# On the structures of a monoid of triangular vector- permutation polynomials,
its group of units and its induced group of permutations

Amr Ali Abdulkader Al-Maktry  Department of Analysis and Number Theory (5010)
Technische Universität Graz
Kopernikusgasse 24/II
8010 Graz, Austria<EMAIL_ADDRESS>
###### Key words and phrases:

vector- polynomial, polynomial function, monoid, polynomial automorphism,
polynomial permutation, unit-valued polynomial, iterated semi-direct product

###### 2010 Mathematics Subject Classification:

Primary 13B25, 20B27, 16W20; Secondary 05A05, 20B25, 13M10, 14R99

## Abstract

Let $n>1$ and let $R$ be a commutative ring with identity the $1\neq 0$. A
vector-polynomials
$\vec{f}=(f_{0},f_{1}+x_{2}g_{1},\ldots,f_{n-1}+x_{n}g_{n-1}),$ where
$f_{0}\in R[x_{1}]$ permutes the elements of $R$ and $f_{i},g_{i}\in
R[x_{1},\ldots,x_{i}]$ with $g_{i}$ mapping $R^{i}$ into the units of $R$
($i=1,\ldots,n-1$); permutes the elements of $R^{n}$. The set of all such
vector-polynomial $\mathcal{MT}_{n}$ is a monoid with respect to composition.
Further, $\vec{f}$ is invertible in $\mathcal{MT}_{n}$ if and only if $f_{0}$
is an $R$ automorphism of $R[x_{1}]$ and $g_{i}$ is invertible in
$R[x_{1},\ldots,x_{i}]$ for $i=1,\ldots,n-1$. When $R$ is finite, the monoid
$\mathcal{MT}_{n}$ induces a finite group of permutations of $R^{n}$.
Moreover, we decompose the $\mathcal{MT}_{n}$ into an iterated semi-direct
product of $n$ monoids. Such a decomposition allows us to obtain similar
decompositions of its group of units, and when $R$ is finite, of its induced
group of permutations. Furthermore, the decomposition of the induced group
helps us to characterize some of its properties.

##  1\. Introduction

Let $R$ be a commutative ring (not necessarily finite) with unity and let
$R[x_{1},\ldots,x_{n}]$ be its polynomial ring in $n$ variables. A vector-
polynomial $\vec{f}$ is an $n$-tuple $(f_{1},\ldots,f_{n})$ of polynomials
over $R$ in the variables $x_{1},\ldots,x_{n}$, i. e. is
$\vec{f}=(f_{1},\ldots,f_{n})$. It is well- known that every vector-polynomial
$\vec{f}$ defines an $R$-endomorphism $\phi_{\vec{f}}$ of
$R[x_{1},\ldots,x_{n}]$ by $\phi_{\vec{f}}(g)=g(f_{1},\ldots,f_{n})$ for any
$g\in R[x_{1},$…$,x_{n}]$. Furthermore, $\vec{f}$ is an $R$-automorphism if
and only it is invertible with respect to composition. The set of all these
these automorphisms is a group, which is denoted by $GA_{n}$. Vector-
polynomials are commonly referred to as polynomial maps in the literature. we
are interested not only in the vector-polynomial (polynomial map) but also in
its induced function (map) on $R^{n}$ via evaluation. This distinction helps
avoid confusion between an $n$-tuple of polynomials and a single polynomial in
this context.

Several results have been achieved concerning these $R$-endomorphisms of
$R[x_{1},\ldots,x_{n}]$, mainly considering the group $GA_{n}$, its subgroups,
and its elements. The developed ideas cover several aspects of investigations,
including the Jacobian Conjecture and equivalent statements (see for example
and the references therein[22]), the inclusion of the tame group (see for
example[9, 24, 20]), and the study of structures of some subgroups of $GA_{n}$
(see [2, 25]).

Previous works have primarily focused on the case where $R$ is an infinite
field with characteristic zero. Nevertheless, researchers have considered also
finite fields to investigating similar questions (see [12, 14, 10]),
disproving some famous results known to be true in the infinite case (see [7,
14]), and reformulating some others (for instance [13]). Notably, the
permutations induced by the elements of $GA_{n}$ on $R^{n}$ have a powerful
impact.

In this paper, we explore the structure of a monoid of triangular vector-
polynomials with a specific property. We consider the monoid of all triangular
vector polynomials of the form
$(f_{0},f_{1}+x_{2}g_{1}\ldots,f_{n-1}+x_{n}g_{n-1})$ with the condition that
$f_{0}$ is a permutation polynomial on $R$ and $g_{i}$ is a unit-valued on $R$
for $i=1,\ldots,n-1$. The elements of this monoid define permutations of
$R^{n}$. Additionally, we observe that the group of units of this monoid
generally contains properly the (lower) Jonquières (classical) triangular
group unless $R$ is a reduced ring. In a reduced ring the invertible
polynomials coincide with the units of the ring $R$, and in this case, the
group of the triangular monoid is just the Jonquières group (see Remark 4.11).

The monoid under considerations can be decomposed into an iterated semi-direct
product of $n$ monoids by iteration (see Theorem 4.5). Surprisingly, this
decomposition admits a similar decomposition of the group of units of the
triangular monoid into an iterated semi-direct product of $n$ groups by
finding the units of those factor monoids in the first decomposition (Theorem
4.10). Further, when $R$ is finite, the triangular monoid induces a group of
permutations on $R^{n}$ which, is its homorphic image via the natural
epimorphism $\pi_{n}$(defined in Remark 2.12). Again, we obtain an iterated
semi-direct product decomposition of the induced group just by taking images
(via $\pi_{n}$) of the factor monoids in that of the triangular monoid
(Theorem 4.14). Such decomposition allows us to characterize the solvability
and the nilpotency of this group restricted (by the Cheiense Reminder
Theorem)to the local case (Theorem 4.20).

It is remarkable that in the case of algebraically closed fields, the
triangular monoid, its group of units, and its induced group all coincide with
the Jonquières triangular group. This is because unit-valued polynomials,
invertible polynomials and the units of $R$ are all the same. This motivates
us to suggest a modification for the tameness problem in the general case by
replacing the Jonquières with our triangular monoid, and then posting some
related questions (see Section6). Having previously worked on polynomial
functions and polynomial permutations over finite commutative rings (in one
variable) rather than polynomial mappings (in several variables), we can
ensure that our motivation to study the triangular monoid emanates from a
simple question on polynomial permutations. In other words, given a finite
ring $R$ and $R_{n}$ is its dual numbers extension in $n$ variables. Then,
does every polynomial permutation on $R_{n}$ can be viewed as a vector-
permutation polynomial on the base ring $R$? More generally, can the group of
permutation polynomials on $R_{n}$ be viewed as a group of permutations of
$R^{n+1}$ induced vector-polynomials in $n+1$ variables? We answer this
question affirmatively and provide more details (see Section 5).

The structure of our paper is as follows: in Section 2, we introduce our
notation and basic definitions, and we recall some facts about polynomials and
polynomial functions. In section 3, we describe the elements of the triangular
monoid and its units and also obtain some simple relations for them. Section 4
consists of three subsections, the first subsection contains the decomposition
of the triangular monoid into an iterated semi-direct product of monoids,
while the other two sections are devoted to the decompositions of its group of
units and its induced group of permutations, respectively. In section 5, we
consider the problem of embedding the monoid of permutation polynomial (in one
variable) of certain ring extensions of the ring $R$ into the triangular
monoid. Finally, in Section 6, we formulate the tameness problem and suggest
some related questions.

##  2\. preliminaries

Throughout this paper, let $R$ be a commutative ring with $1\neq 0$,
$R^{\times}$ its group of units, and whenever $R$ is a finite local ring, $M$
will denote its unique maximal ideal. Also, we use $\mathbb{F}q$ to denote its
residue field $R/M$ or, more generally, a finite field of $q$ elements.

Let $n\geq 1$. By $n$, we denote the number of variables $x_{1},\ldots,x_{n}$;
and to distinguish the elements of $R[x_{1},\ldots,x_{n}]^{n}$ from those of
$R[x_{1},\ldots,x_{n}]$, we use the symbol $\vec{f}$ to indicate an $n$-tuple
$(f_{1},\ldots,f_{n})$ of polynomials $f_{1},\ldots,f_{n}\in
R[x_{1},\ldots,x_{n}]$. In a similar manner we use symbols like $\vec{F}$ to
indicate function $\vec{F}\colon R^{n}\longrightarrow R^{n}$, or equivalently
to indicate an $n$-tuple $(F_{1},\ldots,F_{n})$ functions
$F_{1},\ldots,F_{n}\colon R^{n}\longrightarrow R$, i.e.
$\vec{F}=(F_{1},\ldots,F_{n})$. It is well-known that an vector polynomial
$\vec{f}=(f_{1},\ldots,f_{n})$ defines an $R$\- endomorphism of
$R[x_{1},\ldots,x_{n}]$: $\vec{f}\colon R[x_{1},\ldots,x_{n}]\longrightarrow
R[x_{1},\ldots,x_{n}]$, $\vec{f}(f)=f(f_{1},\ldots,f_{n})$. Such an
endomorphism is called an $R$-automorphism if there is an $n$\- vector
polynomial $\vec{g}(g_{1},\ldots,g_{n})$ such that

$\vec{f}\circ\vec{g}=(g_{1}(f_{1},\ldots,f_{n}),\ldots,g_{n}(f_{1},\ldots,f_{n}))=(x_{1},\ldots,x_{n}).$

Here, we notice that the above operation is anti the normal composition.
However, we will adapt later the normal one (see Definition 2.6). The reason
is that we are also interested in the (vector)-functions induced by
(vector)-polynomials, and we require that the operation on polynomials and
their functions be consistent. Also, it is noticeable here that despite any
one of these operations being under consideration in the monoid
${R[x_{1},\ldots,x_{n}]}^{n}$, the terms invertible, left invertible and right
invertible are equivalent [22].

Furthermore, we note that if $\vec{F}$ is vector function obtained from the
$R$-automorphism $\vec{f}$ via evaluation, that is,

$\vec{F}(a,\ldots,a_{n})=\vec{f}(a_{1},\ldots,a_{n})=(f_{1}(a_{1},\ldots,a_{n}),\ldots,f_{n}(a_{1},\ldots,a_{n})),\text{
for every }(a_{1},\ldots,a_{n})\in R^{n},$

permutes the elements of $R^{n}$. However, the converse is not true. For
example, consider the case $n=1$, $R=\mathbb{F}_{q}$, and
$\vec{f}(x)=f(x)=x^{q}$. Evidently, $\vec{F}=F=id_{R^{n}}$; however, there is
no a polynomial $g$ such that $f\circ g=x$.

######  Notation 2.1.

For $k\geq 1$, let $R_{[k]}$ denote the polynomial ring
$R[x_{1},\ldots,x_{k}]$. But, when $k=1$, we use $R[x]$ or $R[x_{1}]$ to
denote polynomial ring in one variable rather than writing $R_{[1]}$.

######  Definition 2.2.

Let $R$ be a commutative ring, and $f,g\in R_{[n]}$. Then:

  1. (1)

The polynomial $f$ gives rise to a polynomial function $F\colon
R^{n}\longrightarrow R$ by substitution for the variables, that is,
$F(a_{1},\ldots,a_{n})=f(a_{1},\ldots,a_{n})$ for every
$(a_{1},\ldots,a_{n})\in R^{n}$. Further, if $F$ maps $R^{\times}$ into
$R^{\times}$ the group of units of $R$, we call $f$ a unit-valued polynomial
and $F$ is a uni-valued polynomial function (on $R$)

  2. (2)

If $g(a_{1},\ldots,a_{n})=f(a_{1},\ldots,a_{n})$ for every
$(a_{1},\ldots,a_{n})\in R^{n}$, this means that $f$ and $g$ induce the same
function $F$ on $R$ and we write $f\operatorname{{}\triangleq{}}g$ on $R$. In
this case, we say $f$ and are equivalent on $R$.

  3. (3)

We denote by $\mathcal{F}(R^{n})$ the set of all polynomial functions (in $n$
variables) on $R$.

  4. (4)

We denote by $UV(R_{[n]})$ the set of all unit-valued polynomials in $n$
variables.

From Definition2.2, we have easily the following fact.

######  Fact 2.3.

Let $R$ be a commutative ring. Then

  1. (1)

$UV(R_{[n]})$ is a submonoid of $(R_{[n]},``\circ")$;

  2. (2)

$\mathcal{F}(R^{n})$ is a monoid with respect to composition;

  3. (3)

$\mathcal{F}(R^{n})$ is a commutative ring with pointwise operation.
Furthermore, $\mathcal{F}(R^{n})$ is finite whenever $R$ is finite, and in
this case the group of units of $\mathcal{F}(R^{n})^{\times}$ consists of all
unit-valued polynomial functions.

######  Remark 2.4.

  1. (1)

By Definition 2.2, every unit polynomial in a polynomial ring is a unit-valued
polynomial but not vice versa. For example, consider the case $R$ is a finite
local ring and $f(x)=(x^{q}-x)+1$.

  2. (2)

When $R$ is a domain, the invertible polynomials are just the units of $R$.

  3. (3)

When $R$ contains a non-zero nilpotent, it contains a nilpotent $r\neq 0$ such
that $r^{2}=0$. But, then the polynomial $f(x)=rx+1$ admits $rx-1$ as
multiplicative inverse.

  4. (4)

Let $R$ be a finite ring and $f(x)=\sum_{i=0}^{k}a_{i}x^{i}$ is a unit in
$R[x]$ if and only if $a_{0}$ is a unit of $R$ and $a_{i}$ is nilpotent for
$i=1,\ldots k$ [3, Propostion. 1.3.3].

######  Definition 2.5.

Let $R$ be a finite commutative ring. We call the group
$\mathcal{F}(R^{n})^{\times}$ the group unit-valued polynomial functions (in
$n$ variables) on $R$.

######  Definition 2.6.

Let

$\vec{f}=(f_{1},\ldots,f_{n}),\text{ and }\vec{g}=(g_{1},\ldots,g_{n}),\text{{
where} }f_{i},g_{i}\in R_{[n]}\text{ for }i=1,\ldots,n.$

Then we define $f_{i}(\vec{g})=f_{i}(g_{1},\ldots,g_{n})$ for $i=1,\ldots,n$;
and

$\vec{f}\circ\vec{g}=(f_{1}(\vec{g}),\ldots,f_{n}(\vec{g}))=(f_{1}(g_{1},\ldots,g_{n}),\ldots,f_{n}(g_{1},\ldots,g_{n})).$

From the Definition 2.6, one easily prove the following fact.

######  Fact 2.7.

For $=1,\ldots,n$, let $f_{i}\in R_{[i]}$. Then

  1. (1)

$(f_{1},f_{2},\ldots,f_{n})=(f_{1},x_{2},\ldots,x_{n})\circ\cdots\circ(x_{1},x_{2},\ldots,f_{n})$

  2. (2)

$(x_{1},\ldots,x_{i},f_{i+1},\ldots
f_{n})\circ\vec{g}=\vec{g}\circ(x_{1},\ldots,x_{i},f_{i+1}(\vec{g}),\ldots
f_{n}(\vec{g}))$, where

$\vec{g}=(x_{1},\ldots,x_{j-1},f_{j},x_{j+1},\ldots,x_{n})$ and $j\leq i$.

######  Definition 2.8.

Let $R$ be a commutative ring, and $f_{1},\ldots,f_{n};g_{1},\ldots,g_{n}\in
R_{[n]}$. Then:

  1. (1)

The $n$-vector polynomial $\vec{f}=(f_{1},\ldots,f_{n})^{t}$ gives rise to an
$n$-vector polynomial function $\vec{F}\colon R^{n}\longrightarrow R^{n}$ by
substitution for the variables, that is,

$\vec{F}(a_{1},\ldots,a_{n})=\vec{f}(a_{1},\ldots,a_{n})=(f_{1}(a_{1},\ldots,a_{n}),\ldots,f_{n}(a_{1},\ldots,a_{n})),\text{
for every }(a_{1},\ldots,a_{n})\in R^{n}.$

If $\vec{F}$ is bijective, we call $\vec{F}$ a vector-polynomial permutation,
$\vec{f}$ a vector-permutation polynomial and $f_{i}$ a permutation polynomial
for $i=1,\ldots,n$.

  2. (2)

If $\vec{g}=(g_{1},\ldots,g_{n})$ and
$\vec{g}(a_{1},\ldots,a_{n})=\vec{f}(a_{1},\ldots,a_{n})$ for every
$(a_{1},\ldots,a_{n})\in R^{n}$, this means that $\vec{f}$ and $\vec{g}$
induce the same function $F$ on $R^{n}$ and we write
$\vec{f}\operatorname{{}\triangleq{}}\vec{g}$ on $R^{n}$.

  3. (3)

We denote by $\vec{\mathcal{F}}(R^{n})$ the set of all $n$ vector-polynomial
functions.

######  Remark 2.9.

When $n>1$, a permutation polynomial $f\in R_{[n]}$ can not alone induce a
permutation of $R^{n}$ by Definition 2.8. However, when $n=1$, the concept of
permutation polynomial and vector-permutation polynomial coincide. Permutation
polynomials in one variable and their induced permutations play a vital rule
in this context. For this reason we collect the related terminology in the
following definition.

######  Definition 2.10.

Let $R$ be a commutative ring, and let $f\in R[x]$ and $F\colon
R\longrightarrow R$ be its induced function. Then

  1. (1)

$f$ is called a _permutation polynomial_ whenever $F$ is a permutation of $R$.
In this case, $F$ is called _polynomial permutation_.

  2. (2)

By $\mathcal{MP}(R)$, we denote the monoid of permutation polynomials on $R$
consisting of all permutation polynomials.

  3. (3)

The group of units of $\mathcal{MP}(R)$ consists of all $R$-automorphisms of
$R[x]$, which we denote by $Aut_{R}(R[x])$.

  4. (4)

If $R$ is finite, the set of all polynomial permutations on $R$ is a finite
group which we denote by $\mathcal{P}(R)$.

######  Remark 2.11.

In [5], Gilmer showed that a polynomial $f\in R[x]$ is an $R$ automorphism of
$R[x]$ if and only if $f(x)=a_{0}+a_{1}x+\cdots+a_{m}x^{m}$ with $a_{1}$ is
unit and $a_{i}$ is nilpotent for every $i\geq 2$. It follows, when $R$ is
reduced, i.e. $R$ has no non-trivial nilpotents elements, that $Aut_{R}(R[x])$
consists only of invertible linear polynomials.

######  Remark 2.12.

  1. There is a natural epimorphism $\pi_{n}\colon{R_{[n]}}^{n}\longrightarrow\vec{\mathcal{F}}(R^{n})$ with respect to composition which maps every vector-polynomial into its induced vector-polynomial function $\vec{F}$.

######  Lemma 2.13.

Let $x_{1},\ldots,x_{k+1}$ be variables, and let $f,g,h,u\in R_{[k]}$ such
that $h$ and $u$ are unit-valued polynomials. Then
$f+x_{k+1}u\operatorname{{}\triangleq{}}g+x_{k+1}h$ on $R$ if and only if
$f\operatorname{{}\triangleq{}}g$ and $u\operatorname{{}\triangleq{}}h$ on $R$

###### Proof.

Suppose that $f+x_{k+1}u\operatorname{{}\triangleq{}}g+x_{k+1}h$ on $R$. Then
setting $x_{k+1}=0$ yields that $f\operatorname{{}\triangleq{}}g$ on $R$.
Hence, $x_{k+1}u\operatorname{{}\triangleq{}}x_{k+1}h$ on $R$. But, then
$x_{k+1}=1$ implies that $u\operatorname{{}\triangleq{}}h$ on $R$. The other
implication is obvious. ∎

######  Fact 2.14.

Let $R$ be a finite commutative ring. Let $\vec{F}$ be an $n$-vector
polynomial permutation of $R^{n}$ induced by the $n$-vector polynomial
$\vec{f}=(f_{1},\ldots,f_{n})$, where $f_{1},\ldots,f_{n}\in R_{[n]}$. If the
inverse of $\vec{F}$, $\vec{F}^{-1}$, is induced by $n$-vector polynomial
$\vec{g}=(g_{1},\ldots,g_{n})$, where $g_{1},\ldots,g_{n}\in R_{[n]}$, then
$f_{i}\circ g_{i}\operatorname{{}\triangleq{}}x_{i}$ on $R$ for $i=1,\dots,n$.

In the following, we recall the definition of a short exact sequence of groups
and its relation with the semi-direct product of groups. A finite sequence of
groups and homomorphisms

$1\rightarrow A\xrightarrow{\alpha}B\xrightarrow{\beta}C\rightarrow 1$

is a short exact sequence if $\operatorname{Im}\alpha=\ker\beta$. Further, if
$\alpha$ is the inclusion function, $i$, $B$ is an extension of $A$ by $C$.
This allows us to give a definition of the semi-direct product of groups by
means of a short exact sequence.

######  Definition 2.15.

[18, Page 760] An extension

$1\rightarrow A\xrightarrow{i}B\xrightarrow{p}C\rightarrow 1$

is split if there exists a homomorphism $j\colon C\rightarrow B$ such that
$pj=id_{C}$. In this case, the group $B$ is called the semi-direct product of
$A$ by $B$, which is denoted by $A\rtimes C$.

##  3\. The monoid of triangular vector-permutation polynomials

In this section, we give the construction of a monoid of triangular vector-
polynomials. Recall from Definition 2.2 that $UV(R_{[k]})$ stands for the
monoid of unit-valued polynomials in $k$ variables.

######  Lemma 3.1.

Let $R$ be a commutative ring. Let
$\vec{f}=(f_{1},f_{2}+x_{2}g_{2},\ldots,f_{n}+x_{n}g_{n})$ and
$\vec{h}=(h_{1},h_{2}+x_{2}w_{2},\ldots,h_{n}+x_{n}w_{n})$, where
$f_{1},h_{1}\in\mathcal{MP}(R)$, $f_{i},h_{i}\in R_{[i-1]}$ and
$u_{i},w_{i}\in UV(R_{[i-1]})$ with $i=2,\ldots,n-1$. Then
$\vec{f}\circ\vec{h}=(l_{1},l_{2}+x_{2}r_{2},\ldots,l_{n}+x_{n}r_{n}),$ for
some $l_{1}\in\mathcal{MP}(R)$, $l_{i}\in R_{[i-1]}$ and $r_{i}\in
UV(R_{[i-1]})$ for $i=2,\ldots,n$.

###### Proof.

Let $\vec{h}\in\mathcal{MT}_{n}$ and $\vec{f}$ as above. Then

By Definition 2.6,

$\vec{f}\circ\vec{h}=(l_{1},l_{2}+x_{2}r_{2},\ldots,l_{n}+x_{n}r_{n}),$ (1)

where

$\begin{matrix}l_{1}(x_{1})=f_{1}(h_{1}(x_{1}))\\\
l_{i}(x_{1},\ldots,x_{i-1})=f_{i}(\vec{v_{i}})+h_{i}(x_{1},\ldots,x_{i-1})g_{i}(\vec{v_{i}}),\\\
r_{i}(x_{1},\ldots,x_{i-1})=w_{i}(x_{1},\ldots,x_{i-1})g(\vec{v_{i}})\text{
and }\\\
\vec{v_{i}}(x_{1},\ldots,x_{i-1})=(h_{1}(x_{1}),h_{2}(x_{1})+x_{2}w_{1}(x_{1}),\ldots,h_{i-1}(x_{1},\ldots,x_{i-2})+x_{i-1}w_{i-1}(x_{1},\ldots,x_{i-2})),\\\
i=2,\ldots,n.\end{matrix}$ (2)

But, then $l_{1}=f_{1}\circ h_{1}\in\mathcal{MP}(R)$ since
$f_{1},h_{1}\in\mathcal{MP}(R)$; and for $i=2,\ldots,n$,
$r_{i}=w_{i}g(\vec{v_{i}})\in UV(R_{[i-1]})$ since $w_{i},g(\vec{v_{i}})\in
UV(R_{[i-1]})$. ∎

Recall from Remark 2.12 that $\pi_{n}$ stands for the epimorphism assigning
for every vector-polynomial to its induced function on $R^{n}$.

######  Lemma 3.2.

Let $R$ be a commutative ring. Let $f_{1}\in\mathcal{MP}(R)$ and for
$i=2,\ldots,n$ let $f_{i}\in R_{[{i-1}]}$ and $g_{i}\in UV(R_{[{i-1}]]})$ .
Then the vector-polynomial:
$\vec{f}=(f_{1},f_{2}+x_{2}g_{2},\ldots,f_{n}+x_{n}g_{n})$ induces a
permutation of $R^{n}$, that is $\pi_{n}(\vec{f})$ is a permutation of
$R^{n}$.

###### Proof.

Set $\vec{F}=\pi_{n}(\vec{f})$. First, we show that $\vec{F}$ is one to one.
For this, let $a_{i},b_{i}\in R$ for $i=1,\dots,n$ such that

$\vec{F}(a_{1},\ldots,a_{n})=\vec{f}(a_{1},\ldots,a_{n})=\vec{f}(b_{1},\ldots,b_{n})=\vec{F}(b_{1},\ldots,b_{n}).$

Then, we have the following equations:

$\begin{matrix}f_{1}(a_{1})=f_{1}(b_{1})\\\
f_{2}(a_{1})+a_{2}g_{1}(a_{1})=f_{2}(b_{1})+b_{2}g_{1}(b_{1})\\\ \vdots\\\
f_{n}(a_{1},\ldots,a_{n-1})+a_{n}g_{n}(a_{1},\ldots,a_{n-1})=f_{n}(b_{1},\ldots,b_{n-1})+b_{n}g_{n}(b_{1},\ldots,b_{n-1})\end{matrix}$

Thus, from the first equation, since $f_{1}$ is a permutation polynomial on
$R$, we have $a_{1}=b_{1}$. Then, the second equation becomes
$a_{2}g_{2}(a_{1})=b_{2}g_{2}(a_{1})$. But, then $a_{2}=b_{2}$ as
$g_{2}(a_{1})$ is a unit since $g_{2}$ is a unit-valued polynomial on $R$.
Continuing in this manner, we find that $a_{i}=b_{i}$, $1\leq i\leq n-1$ and
the $i+1$th equation is reduced to
$a_{i+1}g_{i+1}(a_{1},\ldots,a_{i})=b_{i+1}g_{i+1}(a_{1},\ldots,a_{i})$.
Hence, $a_{i+1}=b_{i+1}$ since $g_{i+1}$ is a unit-valued polynomial.
Therefore, $a_{i}=b_{i}$ for $i=1,\dots,n$ To show $\vec{F}$ is surjective,
let $(b_{1},\ldots,b_{n})\in R^{n}$ be arbitrary. Then, we solve the equation

$\vec{f}(x_{1},\ldots,x_{n})=(b_{1},\ldots,b_{n}).$

Therefore, we have the following system of $n$ equations:

$\begin{matrix}f_{1}(x_{1})=y_{1}\\\ f_{2}(x_{1})+x_{2}g_{2}(x_{1})=y_{2}\\\
\vdots\\\
f_{n}(x_{1},\ldots,x_{n-1})+x_{n}g_{n}(x_{1},\ldots,x_{n-1})=y_{n}.\end{matrix}$

But, then since $f_{1}$ is a permutation on $R$,
$x_{1}=a_{1}=F_{1}^{-1}(y_{1})$ is the solution of the 1st equation, where
$F_{1}^{-1}$ stands for the inverse permutation of the permutation obtained by
$f_{1}$ on $R$. So, the second equation becomes
$f_{2}(a_{1})+x_{2}g_{2}(a_{1})=y_{2}$. Since $g_{2}$ is a unit-valued on $R$,
$g_{2}(a_{1})$ is invertible in $R$. Thus,
$x_{2}=a_{2}=(y_{2}-f_{2}(a_{1})){g_{2}(a_{1})}^{-1}$. Keeping solving these
equations and the fact that $g_{i}$ is a unit valued on $R$, yields

$x_{i}=a_{i}=(y_{i-1}-f_{i}(a_{1},\ldots,a_{i-1})){g_{i}(a_{1},\ldots,a_{i-1})}^{-1}\text{,
}i=2,\ldots,n.$

∎

######  Theorem 3.3.

Let $R$ be a commutative with unity and $n>1$ an integer. Let
$\mathcal{MT}_{n}$ be the set of all vector-polynomials of the form:

$\vec{f}(x_{1},\ldots,x_{n})=(f_{1}(x_{1}),f_{2}(x_{1})+x_{2}g_{2}(x_{1}),\ldots,f_{n}(x_{1},\ldots,x_{n-1})+x_{n}g_{n}(x_{1},\ldots,x_{n-1})),$
(3)

where $f_{1}$ is a permutation polynomial on $R$, $f_{i}\in R_{[{i-1}]}$ and
$g_{i}\in UV(R_{[{i-1}]]})$ for $i=2,\ldots,n$. Then $\mathcal{MT}_{n}$ is a
monoid of vector-permutation polynomials with respect to composition.
Moreover, if $R$ is finite, then $\pi_{n}(\mathcal{MT}_{n})$ is a group of
permutations of $R^{n}$.

###### Proof.

By Lemma 3.2, $\mathcal{MT}_{n}$ consists of vector-permutation polynomials.
To show $\mathcal{MT}_{n}$ is a monoid with respect to the composition
$``\circ"$ we need only to show $\mathcal{MT}_{n}$ is closed with respect to
$``\circ"$, since the identity $(x_{1},x_{2},\ldots,x_{n})\in\mathcal{MT}_{n}$
and composition is always associative (see Definition 2.6). But, this is the
case by Lemma 3.1.

Moreover, if $S_{R^{n}}$ is the symmetric group on $R^{n}$, then the first
part of the proof shows that $\pi_{n}(\mathcal{MT}_{n})\subseteq S_{R^{n}}$.
Since we consider the case $R$ is finite, $\pi_{n}(\mathcal{MT}_{n})$ is a
finite subset of the finite group $S_{R^{n}}$. Thus, we need only show that
$\pi_{n}(\mathcal{MT}_{n})$ is closed with respect to composition. Now, let
$\vec{F},\vec{G}\in\pi_{n}(\mathcal{MT}_{n})$. Then there exist
$\vec{f},\vec{g}\in\mathcal{MT}_{n}$ such that $\vec{F}=\pi_{n}(\vec{f})$ and
$\vec{G}=\pi_{n}(\vec{g})$. Therefore,

$\vec{F}\circ\vec{G}=\pi_{n}(\vec{f})\circ\pi_{n}(\vec{g})=\pi_{n}(\vec{f}\circ\vec{g})\in\pi_{n}(\mathcal{MT}_{n}).$

∎

As a direct result, we have the following corollary.

######  Corollary 3.4.

Let $R$ be a finite commutative ring. Let $k>1$ and $f,g\in R_{[k]}$. Suppose
that $f$ is a unit-valued. Then
$g(x_{1},\ldots,x_{k})+x_{k+1}f(x_{1},\ldots,x_{k})$ is a permutation
polynomial of $i$ variables for every $i\geq k+1$.

######  Remark 3.5.

The finiteness condition on $R$ in the last part of Theorem 3.3 is sufficient
for $\pi_{n}(\mathcal{MT})_{n}$ to be a group. However, if we consider the
case, $R$ is an infinite field, then
$\pi_{n}(\mathcal{MT}_{n})\cong\mathcal{MT}_{n}$ since if $f\in R_{[k]}$
($k\geq 1$) maps $R^{k}$ onto zero then $f$ is necessary the constant zero
polynomial. Therefore, we notice the following two examples:

  1. (1)

Let $R$ be the real numbers. Then, it is evident that
$(x_{1}^{3},x_{2},x_{3},\ldots,x_{n})$ is a non-invertible element in
$\mathcal{MT}_{n}$. Hence, $\pi_{n}(\mathcal{MT}_{n})$ is not a group.

  2. (2)

If $R$ is an algebraically closed field, then $\pi_{n}(\mathcal{MT}_{n})$ is
an infinite group. Because in this case, $\mathcal{MT}_{n}$ is a group (we see
this later in Remark 4.11).

######  Definition 3.6.

Let $n>1$. We call the monoid $\mathcal{MT}_{n}$ constructed in Theorem 3.3
the monoid of triangular vector-permutation polynomials or more briefly the
triangular monoid.

######  Notation 3.7.

We set $\mathcal{MT}_{1}$ to be the monoid $\mathcal{MP}(R)$ of permutation
polynomials on $R$.

Our next aim is to describe the group of units of the monoid of vector-
permutation polynomials constructed in theorem 3.3. Before doing so, we need
to fix some notation.

######  Notation 3.8.

  1. (1)

Let $(1:f)$ denote $(f(x_{1}),x_{2},\ldots,x_{n})$, where $f\in R[x_{1}]$ is a
permutation polynomial on $R$.

  2. (2)

For $i>1$, let $(i:u;f)$ denote
$(x_{1},\ldots,x_{i-1},f+x_{i}u,x_{i+1},\ldots,x_{n})$, where $f,u\in
R_{[i-1]}$ such that $u\in UV(R_{[i-1]})$ (i.e. $u$ is a unit-valued
polynomial).

  3. (3)

When there is no confusion we write $\vec{f}\vec{g}$ instead of
$\vec{f}\circ\vec{g}$.

######  Remark 3.9.

  1. (1)

Let $\vec{f}\in\mathcal{MT}_{n}$ by definition, $\vec{f}$ has the form (3).
Then in view of Notation 3.8, we have by Fact 2.7,

$\vec{f}=(1:f_{1})\circ(2:u_{i};f_{i})\circ\cdots\circ(n:u_{n};f_{n}),\text{or
simply we write}$ $\vec{f}=(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i}),$ (4)

where $f_{0}\in R[x_{1}]$ is a permutation polynomial and $f_{i},u_{i}\in
R_{[i-1]}$ such that $u_{i}$ is a unit-valued polynomial for $i=2,\ldots,n$.
Clearly, that the expression (4) is uniquely determined by the expression (3)

  2. (2)

Again in view of Notation 3.8 , we see that for $i>1$,
$(1:x_{1})=(i:1;0)=(x_{1},\ldots,x_{n})$ is the identity element of
$\mathcal{MT}_{n}$.

  3. (3)

Let $\vec{g}=(1:g_{1})\prod_{i=2}^{n}(i:w_{i};g_{i})$. To obtain an expression
of $\vec{g}\vec{f}$ similar to expression (4), we adopt Equations (1) and (2)
to get

$\vec{g}\vec{f}=(1:g_{1}\circ
f_{1})\prod_{i=2}^{n}(i:u_{i}w_{i}(\vec{v_{i}});g_{i}(\vec{v_{i}})+w_{i}(\vec{v_{i}})f_{i}),\text{
where }$ (5)

$\vec{v_{i}}(x_{1},\ldots,x_{i-1})=(f_{1}(x_{1}),f_{2}(x_{1})+x_{2}u_{2}(x_{1}),\ldots,f_{i-1}(x_{1},\ldots,x_{i-1})+x_{{i-1}}u_{i-1}(x_{1},\ldots,x_{i-2}))$;
$i=2,\ldots,n$. We notice here that $w_{i}(v_{i})$ is a unit-valued polynomial
since $w_{i}$. So, expression 5 can be written in the expression 4 by choosing
$h_{1}=g_{1}\circ f_{1}$,

The following Lemmas are straightforward.

######  Lemma 3.10.

Let $f,g\in R[x_{1}]$ be permutation polynomials. Then

  1. (1)

$(1:f)(1:g)=(1:f\circ g)$;

  2. (2)

$\vec{f}=(1:f)$ is a unit (invertible) in $\mathcal{MT}_{n}$ if and only if
$f$ is invertible (with respect to the composition of polynomials) if and only
if $f$ is an $R$-automorphism of $R[x_{1}]$. In this case
$\vec{f}^{-1}=(1:f^{-1})$, where $f^{-1}$ is the inverse of $f$ with respect
to composition.

######  Lemma 3.11.

Let $i>1$ and $f,g,u,v\in R_{[i-1]}$ such that $u$ and $v$ are unit-valued
polynomials. Then

  1. (1)

$(i:u;f)(i:v;g)=(i:uv;f+ug)$;

  2. (2)

$\vec{f}=(i\\!:u;f)$ is a unit in $\mathcal{MT}_{n}$ if and only if $u$ is a
unit of the polynomial ring $R_{[i-1]}$ (with respect to multiplication). In
this case, $\vec{f}^{-1}=(i:u^{-1};-u^{-1}f)$, where $u^{-1}$ is the inverse
of $u$ in the ring $R_{[i-1]}$ (with respect to multiplication).

By means of Notation 3.8, Remark 3.9, Definition 2.8 and Lemma 2.13, we get
without effort the following useful lemma.

######  Lemma 3.12.

Let $R$ be a finite commutative ring. Let $f_{1},g_{1}\in R[x_{1}]$ be
permutation polynomials on $R$, and $f_{i},g_{i},v_{i},u_{i}\in R_{[i-1]}$
such that $u_{i}$ and $v_{i}$ are unit-valued polynomials on $R$ for
$i=2,\ldots,n$. The following statements are equivalent

  1. (1)

$(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i})\operatorname{{}\triangleq{}}(1:g_{1})\prod_{i=2}^{n}(i:v_{i};g_{i})$
on $R^{n}$;

  2. (2)

$(1:f_{1})\operatorname{{}\triangleq{}}(1:g_{1})$ and
$(i:u_{i};f_{i})\operatorname{{}\triangleq{}}(i:v_{i};g_{i})$ on $R^{n}$ for
$i=2,\dots,n$;

  3. (3)

$f_{1}\operatorname{{}\triangleq{}}g_{1}$,
$f_{i}\operatorname{{}\triangleq{}}g_{i}$ and
$u_{i}\operatorname{{}\triangleq{}}v_{i}$ on $R$ for $i=2,\dots,n$.

######  Theorem 3.13.

Let $R$ be a commutative ring. Then the group of units of the monoid of
triangular vector-permutation polynomials $\mathcal{MT}_{n}$ consists of all
elements of the form

$\vec{f}=(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i}),$

where $f_{1}\in R[x_{1}]$ is an $R$-automorphism, $f_{i}\in R_{[i-1]}$ and
$u_{i}\in{R_{[i-1]}}^{\times}$ for $i=2,\ldots,n$.

###### Proof.

Let $\vec{f}\in\mathcal{MT}_{n}$. Then, by Remark 3.9,
$\vec{f}=(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i})$ where
$f_{1}\in\mathcal{MP}(R)$, $f_{i}\in R_{[i-1]}$ and $u_{i}\in UV(R_{[i-1]})$.
First, suppose that $\vec{f}$ is a unit in $\mathcal{MT}_{n}$. Then there
exists $\vec{g}=(1:g_{1})\prod_{i=2}^{n}(i:w_{i};g_{i})$, where
$f_{1}\in\mathcal{MP}(R)$, $f_{i}\in R_{[i-1]}$ and $u_{i}\in UV(R_{[i-1]})$
with $\vec{g}\vec{f}=(x_{1},\ldots,x_{n})$. By Remark 3.9 ( Equation (5)),

$\vec{g}\vec{f}=(1:g_{1}\circ
f_{1})\prod_{i=2}^{n}(i:u_{i}w_{i}(\vec{v_{i}});g_{i}(\vec{v_{i}})+w_{i}(\vec{v_{i}})f_{i}),\text{
where }\vec{v_{i}}\in{R_{[i-1]}}^{i-1}\text{ for }i=2,\ldots,n.$

Again by Remark 3.9, we have the following equalities

$\begin{matrix}g_{1}\circ f_{1}=x_{1}\\\
g_{i}(\vec{v_{i}})+w_{i}(\vec{v_{i}})f_{i}x_{i}=x_{i}\text{ for
}i=2,\ldots,n.\end{matrix}$

The first equality implies that $g_{1}\circ f_{1}$ is an $R$-automorphism.
From the second equality, we infer that $w_{i}(\vec{v_{i}})f_{i}=1$ since
$g_{i}(\vec{v_{i}}),w_{i}(\vec{v_{i}})$ and $f_{i}$ are independent from the
variable $x_{i}$ for $i=2,\ldots,n$. Thus, $f_{i}\in{R_{[i-1]}}^{\times}$.
Conversely, assume that $f_{1}\in R[x_{1}]$ is an $R$-automorphism, $f_{i}\in
R_{[i-1]}$ and $u_{i}\in{R_{[i-1]}}^{\times}$ for $i=2,\ldots,n$ Then, by
construction and Lemma 3.10 and Lemma 3.11, $\vec{f}$ is invertible being
product (composition) of units in $\mathcal{MT}_{n}$. ∎

From Theorem 3.13, Lemma 3.10 and Lemma 3.11 we have the following result.

######  Corollary 3.14.

Let $\vec{f}\in\mathcal{MT}_{n}$. Suppose that
$\vec{f}=(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i})$ where
$f_{0}\in\mathcal{MP}(R)$, $f_{i},u_{i}\in R_{[i-1]}$ such that $u_{i}\in
UV(R_{[i-1]})$. Then $\vec{f}$ is invertible if and only if $(1:f_{1})$ and
$(i:u_{i};f_{i})$ are invertible for $i=2,\ldots,n$. Further, the inverse is
given by the relation

$\vec{f}^{-1}=(\prod_{i=n}^{2}(i:u_{i}^{-1};-u_{i}^{-1}f_{i}))(1:f_{1}^{-1}),$

where $f_{1}^{-1}$ is the inverse of $f_{1}$ with respect to composition and
for $i=2,\ldots,n$, $u_{i}^{-1}$ is the inverse of $u_{i}$ with respect to
multiplication.

######  Definition 3.15.

We denote by $TR_{n}$ the group of units of the monoid $\mathcal{MT}_{n}$. We
call $TR_{n}$ the group of triangular vector-permutation polynomials, or
shortly the triangular group.

If $R$ is finite, then necessary $\pi_{n}({\mathcal{MT}_{n}})$ is a group of
permutations of $R^{n}$ by Theorem 3.13. Further, in this case we have also
$\pi_{n}({TR_{n}})$ is a group of permutations of $R^{n}$ (being a closed
subset of the finite group $\pi_{n}({\mathcal{MT}_{n}})$). An interesting
question arises from what already we mentioned: Is it true that
$\pi(\mathcal{MT}_{n})\neq\pi_{n}(TR_{n})$? Fortunately, we can generally
answer this question affirmatively except in the cases $R$ is a finite product
of the field of two elements $\mathbb{F}_{2}$. Without loss of generality, we
consider only the case that $R$ is a finite local ring since the general case
follows by the Chinese Reminder Theorem.

Before doing so, we recall the following celebrated fact which is a special
case of a more general result due to Nöbauer [17, Theorem 2.3]. For a proof of
this fact, we refer the interested reader to a paper by Nechaev [15, Theorem
3].

From now on for a polynomial $f$, let $f^{\prime}$ be its formal derivative.

######  Lemma 3.16.

[17, Theorem. 2.3] Let $R$ be a finite local ring, that is not a field, $M$
its maximal ideal, and $f\in R[x]$.

Then $f$ is a permutation polynomial on $R$ if and only if the following
conditions hold:

  1. (1)

$f$ is a permutation polynomial on $R/M$;

  2. (2)

for all $a\in R$, $f^{\prime}(a)\neq 0\mod{M}$.

######  Lemma 3.17.

Let $R$ be a finite local ring which is not a field with residue field
$\mathbb{F}_{q}$. Then there is no invertible unit-valued polynomial
represents the unit-valued polynomial function induced by $g(x)=(x^{q}-x)+1$,
i.e. $\pi_{1}(g)\in\pi_{1}(UV(R[x])\setminus{R[x]}^{\times})$.

###### Proof.

Assume to the contrary that there is an invertible $f\in R[x]$ such that $f$
and $g$ induce the same function on $R$. Then by remark 2.4, since
$f(0)=g(0)=1$, $f(x)=1+m_{1}x+\cdots+m_{n}x^{n}$ with $m_{1},\ldots,m_{n}\in
M$. But, then, $xf(x)$ and $xg(x)$ induce the same function on $R$. However,
by Lemma 3.16, $xf(x)$ is a permutation polynomial while $xg(x)$ is not (since
the derivative of $xg(x)$ admits a root on $\mathbb{F}_{q}$). This
contradiction implies that such a polynomial $f$ does not exist. ∎

######  Theorem 3.18.

Let $R$ be a finite local ring. Then
$\pi_{n}(TR_{n})=\pi_{n}(\mathcal{MT}_{n})$ if and only if $R=\mathbb{F}_{2}$.

###### Proof.

If $R=\mathbb{F}_{2}$, one see easily $1$ is the only unit of $R$ and hence
every unit-valued polynomials $u(x_{1},$…$,x_{k})$ in $k$ variables (for any
$k\geq 1$) is equivalent to the (invertible) constant polynomial $1$. On the
other hand, there are only two permutations of $R$ that are induced by the
(invertible) permutation polynomials $x$ and $x+1$.

Now, if $\vec{F}\in\pi_{n}(\mathcal{MT}_{n})$, then $\vec{F}=\pi_{n}(\vec{f})$
for some $\vec{f}\in\pi(\mathcal{MT}_{n})$, where

$\vec{f}=(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i}),$

with $f_{0}\in\mathcal{MP}(R)$ and $f_{i},u_{i}\in R_{[i-1]}$ such that
$u_{i}\in UV(R_{[i-1]})$ for $i=2,\ldots,n$. But, the above discussion tell us
that $u_{i}\operatorname{{}\triangleq{}}1$ on $R$ (for $i=2,\ldots n$) and
either $f_{1}\operatorname{{}\triangleq{}}x_{1}$ or $x_{1}+1$ on $R$. Without
loss of generality say $f_{1}\operatorname{{}\triangleq{}}x$. So, if we
consider $\vec{g}=(1:x_{1})\prod_{i=2}^{n}(i:1;f_{i})$, then, by Theorem 3.13
and Definition 2.8, $\vec{g}\in TR_{n}$ and
$\pi_{n}(\vec{g})=\pi_{n}(\vec{f})=\vec{F}$. Thus,
$\vec{F}\in\pi_{n}(TR_{n})$.

For the other direction, we construct a vector-polynomial permutation
$\vec{F}\in\pi_{n}(\mathcal{MT}_{n})\setminus\pi_{n}(TR_{n})$. However, we
distinguish two cases on $R$. First, assume that $R$ is a local ring with
maximal ideal $M\neq\\{0\\}$. Now, consider the unit-valued polynomial
$g(x_{1})=(x_{1}^{q}-x_{1})+1$. Then
$\vec{g}=(2:g;0)\in\mathcal{MT}_{n}\setminus TR_{n}$ and hence
$\pi(\vec{g})\in\pi(\mathcal{MT}_{n})$. We claim that
$\pi(\vec{g})\notin\pi(TR_{n})$. Otherwise, suppose that there is a vector-
permutation polynomial $\vec{f}\in TR_{n}$, such that
$\pi_{n}(\vec{f})=\pi_{n}(\vec{g})$. Then, writing
$\vec{f}=(1:f_{1})\prod_{i=2}^{n}(i:u_{i};f_{i})$ implies that $u_{2}$ is a
unit of $R[x_{1}]$ by Theorem 3.13. Since $\pi(\vec{g})=\pi_{n}(\vec{f})$, we
have by Lemma 3.12, $u_{2}\operatorname{{}\triangleq{}}g$ on $R$. But, such a
unit polynomial $u_{2}$ does not exist by Lemma 3.17. Hence, the claim.

Now, consider the case $\mathbb{F}_{q}$ with $q\neq 2$. Here, one can see
easily, that unit polynomials are only the non zero-constants. Then, since
$q\neq 2$, there is an element $a\in\mathbb{F}_{q}$ such that $a\neq 0,1$ and
we can define a non constant unit-valued function
$F\colon\mathbb{F}_{q}\longrightarrow\mathbb{F}_{q}$,
$F(b)=\begin{cases}a&\textnormal{if }b=a\\\ 1&\textnormal{if }b\neq
a\end{cases}.$ Also, by Lagrange interpolation, we can find a polynomial $g$
represents $F$. Then the same argument as the previous case shows that
$\pi_{n}(2:g;0)\in\pi_{n}(\mathcal{MT}_{n})\setminus\pi_{n}(TR_{n})$.

∎

######  Remark 3.19.

Let $k>n$. Then the map

$e_{n,k}\colon R_{[n]}^{n}\longrightarrow
R_{[k]}^{k},\vec{f}=(f_{2},\ldots,f_{n})\rightarrow(f_{2},\ldots,f_{n},x_{n+1}\ldots,x_{k})$

is well known to be an embedding and by its restrictions on the monoid
$\mathcal{MT}_{n}$ and its group of units $TR_{n}$ respectively, we have the
following embeddings

  1. (1)

$\mathcal{MT}_{n}\hookrightarrow\mathcal{MT}_{k}$; and

  2. (2)

$TR_{n}\hookrightarrow TR_{k}$.

Also, there is an embedding
$\bar{e}_{n,k}\colon\pi_{n}(\mathcal{MT}_{n})\longrightarrow\pi_{k}(\mathcal{MT}_{k}),\pi_{n}(\vec{f})\rightarrow\pi_{k}\circ
e_{n,k}(\vec{f})$.

##  4\. Some structure results

In this section, we investigate the structures of the monoid
$\mathcal{MT}_{n}$, its group of units $TR_{n}$; and when$R$ is finite, its
induced group $\pi_{n}(\mathcal{MT}_{n})$ of permutations on $R^{n}$. In
particular, we see that the structures of $TR_{n}$ and
$\pi_{n}(\mathcal{MT}_{n})$ depend mainly on that of $\mathcal{MT}_{n}$.

From now on, when $H$ is a monoid let $H^{\times}$ denote its group of units.

###  4.1. The structure of the monoid $\mathcal{MT}_{n}$

In this subsection, we see that the triangular monoid $\mathcal{MT}_{n}$ can
be decomposed into an iterated semi-direct product of $n$ monoids.

The concept of the semi-direct product of monoids plays a vital role in this
part of the paper. In the following, we recall its definition. Let $A$ and $B$
be monoids and $End(B)$ be the monoid of endomorphisms of $B$ with respect to
composition. If $\phi\colon A\longrightarrow End(B)$, $a\rightarrow\phi_{a}$,
is a homomorphism then the semi-direct product $B\rtimes_{\phi}A$ (or simply
$B\rtimes A$) is the monoid with elements $\\{(a,b)\colon a\in A,b\in B\\}$
and operation $(a,b)(c,d)=(ac,b\phi_{a}(d))$ (when $A$ acts on $B$ from left)
or $(a,b)(c,d)=(ac,\phi_{c}(b)d)$ (when $A$ acts on $B$ from right)‘(see for
example [16]). We notice here that if we chose $A$ and $B$ to be two groups;
and replace $End(B)$ by the group $Aut(B)$ of automorphisms (i.e. $\phi\colon
A\longrightarrow Aut(B)$), then we just have the semi-direct product of the
group $B$ by $A$ with the same operation as mentioned before. We note here
that $(a,b)^{-1}=(a^{-1},\phi_{a^{-1}}(b^{-1}))$ and
$\phi_{a^{-1}}=\phi_{a}^{-1}$.

######  Lemma 4.1.

Fix $2\leq k\leq n$. Let $\mathcal{ML}^{n}_{k}$ denote the set of vector
polynomials of the form

$(k:u;f)$, where $f,u\in R_{[k-1]}$ such that $u\in UV(R_{[k-1]})$, is a
submonoid of the monoid $\mathcal{MT}_{n}$ which is isomorphic to the semi-
direct product of the additive group of $R_{[k-1]}$ by the monoid
$(UV(R_{[k-1]}),\cdot)$ of unit-valued polynomials. That is,
$\mathcal{ML}^{n}_{k}\cong R_{[k-1]}\rtimes UV(R_{[k-1]})$.

###### Proof.

It is clear that the identity $(k:1;0)\in\mathcal{ML}^{n}_{k}$. Also, by Lemma
3.11, $\mathcal{ML}^{n}_{k}$ is closed with respect to composition, whence
$\mathcal{ML}^{n}_{k}$ is a submonoid of $\mathcal{MT}_{n}$. Now, given a
unit-valued polynomial $u\in UV(R_{[k-1]})$, one can define an endomorphism
$\phi_{u}\colon R_{[k-1]}\longrightarrow R_{[k-1]}$ by $\phi_{u}(f)=uf$ for
every $f\in R_{[k-1]}$. Consider, the map $\phi\colon
UV(R_{[k-1]})\longrightarrow(End(R_{[k-1]}),``\circ")$,
($u\rightarrow\phi_{u}$). We claim that $\phi$ is a homomorphism. To show the
claim, let $u,v\in UV(R_{[k-1]})$ and consider $f\in R_{[k-1]}$. We have,

$\phi_{u}\circ\phi_{v}(f)=\phi_{u}(\phi_{v}(f))=\phi_{u}(vf)=uvf=\phi_{uv}(f).$

Thus $\phi_{u}\circ\phi_{v}=\phi_{uv}.$ This proves the claim. Hence, the
semi-direct product $R_{[k-1]}\rtimes UV(R_{[k-1]})$ is defined with the
following operation

$(u,f)(v,g)=(uv,f+ug),$

where $u,v\in UV(R_{[k-1]})$ and $f,g\in R_{[k-1]}$. The rest is to show that
the map

$\psi\colon R_{[k-1]}\rtimes
UV(R_{[k-1]})\longrightarrow\mathcal{ML}^{n}_{k},\text{
}(u,f)\rightarrow(k:u;f)$

is an isomorphism. We leave this to the reader.

∎

Next, we show that the monoid $\mathcal{MT}_{n}$ can be viewed as the semi-
direct product of the $\mathcal{ML}^{n}_{n}$ (defined in Lemma 4.1) by the
monoid $\mathcal{MT}_{n-1}$. We begin with proving the following technical
Lemma.

######  Lemma 4.2.

Let $n>1$. Then $\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$ is equipped
with the operation

$(\vec{h},((n:u;f))(\vec{l},((n:v;g))=(\vec{h}\circ\vec{l},(n:u(\vec{l})v;f(\vec{l})+u(\vec{l})g)),$
(6)

where $\vec{h},\vec{l}\in\mathcal{MT}_{n-1}$ and
$(n:u;f),(n:v;g)\in\mathcal{ML}^{n}_{n}$

###### Proof.

Let $\vec{h}\in\mathcal{MT}_{n-1}$. For simplicity write
$\vec{h}=(h_{1},\ldots,h_{n-1})$, where $h_{i}\in R_{[i]}$ for
$i=1,\ldots,n-1$ (keeping in mind that $\vec{h}$ satisfies the conditions of
expression (3)). Then define a map
$\phi_{\vec{h}}\colon\mathcal{ML}^{n}_{n}\longrightarrow\mathcal{ML}^{n}_{n}$
by $((n:u;f))\phi_{\vec{h}}=(n:u(\vec{h});f(\vec{h}))$ for
$(n:u;f)\in\mathcal{ML}^{n}_{n}$. Since $u$ is a unit-valued polynomial (in
$n-1$ variables), we have that $u(\vec{h})$ is a unit-valued polynomial (in
$n-1$ variables). This shows $((n:u;f))\phi_{\vec{h}}\in\mathcal{ML}^{n}_{n}$.
Now, let $(n:v;g)\in\mathcal{ML}^{n}_{n}$ and consider

$\displaystyle((n:u;f)(n:v;g))\phi_{\vec{h}}$
$\displaystyle=((n:uv;f+ug))\phi_{\vec{h}}\text{ by Lemma~{}\ref{relat2}}$
$\displaystyle=(n:(uv)(\vec{h});f(\vec{h})+(ug)(\vec{h}))=(n:u(\vec{h})v(\vec{h});f(\vec{h})+u(\vec{h})g(\vec{h}))$
$\displaystyle{}_{=}^{\text{Lemma~{}\ref{relat2}}}(n:u(\vec{h});f(\vec{h}))(n:v(\vec{h});g(\vec{h}))=((n:u;f))\phi_{\vec{h}}((n:v;g))\phi_{\vec{h}}.$

Thus, $\phi_{\vec{h}}\in End(\mathcal{ML}^{n}_{n})$. Then, the map
$\phi\colon\mathcal{MT}_{n-1}\longrightarrow End(\mathcal{ML}^{n}_{n})$,
${\vec{h}}\rightarrow\phi_{\vec{h}}$ is a homomorphism. Indeed, if
$\vec{l}=(l_{1},\ldots,l_{n-1})\in\mathcal{MT}_{n-1}$, we have

$\displaystyle((n:u;f))\phi_{\vec{h}\circ\vec{l}}$
$\displaystyle=(n:u(\vec{h}\circ\vec{l});f(\vec{h}\circ\vec{l}))=(n:u(h_{1}(\vec{l}),\ldots,h_{n-1}(\vec{l}));f(h_{1}(\vec{l}),\ldots,h_{n-1}(\vec{l})))$
$\displaystyle=(n:u(h_{1},\ldots,h_{n-1});f(h_{1},\ldots,h_{n-1}))\phi_{\vec{l}}=((n:u;f))\phi_{\vec{h}})\phi_{\vec{l}}$
$\displaystyle=((n:u;f))\phi_{\vec{h}}\circ\phi_{\vec{l}}.$

Thus, the semi-direct product $\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$
is defined with the following operation

$\displaystyle(\vec{h},((n:u;f))(\vec{l},((n:v;g))$
$\displaystyle=(\vec{h}\circ\vec{l},((n:u;f))\phi_{\vec{l}}(n:v;g))$
$\displaystyle=(\vec{h}\circ\vec{l},(n:u(\vec{l})v;f(\vec{l})+u(\vec{l})g))$

or more explicitly

$(\vec{h},(n:u;f))(\vec{l},(n:v;g))=((h_{1}\circ
l_{1},h_{2}(l_{1},l_{2}),\ldots,h_{n-1}(l_{1},\ldots,l_{n-1})),(n:u(\vec{l})v;f(\vec{l})+u(\vec{l})g)).$
(7)

∎

######  Proposition 4.3.

Let $n>1$ and let $\mathcal{ML}^{n}_{n}$ denote the monoid of vector
polynomials of the form

$(n:u;f)$, where $f,u\in R_{[k-1]}$ and $u\in UV(R_{[k-1]})$. Then
$\mathcal{MT}_{n}\cong\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$.

###### Proof.

In view of Lemma 4.2 the semi-direct product
$\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$ is defined by the operation
given in Equation 6 (more explicit in Equation7). Define a map
$\psi\colon\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}\longrightarrow\mathcal{MT}_{n}$
as the following: if
$(\vec{h},(n:u;f))\in\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$, where
$u\in UV(R_{[n-1]})$ and $\vec{h}=(h_{1},\ldots,h_{n-1})$; $h_{i}\in R_{[i]}$
for $i=1,\ldots,n-1$, then set

$\psi(\vec{h},(n:u;f))=(h_{1},\ldots,h_{n-1},f+ux_{n}).$

Note that, by the definition of $\mathcal{MT}_{n-1}$, $h_{1}$ is a permutation
polynomial on $R$ and $h_{i}=f_{i}+u_{i}x_{i}$ where $f_{i},\in R_{[i-1]}$ and
$u_{i}\in UV(R_{[i-1]})$ for $i=2,\ldots,n-1$. This means that
$\psi(\vec{h},(n:u;f))\in\mathcal{MT}_{n}$. Now, we show that $\psi$ is a
homomorphism. Let $(\vec{h},(n:u;f))$ and
$(\vec{l},(n:v;g))\in\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$, and
consider,

$\psi((\vec{h},(n:u;f))(\vec{l},(n:v;g))){}_{=}^{\text{Equation~{}\ref{explcform}}}\\\
=\psi(((h_{1}\circ
l_{1},h_{2}(l_{1},l_{2}),\ldots,h_{n-1}(l_{1},\ldots,l_{n-1})),(n:u(\vec{l})v;f(\vec{l})+u(\vec{l})g)))\\\
=(h_{1}\circ
l_{1},h_{2}(l_{1},l_{2}),\ldots,h_{n-1}(l_{1},\ldots,l_{n-1})),f(\vec{l})+u(\vec{l})g+u\big{(}\small{\vec{l}}\big{)}vx_{n})\\\
=(h_{1}\circ
l_{1},h_{2}(l_{1},l_{2}),\ldots,h_{n-1}(l_{1},\ldots,l_{n-1})),f(l_{1},\ldots,l_{n-1})+u(l_{1},\ldots,l_{n-1})g+u(l_{1},\ldots,l_{n-1})vx_{n})\\\
=(h_{1},\ldots,h_{n-1},f+ux_{n})(l_{1},\ldots,l_{n-1},g+vx_{n})=\psi((\vec{h},(n:u;f)))\psi((\vec{l},(n:v;g))).$

The hardest part of the proof has been finished, and the remaining details are
straightforward and left to the reader. ∎

We are now in position to decompose the triangular monoid $\mathcal{MT}_{n}$
into an iterated semi-direct product of
$\mathcal{ML}^{n}_{n},\ldots,\mathcal{ML}^{2}_{2}$, and
$\mathcal{MT}_{1}=\mathcal{MP}(R)$ (see Notation 3.7). In fact, we deduce
more.

######  convention 4.4.

From now on, by $A_{1}\rtimes A_{2}\rtimes A_{3}$, we means
$A_{1}\rtimes(A_{2}\rtimes A_{3})$, i.e. we start computing from the left.

######  Theorem 4.5.

Let $n>1$. Then

  1. (1)

$\mathcal{MT}_{n}\cong\mathcal{ML}^{n}_{n}\rtimes\cdots\rtimes\mathcal{ML}^{2}_{2}\rtimes\mathcal{MP}(R)$;

  2. (2)

$\mathcal{MT}_{n}\cong(R_{[n-1]}]\rtimes
UV(R_{[k-1]}))\rtimes\cdots\rtimes(R{[x_{1}]}\rtimes
UV(R{[x_{1}]}))\rtimes\mathcal{MP}(R)$.

###### Proof.

By Proposition 4.3,
$\mathcal{MT}_{n}\cong\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1}$. Then
iterated applications of this relation yield the first expression. Then, we
use Lemma 4.1, to have $\mathcal{ML}^{i}_{i}\cong R_{[i-1]}\rtimes
UV(R_{[i-1]})$ for $i=2,\ldots,n$. Combining this with the first expression
gives the second expression.

∎

###  4.2. The stricture of the triangular group $TR_{n}$

Using the fact that monoid homomorphisms preserve the identity, we will see
that the group units of a semi-direct product of monoids is just the semi-
direct product of their groups of units. Then, we apply this result to the
decomposition of $\mathcal{MT}_{n}$ given in Theorem 4.5 to obtain a
decomposition of the group $TR_{n}$ into an iterated semi-direct product of
groups.

######  Lemma 4.6.

Let $A$ and $B$ be monoids. Consider the homomorphism $\phi\colon
A\longrightarrow End(B)$ ($a\rightarrow\phi_{a}$, $\phi_{a}\colon
B\longrightarrow B$ is a homomorphism ), and let $\psi_{a}$ be the restriction
of $\phi_{a}$ on $B^{\times}$ then

  1. (1)

$\phi_{a}\in Aut(B)$ for every $a\in A^{\times}$;

  2. (2)

$\psi_{a}\in Aut(B^{\times})$ for every $a\in A^{\times}$.

###### Proof.

Consider an element $a\in A^{\times}$, by the definition of $\phi$, $\phi_{a}$
is a homomorphism. So, we need only to show that $\phi_{a}$ is a bijection.
Now,

$\phi_{a}\circ\phi_{a^{-1}}=\phi(aa^{-1})=\phi(1_{A})=\phi_{1_{A}}=\phi(1_{A})=\phi(a^{-1}a)=\phi_{a^{-1}}\circ\phi_{a},$

but, a homomorphism preserves the identity i.e. $\phi(1_{A})=\phi_{1_{A}}$ is
the identity endomorphism of $B$. Hence $\phi_{a}\in Aut(B)$. This proves (1).

(2) Since a homomorphism preserves units and the restriction of a homomorphism
is a homomorphism, we have that $\psi_{a}\colon B^{\times}\longrightarrow
B^{\times}$ is a homomorphism for every $a\in A^{\times}$. Then similar
argument as in (1) shows that $\psi_{a}\in Aut(B^{\times})$ for every $a\in
A^{\times}$.

∎

######  Remark 4.7.

If $A$ and $B$ are monoids and $\phi\colon A\longrightarrow End(B)$
($a\rightarrow\phi_{a}$, $\phi_{a}\colon B\longrightarrow B$ is a homomorphism
) is a homomorphism, then we have a homomorphism $\psi\colon
A^{\times}\longrightarrow Aut(B^{\times})$ ($a\rightarrow\psi_{a}$, where
$\psi_{a}$ is the restriction of $\phi_{a}$ on $B^{\times}$) by Lemma 4.6.
Furthermore, the semi-direct product (of groups)
$B^{\times}\rtimes_{\psi}A^{\times}$ with respect to $\psi$ is defined with
operation inherited from that of the semi-direct product (of monoids)
$B\rtimes_{\phi}A$.

######  Proposition 4.8.

Let $A$ and $B$ be monoids and $\phi\colon A\longrightarrow End(B)$
($a\rightarrow\phi_{a}$, $\phi_{a}\colon B\longrightarrow B$ is a homomorphism
) be a homomorphism. Let $\psi\colon A^{\times}\longrightarrow
Aut(B^{\times})$ be the homomorphism defined by $a\rightarrow\psi_{a}$, where
$\psi_{a}$ is the restriction of $\phi_{a}$ on $A^{\times}$, then

$(B\rtimes_{\phi}A)^{\times}=B^{\times}\rtimes_{\psi}A^{\times}.$

###### Proof.

By the definition of $\psi$, the operation of
$B^{\times}\rtimes_{\psi}A^{\times}$ is just the operation of
$B\rtimes_{\phi}A$ restricted into the set $A^{\times}\bigtimes B^{\times}$.
Thus,
$B^{\times}\rtimes_{\psi}A^{\times}\subseteq(B\rtimes_{\phi}A)^{\times}$, and
hence $B^{\times}\rtimes_{\psi}A^{\times}$ is a subgroup of
$(B\rtimes_{\phi}A)^{\times}$. For the other inclusion, consider
$(a,b)\in(B\rtimes_{\phi}A)^{\times}$. Then there exists
$(a_{1},b_{1})\in(B\rtimes_{\phi}A)^{\times}$ such that

$(a,b)(a_{1},b_{1})=(aa_{1},b\phi_{a}(b_{1}))=(1_{A},1_{B})=(a_{1}a,b_{1}\phi_{a_{1}}(b))=(a_{1},b_{1})(a,b).$

Therefore, we have the following two equalities

$aa_{1}=a_{1}a=1_{A},\text{ and }$
$b\phi_{a}(b_{1})=b_{1}\phi_{a_{1}}(b)=1_{B}.$

The first equality implies that $a\in A^{\times}$ and $a_{1}=a^{-1}$, while
the other implies $b$ is right invertible. Also, applying the automorphism
$\phi_{a}$ to the equality $b_{1}\phi_{a_{1}}(b)=1_{B}$ yields since
$a_{1}=a^{-1}$,

$\phi_{a}(b_{1})b=\phi_{a}(b_{1})\phi_{1_{A}}(b)=\phi_{a}(b_{1})\phi_{aa^{-1}}(b)=\phi_{a}(b_{1})\phi_{a}(\phi_{a_{1}}(b))=\phi_{a}(b_{1}\phi_{a_{1}}(b))=\phi_{a}(1_{B})=1_{B},$

hence $\phi_{a}(b_{1})b=1_{B}$, and $b$ is a left invertible. Thus, $b\in
B^{\times}$ (being left and right invertible). Therefore, $(a,b)\in
B^{\times}\rtimes_{\psi}A^{\times}$.

∎

######  Proposition 4.9.

Fix $2\leq k\leq n$. Let $\mathcal{L}^{n}_{k}$ denote the set of vector
polynomials of the form

$(k:u;f)$, where $f\in R_{[k-1]}$ and $u\in R_{[k-1]}^{\times}$. Then

  1. (1)

$\mathcal{L}^{n}_{k}$ is a subgroup of the group $TR_{n}$;

  2. (2)

$\mathcal{L}^{n}_{k}\cong R_{[k-1]}\rtimes R_{[k-1]}^{\times}$, that is,
$\mathcal{L}^{n}_{k}$ is the semi-direct product of additive group of the
polynomial ring in $k-1$ variables by its group of units;

  3. (3)

$\mathcal{L}^{n}_{k}\lhd TR_{n}$ if and only if $n=k$. In this case,

$TR_{n}\cong\mathcal{L}^{n}_{n}\rtimes TR_{n-1}.$

###### Proof.

(1) Let $\mathcal{ML}^{n}_{k}$ be the monoid defined in Lemma 4.1 containing
all vector polynomials of the form $(k:u;f)$ , where $f\in R_{[k-1]}$ and
$u\in UV(R_{[k-1]})$. Then, by Theorem 3.13

${\mathcal{ML}^{n}_{k}}^{\times}=\mathcal{ML}^{n}_{k}\cap{\mathcal{MT}_{n}}^{\times}=\mathcal{ML}^{n}_{k}\cap
TR_{n}=\mathcal{L}^{n}_{k}.$

(2) By Lemma 4.1, $\mathcal{ML}^{n}_{k}\cong R_{[k-1]}\rtimes UV(R_{[k-1]})$
(semi-direct product of monoids). But, by definition,
$UV(R_{[k-1]})^{\times}={R_{[k-1]}}^{\times}$; and
$(R_{[k-1]},+)^{\times}=R_{[k-1]}$. Thus, the semi-direct product
$R_{[k-1]}\rtimes{R_{[k-1]}}^{\times}$, is defined by Remark 4.7. Now,

$\mathcal{L}^{n}_{k}{}_{=}^{(\ref{1})}{\mathcal{ML}^{n}_{k}}^{\times}{}_{\cong}^{Lemma~{}\ref{semdm}}(R_{[k-1]}\rtimes
UV(R_{[k-1]}))^{\times}{}_{=}^{Propostion~{}\ref{unisemi}}R_{[k-1]}\rtimes{R_{[k-1]}}^{\times}.$

(3) Assume, first that $k=n$. Then, by Theorem 3.3, Remark 4.7 and Proposition
4.8, we have

$TR_{n}={\mathcal{MT}_{n}}^{\times}\cong(\mathcal{ML}^{n}_{n}\rtimes\mathcal{MT}_{n-1})^{\times}=\mathcal{L}^{n}_{n}\rtimes
TR_{n-1}.$

This also, shows that $\mathcal{L}^{n}_{n}\lhd TR_{n}$. Now, assume that
$2\leq k<n$ and let $(k:1;x_{k-1})\in\mathcal{L}^{n}_{k}$. So, if
$(k+1:1;x_{k})\in TR_{n}\setminus\mathcal{L}^{n}_{k}$,
$(k+1:1;x_{k})^{-1}=(k+1:1;-x_{k})$ by Lemma 3.11. Then, we have by direct
calculations (Definition 2.6),

$\displaystyle(k+1:1;x_{k})(k:1;x_{k-1})(k+1:1;-x_{k})$
$\displaystyle=(x_{1},\ldots,x_{k-1},x_{k-1}+x_{k},x_{k-1}+x_{k+1},x_{k+2},\ldots,x_{n})$
$\displaystyle=(k:1;x_{k-1})(k+1:1;x_{k-1})\notin\mathcal{L}^{n}_{k}.$

∎

As a result of Proposition 4.9, we are able to decompose the group $TR_{n}$
into an iterated semi-direct product of
$\mathcal{L}^{n}_{n},\ldots,\mathcal{L}^{2}_{2}$, and $TR_{1}=Aut_{R}(R[x])$.
Alternatively, we can replace $\mathcal{L}^{k}_{k}$ by
$R_{[k-1]}]\rtimes{R_{[k-1]}}^{\times}$ ($k=2,\ldots,n$) to obtain another
equivalent decomposition.

######  Theorem 4.10.

Let $n>1$ and let $TR_{n}$ be the triangular group. Then

  1. (1)

$TR_{n}\cong\mathcal{L}^{n}_{n}\rtimes\cdots\rtimes\mathcal{L}^{2}_{2}\rtimes
Aut_{R}(R[x])$;

  2. (2)

$TR_{n}\cong(R_{[n-1]}]\rtimes{R_{[n-1]}}^{\times})\rtimes\cdots\rtimes(R{[x_{1}]}\rtimes
R{[x_{1}]}^{\times})\rtimes Aut_{R}(R[x])$.

Moreover, $TR_{n}$ is solvable if and only if $Aut_{R}(R[x])$ is solvable . In
particular, if $R$ is a domain, $TR_{n}$ is solvable, and
$TR_{n}\cong(R_{[n-1]}]\rtimes
R^{\times})\rtimes\cdots\rtimes(R{[x_{1}]}\rtimes R^{\times})\rtimes(R\rtimes
R^{\times})$.

###### Proof.

By Proposition 4.9, $TR_{n}\cong\mathcal{L}^{n}_{n}\rtimes TR_{n-1}$. Then
iterated application of this relation yields the first expression. Again, by
Proposition 4.9, we have $\mathcal{L}^{i}_{i}\cong R_{[i-1]}\rtimes
R_{[i-1]}^{\times}$ for $i=2,\ldots,n$. Combining these results yields the
second expression. Moreover, it is well known that Abelian groups are solvable
and extensions of solvable groups are solvable. Therefore, by the second
expression, $TR_{n}$ is solvable if and only if $Aut_{R}(R[x])$ is solvable.
In particular, if $R$ is a domain, then $Aut_{R}(R[x])\cong R\rtimes
R^{\times}$ and for each $i$, ${R_{[i]}}^{\times}=R^{\times}$. Thus, $TR_{n}$
is solvable, and the claim follows. ∎

######  Remark 4.11.

  1. (1)

In algebraic geometry, the Jonquières (classical) triangular group $KTR_{n}$
is defined to be the group of all vector-polynomials of the form

$\vec{f}=(a_{1}x_{1}+b_{0},f_{1}(x_{1})+a_{2}x_{2},\ldots,f_{n-1}(x_{1},\ldots,x_{n-1})+a_{n}x_{n}),$
(8)

where $a_{1},\ldots,a_{n}\in R^{\times}$ and $b\in R$ (see [23]). In general,
we have the inclusions:

$KTR_{n}\subseteq TR_{n}\subseteq\mathcal{MT}_{n}.$ (9)

However, by Theorem 4.10 and Theorem 4.5, one can easily see that

$KTR_{n}=TR_{n}=\mathcal{MT}_{n}$

if and only if:

a):  $UV(R_{[i]})={R_{[i]}}^{\times}=R^{\times}$ b):
$\mathcal{MP}(R)=Aut_{R}(R[x])=\\{ax+b\colon a\in R^{\times}\\}$.

In the following examples, we demonstrate how the validity of the previous two
conditions determines the nature of the inclusions of relation (9).

     * •

When $R$ is a $D$-ring (see for example [11]), $R$ satisfies condition 1a) by
the definition of $D$-rings. Then $KTR_{n}=TR_{n}$ (note that it is enough
here to have $R_{[i]}^{\times}=R^{\times}$). Further, if $R$ is an
algebraically closed field, then

$KTR_{n}=TR_{n}=\mathcal{MT}_{n}.$

Because, in this case condition 1b) is valid in $R$ as well. Indeed, we have
that every vector-permutation polynomial is invertible (see [4, Theorem 2.2]
and [22, Page. 80]).

     * •

If $R$ is a non-algebraically closed field (for example, real numbers and
finite fields),

$KTR_{n}=TR_{n}\subsetneq\mathcal{MT}_{n}.$

In this case, it is clear that ${R_{[i]}}^{\times}=R^{\times}$ (for each $i$)
and $Aut_{R}(R[x])$ consists of all linear polynomials. Also, since $R$ is not
an algebraically closed field, there is irreducible polynomial $f$ of degree
$\geq 2.$ Evidently, $f\in UV(R_{[i]})\setminus{R_{[i]}}^{\times}$

     * •

When $R$ is not reduced (i.e. containing a non-zero nilpotent), we have

$KTR_{n}\subsetneq TR_{n}\subseteq\mathcal{MT}_{n}.$

Here, we have by Remark 2.4, $\in{R_{[i]}}^{\times}\neq R^{\times}$ for any
non-zero nilpotent $r$. In particular, if $R$ is a finite local ring that is
not a field,

$KTR_{n}\subsetneq TR_{n}\subsetneq\mathcal{MT}_{n},$

since in this case we also have $1+(x^{q}-x)\in UV(R[x])\setminus
R[x]^{\times}$.

  2. (2)

In Equation (8), if $a_{1}=\ldots=a_{n}=1$, we get the so called unitriangular
group of automorphisms. In [2], a decomposition of the unitriangular group on
fields of characteristic zero into an iterated semi-product of Abelian groups
is given.

######  Remark 4.12.

Unlike solvability, the decomposition of Theorem 4.10 can not determine
whether the group $TR_{n}$ is nilpotent or not. However, in the following we
notice for a large class of rings that $TR_{n}$ is not nilpotent.

  1. (1)

Let $n>1$ and $R$ is an infinite field. Then $TR_{n}$ is not nilpotent.

  2. (2)

Let $n>2$ and $R$ be a commutative ring with Jacobson radical $J(R)$ such that
$R/J(R)$ is the finite field of $q$ elements $\mathbb{F}_{q}$.

To show these assertions, it is enough to find a subgroup of $TR_{n}$, which
is not nilpotent.

Indeed the first assertion follows from the fact that the unitriangular group
(defined in Remark 4.11) is not nilpotent (see[[2] Theorem 2.1]). To prove the
second assertion, we consider first the case $R=\mathbb{F}_{q}$. But, then the
unitrianguar group is not nilpotent (see [[10] Theorem 3]). For the case
$J(R)\neq\\{0\\}$, we note that there exists a natural epimorphism $\pi\colon
R\longrightarrow R/J(R)$ given by $\pi(a)=a\mod J(R)$. Evidently $\pi$ can be
extended in to an epimorphism of monoids
$\tilde{\pi}\colon{R_{[n]}}^{n}\longrightarrow{{R/J(R)}_{[n]}}^{n}$, and one
easily see that
$\tilde{\pi}(UT_{n}(R))=UT_{n}(R/J(R))=UT_{n}(\mathbb{F}_{q})$, where
$UT_{n}(A)$ stands for the unitriangular group over the ring $A$. Hence
$UT_{n}(R)$ is not nilpotent since $\tilde{\pi}(UT_{n}(R))$ is not nilpotent.
Thus, $TR_{n}$ is not nilpotent.

###  4.3. The structure of the induced permutations group
$\pi_{n}(\mathcal{MT}_{n})$

Recall that $\pi_{n}$ stands for the natural homomorphism assigning for each
$\vec{f}\in{R_{[n]}}^{n}$ its induced vector-function $\vec{F}\colon
R^{n}\longrightarrow R^{n}$.

In this subsection, we explore the structure of the group
$\pi_{n}(\mathcal{MT}_{n})$ of permutations of $R^{n}$ induced by the elements
of the triangular monoid $\mathcal{MT}_{n}$. We see that this group can be
decomposed into an iterated semi-direct product of $n$ groups in a similar
manner to what we have already seen for the group $TR_{n}$.

######  Proposition 4.13.

Fix $2\leq k\leq n$. Let $\mathcal{ML}^{n}_{k}$ be the monoid of vector
permutation polynomials of the form $(k:u;f)$, where $f\in R_{[k-1]}$ and
$u\in UV(R_{[k-1]})$. Then

  1. (1)

$\pi_{n}(\mathcal{ML}^{n}_{k})$ is a subgroup of the group
$\pi_{n}(\mathcal{MT}_{n})$;

  2. (2)

$\pi_{n}(\mathcal{ML}^{n}_{k})\cong\mathcal{F}(R^{k-1})\rtimes\mathcal{F}(R^{k-1})^{\times}$,
that is, $\mathcal{ML}^{n}_{k}$ is the semi-direct product of the additive
group of the ring of polynomial functions (on $R$) in $k-1$ variables by its
group of units;

  3. (3)

$\pi_{n}(\mathcal{ML}^{n}_{k})\lhd\pi_{n}(\mathcal{MT}_{n})$ if and only if
$n=k$. In this case,

$\pi_{n}(\mathcal{MT}_{n})\cong\pi_{n}(\mathcal{ML}^{n}_{n})\rtimes\pi_{n}(\mathcal{MT}_{n-1}).$

###### Proof.

Let $k$ be fixed. (1) Obvious

(2) By Lemma 4.1 , $\mathcal{ML}^{n}_{k}\cong R_{[k-1]}\rtimes UV(R_{[k-1]})$.
Then it is easy to see that the sets sets $A=\\{(k:1;f):f\in R_{[k-1]}\\}$ and
$B=\\{(k:u;0):u\in UV(R_{[k-1]}\\}$ are isomorphic to $R_{[k-1]}$ and
$UV(R_{k-1})$ respectively. Further, $\pi_{n}(A)$ and $\pi_{n}(B)$ are
subgroups of the finite group $\pi_{n}(\mathcal{ML}^{n}_{k})$ (being non empty
and closed with respect to composition).

Now, we claim that $\pi_{n}(A)\cong\mathcal{F}(R^{k-1})$ and
$\pi_{n}(B)\cong\mathcal{F}(R^{k-1})^{\times}$. To show this claim, let
$\pi\colon R_{[k-1]}\longrightarrow\mathcal{F}(R^{k-1})$ be the natural
epimorphism mapping every polynomial to its induced function on $R$. Then
define a map $\Psi\colon\pi_{n}(A)\longrightarrow\mathcal{F}(R^{k-1})$ by
$\Psi(\vec{F})=\pi(f)$, where $f\in R_{[k-1]}$ such that
$\vec{F}=\pi_{n}((k:1;f))$. By Lemmas 2.13 and 3.12, $\Psi$ is well defined.
Let $\vec{G}\in\pi_{n}(A)$. Then $\vec{G}=\pi_{n}((k:1;g))$. Consider,

$\displaystyle\Psi(\vec{F}\vec{G})$
$\displaystyle=\Psi(\pi_{n}((k:1;f))\circ\pi_{n}((k:1;g)))=\Psi(\pi_{n}((k:1;f)(k:1;g)))$
$\displaystyle{}_{=}^{Lemma~{}\ref{relat2}}\Psi(\pi_{n}((k:1;f+g))=\pi_{n}(f+g)=\pi_{n}(f)+\pi_{n}(g).$

Now, by the definitions of the set $A$, and the maps $\Psi$, $\pi_{n}$ and
$\pi$, we have

$\Psi(\pi_{n}(A))=\pi(R_{[k-1]})=\mathcal{F}(R^{k-1}).$

Similarly, one can show that $\pi_{n}(B)\cong\mathcal{F}(R^{k-1})^{\times}$.
Consider, the map
$\Phi\colon\pi_{n}(\mathcal{ML}^{n}_{k})\longrightarrow\pi_{n}(B),(\pi_{n}((k:u;f))\rightarrow\pi_{n}((k:u;0)))$,
it is obvious that $\Phi$ is surjective. So, if
$\pi_{n}((k:u;f)),\pi_{n}((k:v;g))\in\pi_{n}(\mathcal{ML}^{n}_{k})$, we have

$\displaystyle\Phi(\pi_{n}((k:u;f))\circ\pi_{n}((k:v;g)))=\Psi(\pi_{n}((k:u;f)(k:v;g)))$
$\displaystyle=\Phi(\pi_{n}((k:uv;f+ug))=(\pi_{n}((k:uv;0))=\pi_{n}((k:u;0)(k:v;0))=$
$\displaystyle\pi_{n}((k:u;0)\circ\pi_{n}((k:v;0))=\Phi(\pi_{n}((k:u;f)))\circ\Phi(\pi_{n}((k:v;g))).$

Now, if $j\colon\pi_{n}(B)\longrightarrow\pi_{n}(\mathcal{ML}^{n}_{k})$ is the
inclusion, then evidently, $\Phi j=id_{\pi_{n}(B)}$. Therefore, the following
extension splits

$1\rightarrow\pi_{n}(A)\xrightarrow{i}\pi_{n}(\mathcal{ML}^{n}_{k})\xrightarrow{\Phi}\pi_{n}(B)\rightarrow
1.$

Thus,

$\pi_{n}(\mathcal{ML}^{n}_{k})=\pi_{n}(A)\rtimes\pi_{n}(B)\cong\mathcal{F}(R^{k-1})\rtimes\mathcal{F}(R^{k-1})^{\times}.$

(3) Let $\vec{F}\in\pi_{n}(\mathcal{MT}_{n})$. Then $\vec{F}=\pi_{n}(\vec{f})$
for some $\vec{f}=(f_{1},\ldots,f_{n-1},f_{n})$ with $f_{i}\in R_{[i]}$ for
$i=1,\ldots,n$ (by the definition of the triangular monoid
$\mathcal{MT}_{n}$). Define
$P\colon\pi_{n}(\mathcal{MT}_{n})\longrightarrow\pi_{n-1}(\mathcal{MT}_{n-1})$
by letting $P(\vec{F})=\pi_{n-1}((f_{1},\ldots,f_{n-1}))$. Clearly, $P$ is an
epimorphism. Also, the map
$j\colon\pi_{n-1}(\mathcal{MT}_{n-1})\longrightarrow\pi_{n}(\mathcal{MT}_{n})$,
$(\pi_{n-1}((f_{1},\ldots,f_{n-1})))\rightarrow\pi_{n}((f_{1},\ldots,f_{n-1},x_{n}))$
is obviously a monomorphism. Further, $Pj=id_{\pi_{n-1}(\mathcal{MT}_{n-1})}$.
Hence, the following extension splits

$1\rightarrow\pi_{n}(\mathcal{ML}^{n}_{n})\xrightarrow{i}\pi_{n}(\mathcal{MT}_{n})\xrightarrow{P}\pi_{n-1}(\mathcal{MT}_{n-1})\rightarrow
1.$

Therefore,
$\pi_{n}(\mathcal{MT}_{n})\cong\pi_{n}(\mathcal{ML}^{n}_{k})\rtimes\pi_{n-1}(\mathcal{MT}_{n-1}),\text{
and }\pi_{n}(\mathcal{ML}^{n}_{k})\lhd\pi_{n}(\mathcal{MT}_{n}).$

Now, assume that $2\leq k<n$. Then, we adopt the argument given in the last
part of the proof of Proposition 4.9 (but with applying the homomorphism
$\pi_{n}$). So, if
$\pi_{n}((k+1:1;x_{k}))\in\pi_{n}(TR_{n})\subseteq\pi_{n}(\mathcal{MT}_{n})$,
$(\pi_{n}((k+1:1;x_{k})))^{-1}=\pi_{n}((k+1:1;x_{k})^{-1})=\pi_{n}((k+1:1;-x_{k}))$.
Consider an element $\pi_{n}(k:1;x_{k-1})\in\pi_{n}(\mathcal{ML}^{n}_{k})$, we
have since $\pi_{n}$ is a homomorphism,

$\displaystyle\pi_{n}((k+1:1;x_{k})(k:1;x_{k-1})$
$\displaystyle(k+1:1;-x_{k}))=$ $\displaystyle=\pi_{n}((x_{1},\ldots,x_{k-1},$
$\displaystyle x_{k-1}+x_{k},x_{k-1}+x_{k+1},x_{k+2},\ldots,x_{n}))=$
$\displaystyle=\pi_{n}((k:1;x_{k-1}))\pi_{n}((k+1:1;x_{k-1})).$

Thus, since $\pi_{n}((k:1;x_{k-1}))\in\pi_{n}(\mathcal{ML}^{n}_{k})$,
$\pi_{n}((k:1;x_{k-1}))\pi_{n}((k+1:1;x_{k-1}))\notin\pi_{n}(\mathcal{ML}^{n}_{k})$
if and only if $\pi_{n}((k+1:1;x_{k-1}))\notin\pi_{n}(\mathcal{ML}^{n}_{k})$.
Let us assume that $\pi_{n}((k+1:1;x_{k-1}))\in\pi_{n}(\mathcal{ML}^{n}_{k})$.
By the definition of $\pi_{n}(\mathcal{ML}^{n}_{k})$, there exist $\vec{f}\in\
\mathcal{ML}^{n}_{k}$ such that $\pi_{n}(\vec{f})=\pi_{n}((k+1:1;x_{k-1}))$.
Hence, there exist $f\in R_{[k-1]}$ and $u\in UV(R_{[k-1]})$ such that
$\vec{f}=(k:u;f)$ by the definition of the monoid $\mathcal{ML}^{n}_{k}$. But,
then by Lemma 3.12, $(k:u;f)\operatorname{{}\triangleq{}}(k+1:1;x_{k-1})$ on
$R^{n}$, which implies that
$x_{k+1}\operatorname{{}\triangleq{}}x_{k+1}+x_{k-1}$ on $R$ by Lemma 2.13.
Thus, $x_{k-1}\operatorname{{}\triangleq{}}0$ on $R$, that is the polynomial
$x_{k-1}$ maps $R^{n}$ to the zero, which is impossible since $R$ contains
$1$. Therefore, $\pi_{n}((k+1:1;x_{k-1}))\notin\pi_{n}(\mathcal{ML}^{n}_{k})$
and $\pi_{n}(\mathcal{ML}^{n}_{k})$ is not normal subgroup of
$\pi_{n}(\mathcal{MT}_{n})$. ∎

Similar to Theorem 4.10, we view the group $\pi_{n}(\mathcal{MT}_{n})$ as
iterated semi-direct products of groups. Also, we obtain a counting formula
for its order.

######  Theorem 4.14.

Let $n>1$ and let $R$ be a finite commutative ring. Then

  1. (1)

$\pi_{n}(\mathcal{MT}_{n})\cong\pi_{n}(\mathcal{ML}^{n}_{n})\rtimes\cdots\rtimes\pi_{2}(\mathcal{ML}^{2}_{2})\rtimes\mathcal{P}(R)$;

  2. (2)

$\pi_{n}(\mathcal{MT}_{n})\cong(\mathcal{F}(R^{n-1})\rtimes\mathcal{F}(R^{n-1})^{\times})\rtimes\cdots\rtimes(\mathcal{F}(R)\rtimes\mathcal{F}(R)^{\times})\rtimes\mathcal{P}(R)$;

  3. (3)

$|\pi_{n}(\mathcal{MT}_{n})|=|\mathcal{P}(R)|\times\prod_{i=1}^{n-1}|\mathcal{F}(R^{i})||\mathcal{F}(R^{i})^{\times}|.$

###### Proof.

(3) follows from (1). (2) and follows from (1) and Proposition 4.13-(2). By
Proposition 4.13-(3),
$\pi_{n}(\mathcal{MT}_{n})\cong\pi_{n}(\mathcal{ML}^{n}_{n})\rtimes\pi_{n}(\mathcal{MT}_{n-1})$.
Then one can use this relation and induction to prove (1). We note here that
$\mathcal{MT}_{1}=\mathcal{MP}(R)$ and
$\pi_{1}(\mathcal{MP}(R))=\mathcal{P}(R)$. ∎

The second decomposition of the previous theorem will allow obtaining more
information about the group $\pi_{n}(\mathcal{MT}_{n})$. However, without loss
of generality, we restrict ourselves to finite local rings. We will need some
facts about polynomial functions over finite local rings. In the proof of the
following proposition, we use the fact that the order of a finite commutative
is a power of $q$, where $q$ is the number of elements of its residue field
$\mathbb{F}_{q}$.

######  Proposition 4.15.

Let $R$ be a finite local ring and $k\geq 1$. Then the group
$(\mathcal{F}(R^{k}),``+")$ is a $p-$group.

###### Proof.

Let $F\in\mathcal{F}(R^{k})$, then there exists a polynomial $f_{F}\in
R_{[k]}$ of minimal degree representing $F$. Since $\mathcal{F}(R^{k})$ is
finite, there exists a positive number $n$ such that $\deg f_{F}\leq n$ for
every $F\in\mathcal{F}(R^{k})$. Let

$A=\\{g\in R_{[k]}\colon\deg g\leq n\\}.$

Then $A$ is a group with respect to addition. Further,
$|A|=|R|^{1+\sum_{i=1}^{k}l_{i}}=q^{({1+\sum_{i=1}^{k}l_{i}})t}$, where
$l_{i}$ is the number of monomials of degree $i$. Hence, $A$ is a $p-$ group
since $q$ is a power of a prime $p$ (say $q=p^{r}$ for some positive integer).
Now, define a map $\psi\colon A\longrightarrow\mathcal{F}(R^{k})$ as a
follows: if $F$ is the function induced by $f$, we let $\psi(f)=F$. Clearly,
is a homomorphism. It is also a surjective since $f_{F}\in A$ for every
$F\in\mathcal{F}(R^{k})$. Thus, by the First Isomorphism Theorem of groups,

$\raise 2.0pt\hbox{$A$}\big{/}\lower
2.0pt\hbox{$\mathcal{F}(R^{k})$}\;\cong\;\ker\psi.\text{ and
}|\mathcal{F}(R^{k})|=\frac{|A|}{|\ker\psi|}.$

But, $|\ker\psi|=p^{m}$ for some non-negative integer $m$ (being a subgroup of
a $p-$ group). Therefore, $\mathcal{F}(R^{k})$ is a $p-$ group. ∎

######  Theorem 4.16.

Let $R$ be a finite local ring and $k\geq 1$. Then

$\frac{|\mathcal{F}(R^{k})^{\times}|}{|\mathcal{F}(R^{k})|}=\frac{(q-1)^{q^{k}}}{q^{q^{k}}}.$

###### Proof.

Set $A=R/M\cong\mathbb{F}_{q}$. Then, since every function
$F\colon\mathbb{F}_{q}^{k}\longrightarrow\mathbb{F}_{q}$ is a polynomial
function i.e represented by a polynomial in $k$ variables by means of Lagrange
Polynomial in $k$ variables, we have $|\mathcal{F}(A^{k})|=q^{{q^{k}}}$. Also,
since $|A^{\times}|=(q-1)$ we have
$|\mathcal{F}(R^{k})^{\times}|=(q-1)^{q^{k}}$. Since polynomial functions
preserve congruences modulo $M$, there exists a natural ring epimorphism
$\lambda\colon\mathcal{F}(R^{k})\longrightarrow\mathcal{F}(A^{k})$. Thus,

$|\mathcal{F}(R^{k})|=|\ker\lambda||\mathcal{F}(A^{k})|=q^{q^{k}}|\ker\lambda|.$

Now, an $a\in R$ is a unit in $R$ if and only if $a+M$ is a unit in $A$, and
so, $F\in\mathcal{F}(R^{k})^{\times}$ if and only if
$\lambda(F)\in\mathcal{F}(A^{k})^{\times}$. This yields,
$\mathcal{F}(R^{k})^{\times}=\lambda^{-1}(\mathcal{F}(A^{k})^{\times})$,
whence

$|\mathcal{F}(R^{k})^{\times}|=|\ker\lambda||\mathcal{F}(A^{k})^{\times}|=(q-1)^{q^{k}}|\ker\lambda|.$

Therefore,

$\frac{|\mathcal{F}(R^{k})^{\times}|}{|\mathcal{F}(R^{k})|}=\frac{(q-1)^{q^{k}}|\ker\lambda|}{q^{q^{k}}|\ker\lambda|}=\frac{(q-1)^{q^{k}}}{q^{q^{k}}}.$

∎

######  Corollary 4.17.

Let $R$ be a finite local ring that is not a field with residue field
$\mathbb{F}_{2}$. Then $\mathcal{F}(R^{k})^{\times}$ is a $2$-group.

###### Proof.

By Theorem 4.16,

$|\mathcal{F}(R^{k})^{\times}|=\frac{(2-1)^{2^{k}}}{2^{2^{k}}}|\mathcal{F}(R^{k})|=\frac{|\mathcal{F}(R^{k})|}{2^{2^{k}}}.$

Since $R\neq\mathbb{F}_{2}$, the proof of Theorem 4.16 shows that
$|\mathcal{F}(R^{k})|>|\mathcal{F}(\mathbb{F}_{2}^{k})|={2^{2^{k}}}$. But,
then the result follows by Proposition 4.15.

∎

######  Remark 4.18.

Similar to Theorem 4.16, but in one variable, Jiang [8] has obtained the
following relation between the number of polynomial functions and the number
of polynomial permutations on a finite local ring $R$ which is not a field:

$\frac{|\mathcal{F}(R)|}{|\mathcal{P}(R)|}=\frac{q!(q-1)^{q}}{q^{2q}}.$ (10)

######  Lemma 4.19 ([6] Theorem 4.1).

Let $R$ be a finite local commutative ring with a residue field of $q$
elements $\mathbb{F}_{q}$. Let $\mathcal{P}(R)$ be the group of polynomial
permutations. Then

  1. (1)

$\mathcal{P}(R)$ is solvable if and only if $q\leq 4$;

  2. (2)

$\mathcal{P}(R)$ is nilpotent if and only if $q=2$;

  3. (3)

$\mathcal{P}(R)$ is abelian if and only if $R=\mathbb{F}_{2}$.

With these preparations, we are ready to present our main theorem of this
subsection.

######  Theorem 4.20.

Let $n\geq 1$ and let $R$ be a finite local commutative ring with a residue
field of $q$ elements $\mathbb{F}_{q}$. Let $\pi_{n}(\mathcal{MT}_{n})$ be the
group of permutations on $R^{n}$ induced by the monoid $\mathcal{MT}_{n}$.
Then

  1. (1)

$\pi_{n}(\mathcal{MT}_{n})$ is solvable if and only if $q\leq 4$;

  2. (2)

$\pi_{n}(\mathcal{MT}_{n})$ is nilpotent if and only if $q=2$;

  3. (3)

$\pi_{n}(\mathcal{MT}_{n})$ is abelian if and only if $n=1$ and
$R=\mathbb{F}_{2}$

###### Proof.

In the light of Lemma 4.19 we may assume that $n>1$ since
$\pi_{1}(\mathcal{MT}_{1})=\mathcal{P}(R)$.

By Theorem 4.14,

$\pi_{n}(\mathcal{MT}_{n})\cong(\mathcal{F}(R^{n-1})\rtimes\mathcal{F}(R^{n-1})^{\times})\rtimes\cdots\rtimes(\mathcal{F}(R)\rtimes\mathcal{F}(R)^{\times})\rtimes\mathcal{P}(R).$

Now, $\mathcal{F}(R^{i})$ and $\mathcal{F}(R^{i})^{\times}$ are solvable
groups (being abelian groups). Also, extensions of solvable groups are
solvable. Therefore $\pi_{n}(\mathcal{MT}_{n})$ is solvable if and only if
$\mathcal{P}(R)$ is solvable if and only $q\leq 4$ by Lemma 4.19. This proves
(1).

(2) Since every subgroup of a nilpotent group is nilpotent, we need only to
consider the case $q=2$ by Lemma 4.19. Assume that $q=2$ and without loss of
generality we consider the case $R\neq\mathbb{F}_{q}$. We claim that
$\pi_{n}(\mathcal{MT}_{n})$ is a $2$-group, hence it is nilpotent since $p$\-
groups are nilpotents. To show the claim, we notice that by means of the
decomposition of $\pi_{n}(\mathcal{MT}_{n})$, $\pi_{n}(\mathcal{MT}_{n})$ is a
$2$-group if and only if $\mathcal{P}(R)$, $\mathcal{F}(R^{i})$ and
$\mathcal{F}(R^{i})^{\times}$ are 2-groups for $i=1,\ldots,n-1$. Proposition
4.15 shows that $\mathcal{F}(R^{i})$ is a $2-group$, while
$\mathcal{F}(R^{i})^{\times}$ is a $2-group$ by Corollary 4.17 for
$i=1,\ldots,k$. Finally, using Equation (10), we see that $\mathcal{P}(R)$ is
a $2$-group.

To prove (3), we need only to consider the case $R=\mathbb{F}_{2}$ and $n=2$
since $\pi_{2}(\mathcal{MT}_{2})$ is embedded in $\pi_{n}(\mathcal{MT}_{n})$
for every $n>2$ by Remark 3.19. But, then routine calculations show that
$\pi_{2}((x_{1},x_{1}+x_{2}))\circ\pi_{2}((x_{1}+1,x_{1}+x_{2}))\neq\pi_{2}((x_{1}+1,x_{1}+x_{2}))\circ\pi_{2}((x_{1},x_{1}+x_{2}))$.
∎

##  5\. A connection to the group of polynomial permutations of the ring dual
numbers of several variables

For be a comitative ring $R$ (not necessary finite) with unity, the ring of
dual numbers of $n$ variables over $R$ is the quotient ring
$R[y_{1},\ldots,y_{n}]/I$, where $I$ is the ideal generated by the set
$\\{y_{i}y_{j}\mid i,j\in\\{1,\ldots,n\\}\\}$, alternatively, the free
$R$-algebra ${R}[\operatorname{\alpha}_{1},\ldots,\operatorname{\alpha}_{n}]$
with basis
$\\{1,\operatorname{\alpha}_{1},\ldots,\operatorname{\alpha}_{n}\\}$ and
$\operatorname{\alpha}_{i}\operatorname{\alpha}_{j}=0$ for $i,j=1,\ldots,n$.
We write $R_{n}$ instead of
${R}[\operatorname{\alpha}_{1},\ldots,\operatorname{\alpha}_{n}]$ to denote
this ring. In fact,

$R_{n}=\\{r_{0}+\sum\limits_{i=1}^{n}r_{i}\operatorname{\alpha}_{i}\mid
r_{0},r_{i}\in R,\text{ with
}\operatorname{\alpha}_{i}\operatorname{\alpha}_{j}=0\text{ for }1\leq i,j\leq
n\\},$

and it follows that every polynomial $g\in R_{n}[x]$ has a unique
representation $g=g_{0}+\sum\limits_{i=1}^{n}g_{i}\operatorname{\alpha}_{i}$,
where $g_{0},g_{1},\ldots,g_{k}\in R[x]$. Also, it is evident that
$a_{0}+\sum_{i=1}^{n}a_{i}\rightarrow(a_{0},\ldots,a_{n})$ establishes a one
to one correspondence between $R_{n}$ and $R^{n+1}$. This correspondence
motivates us to relate some groups of permutations of $R_{n}$ and $R$ at least
in the finite case.

Throughout this section, since we deal with two different rings, we use the
notation $\mathcal{MT}_{n+1}(R)$ for the triangular mononid over the ring $R$
to not confuse the coefficients ring.

In this section, for a commutative ring $R$, we embed the monoid
$\mathcal{MP}(R_{n})$ of permutation polynomials on $R_{n}$ in the triangular
monoid $\mathcal{MT}_{n+1}(R)$. Moreover, this embedding leads to the
embedding of the group of polynomial permutations $\mathcal{P}(R_{n})$ in the
group of $\pi_{n+1}(\mathcal{MT}_{n+1}(R))$, whenever $R$ is finite.

######  Lemma 5.1.

Let $k>1$, $R$ be a commutative ring and
$g=g_{0}+\sum\limits_{i=1}^{k}g_{i}\operatorname{\alpha}_{i}$, where
$g_{0},\ldots,g_{n}\in R[x]$.

  1. (1)

If $f\in R[x]$, then

$f(g)=f(g_{0})+\sum\limits_{i=1}^{k}g_{i}f^{\prime}(g_{0})\operatorname{\alpha}_{i}.$

  2. (2)

If $f=f_{0}+\sum\limits_{i=1}^{k}f_{i}\operatorname{\alpha}_{i}$, where
$f_{0},\ldots,f_{k}\in R[x]$, then

$f(g)=f_{0}(g_{0})+\sum\limits_{i=1}^{k}(g_{i}f_{0}^{\prime}(g_{0})+f_{i}(g_{0}))\operatorname{\alpha}_{i}.$

###### Proof.

(2) follows from (1). To show (1), write $f(x)=\sum\limits_{j=1}^{m}x^{j}$.
Then expanding $f(g)$ with the fact that
$\operatorname{\alpha}_{i}\operatorname{\alpha}_{j}=0$ for all $i,j$ yield the
result. ∎

We need the following facts from [1].

######  Lemma 5.2.

[1, Theorem 4.1] Let $R$ be a finite ring. Let
$f=f_{0}+\sum\limits_{i=1}^{k}f_{i}\operatorname{\alpha}_{i}$, where
$f_{0},\ldots,f_{k}\in R[x]$. Then $f$ is a permutation polynomial on $R_{k}$
if and only if the following conditions hold:

  1. (1)

$f_{0}$ is a permutation polynomial on $R$;

  2. (2)

$f_{0}$ is a unit-valued polynomial on $R$.

We conclude this section with the following embedding result.

######  Lemma 5.3.

[1, Corollary 3.6]

Let $f=f_{0}+\sum\limits_{i=1}^{k}f_{i}\operatorname{\alpha}_{i}$ and
$g=g_{0}+\sum\limits_{i=1}^{k}g_{i}\operatorname{\alpha}_{i}$, where
$f_{0},\ldots,f_{k},g_{0},\ldots,g_{k}\in R[x]$.

Then $f$ and $g$ induce the same function on $R_{k}$ if and only if the
following conditions hold:

  1. (1)

$f_{i}\operatorname{{}\triangleq{}}g_{i}$ on $R$ for $i=0,\dots,k$;

  2. (2)

$f_{0}^{\prime}\operatorname{{}\triangleq{}}g_{0}^{\prime}$ on $R$.

######  Proposition 5.4.

Let $n>1$ and let $R$ be a commutative ring. Then

  1. (1)

$\mathcal{MP}(R_{n})\hookrightarrow\mathcal{MT}_{n+1}(R)$; and whenever $R$ is
finite

  2. (2)

$\mathcal{P}(R_{n})\hookrightarrow\pi_{n+1}(\mathcal{MT}_{n+1}(R))$.

###### Proof.

(1) Let
$f=f_{0}+\sum\limits_{i=1}^{n}f_{i}\operatorname{\alpha}_{i}\in\mathcal{MP}(R_{n})$,
where $f_{0},\ldots,f_{n}\in R[x]$. Then, $f_{0}$ is a permutation polynomial
and $f_{0}^{\prime}$ is a unit-valued polynomial by Lemma 5.2. Define
$\psi\colon\mathcal{MP}(R_{n})\longrightarrow\mathcal{MT}_{n+1}$ by

$\psi(f)=(f_{0}(x_{1}),f_{1}(x_{1})+x_{2}f_{0}^{\prime}(x_{1}),\ldots,f_{n}(x_{1})+x_{n+1}f_{0}^{\prime}(x_{1})).$

Then, since $f_{0}$ is a permutation polynomial and $f_{0}^{\prime}$ is a
unit-valued polynomial, $\psi(f)$ has the form (3). Thus,
$\psi(f)\in\mathcal{MT}_{n+1}$ and $\psi$ is defined. Now, let
$g=g_{0}+\sum\limits_{i=1}^{k}g_{i}\operatorname{\alpha}_{i}\in\mathcal{MP}(R_{n})$,
where $g_{0},\ldots,g_{k}\in R[x]$. By Lemma 5.1, $f\circ
g(x)=f_{0}(g_{0}(x))+\sum\limits_{i=1}^{k}(g_{i}(x)f_{0}^{\prime}(g_{0}(x))+f_{i}(g_{0}(x)))\operatorname{\alpha}_{i}$.

Thus, by the definition of $\psi$ and the definition of composition,

$\displaystyle\psi(f\circ g)$
$\displaystyle=(f_{0}(g_{0}(x_{1})),(g_{1}(x_{1})f_{0}^{\prime}(g_{1}(x_{1}))+f_{1}(g_{0}(x_{1})))+x_{2}g_{0}^{\prime}(x_{1})f_{0}^{\prime}(g_{0}(x_{1})),$
$\displaystyle\ldots,$
$\displaystyle(g_{n}(x_{1})f_{0}^{\prime}(g_{0}(x_{1}))+f_{n}(g_{0}(x_{1})))+x_{n+1}g_{0}^{\prime}(x_{1})f_{0}^{\prime}(g_{0}(x_{1})))=$
$\displaystyle=\psi(f)\circ\psi(g).$

Thus, $\psi$ is a homomorphism. Evidently, $\psi$ is one-to-one. (2) Define
$\phi\colon\mathcal{P}(R_{n})\longrightarrow\pi_{n+1}(\mathcal{MT}_{n+1})$ by
$\phi(F)=\pi_{n+1}(\psi(f))$, where
$f=f_{0}+\sum\limits_{i=1}^{n}f_{i}\operatorname{\alpha}_{i}\in\mathcal{MP}(R_{n})$
is a permutation polynomial inducing $F$ on $R_{n}$ , with
$f_{0},\ldots,f_{n}\in R[x]$. Then, by Lemma 5.3 and Lemma 3.12, $\phi$ is
well defined. Also, it is a homomorphism being a composition of homomorphisms.
To show $\phi$ is injective, consider $G\in\mathcal{P}(R_{n})$ such that
$\phi(F)=\phi(G)$. So, there exists
$g=g_{0}+\sum\limits_{i=1}^{k}g_{i}\operatorname{\alpha}_{i}\in\mathcal{MP}(R_{n})$,
where $g_{0},\ldots,g_{k}\in R[x]$ such that $G$ is induced by $g$. Then
$\phi(F)=\phi(G)$ implies that $\pi_{n+1}(\psi(g))\pi_{n+1}\psi(f)$, that is

$\displaystyle\pi_{n+1}((f_{0}(x_{1}),f_{1}(x_{1})+x_{2}f_{0}^{\prime}(x_{1})$
$\displaystyle,\ldots,f_{n}(x_{1})+x_{n+1}f_{0}^{\prime}(x_{1})))=$
$\displaystyle=\pi_{n+1}((g_{0}(x_{1}),g_{1}(x_{1})+x_{2}g_{0}^{\prime}(x_{1}),\ldots,g_{n}(x_{1})+x_{n+1}g_{0}^{\prime}(x_{1}))).$

Thus, by Lemma 3.12,

$\begin{matrix}f_{0}(x_{1})\operatorname{{}\triangleq{}}g_{0}(x_{1})\text{ on
}R\\\
f_{i}(x_{1})+x_{i+1}f_{0}^{\prime}(x_{1})\operatorname{{}\triangleq{}}g_{i}(x_{1})+x_{i+1}g_{0}^{\prime}(x_{1})\text{
on }R\text{ for }i=1,\dots,n.\end{matrix}$

But, by Lemma 5.3, this means $f$ ang $g$ induce the same function on $R_{n}$,
that is $F=G$. This finishes the proof. ∎

##  6\. The Tame Monoid and related questions

Recall from Remark 4.11 that the classical triangular group $KTR_{n}$ consists
of all invertible vector polynomials of the form
$(a_{1}x_{1}+a_{1},f_{1}(x_{1})+a_{2}x_{2},\ldots,f_{n-1}(x_{1},\ldots,x_{n-1})+ax_{n})$,
where $a_{0}\in R$ and $a_{1},\ldots,a_{n}\in R^{\times}$. It is well-known
that the group $KTR_{n}$ and the affine Group $Aff_{n}$ consisting of all
invertible linear vector-polynomial generate the so-called tame group
$Tm_{n}$. An element of $Tm_{n}$ is called tame, and the tame question asks
(when $R$ is a field and $n\geq 2$) whether the group $GA_{n}$ of all
invertible vector polynomials (all R -automorphism of ${R_{[n]}})$ consists
only of tame elements, that is, whether $GA_{n}=Tm_{n}$ or not ? When $n=2$,
the equality always holds; this was proved by Jung [9] for fields of
characteristic zero, and was shown for positive characteristic by Van der Kulk
[24]. For the case, $n=3$ and $R$ is a field of characteristic zero Shestakov
and Umirbaiev [19] showed that $GA_{3}\neq Tm_{3}$ by showing that the Nagata
automorphism is not in $Tm_{3}$. While the case $n=3$ for positive
characteristic fields and the general case $n>3$ for any field are still open.
Unfortunately, the embedding of the Nagata automorphism (by adding the
variables $x_{4},\ldots,x_{n}$) in the group $GA_{n}$ is shown by Smith [21]to
be an element of the tame group $Tm_{n}$ for $n\geq 4$. We notice here that
Maubach [12] showed that $\pi_{n}(Tm_{n})$ is just the group of all even
permutations of $F_{2^{r}}$ with $r>1$, and noticed that to see $GA_{n}\neq
Tm_{n}$ is equivalent to find an element $\vec{f}\in GA_{n}$ with
$\pi_{n}(\vec{f})$ is an odd permutation.

The case $n=1$ is much easier as in this case
$Aut_{R}(R[x])=GA_{1}=Tm_{1}=\\{ax+b:a,b\in R,a\neq 0\\}$. However, when $R$
is not reduced, we have $Tm_{1}\neq GA_{1}$ as for example $x+rx^{2}\in
GA_{1}\setminus Tm_{1}$ for every non-zero nilpotent $r$. Then may be the
tameness problem is more trivial on non- reduced rings and always given an $n$
there exists a non-tame invertible element. But, we could not find a reference
discussing the tameness problem on non-reduced rings. Nevertheless, this does
not dampen our enthusiasm for suggesting a more general tame structure rather
than the tame group. This generalized structure allows us to modify the
tameness problem and to ask some related questions.

In the following, we define the tame monoid and give our general definition
for tame elements.

######  Definition 6.1.

Let $R$ be a commutative ring with unity. We call the monoid generated by the
triangular monoid $\mathcal{MT}_{n}$ and the affine group $Aff_{n}$, the tame
monoid which we denote by $\mathcal{TAM}_{n}$. We call every element of
$\mathcal{TAM}_{n}$ tame.

Let $\mathcal{MPP}_{n}$ denote the monoid of vector-permutation polynomials.
Then, by construction, $\mathcal{MPP}_{n}$ contains $\mathcal{TAM}_{n}$. Now,
we have the following general form of the tameness problem.

######  Question 6.2.

Let $n>1$, and let $R$ be a commutative ring with unity. Is it true that
$\mathcal{MPP}_{n}=\mathcal{TAM}_{n}$?

######  Remark 6.3.

  1. (1)

If $n=1$, the we have always
$\mathcal{MPP}_{1}=\mathcal{TAM}_{1}=\mathcal{MP}(R)$, as in this case
$\mathcal{MPP}_{1}=\mathcal{MT}_{1}$ and $Aff_{1}\subseteq\mathcal{MT}_{1}$.

  2. (2)

If $R$ is an algebraically closed field, we have, by Remark 4.11,
$\mathcal{MT}_{n}=KTR_{n}$ and $\mathcal{MPP}_{n}=GA_{n}$. But, then Question
6.2 is reduced to the classical problem of the tame group.

We close our paper with two related questions that can be more sensible in the
case of finite fields or non-reduced rings.

######  Question 6.4.

Let $\vec{f}\in\mathcal{MPP}_{n}$. Then does
$\pi_{n}(\vec{f})\in\pi_{n}(\mathcal{TAM}_{n})$ implies that
$\vec{f}\in\mathcal{TAM}_{n}$?

Another result of Maubach [12] over the finite field $\mathbb{F}_{q}$ with
$q=2$ or $q$ is odd, shows that $\pi_{n}(Tm_{n})$ is the symmetric group on
the elements of $\mathbb{F}_{q}$. At least in this case we know the answer to
the following question to be true.

######  Question 6.5.

Is every vector-polynomial permutation only tamely represented? In terms of
symbols: Let $\vec{F}\in\pi_{n}(\mathcal{MPP}_{n})$. Then does there exist
$\vec{f}\in\mathcal{TAM}_{n}$ such that $\pi_{n}(\vec{f})=\vec{F}$?

Acknowledgment. The author is supported by the Austrian Science Fund (FWF):P
35788-N. The author would like to thank Amran A. Al-Aghbari for proof reading.

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