# Non-commutative factorizations and finite-dimensional representations of
free algebras

P. N. Ánh  Hun-Ren Rényi Institute of Mathematics, 1364 Budapest, Pf. 127
Hungary<EMAIL_ADDRESS>and  F. Mantese  Department of Computer Science,
University of Verona, Strada le Grazie 15, 37134 Italy
<EMAIL_ADDRESS>
###### Abstract.

A very first step to develop non-commutative algebraic geometry is the
arithmetic of polynomials in non-commuting variables over a commutative field,
that is, the study of elements in free associative algebras. This
investigation is presented as a natural extension of the classical theory in
one variable by using Leavitt algebras, which are localizations of free
algebras with respect to the Gabriel topology defined by an ideal of
codimension 1. In particular to any polynomial in $n$ non-commuting variables
with non-zero constant term we associate a finite-dimensional module over the
free algebra of rank $n$, which turns out to be simple if and only if the
polynomial is irreducible. This approach leads to new insights in the study of
the factorization of polynomials into irreducible ones and other related
topics, such as an algorithm to divide polynomials or to compute the greatest
common divisor between them, or a description of similar polynomials. The case
of polynomials with zero constant term reduces to an open question whether the
intersection of nonunital subalgebras of codimension 1 in a free associative
algebra is trivial, that is, 0.

###### Key words and phrases:

free algebras, irreducibile polynomials, atomic factorizations, Leavitt
algebras, finitely presented modules, flat rings of quotients

###### 2020 Mathematics Subject Classification:

Primary 16S88, 16S90; secondary 16G20, 16P50.

The first author was partially supported by National Research, Development and
Innovation Office NKHIH K138828 and K132951, by both Vietnam Institute for
Advanced Study in Mathematics (VIASM) and Vietnamese Institute of Mathematics.
The second author was partially supported by the National Recovery and
Resilience Plan (NRRP), Mission 4 Component 2 Investment 1.1 - Call PRIN 2022
No. 104 of February 2, 2022 of Italian Ministry of University and Research;
Project 2022S97PMY (subject area: PE - Physical Sciences and Engineering)
”Structures for Quivers, Algebras and Representations (SQUARE)”. She is
moreover member of INDAM - GNSAGA

##  1\. Introduction

The classical theory of polynomials in one variable over a commutative field
$K$ assigns to every irreducible polynomial its residue field, i.e., its
finite-dimensional field extension of remainders by Euclidean division, and
shows that every polynomial is uniquely a product of irreducible ones up to
permutation and units. The division algorithm to compute the greatest common
divisor of polynomials leads to a complete picture of the lattice of the
ideals of $K[x]$. Moreover, the irreducible representations of $K[x]$ are
explicitly constructed from the irreducible polynomials in $K[x]$.

Analogous results for polynomials in several commutative variables are no
longer true. For instance, the factor algebra of an irreducible polynomial in
$K[x_{1},\cdots,x_{n}]$ for $n\geq 2$ is not a field extension of $K$ and is
never of finite dimension.

For $n\geq 2$ polynomials in non-commuting variables are elements of the free
associative algebra $K\langle x_{1},\cdots,x_{n}\rangle$. Factorization
problems in $K\langle x_{1},\cdots,x_{n}\rangle$ have been deeply studied by
Cohn in [7], [9], [11], [10], where he showed that free algebras have a weak
division algorithm, they are weak Bézout free ideal rings, and they are unique
factorization rings, i.e any element is the product of irreducible ones in a
unique way up to units and similarity. Roughly speaking, weaker versions of
the relevant properties of $K[x]$ to be a Euclidean domain, a principal ideal
domain and a unique factorization domain hold for the free algebra $K\langle
x_{1},\cdots,x_{n}\rangle$. Moreover, by considering the category of strictly
cyclic modules or more generally of (specific) torsion modules over free ideal
rings, Cohn succeeded to show a categorical simplicity of factor modules of
the free algebra by irreducible polynomials. However, Cohn’s approach leaves
open the problem to parametrize the irreducible representations of $K\langle
x_{1},\cdots,x_{n}\rangle$ by irreducible polynomials, as in the classical
case of one variable.

The aim of our work is to present a theory of polynomials which coincides with
the classical one in case of one variable. The starting point is the Cohn’s
unique normal form [8]

$\lambda=k+\sum\limits_{i=1}^{n}x_{i}\lambda_{x_{i}}\quad(k\in
K;\,\lambda_{x_{i}}\in K\langle x_{1},\cdots,x_{n}\rangle)$

of any polynomial $\lambda\in K\langle x_{1},\cdots,x_{n}\rangle$, where
$\lambda_{x_{i}}=\lambda_{i}$ is called the _cofactor_ with respect to the
$i$-th axis (i.e., variable) $x_{i}$. Hence taking cofactors behaves like
taking generalized partial differentiation or truncated partial inverse along
axes. One can iterate this process to obtain further cofactors of $\lambda$
until getting nonzero constants. For any polynomial $\lambda$ in $K\langle
x_{1},\cdots,x_{n}\rangle$ with nonzero constant term we define an easily
handled finite-dimensional vector space $V_{\lambda}$, spanned by all nonzero
iterated cofactors of $\lambda$, which admits a natural left module structure
over the free algebra. This new invariant becomes a good tool in the study of
the polynomial $\lambda$ itself, independent on $K\langle
x_{1},\cdots,x_{n}\rangle$. For instance, the module $V_{\lambda}$ helps to
decide the irreducibility of $\lambda$, to compute its similar polynomials and
its irreducible factors or indecomposable factors, i.e., primary
decompositions, and to compute other invariants such as endomorphism rings,
submodule lattices, etc. In particular, $V_{\lambda}$ is simple if and only
$\lambda$ is irreducible. More generally, if $\lambda=\pi_{1}\cdots\pi_{l}$ is
a factorization of $\lambda$ as a product of irreducible polynomials
$\pi_{1},\cdots,\pi_{l}$, then $l$ is the length of $V_{\lambda}$ and
$\\{V_{\pi_{1}},\cdots,V_{\pi_{l}}\\}$ is the set of composition factors of
$V_{\lambda}$. If $\gamma$ is another polynomial with nonzero constant term,
then $V_{\lambda}$ and $V_{\gamma}$ are isomorphic $K\langle
x_{1},\cdots,x_{n}\rangle$-modules if and only if the $K\langle
x_{1},\cdots,x_{n}\rangle$-modules $K\langle
x_{1},\cdots,x_{n}\rangle/K\langle x_{1},\cdots,x_{n}\rangle\lambda$ and
$K\langle x_{1},\cdots,x_{n}\rangle/K\langle x_{1},\cdots,x_{n}\rangle\gamma$
are such. Therefore $V_{\lambda}$ encodes the structure of $\lambda$.

The module structure of $V_{\lambda}$ is inspired by ideas coming from Leavitt
algebras, which are universal with respect to an isomorphism property between
finite rank free modules [13]. This is not surprising, since free algebras and
Leavitt algebras are tightly connected by the concept of localization (see
Section 3). This means that even if these two algebraic structures are deeply
different in view of their ring properties (for instance Leavitt algebras are
far from being domains), their module categories are closely related.
Moreover, any free algebra can be seen naturally in two ways as a subalgebra
of a Leavitt algebra. This fact reflects the decisive role of Leavitt algebras
in the investigation of both right and left factorizations of polynomials. By
symmetry it is enough to study right factorizations. Since Leavitt algebras
are one-sided Bézout rings, i.e. any finitely generated right (or left) ideal
is principal, moving to this larger setting it is possible to define the
_generalized greatest common divisor_ , shortly, the _greatest common divisor_
, for finitely many polynomials $\lambda_{i}$ in $K\langle
x_{1},\cdots,x_{n}\rangle$: it is the uniquely determined, up to scalars,
polynomial with nonzero constant term which is a generator of the left ideal
of the extended Leavitt algebra generated by the $\lambda_{i}$’s.

In this manner we succeed to have an algorithm computing the greatest common
divisor of polynomials and to divide polynomials in $K\langle
x_{1},\cdots,x_{n}\rangle$ analogous to the Euclidean algorithm for $K[x]$.
This will be a crucial tool for our theory. Another basic notion in our
treatment is that of similarity, which is naturally inspired by the similarity
of linear transformations. Namely, two elements are _similar_ if the
corresponding factors by the left (right) ideals generated by them are
isomorphic (see [10]). We will show that two elements $\lambda$ and $\gamma$
with nonzero constant term are similar if and only if their associated finite-
dimensional modules $V_{\lambda}$ and $V_{\gamma}$ are isomorphic. The article
is organized as follows. In Section 2 we state the main result, Theorem 2.3,
that every irreducible polynomial with constant term in $n$ non-commutative
variables defines a simple finite-dimensional module over the free algebra of
rank $n$, whence a finite-dimensional division algebra, opening several
exciting related questions. Consequently, every polynomial with nonzero
constant term factors uniquely into products of irreducible up to units and
similarity, in the sense that if
$\lambda=\delta_{1}\cdots\delta_{s}=\pi_{1}\cdots\pi_{t}$ where the
$\delta_{i}$’s and the $\pi_{j}$’s are irreducible, then $s=t$ and there
exists a permutation $\sigma$ such that $\delta_{i}$ is similar to
$\pi_{\sigma(i)}$, even if $\lambda$ is no longer equal to a permutated
product $\prod\limits_{i=1}^{s}\pi_{\sigma(i)}$. Since irreducibilty does not
change under automorphisms, polynomials with zero constant term may become
ones with nonzero constant after suitable substitutions. The feasibility of
this trick is an open question whether or not the intersection of nonunital
subalgebras of codimension 1 is 0.

Section 3 develops shortly the theory of Leavitt algebras necessary for
Theorem 2.3 and presents the interaction between Leavitt algebras and free
algebras in terms of their module categories. The importance of Leavitt
algebras is underlined by the fact that they contain the pair of free algebras
$K\langle x_{1},\cdots,x_{n}\rangle$ and $K\langle
x^{*}_{1},\cdots,x^{*}_{n}\rangle$ together with both the isomorphism and
anti-isomorphism induced canonically by the assignment $x_{i}\mapsto
x^{*}_{i}\,(i=1,\cdots,n)$. This parallel reflects precisely both the left and
the right factorizations of polynomials. Hence we deal mainly with the right
factorizations of polynomials.

Section 4 is devoted to prove Theorem 2.3 with additional results and
consequences on the arithmetic of polynomials in non-commutative variables.

Section 5 presents applications of Theorem 2.3 to companion structures of
polynomials with nonzero constant term, like their similarity classes, atomic
factorizations etc. Some important problems concerning irreducible polynomials
and finite-dimensional non-commutative division algebras are stated.

Section 6 contains further problems, discusses primary decompositions of a
polynomial as indecomposable decompositions of its module of remainders and
shows that Theorem 2.3 remains true when the investigated polynomial is
considered as an element of a free algebra of an arbitrary rank. Hence the
study of an individual polynomial via the modules of remainders is independent
of the free algebra which contains it.

##  2\. Atomic factorizations in free algebras

Throughout this work $K$ denotes an arbitrary commutative field, rings are
unital $K$-algebras, and modules are unitary whence $K$-spaces, too. When we
discuss to _nonunital subalgebras_ of a given $K$-algebra $A$, we mean
multiplicatively closed subspaces without identity. A ring $A$ is _projective-
free_ if projective modules are free. It is worth to note that being
projective-free is left-right symmetric. Two elements $a,b\in A$ are _similar_
(over $A$) if the left $A$-modules $A/Aa,A/Ab$ are isomorphic. The
_codimension_ of a subspace is the dimension of its direct complement. The
_colength_ of a submodule is the length of its factor module.

A non-zero element of a ring $A$ is called _irreducible_ , or _an atom_ in
Cohn’s terminology [7], if it cannot be written as a product of two nonzero
non-invertible elements. This definition is clearly preserved by
automorphisms. A factorization of an element as a product of irreducible ones
is called an _atomic factorization_. Notice that in the non commutative
setting irreducible and prime elements may be different. For example, the
equality
$\lambda=(x_{1}x_{2}x_{3}+x_{1}+x_{3})(1+x_{2}x_{1})=(1+x_{1}x_{2})(x_{3}x_{2}x_{1}+x_{1}+x_{3})$
analyzed in [16] shows that the irreducible element $1+x_{1}x_{2}$ divides
$\lambda$ but it divides neither $1+x_{2}x_{1}$ nor
$x_{1}x_{2}x_{3}+x_{1}+x_{3}$. Indeed the non-commutativity requires to take
into account both the left and the right division in the definition of
primeness. Moreover, although prime elements are trivially irreducible, their
existence is not necessarily guaranteed. Conversely, even if irreducible
elements are not necessary prime, their existence is ensured for instance by
the existence of a degree function, as in the case of polynomials in non-
commuting variables, where any non-invertible element admits at least one
atomic factorization. Thus in general it makes more sense to look for atomic
than prime factorizations. A free $K$-algebra $\Lambda=\Lambda_{n}$ of rank
$n$ determines uniquely $K$ via $K=U(\Lambda)\cup\\{0\\}$ and $n={\rm
gldim}\,(\Lambda/[\Lambda,\Lambda])$, where $U(A)$ is the unit group of the
ring $A$ and $[A,A]$ is the commutator ideal. The equality $n={\rm
gldim}\,(\Lambda/[\Lambda,\Lambda])$ is the famous syzygy theorem of Hilbert
computing the global dimension of a commutative polynomial algebra in $n$
variables. However, ${\rm gldim}\,\Lambda_{1}=1$ holds obviously. If a free
generating set $\\{x_{1},\cdots,x_{n}\\}$ is designated, i.e $\Lambda=K\langle
x_{1},\cdots,x_{n}\rangle$, one gets a normal form for the elements in
$\Lambda$. In other words, a free generating set gives a polynomial
description for the elements of $\Lambda$. The two-sided ideal
$I=\sum\limits_{i=1}^{n}x_{i}\Lambda=\sum\limits_{i=1}^{n}\Lambda x_{i}$ is a
nonunital free algebra on $\\{x_{1},\cdots,x_{n}\\}$ which is also a free
$\Lambda$-module of rank $n$ on each side. However, neither the set
$\\{x_{1},\cdots,x_{n}\\}$ nor $I$ can be recovered from $\Lambda$. In fact,
any free generating set describes an automorphism of $\Lambda$ up to a
permutation. As we will see in the sequel, for the purpose of our work the
question whether there is an element of $\Lambda$ whose normal form with
respect to any free generating set is always a polynomial with zero constant
term is crucial. Notice that this is equivalent to the question whether the
intersection of two-sided ideals of codimension 1 in $\Lambda$ is trivial,
that is, 0. Maximal two-sided ideals of codimension 1 in $\Lambda$ are exactly
nonunital subalgebras of codimension 1, as it is easy to see. Hence it is an
open, interesting question whether the intersection of nonunital subalgebras
of codimension 1 in $\Lambda$ is 0.

In the sequel, $\Lambda^{*}=\Lambda^{*}_{n}$ will denote the free algebra on
the generating set $\\{x^{*}_{i},\cdots,x^{*}_{n}\\}$, so that
$\Lambda=K\langle x_{1},\cdots,x_{n}\rangle$ and $\Lambda^{*}=K\langle
x^{*}_{1},\cdots,x^{*}_{n}\rangle$ with
$I^{*}=\sum\limits_{i=1}^{n}\Lambda^{*}x^{*}_{i}=\sum\limits_{i=1}^{n}x^{*}_{i}\Lambda^{*}$.
Both the isomorphism and anti-isomorphism from $\Lambda$ onto $\Lambda^{*}$
induced by $x_{i}\mapsto x^{*}_{i}\,\,(i=1,\cdots,n)$ are called _canonical_ ,
making the presentation lucidly. Therefore an arbitrary monomial
$x_{i_{1}}\cdots x_{i_{l}}$ goes to $x^{*}_{i_{1}}\cdots x^{*}_{i_{l}}$
$(x^{*}_{i_{l}}\cdots x^{*}_{i_{1}})$ under this canonical (anti)isomorphism.

The free monoid $\bf B$ on $\\{x_{1},\cdots,x_{n}\\}$ provides the most
natural $K$-basis of $\Lambda$. For $b=x_{i_{1}}\cdots x_{i_{m}}\in\bf B$ of
_length_ or _degree_ $|b|=m$, monomials $h_{b}(l)=x_{i_{1}}\cdots x_{i_{l}}$
and $t_{b}(l)=x_{i_{l+1}}\cdots x_{i_{m}}$ are called the _head_ of length $l$
or the _tail_ of _colength_ (_codegree_) $l$ of $b$, respectively. We define
$h_{b}(0)=1=t_{b}(|b|)$ and $h_{b}(|b|)=b=t_{b}(0)$. Thus $1\in\Lambda$ is an
empty monomial of length 0. The _transpose_ ${{}^{\tau}}b$ of $b$ is
$x_{i_{m}}\cdots x_{i_{1}}$. The _(total) degree_ $\deg\lambda$ of
$\lambda\in\Lambda$, denoted also by $|\lambda|$, is the greatest length of
its monomials. Hence $|0|=-\infty$ and $|k|=0$ for $0\neq k\in K$. One can
write each $\lambda\in\Lambda$ uniquely in the form

(1)
$\displaystyle\lambda=k+\sum\limits_{i=1}^{n}x_{i}\lambda_{x_{i}}\quad(k\in
K;\,\lambda_{x_{i}}\in\Lambda).$

The polynomial $\sum x_{i}\lambda_{x_{i}}=\underline{\lambda}$ is called the
_main part_ of $\lambda$ while, according to Cohn [8], $\lambda_{x_{i}}$ is
the _right cofactor_ of _colength_ 1 with respect to $x_{i}\,(i=1,\cdots,n)$.
Moreover, a monomial $b$ in the normal form (1) of $\lambda$ is called a
_maximal monomial of_ $\lambda$ if it is not a proper head of another monomial
of $\lambda$ in (1). The polynomial $\lambda$ is _with_ or _without constant_
according to $k\neq 0$ or $k=0$, respectively. An element $\lambda$ with $k=1$
is also called a _comonic polynomial_.

Right cofactors $(\lambda_{x_{i}})_{x_{j}}$ of $\lambda_{x_{i}}$, denoted as
$\lambda_{x_{i}x_{j}}$, are called _cofactors of colength_ 2 of $\lambda$, and
so on until getting nonzero constants. In other words, if $b=x_{i_{1}}\cdots
x_{i_{l}}$ is a monomial and if $b_{1},\cdots,b_{m}$ are the monomials of
$\lambda$ having the head $b$ with coefficient $k_{1},\cdots,k_{m}$,
respectively, then the right cofactor $\lambda_{b}$ of colength $l=|b|$
defined by $b$ is

(2)
$\displaystyle\lambda_{b}=\left(\lambda_{x_{i_{1}}}\right)_{x_{i_{2}}\cdots
x_{i_{l}}}=\cdots=\sum\limits_{i=1}^{m}k_{i}t_{b_{i}}(|b|).$

Cofactors $\lambda_{b}$ are defined only for heads $b$ of maximal monomials in
$\lambda$ and their constant term may be 0. Hence for convenience we make the
convention that cofactors of constants are 0 and put $\lambda_{b}=0$ if $b$ is
not a head of any maximal monomials of $\lambda$. Consequently, forming the
cofactor with respect to $x_{i}$ provides a $\star$-action of $\Lambda$ on
itself, that is, $x_{i}\star\lambda=\lambda_{i}$ for every
$\lambda\in\Lambda$, which behaves like a generalized differentiation on the
axis $x_{i}$ or a truncated partial inverse to the direction $x_{i}$. In fact,
this action makes $\Lambda$ a left $\Lambda$-module which is an essential
extension of its unique minimal trivial $\Lambda$-submodule $K$. Hence we have

######  Proposition 2.1.

Forming right cofactors with respect to variables $x_{i}\,(i=1,\cdots,n)$
makes $\Lambda$ a left module over itself, denoted as $\Lambda_{\partial}$,
which is an essential extension of its unique minimal trivial submodule $K$.

###### Proof.

Let $\lambda\in\Lambda$ be an arbitrary proper polynomial, i.e.,
$|\lambda|\geq 1$. Then for a maximal monomial $b$ of $\lambda$ which has a
nonzero coefficient $k_{b}\in K$, one has
${}^{\tau}b\star\lambda=\lambda_{b}=k_{b}\neq 0$. This shows that $K$ is the
smallest essential submodule of $\Lambda_{\partial}$, completing the proof. ∎

Notice that Proposition 2.1 could be a starting point in the investigation of
injective envelopes of finite-dimensional modules over free algebras.

For example, Proposition 2.1 shows that when $\Lambda=\Lambda_{1}=K[x]$,
forming cofactors defines a module structure on $K[x]$, namely, the module
$K[x]_{\partial}=\Lambda_{\partial}$ which is the injective hull of the simple
trivial module $K$ independently on the characteristic of $K$. By this reason
$\Lambda_{\partial}$ can be called _generalized directional Leavitt module_.

It turns out that given any polynomial $\lambda\in\Lambda$ with nonzero
constant term, one can modify the $\star$-action defined above to obtain a
powerful tool for the study of $\lambda$ itself. Namely, any polynomial
$\lambda=k_{\lambda}+\underline{\lambda}$ such that $0\neq k_{\lambda}\in K$,
makes $\Lambda$ a left module over itself, denoted by $\Lambda_{\lambda}$, via
linear transformations induced by putting

(3) $\displaystyle
x_{i}\star_{\lambda}1=-k^{-1}_{\lambda}\lambda_{x_{i}}\quad\&\quad
x_{i}\star_{\lambda}b=\begin{cases}b_{x_{i}}=t_{b}(1)&\text{if}\qquad
x_{i}=h_{b}(1),\\\ 0&\text{if}\qquad x_{i}\neq h_{b}(1)\end{cases}$

for $i=1,\cdots,n$ and $1\neq b\in\bf B$. Thus $x_{i}\star_{\lambda}1=0$ holds
if $x_{i}$ is not a head of any monomial of $\lambda$, or equivalently,
$\lambda_{x_{i}}$ is not defined, that is, $\lambda_{x_{i}}=0$ by our
convention. Since
$x_{i}\star_{\lambda}\lambda=x_{i}\star_{\lambda}k_{\lambda}+\lambda_{i}=0$
holds for all indices $i$, $K\lambda$ is isomorphic to the trivial
$\Lambda$-module. Furthermore, for any
$\gamma=k_{\gamma}+\underline{\gamma}\in\Lambda$ one has
$x_{i}\star_{\lambda}\gamma\lambda=x_{i}\star_{\gamma}(k_{\gamma}\lambda+\underline{\gamma}\lambda)=\gamma_{i}\lambda$
whence $\Lambda\lambda$ is a submodule of $\Lambda_{\lambda}$. In fact,
$\Lambda\lambda$ contains the smallest trivial submodule $K\lambda$ and it is
isomorphic to the generalized directional Leavitt module $\Lambda_{\partial}$
via the map $\gamma\lambda\mapsto\gamma$.

Notice that if $\lambda$ is comonic then for any $\gamma=k+\sum
x_{i}\gamma_{x_{i}}\in\Lambda$, where $k\in K$, we have
$x_{j}\star\gamma=-k\lambda_{x_{j}}+\gamma_{x_{j}}$.

We are now in position to define one of the most important objects in our
investigation.

######  Definition 2.1.

Let $\lambda\in\Lambda$ be a polynomial with nonzero constant term. Then
$V_{\lambda}$ is the $K$-subspace of $\Lambda$ generated by all cofactors of
$\lambda$.

In particular, $0\neq\lambda\in K$ implies $V_{\lambda}=0$ as well as
$|\lambda|=1$ implies $V_{\lambda}=K$. If $|\lambda|\geq 1$, then
$V_{\lambda}$ has finite dimension. For example, if
$\lambda=1+x^{2}_{1}+x_{1}x_{2}+x^{2}_{2}x_{3}+x_{3}x_{2}$, then
$1,x_{1}+x_{2},x_{2}x_{3},x_{2},x_{3}$ are cofactors of $\lambda$ and so
$V_{\lambda}$ has dimension 5 with a basis
$\\{1,x_{1},x_{2},x_{2}x_{3},x_{3}\\}$. For
$\lambda=1+x_{1}+x_{2}+x^{2}_{1}+x_{1}x_{2}+x_{2}x_{1}+x^{2}_{2}$, then
$\\{1,1+x_{1}+x_{2}\\}$ is the set of all cofactors of $\lambda$ and
$V_{\lambda}$ has dimension 2 with a basis $\\{1,x_{1}+x_{2}\\}$. It is a
challenging and exciting problem to determine or to estimate the dimension of
$V_{\lambda}$.

######  Proposition 2.2.

Let $\lambda\in\Lambda$ be a polynomial with nonzero constant term. Then
$V_{\lambda}$ is a left $\Lambda$-submodule of $\Lambda_{\lambda}$.

###### Proof.

It follows directly from the definition of the $\star_{\lambda}$-action, since
if $\lambda_{b}$ is a cofactor of $\lambda$, then
$x_{i}\star_{\lambda}\lambda_{b}$ is a linear combination of cofactors of
$\lambda$. ∎

For instance, consider
$\lambda=1+x^{2}_{1}+x_{1}x_{2}+x^{2}_{2}x_{3}+x_{3}x_{2}$ and
$V_{\lambda}=\langle 1,x_{1}+x_{2},x_{2}x_{3},x_{2},x_{3}\rangle$, as before.
Then $x_{1}\star_{\lambda}1=-x_{1}-x_{2}$,
$x_{1}\star_{\lambda}(x_{1}+x_{2})=1$, $x_{1}\star_{\lambda}x_{2}x_{3}=0$,
$x_{1}\star_{\lambda}x_{2}=0$, $x_{1}\star_{\lambda}x_{3}=0$. Similar
computations for $x_{2}$ and $x_{3}$ instead of $x_{1}$.

Finally notice that $V_{\lambda}$ and $\Lambda\lambda$ are both submodules of
$\Lambda_{\lambda}$ with trivial intersection, and it is straightforward to
see that if $\gamma\in V_{\lambda}$ then $V_{\gamma}\subseteq V_{\lambda}$.

######  Remark 2.2.

One can define dually _left cofactors_ of $\lambda\in\Lambda$, and so left
cofactors of higher colength, by using the normal form

(4)
$\displaystyle\lambda=k+\sum\limits_{i=1}^{n}{{}_{x_{i}}\lambda}x_{i}\quad(k\in
K;\,_{x_{i}}\lambda\in\Lambda).$

Consequently, one obtains that $\Lambda$ admits a right $\Lambda$-module
structure with respect to the right $\star$-action by forming left cofactors,
that is, for every $b\in\bf B$ and $x_{i}\,(i\in\\{1,2,\cdots,n\\})$ one has

(5) $\displaystyle b\star x_{i}=\begin{cases}h_{b}(|b|-1)&\text{if}\qquad
x_{i}=t_{b}(|b|-1),\\\ 0&\text{if}\qquad x_{i}\neq t_{b}(|b|-1)\end{cases}.$

In particular, $x_{i}\star k=0$ for every scalar $k\in K$. We denote this
module structure of $\Lambda$ as ${}_{\partial}\Lambda$.

Furthermore, any polynomial
$\lambda=k_{\lambda}+\sum\limits_{i=1}^{n}{{}_{x_{i}}}\lambda x_{i},\,0\neq
k_{\lambda}\in K$, makes $\Lambda$ a right $\Lambda$-module, denoted by
${}_{\lambda}\Lambda$, by putting

(6) $\displaystyle 1\star_{\lambda}x_{1}=-{{}_{x_{i}}\lambda
k^{-1}_{\lambda}}\,\,\,\,\&\,\,\,\,b\star_{\lambda}x_{1}=\begin{cases}h_{b}(|b|-1)&\text{if}\quad
x_{i}=t_{b}(|b|-1),\\\ 0&\text{if}\quad x_{i}\neq t_{b}(|b|-1)\end{cases}$

for $i=1,\cdots,n$ and $1\neq b\in\bf B$. The vector subspace $W_{\lambda}$
generated by the right cofactors of $\lambda$ becomes a finite-dimensional
right $\Lambda$-submodule of ${}_{\lambda}\Lambda$. We will need both
$V_{\lambda}$ and $W_{\lambda}$ to compute atomic factorizations of $\lambda$
in Subsection 5.2.

With the notation above, the main goal of this work is

######  Theorem 2.3.

Let $\gamma,\lambda\in\Lambda$ be comonic polynomials of positive degree and
$V_{\gamma}$ and $V_{\lambda}$ the finite-dimensional left $\Lambda$-modules
with respect to the $\star$-action defined by $\gamma$ and $\lambda$,
respectively. Then:

  1. (1)

$\lambda$ is an irreducible polynomial if and only if $V_{\lambda}$ is a
simple $\Lambda$-module.

  2. (2)

If $\lambda=\pi_{1}\cdots\pi_{m}$ is a factorization of $\lambda$ into a
product of irreducible polynomials, then $m$ is the length of $V_{\lambda}$
and its composition factors are isomorphic to the simple modules
$V_{\pi_{i}}$, $i=1,\cdots m$. In particular $m$ is an invariant of $\lambda$.

  3. (3)

$V_{\lambda}\cong V_{\gamma}$ if and only if
$\Lambda/\Lambda\gamma\cong\Lambda/\Lambda\lambda$, that is, $\gamma$ and
$\lambda$ are similar over $\Lambda$.

$\operatorname{End}(V_{\pi})$ is called the _(division) finite-dimensional
algebra_ associated to a comonic (irreducible) polynomial $\pi\in\Lambda$.
Theorem 2.3 suggests a question of central importance, namely, to characterize
comonic irreducible polynomials $\lambda$ such that
$\operatorname{End}_{\Lambda}(V_{\lambda})$ is non-commutative. We shall
discuss polynomials with zero constant term in Sections 4 and 5.

Theorem 2.3 suggests how to extend the concept ”divisibility” and hence
”primeness” in the non-commutative setting. A nonzero, non-invertible element
$\delta$ _divides_ another element $\lambda$ if $\lambda$ admits a
factorization $\lambda=\alpha\beta\gamma$ such that $\delta$ is one of the
$\alpha$, $\beta$ or $\gamma$. A nonzero, non-invertible element $\pi$ is
called _prime_ if $\pi$ has the property: $\pi$ divides a product
$\alpha\beta$ if and only if there is a polynomial $\delta$ such that $\delta$
divides either $\alpha$ or $\beta$ and $\delta$ is in addition similar to
$\pi$. Then Cohn’s Theorem III.3.2 [7, 1st ed.] shows for polynomials, even
more generally, for elements in atomic 2-firs, that $\pi$ is prime if and only
if it is irreducible.

######  Remark 2.3.

As we will see in Section 4, the definition of the $\star$-action in
$\Lambda=\Lambda_{n}$ is inspired by the multiplication by the elements
$x_{i}^{*}\,(i=1,\cdots,n)$ in the associated Leavitt algebra $L(1,n)$. In
case of $n=1$, so that $\Lambda_{1}=K[x]$, the extended Leavitt algebra is the
algebra of Laurent polynomials $K[x,x^{-1}]$ and $x^{*}$ can be seen as
$x^{-1}$. Assume $\lambda=1+k_{1}x+\cdots+k_{l}x^{l}$ is a comonic polynomial
in $K[x]$, so that $\lambda=1+x\lambda_{x}$ and $1=-x\lambda_{x}+\lambda$. If
$g=k+xg_{x}$ is any polynomial in $K[x]$, then
$x\star_{\lambda}g=-k\lambda_{x}+g_{x}=x^{-1}\cdot g-kx^{-1}\lambda$, and
hence the $\star_{\lambda}$-action coincides with the multiplication by
$x^{-1}$ modulo $K[x,x^{-1}]\lambda$. Consequently, using the
$\star_{\lambda}$-action on $K[x]$, one can see that $g$ is a multiple of
$\lambda$, i.e. $g\in K[x]\lambda$, if and only if
$x^{|g|}\star_{\lambda}g=0$. Moreover, denoting by

$\gamma=x^{l}+k_{1}x^{l-1}+\cdots+k_{l}=x^{l}\lambda(x^{-1})$

the reciprocal polynomial of $\lambda$, we get
$K[x,x^{-1}]\lambda=K[x,x^{-1}]\gamma(x^{-1})$ establishing a one-to-one
correspondence between remainders modulo $\lambda$ in $K[x]$ and
$\gamma(x^{-1})$ in $K[x^{-1}]$, respectively.

$V_{\lambda}$ is clearly the $K$-subspace of $K[x]$ spanned by
$\\{1,x,\cdots,x^{l-1}\\}$ and it is clearly closed under the
$\star_{\lambda}$-action induced by the rule $x\star_{\lambda}1=-\lambda_{x}$
(so that, for instance, $x\star_{\lambda}x=1$). In particular, $V_{\lambda}$
is isomorphic to the $K[x^{-1}]$-module $K[x^{-1}]/K[x^{-1}]\gamma(x^{-1})$
with respect to the $K$-algebra isomorphism $K[x]\rightarrow
K[x^{-1}]:x\mapsto x^{-1}$. So $V_{\lambda}=\Lambda\star_{\lambda}1$ is a
cyclic module and the annihilator of $1\in V_{\lambda}$ is the ideal
$K[x]\gamma$. Indeed, from $x\star_{\lambda}x=1$ it follows that
$k_{i}x^{l-i+1}\star_{\lambda}1=x^{l-1}\star_{\lambda}k_{i}x^{i-1}$ and

$\gamma\star_{\lambda}1=(x^{l}+k_{1}x^{l-1}+\cdots+k_{l})\star_{\lambda}1=x^{l}\star_{\lambda}1+x^{l-1}\star_{\lambda}\lambda_{x}=$
$=x^{l-1}\star_{\lambda}(-\lambda_{x}+\lambda_{x})=x^{l-1}\star_{\lambda}0=0.$

Hence in general $V_{\lambda}$ is not isomorphic to $K[x]/K[x]\lambda$ even
when $\lambda$ is an irreducible polynomial.

In case $\lambda$ is irreducible, the ideal $K[x]\gamma$ is maximal, and hence
$V_{\lambda}$ is isomorphic to the irreducible representation
$K[x]/K[x]\gamma$. Therefore, in case of one variable, Theorem 2.3 provides a
parameterization of the irreducible representations of $K[x]$ alternative to
the classical one. Namely, one uses the reciprocical $\gamma$ of an
irreducible polynomial $\lambda$ for this job.

To conclude Remark 2.3 notice that $V_{\lambda}$ can be identified with the
space of remainders by the Euclidean division by $\lambda$. In the classical
approach using the factor algebra $K[x]/K[x]\lambda$, in order to compute the
remainder of $g$ by $\lambda$ without carrying over the division, one has to
substitute $x^{l}$ by $(-k_{0}-k_{1}x-\cdots-k_{l-1}x^{l-1})$. Unfortunately
this approach is not extendable to the case of several non-commutative
variables. However, using the module $V_{\lambda}$ and the $\star_{\lambda}$
action, in order to find the remainder of $g$ by $\lambda$, we can compute
recursively $x\star_{\lambda}g$ until we get an element in $V_{\lambda}$. As
we will see in Section 4 where the proof of Theorem 2.3 is presented, this
approach can be generalized for several non-commutative variables and is, in
fact, the crucial idea of the proof.

From now on, let us assume $n\geq 2$. Then Theorem 2.3 shows a relatively
transparent and simple way assigning to every proper and comonic polynomial
$\lambda$ a finite-dimensional $\Lambda$-module $V_{\lambda}$ whose
endomorphism ring is a finite-dimensional algebra. $V_{\lambda}$ plays the
role of ”remainders” in a generalized Euclidean division algorithm by the
comonic polynomial $\lambda$. This approach provides also an easy algorithm to
compute the remainder of a polynomial $\mu$ at the division by $\lambda$, and
then to compute the unique expression $\mu=\eta\lambda+\rho$ with $\rho\in
V_{\lambda}$, as we will see in Section 4. Our treatment simplifies and
sharpens Cohn’s approach of atomic factorizations discussed in [7], [9].

##  3\. The tool: Leavitt algebras

By definition (cf. [13]) the _module type_ of a ring is either _IBN_ , an
abbreviation for _invariant basis number_ , that is, an isomorphism between
modules generated freely by $m$ and $n$ elements, respectively, implies $m=n$;
or the first integral pair $(m,n)\,(1\leq m\leq n-1)$ such that modules
generated freely by $m$ and $n$ elements, respectively, are isomorphic. This
definition implies immediately the following obvious but important property of
rings without IBN.

######  Proposition 3.1.

If free modules of rank $m$ and $n\geq m+1$, respectively, are isomorphic,
then any finitely generated module can be generated by at most $m$ elements.
In particular, a ring of module type $(1,n),\,\,n\geq 2$ is a Bezout ring,
i.e., every finitely generated one-sided ideal is principal.

For the existence of rings of arbitrary module type $(m,n)$, Leavitt invented
the algebra $\mathbb{L}_{K}(m,n)$ by inverting the matrix $(x_{ij})$ over the
free algebra $\Lambda=K\langle x_{ij}\rangle\,(i=1,\cdots,m;j=1,\cdots,n)$
[13]. Although $\mathbb{L}_{K}(1,n)$ is now somewhat well-understood,
$\mathbb{L}_{K}(m,n)\,(2\leq m\leq n-1)$ is still a big puzzle.

We present below an elementary and transparent approach to
$\mathbb{L}_{K}(1,n)$, shortly $\mathbb{L}$, by listing results in such an
order that their proof is immediate. For the details we refer to [4].

By definition $\mathbb{L}$ is the localization of $\Lambda=K\langle
x_{1},\cdots,x_{n}\rangle=\Lambda_{n}$ inverting the row
$(x_{1},\cdots,x_{n})$, i.e., $\mathbb{L}$ is generated by elements $x_{i}$
and $x^{*}_{i}\,(i=1,\cdots,n)$ subject to relations

(7) $\displaystyle\mathop{\sum}\limits_{i=1}^{n}x_{i}x^{*}_{i}=1\qquad\&\qquad
x^{*}_{j}x_{i}=\begin{cases}1&\text{if}\qquad j=i,\\\ 0&\text{if}\qquad j\neq
i\end{cases}.$

The images of $x_{i},x^{*}_{i}$ in $\mathbb{L}$ are also denoted by
$x_{i},x^{*}_{i}$ themselves by convenience and by the fact that free algebras
$\Lambda$ and $\Lambda^{*}$ are isomorphic canonically to the $K$-subalgebras
$\bar{\Lambda}$ and ${\bar{\Lambda}}^{*}$ of $\mathbb{L}$ generated by
$\\{x_{1},\cdots,x_{n}\\}$ and $\\{x^{*}_{1},\cdots x^{*}_{n}\\}$,
respectively. This result is immediate by Theorem 3.3 from which almost all
basic properties of $\mathbb{L}$ can be read off easily. By putting
$b^{*}=x^{*}_{i_{l}}\cdots x^{*}_{i_{1}}$ for $b=x_{i_{1}}\cdots
x_{i_{l}}\in\bf B$, every element $r\in\mathbb{L}$ can be expressed (not
necessarily uniquely) as a linear combination $\sum
k_{i}b_{i}c^{*}_{i}\,(0\neq k_{i}\in K;b_{i},c_{i}\in\bf B)$. To verify
Theorem 3.3 we need first

######  Lemma 3.2.

For any $r\in\mathbb{L}$, $r\in({\bar{\Lambda}}\cap r\mathbb{L})\mathbb{L}$
and ${\bar{\Lambda}}\cap r\mathbb{L}$ is a finitely generated right ideal of
$\bar{\Lambda}$. In particular, if $\mathfrak{R}$ is a right ideal of
$\mathbb{L}$ and $R=\bar{\Lambda}\cap\mathfrak{R}$, then
$\mathfrak{R}=R\mathbb{L}$. Moreover $R$ is a finitely generated right ideal
of $\bar{\Lambda}$ if and only if $\mathfrak{R}$ is principal. By symmetry,
$\mathfrak{L}=\mathbb{L}(\bar{\Lambda}^{*}\cap\mathfrak{L})=\mathbb{L}L$ holds
for every left ideal $\mathfrak{L}$ of $\mathbb{L}$ and $L$ is a finitely
generated left ideal of ${\bar{\Lambda}^{*}}$ if and only if ${\mathfrak{L}}$
is principal.

###### Proof.

The equality in $\mathbb{L}$

$1=\sum\limits_{i=1}^{n}x_{i}x^{*}_{i}=\sum\limits_{i=1}^{n}x_{i}\left(\sum\limits_{j=1}^{n}x_{j}x^{*}_{j}\right)x^{*}_{i}=\cdots=\sum\limits_{b,c\in\bf
B}^{|b|=|c|=l}bc^{*}\quad(l\in\mathbb{N})$

and $r\cdot 1=r=1\cdot r$ imply

(8) $\displaystyle r=\sum\limits_{b,c\in\bf
B}^{|b|=|c|=l}(rb)c^{*}=\sum\limits_{b,c\in\bf
B}^{|b|=|c|=l}b(c^{*}r)\quad(l\in\mathbb{N})$

with $rb\in\bar{\Lambda}$ and $c^{*}r\in{\bar{\Lambda}}^{*}$ for almost all
$l$. Consequently, if $r$ is represented as a linear combination
$r=\sum\limits_{i=1}^{l}b_{i}c^{*}_{i}$ with the least number $l$ of non-zero
coefficients, setting $d=d_{r}=\max\\{|c_{i}|\,|\,i=1,\cdots,l\\}$, then for
$m=d+1$ one has $rb\in r\mathbb{L}\cap\Lambda$ for all monomials $b$ of degree
$m$. Thus $r\in(r\mathbb{L}\cap\bar{\Lambda})\mathbb{L}$ holds by (8) and
hence $\mathfrak{R}=R\mathbb{L}$. Moreover, $r\mathbb{L}\cap\bar{\Lambda}$ is
trivially generated by the finite set $\\{rb\,|\,\deg b=m\\}$, hence the proof
is complete by Proposition 3.1. ∎

######  Remark 3.1.

Recall that $I$ resp., $I^{*}$ is the two-sided ideal of $\Lambda$ resp.,
$\Lambda^{*}$ defined by the set $\\{x_{1},\cdots,x_{n}\\}$ resp.,
$\\{x^{*}_{1},\cdots,x^{*}_{n}\\}$ of variables. The corresponding ideal
topology ${\mathfrak{T}}\,(resp.,{\mathfrak{T}}^{*})$ is a right (resp., left)
Gabriel topology of $\Lambda$ (resp., $\Lambda^{*}$). By [15, Lemma IX.1.6],
$Q_{I}(\Lambda)=\underrightarrow{\rm
lim}_{l\in\mathbb{N}}\operatorname{Hom}_{\Lambda}(I^{l},\Lambda_{\Lambda})$ is
a ring with respect to the multiplication induced by partial composition of
maps. Another equivalent way is to identify $Q_{I}(\Lambda)$ with the subring
of the maximal ring of right quotients of $\Lambda$ generated by $\Lambda$ and
$\operatorname{Hom}_{\Lambda}(I_{\Lambda},\Lambda)$. By definition
$Q_{I}(\Lambda)$ is the ring of right quotients of $\Lambda$ by the Gabriel
topology $\mathfrak{T}$.

######  Theorem 3.3.

$\mathbb{L}$ is naturally isomorphic to the ring $Q_{I}(\Lambda)$ of right
quotients of $\Lambda$ by the ideal topology
${\mathfrak{T}}=\\{I^{l}\,|\,l\in\mathbb{N}\\}$ hence it is a flat left module
over $\Lambda$. By symmetry, $\mathbb{L}$ is also isomorphic to the ring
$Q_{I^{*}}(\Lambda^{*})$ of left quotients of $\Lambda^{*}$ by the ideal
topology ${\mathfrak{T}}^{*}=\\{(I^{*})^{l}\,|\,l\in\mathbb{N}\\}$ hence it is
a flat right module over $\Lambda^{*}$.

###### Proof.

Since the $I^{l}_{\Lambda}$’s are freely generated by $n^{l}$ monomials of
degree $l$, $\mathfrak{T}$ is a perfect Gabriel topology by [15, Propositions
VI.6.10, XI.3.3 and XI.3.4(d)]. Hence $Q_{I}(\Lambda)={\underrightarrow{\rm
lim}}_{l\in\mathbb{N}}\operatorname{Hom}_{\Lambda}(I^{l},\Lambda_{\Lambda})$
is a $K$-algebra, called the _ring of right quotients_ of $\Lambda$ by
$\mathfrak{T}$, as we already mentioned. For every monomial $b\in{\bf
B}\,(|b|=l\geq 1)$ of positive degree $l$ let $b^{\star}$ be the right
$\Lambda$-module homomorphism $b^{\star}:I^{l}\rightarrow\Lambda_{\Lambda}$
such that

$c\in I^{l}\cap{\bf B}\mapsto b^{\star}(c)=\begin{cases}1&\text{if}\qquad
c=b,\\\ 0&\text{if}\qquad c\neq b\end{cases}.$

In view of the equalities

$I^{l}=\bigoplus\limits_{b\in\bf
B}^{|b|=l}b\Lambda\qquad\&\qquad\operatorname{Hom}_{\Lambda}(I^{l}_{\Lambda},\Lambda)=\bigoplus\limits_{b\in\bf
B}^{|b|=l}{{}_{\Lambda}\Lambda}b^{\star}$

and the fact that all
$\operatorname{Hom}_{\Lambda}(I^{l},\Lambda_{\Lambda})\,(l\in\mathbb{N})$ are
free left $\Lambda$-modules of rank $n^{l}$ one obtains that
${{}_{\Lambda}Q_{I}}(\Lambda)$ is flat. For $b=b_{i_{1}}\cdots b_{i_{l}}$ it
is easy to check the equality $b^{\star}=b^{\star}_{i_{l}}\cdots
b^{\star}_{i_{1}}$ in $Q_{I}(\Lambda)$. Hence $\Lambda$ and the
$\,x^{\star}_{i}$’s generate $Q_{I}(\Lambda)$ and the row
$(x_{1},\cdots,x_{n})$ is the inverse of the column
${}^{t}(x^{\star}_{1},\cdots,x^{\star}_{n})$. Therefore, from the definition
of $\mathbb{L}$, there is a canonical map $\phi:\mathbb{L}\rightarrow
Q_{I}(\Lambda)$ induced by the identity map on $\Lambda$ and the rule
$x^{*}_{i}\mapsto x^{\star}_{i}$ for $i=1,\cdots,n$. Consequently, $\phi$ is
surjective and restricts to the identity on $\Lambda$ whence $\Lambda$ is a
subalgebra of $\mathbb{L}$.

For the injectivity of $\phi$, assume $\phi(r)=0$ for some $r\in\mathbb{L}$.
Then, by Lemma 3.2, $r\neq 0$ implies $0\neq r\mu\in\Lambda$ for some monomial
$\mu$ and so $0\neq r\mu=\phi(r\mu)=\phi(r)\phi(\mu)=0$, a contradiction.
Consequently, $\phi$ is an isomorphism whence $\mathbb{L}$ is isomorphic to
$Q_{I}(\Lambda)$. ∎

Interchanging the role of $x_{i}$ and $x^{*}_{i}$ one obtains

######  Corollary 3.4.

The free algebra $\Lambda^{*}=K\langle x^{*}_{1},\cdots,x^{*}_{n}\rangle$ is
naturally isomorphic to the subalgebra ${\bar{\Lambda}}^{*}$ of $\mathbb{L}$.
The rule $b\in{\bf B}\mapsto b^{*}$ induces an involution of $\mathbb{L}$
which restricts to the canonical anti-isomorphism from $\Lambda$ to
$\Lambda^{*}$.

Hence $\Lambda$ and $\Lambda^{*}$ are viewed naturally as $K$-subalgebras of
$\mathbb{L}$. Moreover, $\Lambda^{*}$ and $\Lambda$ are isomorphic free
algebras, for example, under the star-forgetting rule
$b^{*}\mapsto{{}^{\tau}b}\,(b\in\bf B)$.

Since a nonunital subalgebra $A\subseteq\Lambda$ of codimension 1 is both a
two-sided ideal and hence a free nonunital algebra of rank $n$, as it is
straightforward to verify, Theorem 3.3 provides a variable-free definition of
$\mathbb{L}$ and sheds a light on ”canonical” involutions of $\mathbb{L}$ and
a close connection to the automorphism groups of free algebras.

######  Corollary 3.5.

If $A\subseteq\Lambda$ is a nonunital subalgebra of codimension 1, then $A$ is
a two-sided ideal of $\Lambda$ and $\mathbb{L}\cong Q_{A}(\Lambda)$, the ring
of right quotients with respect to the ideal topology
$\\{A^{l}\,|\,l\in\mathbb{N}\\}$.

As a ring of partial maps, every element $r\in\mathbb{L}$ is represented by a
$\Lambda$-homomorphism $r:S_{r}\rightarrow R_{r}$ from the right ideal
$S_{r}=\\{\lambda\in\Lambda\,|\,r\lambda\in\Lambda\\}$, the _domain_ , onto
$R_{r}=\mathbb{L}r\cap\Lambda$, the _range_ of $r$. Hence $S_{r}$ has finite
codimension because it contains some power of $I$ and $R_{r}$ is finitely
generated. However $R_{r}$ has not, in general, finite codimension. As a first
obvious consequence, by comparing degrees of $\lambda(b)=\lambda b$ and
$r(b)=rb$ for $\lambda\in\Lambda,\,r\in\Lambda^{*}$ and $b\in\bf B$ a monomial
contained in the range $R_{r}$ of $r$, we have clearly the following well-
known special case of [2, Corollary 1.5.7]

######  Corollary 3.6.

$\Lambda\cap\Lambda^{*}=K$ as subalgebras of $\mathbb{L}$.

######  Proposition 3.7.

If $n\geq 2$, then $\mathbb{L}$ is a purely infinite simple $K$-algebra, i.e.,
for every non-zero $r\in\mathbb{L}$ one has $srt=1$ for suitable elements
$s,t\in\mathbb{L}$

###### Proof.

Assume $n\geq 2$. For the simplicity of $\mathbb{L}$, let $0\neq
r\in\mathbb{L}$ and consider the elements
$0\neq\lambda=\sum\limits_{i=1}^{l}k_{i}b_{i}\in r\mathbb{L}\cap\Lambda$ with
smallest $l$. Among them, choose an element $\bar{\lambda}$ with smallest
degree. Then first by multiplying an appropriate $x_{i}$ on the right if
$\bar{\lambda}$ has non-zero constant and then on the left by another
$x^{*}_{j}\neq x^{*}_{i}$, which is possible since $n\geq 2$, one can see that
all $b_{i}$ must start with a same variable, say $x$, and $\lambda$ is without
constant if it is not a non-zero scalar. Hence $x^{*}\lambda$ has lower degree
and so $\lambda$ must be a nonzero constant. This implies the purely infinite
simplicity of $\mathbb{L}$. ∎

In the following corollary we list some basic properties of $\mathbb{L}$ that
can be deduced from Theorem 3.3. In particular Claim (3) describes the open
left ideals of $\mathbb{L}$.

######  Corollary 3.8.

Let $\mathfrak{L}$ a left ideal of $\mathbb{L}$ and
$L=\Lambda^{*}\cap\mathfrak{L}$. Then:

  1. (1)

$\mathfrak{L}=\mathbb{L}L$ holds for every left ideal $\mathfrak{L}$ of
$\mathbb{L}$, so $L$ is a finitely generated left ideal of $\Lambda^{*}$ if
and only if $\mathfrak{L}$ is finitely generated.

  2. (2)

The algebra $\mathbb{L}$ is hereditary and projective-free.

  3. (3)

A left ideal $N$ of $\Lambda^{*}$ satisfies $\mathbb{L}N=\mathbb{L}$ iff $N$
contains a power of $I^{*}$.

  4. (4)

$\mathfrak{L}$ is finitely generated of finite colength if and only if $L$ is
of finite codimension.

###### Proof.

Claim $(1)$ is given by Lemma 3.2.

Claim (2) follows from the fact that
$\mathfrak{L}=\mathbb{L}L=\mathbb{L}\otimes_{\Lambda^{*}}L$ by the flatness of
$\mathbb{L}_{\Lambda^{*}}$ ensured in Theorem 3.3. Thus we conclude since free
algebras, being free ideal rings, are hereditary and projective-free.

If $N$ contains $I^{l}$ for a suitable $l\geq 0$, then from the identity
$1=\sum\limits_{b,c\in\bf B}^{|b|=|c|=l}bc^{*}$ we get
$\mathbb{L}N=\mathbb{L}$. Conversely, if
$1=\sum\limits_{i=1}r_{i}\lambda^{*}_{i}$ for appropriate $r_{i}\in\mathbb{L}$
and $\lambda^{*}_{i}\in N$, we can choose an integer $l$ big enough such that
$b^{*}r_{i}\in\Lambda^{*}$ for any $b\in\bf B$ with $|b|=l$ and any $r_{i}$.
So we get that $I^{l}\leq N$. This shows Claim (3).

Since $\mathbb{L}$ is projective free, any principal left ideal is free and of
infinite length. It follows easily that if $\mathfrak{L}$ has finite colength,
then it is essential. From the essentiality of $\mathfrak{L}$ we get that also
$L$ is essential in $\Lambda^{*}$. By the result of Rosenmann and Rosset in
[14, Theorem 3.3], we conclude that $L$ has finite codimension since it is
finitely generated and essential. To conclude Claim $(4)$ assume $L$ has
finite codimension. In particular, again by [14, Theorem 3.3], it is finitely
generated and hence $\mathfrak{L}$ is finitely generated. Notice that if
$\mathfrak{L}<\mathfrak{M}<\mathbb{L}$, then
$L<\mathfrak{M}\cap\Lambda^{*}<\Lambda^{*}$, so $\mathfrak{L}$ is of finite
colength since $L$ has finite colength. ∎

In the notation of the previous corollary, if $L$ is an arbitrary maximal left
ideal of $\Lambda^{*}$ such that the associated primitive ideal
$J=\operatorname{ann}_{\Lambda^{*}}\Lambda^{*}/L$ is not contained in $I^{*}$,
then $L$ cannot contain any power $(I^{*})^{l}$. For, $(I^{*})^{l}\subseteq L$
implies $J+(I^{*})^{l}\subseteq L$. However, $J+I^{*}=\Lambda^{*}$ implies
$J=J\Lambda^{*}=J^{2}+JI^{*}$ and so
$\Lambda^{*}=(J+I^{*})^{2}=J+(I^{*})^{2}=\cdots=J+(I^{*})^{l}=J^{2}+(I^{*})^{l}=\cdots=J^{m}+(I^{*})^{l}$
for any $l,m\in\mathbb{N}$. Consequently, one has $L=\Lambda^{*}$, which is
impossible. Hence $\mathbb{L}L\neq\mathbb{L}$ by Corollary 3.8(3). Since $L$
is maximal, Corollary 3.8(1) implies that $\mathbb{L}L$ is a maximal left
ideal of $\mathbb{L}$. Obvious examples for such maximal left ideals are ones
which contain maximal two-sided ideals of the free algebra $\Lambda^{*}$
different from $I^{*}$. In this way one obtains a nice large class of new
simple modules over $\mathbb{L}$ with interesting endomorphism algebras. The
subclass induced by maximal left ideals of $\Lambda^{*}$ which have finite
codimension and do not coincide with $I^{*}$ is implicitly found in Ara’s
Proposition 3.3 and Theorem 5.1 [5]. More generally, the preceding argument
shows that for pairwise different maximal two-sided ideals $J_{i}$ of
$\Lambda^{*}$ together with positive integers $l_{i}\in\mathbb{N}$, we have
$J^{l_{1}}_{1}+J^{l_{2}}_{2}\cdots J^{l_{m}}_{m}=\Lambda^{*}$. Hence if $L$ is
a finite-codimensional left ideal and $J$ is the maximal two-sided ideal
contained in $L$ such that inverse images of maximal two-sided ideals in the
factor algebra $\Lambda^{*}/J$ are different from $I^{*}$, then the extended
principal left ideal $\mathbb{L}L$ satisfies $L=\Lambda^{*}\cap\mathbb{L}L$
and the modules $\Lambda^{*}/L$ and $\mathbb{L}/\mathbb{L}L$ have the same
length. On the other hand, by Corollary 3.8(4) every finitely presented left
$\mathbb{L}$-module of finite length is isomorphic to
$\mathbb{L}/\mathfrak{L}$ such that $L=\Lambda^{*}\cap\mathfrak{L}$ is finite-
codimensional. Assume that $\mathfrak{L}$ is a maximal left ideal of
$\mathbb{L}$. Then the intersection $L=\mathfrak{L}\cap\Lambda^{*}$ is not
necessarily a maximal left ideal of $\Lambda^{*}$. $L$ is maximal only in the
set of left ideals $M$ of $\Lambda^{*}$ satisfying
$M=(\mathbb{L}M)\cap\Lambda^{*}$. Since $L=\Lambda^{*}\cap\mathfrak{L}$ is
finite-codimensional there is $L=L_{0}\subseteq L_{1}\subseteq\cdots\subseteq
L_{l}=\Lambda^{*}$ is a maximal ascending chain of left ideals with simple
factors $L_{i+1}/L_{i}\,(i=0,\cdots,l-1)$. Moreover,
$\mathbb{L}L_{i}=\mathbb{L}$ holds for all $i\geq 1$ and $L_{1}/L$ is a simple
finite-dimensional $\Lambda^{*}$-module such that the associated primitive
ideal is a finite-dimensional maximal two-sided ideal $J\neq I^{*}$.
Consequently, if $M$ is a maximal left ideal of $\Lambda^{*}$ with
$\Lambda^{*}/M\cong L_{1}/L$ one gets
$\mathbb{L}/\mathbb{L}M\cong\mathbb{L}/\mathfrak{L}$. Moreover, one obtains
that all the other possible composition factors of $\Lambda^{*}/L$ are
isomorphic to the trivial module $K=\Lambda^{*}/I^{*}$. This establishes an
one-to-one correspondence between isomorphism classes of finitely presented
irreducible representations of the Leavitt algebra $\mathbb{L}$ and ones of
finite-dimensional non-trivial (i.e., not isomorphic to the trivial
$\Lambda^{*}$-module $K$) representations of $\Lambda^{*}$. In this manner we
obtain explicitly the above mentioned results in [5]. Localization theory
shows that the torsion modules defined by the Gabriel topology
${\mathfrak{T}}^{*}=\\{(I^{*})^{l}\,|\,l\in\mathbb{N}\\}$ form a Serre
localizing subcategory of $\Lambda^{*}$-Mod and $\mathbb{L}$-Mod is the
quotient category of $\Lambda^{*}$-Mod by this Serre category. Hence Ara’s
Proposition 3.3 and Theorem 5.1 [5] are consequences of Theorem 3.3 and the
Gabriel’s localization theory. However, one need not to use sledge hammer to
crack nuts and to both readers’ benefit and the sake of the elementary self-
containedness we present the special case of finitely presented modules of
finite length.

Let $\mathfrak{FD}$ be the full subcategory of finite-dimensional left
$\Lambda^{*}$-modules. A finite-dimensional $\Lambda^{*}$-module is called a
_Leavitt_ module if none of its simple subfactors is isomorphic to the trivial
module $K$, that is, its annihilator ideal is a product of maximal two-sided
finite-codimensional ideals different from $I^{*}$. The full subcategory of
$\Lambda^{*}$-Mod consisting of all Leavitt modules is called the _Leavitt
category_. A finite-dimensional module is called a _Leavitt-torsion_ module if
its simple subfactors are isomorphic to the trivial module $K$. Any
$M\in\mathfrak{FD}$ contains the largest submodule $T(M)$ which has all simple
subfactors isomorphic to the trivial module $K$. Then $T(M/T(M))=0$ holds
clearly. $T(M)$ is called the _Leavitt torsion submodule_ or simply _torsion
submodule_ of $M$. If $T(M)=0$, then $M$ is a _(Leavitt) torsionfree module_.
Leavitt modules are clearly torsionfree but torsionfree modules are not
necessarily Leavitt modules since they may contain simple subfactors which are
not contained in their socle and isomorphic to the simple trivial module $K$.
The full subcategory $\mathfrak{LT}$ of Leavitt-torsion modules is clearly a
Serre subcategory of $\mathfrak{FD}$. If $M$ is a Leavitt module of length
$l$, then the flatness of $\mathbb{L}_{\Lambda^{*}}$ implies that
${\mathfrak{M}}=\mathbb{L}\otimes_{\Lambda^{*}}M$ is finitely presented and
has the same length $l$. Namely, if $0=M_{0}\subseteq
M_{1}\subseteq\cdots\subseteq M_{l}=M$ is a composition chain of $M$, then the
flatness of $\mathbb{L}_{\Lambda^{*}}$ implies that
$\mathbb{L}\otimes_{\Lambda^{*}}M_{i}$ $(i=0,1,\cdots,l)$ form a corresponding
composition chain of $\mathfrak{M}$. Hence by comparing lengths and by the now
routine use of the equality (8) together with similar reasoning for Corollary
3.8, one can see that $M$ is isomorphic to $\\{1\otimes m\,|\,m\in
M\\}\subseteq\mathfrak{M}$ as left modules over $\Lambda^{*}$. After this
identification we have the lattice isomorphism between the submodule lattices
of $\mathfrak{M}$ and $M$ induced by $N\subseteq
M\mapsto{\mathfrak{N}}=\mathbb{L}\otimes_{\Lambda^{*}}N=\mathbb{L}N\mapsto{\mathfrak{N}}\cap
M$, respectively. Any module homomorphism $f\colon M\rightarrow N$ between
Leavitt modules induces clearly the homomorphism $\mathbb{L}\otimes
f:{\mathfrak{M}}=\mathbb{L}\otimes_{\Lambda^{*}}M\rightarrow{\mathfrak{N}}=\mathbb{L}\otimes_{\Lambda^{*}}N$.
Furthermore, any homomorphism
${\mathfrak{f}}:{\mathfrak{M}}\rightarrow\mathfrak{N}$ is induced by some
homomorphism $\bar{f}\colon\bar{M}\rightarrow\bar{N}$ between submodules
$\bar{M},\bar{N}$ of $M,\,N$, respectively. The equality
${\mathfrak{f}}=\mathbb{L}\otimes\bar{f}$ together with the above deduced
lattice isomorphism of submodule lattices show $\bar{M}=M$ and $\bar{N}=N$.
Hence the Leavitt category of $\Lambda^{*}$ can be considered as a full and
faithful subcategory of $\mathbb{L}$-Mod. On the other hand, if $\mathfrak{M}$
is any finitely presented left $\mathbb{L}$-module of finite length, then
$\mathfrak{M}\cong\mathbb{L}/\mathfrak{L}$ by Proposition 3.1 and Corollary
3.8(4) where $\mathfrak{L}$ is a principal ideal of $\mathbb{L}$ of finite
colength. Since any homomorphism
$f\colon\mathbb{L}/\mathfrak{L}_{1}=\mathfrak{M}_{1}\rightarrow\mathbb{L}/\mathfrak{L}_{2}=\mathfrak{M}_{2}$
between finitely presented modules of finite length is uniquely determined by
the image $(1+\mathfrak{L}_{1})f\equiv
r\mod\mathfrak{L}_{2}\,(r\in\mathbb{L})$, the equality (8) shows that $f$ is
uniquely determined by the images $(b^{*}+\mathfrak{L}_{1})f\equiv
b^{*}r\mod\mathfrak{L}_{2}$ with all $b^{*}r\in\Lambda^{*}$, where $b$ runs
over all monomials of an appropriate degree $l$ big enough. Therefore $f$ is
uniquely determined by a homomorphism
${\overline{f}}\colon\sum\limits_{|b|=l}\Lambda^{*}b^{*}/L_{1}={\overline{M}}_{1}\rightarrow\sum\limits_{|b|=l}\Lambda^{*}b^{*}r/L_{2}={\overline{M}}_{2}$
where $L_{i}=\Lambda^{*}\cap\mathfrak{L}_{i}$ and $M_{i}=\Lambda^{*}/L_{i}$
contains $\overline{M}_{i}\,(i=1,2)$. Corollary 3.8(3) implies that
$M_{1}/{\overline{M}_{1}}$ is a torsion module. Note the fact that both
$M_{1}$ and $M_{2}$ are torsion free. Furthermore, any homomorphism $g\colon
M\subseteq M_{1}\rightarrow M_{2}$ such that $M_{1}/M$ is a torsion module,
defines a homomorphism $1\otimes g\colon\mathfrak{M}_{1}=\mathbb{L}\otimes
M\rightarrow\mathfrak{M}_{2}$ and $\overline{1\otimes g}$ coincides with $g$
on $M$.

Take now an arbitrary $M\in\mathfrak{FD}$. Then $M$ maps to a left
$\Lambda^{*}$-submodule $\overline{M}=\\{1\otimes m\,\,\,|\,\,m\in M\\}$ of
$\mathfrak{M}=\mathbb{L}\otimes_{\Lambda^{*}}M$ whose kernel is exactly
$T(M)$. If $f\colon\mathfrak{M}\rightarrow\mathfrak{N}=\mathbb{L}\otimes
N\,\,(N\in\mathfrak{FD})$ is any $\mathbb{L}$-homomorphism, then there is
$l\in\mathbb{N}$ such that $b^{*}(1\otimes m)f\in{\overline{N}}=\\{1\otimes
n\,\,\,|\,\,n\in N\\}\subseteq\mathfrak{N}$ holds for all $b\in\bf B$ with
$|b|=l$. Hence $f$ is induced by a $\Lambda^{*}$-homomorphism from a submodule
$\underline{M}$ of $M$ with $T(M/\underline{M})=M/\underline{M}$ into a
submodule $\underline{N}$ of $N/T(N)$. This implies Theorem 5.1 of Ara [5]
that the category $\mathfrak{FPL}$ of finitely presented left
$\mathbb{L}$-modules of finite length is the quotient category of
$\mathfrak{FD}$ by the Serre subcategory $\mathfrak{TL}$. In particular, the
Leavitt category is equivalent to a full subcategory of $\mathfrak{FPL}$.
Notice that although the Leavitt category seems to be equivalent to the
category $\mathfrak{FPL}$ of finitely presented $\mathbb{L}$-modules of finite
length, it is an open and important question whether they are actually
equivalent. To conclude we have obtained a refinement of Theorem 5.1 of Ara
[5] as

######  Corollary 3.9.

Finitely presented $\mathbb{L}$-modules of finite length form the category
$\mathfrak{FPL}$ which is the quotient category of finite-dimensional
$\Lambda^{*}$-modules by one of Leavitt-torsion modules, whence the Leavitt
category is fully embedded in $\mathfrak{FPL}$ via the functor
$M\mapsto\mathbb{L}\otimes_{\Lambda^{*}}M=\mathfrak{M}$ with
$\operatorname{End}(\mathfrak{M})\cong\operatorname{End}(M)$. Hence for any
finite-dimensional factor algebra $A$ of $\Lambda^{*}$ such that $I^{*}$ maps
to $A$, finite-dimensional left $A$-modules are fully embedded in
$\mathfrak{FDL}$.

######  Remarks 3.2.

  1. (1)

Although finitely generated $\mathbb{L}$-modules are cyclic by Proposition
3.1, finitely generated Leavitt $\Lambda^{*}$-modules are not necessarily
cyclic. For example, if $J\neq I^{*}$ is a maximal finite-codimensional two-
sided ideal of $\Lambda^{*}$ and $M=(\Lambda^{*}/J)^{l}$ with $l\geq 2$, then
$M$ is clearly a finitely generated non-cyclic Leavitt $\Lambda^{*}$-module.
The submodule lattice of $M$ is isomorphic to that of the $\mathbb{L}$-module
$\mathfrak{M}=\mathbb{L}\otimes_{\Lambda^{*}}M$, which is a cyclic module. So
$\mathfrak{M}$ is isomorphic to $\mathbb{L}/\mathfrak{L}$ for a suitable
principal left ideal $\mathfrak{L}$, by the fact that $M$ is finitely
presented. However, $M$ and $\Lambda^{*}/L,\,(L=\Lambda^{*}\cap\mathfrak{L})$
are not isomorphic $\Lambda^{*}$-modules.

  2. (2)

There are several ways to realize a particular finite dimensional $K$-algebra
as a factor algebra of $\Lambda^{*}$. For example, to every point
$\underline{k}=(k_{1},\cdots,k_{n})$ of the affine space $K^{n}$ the factor
algebra
$\Lambda^{*}/I_{\underline{k}},\,\,I_{\underline{k}}=\sum\Lambda(x^{*}_{i}-k_{i})$
is isomorphic to $K$.

  3. (3)

There are left ideals $L\subseteq\Lambda^{*}$ of finite codimension not
containing any power of $I^{*}$ satisfying $L\neq\mathbb{L}L\cap\Lambda^{*}$
which are maximal left ideals. For example,
$I^{*}\cap\sum\limits_{i=1}^{n}\Lambda^{*}(x^{*}_{i}-1)=L\neq\mathbb{L}L\cap\Lambda^{*}=\sum\limits_{i=1}^{n}\Lambda^{*}(x^{*}_{i}-1)$.

Corollary 3.9 raises an important but vague question about how Leavitt modules
can be constructed and classified. It is a big surprise that finite-
dimensional $\Lambda$-modules $V_{\lambda}$ induced by noncommutative
polynomials $\lambda\in\Lambda$ provide a simple transparent partial answer
with remarkable consequences via Theorem 2.3.

We conclude this section with the following easy result which will be very
useful in the sequel.

######  Lemma 3.10.

If $\lambda\in\Lambda$ is a polynomial without constant, then
$\mathbb{L}\lambda=\sum\mathbb{L}{\gamma_{i}}$ for suitable polynomials
$\gamma_{i}$’s with constant and of degree lower than $\lambda$.

###### Proof.

Assume $\lambda=\sum\limits_{i=1}^{n}x_{i}\lambda_{x_{i}}$, where the
$\lambda_{x_{i}}$’s are the cofactors of $\lambda$ defined in the previous
section. Then
$\mathbb{L}\lambda=\sum\limits_{i=1}^{n}\mathbb{L}\lambda_{x_{i}}$ and we can
repeat the argument until we get
$\mathbb{L}\lambda=\sum\limits_{j=1}^{l}\mathbb{L}\lambda_{b_{j}}$ for the
$b_{j}$’s monomials of $\lambda$ of minimal degree. ∎

##  4\. The proof

In this section we are going to prove Theorem 2.3, by means of the results on
the Leavitt algebra $\mathbb{L}$ previously presented. Recall in particular
that both $\Lambda=K\langle x_{1},\cdots,x_{n}\rangle$ and
$\Lambda^{*}=K\langle x^{*}_{1},\cdots,x^{*}_{n}\rangle$ are free subalgebras
of $\mathbb{L}$. Without loss of generality $\lambda\in\Lambda$ is assumed
comonic and $|\lambda|\geq 1$, i.e., $\lambda=1+\sum
x_{i}\lambda_{x_{i}}=1+{\underline{\lambda}}\neq 1$. Once $\lambda$ is fixed,
we set $L_{\lambda}=\Lambda^{*}\cap\mathbb{L}\lambda$ and
${M_{\lambda}}=\mathbb{L}/\mathbb{L}\lambda$.

######  Lemma 4.1.

Let $\lambda\in\Lambda$ be a comonic polynomial. If
$\alpha\in\mathbb{L}\lambda\cap\Lambda$, then $\alpha\in\Lambda\lambda$. Hence
$\mathbb{L}\lambda\cap\Lambda=\Lambda\lambda$.

###### Proof.

Put ${\underline{\lambda}}=-\delta$ for convenience. For $r\in\mathbb{L}$ with
$\alpha=r\lambda$ one has $\alpha=r-r\delta$, so that
$r=\alpha+r\delta=\alpha+(\alpha+r\delta)\delta=\cdots=\alpha+\alpha\delta+\cdots+\alpha\delta^{l}+r\delta^{l+1}$
for all $l\in\mathbb{N}$. Since there exists $l$ large enough such that
$r\delta^{l+1}\in\Lambda$, we get $r\in\Lambda$ and so
$\alpha\in\Lambda\lambda$. ∎

######  Corollary 4.2.

If $\lambda\in\Lambda$ is comonic and $|\lambda|\geq 1$, then
$\mathbb{L}\lambda$ is a proper left ideal of $\mathbb{L}$ and $L_{\lambda}$
is a proper left ideal of $\Lambda^{*}$. In particular, if
$\mathbb{L}\gamma=\mathbb{L}\gamma$ for nonzero polynomials
$\delta,\gamma\in\Lambda$ with constant, then $\gamma$ is a scalar multiple of
$\delta$.

###### Proof.

By the previous lemma, polynomials in $\Lambda$ with constant of degree
$\leq|\lambda|-1$ do not belong to $\Lambda\lambda$ and also not to
$\mathbb{L}\lambda$. Hence $\mathbb{L}\lambda$ is a proper left ideal of
$\mathbb{L}$. Moroever, $L_{\lambda}$ is a proper left ideal of $\Lambda^{*}$
by Corollary 3.8(1). ∎

######  Lemma 4.3.

If $\lambda$ is comonic, then $\Lambda+\mathbb{L}\lambda=\mathbb{L}$, whence
$\Lambda/\Lambda\lambda\cong\mathbb{L}/\mathbb{L}\lambda$ as
$\Lambda$-modules.

###### Proof.

For ${\underline{\lambda}}=-\delta$, the congruence
$1\equiv\delta\mod\mathbb{L}\lambda$ implies
$1\equiv\delta^{l}\mod\mathbb{L}\lambda$ and hence $r\equiv
r\delta^{l}\mod\mathbb{L}\lambda$ for any $l\in\mathbb{N}$ and any
$r\in\mathbb{L}$. Consequently, choosing $l$ large enough such that
$r\delta^{l}\in\Lambda$, we get
$(r+\mathbb{L}\lambda)\cap\Lambda\neq\emptyset$ for all $r\in\mathbb{L}$.
Hence $\Lambda+\mathbb{L}\lambda=\mathbb{L}$ holds and
$\Lambda/\Lambda\lambda\cong\mathbb{L}/\mathbb{L}\lambda$ as $\Lambda$-modules
by the equality $\mathbb{L}\lambda\cap\Lambda=\Lambda\lambda$ of Lemma 4.1 and
the basic isomorphism theorem of Noether. ∎

######  Remark 4.1.

Notice that $\Lambda/\Lambda\lambda$ becomes an $\mathbb{L}$-module by putting

$x^{*}_{j}(\gamma+\Lambda\lambda)=(x_{j}\star_{\lambda}\gamma)+\Lambda\lambda=-k\lambda_{j}+\gamma_{j}+\Lambda\lambda$

for every $\gamma=k+\sum x_{i}\gamma_{i}\in\Lambda$, where $\gamma_{j}$ and
$\lambda_{j}$ are the $x_{j}$-th cofactors of $\gamma$ and $\lambda$,
respectively. This definition is well-posed because
$x_{i}\star_{\lambda}\lambda=0$ and
$x_{i}\star_{\lambda}\Lambda\lambda\subseteq\Lambda\lambda$ hold for all
$i=1,\cdots,n$, as it is routine to check. In fact, the Leavitt algebra
$\mathbb{L}$ with Lemma 4.3 is a proper motivation for the
$\star_{\lambda}$-action defined in Section 2.

Notice that the $\star_{\lambda}$-action endows $\Lambda\lambda$ and $\Lambda$
with a structure of left $\Lambda^{*}$-modules where multiplication by
$x^{*}_{i}$ is exactly the $\star_{\lambda}$ action of $x_{i}$, that is,

$x_{i}^{*}\times r=x_{i}\star_{\lambda}r$

for any $r\in\Lambda$. However they are not modules over $\mathbb{L}$ because
$\sum x_{i}x^{*}_{i}$ is not the identity on them.

Facts on the $\star_{\lambda}$-action listed in Section 2 which are useful in
studying injective envelopes of finite-dimensional $\Lambda$-modules are
summarized as

######  Corollary 4.4.

Let $\lambda=1+\sum x_{i}\lambda_{i}$. Then $\Lambda$ is a left
$\Lambda$-module via the $\star_{\lambda}$-action
$x_{j}\star_{\lambda}\gamma=-k\lambda_{j}+\gamma_{j}$ for $\gamma=k+\sum
x_{i}\gamma_{i}\in\Lambda$. If $|\lambda|\geq 1$, then $V_{\lambda}\neq 0$ is
essentially embedded in $\Lambda/\Lambda\lambda$.

The last claim of Corollary 4.4 follows easily since
$b\star_{\lambda}\gamma\in V_{\lambda}$ for any monomial $b\in\bf B$ of length
$|\gamma|$.

######  Lemma 4.5.

In $\mathbb{L}$ the equalities
$V_{\lambda}+\mathbb{L}\lambda=V_{\lambda}\oplus\mathbb{L}\lambda=\Lambda^{*}+\mathbb{L}\lambda$
hold. Moreover $L_{\lambda}$ is a finite codimensional left ideal of
$\Lambda^{*}$ and
$V_{\lambda}\cong\Lambda^{*}/(\mathbb{L}\lambda\cap\Lambda^{*})=\Lambda^{*}/L_{\lambda}$.
In particular, $\operatorname{codim}L_{\lambda}=\dim V_{\lambda}$.

###### Proof.

Since $-1\equiv{\underline{\lambda}}\mod\mathbb{L}\lambda$, for any
$b^{*}\,(b\in\bf B)$ we have $-b^{*}\equiv
b^{*}{\underline{\lambda}}\mod\mathbb{L}\lambda$. Hence any $b^{*}\,(b\in\bf
B)$ with $|b|\geq|\lambda|$ is either in $L_{\lambda}$ or congruent modulo
$L_{\lambda}$ to linear combinations of $c^{*}\,(c\in\bf B)$ with
$|c|\leq|\lambda|-1$. Consequently, $\Lambda^{*}$ is the sum of $L_{\lambda}$
and the $K$-space generated by the finite set
$\\{b^{*}\,|\,|b|\leq|\lambda|-1,b\in\bf B\\}$ and hence $L_{\lambda}$ is a
finite codimensional left ideal of $\Lambda^{*}$. In particular, if
$|\lambda|=1$ then $L_{\lambda}$ and hence $\mathbb{L}\lambda$ (by Corollary
3.9) are maximal left ideals of $\Lambda^{*}$ and $\mathbb{L}$, respectively.

Assume now that a cofactor $\lambda_{b}$ of $\lambda$ is not zero for some
monomial $b\in\bf B$, that is, $b$ is the head of some maximal monomial in
$\lambda$. Then

(9) $\displaystyle-b^{*}\equiv b^{*}{\underline{\lambda}}=\sum
k_{j}t_{b}(j)^{*}+\lambda_{b}\Longrightarrow-b^{*}-\sum
k_{j}t_{b}(j)^{*}\equiv\lambda_{b}\mod\mathbb{L}\lambda,$

where $k_{j}$ is the coefficient of a proper head
$h_{b}(j)\,(j\in\\{1,\cdots,|b|-1\\})$ of $b$ which is also a monomial in
$\lambda$. This implies $\lambda_{b}\in\Lambda^{*}+\mathbb{L}\lambda$. Since
by Lemma 4.1 we get $\lambda_{b}\notin\mathbb{L}\lambda$, we conclude that
$V_{\lambda}$ is canonically embedded into $\Lambda^{*}/L_{\lambda}$. On the
other hand, the congruences $-x^{*}_{i}\equiv\lambda_{x_{i}}$ modulo
$\mathbb{L}\lambda$ and an obvious induction based on congruences (9), imply
that any monomial $c^{*}$ of length $\leq|\lambda|$ is a linear combination of
cofactors $\lambda_{b}$ of $\lambda$ modulo $\mathbb{L}\lambda$, so that
$V_{\lambda}$ maps surjectively onto $\Lambda^{*}/L_{\lambda}$. So we have
that the $K$-vector spaces $V_{\lambda}$ and $\Lambda^{*}/L_{\lambda}$ are
isomorphic, and hence $\operatorname{codim}L_{\lambda}=\dim V_{\lambda}$.
Finally, by congruences (9), we easily get that
$V_{\lambda}+\mathbb{L}{\lambda}=V_{\lambda}\oplus\mathbb{L}\lambda=\Lambda^{*}+\mathbb{L}\lambda$.
∎

Notice that the proof of Lemma 4.5 implies the following important equality

$\Lambda^{*}+\Lambda^{*}\lambda=V_{\lambda}+\Lambda^{*}\lambda=V_{\lambda}\oplus\Lambda^{*}\lambda.$

Together with the algorithm described in the proof of [14, Lemma 3.2], the
proof of Lemma 4.5 yields an algorithm to write down a strong Schreier basis
of $\Lambda^{*}/L_{\lambda}$ which generates a direct complement $V$ of
$L_{\lambda}$ with respect to $\Lambda^{*}$. Then one can use the projection
of $\Lambda^{*}$ onto $V$ along $L_{\lambda}$ to write down a basis of
$L_{\lambda}$. This algorithm for getting bases of finitely-codimensional left
ideals plays a crucial role in finding polynomials similar to one with nonzero
constant term, for instance see Examples 5.2 and 5.3.

As a direct consequence of the proof to Lemma 4.3 we have

######  Corollary 4.6.

The ${\Lambda}$-module $V_{\lambda}$ and the $\Lambda^{*}$-module
${\Lambda^{*}}/{L_{\lambda}}$ are isomorphic with respect to the canonical
ring isomorphism $\Lambda^{*}\cong\Lambda$. Moreover, if $\lambda$ is linear,
then $\operatorname{codim}{L_{\lambda}}=1$ and $\mathbb{L}/\mathbb{L}\lambda$
is a simple $\mathbb{L}$-module whose endomorphism ring is $K$.

###### Proof.

It is straightforward to verify that the $K$-isomorphism between $V_{\lambda}$
and $\Lambda^{*}/L_{\lambda}$ given in Lemma 4.5 is $\Lambda$-linear with
respect to the canonical ring isomorphism $\Lambda\cong\Lambda^{*}$. The last
statement follows by Corollary 3.8. ∎

By the results in Section 3, we immediately get the following

######  Corollary 4.7.

The $\mathbb{L}$-module $M_{\lambda}=\mathbb{L}/\mathbb{L}\lambda$ is finitely
presented and of finite length.

It is time to better understand why and how interrelations between
$\Lambda,\Lambda^{*}$ and the modules
$\mathbb{L}/\mathbb{L}\lambda=M_{\lambda}$, $V_{\lambda}$,
$\Lambda^{*}/L_{\lambda}$ are crucial for the aim of this work. First, as we
already mentioned, the rule $x_{i}\mapsto x^{*}_{i}$ induces both the natural
canonical algebra isomorphism and the canonical anti-isomorphism between
$\Lambda$ and $\Lambda^{*}$. The canonical isomorphism induces the module
isomorphism between $V_{\Lambda}$ and ${\Lambda^{*}}/L_{\lambda}$ as we
already mentioned in Remark 2.3 in the case of one variable . Moreover, thanks
to results in Section 3, the $\Lambda^{*}$-module $\Lambda^{*}/L_{\lambda}$
and the $\mathbb{L}$-module $\mathbb{L}/\mathbb{L}\lambda$ share important
features via localization, and also in terms of finiteness or dimensions. Thus
investigating the $\Lambda$-module $V_{\lambda}$ or the $\mathbb{L}$-module
$\mathbb{L}/\mathbb{L}\lambda$ leads to a deeper study of $\lambda$ and turns
out to be a direct, more appropriate way to study the factorizations of
$\lambda$ than what is proposed by Cohn in [7], [9]. In fact, it provides a
way to find polynomials similar to $\lambda$, to check irreducibility to
$\lambda$ as well as its factorization. The canonical anti-isomorphism induces
the symmetry between left and right modules over free algebras, so that
results on factorization and divisibility we will get on the right can be
stated also on the left. As a first application of these ideas we have an
easily used criterion for divisibility.

######  Theorem 4.8.

If $\lambda$ is a comonic polynomial and $\gamma$ is an arbitrary polynomial
in $\Lambda$, then $\lambda$ is a right factor of $\gamma$ if and only if
$b\star_{\lambda}\gamma=0$ for all monomials $b\in\bf B$ of degree $|\gamma|$.
There is an algorithm to find the left quotient $\delta$ of $\gamma$
satsisfying $\gamma=\delta\lambda$. Symmetrically, $\lambda$ is a left factor
of $\gamma$ if and only if $\gamma\star_{\lambda}b=0$ for all monomials
$b\in\bf B$ of degree $|\gamma|$ and one can find successively the right
quotient $\delta$ of $\gamma$ satisfying $\gamma=\lambda\delta$.

###### Proof.

By Lemma 4.1, $\lambda$ is a right factor of $\gamma$ if and only if
$\gamma\in\mathbb{L}\lambda$. In view of the equality (8),
$\gamma\in\mathbb{L}\lambda$ holds if and only if
$b^{*}\gamma\in\mathbb{L}\lambda$ for all monomials $b\in\bf B$ of degree $l$,
where $l$ is an arbitrary positive integer. However $b^{*}\gamma$ is clearly
congruent to ${}^{\tau}b\star_{\lambda}\gamma$ modulo $\mathbb{L}\lambda$, as
it follows by Lemma 4.5 and the obvious induction. Therefore $\lambda$ is a
right factor of $\gamma$ if and only if
$b\star_{\lambda}\gamma\in\mathbb{L}\lambda$ for all monomials $b$ with
$|b|=|\gamma|$. Since the degree of $b\star_{\lambda}\gamma$ is at most
$|\lambda|-1$, Lemma 4.1 implies $b\star_{\lambda}\gamma=0$ if $|b|=|\gamma|$.
On the other hand, the equality $\gamma=\delta\lambda$ implies
$x_{i}\star_{\lambda}\gamma=\gamma_{i}\lambda$ as we have already seen. This
reduces the computation of the left quotient $\delta$ of $\gamma$ of one of
lower degree. Thus the proof is complete. ∎

The following theorem is the key tool in proving Theorem 2.3 and it is, in
fact, an algorithm to find a greatest common divisor of two comonic
polynomials in $\Lambda$.

######  Theorem 4.9.

Let $\gamma$ and $\delta$ be comonic polynomials in $\Lambda$. Then there is a
comonic polynomial $\lambda\in\Lambda$ such that
$\mathbb{L}\delta+\mathbb{L}\gamma=\mathbb{L}\lambda$.

###### Proof.

Without loss of generality one can assume that $\delta$ and $\gamma$ are
proper polynomials with $|\delta|\leq|\gamma|=m$ and none of them is contained
in the left ideal of $\mathbb{L}$ generated by the other. We use induction on
$m$. The claim is obvious in view of Corollary 4.6 if $m=1$. So assume $m>1$
and put $\mathfrak{L}=\mathbb{L}\delta+\mathbb{L}\gamma$. If
$\mathfrak{L}=\mathbb{L}$, one takes $\lambda=1$. Hence one can assume
$\mathfrak{L}\neq\mathbb{L}$. Since $\gamma$ does not belong to
$\mathbb{L}\delta$, by Theorem 4.8 there exits a monomial $b_{1}\in\bf B$ with
$|b_{1}|=\deg\gamma$ such that $0\neq b_{1}\star_{\delta}\gamma=\delta_{1}\in
V_{\delta}$ and $|\delta_{1}|\leq|\delta|-1$. Since
$\delta_{1}=b_{1}\star_{\delta}\gamma\in b^{*}_{1}\delta+\mathbb{L}\delta$ one
has
$\mathbb{L}\delta_{1}\subseteq\mathbb{L}\gamma+\mathbb{L}\delta=\mathfrak{L}$,
and so $\delta_{1}\in\mathfrak{L}$. Hence we conclude
$\mathbb{L}\delta_{1}+\mathbb{L}\delta\leq\mathfrak{L}$. It is possible that
$\delta_{1}$ is a polynomial without constant. However, by Lemma 3.10 one can
replace $\delta_{1}$ by a comonic polynomial in
$\mathbb{L}\delta_{1}\cap\Lambda$. Thus one can assume $\delta_{1}$ comonic.
If $\gamma\notin\Lambda\delta_{1}$, then one obtains in the same manner $0\neq
b_{2}\star_{\delta_{1}}\gamma=\delta_{2}\in V_{\delta_{1}}\cap\mathfrak{L}$
comonic, $|\delta_{2}|\leq|\delta_{1}|-1\leq|\delta|-2$ and
$\mathbb{L}\delta_{2}+\mathbb{L}\delta\leq\mathfrak{L}$. If
$\gamma\notin\mathbb{L}\delta_{2}$ we continue this process and after at most
$|\gamma|-1$ steps we find $\mu\in\Lambda\cap\mathfrak{L}$ with constant such
that $|\mu|\leq|\delta|-1$ with $\gamma\in\Lambda\mu$, so that
$\mathbb{L}\mu+\mathbb{L}\delta=\mathfrak{L}$. If $|\delta|\leq m-1$, then the
theorem follows from the induction hypothesis. If $|\delta|=m$, then changing
the role of $\delta$ with $\mu$ and $\gamma$ with $\delta$ we will obtain
again the claim after repeating the above argument. Therefore the proof is
complete. ∎

Thanks to Theorem 4.9, one obtains immediately the first claim of Theorem 2.3
describing comonic irreducible polynomials in terms of simple modules, as for
one variable.

######  Corollary 4.10.

Let $\lambda$ be a comonic polynomial in $\Lambda$.

  1. (1)

$\lambda$ is irreducible if and only if $\mathbb{L}/\mathbb{L}\lambda$ is a
simple left $\mathbb{L}$-module.

  2. (2)

$\lambda$ is irreducible if and only if $\Lambda^{*}/L_{\lambda}$ is a simple
$\Lambda^{*}$-module.

  3. (3)

$\lambda$ is irreducible if and only if $V_{\lambda}$ is a simple
$\Lambda$-module.

In particular, if $\lambda$ is irreducible, then
$D=\operatorname{End}(M_{\lambda})\cong\operatorname{End}(\Lambda^{*}/L_{\lambda})\cong\operatorname{End}(V_{\lambda})$
is a finite-dimensional division $K$-algebra, hence $D\cong K$ if $K$ is
algebraically closed.

###### Proof.

$1)$ Assume $\lambda$ irreducible. In order to show that $\mathbb{L}\lambda$
is maximal we have to see that $\mathbb{L}\lambda+\mathbb{L}l=\mathbb{L}$ for
any $l\in\mathbb{L}$. By Lemma 4.3 we can reduce to show that
$\mathbb{L}\lambda+\mathbb{L}\alpha=\mathbb{L}$ for any $\alpha\in\Lambda$. By
Lemma 3.10 we can assume $\alpha$ comonic, and so the result follow from
Theorem 4.9.

Conversely, if $\lambda$ is reducible, then $\lambda=\gamma_{1}\gamma_{2}$,
with $\gamma_{i}$ comonic and of degrees strictly less than $\lambda$. Hence
$\mathbb{L}\lambda\leq\mathbb{L}\gamma_{2}\leq\mathbb{L}$ and, by Corollary
4.2, the inclusions are proper since $\gamma_{2}\notin\Lambda\lambda$. So, if
$\mathbb{L}\lambda$ is a maximal left ideal, then $f$ has to be irreducible.

$2)$ Assume $\lambda$ irreducibile. Let $J$ be any proper left ideal of
$\Lambda^{*}$ strictly containing $L_{\lambda}$. Then $\mathbb{L}J=\mathbb{L}$
holds because $\mathbb{L}\lambda$ is a maximal left ideal of $\mathbb{L}$ by
the previous point. Consequently, Corollary 3.8(3) shows that $J$ contains a
power $(I^{*})^{l}$ for some $l\in\mathbb{N}$. This means that
${\overline{J}}=J/(I^{*})^{l}$ is a left ideal of $A=\Lambda^{*}/(I^{*})^{l}$
which is a finite-dimensional local ring. However, since by assumption
$\lambda$ is comonic, its constant term is $1$ and hence for a maximal
monomial $b$ of $\lambda$ we get $b^{*}\lambda=b^{*}+\cdots+k_{b}$ where
$k_{b}$ is the non-zero coefficient of $b$ in $\lambda$. So $b^{*}\lambda$ is
in $L_{\lambda}$ and its image in $A$ is a unit of $A$. Thus
${\overline{J}}=A$ holds. This implies $J=\Lambda^{*}$, proving that
$L_{\lambda}$ is maximal in $\Lambda^{*}$.

Conversely, if $L_{\lambda}$ is maximal in $\Lambda^{*}$, then
$\mathbb{L}\lambda$ is maximal by results in Section 3 and Corollary 4.2.

$3)$ This follows immediately since $V_{\delta}$ and $\Lambda^{*}/L_{\delta}$
are isomorphic with respect to the $\ast$-omitting ring isomorphism

∎

We note that when $n=1$, then the field $D$ of Corollary 4.10 is precisely the
field extension $K[x]/K[x]\lambda$ of $K$ by $\lambda$.

######  Corollary 4.11.

If a left ideal $\mathfrak{L}$ of $\mathbb{L}$ contains some
$\gamma\in\Lambda$, then $\mathfrak{L}=\mathbb{L}\lambda$ for some comonic
$\lambda\in\Lambda$. In particular, it has finite colength and
$\operatorname{End}(\mathbb{L}/\mathfrak{L})$ is finite dimensional.

###### Proof.

By assumption $\mathfrak{L}\cap\Lambda^{*}$ contains
$L_{\gamma}=\mathbb{L}\gamma\cap\Lambda^{*}$. By Lemma 3.10 we can assume
$\gamma$ comonic. By Lemma 4.5, applying the modular law we get
$\mathfrak{L}\cap(\Lambda^{*}+\mathbb{L}\gamma)=\mathfrak{L}\cap(V_{\gamma}+\mathbb{L}\gamma)=W+\mathbb{L}\gamma$
where $W=V_{\gamma}\cap\mathfrak{L}$. Hence Corollary 3.8(1) yields
$\mathfrak{L}=\mathbb{L}\gamma+\mathbb{L}W$. Since $W$ is a finite-dimensional
subspace of $\Lambda$ the proof is immediately completed by successive
applications of Theorem 4.9. ∎

The previous result has a nice important consequence.

######  Corollary 4.12.

If $\lambda\in\Lambda$ is comonic, then there is a bijection between right
factors of $\lambda$ (up to a scalar) and left ideals of $\mathbb{L}$
containing $\lambda$. Hence the submodule lattice of
$\mathbb{L}/\mathbb{L}\lambda$ is isomorphic to the distributive lattice ${\bf
L}(\Lambda\lambda,\Lambda)$ of principal left ideals of $\Lambda$ containing
$\lambda$.

###### Proof.

If $\mathfrak{L}$ is a left ideal of $\mathbb{L}$ containing
$\mathbb{L}\lambda$, then by Corollary 4.11 there exists $\delta\in\Lambda$
such that $\mathfrak{L}=\mathbb{L}\delta$, so that $\delta$ is a right divisor
of $\lambda$. ∎

Note that the distributivity of the lattice ${\bf L}(\Lambda\lambda,\Lambda)$
is a deep result of Cohn [7, Theorem 4.3.3].

The next corollary shows that irreducible polynomials without constant
generate $\mathbb{L}$.

######  Corollary 4.13.

If $\pi\in\Lambda$ is irreducible without constant, then
$\mathbb{L}\pi=\mathbb{L}$. In particular, $\mathbb{L}\lambda=\mathbb{L}$ if
$\lambda$ is a product of irreducible polynomials without constant.

###### Proof.

Let $\pi\in I$ be irreducible. By Corollary 4.11,
$\mathbb{L}\pi=\mathbb{L}\lambda$ for a comonic polynomial $\lambda\notin I$.
By Lemma 4.1 we get that $\pi$ is a multiple of $\lambda$. Thus the
irreducibility of $\pi$ implies $\lambda=1$. ∎

As a next step to the proof Theorem 2.3, we show the equivalence of the
similarity over $\mathbb{L}$ and $\Lambda$.

######  Corollary 4.14.

Let $\lambda\in\Lambda$ be comonic and $\lambda=\delta_{1}\cdots\delta_{m}$ be
a factorization of $\lambda$ into irreducible factors. Then:

  1. (1)

The $\mathbb{L}$-module $M_{\lambda}=\mathbb{L}/\mathbb{L}\lambda$ has a
composition series with composition factors the simple modules
$\mathbb{L}/\mathbb{L}\delta_{i}$, for $i=1,\cdots,m$.

  2. (2)

The composition factors of $\Lambda$-module $V_{\lambda}$ contains the simple
modules $V_{\delta_{i}}$, for $i=1,\cdots,m$.

  3. (3)

Given a monic polynomial $\gamma\in\Lambda$, $\gamma$ and $\lambda$ are
similar over $\Lambda$ if and only if they are similar over $\mathbb{L}$ .

######  Remark 4.2.

Together with Corollary 3.8(3), Claim (3) of Corollary 4.14 says that
$V_{\lambda}$ has length $\geq m$ and all the other possible composition
factors are isomorphic to the trivial module $K$. We will show that there are
no more composition factors and so $|V_{\lambda}|=m$ in Corollary 4.15.

###### Proof.

$(1)$ Since $\lambda$ is comonic, all the $\delta_{i}$’s have constant term
and we can assume they are comonic. By Lemma 4.1, $\mathbb{L}$ admits a
descending chain
$\mathbb{L}\supset\mathbb{L}\delta_{l}\supset\mathbb{L}\delta_{l-1}\delta_{l}\supset\cdots\supset\mathbb{L}\delta_{2}\cdots\delta_{l}\supset\mathbb{L}\lambda$
of pairwise different left ideals. For $i=1,\cdots,m$ let
$\gamma_{i}=\delta_{i}\cdots\delta_{l}$, so
$\gamma_{1}=\lambda,\,\gamma_{l}=\delta_{l}$ and $\gamma_{l+1}=1$. For any
$i=1,\cdots,m$ let $T_{i}=\mathbb{L}\gamma_{i+1}/\mathbb{L}\gamma_{i}$. Then
$T_{i}$ is a non-zero left $\mathbb{L}$-module and if we denote by
$\alpha_{i+1}$ the image of $\gamma_{i+1}$ in $T_{i}$, then
$T_{i}=\mathbb{L}a_{i+1}$ and $\delta_{i}a_{i+1}=0$ holds. So we get
$T_{i}\cong\mathbb{L}/\mathbb{L}{\delta_{i}}$, and
$\mathbb{L}/\mathbb{L}\delta_{i}$ is simple since $\delta_{i}$ is irreducible.

$(2)$ Put $L_{i}=\Lambda^{*}\cap\mathbb{L}\gamma_{i}\,(i=1,\cdots,l)$. By
Corollary 4.6 $L_{l}$ is a maximal left ideal of $\Lambda^{*}$ and
$V_{\delta_{l}}\cong\Lambda^{*}/L_{l}$ is a simple composition factor of
$V_{\lambda}$. From Claim $(1)$ and by Corollary 3.8 together with Corollary
4.6 again, for each $i=1,\cdots,l-1$ the $\Lambda^{*}$-module $L_{i+1}/L_{i}$
has one composition factor isomorphic to $V_{i}$. Furthermore by the comments
after Corollary 3.8 all possible other composition factors of all
$L_{i+1}/L_{i}\,(i=1,\cdots,l-1)$ are the trivial $\Lambda^{*}$-module $K$.
Consequently, Claim $(2)$ is verified.

$(3)$ By Proposition 3.3.4 [7] the similarity of $\gamma$ and $\lambda$ over
$\Lambda$ implies an equality $\gamma\alpha=\beta\lambda$ for some
$\alpha,\beta\in\Lambda$ such that $\gamma,\beta$ have no common left factor
while $\alpha,\lambda$ have no common right factor. By
$\Lambda\alpha\cap\Lambda\lambda\neq 0$ one sees that
$\Lambda\alpha+\Lambda\lambda=\Lambda\mu$ for some $\mu\in\Lambda$ whence
$\mu$ must be a unit because $\alpha$ and $\lambda$ have no common right
factor. Consequently, $\Lambda\alpha+\Lambda\lambda=\Lambda$ and hence the
endomorphism $1\mapsto\alpha$ of ${}_{\mathbb{L}}\mathbb{L}$ induces a
surjective homomorphism $f:{M_{\gamma}}\rightarrow{M_{\lambda}}$ by the
equality
$(\mathbb{L}\gamma)\alpha=\mathbb{L}(\gamma\alpha)=\mathbb{L}(\beta\lambda)\subseteq\mathbb{L}\lambda$.
Since both $M_{\gamma}$ and $M_{\lambda}$ have finite length one obtains
$\lg({M_{\lambda}})\leq\lg({M_{\gamma}})$ whence
$\lg({M_{\lambda}})=\lg({M_{\gamma}})$ by symmetry. This shows that $f$ is an
isomorphism and so $\gamma$ and $\lambda$ are similar over $\mathbb{L}$. On
the other hand, an isomorphism between $\mathbb{L}$-modules $M_{\gamma}$ and
$M_{\lambda}$ is clearly a $\Lambda$-isomorphism between
$\Lambda/\Lambda\gamma$ and $\Lambda/\Lambda\lambda$ in view of Lemma 4.3,
whence the similarity over $\mathbb{L}$ implies the one over $\Lambda$. This
completes the proof. ∎

######  Remark 4.3.

It is important to see clearly the role of $\lambda$ and $V_{\lambda}$ even in
the case $n=1$. As we have already seen in Remark 2.3, for a comonic
polynomial $\lambda$ and its reciprocal polynomial $\gamma$, the module
$V_{\lambda}\cong K[x]/K[x]\gamma$ is not in general isomorphic to
$K[x]/K[x]\lambda$. Hence passing from $\Lambda=K[x]$ to
$\mathbb{L}=K[x,x^{-1}]$ we have the parallel appearance of both $\lambda$ and
$\gamma$ but they are still not similar. To make them similar, or more
generally to make all irreducible polynomials defining the same field
extension of $K$ similar, which is reasonable, one needs a more general notion
of similarity of linear transformations. By passing to the corresponding
companion matrix, it is proved in [1, Proposition 1.6 (2)] that they, i.e.,
the associated companion matrices, are indeed similar matrices.

By Corollaries 4.10 and 4.14, the proof of Theorem 2.3 is complete by the
verification of

######  Corollary 4.15.

Two comonic polynomials $\gamma$ and $\lambda\in\Lambda$ are similar if and
only if the $\Lambda$-modules $V_{\gamma}$ and $V_{\lambda}$ are isomorphic.

###### Proof.

As usual, let $L_{\gamma}=\Lambda^{*}\cap\mathbb{L}\gamma$ and
$L_{\lambda}=\Lambda^{*}\cap\mathbb{L}\lambda$. By Corollary 4.6, we show that
$\gamma,\lambda$ are similar if and only if
$N_{\gamma}=\Lambda^{*}/L_{\gamma},N_{\lambda}=\Lambda^{*}/L_{\lambda}$ are
isomorphic $\Lambda^{*}$-modules. Since the sufficiency is immediate, we need
only to show the necessity, that is, $N_{\gamma}$ and $N_{\lambda}$ are
isomorphic if $M_{\gamma}$ and $M_{\lambda}$ are isomorphic
$\mathbb{L}$-modules. In view of Corollary 3.9 it is enough to see that the
length of $N_{\lambda}$ is $l$ if $\lambda=\pi_{1}\cdots\pi_{l}$ is a product
of $l$ irreducible polynomials. We use induction on $l$. The claim for $l=1$
is exactly Corollary 4.10. Assume the claim for $l-1\geq 1$ and put
$\lambda=\pi_{1}\cdots\pi_{l}=\pi_{1}{\overline{\lambda}}$. Denoting by
$L_{\overline{\lambda}}=\Lambda^{*}\cap\mathbb{L}{\overline{\lambda}}$, by
induction $N_{\overline{\lambda}}=\Lambda^{*}/L_{\overline{\lambda}}$ has
length $l-1$ . To complete the induction it is enough to see that
$L_{\overline{\lambda}}/L_{\lambda}$ is a simple $\Lambda^{*}$-module. First
notice that $L_{\overline{\lambda}}/L_{\lambda}$ is not the zero module, since
$L_{\lambda}\neq L{\overline{\lambda}}$ by Corollary 3.8. Consider the right
multiplication by $\overline{\lambda}$ in $\mathbb{L}$. It induces an
isomorphism between $\mathbb{L}/\mathbb{L}\pi_{1}$ and
$\mathbb{L}\overline{\lambda}/\mathbb{L}\lambda$. Thanks to Corollary 3.9 and
Corollary 4.10, we get an isomorphism between the simple module
$\Lambda^{*}/L_{\pi_{1}}$ and $L_{\overline{\lambda}}/L_{\lambda}$, from which
we deduce that there exists a $y\in L_{\overline{\lambda}}$ such that its
annihilator is $L_{\pi_{1}}$. Assume now $L_{\overline{\lambda}}/L_{\lambda}$
is not simple and let us conclude obtaining a contradiction. Indeed, for any
ideal $L$ properly between $L_{\lambda}$ and $L_{\overline{\lambda}}$,
$\mathbb{L}L=\mathbb{L}\overline{\lambda}$ by Corollary 3.8. Since
$L_{\overline{\lambda}}/L_{\lambda}$ is finite dimensional, this means that
its annihilator ideal contains a power $(I^{*})^{m}$ for a suitable $m>0$.
Consequently, $(I^{*})^{m}y\in L_{\lambda}$, so that $(I^{*})^{m}\leq
L_{\pi_{1}}$. Then, arguing as in Corollary 4.10(2), in the local ring
$\Lambda^{*}/(I^{*})^{m}$ the left ideal $L_{\pi_{1}}/(I^{*})^{m}$ contains a
unit and hence $\Lambda^{*}=L_{\pi_{1}}$. This is contradiction since
$\pi_{1}$ is comonic and irreducible, so we conclude the proof. ∎

We conclude this section emphasizing how our approach of considering
polynomials in $\Lambda$ as elements of $\mathbb{L}$ and studying their
associated modules could be an effective tool for the study of the arithmetic
of polynomials in noncommutating variables. First of all, given any finitely
many arbitrary polynomials $\lambda_{1},\cdots,\lambda_{l}\in K\langle
x_{1},\cdots x_{n}\rangle$, we can define their _generalized greatest common
divisor_ , shortly _greatest common divisor_ , as the polynomial with constant
$\delta\in K\langle x_{1},\cdots,x_{n}\rangle$ which is a generator of the
left ideal $\sum\limits_{i=1}^{l}\mathbb{L}\lambda_{i}$. As an immediate
consequence of Corollary 4.11, this greatest common divisor $\delta$ exists
and is unique up to a scalar, even in the case that the ordinary least common
multiple of the $\lambda_{i}$’s is $0$. It is important to note the following:
while the greatest common divisor of a polynomial with constant is itself up
to a scalar, they are different in the case of a polynomial without constant,
since the greatest common divisor has constant by definition. For example, the
equality
$\mathbb{L}(x_{1}+x^{3}_{1}+x_{1}x^{2}_{2}+x_{2}+x_{2}x^{2}_{1}+x^{3}_{2})=\mathbb{L}(1+x^{2}_{1}+x^{2}_{2})$
shows by the definition that the greatest common divisor of the polynomial
$\lambda=x_{1}+x^{3}_{1}+x_{1}x^{2}_{2}+x_{2}+x_{2}x^{2}_{1}+x^{3}_{2}$ is the
comonic polynomial $1+x^{2}_{1}+x^{2}_{2}$. Theorem 4.9 offers also an
algorithm to find $\lambda$. Because of its importance, we rephrase this fact
together with Theorem 4.9 as

######  Theorem 4.16.

Given $\lambda$ and $\gamma$ monic polynomials in $\Lambda$, then there exists
a unique monic polynomial $\delta$ in $\Lambda$ such that $\delta$ is a right
divisor of $\lambda$ and $\gamma$, and any other common right divisor is a
right divisor of $\delta$.

For example, let $\lambda=1+x^{2}_{1}+x^{2}_{2}$. Then
$V_{\lambda}=K+Kx_{1}+Kx_{2}$ has dimension 3. It is easy to see that
$V_{\lambda}$ is a simple $\Lambda$-module with $x_{1}\star_{\lambda}1=-x_{1}$
and $x_{2}\star_{\lambda}1=-x_{2}$. Hence $\lambda$ is irreducible. Let
$\gamma=1+x_{1}+x^{2}_{1}$. Then
$x^{2}_{1}\star_{\lambda}\gamma=(x^{2}_{1}\star_{\lambda}1)+(x^{2}_{1}\star_{\lambda}x_{1})+(x^{2}_{1}\star_{\lambda}x^{2}_{1})=-1-x_{1}+1=-x_{1}\neq
0$ implies that the greates common divisor of $\lambda$ and $\gamma$ is 1. If
$\alpha=1+x^{2}_{1}+x_{2}+x_{2}x^{2}_{1}+x^{2}_{2}+x^{3}_{2}$, then one can
check routinely that $b\star_{\lambda}\alpha=0$ for all monomials $b$ in
$x_{1},x_{2}$ of degree 3. Hence $\lambda$ is the greatest common divisor of
$\alpha,\lambda$. Actually, $\alpha=(1+x_{2})\lambda$ holds.

Moreover one can solve the so-called ”membership problem” in $\Lambda$ by a
relatively simple and quick algorithm as an easy consequence of Theorem 4.8.
Namely, let $\gamma=1+\sum x_{i}\gamma_{i}=1+\underline{\gamma}$ be a comonic
polynomial in $\Lambda$ and $\lambda=k_{\lambda}+\sum x_{i}\lambda_{i}$ an
arbitrary polynomial of degree $m$. Let
$R_{\lambda}=\\{\delta_{b}=b\star_{\gamma}\lambda\neq 0\ \,\,|b\in\bf
B,\,|b|=|\lambda|\\}\subseteq V_{\gamma}$. According to Theorem 4.8 $\gamma$
is a right factor of $\lambda$ if and only if this set $R_{\lambda}$ is empty.
Hence $R_{\lambda}$ behaves like a set of ”generalized remainders of $\lambda$
at division on the left by $\gamma$. If $R_{\lambda}$ is not empty, then

$\mathbb{L}\gamma+\mathbb{L}\lambda=\mathbb{L}\gamma+\sum\limits_{\delta_{b}\in
R_{\lambda}}\delta_{b}$

and for $\delta_{\lambda}=\sum\limits_{\delta_{b}\in R_{\lambda}}b\delta_{b}$
one can find algorithmically $\rho_{\lambda}\in\Lambda$ such that
$\lambda=\rho_{\lambda}\gamma+\delta_{\lambda}$.

Hence we have an expression $\lambda=\rho_{\lambda}\gamma+\delta_{\lambda}$ as
a final result of dividing $\lambda$ by $\gamma$. We summarize these results
in the following theorem.

######  Theorem 4.17.

Let $\gamma\in\Lambda$ be a comonic polynomial and $\lambda\in\Lambda$ of
degree $m$. Then $\lambda\in\Lambda\gamma$ if and only if
$b\star_{\gamma}\lambda=0$ holds for every monomial $b\in\bf B$ of degree $m$.
Moreover, if $R_{\lambda}=\\{\delta_{b}=b\star_{\gamma}\lambda\neq
0\,|\,|b|=m\\}$, then there is a unique polynomial $\rho_{\lambda}\in\Lambda$
satisfying $\lambda=\rho_{\lambda}\gamma+\delta_{\lambda}$ where
$\delta_{\lambda}=\sum\limits_{\delta_{b}\in R_{\lambda}}b\delta_{b}$. By
symmetry, one has an algorithm to determine the right quotient
${}_{\lambda}\rho$ and the set of remainders
${{}_{\lambda}}R=\\{_{b}\delta=\lambda\star_{\gamma}b\neq 0\,|\,|b|=m\\}$
together with the expression
$\lambda=\gamma({{}_{\lambda}\rho})+_{\lambda}\delta$ where
${}_{\lambda}\delta=\sum\limits_{{}_{b}\delta\in{{}_{\lambda}}R}{{}_{b}\delta}b$.

######  Remarks 4.4.

  1. (1)

For the particular computation it is worth to note that it is not necessarily
to compute all $b\star_{\gamma}\lambda$ with $|b|=m=|\lambda|$. It is enough
to take such monomials $b$ of a minimal degree with $b\star_{\gamma}\lambda=0$
or $b\star_{\gamma}\lambda\in V_{\gamma}$, respectively. The constraint
$|b|=m$ is required only to grant that $b\star_{\gamma}\lambda$ is an element
of $V_{\gamma}$ when it is not zero.

  2. (2)

Although the ”membership problem” is solved, the division algorithm is
completely different from the case of polynomials in commutating variables
despite of the fact that it looks formally almost the same. First the
”generalized remainder” $\delta_{\lambda}$ by dividing $\lambda$ by $\gamma$
is, in general, not an element of $V_{\gamma}$. Secondly, the degree of
$\delta_{\lambda}$ can be arbitrary. In fact, we have no idea to describe the
set $\\{\delta_{\lambda}=\sum b\delta_{b}\,|\,\lambda\in\Lambda\\}$. It is
clear that every polynomial $\mu\in\Lambda$ with $|\mu|\leq|\gamma|$ can
appear as a remainder at a division by $\gamma$ if $\mu$ is not a scalar
multiple of $\gamma$ in case of equal degree. For example, if
$\gamma=1-x_{1}$, then $\mu=1-x_{2}\notin\Lambda\gamma$ but its ”strict
remainder” in our sense is $x^{*}_{1}\mu=x^{*}_{1}\equiv 1$ mod
$\mathbb{L}\gamma$. Even for $m=1-x^{l}_{1}x_{2}$ we have
$(x^{*}_{1})^{l}\mu=(x^{*}_{1})^{l}-x_{2}\equiv 1-x_{2}\equiv 1$ mod
$\mathbb{L}\gamma$. Therefore for the particular use it is reasonable to
determine only the set $R_{\lambda}$.

  3. (3)

To check the simplicity of $\mathbb{L}/\mathbb{L}\pi$ where $\pi$ has
constant, one needs compute $\mathbb{L}\pi+\mathbb{L}\alpha$ where $\alpha$
runs over all nonzero elements of $V_{\pi}$. This is a definitely harder,
exhaustive job than the memership problem. According to Theorem 4.9 it can be
verified in finitely many steps for each element $\alpha\in V_{\pi}$. Hence
the exercise can be carried out in finitely many steps when the base field $K$
is finite. If $K$ is infinite, then $V_{\pi}$ has unfortunately infinitely
many elements although it has a finite dimension. Consequently, finding right
factors and irreducible factorizations of a polynomial is more difficult job.

Finally, by Theorem 2.3 we know that, given $\lambda\in\Lambda$, its
irreducible decompositions are unique up to similarity. For some particular
polynomials we can improve this result obtaining the uniqueness of the
irreducible factorization in the classical sense.

######  Corollary 4.18.

If $\lambda=\pi_{1}\pi_{2}\cdots\pi_{l}$ is a product of polynomials $\pi_{i}$
similar to an irreduclible polynomial $\pi$, then this factorization of
$\lambda$ is unique up to a scalar, that is, if
$\lambda=\delta_{1}\cdots\delta_{l}$ is another prime factorization of
$\lambda$, then each $\delta_{j}$ is a scalar multiple of $\pi_{j}$ for
$j=1,\cdots,l$.

###### Proof.

Since $V_{\lambda}$ has the distributive submodule lattice by Corollary 4.12,
one obtains immediately that $V_{\lambda}$ and $\mathbb{L}/\mathbb{L}\lambda$
are uniserial modules, that is, their submodules form a unique finite chain if
all of their simple subfactors are isomorphic. We then conclude the proof
recalling that if $\mathbb{L}\delta=\mathbb{L}\gamma$ for $\gamma$ and
$\delta$ in $\Lambda$ with constant, then by Corollary 4.2 we get that
$\delta$ is a scalar multiple of $\gamma$ ∎

##  5\. Applications to factorizations of polynomials

In the next two sections we give some hints for future directions that can be
explored. As shown by results and techniques exposed in the previous sections,
free algebras and Leavitt algebras are deeply interrelated. Facts in free
algebras are used to investigate Leavitt algebras and vice versa. However,
there is sometime the need to treat them separately in their own language, to
better understand their applications and their relevance in the setting we
choose to focus on. For instance, from this point of view, results of Section
4 provide an algorithm to find factorizations of $\lambda\in\Lambda$ starting
from the right. By symmetry or by using the canonical ∗ involution of
$\mathbb{L}$ and the anti-isomorphism between $\Lambda$ and $\Lambda^{*}$,
they yield also a way to find factorizations of $\lambda$ starting from the
left. Moreover, given a comonic polynomial $\lambda\in\Lambda$, instead of
$\mathbb{L}/\mathbb{L}\lambda$ and $\Lambda^{*}/L$ one can use $V_{\lambda}$
and the algorithm described in proof of Theorem 4.9 to decide irreducibility,
similarity, factorization, and more generally, to do standard arithmetic of
non-commutative polynomials. Since $\operatorname{End}_{\Lambda}(V_{\lambda})$
is a finite-dimensional division $K$-algebra and $V_{\lambda}$ has a quite
easily handled $K$-basis it is interesting to closely study
$\operatorname{End}_{\Lambda}(V_{\lambda})$. The first question arising is
which non-algebraically closed field $K$ has a comonic irreducible polynomial
$\lambda\in\Lambda$ such that $\operatorname{End}_{\Lambda}(V_{\lambda})$ is
non-commutative. The next one is to determine non-commutative finite-
dimensional division $K$-algebras which are not representable as
$\operatorname{End}_{\Lambda}(V_{\lambda})$ of irreducible polynomials
$\lambda$, and so on.

###  5.1. Companions to a polynomial: similarity, remainders etc.

Polynomials in not necessarily commuting variables form perhaps the most
classical and quite hard subject of algebra. By definition, a polynomial
$\lambda$ is dependent only on variables appearing in its expression. Hence it
does not depend on the carrying free algebra $\Lambda$, which may have even
infinitely many variables. From this point of view the finite-dimensional
space $V_{\lambda}$ seems to be a good tool, a companion data structure as a
simple combinatorial invariant. It plays the role of the module of remainders
and it is easy to describe. As we have already noticed, it can be used to
determine whether $\lambda$ is irreducible and then to find, at least
theoretically, its factorizations together with other related structures like
its endomorphism ring and similar polynomials. By definition, to find a
polynomial $\gamma$ similar to $\lambda$ one needs first to compute the factor
module
$M{{}_{\lambda}}=\Lambda/\Lambda\lambda\cong\mathbb{L}/\mathbb{L}\lambda$,
then find another generator $m$ of $M_{\lambda}$ such that
$\operatorname{ann}_{\mathbb{L}}m=\\{r\in\mathbb{L}\,|\,rm=0\in
M_{\lambda}\\}\neq\mathbb{L}\lambda$, and finally look for a polynomial
$\gamma\in\Lambda$ such that
$\operatorname{ann}_{\mathbb{L}}m=\mathbb{L}\gamma$. Hence $\gamma$ is similar
to $\lambda$.

However, this direct approach with a few exception is not feasible.
Fortunately, for a comonic polynomial $\lambda$ one can use $V_{\lambda}$ to
provide a way to compute all polynomials similar to $\lambda$, that is, to
determine the similarity class $\mathfrak{S}\lambda$ of $\lambda$. Since
$V_{\lambda}$ is finite-dimensional, it is theoretically feasible to find all
the generators of $V_{\lambda}$ and then to any generator $m\in V_{\lambda}$
the unique comonic polynomial $\gamma_{m}$ similar to $\lambda$ satisfying
$\mathbb{L}\gamma_{m}=\mathbb{L}L_{m}$, where $L_{m}$ is the image of
$\operatorname{ann}_{\Lambda}m=\\{\mu\in\Lambda\,\,|\,\,\mu\star m=0\\}$ under
the canonical isomorphism from $\Lambda$ onto $\Lambda^{*}$, provided that
such a $\gamma_{m}$ exists. These results are presented in the next claim.

######  Proposition 5.1.

Let $\lambda\in\Lambda$ be a comonic polynomial. Then, up to non-zero scalars
and automorphisms of $V_{\lambda}$, every comonic polynomial $\gamma$ similar
to $\lambda$ determines uniquely a generator $m_{\gamma}$ of $V_{\lambda}$.

###### Proof.

By Corollary 4.15, the $\Lambda$-modules $V_{\gamma}$ and $V_{\lambda}$ are
isomorphic, whence $\gamma$ determines uniquely a generator $m_{\gamma}$ of
$V_{\lambda}$ up to automorphisms of $V_{\lambda}$, where the automorphism
group of $V_{\lambda}$ is isomorphic to that of the $\mathbb{L}$-module
${\mathbb{L}}/{\mathbb{L}\lambda}$. ∎

######  Remark 5.1.

This assertion can be considered as a starting step describing similarity
classes of polynomials. As the first trivial application one sees immediately
that the similarity class of a linear polynomial with constant has exactly one
element up to scalars. In view of [7, 1st ed. Proposition 3.3.4] similar
polynomials provide a method to find different not permutated factorizations.
In particular, until Proposition 5.1 there is no practical method finding
similar polynomials. A thorough study of similar polynomials will be a subject
of forthcoming works.

######  Example 5.2.

In commutative algebra it is well-known that the maximal ideals of the
polynomial algebra $A=K[x_{1},\cdots,x_{n}]$ of codimension 1, are the ideals
$\sum\limits_{i=1}^{n}A(x_{i}-k_{i})$ where
${\mathfrak{k}}=(k_{1},\cdots,k_{n})\in K^{n}$ runs over the points in the
$n$-dimensional affine space $K^{n}$. For free associative algebra
$\Lambda_{n}$, by passing to the corresponding factors, non-trivial one-
dimensional $\Lambda_{n}$-modules are clearly parameterized by the polynomials
$1-\sum\limits_{k_{i}\neq 0}k_{i}x_{i}$ for points
${\mathfrak{k}}=(k_{1},\cdots,k_{n})\in K^{n}$.

######  Example 5.3.

Let $\lambda=1+x_{1}x_{2}\in\Lambda=\Lambda_{2}$. We compute the similarity
class of $\lambda$. By definition the 2-dimensional $K$-space
$V_{\lambda}=\\{1,x_{2}\\}$ is a $\Lambda_{2}$-module with respect to
$x_{1}\star 1=-x_{2}$ and $x_{2}\star 1=0$. One can see that $\lambda$ is
irreducible and $\\{1+kx_{2},\,x_{2}\,\,|\,\,k\in K\\}$ is the set of all
generators of $V_{\lambda}$. Put $v=1\in V_{\lambda}$, then
$\operatorname{ann}_{\Lambda}v$ is a $\Lambda_{2}$-module freely generated by
$\\{x_{2},x^{2}_{1},1+x_{2}x_{1}\\}$ whence $L_{v}$ is freely generated by
$\\{x^{*}_{2},(x^{*}_{1})^{2},\,1+x^{*}_{2}x^{*}_{1}\\}$. Consequently,
$\mathbb{L}L_{v}=\mathbb{L}\lambda$ because
$\lambda=x_{2}x^{*}_{2}+x^{2}_{1}(x^{*}_{1})^{2}+x_{1}x_{2}(1+x^{*}_{2}x^{*}_{1})$.
For $w=x_{2}\in V_{\lambda}$ we see that $\operatorname{ann}w$ is freely
generated by $\\{x_{1},x^{2}_{2},1+x_{1}x_{2}\\}$ and so by symmetry
$\mathbb{L}L_{w}=\mathbb{L}(1+x_{2}x_{1})$, that is, $\lambda$ is similar to
$1+x_{2}x_{1}$.

To describe the similarity class of $\lambda$, for any $0\neq k\in K$ let
$v_{k}=1+kx_{2}$ be an arbitrary generator of $V_{\lambda}$ different from $1$
and $x_{2}$. We intend to look for the unique comonic polynomial
$\gamma_{k}\in\Lambda$ such that $\mathbb{L}\gamma_{k}=\mathbb{L}L_{k}$, where
$L_{k}$ is the image of $\operatorname{ann}v_{k}$ in $\Lambda^{*}$ with
respect to the canonical isomorphism from $\Lambda$ onto $\Lambda^{*}$. If
such a $\gamma_{k}$ exists, since $V_{\gamma_{k}}$ contains 1 and has
$K$-dimension 2, $\gamma_{k}$ must have degree 2 in view of the definition of
$V_{\gamma_{k}}$. Moreover, $V_{\gamma_{k}}$ has a $K$-basis
$\\{1,u=k_{1}x_{1}+k_{2}x_{2}\neq 0\\}$ whence at least one of the
$k_{1},k_{2}\in K$ must be not zero. Furthermore
$\gamma_{k}=1+x_{1}\gamma_{1}+x_{2}\gamma_{2}$ where
$\gamma_{1}=l_{1}+l_{2}u,\gamma_{2}=m_{1}+m_{2}u$ with appropriate
coefficients $l_{1},l_{2}$ and $m_{1},m_{2}$ from $K$ and at least one of the
coefficients $l_{2},m_{2}$ is not zero. Expressing

$\gamma_{k}=\gamma=1+x_{1}[l_{1}+l_{2}(k_{1}x_{1}+k_{2}x_{2})]+x_{2}[m_{1}+m_{2}(k_{1}x_{1}+k_{2}x_{2})]$

we have

$\gamma_{k}=\gamma=1+l_{1}x_{1}+k_{1}l_{2}x^{2}_{1}+k_{2}l_{2}x_{1}x_{2}+m_{1}x_{2}+k_{1}m_{2}x_{2}x_{1}+k_{2}m_{2}x^{2}_{2}.$

Hence $\gamma$ generates $\mathbb{L}L_{k}$ if
$\Lambda^{*}\cap\mathbb{L}\gamma\subseteq L_{k}$, i.e.,
$\\{x^{*}_{i}x^{*}_{j}\gamma\,\,|\,i,j=1,2\\}\subseteq L_{k}$. In other words,
we have to see whether

$\\{(x^{*}_{1})^{2}+l_{1}x^{*}_{1}+l_{1}l_{2},(x^{*}_{2})^{2}+m_{1}x^{*}_{2}+m_{2}k_{2},x^{*}_{1}x^{*}_{2}+m_{1}x^{*}_{1}+k_{1}m_{2},x^{*}_{2}x^{*}_{1}+l_{1}x^{*}_{2}+k_{2}l_{2}\\}$

is contained in $L_{k}$. This means whether

$\\{x^{2}_{1}+l_{1}x_{1}+l_{1}l_{2},x^{2}_{2}+m_{1}x_{2}+m_{2}k_{2},x_{1}x_{2}+m_{1}x_{1}+k_{1}m_{2},x_{2}x_{1}+l_{1}x_{2}+k_{2}l_{2}\\}$

annihilates $v_{k}\in V_{\lambda}$. By $u\star
v_{k}=(k_{1}x_{1}+k_{2}x_{2})\star(1+kx_{2})=-k_{1}x_{2}+kk_{2}$, the equality
$0=(x^{2}_{1}+l_{1}x_{1}+l_{2}l_{1})\star v_{k}=(-l_{1}+k)x_{2}+l_{1}l_{2}$
implies $l_{1}=k\neq 0$ and $l_{2}=0$. Furthermore
$0=(x^{2}_{2}+m_{1}x_{2}+m_{2}k_{2})\star
v_{k}=m_{1}k+m_{2}k_{2}+kk_{2}m_{2}x_{2}$ implies $k_{2}m_{2}=0$ whence
$m_{1}k=0$ and so $m_{1}=0$ follows. Consequently,
$0=(x_{1}x_{2}+m_{1}x_{1}+k_{1}m_{2})\star
v_{k}=k_{1}m_{2}+(kk_{1}m_{2}-k)x_{2}$ implies $k=0$ which contradicts the
assumption $k\neq 0$. This shows that $\mathbb{L}L_{k}$ cannot be generated by
a polynomial from $\Lambda$ and so the similarity class of $\lambda$ has just
two comonic elements, namely, $\lambda$ and $1+x_{2}x_{1}$.

This example shows two important facts. First not every principal left ideal
of $\mathbb{L}$ whose factor module is isomorphic to some $V_{\lambda}$ of a
polynomial $\lambda\in\Lambda$ with constant (consider for instance $L_{k}$ as
above), has a generator which is a polynomial with constant from $\Lambda$.
Consequently, not every finite-codimensional maximal left ideal of
$\Lambda^{*}$ can be represented as
$V_{\lambda}=\Lambda^{*}\cap\mathbb{L}\lambda$ for an appropriate comonic
polynomial $\lambda$ in $\Lambda$. The second fact is that it provides a way
to determine the similarity class of polynomials with constant. Last but not
least this example suggests a project to classify generators $v$ of
$V_{\lambda}$ satisfying $\mathbb{L}L_{v}=\mathbb{L}\gamma$ for an appropriate
$\gamma\in\Lambda$, where $L_{v}$ denotes the image of $\operatorname{ann}v$
in $\Lambda^{*}$ under the canonical isomorphism from $\Lambda$ to
$\Lambda^{*}$. In the same manner one can decide which 2-dimensional
$\Lambda_{2}$-module can be described as $V_{\lambda}$ for an appropriate
comonic polynomial of degree 2, and one can ask the same question for small
dimensions.

###  5.2. Factorizations of polynomials

Using Leavitt algebras to investigate polynomials requires that the studied
polynomials must be with constant. In view of Lemma 3.10, a polynomial
$\lambda\in\Lambda$ without constant induces a finite set $\\{\lambda_{b}\\}$
of its cofactors, where the $\lambda_{b}$’s run over monomials of minimal
degree in $\lambda$ and they are with constant. So the $\lambda_{b}$’s admit a
greatest common divisor $\delta$ by Theorem 4.9. Hence $\delta$ divides
$\lambda$ if $\delta$ is not a constant. In this way one can find all right
factors of $\lambda$ with constant and by symmetry left factors with constant,
too. However, this argument does not work in case $\delta\in K$. One can avoid
this restriction by an appropriate substitution. For example, one can pass
from the polynomial $x^{2}_{1}x^{3}_{2}$ to the polynomial
$(x_{1}+1)^{2}(x_{2}+1)^{3}$. The feasibility of this trick depends on a
problem whether there exists a polynomial $\lambda$ with zero constant which
remains with zero constant term at any possible substitution of variables.This
opens the question whether the intersection of nonunital subalgebras of
codimension 1 is zero as we have already pointed out.

In any case, using Leavitt algebra may help to find factors with constant of a
polynomial $\lambda$ without constant when $\mathbb{L}\lambda\neq\mathbb{L}$.
For example, let

$\lambda=x_{1}x_{2}x_{3}x_{2}x_{1}+x_{1}x_{2}x_{3}+x_{1}x_{2}x_{1}+x_{1}+x_{3}x_{2}x_{1}+x_{3}$

be the polynomial considered by Volčič [16]. Then by Corollary 3.10
$\mathbb{L}\lambda=\mathbb{L}\lambda_{1}+\mathbb{L}\lambda_{3}=\mathbb{L}\lambda_{3}$
by $\lambda_{1}=\lambda_{3}+x_{2}x_{3}\lambda_{3}$, where
$\lambda_{1}=x_{2}(x_{3}x_{2}x_{1}+x_{1}+x_{3})+1$ and
$\lambda_{3}=x_{2}x_{1}+1$. Consequently, $\lambda_{3}$ is a right factor of
$\lambda=(x_{1}x_{2}x_{3}+x_{1}+x_{3})\lambda_{3}$. On the other hand, one can
use left factors of $\lambda$, that is, left factors of the image of $\lambda$
with respect to the $\ast$-adding isomorphism to obtain another factorization
of $\lambda=(x_{1}x_{2}+1)(x_{3}x_{2}x_{1}+x_{3}+x_{1})$. Iterating the
argument of this example one obtains immediately together with Cohn’s
characterization of similarity [7, Proposition 3.3.4] the following result on
factorizations of polynomials without constant.

######  Proposition 5.2.

Let $\lambda\in\Lambda$ be a polynomial without constant and
$\overline{\lambda}$ the image of $\lambda$ by the canonical isomorphism from
$\Lambda$ onto $\Lambda^{*}$. If
$\mathbb{L}\lambda\neq\mathbb{L}\neq{\overline{\lambda}}\mathbb{L}$, then
$\lambda$ has two different factorizations $\lambda=\alpha\beta=\gamma\delta$
such that $\alpha,\delta$ have constant and $\beta,\gamma$ have no constant.

###### Proof.

The assumption that $\mathbb{L}\lambda\neq\mathbb{L}$ implies that $\lambda$
has a factorization $\lambda=\gamma\delta$ such that $\delta$ has constant and
$\gamma$ has no constant. The condition
${\overline{\lambda}}\mathbb{L}\neq\mathbb{L}$ implies by symmetry that
${\overline{\lambda}}={\overline{a}}{\overline{\beta}}$, that is, $\lambda$
has a factorization $\lambda=\alpha\beta$ where $\alpha$ has constant but
$\beta$ has no constant. ∎

Therefore we have a way to produce polynomials having different
factorizations. Just take a product $\lambda=\alpha\beta$ such that $\alpha$
is a product of irreducible polynomials with constant and $\beta$ is a product
of irreducible ones without constant. If $\mathbb{L}\lambda\neq\mathbb{L}$,
then $\lambda$ admits another factorization $\lambda=\gamma\delta$ where
$\delta$ has constant. In view of the results of Section 4, it is a routine
matter to check whether $\mathbb{L}\lambda=\mathbb{L}$ holds. One can start
with any polynomial $\lambda$ without constant and then use its left or right
cofactors to check the inequalities
$\mathbb{L}\lambda\neq\mathbb{L}\neq{\overline{\lambda}}\mathbb{L}$ where
$\overline{\lambda}$ is the image of $\lambda$ by the canonical isomorphism in
$\Lambda^{*}$. This argument can not be applied for polynomials $\lambda$
without constant satisfying $\mathbb{L}\lambda=\mathbb{L}=\lambda\mathbb{L}$
such that $\lambda$ remains without constant under any variables substitution,
even in the case when $\lambda$ has a factorization
$\lambda=\alpha\beta\gamma$ where $\alpha$ and $\gamma$ are polynomials
without constant and $\beta$ is one with constant.

In order to find atomic factorizations of a comonic polynomial
$\lambda\in\Lambda$, we focus first to a right comonic factor
$\gamma\in\Lambda$ of $\lambda$. This means exactly that there is a positive
integer $l\in\mathbb{N}$ such that $b\star_{\gamma}\lambda=0$ for all
monomials $b\in\bf B$ of degree $l$. So this is a routine calculation to
verify whether $\gamma$ is a right factor of another comonic polynomial. To
compute the polynomial $\delta$ satisfying $\lambda=\delta\gamma$ we use the
equality for each index $i\in\\{1,2,\cdots,n\\}$

$x_{i}\star_{\gamma}\lambda=x_{i}\star_{\gamma}\delta\gamma=x_{i}\star_{\gamma}(\gamma+{\bar{\delta}}\gamma)=\delta_{i}\gamma.$

Taking in account that
$\delta=1+{\bar{\delta}}=1+\mathop{\sum}\limits_{i=1}^{n}x_{i}\delta_{i}=1+\mathop{\sum}\limits_{i=1}^{n}x_{i}\delta_{x_{i}}$
the determination of $\delta$ reduces to compute now a not necessarily comonic
polynomial $\delta_{1}$ with constant satisfying
$\delta_{i}\gamma=x_{i}\star_{\gamma}\lambda$ where for each
$i\in\\{1,2,\cdots,n\\}$ the polynomial $x_{i}\star_{\gamma}\lambda$ has lower
degree. Thus in principle one can compute the left quotient $\delta$ of
$\lambda$ at the right division by $\gamma$.

We are now in position to find atomic factorizations of $\lambda$. It is
routine to compute $V_{\lambda}$. Then one can continue as for integers. Take
an arbitrary comonic polynomial $\gamma\in V_{\lambda}$ and verify if $\gamma$
does divide $\lambda$ by computing $b\star_{\gamma}\lambda$ for all monomials
$b\in\bf B$ with $|b|=|\lambda|$. In the positive case one finds further a
factorization $\lambda=\delta\gamma$, and the problem of factorization reduces
to one of lower degree. In principle this can be carry out without difficulty
in case of a finite basic field $K$. However, there is a better way with
possibly less computations as follows. By Corollary 4.14, irreducible left
factors of $\lambda$ correspond to minimal submodules of $V_{\lambda}$. So one
can look for a minimal submodule $S$ of $V_{\lambda}$, then to a generator $m$
of $S$ and finally to an irreducible polynomial $\pi$ determining $m$ as in
5.1. Next step is to check whether $\pi$ divides $\lambda$ from the left. To
check this matter, we need to use $W_{\pi}$ which is the simple finite-
dimensional right $\Lambda$-module associated to $\pi$ defined in Remark 2.2,
and verify whether $\lambda\star_{\pi}b=0$ for all monomials $b\in\bf B$ of
degree $|\lambda|$. If $\pi$ divides $\lambda$, then one can use the right
symmetric version of the above described process to compute the right quotient
$\lambda_{1}$ of $\lambda$ at the left division by $\pi$, i.e.,
$\lambda=\pi\lambda_{1}$ and continue in the same way for $\lambda_{1}$
instead of $\lambda$. If $\pi$ does not divide $\lambda$, then look for the
other generators of $S$, until we find a polynomial similar to $\pi$ which
divides $\lambda$. This method can be carried out step by step until an atomic
factorization of $\lambda$ is found. This is a method to find atomic
factorizations of $\lambda$ from the left hand side. To find right irreducible
factors of $\lambda$, and then atomic factorizations of $\lambda$ starting
from the right hand side, one need first to compute $W_{\lambda}$, generators
of the simple submodules of $W_{\lambda}$ and then irreducible polynomials
$\pi$ representing them which become the right factors of $\lambda$. Of course
the process can be very hard, complicated from a computational point of view.
It is also worth to note that we have to use here both the left and right
version of Theorem 4.9 to get all possible factorizations of $\lambda$.

##  6\. Final remarks and further open questions

In this last section we continue a discussion of an individual polynomial in
noncommuting variables. One companion structure of a polynomial is a primary
decomposition. A polynomial $\lambda$ is called _primary_ if the associated
module $V_{\lambda}$ is indecomposable. In contrast to the case of polynomials
in commuting variables, primary polynomials in noncommuting variables have a
great diversity. For example, $x_{1}x_{2}x_{3}$ and its permutated polynomials
$x_{i}x_{j}x_{k}$ are all primary and none of them is similar to the other.
Since $V_{\lambda}$ is finite-dimensional, it is a direct sum of its
indecomposable submodules and this decomposition is unique up to isomorphism
by Krull-Schmidt theorem. A challenging problem is to find factorizations of a
comonic polynomial $\lambda$ into products of primary polynomials.

###  6.1. Primary decomposition of a polynomial

For a comonic polynomial $\lambda\,(|\lambda|\geq 1)$ the $K$-vector space
$V_{\lambda}$ is a finite-dimensional $\Lambda$-module. Hence it decomposes
uniquely as a direct sum of indecomposable submodules $V_{j}$ up to
permutations. Indecomposable direct summands of $V_{\lambda}$ are called
_primary components_ of $V_{\lambda}$. They are determined by right factors
$\pi_{j}$ of $\lambda$ and can be considered as invariants of $\lambda$.
However, we know almost nothing about them. For example we do not know whether
$\lambda$ is their product. Since $V_{\lambda}$ has a distributive submodule
lattice, Corollary 4.18 shows that, if $\lambda=\pi_{1}\pi_{2}\cdots\pi_{l}$
is a product of polynomials $\pi_{i}$ similar to the same irreducible
polynomial $\pi$, then $V_{\lambda}$ is uniserial and hence it has a unique
primary component.

It is an interesting project to study primary decompositions closely. For
instance, in contrast to the commutative case, there are polynomials $\lambda$
such that $V_{\lambda}$ is indecomposable and their simple subfactors are
pairwise non-isomorphic. For example $x_{1}x_{2}$ is a product of two non-
similar irreducible factors and $\Lambda/\Lambda x_{1}x_{2}$ is
indecomposable. Substituting $x_{1}+1,x_{2}+1$ for $x_{1}.x_{2}$,
respectively, one gets $\lambda=(1+x_{1})(1+x_{2})=1+x_{1}(1+x_{2})+x_{2}$, a
polynomial with constant. By definition $V_{\lambda}$ is a $K$-space generated
by $1,1+x_{2}$. It is now obvious by immediate calculation that $V_{\lambda}$
is indecomposable of length 2 with endomorphism ring isomorphic to $K$.

Finally, one can use the distributivity of the submodule lattice of
$V_{\lambda}$ to show that not every finite-codimensional left ideal of
$\Lambda^{*}$ can be represented as
$L_{\lambda}=\mathbb{L}\lambda\cap\Lambda^{*}$ for some $\lambda\in\Lambda$,
as we already saw in Subsection 5.1. Namely, let $L_{1},L_{2}$ be different
finite-codimensional maximal left ideals of $\Lambda^{*}$ such that their
associated primitive ideal is not $I^{*}$ and such that the factors
$\Lambda^{*}/L_{i}\,(i=1,2)$ are isomorphic. There are a great number of
choices for such maximal left ideals, for example by $L_{\lambda}$ where
$\lambda$ runs over similar irreducible polynomials with constant in
$\Lambda$. Put $L=L_{1}\cap L_{2}$. Then $\Lambda^{*}/L=N$ and hence
$\mathfrak{N}=\mathbb{L}\otimes_{\Lambda^{*}}N$ is a direct sum of two
isomorphic simple modules. Therefore their submodule lattice is not
distributive whence $L$ can not be represented as $L_{\lambda}$ for some
$\lambda\in\Lambda$ with constant.

In fact, there are even maximal two-sided ideals of $\Lambda^{*}$ with
division factor algebras which cannot be represented as $V_{\lambda}$ for some
comonic polynomial $\lambda\in\Lambda$. This claim is obvious for infinite-
dimensional division factor algebras of $\Lambda^{*}$ because all
$V_{\lambda}$ are finite-dimensional. Furthermore, if $K$ is the field
$\mathbb{R}$ of the reals and $\mathbb{H}$ is the field of the real
quaternions given by the two-sided ideal $\langle{x_{1}^{*}}^{2}+1,\
{x_{2}^{*}}^{2}+1,\ x^{*}_{1}x^{*}_{2}+x^{*}_{2}x^{*}_{1}\rangle$, then
$\mathbb{H}$ cannot be represented as $V_{\lambda}$ for any appropriate
comonic polynomial $\lambda\in\mathbb{R}\langle x_{1},x_{2}\rangle$.

###  6.2. Free algebra of arbitrary rank

It is clear that any element $\lambda$ of a free algebra contains only
finitely many variables. Hence its study does not essentially depend on the
other variables. Therefore from now on it is more natural to assume that
$\lambda$ is a polynomial in a free algebra $K\langle X\rangle$ on an
arbitrary possibly infinite variable set $X$. In this case, $V_{\lambda}$
remains a finite-dimensional $\Lambda$-module but the equality
$\mathbb{L}(X)\lambda=\mathbb{L}(X)(\Lambda^{*}\cap\mathbb{L}\lambda)$ does
not hold. Here $\mathbb{L}(X)$ denote the Leavitt algebra by adding only a
right inverse of the row matrix $X$, that is, it is generated by
$\\{x,x^{*}\,|\,x\in X\\}$ subject to

$y^{*}x=\begin{cases}1&\text{if}\qquad x=y,\\\ 0&\text{if}\qquad x\neq
y\end{cases}$

for any two $x,y\in X$ when $X$ is infinite. For example, for infinite $X$ the
right ideal
$\mathbb{L}(X)(1-x)\cap\Lambda^{*}=\Lambda^{*}(1-x^{*})+\sum\limits_{y\neq
x}\Lambda^{*}y^{*}=L$ is still maximal but
$\mathbb{L}(X)(1-x)\neq\mathbb{L}L$. However, $\mathbb{L}(X)\lambda$ is still
a left ideal of colength $l$ if $\lambda=\pi_{1}\cdots\pi_{l}$ is a
factorization of $\lambda$ into $l$ irreducible factors. To verify this result
we need first an infinite version of Theorem 4.9.

######  Theorem 6.1.

Let $\Lambda=K\langle X\rangle$ be the free algebra on an infinite set $X$ and
$\mathbb{L}=\mathbb{L}(X)$ the corresponding Leavitt algebra. Then for any two
polynomials $\delta,\gamma\in\Lambda$ there is a comonic polynomial
$\lambda\in\Lambda$ with
$\mathbb{L}\delta+\mathbb{L}\gamma=\mathbb{L}\lambda$. In particular, any
submodule of the $\mathbb{L}$-module
${}_{\lambda}M=\mathbb{L}/\mathbb{L}\lambda\cong\Lambda/\Lambda\lambda$ is
cyclic and can be generated by some comonic $\gamma\in\Lambda$.

###### Proof.

From the definition one sees immediately that Lemma 3.10 remains true whence
by a trivial induction one can assume without loss of generality that both
$\delta,\lambda$ are comonic. Moreover, one can define a
$\star_{\lambda}$-action of $\Lambda$ on itself for any comonic polynomial
$\lambda\in\Lambda$ in the same way as it was defined before Theorem 2.3
whence $\Lambda/\Lambda\lambda$ becomes naturally a left $\mathbb{L}$-module
isomorphic to $M_{\lambda}=\mathbb{L}/\mathbb{L}\lambda$. Therefore one has
$x\star_{\lambda}\alpha=-k_{\alpha}k^{-1}_{\lambda}\lambda_{x}+\alpha_{x}$ if
$\lambda=k_{\lambda}+\sum\limits_{x\in
X}x\lambda_{x},\,\,\alpha=k_{\alpha}+\sum\limits_{x\in X}x\alpha_{x},\,(0\neq
k_{\lambda},k_{\alpha}\in K)$ where $\beta_{x}$ denotes the cofactor of
$\beta\in\Lambda$ with respect to $x\in X$. If we denote by $X_{\alpha}$ the
subset of variables appearing in $\alpha$, then $X_{\alpha}$ is finite and
$\sum\limits_{x\in X_{\alpha}}x^{*}$ acts as the identity on $\alpha$ mod
$\mathbb{L}\lambda$, that is,

$\sum\limits_{x\in
X_{\alpha}}x(x\star_{\lambda}\alpha)\equiv\alpha\quad\mod\Lambda\lambda.$

Consequently, if $b$ runs over the finite set of all monomials in $x\in
X_{\alpha}$ of the degree $\alpha$, then at least one of these
$b\star_{\lambda}\alpha$ is a non-zero element of $V_{\lambda}$, hence of
degree $\leq|\lambda|-1$. These observations show that almost all results of
Sections 3 and 4 together with their proof remain true if they are formulated
by using $\mathbb{L}\lambda,V_{\lambda}$ and
$M_{\lambda}(\cong\Lambda/\Lambda\lambda)$ and not involving
$L_{\lambda}=\Lambda^{*}\cap\mathbb{L}\lambda,\,N_{\lambda}=\Lambda^{*}/L_{\lambda}$.
Thus we have

######  Corollary 6.2.

Let $\lambda\in\Lambda=K\langle X\rangle\,(|X|=\infty,\,\,|\lambda|\geq 1)$
comonic. Then $V_{\lambda}$ is a finite-dimensional $\Lambda$-module with
respect to the $\star_{\lambda}$-action which embeds essentially in
$\Lambda/\Lambda\lambda\cong M_{\lambda}=\mathbb{L}/\mathbb{L}\lambda$. Both
$V_{\lambda}$ and the $\mathbb{L}$-module $M_{\lambda}$ are simple with
endomorphism ring isomorphic to $K$ if $\lambda$ is linear. For any
$\alpha\in\Lambda$ not contained in $\Lambda\lambda$ there is a monomial $b$
in variables of $\alpha$ of degree $|\alpha|$ with $0\neq
b\star_{\lambda}\alpha\in V_{\lambda}$.

In view of Corollary 6.2 the proof of Theorem 4.9 works also in this case, and
so Theorem 6.1 is verified. ∎

In view of Corollary 6.2 and Theorem 6.1 the same argument in the last part of
Section 4 implies the more general form of Theorem 2.3 as follows.

######  Theorem 6.3.

Let $\gamma,\lambda\in\Lambda=K\langle X\rangle\,$ be comonic polynomials of
positive degree where $X$ has an arbitrary rank, together with finite-
dimensional left $\Lambda$-modules $V_{\gamma},V_{\lambda}$ with respect to
the $\star$-action defined by $\gamma,\lambda$, respectively. Let
$\mathbb{L}(X)$ be the corresponding Leavitt algebra defined by $\Lambda$.

  1. (1)

$\lambda$ is an irreducible polynomial if and only if
$M_{\lambda}=\mathbb{L}(X)/\mathbb{L}(X)\lambda$ and $V_{\lambda}$ are simple
modules over $\mathbb{L}(X)$ and $\Lambda$, respectively.

  2. (2)

If $\lambda=\pi_{1}\cdots\pi_{m}$ is a factorization of $\lambda$ into a
product of irreducible polynomials, then $m$ is the length of both
$M_{\lambda}$ and ${{}_{\Lambda}}V_{\lambda}$ whence $m$ is an invariant of
$\lambda$.

  3. (3)

$V_{\lambda}\cong
V_{\gamma}\Longleftrightarrow\Lambda/\Lambda\gamma\cong\Lambda/\Lambda\lambda\Longleftrightarrow
M_{\gamma}\cong M_{\lambda}$, that is, $\gamma$ and $\lambda$ are similar over
$\Lambda$ iff they are such over $\mathbb{L}(X)$.

Acknowledgement. The authors express their gratitude to Professor Gene Abrams
for his critical comments, questions and noble English assistance making this
work transparent and enjoyable to the readers.

## References

  * [1] G. Abrams, P. N. Ánh, A matrix viewpoint for various algebraic extensions, _Elemente der Mathematik_ 75 (2020), 1-17, https://doi.org/1013-6018/20/020001-17
  * [2] G. Abrams, P. Ara, M. Siles Molina, _Leavitt path algebras_ , Lecture Notes in Mathematics 2191, Springer 2018.
  * [3] P. N. Ánh, Skew polynomial rings: the Schreier technique, _Acta Math. Vietnamica_ , https://doi.org/10.1007/s40206-021-00466-7
  * [4] P.N. Ánh, M.Siddoway, Leavitt path algebras are flat rings of quotients. Arxiv version: arXiv:2108.11987
  * [5] P. Ara, Finitely presented modules over Leavitt algebras, _J. Pure Appl. Algebra_ 191(1-2)(2004), 1 – 21.
  * [6] P. Ara, M. Brustenga, Module theory over Leavitt path algebras and K-theory, _J. Pure Appl. Algebra_ 214 (2010), 1131 – 1151.
  * [7] P. M. Cohn, _Free rings and their relations_ , AP 1. ed. 1971, 2. ed. 1985.
  * [8] P. M. Cohn, _An introduction to ring theory_ , Springer UMS, 1999.
  * [9] P. M. Cohn, _Free ideal rings and localization in general rings_ , Cambridge University Press 2006.
  * [10] P.M. Cohn, Non commutative unique factorization domains, _Trans. Amer. Math. Soc_ , 109 (1963), 313-331.
  * [11] P.M. Cohn, Rings with a weak algortithm, _Trans. Amer. Math. Soc_ , 109 (1963), 332-356.
  * [12] J. Lewin, Free modules over free algebras and free group algebras: The Schreier technique, _Trans. Amer. Math. Soc._ 145(1969), 455-465.
  * [13] W. G. Leavitt, The module type of a ring, _Trans. Amer. math. Soc._ 103(1)(1962), 113 – 130.
  * [14] A. Rosenmann, S. Rosset, Ideals of finite codimension in free algebras and the fc-localization, _Pacific J. Math._ 162(2)(1994), 351 - 371.
  * [15] Bo Stenström, _Rings of quotients_ , Grundl math. Wiss. 217, Springer 1975.
  * [16] J. Volčič, Determinantal zeros and factorization of noncommutative polynomials, talk on the Graz’s conference _Rings and Factorizations_ , July 10 – 14, 2023.