{n}(L)$ for some $L\equiv F$, $H\cong
PT_{n}(L)\times Z(H)$ as desired.

If $F$ is real closed, then $F^{\times}\cong\\{\pm\textbf{1}\\}\times B(G)$,
where $B(G)$ is the multiplicative group of positive elements of $F$, and
therefore definable in $F^{\times}\equiv Z(G)$ by:

$\Phi(x)=\exists y(y^{2}=x).$

Now $G^{\prime}\cdot\langle d_{i}:i=1,\ldots n\rangle$ is absolutely definable
in $G$ by $\Psi(x)=\exists x(x^{2}\in G^{\prime})$. This subgroup is evidently
regularly bi-interpretable with $F$. As the absolutely definable subgroup
$\\{\pm\textbf{1}\\}$ splits from $G^{\prime}$, in $H$ the corresponding
subgroup splits from $H^{\prime}$. Therefore, $Z(H)=\\{\pm\textbf{1}\\}\times
B(H)$, where $B(H)\equiv B(G)$, splits from $H$.

∎

###  5.5 Abelian deformations of $T_{n}(\mathbb{Q})$ which are not isomorphic
to any $T_{n}(F)$ for any field $F$

Let $G=\textrm{T}_{n}(F)$. Note that since $G\cong(\textrm{UT}_{n}(F)\rtimes
B_{n}(F))\times F^{\times}$, any group of the form
$H=(\textrm{UT}_{n}(L)\rtimes B_{n}(L))\times A$, where $L\equiv F$ and
$A\equiv F^{\times}\equiv Z(G)$ is elementarily equivalent to
$\textrm{T}_{n}(F)$ by a combination of the Feferman-Vaught Theorem on
elementary equivalence of direct products, a simple ultraproduct argument, and
Keisler-Shelah’s Theorem. This is one source of elementsarily equivalent
groups to $\textrm{T}_{n}(F)$, but not isomorphic to any $T_{n}$. In the
following, we shall construct a different type of example.

######  Lemma 5.14.

There exists a countable field $K$, such that $\mathbb{Q}\equiv F$ and also
$Ext(F^{\times},F^{\times})\neq 1$.

###### Proof.

By ([5], Proposition 2.2.16) there exists a countable ring $R$ where
$R\equiv\mathbb{Z}$ and $Ext(R^{+},R^{+})\neq 0$. In fact, for such a ring
$R$, $R^{+}=A\oplus D$ where $A\neq 0$ is a reduced abelian group and $D\neq
0$ is a countable torsion-free divisible abelian group. Then by ([19], 54.5)
$Ext(D,A)\neq 0$, which implies that $Ext(R^{+},R^{+})\neq 0$.

Claim: Let $F$ be the field of fraction of the integral domain $R$ discussed
above. Then $F^{\times}\cong\\{\pm 1\\}\times A\times D$ where $A$ and $D$ are
both nontrivial, $A$ is reduced and $D$ is countable, torsion-free, and
divisible.

Proof of the claim. We know that $\mathbb{Q}^{\times}=\\{\pm 1\\}\times C$
where $C\cong\prod_{i\in\omega}C_{\infty}$, and $C_{\infty}$ is the infinite
cyclic group. First, we show that $C$ is definable in $\mathbb{Q}^{\times}$
uniformly with respect to $Th(\mathbb{Q})$. By a theorem of [36] the ring
$\mathbb{Z}$ is definable in $\mathbb{Q}$. Arithmetic
$\mathbb{N}=\langle|\mathbb{N}|,+,\times,0,1\rangle$ is definable in
$\mathbb{Z}$. In particular, the natural order on $\mathbb{Z}$ is definable in
$\mathbb{Z}$. Since $\mathbb{Z}$ is definable in $\mathbb{Q}$, it is easy to
see that natural order on $\mathbb{Q}$ is definable in $\mathbb{Q}$. Now
$\mathbb{Q}^{\times}$ is definable in $\mathbb{Q}$ as the set of invertible
elements. The subgroup $C$ is definable in $\mathbb{Q}^{\times}$ as the set of
all positive elements of $\mathbb{Q}$. The subgroup $\\{\pm 1\\}$ is obviously
definable in $\mathbb{Q}^{\times}$.

We note that $\mathbb{Q}$ is interpretable in $\mathbb{Z}$ and the same
formulas that interpret $\mathbb{Q}$ in $\mathbb{Z}$ interpret $F$ in $R$.
Thus, indeed $\mathbb{Q}\equiv F$. The formulas that define $\mathbb{Z}$ in
$\mathbb{Q}$ define a subring $R$ of $K$ in $K$ that is elementarily
equivalent to $\mathbb{Z}$. Again, the formulas that define $C$ in
$\mathbb{Q}$ define a subgroup $E$ of the multiplicative group $F^{\times}$ of
$F$ in $F$. So, clearly $F^{\times}=\\{\pm 1\\}\times E$. Next, we will show
that $E$ has a direct factor which is isomorphic as an abelian group to
$R^{+}$. By Gödel’s work all primitive recursive functions on $\mathbb{N}$ are
arithmetically definable, in particular the predicate $z=2^{x}$ is
arithmetically definable. That is, there is a first-order formula $exp(z,y,x)$
of the language of arithmetic such that

$z=2^{x}\Leftrightarrow\mathbb{N}\models exp(z,1+1,x).$

In particular, it can be seen that the set
$2^{\mathbb{N}}=\\{2^{n}:n\in\mathbb{N}\\}$ has a definable arithmetic
structure, also denoted by $2^{\mathbb{N}}$. Since $\mathbb{N}$ is definable
in $\mathbb{Z}$, so is $2^{\mathbb{N}}$. This implies that $2^{\mathbb{Z}}$ is
definable in $\mathbb{Q}$ and also carries a ring structure interpretable in
$\mathbb{Q}$ and isomorphic to the ring of integers $\mathbb{Z}$. On the other
hand, $2^{\mathbb{Z}}\subset\mathbb{Q}^{\times}$. Formulas that define an
arithmetic structure on $2^{\mathbb{Z}}$ define in $K$ a subgroup $2^{R}\leq
E<K^{\times}$ of $K$ with a definable ring structure isomorphic to $R$. Now
$R^{+}\cong 2^{R}\cong A^{\prime}\times D^{\prime}$ where both $A^{\prime}$
and $D^{\prime}$ are non-empty and $D^{\prime}$ is torsion-free divisible.
This means that $C=A\times D$ where both $A$ and $D$ are nonempty, $A$ is
reduced torsion-free and $D$ is countable, torsion-free and divisible. This
ends the proof of the claim.

As a corollary of the claim and the fact $Ext(D,A)\neq 1$, we can conclude
that $Ext(F^{\times},F^{\times})\neq 1$. ∎

######  Theorem 5.15.

There exists a field $F\equiv\mathbb{Q}$ and a 2-cocycle $f\in
S^{2}((F^{\times})^{n-1},F^{\times})$, such that $T_{n}(F,f)\equiv
T_{n}(\mathbb{Q})=G$, but $T_{n}(F,f)\ncong T_{n}(L)$, for any field $L$.

###### Proof.

Pick the field $F$ found in Lemma 5.14. We can clearly find a CoT 2-cocycle
$f\in S^{2}(F^{\times},F^{\times})$ which is not a 2-coboundary. Set, say,
$f_{1}=f$ and the let the rest of the $f_{i}$ be trivial and form
$H=T_{n}(K,f)$. By Theorem 5.12, we have $H\equiv T_{n}(\mathbb{Q})$. If
$\phi:H\to G$, where $G=T_{n}(L)$ is an isomorphism of groups, then
$\textrm{UT}_{n}(F)\cong\phi(H^{\prime})\cong
G^{\prime}\cong\textrm{UT}_{n}(L)$. This in fact shows that $F\cong L$. Now,
passing to the abelianizations, $\phi$ induces an isomorphism of
$D_{n}(F,f,Z(H))$ onto $D_{n}(L,Z(G))$, mapping $Z(H)$ onto $Z(G)$. This is
impossible, otherwise $f$ would be a 2-coboundary.∎

##  6  Groups first-order equivalent to $\textrm{GL}_{n}(F)$, $n\geq 3$

In this section $F$ is an infinite field of characteristic $\neq 2$. The case
of $\textrm{GL}_{n}(F)$ is similar to the case of $\textrm{T}_{n}(F)$, but
with more subtlety. We know that

$G=\textrm{GL}_{n}(F)\cong\textrm{SL}_{n}(F)\rtimes d_{1}(F^{\times})$

Hence, $\textrm{SL}_{n}(F)=G^{\prime}$, while there is a non-trivial
intersection between $D_{n}(F)$ and $\textrm{SL}_{n}(F)$. In fact, for any
$\alpha\in F^{\times}$

$d_{1}(\alpha^{n})=\textrm{diag}(\alpha)\prod_{i=2}^{n}d_{1}(\alpha)d_{i}(\alpha^{-1})$
(6.1)

while $\prod_{i=2}^{n}d_{1}(\alpha)d_{i}(\alpha^{-1})\in\textrm{SL}_{n}(F)$
and $\textrm{diag}(\alpha)\in Z(G)$. On the other hand, for
$G=\textrm{GL}_{n}(F)$, since $G^{\prime}=\textrm{SL}_{n}(F)$ and
$\textrm{SL}_{n}(F)$ is boundedly generated by commutators, $G^{\prime}$ is
absolutely and uniformly definable in $G$. By Theorem 4.6 if $H\equiv G$, then
$H^{\prime}\cong\textrm{SL}_{n}(L)$ for some field $L\equiv F$, $Z(H)\equiv
F^{\times}$ and

$G/(G^{\prime}\cdot Z(G))\cong F^{\times}/(F^{\times})^{n}$

In fact we have proven the following statement.

Theorem A4. If $H$ is any group such that $H\equiv\textrm{GL}_{n}(F)$, then
$H$ fits into the short exact sequence:

$1\rightarrow H^{\prime}\cdot Z(H)\rightarrow
H\rightarrow\frac{L^{\times}}{(L^{\times})^{n}}\rightarrow 1$

where $H^{\prime}\cong\textrm{SL}_{n}(L)$, $L\equiv F$, as fields and
$Z(H)\equiv F^{\times}$.

The above enables us to characterize some important cases.

Define the _isolator_ of $M$ in $G$ by

$Is_{G}(M)=\\{g\in G|x^{n}\in M,\text{ for some $n\in\mathbb{N}$}\\}$

Clearly $Is_{G}(M)$ is a normal subgroup of $G$. Now consider the following
sequence of subgroups of $G=\textrm{GL}_{n}(F)$

$G\geq Is(G^{\prime}\cdot Z(G))\geq Is(G^{\prime})\cdot Z(G)\geq
Is(G^{\prime})\geq G^{\prime}$

Now let $G=\textrm{GL}_{n}(F)$ or, in fact, a group
$G\equiv\textrm{GL}_{n}(F)$. By Theorem A4., the sequence above becomes

$G=Is(G^{\prime}\cdot Z(G))\geq Is(G^{\prime})\cdot Z(G)\geq
Is(G^{\prime})\geq G^{\prime}\geq 1$ (6.2)

We shall refer to this sequence in the section, when we try to characterize
groups elementarily equivalent to $\textrm{GL}_{n}(F)$ for some classes of
fields.

###  6.1 Algebraically closed fields

######  Theorem 6.1.

Let $F$ be a characteristic zero algebraically closed field. Then the
following are equivalent:

  1. 1.

$H\equiv\textrm{GL}_{n}(F)$,

  2. 2.

$H\cong(H^{\prime}\times Z(H))/\mu_{n}$, $H^{\prime}\cong\textrm{SL}_{n}(L)$,
$Z(H)\equiv\mathbb{C}^{\times}$, $\mu_{n}=ker\phi$, $\phi:H^{\prime}\times
Z(H)\to H,(h,z)\mapsto hz$.

###### Proof.

$(1.)\Rightarrow(2.)$ Going back to the sequence in Theorem A4.,
$F^{\times}=(F^{\times})^{n}$. As both $G^{\prime}$ and $Z(G)$ are absolutely
and uniformly definable in $G$, the homomorphism
$\phi:\textrm{SL}_{n}(F)\times Z(G)\to G$, $(g,a)\mapsto ga$ is absolutely and
uniformly definable in $G$, as is its finite kernel $\mu_{n}$. Therefore, the
induced isomorphism, say $\tilde{\phi}$ is absolutely and uniformly
interpretable in $G$. The same formulas set an isomorphism as in (2.) in $H$.

$G=G^{\prime}Z(G)$, we have $H=H^{\prime}Z(H)$, while
$Z(H)\equiv\mathbb{C}^{\times}$ and $H^{\prime}\cong\textrm{SL}_{n}(L)$ for
some $L\equiv F$. Note that the subgroup $\mu_{n}$ is isomorphic to
$\\{(\omega,\omega^{-1}):\omega^{n}=1,\omega\in\mathbb{C}\\}$ or just the
group of $n$-th roots of unity. One might call the homomorphism $\phi$ a _non-
standard central isogeny_.

$(2.)\Rightarrow(1.)$ By assumption and the classical Fefferman and Vaught
Theorem $G^{\prime}\times Z(G)\equiv H^{\prime}\times Z(H)$. Now, the same
formulas that interpret $(H^{\prime}\times Z(H))/\mu_{n}$ in $H$, interpret
$(G^{\prime}\times Z(G))/\mu_{n}$ in $G$, where
$G^{\prime}=\textrm{SL}_{n}(F)$. Therefore, $H\equiv\textrm{GL}_{n}(F)$.

∎

###  6.2 Real closed fields

######  Lemma 6.2.

Let $G$, be a group that fits into the sequence of Theorem A4. where

  1. 1.

$Z=Z(G)$ splits over its torsion subgroup $T(Z)$, that is, there exists a
subgroup $B$ of $Z$, such that $Z=B\times T$

  2. 2.

$A$ is a radicable (multiplicatively divisible) group.

Then

$G\cong Is_{G}(G^{\prime})\times B$

Where $G^{\prime}\cong\textrm{SL}_{n}(L)$

###### Proof.

Clearly, in the situation above the sequence in (6.2) becomes:

$G=Is(G^{\prime})\cdot B\geq Is(G^{\prime})\geq\textbf{1}$

Where, $[B,Is(G^{\prime})]=\textbf{1}$, both $B$ and $Is(G^{\prime})$ are
normal in $G$ and $B\cap Is(G^{\prime})=\textbf{1}$. ∎

######  Theorem 6.3.

Let $F$ be a real closed field, that is, $F\equiv\mathbb{R}$, $\mathbb{R}$ the
field of real numbers, and let $H$ be a group. The following are equivalent:

  1. 1.

$H\equiv\textrm{GL}_{n}(\mathbb{R})$,

  2. 2.

$H\cong\left(\textrm{SL}_{n}(L)\cdot\langle d_{1}\rangle\right)\times B(H)$,
where $L\equiv\mathbb{R}$ and $B(H)\equiv(\mathbb{R}_{+})^{\times}$, where
$(\mathbb{R}_{+})^{\times}$ is the multiplicative group of positive reals.

###### Proof.

(1.) $\Rightarrow$ (2.): Clearly, $G=\textrm{GL}_{n}(\mathbb{R})$ satisfies
the conditions of Lemma 6.2, so $\textrm{GL}_{n}(\mathbb{R})\cong
Is_{G}(\textrm{SL}_{n}(\mathbb{R}))\times B(G)$ as described while
$Is_{G}(G^{\prime})=Is_{G}(\textrm{SL}_{n}(\mathbb{R}))=\textrm{SL}_{n}(L)\cdot\langle
d_{1}\rangle$. This latter subgroup is an absolutely and uniformly definable
subgroup of $G$, which is regularly bi-interpretable with $F$. Therefore,
$Is(H^{\prime})\cong\textrm{SL}_{n}(\mathbb{R})\cdot\langle d_{1}\rangle$. The
subgroup $B(G)$ is definable in $Z(G)$, by the equation

$\Phi(x)=\exists y(y^{2}=x)$

This subgroup is both torsion-free and radicable, both properties
axiomatizable with a set of first-order sentences. Hence,
$B(H)\equiv(\mathbb{R}_{+})^{\times}$ splits as a direct summand from the
group $H$, the complement being $Is_{H}(H^{\prime})$, whose structure has been
described above.

(2.) $\Rightarrow(1.)$: If $L\equiv\mathbb{R}$, then
$\textrm{SL}_{n}(L)\cdot\langle
d_{1}\rangle\equiv\textrm{SL}_{n}(\mathbb{R})\cdot\langle d_{1}\rangle$ by an
ultrapower argument and the fact that the action of $d_{1}\in\textrm{GL}_{n}$
on $\textrm{SL}_{n}$ in both cases is the same. The rest follows from the
Feferman-Vaught Theorem. ∎

###  6.3 NFT-fields

Let $F$ be an NFT field; then the sequence in (6.2),
$Is(G^{\prime})=\textrm{SL}_{n}(F)\rtimes T(d_{1}(F^{\times}))$ is uniformly
definable in $G=\textrm{GL}_{n}(F)$ and clearly regularly bi-interpretable
with $F$. Let $T$ be the torsion subgroup of $Z(G)$ and $G_{0}\leq Z(G)$ its
complement. We call any subgroup $G_{0}$ with this description an _addition of
$G$_ and the quotient $G_{f}=G/G_{0}$ a _foundation of $G$_. Since $T$ is a
direct product of cyclic groups, such a subgroup $G_{0}$ always exists. Hence,
$G$ is a central extension of $G_{0}$ by $G_{f}$. Note that $G_{f}=G/G_{0}$
contains a copy of $\textrm{SL}_{n}(F)$, since $G^{\prime}\cap G_{0}=1$.

######  Lemma 6.4.

Let $F$ be an NFT-field. Then, for any two additions $A_{1}$ and $A_{2}$ of
$G=\textrm{GL}_{n}(F)$, there exists an automorphism $\phi$ of
$\textrm{GL}_{n}(F)$ such that $\phi(A_{1})=A_{2}$.

###### Proof.

We recall that if $\eta_{i}:F^{\times}/T\to F^{\times}$ are the corresponding
transversals of $A_{i}$’s, then $\mu:F^{\times}/T\to T$ defined by
$\mu(\alpha)=\eta_{1}(\alpha)\eta_{2}^{-1}(\alpha)$ is indeed a homomorphism.
In fact, since $F^{\times}\cong T\times F^{\times}/T$, we can just assume that
$\mu:F^{\times}\to T$ is defined as $\mu(T)=1$ and
$\mu|_{F^{\times}/T}=\eta_{1}\eta_{2}^{-1}$. So, to prove the statement, we
need to show that there exists an automorphism $\phi$ of $\textrm{GL}_{n}(F)$
such that $\phi(\textrm{diag}(\alpha))=\textrm{diag}(\alpha\cdot\mu(\alpha))$.
To this end, consider the subgroup $H$ of $\textrm{GL}_{n}(F)$ generated by
$\textrm{SL}_{n}(F)$ and the subgroup
$H_{1}=\\{d_{1}(\alpha)\textrm{diag}(\phi(\alpha))|\alpha\in F^{\times}\\}$.
In fact, $H=\textrm{SL}_{n}(F)\cdot H_{1}$. All we really need to show is that
if $x\in\textrm{SL}_{n}(F)\cap H_{1}$, then $x=1$. So, assume that $x$ is such
an element. Then $\alpha\mu^{n}(\alpha)=det(x)=1$. Let $\alpha=\beta\cdot t$
where $\beta$ is of infinite order and $t$ is torsion. Clearly, we must have
$\beta=1$, which implies that $\mu(\alpha)=\mu(\beta)=1$. Hence $t=1$ and
$x=1$.

Next, we claim that $H=\textrm{GL}_{n}(F)$. We already know that
$\textrm{SL}_{n}(F)\leq H$. Now we need to prove that $d_{1}(\alpha)\in H$ for
all $\alpha\in F^{\times}$. Write again $\alpha=\beta\cdot t$ as above. Note
that if $\beta=1$, then $d_{1}(\alpha)diag(\mu(\alpha))=d_{1}(t)$. So, in
fact, for all $t\in T(F^{\times})$, $d_{1}(t)\in H$. Hence
$d_{i}(t)=d_{1}(t)d_{1}(t^{-1})d_{i}(t)\in H$. So for all such $t$,
$diag(t)\in H$. Since $im(\mu)\leq T$, all $\textrm{diag}(\mu(\alpha))\in H$,
and hence $d_{1}(\alpha)\in H$ for all $\alpha\in F^{\times}$.

Next, define

$\phi:\textrm{GL}_{n}(F)\to\textrm{GL}_{n}(F),\quad(g,d_{1}(\alpha))\mapsto(g,d_{1}(\alpha)\textrm{diag}(\mu(\alpha)))$

The map $\phi$ is a homomorphism:

$\displaystyle\phi((g_{1},d_{1}(\alpha_{1}))\cdot(g_{2},d_{1}(\alpha_{2})))$
$\displaystyle=\phi((g_{1}g_{2}^{d_{1}(\alpha_{1})},d_{1}(\alpha_{1})d_{1}(\alpha_{2}))$
$\displaystyle=(g_{1}g_{2}^{d_{1}(\alpha_{1})},d_{1}(\alpha_{1}\alpha_{2}diag(\mu(\alpha_{1}\alpha_{2})))$
$\displaystyle=(g_{1}g_{2}^{d_{1}(\alpha_{1})\textrm{diag}(\mu(\alpha_{1})},d_{1}(\alpha_{1}\alpha_{2}\textrm{diag}(\mu(\alpha_{1}\alpha_{2})))$
$\displaystyle=(g_{1},d_{1}(\alpha_{1})\textrm{diag}(\mu(\alpha_{1})))\cdot(g_{2},d_{1}(\alpha_{2})\textrm{diag}(\mu(\alpha_{2}))$

It is also injective. Again, if $\alpha=\beta\cdot t$ with $\beta$ of infinite
order and $t$ torsion and $d_{1}(\alpha)\textrm{diag}(\phi(\alpha))=1$, we
immediately conclude that $\beta=1$ and $\mu(\alpha)=1$, implying that $t=1$,
hence $\alpha=1$. This implies that $\phi(g,d_{1}(\alpha))=1$,
$g=d_{1}(\alpha)=1$.

By the above, it is also surjective, and

$\phi(\textrm{diag}(\alpha))=\textrm{diag}(\alpha\cdot\mu(\alpha))$

∎

######  Lemma 6.5.

Let $F$ be an NFT-field, $G=\textrm{GL}_{n}(F)$ with an addition $G_{0}$ and
$H\equiv G$. Then there exists a subgroup $H_{0}\leq Z(H)$ such that
$Is(H^{\prime})\cdot Z(H)=Is(H^{\prime})\times H_{0}$ and $G_{f}\equiv
H/H_{0}$.

###### Proof.

We proceed with an ultrapower argument. Let $D$ be an ultrafilter $I$ such
that $G^{I}/D\cong H^{I}/D$. We observe that $H^{*}=H^{I}/D$ is isomorphic to
$\textrm{GL}_{n}(F^{I}/D)$. Set $L=F^{I}/D$ and assume
$H^{*}=\textrm{GL}_{n}(L)$. Since $G^{\prime}$ is of finite width, it is not
very hard to prove $(G^{\prime})^{*}=(G^{*})^{\prime}$ and then to conclude
that $Is((G^{\prime})^{*})=(Is(G^{\prime}))^{*}$. Therefore, if $G_{0}$ is an
addition to $G$, then $G^{*}_{0}=G_{0}^{I}/D$ is an addition to
$G^{*}=G^{I}/D$. Assume that $A_{1}$ is the image of $G^{*}_{0}$ in $H^{*}$
under the assumed isomorphism. By Lemma 6.4 there exists an automorphism of
$H^{*}$ mapping $A$ onto $H_{0}^{*}$. Hence,

$(G_{f})^{*}\cong G^{*}/G^{*}_{0}\cong H^{*}/H_{0}^{*}\cong(H_{f})^{*}$

implying that $G_{f}\equiv H/H_{0}$. ∎

######  Proposition 6.6.

Let $F$ be an NFT-field, $G=\textrm{GL}_{n}(F)$ and $H\equiv G$. Then there
exists an addition $H_{0}$ of $H$, that is, a subgroup $H_{0}\leq Z(H)$, such
that $Is(H^{\prime})Z(H)=Is(G^{\prime})\times H_{0}$, a field $L\equiv F$, and
an addition $K_{0}$ of $K=\textrm{GL}_{n}(L)$ such that $H/H_{0}\cong K_{f}$.

###### Proof.

Since $G^{\prime}=\textrm{SL}_{n}(F)$ and $G^{\prime}\cap G_{0}=1$ and
$G_{0}\leq Z(G)$, $(G_{f})^{\prime}\cong\textrm{SL}_{n}(F)$. Therefore
$(G_{f})^{\prime}$ is regularly bi-interpretable with $F$ with respect to the
images of the constants $\mathcal{T}$ in $G_{f}$. We shall identify
$(G_{f})^{\prime}$ with $\textrm{SL}_{n}(F)$ and note that the image of
$Is(G^{\prime})$ in $G_{f}$ is also isomorphic to $Is(G^{\prime})$. Consider
an addition $H_{0}$ of $H$ as in Lemma 6.5, and let $H_{f}=H/H_{0}$. We can
now conclude that $(H_{f})^{\prime}\cong\textrm{SL}_{n}(L)$ for some field
$L\equiv F$. Again, to keep the notation light we identify $(H_{f})^{\prime}$
with the copy of $SL_{n}(L)$ in $H_{f}$. From this point on, we also set
$K=\textrm{GL}_{n}(L)$ and pick an addition $K_{0}$ of $K$.

Consider the first-order formula, say $\Phi(x)$, resulting from the
conjunction of all the following:

$\exists y\in T_{ij}(xt_{ij}x^{-1}=y),\text{ if }i=1\text{ or }j=1$
$xt_{ij}x^{-1}=1,\text{ if }i\neq 1\text{ and }j\neq 1.$

The formula defines a subgroup $\Delta_{1}(G_{f})$. Furthermore,
$G_{f}\models\Theta\wedge\Psi$, where

$\Theta:\forall y((y\in T_{12}\wedge y\neq 1)\to\exists x(\Phi(x)\wedge
xt_{12}x^{-1}=y)$

and

$\Psi:\forall y_{1},y_{2}((\Phi(y_{1})\wedge\Phi(y_{2})\wedge
y_{1}t_{12}y_{1}^{-1}=y_{2}t_{12}y_{2}^{-1})\to y_{1}y_{2}^{-1}\in Z(G_{f}))$

Hence $\Delta_{1}(G_{f})/Z(G_{f})$ is definably isomorphic to $F^{\times}$ via
an isomorphism, say $\psi_{G}:\Delta_{1}/Z(G_{f})\to F^{\times}$. The same
formula $\Phi$ defines in $H$ a subgroup $\Delta_{1}(H)$ isomorphic to
$L^{\times}\cdot Z(H_{f})$ where $\Delta_{1}(H_{f})/Z(H_{f})$ is isomorphic to
$L^{\times}$ via an isomorphism, say $\psi_{H}$ and $H_{f}$ is generated by
$\Delta_{1}(H_{f})$ and $(H_{f})^{\prime}$. Since $Z(G_{f})$ is finite,
$(G_{f})^{\prime}\cdot T(\Delta_{1}(F))$ is definable in $G_{f}$ and regularly
bi-interpretable with $F$. Hence, we might identify $H_{f}\cdot Z(H_{f})$ with
the copy of $SL_{n}(L)\cdot Z(K_{f})$ in $K_{f}$.

In $G=\textrm{GL}_{n}(F)$, we have that for any $\alpha\in F^{\times}$,
Equation (6.1) holds in $G$. Therefore, the following holds in $G_{f}$

$G_{f}\models\forall
x\in\Delta_{1}(G_{f})\left((xZ)^{n}=\prod_{i=2}^{n}d_{1}(\psi_{G}(xZ))d_{i}((\psi_{G}(xZ))^{-1})Z\right)$
(6.3)

The same holds for $H_{f}$ with obvious substitutions. Now, let
$\mu:L^{\times}\to T(L^{\times})$ be the defining transversal (homomorphism)
of the addition $K_{0}$. Define a map $\phi:H_{f}\to K_{f}$ such that

  * •

$\phi=Id$ on $H_{f}^{\prime}\cdot Z(H_{f})$

  * •

$\phi(\alpha)=d_{1}(\psi(\alpha)[\mu(\psi(\alpha))]^{-1})$, where
$\alpha\in\Delta_{1}$, ranges over all representatives of the cosets of
$\Delta_{1}/Z$. Note that $\Delta_{1}$ splits over $Z$ and $\psi$ depends only
on the cosets.

For $\alpha$ as above, $\psi(\alpha)\in L^{\times}$ can be written as
$t\cdot\beta$, where $t$ is torsion and $\beta$ is torsion-free. The element
$t$ is uniquely determined.

Let us agree that in this context diag is understood as an isomorphism
$\textrm{diag}:T(L^{\times})\to Z(K_{f})$.

Again identifying $\textrm{SL}_{n}(L)$ with its copy in $K_{f}$ we can verify
that

$K_{f}\cong\left[\left(\textrm{SL}_{n}(L)\cdot Z(G_{f})\right)\rtimes
L^{\times}\right]/N$ (6.4)

where $L^{\times}$ acts on $\textrm{SL}_{n}(L)$ as $d_{1}(L^{\times})$ and $N$
is the normal subgroup generated by all the following elements as $\alpha$
ranges over $L^{\times}$:

$\left(\prod_{i=2}^{n}(d_{1}(\alpha)d_{i}(\alpha)^{-1})\textrm{diag}(t\left[\mu(\alpha))\right]^{-1},\alpha^{-n}\right)$
(6.5)

Now, since (6.3) also holds in $H_{f}$ with obvious substitutions, we observe
that for $\alpha\in\Delta_{1}$ as described,

$\phi^{n}(\alpha)=\textrm{diag}\left(t[\mu(\psi(\alpha)]^{-1}\right)\prod_{i=2}^{n}d_{1}(\psi(\alpha))d_{i}(\psi(\alpha))^{-1}$

Hence, we are guaranteed that $\phi$ is well defined on
$\Delta_{1}\cap[(H_{f})^{\prime}\cdot Z(H_{f})]$. This finishes the proof.∎

####  6.3.1 Abelian Deformations of $\textrm{GL}_{n}$ over NFT-fields

Let $L$ be an NFT-field, and let $H=\textrm{GL}_{n}(L)$. Consider an addition
$H_{0}$ and the corresponding foundation $H_{f}$. Identify $(H_{f})^{\prime}$
with $\textrm{SL}_{n}(F)$ and let:

  1. 1.

$\Sigma_{1}(H_{f})=\\{(\prod_{i=2}^{n}(d_{1}(\alpha)d_{i}(\alpha)^{-1})|\alpha\in
L^{\times}\\}$;

  2. 2.

$\Delta_{1}(H_{f})=Is_{H_{f}}(\Sigma_{1}(H_{f})\cdot Z(H_{f}))$ [Note that
this is the same subgroup as the one defined in the proof of 6.6.]

  3. 3.

Let $\pi:H\to H_{f}$ be the canonical epimorphism and let
$H_{1}=(H_{f})^{\prime}\cdot Z(H_{f})$. Observe that
$\pi(Is_{H}(H^{\prime}))=H_{1}$. Since $Is_{H}(H^{\prime})\cap H_{0}=1$, $\pi$
maps $Is_{H}(H^{\prime})$ isomorphically onto $H_{1}$

We note that $H_{1}\cap\Delta_{1}=\Sigma_{1}\cdot Z(H_{f})$. By (6.4) and
(6.5) we might represent elements of $H_{f}$ as pairs $(c,a)$ where $c\in
H_{1}$, $a\in\Delta_{1}$. Furthermore, there exists a symmetric 2-cocycle
$p:A/A^{n}\times A/A^{n}\to\Sigma_{1}\cdot Z(H_{f})$, where the product in
$H_{f}$ is given by:

$(c_{1},a_{1})(c_{2},a_{2})=(c_{1}c_{2}^{a_{1}}p(a_{1}\Sigma_{1},a_{2}\Sigma_{2}),a_{1}a_{2}).$

######  Definition 6.7.

Let $L$ be an NFT-field, $L^{\times}=A(L)\times T(L)$, where $A=A(L)$ is
torsion-free and $T=T(L)$ is torsion, $H=\textrm{GL}_{n}(L)$ and
$H_{f}=G/G_{0}$ for some addition $H_{0}$ of $H$. Consider $B\equiv A$ and
$h\in S^{2}(A/A^{n},B)$. Define $E=GL_{n}(L,h,B)$ as the central extension:

$1\to B\to E\to H_{f}\to 1$

where the group multiplication is given by:

$(b_{1},c_{1}a_{1})(b_{2},c_{2}a_{2})=(b_{1}b_{2}h(a_{1}\Sigma_{1},a_{2}\Sigma_{1}),c_{1}c_{2}^{a_{1}}p(a_{1}\Sigma_{1},a_{2}\Sigma_{1})a_{1}a_{2})$

for $b_{i}\in B$, $c_{i}\in H_{1}$ and $a_{i}\in\Delta_{1}(H_{f})$.

######  Theorem 6.8.

Let $F$ be an NFT-field. If $H\equiv\textrm{GL}_{n}(F)$, then $H\cong
GL_{n}(L,h,B(H))$, as in Definition 6.7, for some $L\equiv F$, $h\in
S^{2}(A/A^{n},B)$, and $B\equiv F^{\times}/T$.

###### Proof.

Let $G=\textrm{GL}_{n}(F)$. Pick any addition $G_{0}$ of $G$. Although it may
not be definable in $G$, it is absolutely interpretable in $G$ as
$Z(G)/T(Z(G))$. Therefore, in $H$ any addition $H_{0}$ is elementarily
equivalent to $G_{0}$. The existence of $K=\textrm{GL}_{n}(L)$ such that
$H_{f}\cong K_{f}$ was established in Proposition 6.6. Therefore, $H$ fits in
a central extension:

$1\to H_{0}\to H\xrightarrow{\pi}K_{f}\to 1.$

We shall identify $H_{f}$ with $K_{f}$, and take $\pi:H\to H_{f}$ to be the
canonical epimorphism. The restriction of $\pi$ to $Is(H^{\prime})$ is an
isomorphism. We identify $SL_{n}(L)\cong H^{\prime}$ with its copy in $H$.
Furthermore, the subgroup $\pi^{-1}(\Sigma_{1}(H_{f}))$ can also be
isomorphically identified with a subgroup of $Is_{H}(H^{\prime})$ and we
denote it by $\Sigma_{1}(H)$. Now, $H$ is generated by $Is(H^{\prime})$ and
$\Pi_{1}(H)=Is_{H}(H_{0}\cdot\Sigma_{1}(H))$. Therefore, the product in $H$
will be completely determined by the products in $Is(H^{\prime})$,
$\Pi_{1}(H)=Is_{H}(Z(H)\cdot\Sigma_{1}(H))$ and the conjugate action of the
latter on the former. The conjugate action is completely determined by the
structure of $H_{f}$, which is known up isomorphism. We fully understand the
structure of $Is_{H}(H^{\prime})$. The subgroup $\Pi_{1}(H)$ is an abelian
group that fits into

$1\to Z(H)\times\Sigma_{1}(H)\to\Pi_{1}(H)\to A/A^{n}\to 1.$

In fact, $\Pi_{1}(G)$ is a definable abelian subgroup of $G$. Note that both
$Z(G)$ and $\Sigma_{1}(G)$ are definable in $G$ with parameters $\mathcal{T}$
and $Is_{G}(Z(G)\cdot\Sigma_{1}(G))=\\{x\in G|x^{n}\in
Z(G)\cdot\Sigma_{1}(G)\\}$. Hence, the same formulas that define $\Pi_{1}(G)$
in $G$ define $\Pi_{1}(H)$ in $H$. Now, since

$Ext(A/A^{n},\Pi_{1}(H)\times H_{0})\cong Ext(A/A^{n},\Pi_{1}(H))\times
T(Z(H)))\times Ext(A/A^{n},H_{0})$

while the extension of $\Pi_{1}(H)\times T(Z(H))$ by $A/A^{n}$ is completely
determined within $H_{f}$ by the 2-cocycle $p$ and the conjugate action of
elements of $\Pi_{1}(H)$ on $H_{1}$. What is left to completely fix the
product in $H$ is determined by a 2-cocycle $h\in S^{2}(A/A^{n},H_{0})$, which
cannot necessarily be recovered from the first-order theory of $G$. ∎

####  6.3.2 A group elementarily equivalent to $\textrm{GL}_{n}(\mathbb{Q})$
but not isomorphic to any $\textrm{GL}_{n}$

Let $L$ be a countable non-standard model of the field $\mathbb{Q}$ and
$H=\textrm{GL}_{n}(L)$. Then by an argument similar to the one in the proof of
Theorem 5.14, $L^{\times}$ contains a countable torsion-free divisible
(radicable) subgroup $D$. Identify $D$ with a central subgroup of
$\textrm{GL}_{n}(L)$. Since $D$ is divisible, it will split from the center
and from the group. Therefore,

$\textrm{GL}_{n}(\mathbb{Q})\equiv\textrm{GL}_{n}(L)\cong\textrm{GL}_{n}(L)\times
D\equiv\textrm{GL}_{n}(\mathbb{Q})\times D$

First, $\textrm{GL}_{n}(\mathbb{Q})\ncong\textrm{GL}_{n}(\mathbb{Q})\times D$,
since $Z(\textrm{GL}_{n}(\mathbb{Q}))\cong\mathbb{Q}^{\times}$ has no
nontrivial divisible (radicable) subgroups. Now, assume
$H=\textrm{GL}_{n}(F)\cong\textrm{GL}_{n}(\mathbb{Q})\times D=G$ for some
field $F$. Then $H^{\prime}=\textrm{SL}_{n}(F)$ and
$G^{\prime}=\textrm{SL}_{n}(\mathbb{Q})$. The isomorphism must map
$H^{\prime}$ onto $G^{\prime}$, while by the abstract isomorphism theorem for
$\textrm{SL}_{n}$, $F\cong\mathbb{Q}$. Contradiction!∎

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