# Finitely Star Regular Domains (⋆)

A. Mimouni  Department of Mathematics, KFUPM, Dhahran 31261, KSA
<EMAIL_ADDRESS>
###### Abstract.

Let $R$ be an integral domain, $Star(R)$ the set of all star operations on $R$
and $StarFC(R)$ the set of all star operations of finite type on $R$. Then $R$
is said to be star regular if $|Star(T)|\leq|Star(R)|$ for every overring $T$
of $R$. In this paper we introduce the notion of finitely star regular domain
as an integral domain $R$ such that $|StarFC(T)|\leq|StarFC(R)|$ for each
overring $T$ of $R$. First, we show that the notions of star regular and
finitely star regular domains are completely different and do not imply each
other. Next, we extend/generalize well-known results on star regularity in
Noetherian and Prüfer contexts to finitely star regularity. Also we handle the
finite star regular domains issued from classical pullback constructions to
construct finitely star regular domains that are not star regular and enriches
the literature with a such class of domains.

###### 2010 Mathematics Subject Classification:

13A15, 13A18, 13F05, 13G05, 13C20

(⋆) Supported by KFUPM under DSR Research Grant #: RG161001.

##  1\. Introduction

Let $R$ be an integral domain with quotient field $L$, $F(R)$ the set of
nonzero fractional ideals of $R$ and $f(R)$ the set of nonzero finitely
generated fractional ideals of $R$. A mapping $*:F(R)\rightarrow F(R)$,
$E\mapsto E^{*}$, is called a star operation on $R$ if the following
conditions hold for all $a\in L\setminus\\{0\\}$ and $E,F\in F(R)$:

  1. (I)

$(aE)^{*}=aE^{*}$;

  2. (II)

$E\subseteq E^{*}$; if $E\subseteq F$, then $E^{*}\subseteq F^{*}$; and

  3. (III)

$(E^{*})^{*}=E^{*}$.

The simplest star operations are the $d$-operation defined by $E^{d}=E$ for
every $E\in F(R)$, and the $v$-operation defined by $E^{v}=(E^{-1})^{-1}$
(where $E^{-1}=(R:E)=\\{x\in L|xE\subseteq R\\}$) for every $E\in F(R)$. A
star operation $*$ is said to be of finite type (or of finite character) if
for each nonzero (fractional) ideal $E$ of $R$, $E^{*}=\bigcup F^{*}$ where
the union is taken over all nonzero finitely generated subideals $F$ of $E$.
Also a star operation is stable if $(E\cap F)^{*}=E^{*}\cap F^{*}$ for each
$E,F\in F(R)$. To any star operation $*$ on $R$, we associate a star operation
of finite type $*_{f}$ and a stable star operation of finite type $*_{w}$ by
setting respectively $E^{*_{f}}=\bigcup\\{F^{*}|F\in f(R),F\subseteq E\\}$ and
$E^{*_{w}}=\bigcup\\{(E:F)|F\in f(R),F^{*}=R\\}$. Notice that $v_{f}=t$ and
$t_{w}=w$. For star operations $*$ and $*^{\prime}$ on $R$, $*\leq*^{\prime}$
provided that $E^{*}\subseteq E^{*^{\prime}}$ for every $E\in F(R)$. Clearly
$d\leq w\leq t\leq v$ and for every star operation $*$ on $R$, $d\leq*\leq v$
and $d\leq*_{f}\leq t$. We denote by $\text{Star}(R)$ the set of all star
operations on $R$ and $StarFC(R)$ the set of all star operations of finite
type on $R$.
Recently, motivated by well-known characterizations of integrally closed and
Noetherian divisorial domains [22, 38], the author of this paper, together
with E. Houston and M. H. Park, started a long and deep study of some ring-
theoretic properties of integral domains having only finitely many star
operations in different contexts of integral domains. Namely, complete
characterizations are given in the cases of local Notherian domains with
infinite residue field and integrally closed domains see [25, 26, 27, 28, 29]
(see also [11, 30, 24, 44], and [46, 47]).
In [28], the authors studied, for a Noetherian domain $R$, how $|Star(R)|$
affects $|Star(T)|$ for each proper overing $T$ of $R$ with the emphasis on
the case where $Star(R)$ is finite. They introduced the notion of a star
regular domain as a domain $R$ such that $|Star(T)|\leq|Star(R)|$ for each
overring $T$ of $R$. Notice that a Noetherian domain $R$ (which is not a
field) with finitely many star operations has Krull dimension one. The authors
constructed a Noetherian domain $R$ with $|Star(R)|=1$ (equivalently, $R$ is a
divisorial domain), but having an overring $T$ with $|Star(T)|=\infty$. Next,
they showed that for a one-dimensional Noetherian domain $R$, if $R$ is
locally star regular, then it is star regular, and the converse holds if
$Star(R)$ is finte. They conjectured that “if $R$ is a local Noetherian domain
with $1<|Star(R)|<\infty$, then $R$ is star regular”, and proved that this
conjecture holds if $R$ has infinite residue field. They also considered the
question of whether finiteness of $Star(T)$ for each proper overring of a
Noetherian domain $R$ implies finiteness of $Star(R)$, and showed that this
occurred when R is non-local.
In [29], the authors investigated star regular domains in the context of
Prüfer domains. They showed that star regularity for Prüfer domains with only
finitely many star operations reduces to star regularity of Prüfer domains $R$
possessing a nonzero prime ideal $P$ contained in the Jacobson radical of $R$
such that $Spec(R/P)$ is finite ([29, Theorem 3.1]). More precisely, they
proved that if $R$ is a semi-local Prüfer domain with more than one maximal
ideal such that $R/P$ is strongly discrete (where $P$ is the largest prime
ideal contained in the Jacobson radical of $R$), then $R$ is star regular
([29, Theorem 3.11]).
In [34], the authors investigated some ring-theoretic properties of certain
classes of integral domains with only finitely many star operations of finite
type. Several generalizations/analogues of well-known results on integral
domains with finitely many star operations were extended to integral domains
with finitely many star operations of finite type. Namely in Noetherian-like
settings such us Mori domains, pullback constructions and more.

The purpose of this paper is to introduce and study the notion of finitely
star regular domains, that is, integral domains $R$ such that
$|StarFC(T)|\leq|StarFC(R)|$ for each overring $T$ of $R$. The class of
finitely star regular domains includes the class of Prüfer domains (since
$|StarFC(R)|=1$ for every Prüfer domain $R$), and coincides with the class of
star domains in the Neotherian context. Motivated by the fact that Prüfer
domains are finitely star regular domains, but not always star regular, in
section 2, we start by showing that the two notions of star and finitely star
regular domains are completely different and do not imply each other, (see
Example 2.3, Example 2.4, and Example 2.5). Next, we extend [26, Theorem 2.3]
to a one-dimensional domain of finite character (i.e., every nonzero non-unit
element is contained in a finitely many maximal ideals). That is, for a one-
dimensional domain $R$ of finite character such that $StarFC(R)$ is finite,
$|StarFC(R)|=\displaystyle\prod_{M\in Max(R)}|StarFC(R_{M})|$ (Theorem 2.10).
The second main theorem asserts that if $R$ is a one-dimensional quasi-Prüfer
domain (i.e. its integral closure is a Prüfer domain) such that $StarFC(R)$ is
finite, then $R$ is finitely star regular if and only if $R_{M}$ is finitely
star regular for every maximal ideal $M$ of $R$ (Theorem 2.11). The third
section deals with certain pullbacks that are finitely star regular. The main
result asserts that a large class of integral domains related to valuation
domains, namely the class of pseudo-valuation domains ($PVD$ for short), are
always finitely star regular, but not star regular in general. Next, we deal
with the classical pullbacks issued from valuation domains in order to
enriches the literature with such a class of integral domains.

Throughout, $R$ denotes an integral domain (which is not a field),
$R^{\prime}$ its integral closure and $\overline{R}$ its complete integral
closure. The set of all maximal ideals of $R$ is denoted by $Max(R)$, and if
$S$ is an overring of $R$, $[R,S]$ denotes the set of all intermediate rings
between $R$ and $S$ (i.e. rings $T$ such that $R\subseteq T\subseteq S$).

##  2\. General Results

Recall that an integral domain $R$ is said to be a star regular domain if
$|Star(T)|\leq|Star(R)|$ for every overring of $R$. Next, we introduce the
notion of finitely star regular domains.

######  Definition 2.1.

Let $R$ be an integral domain. We say that $R$ is finitely star regular if
$|StarFC(T)|\leq|StarFC(R)|$ for each overring $T$ of $R$.

The next proposition deals with the class of integrally closed domains and
shows that it is an important class of finitely star regular domains that are
not necessarily star regular.

######  Proposition 2.2.

Let $R$ be an integrally closed domain. Then $R$ is finitely star regular.

###### Proof.

Assume that $R$ is integrally closed. If $|StarFC(R)|=\infty$, we are done.
Assume that $StarFC(R)$ is finite. Then by [27, Theorem 3.1], $R$ is a Prüfer
domain. Thus every overring $T$ of $R$ is Prüfer and so
$|StarFC(T)|=|StarFC(R)|=1$, as desired. ∎

Our next three examples show that the notions of star regular domain and
finitely star regular domain do not implies each other. The first one is an
example of a finitely star regular domain $R$ with $Star(T)$ finite for every
overring $T$ of $R$, but which is not star regular (as it has an overring $T$
with $|Star(R)|<|Star(T)|$). The second one is an example of a star regular
domain which is not finitely star regular with $StarFC(R)$ finite but
$Star(R)$ is infinite. In the third example, $R$ is finitely star regular but
not star regular with $Star(R)$ finite and having an overring $T$ with
$|Star(T)|=\infty$. Notice that if $V$ is a valuation domain, then
$|Star(V)|\leq 2$.

######  Example 2.3.

Let $V$ be a valuation domain with a principal maximal ideal $M$ and suppose
that $V$ has a nonzero non-maximal ideal $P$ such that $P=P^{2}$. Clearly for
every overring $T$ of $V$, $|StarFC(T)|=|StarFC(V)|=1$. Thus $V$ is a finitely
star regular domain. However, $|Star(V_{P})|=2$ while $|Star(V)|=1$. Hence $V$
is not star regular.

######  Example 2.4.

The following is an example of a star regular domain which is not finitely
star regular. Let $k$ be an infinite field, $X$ and $Y$ indeterminates over
$k$ and set $D=k[X^{3},X^{7}]$ and $D_{1}=k[X^{3},X^{7},X^{8}]$. By [41,
Theorem 2.2], $D$ is a divisorial Noetherian domain and so
$|Star(D)|=|StarFC(D)|=1$, while $D_{1}$ is a Noetherian overring of $D$ with
$|Star(D_{1})|=|StarFC(D_{1})|=\infty$. Indeed, let
$M_{1}=(X^{3},X^{7},X^{8})$. Then $M_{1}$ is a maximal ideal of $D_{1}$ with
$(D_{1}:M_{1})=k[X^{3},X^{4},X^{5}]$ and so $(D_{1})_{M_{1}}$ is a Noetherian
local domain satisfying the conditions of [26, Theorem 3.9]. Thus
$|StarFC((D_{1})_{M_{1}})|=|Star((D_{1})_{M_{1}})|=\infty$. By [26, Theorem
2.3], $|StarFC(D_{1})|=|Star(D_{1})|=\infty$. Now set
$V=k(X)[[Y]]=k(X)+Yk(X)[[Y]]$, $R=D+Yk(X)[[Y]]=D+M$ and $T=D_{1}+M$. By [34,
Theorem 4.4], $|StarFC(R)|=|StarFC(D)|=1$, but
$|StarFC(T)|=|StarFC(D_{1})|=\infty$. Thus $R$ is not finitely star regular.
However, $|Star(R)|=\infty$ and so $R$ is star regular. Indeed, for every
maximal ideal $P$ of $k[X]$, set $S_{P}=k[X]_{P}+M$. Then $S_{P}$ is a
fractional overring of $R$, and $(R:S_{P})=M$, so that $(S_{P})_{v}=(R:M)=V$.
Now, by [25, Proposition 2.7], $*_{P}=\delta(d_{S_{P}},v)$ defined by
$E^{*_{P}}=ES_{P}\cap E_{v}$ is a star operation on $R$. Moreover, for
$P\not=Q$ maximal ideals of $k[X]$, $(S_{P})^{*_{P}}=S_{P}\cap V=S_{P}$ and
$(S_{P})^{*_{Q}}=S_{Q}S_{P}\cap V=V\cap V=V$. Hence $*_{P}\not=*_{Q}$. Since
$k[X]$ has infinitely many maximal ideals that induce different star
operations on $R$, $R$ has infinitely many star operations.

The following is an example of a non star regular Prüfer domain $R$ with only
finitely many star operations having an overring $T$ with $|Star(T)|=\infty$.
The Example is given in [29, Example 2.4], and for the convenience of the
reader, we include it here.

######  Example 2.5.

The following is an example of a finitely star regular which is not star
regular. Let $A$ be the direct sum of countably infinitely many copies of $G$,
where $G$ is the totally ordered group $\mathbb{R}\bigoplus\mathbb{Z}$ with
lexicographic order (i.e., for
$(a,b),(a^{\prime},b^{\prime})\in\mathbb{R}\bigoplus\mathbb{Z}$,
$(a,b)\geq(a^{\prime},b^{\prime})$ if $a>a^{\prime}$, or $a=a^{\prime}$ and
$b\geq b^{\prime}$). For $(a_{i})_{i=1}^{\infty}\in A$, define $(a_{i})\geq 0$
if $a_{i}\geq 0$ for all $i\geq 1$. Then $A$ is a lattice ordered group.
According to Jaffard and Ohm [33, 43], every lattice ordered group is a group
of divisibility of a Bézout domain. Thus there is a Bézout domain $R$ with
$K^{*}/U(R)\cong A$, where $U(R)$ is the set of units of $R$. Since $A$ is the
weak direct product of $G$’s, $R$ has finite character. It follows easily from
[37, Theorem 3.2] and [24, Lemma 3.4] that $R$ is an $h$-local Prüfer domain
with $\text{dim}\,R=2$. Thus each overring $T$ of $R$ is a Pruüfer domain and
so $|StarFC(T)|=|StarFC(R)|=1$. Hence $R$ is finitely star regular. Since for
each maximal ideal $M$ of $R$, $R_{M}$ has value group $G$, $MD_{M}$ is
principal. Since $R$ has finite character, each $M$ is invertible. Hence $R$
is a divisorial domain [22, Theorem 5.1]. Let $M$ be a maximal ideal of $R$,
and let $P$ be the nonzero prime contained in $M$. Since the value group of
$R_{P}$ is equal to $\mathbb{R}$, $PR_{P}$ is not principal. Hence $P=P^{2}$,
and so $R_{M}$ is not strongly discrete. By [29, Theorem 2.3], there is an
overring $T$ of $R$ with $|Star(T)|=\infty$ (and we may take $T=\bigcap
R_{P}$, where the intersection is taken over the height-one primes of $R$).
Notice that, for each positive integer $n$, if in the definition of $A$, we
replace $n$ of the $G$ by $\mathbb{R}\bigoplus\mathbb{R}$, then the resulting
$R$ satisfies $|Star(R)|=2^{n}$ by [25, Theorem 3.1], and $R$ has an overring
$T$ with $|Star(T)|=\infty$ ([29, Remark 2.5]).

The next two lemmas are crucial. The first one is [25, Proposition 2.7] and we
shall use it whenever we consider a proper overring $T$ of the base ring $R$
and a star operation on $T$. The second one is a direct combination of [25,
Proposition 2.4 and Proposition 4.6] and we shall use it whenever we consider
the particular overring $T=R_{M}$ for some maximal ideal $M$.

######  Lemma 2.6.

([25, Proposition 2.7]) Let $R$ be an integral domain, $T$ an overring of $R$,
$*$ a star operation on $R$ and $*^{\prime}$ a star operation on $T$. Then the
map $\delta(*^{\prime},*):F(R)\rightarrow F(R)$, $E\mapsto
E^{\delta(*^{\prime},*)}:=(ET)^{*^{\prime}}\cap E^{*}$, defines a star
operation on $R$. Moreover, if $*$ and $*^{\prime}$ are of finite type, then
so is $\delta(*^{\prime},*)$.

######  Lemma 2.7.

Let $R$ be an integral domain, $M$ a maximal ideal of $R$ and
$*\in\text{StarFC}(R)$. Then $*(M)$ defined on $R_{M}$ by
$(AR_{M})^{*(M)}=A^{*}R_{M}$ for every $A\in f(R)$ is a star operation on
$R_{M}$. Moreover, if $R$ is $v$-coherent, then every star operation on
$R_{M}$ is of this form, that is, if $*^{\prime}\in\text{StarFC}(R_{M})$, then
$*^{\prime}=*(M)$ for some $*\in\text{StarFC}(R)$.

######  Theorem 2.8.

Let $R$ be an integral domain such that $(R:\overline{R})=M$ is a maximal
ideal of $R$ and suppose that $\overline{R}$ is a $PID$. Then for every
$T\in[R,\overline{R}]$, $|Star(T)|\leq|Star(R)|$ and
$|StarFC(T)|\leq|StarFC(R)|$.

###### Proof.

First notice that $M^{-1}=(M:M)=\overline{R}$ and so $M$ is a divisorial ideal
of $R$. Let $T\in[R,\overline{R}]$. Since $R$ is a $PID$,
$|StarFC(\overline{R})|=1\leq|StarFC(R)|$, so without loss of generality we
may assume that $R\subsetneqq T\subsetneqq\overline{R}$. Thus $(R:T)=M$ and so
$M$ is an ideal of $T$ and $T_{v}=M^{-1}=\overline{R}$. Notice that if $S$ is
a fractional overring of $T$, then
$S\subseteq\overline{S}=\overline{T}=\overline{R}=M^{-1}$. Since
$(M^{-1})^{*}$ is a fractional overring of $T$, then $(M^{-1})^{*}=M^{-1}$ for
every $*\in Star(T)$. Thus for every nonzero fractional ideal $I$ of $T$ with
$IM^{-1}=xM^{-1}$,
$I^{*}\subseteq(IM^{-1})^{*}=(xM^{-1})^{*}=xM^{-1}=IM^{-1}$. Now let
$*_{1}\not=*_{2}$ be star operations (resp. star operations of finite type) on
$T$ and let $A$ be a fractional ideal (resp. a finitely generated fractional
ideal) of $T$ such that $A^{*_{1}}\not=A^{*_{2}}$. Without loss of generality,
we may assume that $A\subseteq M$ (for if $0\not=d\in T$ is such that
$dA\subseteq T$ and $0\not=m\in M$, $mdA\subseteq mT\subseteq M$ and
$(mdA)^{*_{1}}=mdA^{*_{1}}\not=mdA^{*_{2}}=(mdA)^{*_{2}}$). Notice that $A$ is
not a divisorial ideal of $T$ and so $A$ is not an invertible ideal of $T$.
Thus $A(R:A)\not=R$. Now set $AM^{-1}=aM^{-1}$. Then $AM=AMM^{-1}=aMM^{-1}=aM$
and
$(M:A)=(M_{v}:A)=((R:M^{-1}):A)=(R:AM^{-1})=(R:aM^{-1})=a^{-1}M_{v}=a^{-1}M$.
Thus $A(M:A)=a^{-1}AM=M$. Hence $M=A(M:A)\subseteq A(R:A)\subsetneqq R$ and by
maximality of $M$, $M=A(M:A)=A(R:A)$. Thus $(R:A)=(M:A)=a^{-1}M$ and so
$A_{v}=aM^{-1}=AM^{-1}$. Now, by Lemma 2.6, $A^{\delta(*_{1},v)}=A^{*_{1}}\cap
A_{v}=A^{*_{1}}\cap AM^{-1}=A^{*_{1}}$ and $A^{\delta(*_{2},v)}=A^{*_{2}}\cap
A_{v}=A^{*_{2}}\cap AM^{-1}=A^{*_{2}}$. Thus
$\delta(*_{1},v)\not=\delta(*_{2},v)$.
For star operations of finite type $*_{1}\not=*_{2}$ and $A$ a finitely
generated ideal of $T$ ($A\subseteq M$) with $A^{*_{1}}\not=A^{*_{2}}$, let
$B$ be a finitely generated ideal of $R$ such that $BT=A$ (for
$A=\displaystyle\sum_{i=1}^{i=n}a_{i}T$, just take
$B=\displaystyle\sum_{i=1}^{i=n}a_{i}R$). If since $B(R:B)\subseteq
A(T:A)\subsetneqq T$, $B(R:B)\subsetneqq R$. Moreover,
$BM^{-1}=AM^{-1}=aM^{-1}$. Thus
$(M:B)=(M_{v}:B)=((R:M^{-1}):B)=(R:BM^{-1})=(R:aM^{-1})=a^{-1}M_{v}=a^{-1}M$.
Hence $B(M:B)=a^{-1}BM=M$ and so $M=B(M:B)\subseteq B(R:B)\subsetneqq R$. Thus
$M=M(B:M)=M(R:M)$ and so $(R:B)=(M:B)=a^{-1}M$. Hence
$B_{t}=B_{v}=aM^{-1}=BM^{-1}=AM^{-1}$. Finally,
$B^{\delta(*_{1},t)}=(BT)^{*_{1}}\cap B_{t}=A^{*_{1}}\cap AM^{-1}=A^{*_{1}}$
and $B^{\delta(*_{2},t)}=(BT)^{*_{2}}\cap B_{t}=A^{*_{2}}\cap
AM^{-1}=A^{*_{2}}$. It follows that $\delta(*_{1},t)\not=\delta(*_{2},t)$. In
both cases, the map $\delta(-,v):Star(T)\longrightarrow Star(R)$,
$*\longmapsto\delta(*,v)$ (resp. $\delta(-,t):StarFC(T)\longrightarrow
StarFC(R)$, $*\longmapsto\delta(*,t)$) is one-to-one. Hence
$|Star(T)|\leq|Star(R)|$ (resp. $|StarFC(T)|\leq|StarFC(R)|$), as desired. ∎

Recall that an integral domain $R$ is said to be conducive if the conductor
$(R:T)\not=0$ for every overring $T$ of $R$ with $T\subsetneqq qf(R)$,
equivalently $(R:V)\not=0$ for some valuation overring $V$ of $R$. The next
corollary generalizes [28, Proposition 1.10].

######  Corollary 2.9.

([28, Proposition 1.10]) Let $(R,M)$ be a local Noetherian domain such that
$M^{-1}$ is a valuation domain. Then $R$ is star regular.

###### Proof.

Since $M^{-1}$ is a valuation domain, $R$ is a conducive domain and clearly
$M^{-1}=(M:M)=\overline{R}$. Thus $[R,L]=[R,\overline{R}]\cup\\{L\\}$. By
Theorem 2.8, $R$ is star regular. ∎

It is well-known that if $R$ is a Noetherian domain (which is not a field)
with $Star(R)=StarFC(R)$ finite, then $R$ has Krull dimension $dim(R)=1$ ([26,
Theorem 2.1]). Therefore $R$ is of finite character, that is, each nonzero
nonunit element is contained in only finitely many maximal ideals. In [26,
Theorem 2.3], it was proved that for a Noetherian domain $R$,
$|StarFC(R)|=|Star(R)|=\displaystyle\prod_{M\in Max(R)}|Star(R_{M})|$. Our
first theorem generalizes this result to one-dimensional domains of finite
character.

######  Theorem 2.10.

Let $R$ be a one-dimensional domain of finite character such that $StarFC(R)$
is finite. Then $|StarFC(R)|=\displaystyle\prod_{M\in Max(R)}|StarFC(R_{M})|$.

###### Proof.

First set $\mathcal{M}_{1}:=\\{M\in Max(R):|StarFC(R_{M}|=1\\}$ and
$\mathcal{M}_{\geq 2}:=\\{M\in Max(R):|StarFC(R_{M}|\geq 2\\}$. Clearly
$Max(R)$ is the disjoint union of $\mathcal{M}_{1}$ and $\mathcal{M}_{\geq
2}$. Now if $\mathcal{M}_{\geq 2}=\emptyset$, then $|StacFC(R_{M})|=1$ and so
$d_{M}=t({M})$ for every $M\in Max(R)$. Let $*\in StarFC(R)$ and let $I$ be a
finitely generated ideal of $R$. Then for every $M\in Max(R)$,
$I_{t}R_{M}=(IR_{M})^{t(M)}=(IR_{M})^{d_{M}}=IR_{M}$. Hence $I_{t}=I$ and so
$t=d$. Thus $StarFc(R)=\\{d\\}$ and therefore
$|StarFC(R)|=1=\displaystyle\prod_{M\in Max(R)}|StarFC(R_{M})|$. Assume that
$\mathcal{M}_{\geq 2}\not=\emptyset$.
Claim: $\mathcal{M}_{\geq 2}$ is finite. By way of contradiction suppose that
$\mathcal{M}_{\geq 2}$ is infinite. Then for every positive integer $n$
consider a subset $\\{M_{1},\dots,M_{n}\\}\subseteq\mathcal{M}_{\geq 2}$, and
consider the map
$\phi:\displaystyle\prod_{i=1}^{n}StarFC(R_{M_{i}})\longrightarrow StarFC(R)$,
$\star=(*_{i})_{i=1}^{n}\mapsto\phi(\star)=\star_{\phi}$ where $\star_{\phi}$
is defined by
$A^{\star_{\phi}}=\displaystyle\bigcap_{i=1}^{n}(AR_{M_{i}})^{*_{i}}\cap\displaystyle\bigcap_{M\in
Max(R):M\not=M_{i}}(AR_{M})$. By [1, Theorem 2], $\star_{\phi}\in StarFC(R)$
and so $\phi$ is well-defined. Now, let
$\star=(*_{i})_{i=1}^{n}\not=(*^{\prime}_{i})_{i=1}^{n}=\star^{\prime}$. Then
$*_{j}\not=*^{\prime}_{j}$ for some $j\in\\{1,\dots,n\\}$. Let $E$ be a
finitely generated integral ideal of $R_{M_{j}}$ such that
$E^{*_{j}}\not=E^{*^{\prime}_{j}}$ and let $A=E\cap R$. Since
$dim(R_{M_{j}})=1$, $E$ is $M_{j}R_{M_{j}}$-primary and so $A$ is
$M_{j}$-primary. Hence $Max(R,A)=\\{M_{j}\\}$ and so $AR_{M}=R_{M}$ for every
$M\in Max(R)-\\{M_{j}\\}$. So
$A^{\star_{\phi}}=(AR_{M_{j}})^{*_{j}}\cap\displaystyle\bigcap_{M\in
Max(R):M\not=M_{j}}R_{M}=E^{*_{j}}\cap\displaystyle\bigcap_{M\in
Max(R):M\not=M_{j}}R_{M}$. Thus $A^{\star_{\phi}}\cap R_{M_{j}}=E^{*_{j}}\cap
R$. Similarly $A^{\star^{\prime}_{\phi}}=E^{*^{\prime}_{j}}\cap R$. So if
$A^{\star_{\phi}}=A^{\star^{\prime}_{\phi}}$, then $E^{*_{j}}\cap
R=A^{\star_{\phi}}\cap R_{M_{j}}=A^{\star^{\prime}_{\phi}}\cap
R_{M_{j}}=E^{*^{\prime}_{j}}\cap R$. Hence $E^{*{j}}=(E^{*_{j}}\cap
R)R_{M_{j}}=(E^{*^{\prime}_{j}}\cap R)R_{M_{j}}=E^{*^{\prime}_{j}}$, which is
absurd. Hence $A^{\star_{\phi}}\not=A^{\star^{\prime}_{\phi}}$ and so
$\phi(\star)\not=\phi(\star^{\prime})$. Thus $\phi$ is a one-to-one and
therefore $\displaystyle\prod_{i=1}^{n}|StarFC(R_{M_{i}})|\leq|StarFC(R)|$,
for every positive integer $n$. So $|StarFC(R)|=\infty$, which is a
contradiction, completing the proof of the claim.
Now assume that $\mathcal{M}_{\geq 2}=\\{M_{1},\dots,M_{r}\\}$. Let
$\varphi:StarFC(R)\longrightarrow\prod_{M\in Max(R)}StarFC(R_{M})$ defined by
$\varphi(*)=(*(M))_{M\in Max(R)}$. Clearly $\varphi$ is one-to-one and so
$|StarFC(R)|\leq\displaystyle\prod_{M\in
Max(R)}|StarFC(R_{M})|=\displaystyle\prod_{i=1}^{r}|StarFC(R_{M_{i}})|\displaystyle\prod_{M\in\mathcal{M}_{1}}|StarFC(R_{M})|=\\\
\displaystyle\prod_{i=1}^{r}|StarFC(R_{M_{i}})|\leq|StarFC(R)|$ (since
$\displaystyle\prod_{M\in\mathcal{M}_{1}}|StarFC(R_{M})|=1$ and
$\displaystyle\prod_{i=1}^{r}|StarFC(R_{M_{i}})|\leq|StarFC(R)|$ by the proof
of the claim). It follows that $|StarFC(R)|=\displaystyle\prod_{M\in
Max(R)}|StarFC(R_{M})|$. ∎

In [28, Theorem 1.5] it was proved that for a Noetherian domain $R$ with
$Star(R)$ finite, $R$ is star regular if and only if $R_{M}$ is star regular
for every maximal ideal $M$ of $R$. Our next theorem generalizes this result
to one-dimensional quasi-Prüfer domain such that $StarFC(R)$ is finite. Notice
that, for a Noetherian domain with $Star(R)$ finite, $dim(R)=1$ and so
$R^{\prime}$ is a Dedekind domain and so $R$ is a quasi-Prüfer domain.

######  Theorem 2.11.

Let $R$ be a one-dimensional quasi-Prüfer domain such that $StarFC(R)$ is
finite. Then $R$ is finitely star regular if and only if $R_{M}$ is finitely
star regular for every maximal ideal $M$ of $R$.

###### Proof.

Notice that each overring $T\subsetneqq qf(R)$ of $R$ is a one-dimensional
quasi-Prüfer domain.
$\Longleftarrow)$ Assume that $R_{M}$ is finitely star regular for every $M\in
Max(R)$. Let $T$ be a proper overring of $R$. By Theorem 2.10,
$|StarFC(T)|=\displaystyle\prod_{N\in
Max(T)}|StarFC(T_{N})|\leq\displaystyle\prod_{Q=N\cap R:N\in
Max(T)}|StarFC(R_{Q})|\leq\displaystyle\prod_{M\in
Max(R)}|StarFC(R_{M})|=|StarFC(R)|$. Thus $R$ is finitely star regular.
$\Longrightarrow)$ We mimic the proof of [28, Theorem 1.5]. Assume that $R$ is
finitely star regular and let $M$ be a maximal ideal of $R$ and suppose that
there is an overring $T$ of $R_{M}$ such that $|StarFC(T)|>|StarFC(R_{M})|$.
Set $B:=\displaystyle\bigcap_{Q\in Max(R)\setminus\\{M\\}}R_{Q}$ and $S=T\cap
B$. We first note that for every maximal ideal $N$ of $T$, $N\cap
R_{M}=MR_{M}$ since $R_{M}$ is one-dimensional local domain. So $N\cap R=M$.
But since $T$ is one-dimensional quasi-Prüfer domain, $T$ is $h$-local and so
$T$ has only finitely many maximal ideals, say $N_{1},\ldots,N_{r}$. Since $R$
is a one-dimensional quasi-Prüfer domain, it is $h$-local, and hence
$B_{R\setminus M}=K$ by [39, Theorem 22]. This yields $S_{R\setminus
M}=T_{R\setminus M}\cap B_{R\setminus M}=T\cap K=T$. It follows that
$N_{i}\cap S\neq N_{j}\cap S$ for $i\neq j$. It is clear that if $Q$ is a
maximal ideal of $S$ different from the $N_{i}\cap S$, then $Q\cap R=P$ for
some maximal ideal $P$ of $R$ distinct from $M$ and hence $P=PR_{P}\cap S$ and
$S_{Q}=R_{P}$. We then have $Max(S)=\\{N_{i}\cap S\mid
i=1,\ldots,n\\}\cup\\{PR_{P}\cap S\mid P\not=M\\}$. Therefore,

$\displaystyle|StarFC(S)|$ $\displaystyle=\prod_{i=1}^{n}|StarFC(S_{Q_{i}\cap
S})|\cdot\prod_{M\not=N}|StarFC(R_{M})|$
$\displaystyle=\prod_{i=1}^{n}|StarFC(T_{Q_{i}})|\cdot\prod_{M\neq
N}|StarFC(R_{M})|$ $\displaystyle=|StarFC(T)|\cdot\prod_{M\neq
N}|StarFC(R_{M})|$ $\displaystyle>|StarFC(R_{N})|\cdot\prod_{M\neq
N}|StarFC(R_{M})|$ $\displaystyle=|StarFC(R)|.$

Therefore, $R$ is not star regular. ∎

######  Corollary 2.12.

Let $R$ be a domain with $StarFC(R)$ is finite and which satisfies one of the
following conditions:
(1) Each proper overring of $R$ is Archimedean;
(2) Each proper valuation overring of $R$ satisfies the $accp$;
(3) Each proper overring of $R$ is a Mori domain.
Then $R$ is finitely star regular if and only if $R_{M}$ is finitely star
regular for every maximal ideal $M$ of $R$.

###### Proof.

By [6, Proposition 3.1(b), Lemma 3.2 and Corollary 3.9], either $R$ is a
valuation domain or $dim(R^{\prime})=1$ and $R^{\prime}$ is Prüfer. If $R=V$
is a valuation domain, then $R$ is finitely star regular. Assume that
$dim(R^{\prime})=1$ and $R^{\prime}$ is Prüfer. Then $R$ has the finite
character and for each overring $T$ of $R$, either $T$ is a valuation or
$dim(T)=1$ and $T^{\prime}$ is Prüfer. If $T$ is valuation,
$|StarFC(T)|=1\leq|StarFC(R)|$. Assume that $dim(T)=1$ and $T^{\prime}$ is
Prüfer. By Theorem 2.10, $|StarFC(T)|=\displaystyle\prod_{N\in
Max(T)}|StarFC(T_{N})|\leq\displaystyle\prod_{Q=N\cap R,N\in
Max(T)}|StarFC(R_{Q})|\leq\displaystyle\prod_{M\in
Max(R)}|StarFC(R_{M})|=|StarFC(R)|$ as desired. ∎

######  Corollary 2.13.

([28, Theorem 1.5]) Let $R$ be a Noetherian domain with $Star(R)$ finite.
Then $R$ is star regular if and only if $R_{M}$ is star regular for every
maximal ideal $M$ of $R$.

###### Proof.

Notice that $Star(R)=StarFC(R)$ and $dimR=1$. Since $R^{\prime}=\overline{R}$
is a Krull domain, $R^{\prime}$ is a Dedekind domain. The conclusion follows
now from Theorem 2.11. ∎

##  3\. pullback constructions

Let $T$ be a domain, $M$ a maximal ideal of $T$, $K$ its residue field,
$\phi:T\longrightarrow K$ the canonical surjection, $D$ a proper subring of
$K$, and $k:=qf(D)$. Let $R$ be the pullback issued from the following diagram
of canonical homomorphisms:

$\begin{array}[]{cccl}&R:=\phi^{-1}(D)&\longrightarrow&D\\\ (\ \square\
)&\downarrow&&\downarrow\\\
&T&\stackrel{{\scriptstyle\phi}}{{\longrightarrow}}&K=T/M.\end{array}$

Clearly, $M=(R:T)$ and $D\cong R/M$. For ample details on the ideal structure
of $R$ and its ring-theoretic properties, we refer the reader to [2, 3, 9, 10,
13, 14, 19]. The case where $T=V$ is a valuation domain is crucial and we will
refer to this case as a classical diagram of type $(\square)$. Notice that for
the classical diagram, if $I$ is a (fractional) ideal of $R$, then either
$I=\phi^{-1}(E)$ for some $E\in F(D)$ if $M\subset I$, or $I$ is an ideal of
$V$ or $I=a\phi^{-1}(E)$ for some $0\not=a\in M$ and $E$ a $D$-submodule of
$K$ with $D\subseteq E\subset K$ if $I$ is not an ideal of $V$ (the proof
similar to that of [9, Theorem 2.1]). For more on star operations on
pullbacks, see [15, 16].

######  Theorem 3.1.

For the classical diagram of type$(\square)$, assume that $qf(D)=K$. Then $R$
is finitely star regular if and only if $D$ is finitely star regular.

###### Proof.

Assume that $D$ is finitely star regular and let $T$ be an overring of $R$. If
$V\subseteq T$, then $T$ is a valuation domain and so
$|StarFC(T)|=1\leq|StarFC(R)|$. Assume that $R\subseteq T\subsetneqq V$. Then
$T=\phi^{-1}(D_{1})$ where $D_{1}$ is an overring of $D$. Now by [34, Theorem
4.4], $|StarFC(T)|=|StarFC(D_{1})|\leq|StarFC(D)|=|StarFC(R)|$, and therefore
$R$ is finitely star regular.
Conversely, assume that $R$ is finitely star regular and let $D_{1}$ be an
overring of $D$. Set $T=\phi^{-1}(D_{1})$. Then $T$ is an overring of $R$ and
again by [34, Theorem 4.4],
$|StarFC(D_{1})|=|StarFC(T)|\leq|StarFC(R)|=|StarFC(D)|$, and therefore $D$ is
finitely star regular. ∎

######  Example 3.2.

Let $k$ be a field , $X$ an indeterminate over $k$ and set
$D=k[[X^{3},X^{4},X^{5}]]$, $V=k((X))[[Y]]=k((X))+M$ and $R=D+M$. Clearly $R$
is neither integrally closed nor Noetherian domain. By [26, Theorem 3.8],
$|StarFC(D)|=|Star(D)|=3$. Since the only overrings of $D$ are
$D_{1}=k[[X^{2},X^{3}]]$, $D_{2}=k[[X]]$ and $qf(D)=k((X))$ and since $D_{1}$
and $D_{2}$ are Noetherian divisorial domains, $D$ is finitely star regular.
Hence $R$ is finitely star regular by Theorem 3.1. In fact the only proper
overrings of $R$ are $T_{1}=D_{1}+M$, $T_{2}=D_{2}+M$ and $V$. By [34, Theorem
4.4], $|StarFC(T_{1})|=|StarFC(T_{2})|=|StarFC(V)|=1$ while $|StarFC(R)|=3$.

Our next theorem deals with an important of class of finitely star regular
domains that are not in general star regular. It shows that any $PVD$ is a
finitely star regular domain. Recall from Hedstrom and Houston ([21]) that a
domain $R$ is pseudo-valuation domain if it is quasilocal and shares its
maximal ideal with a valuation domain which necessarily must contain $R$ and
be unique. In terms of pullbacks, according to [3, Proposition 2.6], $R$ is a
pseudo-valuation domain if and only if there is a valuation domain $V$ with
maximal ideal $M$ and a subfield $k$ of $V/M=K$ such that $R$ is the pullback
in the following diagram

$\begin{array}[]{ccl}R&\longrightarrow&k\\\ \downarrow&&\downarrow\\\
V&\stackrel{{\scriptstyle\phi}}{{\longrightarrow}}&K=V/M\end{array}$

Notice that a $PVD$ which is not a valuation domain is a $TV$-domain, that is,
the $t$\- and $v$-operations are the same ([31, Proposition 4.3]). We start
with the following useful lemma.

######  Lemma 3.3.

Let $R$ be a $PVD$ (which is not a valuation domain), $V$ its associated
valuation overring, $M$ its maximal ideal, $k=R/M$ and $K=V/M$. If $StarFC(R)$
is finite, then $K$ is algebraic over $k$.

###### Proof.

Assume that $StarFC(R)$ is finite and suppose that $K$ is transcendental over
$k$. Let $\lambda\in K$ transcendental over $k$ and set
$T=\phi^{-1}(k[\lambda])$. Since $(k:k[\lambda])=(0)$,
$T^{-1}=(R:T)=\phi^{-1}(k:k[\lambda])=\phi^{-1}(0)=M$, and so $T_{t}=T_{v}=V$.
Now, for every nonzero prime ideals $p\not=q$ of $k[\lambda]$, set
$T_{P}=\phi^{-1}(k[\lambda]_{P})$ and $T_{q}=\phi^{-1}(k[\lambda]_{q})$. By
Lemma 2.6, $\delta(d_{T_{P}},t)$ and $\delta(d_{T_{q}},t)$ are star operations
on $R$ of finite type and $T^{\delta(d_{T_{P}},t)}=TT_{P}\cap T_{t}=T_{P}\cap
V=T_{P}$ and $T^{\delta(d_{T_{q}},t)}=TT_{q}\cap T_{t}=T_{q}\cap V=T_{q}$.
Thus $\delta(d_{T_{P}},t)\not=\delta(d_{T_{q}},t)$. As $Spec((k[\lambda])$ is
inifinte, $starFC(R)$ would be infinite, which is absurd. It follows that $K$
is algebraic over $k$. ∎

######  Theorem 3.4.

Any $PVD$ is finitely star regular.

###### Proof.

First, if $R$ is a valuation domain, then for every overring $T$ of $R$,
$|StarFC(T)|=|StarFC(R)|=1$. So, without loss of generality, we may assume
that $R$ is not a valuation domain and $StarFC(R)$ is finite. Let $V$ be the
associated valuation of $R$, $M$ its maximal ideal, $k=R/M$ and $K=V/M$. By
Lemma 3.3, $K$ is algebraic over $k$. Now, let $T$ be a proper overring of
$R$. If $V\subseteq T$, then $T$ is a valuation domain and so
$|StarFC(T)|=1\leq|StarFC(R)|$, as desired. Assume that $R\subsetneq
T\subsetneq V$. Then $T=\phi^{-1}(F)$ where $F$ is a subfield of $K$ with
$k\subsetneqq F\subsetneqq K$. We claim that
$\delta(-,t):StarFC(T)\longrightarrow StarFC(R),*\mapsto\delta(*,t)$ is a one-
to-one map. Indeed, let $*\not=*^{\prime}\in StarFC(T)$ and let $A$ be a
finitely generated integral ideal of $T$ such that $A^{*}\not=A^{*^{\prime}}$.
Necessarily $A\subsetneqq M$ and $A$ is not an ideal of $V$. Then
$A=a\phi^{-1}(W)$ where $F\subsetneqq W\subsetneqq K$ is a finite dimensional
$k$-subspace of $K$. Set $W=\displaystyle\sum_{i=1}^{i=r}F\lambda_{i}$ and let
$H=\displaystyle\sum_{i=1}^{i=r}k\lambda_{i}$ and $B=a\phi^{-1}(H)$. Then $B$
is a finitely generated ideal of $R$, $BT=A$ and $B_{t}=B_{v}=A_{v_{T}}=aV$.
Thus $B^{\delta(*,t)}=(BT)^{*}\cap B_{t}=A^{*}\cap A_{v_{T}}=A^{*}$.
Similarly,$B^{\delta(*^{\prime},t)}=(BT)^{*^{\prime}}\cap
B_{t}=A^{*^{\prime}}\cap A_{v_{T}}=A^{*^{\prime}}$. Hence
$B^{\delta(*,t)}\not=B^{\delta(*^{\prime},t)}$ and therefore $\delta(-,t)$ is
one-to-one. It follows that $|StarFC(T)|\leq|StarFC(R)|$ and therefore $R$ is
finitely star regular.
∎

Notice that a $PVD$ is not necessarily a star regular domain as shown by the
following example.

######  Example 3.5.

Let $V$ be a valuation domain with a principal maximal ideal $M$ and a non-
maximal prime ideal $P$ such that $P=P^{2}$. Suppose that $K=V/M$ is a
quadratic extension of a field $k$ and let $R$ be the $PVD$ arising from the
diagram:

$\begin{array}[]{cccl}&R:=\phi^{-1}(k)&\longrightarrow&k\\\ (\ \square\
)&\downarrow&&\downarrow\\\
&V&\stackrel{{\scriptstyle\phi}}{{\longrightarrow}}&K=V/M.\end{array}$

Since $[K:k]=2$, $R$ is a divisorial domain and so $|Star(R)|=1$. However,
$V_{P}$ is an overring of $R$ and $|Star(V_{P})|=2$. Hence $R$ is not star
regular.

######  Example 3.6.

Let $k=\mathbb{Z}_{2}$ and $K$ an extension of $k$ with $[K:k]=4$ (for
instance, let $x$ be a root of the irreducible polynomial
$f(Y)=Y^{4}+Y^{3}+1\in k[Y]$ and $K=k(x)$). Let $X$ be an indeterminate over
$k$ and set $V=K[[X]]=K+M$ and $R=k+M$. Let $T$ be a proper overring of $R$.
If $V\subseteq T$, $T$ is a valuation domain and so
$|StarFC(T)|=1\leq|StarFC(R)|$. If $R\subsetneqq T\subsetneqq V$, then
$T=\phi^{-1}(F)$ where $k\subsetneqq F\subsetneqq K$ is a subfield of $K$.
Necessarily $[K:F]=2$ and so $T$ is a divisorial $PVD$. Hence
$|StarFC(T)|=|Star(T)|=1\leq|Star(R)|=|StarFC(R)|=9$ by [44].

Recall that for the general pullback of type $(\square)$, every star operation
$*$ on $R$ induces a star operation $*_{\phi}$on $D$ defined by
$J^{*_{\phi}}=\phi((\phi^{-1}(J))^{*})$ for every $J\in F(D)$ ([15,
Proposition 2.7 and Proposition 2.6]. In this context, it is easy to check
that if $*$ is of finite type on $R$, then $*_{\phi}$ on $D$ is of finite type
on $D$.

######  Theorem 3.7.

For the classical pullback diagram of type $(\square)$, let
$T=\phi^{-1}(D_{1})$ be an overring of $R$. Then
$|StarFC(T)|\leq|StarFC(D_{1})||StarFC(R)|$. In particular, if
$|StarFC(D_{1})|=1$ for every $D_{1}\in[D,K]$, then $R$ is finitely star
regular.

###### Proof.

Set $T=\phi^{-1}(D_{1})$ and consider the map $\delta:StarFC(T)\longrightarrow
StarFC(D_{1})\times StarFC(R)$,
$*\mapsto(*_{\phi},\delta(*,t))$. We claim that $\delta$ is one-to-one.
Indeed, let $*\not=*^{\prime}\in StarFC(T)$ and let $A$ be a finitely
generated integral ideal of $T$ such that $A^{*}\not=A^{*^{\prime}}$.
Necessarily $A$ is not an ideal of $V$. If $M\subsetneqq A$, then
$A=\phi^{-1}(J)$ for some finitely ideal $J$ of $D_{1}$. In this case
$A^{*}=\phi^{-1}(J^{*_{\phi}})$ and
$A^{*^{\prime}}=\phi^{-1}(J^{*^{\prime}_{\phi}})$. Thus
$J^{*_{\phi}}\not=J^{*^{\prime}_{\phi}}$ and so
$*_{\phi}\not=*^{\prime}_{\phi}$. Assume that $A\subsetneqq M$ and set
$A=a\phi^{-1}(W)$ where $D_{1}\subsetneqq W\subsetneqq K$ is a finitely
generated $D_{1}$-module. If $(D_{1}:W)\not=0$, then $W\subseteq qf(D_{1})$
and so $W$ would be a finitely generated fractional ideal of $D_{1}$. Thus
$A^{*}=a\phi^{-1}(W^{*_{\phi}})$ and
$A^{*^{\prime}}=a\phi^{-1}(W^{*^{\prime}_{\phi}})$. Thus
$W^{*_{\phi}}\not=W^{*^{\prime}_{\phi}}$ and so
$*_{\phi}\not=*^{\prime}_{\phi}$. Assume that $(D_{1}:W)=0$. Set
$W=\displaystyle\sum_{i=1}^{i=r}D_{1}\lambda_{i}$ and let
$H=\displaystyle\sum_{i=1}^{i=r}D\lambda_{i}$ and $B=a\phi^{-1}(H)$. Then $B$
is a finitely generated ideal of $R$, $BT=A$ and $B_{t}=B_{v}=A_{v_{T}}=aV$.
Thus $B^{\delta(*,t)}=(BT)^{*}\cap B_{t}=A^{*}\cap A_{v_{T}}=A^{*}$.
Similarly,$B^{\delta(*^{\prime},t)}=(BT)^{*^{\prime}}\cap
B_{t}=A^{*^{\prime}}\cap A_{v_{T}}=A^{*^{\prime}}$. Hence
$B^{\delta(*,t)}\not=B^{\delta(*^{\prime},t)}$. Thus
$\delta(*)\not=\delta(*^{\prime})$ and hence $\delta$ is one-to-one. It
follows that $|StarFC(T)|\leq|StarFC(D_{1}||StarFC(R)|$.
Now assume that $|StarFC(D_{1}|=1$ for every $D_{1}\in[D,K]$ and let $T$ be an
overring of $R$. If $V\subseteq T$, then $|StarFC(T)|=1\leq|StarFC(R)$. Let
$T$ be a proper overring of $R$. Assume that $R\subsetneqq T\subsetneq V$ and
$T=\phi^{-1}(D_{1})$ where $D\subsetneqq D_{1}\subsetneqq K$. Then
$|StarFC(T)|\leq|StarFC(D_{1})||StarFC(R)|=|StarFC(R)|$ and therefore $R$ is
finitely star regular. ∎

######  Example 3.8.

Let $\mathbb{Q}$ be the field of rational numbers, and $X$ and $Y$
indeterminates over $\mathbb{Q}$. Set $D=\mathbb{Q}[[X^{2},X^{3}]]$,
$V=\mathbb{Q}(\sqrt{2})((X))[[Y]]=\mathbb{Q}(\sqrt{2})((X))+M$ and $R=D+M$.
Clearly
$[D,K]=\\{D,\mathbb{Q}[[X]],\mathbb{Q}((X)),\mathbb{Q}(\sqrt{2})[[X^{2},X^{3}]],\mathbb{Q}(\sqrt{2})[[X]],K\\}$
and every $D_{1}\in[D,K]$ is divisorial. Thus $R$ is finitely star regular.

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# Applications of infinite lower triangular matrices and their group structure
in combinatorics and the theory of orthogonal polynomials.

Paweł J. Szabłowski  Department of Mathematics and Information Sciences,
Warsaw University of Technology
ul Koszykowa 75, 00-662 Warsaw, Polandg<EMAIL_ADDRESS>
(Date: January, 2023)

###### Abstract.

We define important properties of the group of infinite lower-triangular
matrices. In particular, we define several subgroups of this group including
subgroups of the set of Riordan matrices. We also present many examples (in
fact around 20) of such matrices composed of the sequences of important
families of polynomials or numbers as entries of certain lower-triangular
infinite matrices. By applying the well-known matrix operations mostly such as
multiplication and calculating inverse, we see that such representation can be
and is a source of getting new, important relationships between these families

###### Key words and phrases:

lower triangular matrices, Riordan matrices, Appell, Bell, associate, Pascal
subgroups, Bernoulli and Euler polynomials and numbers, orthogonal
polynomials, raising factorials, $q-$series theory, $q-$Hermite, Al-
Salam–Chihara, Rogers-Szegö polynomials.

###### 2020 Mathematics Subject Classification:

Primary 05C198, 33C45A15; Secondary 20H25, 16S15

##  1\. Introduction and notation

Let
$\mathbf{A}\allowbreak\overset{df}{=}\allowbreak\left[a_{n,j}\right]_{n,j\geq
0,}$ with $a_{nj}\allowbreak=\allowbreak 0$ for all $j>n\geq 0$, be a lower
triangular infinite matrix with entries belonging to $\mathbb{C}$ in general.
In the sequel we will consider also some sequences and their generating
functions. In general, these sequences as well as variables in the formal
power series being the expansion of these GF can be complex. However, in
almost all cases we will work with real sequences as well as real GF’s. That
is why we will assume that all variables and numbers will be real. We will
also be aware that without any difficulty the results can be extended to the
complex case. For certain reasons that will be obvious in the sequel, let us
agree that both index entries will start ranging from $0$. It is elementary to
notice that such matrices form a linear space but also more importantly the
non-commutative ring with the ordinary matrix addition and multiplication as
the two ring operations. More precisely by $\mathbf{A\allowbreak+\allowbreak
B\allowbreak=\allowbreak}\left[a_{n,j}+b_{n,j}\right]$ if
$\mathbf{A\allowbreak=\allowbreak[}a_{n,j}]$ and
$\mathbf{B}\allowbreak=\allowbreak$ $\left[b_{n,j}\right]$ and
$\mathbf{AB}\allowbreak=\allowbreak\left[\sum_{k=j}^{n}a_{n,k}b_{k,j}\right]$.
Let us notice that the set of diagonal matrices make a commutative subring of
our ring. Let us agree that diagonal matrices will be denoted in the following
way. Namely, $\left[\left\\{\beta_{n}\right\\}\right]$ will denote diagonal
matrix with $\beta_{n}$ as its ($n+1)\times(n+1)$ entry. Among these diagonal
matrices the one with all entries equal to zero serves as the zero element of
the ring and the diagonal matrix $\left[\left\\{1\right\\}\right]$ (i.e. with
all diagonal elements equal to $1)$ as the ring’s ”one”. These special
matrices will be denoted respectively $\mathbf{0}$ and $\mathbf{1}$. Let us
denote by $\mathcal{S}$ the whole ring of lower triangular matrices and by
$\mathcal{D}$ the subring of all diagonal matrices.

As noticed in [7], every such infinite matrix can be understood as the
sequence of $n\times n$ lower triangular matrices organized in such a way that
the matrix $\mathbf{A}_{n}$ is a submatrix of $\mathbf{A}_{n+1}$ and we have

$\mathbf{A}_{n+1}\allowbreak=\allowbreak\begin{bmatrix}\mathbf{A}_{n}&0\\\
\mathbf{a}_{n}^{T}&\alpha_{n,n}\end{bmatrix},$

where $\mathbf{c}_{n}^{T}$ is certain row vector of dimension $n$ and
$\alpha_{n,n}\neq 0$ certain real number. We will denote such sequences of
matrices by $\left\\{\mathbf{A}_{n}\right\\}$ meaning that we deal with a
lower triangular infinite matrix whose $n\times n$ matrices in their upper
left corners are matrices $\mathbf{A}_{n}$, $n\allowbreak=\allowbreak
1,\ldots.$ Notice also that in accordance with our index convention we have
$\mathbf{A}_{1}\allowbreak=\allowbreak\alpha_{0,0}\allowbreak\overset{df}{=}\allowbreak\alpha_{0}.$
Let us extend this convention and denote all diagonal elements $a_{n,n}$ by
single index that is $a_{n}$ if of course
$\mathcal{S\allowbreak\ni}\allowbreak\mathbf{A\allowbreak=\allowbreak}\left[a_{k,j}\right]$.
In the sequel, we will us convention:
$\mathbf{A}_{n}\allowbreak=\allowbreak[a_{k,j}]_{n}$.

Using this observation and the well-known formulae for the block
multiplication of matrices we see that

(1.1) $\mathbf{A}_{n+1}\mathbf{B}_{n+1}=\begin{bmatrix}\mathbf{A}_{n}&0\\\
\mathbf{a}_{n}^{T}&\alpha_{n}\end{bmatrix}\begin{bmatrix}\mathbf{B}_{n}&0\\\
\mathbf{b}_{n}^{T}&\beta_{n}\end{bmatrix}=\begin{bmatrix}\mathbf{A}_{n}\mathbf{B}_{n}&0\\\
\mathbf{a}_{n}^{T}\mathbf{B}_{n}+\alpha_{n}\mathbf{b}_{n}^{T}&\alpha_{n}\beta_{n}\end{bmatrix}.$

It is also elementary to notice that this ring has divisors of zero. The
following two matrices are the examples of the left and right divisors of zero
in this ring with nonzero elements on the diagonal:

$\begin{bmatrix}a_{0}&0&0\\\ a_{1}&0&0\\\
a_{2}&0&0\end{bmatrix},~{}~{}\begin{bmatrix}0&0&0\\\ 0&b_{1}&0\\\
b_{2}&b_{3}&b_{4}\end{bmatrix}.$

On the other hand matrices having a nonzero element on the $(0,0)$ position ,
i.e., in the upper left corner cannot be right divisors of zero.

Notice also that diagonal matrices with the same number, say $\alpha,$ on the
diagonal , i.e., $\left[\left\\{\alpha\right\\}\right]$ can be identified with
this number since from the formula (1.1) it follows that

$[a_{n,j}][\left\\{\alpha\right\\}]=[\alpha a_{n,j}],$

Now let us discuss the set of matrices $\mathbf{A}$ that are invertible ,
i.e., such matrices for which there exists a matrix (called inverse and
denoted by $\mathbf{A}^{-1}$ such that

$\mathbf{AA}^{-1}\allowbreak=\allowbreak\mathbf{A}^{-1}\mathbf{A\allowbreak=\allowbreak
1.}$

Now notice that if a matrix form $\mathcal{S}$, say $\mathbf{A}$ has
representation $\left\\{\mathbf{A}_{n}\right\\}$ then the diagonal elements of
each of the matrices $\mathbf{A}_{n}$ are their eigenvalues. Consequently,
each matrix $\mathbf{A}_{n}$ having nonzero elements on the diagonal is
invertible. So the set of invertible elements of $\mathbf{\mathcal{S}}$
defined by the conditions: $\mathbf{A\allowbreak=\allowbreak[}a_{n,j}]$ is
invertible iff

$a_{n,j}=0\text{ if }j>n\text{ and }a_{j,j}\neq 0.$

Let us denote by $\mathcal{I}$ the set of invertible matrices. It is
elementary to notice that set of all invertible matrices forms a linear cone ,
i.e., if $\mathbf{A}$ is in $\mathcal{I}$ then all matrices of a form
$\mathbf{A[}\beta]\in\mathcal{I}$ for all $\beta\neq 0$. The family of such
matrices forms a field with the the field operations being the ordinary matrix
addition and multiplication.

The other properties of elements of $\mathcal{S}$ considered as linear
operators, i.e. their eigenvectors and some decompositions at least for the
special, more precisely Riordan matrices, are presented in a recently
published paper [2].

As it follows from, say works T. Banachiewicz or H. Bolt, the formula for
inversion of block matrices we have

(1.2) $\mathbf{A}_{n+1}^{-1}=\begin{bmatrix}\mathbf{A}_{n}^{-1}&0\\\
-\mathbf{a}_{n}^{T}\mathbf{A}_{n}^{-1}/\alpha_{n}&\alpha_{n}^{-1}\end{bmatrix}.$

As a corollary we have the following observation.

######  Remark 1.

1\. If matrix $\mathbf{A}$ has all integer entries and moreover it has $1$ on
its diagonal, then matrix $\mathbf{A}^{-1}$ has also integer entries and
$1^{\prime}s$ on its diagonal.

2\. If matrix $\mathbf{A}$ has polynomial entries except for the diagonal
whose entries are real, then its inverse also has polynomial entries.

3\. If $\mathbf{A\allowbreak=\allowbreak}\left[a_{nj}\right]$ and
$\mathbf{A}^{-1}\allowbreak=\allowbreak\left[b_{nj}\right],$ then also the
following two matrices are inverses of one another:

3a.
$\forall\lambda\in\mathbb{C}:\mathbf{F(}\lambda)\allowbreak\mathbf{=\allowbreak}\left[a_{nj}\lambda^{n-j}\right]$
and
$\mathbf{B(}\lambda\mathbf{)}\allowbreak=\allowbreak\left[b_{nj}\lambda^{n-j}\right].$

3b. Suppose $\left\\{\alpha_{n}\right\\}_{n\geq 1}$, $\alpha_{n}\neq
0:\mathbf{\tilde{A}\allowbreak=\allowbreak}\left[\alpha_{n}a_{n,j}\right]$ and
$\mathbf{\tilde{B}\allowbreak=\allowbreak}\left[b_{n,j}/\alpha_{j}\right].$

4.

$\left[\left\\{\alpha_{n}\right\\}\right]\left[a_{n,j}\right]\left[\left\\{\beta_{j}\right\\}\right]=\left[\alpha_{n}a_{n,j}\beta_{j}\right],$

In particular

(1.3)
$\left[\left\\{\alpha_{n}\right\\}\right]\left[a_{n,j}\right]\left[\left\\{\alpha_{j}\right\\}\right]^{-1}=\left[\alpha_{n}a_{n,j}/\alpha_{j}\right].$

5\. Suppose $\mathbf{A\allowbreak=\allowbreak}\left[d_{n-j}\right],$ for some
sequence $\left\\{d_{j}\right\\},$ with $d_{0}\allowbreak=\allowbreak 1,$ then
$\mathbf{A}^{-1}\allowbreak=\allowbreak\left[e_{n-j}\right],$ where sequence
$\left\\{e_{j}\right\\}$ is defined by the recursion:

$e_{0}=1,~{}~{}e_{k}=-\sum_{s=0}^{k-1}d_{k-s}e_{s}.$

Naturally $\mathcal{S}$ is also a group if we confine ourselves to
multiplication. We will retain the name $\mathcal{S}$ for the group.

Let us notice that for every lower triangular matrix, say
$\mathbf{A}\allowbreak=\allowbreak[a_{n,i}]\mathbf{\in}S$ we can define a
formal power series

$\mathcal{A}(x,y)=\sum_{i=0}^{\infty}\sum_{n=i}^{\infty}x^{i}y^{n}a_{n,i}=\sum_{i=0}^{\infty}x^{i}g_{i}(y),$

where $g_{i}(y)\allowbreak=\allowbreak\sum_{n\geq i}y^{n}a_{n,i}$ that we call
GF of the matrix $\mathbf{A}$. We will use in the sequel the following
notation used in the theory of formal power series (FPS). Namely, if
$p(x)\allowbreak=\allowbreak\sum_{n\geq 0}p_{n}x^{n}$ is a FPS, then
$p_{n}\allowbreak$ is often denoted by :
$\left[\left[x^{n}\right]\right]p(x).$

##  2\. Riordan arrays

One of the most important subgroups of this group is the so-called Riordan
group $\mathcal{R}$ defined in many positions of literature through by the
properties of generating functions (GF) of the entries of elements of
$\mathcal{S}$.

The Riordan group is characterized by the fact that $\forall i\geq
0:g_{i}(y)\allowbreak=\allowbreak f(y)h^{i}(y),$ for some formal power series
$f(y)\allowbreak=\allowbreak f_{0}+\sum_{j\geq 1}f_{j}y^{j}$ and
$h(y)\allowbreak=\allowbreak\sum_{j\geq 1}h_{j}y^{j},$ with $f_{0},h_{1}\neq
0.$ For an element of the Riordan group, we thus have

$\mathcal{A}(x,y)=f(y)\sum_{i=0}^{\infty}x^{i}h^{i}(y)=\frac{f(y)}{1-xh(y)}.$

Hence, the Riordan matrix characterized by the two generating functions $f$
and $h$ is traditionally denoted as $\left(f,h\right).$ It is well-known that
the product of two Riordan matrices, is a Riordan matrix. Hence, the Riordan
matrices form a subgroup $\mathcal{R}$ of the group $\mathcal{S}$.
Unfortunately $\mathcal{R}$ is not a ring since in general a sum of two
Riordan matrices is not Riordan. There appeared many papers concerning Riordan
matrices through the years. They presented many features of this group. We
will recall some of these results concerning mostly the group structure of the
Riordan group for the sake of completeness of the paper.

We start with the following well-known, old result:

######  Theorem 1 (Roman).

Suppose we have two Riordan matrices: $(a(x),b(x))$ and $(c(x),d(x))$ where
functions $a,b,c,d$ are such that $a(0)\allowbreak\neq\allowbreak
0\neq\allowbreak c(0),$ $b(0)\allowbreak=\allowbreak
d(x)\allowbreak=\allowbreak 0$ and $b(0)\neq\allowbreak
0\allowbreak\neq\allowbreak d(0)$. Then their product is a Riordan matrix
$(a(x)c(b(x)),d(b(x))$ and its inverse the Riordan matrix $(a(x),b(x))$ is
equal to $\left(1/a(\bar{b}(x),\bar{b}(x)\right),$ where the function
$\bar{b}$ is defined by the relationship:
$\bar{b}(b(x))\allowbreak=\allowbreak
b\left(\bar{b}\left(x\right)\right)\allowbreak=\allowbreak x$.

###### Proof.

Following [8] the proof is presented in [4], p.43.

We have the following couple of simple observations concerning subgroups of
$\mathcal{R}$.

######  Lemma 1.

1\. Riordan matrices of the form $\left(p(x),\beta x/(1-\alpha x)\right),$ for
some reals $\alpha$, $\beta$ and a formal power series $p\left(x\right)$ such
that $p(0)\neq 0$ forms a subgroup of $\mathcal{R}$. The multiplication rule
within this subgroup is the following:

$\left(p_{1}(x),\frac{\beta_{1}x}{(1-\alpha_{1}x)}\right)\left(p_{2}(x),\frac{\beta_{2}x}{(1-\alpha_{2}x)}\right)=\left(p_{1}(x)p_{2}\left(\frac{\beta_{1}x}{(1-\alpha_{1}x)}\right),\frac{\beta_{1}\beta_{2}x}{(1-(\alpha_{1}+\beta_{1}\alpha_{2})x)}\right).$

We will denote this subgroup by $\mathcal{I}(p,\beta,\alpha)$.

2\. The Riordan matrices of the form $\left(p(x),ax\right),$ for some real $a$
and a formal power series $p(x)$ such that $p(0)\neq 0$ forms a subgroup of
$\mathcal{I}(p,a,\alpha).$ The multiplication rule within this subgroup is the
following:

$(p_{1}(x),a_{1}x)(p_{2}(x),a_{2}x)=\left(p_{1}(x)p_{2}(a_{1}x\right),a_{1}a_{2}x).$

We will denote this subgroup by $\mathcal{L(}p(x),a).$ It has a subgroup
$\mathcal{L(}p(x),1)$ which we will denote by $\mathcal{A}$ (called Appell
subgroup) or $\mathcal{A(}p)$ if one wants to underline the dependence on the
formal power series $p$. Moreover, $\mathcal{A}$ has a subgroup consisting of
elements such that $p(0)\allowbreak=\allowbreak 1.$ Let us denote it by
$\mathcal{O}$.

3\. Riordan matrices of the form $\left(p(x),x/(1-\alpha x)\right),$ for some
real $\alpha$, and a formal power series $p\left(x\right)$ such that $p(0)\neq
0$ forms a subgroup of $\mathcal{R}$. The multiplication rule within this
subgroup is the following:

$\left(p_{1}(x),\frac{x}{(1-\alpha_{1}x)}\right)\left(p_{2}(x),\frac{x}{(1-\alpha_{2}x)}\right)=\left(p_{1}(x)p_{2}\left(\frac{x}{(1-\alpha_{1}x)}\right)\right),\frac{x}{(1-(\alpha_{1}+\alpha_{2})x)}).$

We will denote this subgroup by $\mathcal{P}(p,\alpha)$ and call matrices of
this form a generalized Pascal matrices (briefly GP).

4\. Riordan matrices of the form $\left(1,p(x)\right)$ make another subgroup
of $\mathcal{R}$ ( called associated subgroup, denoted by $\mathcal{C}$) with
the following multiplication rule:

$(1,p_{1}(x))(1,p_{2}(x))=(1,p_{2}(p_{1}(x))).$

5\. Riordan matrices of the form $\left(g(x),xg(x)\right)$ (equivalently
$(f(x)/x,f(x))$) make another subgroup (called Bell subgroup, denoted by
$\mathcal{B}$) with the following multiplication rule:

$\left(g_{1}(x),xg_{1}(x)\right)(g_{2}(x),xg_{2}(x))=(g_{1}(x)g_{2}(xg_{1}(x)),xg_{1}(x)g_{2}(xg_{1}(x))).$

6\. We have:

$\mathcal{S\supset
R\supset\mathcal{I}}(p,a,\alpha)\mathcal{\supset\mathcal{P}}(p,\alpha)\mathcal{\supset
A}(p)\mathcal{\supset O}(p),\text{~{}}\mathcal{R\supset C}.$

Notice also that in general $\mathcal{O\nsupseteqq D}$ unless elements of
$\mathcal{D}$ are of the form $\left[\left\\{r^{n}\right\\}\right]$ for some
$r.$

###### Proof.

The proof is elementary, based entirely on the Theorem 1.

######  Remark 2.

All subgroups mentioned in Lemma 1 except for general Pascal subgroup are
known and mentioned , e.g., in [3] or even earlier, like, e.g., the associated
subgroup is mentioned in [8].

In [3], it was shown that Appell subgroup is normal in $\mathcal{R}$.

######  Remark 3.

Notice also, following [3], that every Riordan matrix $(g(x),f(x))$ can be
presented as the product of an Appell matrix and a Bell one. Namely, we have

$(xg(x)/f(x),x)(f(x)/x,f(x))=(g(x),f(x)).$

######  Remark 4.

Since the matrix $\left(1,x\right)$ plays the rôle of a neutral element in the
Riordan group we see that the inverse of $\left(p(x),x/(1-\alpha
x)\right)^{-1}$ is equal to $\left(p_{1}(x),x/(1+\alpha x)\right)$ where
$p_{1}(x)$ is such that $p_{1}(x/(1-\alpha x))\allowbreak=\allowbreak 1/p(x).$

Hence, let us analyze subgroup $\mathcal{P}(p,\alpha)$ in more detail.

######  Proposition 1.

Let us consider a generalized Pascal matrix $\left(p(x),x/(1-\alpha
x)\right).$ Let us assume $\alpha\in\mathbb{R}$ and that has the following
expansion $p(x)\allowbreak=\allowbreak\sum_{n\geq 0}p_{n}x^{n}$ with
$p_{0}\neq 0$. We have:

1.

$\left[\left[x^{n}\right]\right]p(x)\left(x/(1-\alpha
x)\right)^{k}=\sum_{j=0}^{n-k}p_{j}\alpha^{n-k-j}\binom{n-j-1}{k-1}.$

2.

$p(x/(1-\alpha
x))=p_{0}+\sum_{s=1}^{\infty}x^{s}\sum_{j=0}^{s-1}p_{j+1}\alpha^{s-1-j}\binom{s-1}{j}.$

###### Proof.

1\. Let us find $\left[\left[x^{n}\right]\right]p(x)\left(x/(1-\alpha
x)\right)^{k}.$ To do this, let us expand $\left(x/(1-\alpha x)\right)^{k}$ as
the power series. It is easy, noticing that for $\alpha\allowbreak=\allowbreak
0$ the series is finite and consists of only one element, while for
$\alpha\neq 0$ we have:

$\left(x/(1-\alpha x)\right)^{k}=\frac{1}{\alpha^{k}}\left(\frac{\alpha
x}{1-\alpha x}\right)^{k}\allowbreak=\allowbreak\frac{1}{\alpha^{k}}(1-\alpha
x)\sum_{n\geq k}\binom{n}{k}\left(\alpha x\right)^{n},$

by the well-known binomial theorem. Now applying Cauchy multiplication of
series, we get:

$\displaystyle p(x)\left(x/(1-\alpha
x)\right)^{k}=\left(\frac{1}{\alpha^{k}}(1-\alpha x)\sum_{n\geq
0}\binom{n}{k}\left(\alpha x\right)^{n}\right)$
$\displaystyle=\left(\sum_{j\geq 0}p_{j}x^{j}\right)\frac{1-\alpha
x}{\alpha^{k}}\alpha^{k}x^{k}\sum_{n\geq k}x^{n-k}\alpha^{n-k}\binom{n}{k}$
$\displaystyle=x^{k}\left(1-\alpha x\right)\left(\sum_{j\geq
0}p_{j}x^{j}\right)\sum_{m=0}^{\infty}\alpha^{m}x^{m}\binom{m+k}{k}$
$\displaystyle=(1-\alpha
x)x^{k}\sum_{s=0}^{\infty}x^{s}\sum_{j=0}^{s}p_{j}\alpha^{s-j}\binom{s-j+k}{k}.$

Hence, further we get

$\displaystyle\left[\left[x^{n}\right]\right]p(x)\left(x/(1-\alpha
x)\right)^{k}=\sum_{j=0}^{n-k}p_{j}\alpha^{n-k-j}\binom{n-k-j+k}{k}$
$\displaystyle-\alpha\sum_{j=0}^{n-k-1}p_{j}\alpha^{n-k-1-j}\binom{n-k-1-j+k}{k}$
$\displaystyle=\sum_{j=0}^{n-k}p_{j}\alpha^{n-k-j}\binom{n-j}{k}-\sum_{j=0}^{n-k-1}p_{j}\alpha^{n-k-j}\binom{n-1-j}{k}$
$\displaystyle=p_{n-k}+\sum_{j=0}^{n-k-1}p_{j}\alpha^{n-k-j}\binom{n-j-1}{k-1}=\sum_{j=0}^{n-k}p_{j}\alpha^{n-k-j}\binom{n-j-1}{k-1}$

2\. Let us analyze also what is the formal power series (FPS) of
$p(x/(1-\alpha x))$ provided $FPS$ of $p(x)$ is given by: $\sum_{j\geq
0}p_{j}x^{j}.$

We have:

$\displaystyle\sum_{j=0}^{\infty}p_{j}\alpha^{-j}\left(\frac{\alpha
x}{1-\alpha x}\right)^{j}=(1-\alpha
x)\sum_{j=0}^{\infty}p_{j}\alpha^{-j}\sum_{n\geq
j}^{\infty}\alpha^{n}x^{n}\binom{n}{j}$ $\displaystyle=(1-\alpha
x)\sum_{j=0}^{\infty}p_{j}x^{j}\sum_{n\geq
j}^{\infty}\alpha^{n-j}x^{n-j}\binom{n}{j}=(1-\alpha
x)\sum_{j=0}^{\infty}p_{j}x^{j}\sum_{k=0}^{\infty}x^{k}\alpha^{k}\binom{j+k}{k}$
$\displaystyle=(1-\alpha
x)\sum_{s=0}^{\infty}x^{s}\sum_{j=0}^{s}p_{j}\alpha^{s-j}\binom{s}{s-j}$
$\displaystyle=p_{0}+\sum_{s=1}^{\infty}x^{s}\sum_{j=0}^{s}p_{j}\alpha^{s-j}\binom{s}{s-j}-\alpha\sum_{s=0}^{\infty}x^{s+1}\sum_{j=0}^{s}p_{j}\alpha^{s-j}\binom{s}{s-j}.$

Further, we get:

$\displaystyle p(x/1-\alpha
x))=p_{0}+\sum_{s=1}^{\infty}x^{s}\sum_{j=0}^{s}p_{j}\alpha^{s-j}\binom{s}{j}-\sum_{m=1}^{\infty}x^{m}\sum_{j=0}^{m-1}p_{j}\alpha^{m-j}\binom{m-1}{j}$
$\displaystyle=p_{0}+\sum_{s=1}^{\infty}x^{s}\left(p_{s}+\sum_{j=1}^{s-1}p_{j}\alpha^{s-j}\binom{s-1}{j-1}\right)$
$\displaystyle=p_{0}+\sum_{s=1}^{\infty}x^{s}\left(p_{s}+\sum_{j=1}^{s-1}p_{j}\alpha^{s-j}\binom{s-1}{j-1}\right)=p_{0}+\sum_{s=1}^{\infty}x^{s}\sum_{j=0}^{s-1}p_{j+1}\alpha^{s-1-j}\binom{s-1}{j}.$

We have also the following immediate observations:

######  Remark 5.

1\. If we set $\alpha\allowbreak=\allowbreak 0$ in assertion 1\. of the
Proposition 1, then obviously
$\left[\left[x^{n}\right]\right]p(x)x^{k}\allowbreak=\allowbreak p_{n-k}.$
Further, if we set $\alpha\allowbreak=\allowbreak 1\allowbreak=\allowbreak
p_{j}$ for all $j\geq 0$ then
$\left[\left[x^{n}\right]\right]p(x)\left(x/(1-x)\right)^{k}\allowbreak=\allowbreak\sum_{j=0}^{n-k}\binom{n-j-1}{k-1}\allowbreak=\allowbreak\binom{n}{k}$
, i.e., we deal with the so-called Pascal matrix. That’s why we call the
Riordan matrix $\left(p(x),x/(1-\alpha x)\right)$ a generalized Pascal matrix.

2\. Given an infinite sequence $\left\\{p_{j}\right\\}_{j\geq 0}$ the
following infinite sequence
$\left\\{\sum_{j=0}^{n}\binom{n}{j}p_{j}\alpha^{n-j}\right\\}_{n\geq 0}$ is
called the sequence of generalized binomial transforms (GBT($\alpha$)) of the
sequence $\left\\{p_{j}\right\\}_{j\geq 0}.$ In the case of
$\alpha\allowbreak=\allowbreak 1,$ simply the binomial transform. It is well
known that GBT($\alpha$) and GBT($-\alpha$) are mutually inverse, that is
GBT($-\alpha$) applied to GBT($\alpha$) of a sequence $\left\\{p_{j}\right\\}$
recovers this sequence. Besides GF of the sequence
$\left\\{\hat{p}_{n}\overset{def}{=}\sum_{j=0}^{n}\binom{n}{j}p_{j}\alpha^{n-j}\right\\}_{n\geq
0}$ is equal to

$\frac{1}{1-\alpha x}p(\frac{x}{1-\alpha x}),$

hence

$p(\frac{x}{1-\alpha x})=(1-\alpha x)\sum_{j\geq 0}x^{j}\hat{p}_{j},$

and consequently

$[[x^{n}]]p(\frac{x}{1-\alpha x})=\hat{p}_{n}-\alpha\hat{p}_{n-1}.$

As a corollary, we have the following observation:

######  Corollary 1.

1\. Suppose we have two Riordan matrices $(a(x),b(x))$ and $(d(x),x).$ Then we
have

$(a(x),b(x))(d(x),x)\allowbreak=\allowbreak\left(a(x)d((b(x)),b(x)\right)$

and

$\left(d(x),x)\right)(a(x),b(x))\allowbreak=\allowbreak\left(d(x)a(x),b(x)\right).$

In particular, we have the following form of multiplication by the Pascal
matrix:

(2.1) $\displaystyle\left(\frac{1}{1-\alpha x},\frac{x}{1-\alpha
x}\right)(d(x),x)$ $\displaystyle=$
$\displaystyle\left(D(x,\alpha),\frac{x}{1-\alpha x}\right),$ (2.2)
$\displaystyle(d(x),x)\left(\frac{1}{1-\alpha x},\frac{x}{1-x}\right)$
$\displaystyle=$ $\displaystyle\left(\frac{1}{1-\alpha
x}d(x),\frac{x}{1-x}\right).$

where we denoted
$\hat{d}_{n}\left(\alpha\right)\allowbreak=\allowbreak\sum_{j=0}^{n}\binom{n}{j}\alpha^{n-j}d_{j},$
$D(x,\alpha)\allowbreak=\allowbreak\sum_{j=0}^{\infty}\hat{d}_{j}\left(\alpha\right)x^{j}.$

2\. For $\alpha\in\mathbb{R}$:

$\left(\frac{1}{1-\alpha x},\frac{x}{1-\alpha
x}\right)\left(d(x),\frac{x}{1+\alpha x}\right)=\left(D(x,\alpha),x\right),$

where $D(x)$ is defined as above.

###### Proof.

The only thing that requires justification is (2.1). Following Theorem of
Roman, we have:

$\left(\frac{1}{1-x},\frac{x}{1-x}\right)(d(x),x)=\left(\frac{1}{1-x}d(\frac{x}{1-x}),\frac{x}{1-x}\right).$

Now, let us recall that

$\sum_{n=0}^{\infty}x^{n}\binom{n}{j}\allowbreak=\allowbreak\frac{x^{j}}{(1-x)^{j+1}},$

consequently we have:

$\displaystyle d(\frac{x}{1-x})$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}d_{j}\left(\frac{x}{1-x}\right)^{j}=\sum_{j=0}^{\infty}d_{j}(1-x)\sum_{n=j}^{\infty}x^{n}\binom{n}{j}$
$\displaystyle=$
$\displaystyle(1-x)\sum_{n=0}^{\infty}x^{n}\sum_{j=0}^{n}d_{j}\binom{n}{j}.$

Now it is elementary to notice that
$\frac{1}{1-x}d(\frac{x}{1-x})\allowbreak=\allowbreak\sum_{n=0}^{\infty}x^{n}\hat{d}_{n}\overset{def}{=}D(x)$.

######  Remark 6.

As a byproduct of assertion 2. of the Lemma 1 we see that every Appell matrix
can be decomposed as the product of two generalized Pascal matrices. Namely,
we have for all $\alpha.$

$\left(d(x),\frac{x}{1-\alpha x}\right)\left(1,\frac{x}{1+\alpha
x}\right)=\left(d(x),x\right).$

Why do we study the lower triangular representation of some sequences? Well,
to get, for example some nontrivial identities.

To see what we mean let us consider several examples. Most of them would
concern Riordan matrices of the form $\left(d(x),x\right)$, , i.e., Appell
matrices, since these matrices have form easy to deal with. Besides, there
exist in the literature many examples of matrices of this form whose entries
are polynomials. Moreover, there exist also ready to use formulae for the
inverses of such matrices. In the sequel, we often use matrix operation as
defined by (1.3). We we will denote the result of such operation on the
matrix, say $\mathbf{A}$ as $\left[\left\\{\alpha_{n}\right\\}\right]$
transform of $\mathbf{A.}$

##  3\. Examples

Most of the examples will be Riordan matrices of the form
$\left(d(x,y\right),y)$, with $d(x,y)$ being a GF of the sequence
$\left\\{d_{n}(x)\right\\}_{n\geq 0}$ consisting either of numbers or
polynomials in $x$. By assertion 1. of the Remark 5 we see that
$\left[d_{n-j}(x)\right]$ is the Riordan matrix $\left(d(x,y),y\right).$

Since we intend to present quite a few different examples coming from
different sources, so in order to avoid confusion, let us group, these
examples according to the source they come from.

###  3.1. Bernoulli and Euler polynomials and numbers

####  3.1.1. Polynomials

Let us recall that Bernoulli and Euler polynomials (denoted respectively
$B_{n}(x)$ and $E_{n}(x)$ , for $n-$th polynomial) are defined respectively by
the expansions:

$\displaystyle t\exp(xt)/(\exp(t)-1)$ $\displaystyle=$
$\displaystyle\sum_{n\geq 0}\frac{t^{n}}{n!}B_{n}(x),$ $\displaystyle
2\exp(xt)/(\exp(t)+1)$ $\displaystyle=$ $\displaystyle\sum_{n\geq
0}\frac{t^{n}}{n!}E_{n}(x).$

Hence, we have the following identity

$\displaystyle\sum_{n\geq
0}\frac{2^{n}t^{n}}{n!}B_{n}(x)=2t\exp(2xt)/(\exp(2t)-1)=\left(\sum_{n\geq
0}\frac{t^{n}}{n!}B_{n}(x)\right)\left(\sum_{n\geq
0}\frac{t^{n}}{n!}E_{n}(x)\right)$ $\displaystyle\sum_{k\geq
0}\frac{t^{k}}{k!}\sum_{n=0}^{k}\binom{k}{n}B_{n}(x)E_{k-n}(x).$

Comparing polynomials by $t^{n}$ and dividing both sides by $k!$ we end up
with the identity.

$2^{k}B_{k}(x)/k!=\sum_{n=0}^{k}\left(B_{n}(x)/n!\right)\left(E_{k-n}(x)/(k-n)!\right).$

Consequently, we have the following observation:

(3.1)
$\left[B_{n-j}(x)/(n-j)!\right]\left[E_{n-j}(x)/(n-j)!\right]=\left[\left\\{2^{n}\right\\}\right]\left[B_{n-j}(x)/(n-j)!\right]\left[\left\\{2^{-j}\right\\}\right].$

####  3.1.2. Numbers

We will also use the following notation:

$\displaystyle\varepsilon(n)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ccc}0&\text{if }&n\text{ is odd}\\\
1&\text{if}&n\text{ is
even}\end{array}\right.,~{}H(x)=\left\\{\begin{array}[]{ccc}0&\text{if}&x<0\\\
1&\text{if}&x\geq 0\end{array}\right.,$ $\displaystyle H2(x)$ $\displaystyle=$
$\displaystyle
H(x)+H(x-2)-2H(x-1)=\left\\{\begin{array}[]{ccc}1&\text{if}&x=0\\\
-1&\text{if}&x=1\\\ 0&\text{if}&\text{otherwise}\end{array}\right.,$
$\displaystyle H3(x)$ $\displaystyle=$ $\displaystyle
H(x)-H(x-1)-H(x-2)+H(x-3).$

and $\left\\{E_{n}\right\\},$ and $\left\\{B_{n}\right\\}$ denote respectively
$n-$th Euler and Bernoulli numbers.

We have

$\displaystyle\left[\binom{n}{j}\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[(-1)^{n-j}\binom{n}{j}\right],~{}\left[\lambda^{n-j}\binom{n}{j}\right]^{-1}=\left[(-\lambda)^{n-j}\binom{j}{n}\right],$
$\displaystyle\left[H(n-j)\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[H2(n-j)\right],~{}\left[\varepsilon(n-j)H(n-j)\right]^{-1}=\left[H3(n-j)\right],$

for some real $\lambda.$ Let us recall that matrix $\left[\binom{n}{j}\right]$
it is a well know Pascal matrix mentioned already above. Hence, we have for
example, for all $n>j$

$\sum_{k=j}^{n}(-1)^{k}\binom{n}{k}\binom{k}{j}=0.$

2\. As shown in [7] we have

(3.4)
$\left[\binom{n}{j}\frac{1}{n-j+1}\right]^{-1}=\left[\binom{n}{j}B_{n-j}\right],~{}\left[\varepsilon(n-j)\binom{n}{j}\right]^{-1}=\left[\binom{n}{j}E_{n-j}\right].$

Thus we have for all $n>j:$

$\displaystyle\sum_{k=j}^{n}\binom{n}{k}\binom{k}{j}\frac{B_{k-j}}{n-k+1}$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\sum_{k=j}^{n}e(n-k)\binom{n}{k}\binom{k}{j}E_{k-j}$
$\displaystyle=$ $\displaystyle 0.$

Notice that performing the following matrix multiplications on the first of
the identities (3.4):

$\left[\left\\{1/n!\right\\}\right]\left[\binom{n}{j}B_{n-j}\right]\left[\left\\{j!\right\\}\right],~{}\left[\left\\{1/n!\right\\}\right]\left[\binom{n}{j}\frac{1}{n-j+1}\right]^{-1}\left[\left\\{j!\right\\}\right],$

we end up with the rather unexpected identities:

(3.5) $\displaystyle\left[1/(n-j)!\right]$ $\displaystyle=$
$\displaystyle\left[(-1)^{n-j}/(n-j)!\right]$ (3.6)
$\displaystyle\left[1/(n-j+1)!\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[B_{n-j}/(n-j)!\right],$ (3.7)
$\displaystyle\left[\varepsilon(j-i)/(j-i)!\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[E_{j-i}/(j-i)!\right].$

3\. As shown also in [7] we have further

$\displaystyle\left[\varepsilon(n-j)\binom{n}{j}\frac{1}{n-j+1}\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\binom{n}{j}\sum_{k=0}^{n-j}\binom{n-j}{k}2^{k}B_{k}\right],$
$\displaystyle~{}\left[\varepsilon(n-j)\binom{n}{j}\right]^{-1}$
$\displaystyle=$ $\displaystyle\left[\binom{n}{j}E_{n-j}\right],$
$\displaystyle\left[\binom{2n}{2j}\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[\binom{2n}{2j}E_{2(n-j)}\right],$
$\displaystyle~{}~{}\left[\binom{2n}{2j}\frac{1}{2(n-j)+1}\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\binom{2n}{2j}\sum_{k=0}^{2(n-j)}\binom{2n-2j}{k}2^{k}B_{k}\right].$

Hence in particular we have $\forall s\geq 1:$

$\displaystyle\sum_{k=0}^{s}\binom{2s}{2k}E_{2k}=0,~{}$
$\displaystyle\sum_{k=0}^{2s}\binom{2s}{k}B_{k}/(2s-k+1)=0.$

Notice that we have changed the order of summation in order to get the second
identity and use the following, elementary to prove, identity:

$\sum_{m=\left\lfloor
l/2\right\rfloor}^{s}\binom{2s-l}{2m-l}/(2s-2m+1)=\frac{2^{2s-l}}{2s-l+1}.$

Again, performing right hand side ad left-hand side multiplications by
matrices $\left[\left\\{1/n!\right\\}\right]$ and its inverse we get the
following also unexpected identities:

(3.8) $\displaystyle\left[\varepsilon(j-i)/(j-i+1)!\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\sum_{k=0}^{j-i}2^{k}B_{j-i-k}/(k!(j-i-k)!)\right],$ (3.9)
$\displaystyle\left[1/(2j-2i)!\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[E_{2j-2i}/(2j-2i)!\right],$ (3.10)
$\displaystyle\left[1/(2j-2i+1)!\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[\sum_{k=0}^{2j-2i}2^{k}B_{k}/(k!(2j-2i-k)!\right].$

###  3.2. Pochhammer symbol - raising factorials

We will be using the following notation. For $x\in\mathbb{C}$ let us denote:

(3.11) $(x)_{(n)}\allowbreak=\allowbreak x(x-1)\ldots(x-n+1).$

This polynomial in $x$ will be called falling factorial while the following
polynomial

(3.12) $(x)^{(n)}=x(x+1)\ldots(x+n-1),$

will be called raising factorial. The $n-$th raising factorial is also called
the Pochhammer symbol. In both cases we set $1$ when $n\allowbreak=\allowbreak
0$.

It is also well-know that

(3.13) $(x)_{(n)}=(-1)^{n}(-x)^{(n)},\text{ and
}(x)^{(n)}=(-1)^{n}(-x)_{(n)}.$

Recall, that we have also the so-called binomial theorem stating that for all
complex $\left|x\right|<1$ we have:

(3.14) $(1-x)^{\alpha}\allowbreak=\allowbreak\sum_{j\geq
0}(-x)^{j}(\alpha)_{(j)}/j!=\sum_{j\geq 0}x^{j}(-\alpha)^{(j)}/j!.$

Notice that from this expansion, by the (so-called exponential) generating
function method we get the following identity: true for all
$\alpha,\beta\in\mathbb{C}$:

(3.15) $\displaystyle\left(\alpha+\beta\right)^{(n)}$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{n}\binom{n}{j}\left(\alpha\right)^{(n-j)}\left(\beta\right)^{\left(j\right)},$
(3.16) $\displaystyle\left(\alpha+\beta\right)_{\left(n\right)}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{n}\binom{n}{j}\left(\alpha\right)_{(n-j)}\left(\beta\right)_{\left(j\right)}.$

As the first observation we have the following one:

######  Proposition 2.

$\forall\mathbb{C}\ni x\neq 0:$

(3.17)
$\left[(-1)^{j}\binom{n}{j}\frac{\left(x\right)^{(n)}}{\left(x\right)^{\left(j\right)}}\right]^{-1}=\left[(-1)^{j}\binom{n}{j}\frac{\left(x\right)^{(n)}}{\left(x\right)^{\left(j\right)}}\right].$

###### Proof.

In order to prove this identity we have to show that $\forall n>j:$

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\sum_{k=j}^{n}(-1)^{k}\binom{n}{k}\frac{\left(x\right)^{(n)}}{\left(x\right)^{\left(k\right)}}(-1)^{j}\binom{k}{j}\frac{\left(x\right)^{(k)}}{\left(x\right)^{\left(j\right)}}$
$\displaystyle=$
$\displaystyle\binom{n}{j}\frac{\left(x\right)^{(n)}}{\left(x\right)^{\left(j\right)}}\sum_{k=j}^{n}(-1)^{k-j}\frac{(n-j)!}{(n-k)!(k-j)!}.$

But this is obvious in face of the above calculations.

Now, taking $\alpha\allowbreak=\allowbreak x$ and
$\beta\allowbreak=\allowbreak-x$ and using the standard trick with
multiplication by the diagonal matrix $\left[\left\\{n!\right\\}\right]$ and
its inverse we arrive at the following identity

$\left[\left(x\right)^{\left(n-j\right)}/(n-j)!\right]^{-1}=\left[\left(-x\right)^{\left(n-j\right)}/(n-j)!\right].$

Similarly, setting $x\allowbreak=\allowbreak 1$ in (3.17), we get:

$\left[(-1)^{n}(n-j)!\binom{n}{j}^{2}\right]^{-1}=\left[(-1)^{n}(n-j)!\binom{n}{j}^{2}\right],$

or

$\left[(n-j)!\binom{n}{j}^{2}\right]^{-1}=\left[(-1)^{n-j}(n-j)!\binom{n}{j}^{2}\right],$

Further, following (3.15), we have for all $x\in\mathbb{C}$:

$\left[\left(x\right)^{\left(n-j\right)}/\left(n-j\right)!\right]^{-1}=\left[\left(-x\right)^{\left(n-j\right)}/(n-j)!\right],$

and following (3.16), we have for all $x\in\mathbb{C}$:

$\left[\left(x\right)_{\left(n-j\right)}/\left(n-j\right)!\right]^{-1}=\left[\left(-x\right)_{\left(n-j\right)}/(n-j)!\right].$

###  3.3. Examples coming from $q-$series theory

####  3.3.1. Introduction and notation

Let us introduce a few elementary notions of the so-called $q-$series theory.

$q$ is a parameter. We will assume that $-1<q\leq 1$ unless otherwise stated.
The case $q\allowbreak=\allowbreak 1$ may not always be considered directly,
but sometimes as a left-hand side limit ( i.e.,$q\longrightarrow 1^{-}$). We
will point out these cases.

We will use traditional notations of the $q-$series theory i.e.,

$\left[0\right]_{q}\allowbreak=\allowbreak
0,~{}\left[n\right]_{q}\allowbreak=\allowbreak
1+q+\ldots+q^{n-1}\allowbreak,\left[n\right]_{q}!\allowbreak=\allowbreak\prod_{j=1}^{n}\left[j\right]_{q},\text{with
}\left[0\right]_{q}!\allowbreak=1\text{,}$
$\QATOPD[]{n}{k}_{q}=\left\\{\begin{array}[]{ccc}\frac{\left[n\right]_{q}!}{\left[n-k\right]_{q}!\left[k\right]_{q}!}&,&n\geq
k\geq 0\\\ 0&,&\text{otherwise}\end{array}\right.\text{.}$

$\binom{n}{k}$ will denote the ordinary, well known binomial coefficient. It
turns out that $\QATOPD[]{n}{k}_{q}$ are polynomials in $q$ (called Gauss
polynomials).

It is useful to use the so-called $q-$Pochhammer symbol for $n\geq 1:$

$\left(a|q\right)_{n}=\prod_{j=0}^{n-1}\left(1-aq^{j}\right),~{}~{}\left(a_{1},a_{2},\ldots,a_{k}|q\right)_{n}\allowbreak=\allowbreak\prod_{j=1}^{k}\left(a_{j}|q\right)_{n}\text{,}$

with $\left(a|q\right)_{0}\allowbreak=\allowbreak 1$.

Although the formula below was known much earlier we cite [10] because of its
nice proof. Namely, we have:

(3.18)
$\left(a|q\right)_{n}=\sum_{j=0}^{n}(-a)^{j}q^{\binom{j}{2}}\QATOPD[]{n}{j}_{q}.$

This formula is often referred to as a finite $q-$binomial formula, that will
be generalized just below.

Often $\left(a|q\right)_{n}$ as well as
$\left(a_{1},a_{2},\ldots,a_{k}|q\right)_{n}$ will be abbreviated to
$\left(a\right)_{n}$ (not to be confused with the falling factorial defined
above) and
$\left(a_{1},a_{2},\ldots,a_{k}\right)_{n},$ if it will not cause
misunderstanding.

We will also use the following symbol $\left\lfloor n\right\rfloor$ to denote
the largest integer not exceeding $n$.

It is worth to mention the following two formulae, that are well known.
Namely, the following formulae are true for $\left|t\right|<1,$
$\left|q\right|<1$ (derived already by Euler, see [1] Corollary 10.2.2)

(3.19) $\displaystyle\frac{1}{(t)_{\infty}}\allowbreak$
$\displaystyle=\allowbreak\sum_{k\geq 0}\frac{t^{k}}{(q)_{k}}\text{,}$ (3.20)
$\displaystyle(t)_{\infty}\allowbreak$ $\displaystyle=\allowbreak\sum_{k\geq
0}(-1)^{k}q^{\binom{k}{2}}\frac{t^{k}}{(q)_{k}}\text{.}$

It is easy to notice that
$\left(q\right)_{n}=\left(1-q\right)^{n}\left[n\right]_{q}!$ and that

(3.21)
$\QATOPD[]{n}{k}_{q}\allowbreak=\allowbreak\allowbreak\left\\{\begin{array}[]{ccc}\frac{\left(q\right)_{n}}{\left(q\right)_{n-k}\left(q\right)_{k}}&,&n\geq
k\geq 0\\\ 0&,&\text{otherwise}\end{array}\right.\text{.}$

The above-mentioned formula is just an example where direct setting
$q\allowbreak=\allowbreak 1$ is senseless however, the passage to the limit
$q\longrightarrow 1^{-}$ makes sense.

Notice that, in particular,

(3.22) $\left[n\right]_{1}\allowbreak=\allowbreak
n,~{}\left[n\right]_{1}!\allowbreak=\allowbreak
n!,~{}\QATOPD[]{n}{k}_{1}\allowbreak=\allowbreak\binom{n}{k},~{}(a)_{1}\allowbreak=\allowbreak
1-a,~{}\left(a|1\right)_{n}\allowbreak=\allowbreak\left(1-a\right)^{n}$

and

(3.23)
$\left[n\right]_{0}\allowbreak=\allowbreak\left\\{\begin{array}[]{ccc}1&\text{if}&n\geq
1\\\
0&\text{if}&n=0\end{array}\right.,~{}\left[n\right]_{0}!\allowbreak=\allowbreak
1,~{}\QATOPD[]{n}{k}_{0}\allowbreak=\allowbreak
1,~{}\left(a|0\right)_{n}\allowbreak=\allowbreak\left\\{\begin{array}[]{ccc}1&\text{if}&n=0\\\
1-a&\text{if}&n\geq 1\end{array}\right..$

####  3.3.2. Examples concerning important in $q-$series theory families of
polynomials

In order to proceed further let us prove some auxiliary results.

######  Proposition 3.

1\. $\forall x,y,q\in\mathbb{C},\left|q\right|,\left|y\right|<1:$

$\sum_{j=0}^{\infty}\frac{y^{j}}{\left(q\right)_{j}}\left(x\right)_{j}=\frac{\left(yx\right)_{\infty}}{\left(y\right)_{\infty}},$

2\. $\forall x,q\in\mathbb{C},x\neq 0,\left|q\right|<1,n\geq 1$

(3.24)
$\sum_{j=0}^{n}\QATOPD[]{n}{j}_{q}\left(x\right)_{n-j}x^{j}\left(1/x\right)_{j}=0.$

###### Proof.

1\. Using (3.18), we get

$\displaystyle\sum_{j=0}^{\infty}\frac{y^{j}}{\left(q\right)_{j}}\left(x\right)_{j}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\frac{y^{j}}{\left(q\right)_{j}}\sum_{k=0}^{j}\QATOPD[]{j}{k}_{q}q^{\binom{k}{2}}(-x)^{k}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\frac{(yx)^{k}}{\left(q\right)_{k}}(-1)^{k}q^{\binom{k}{2}}\sum_{j=k}^{\infty}\frac{y^{j-k}}{\left(q\right)_{j-k}}=\frac{\left(yx\right)_{\infty}}{\left(y\right)_{\infty}},$

by (3.19) and (3.20).

2\. Now notice that using $1.$ with $y$ substituted by $xy$ we get

$\frac{\left(y\right)_{\infty}}{\left(yx\right)_{\infty}}=\frac{\left(yx/x\right)_{\infty}}{\left(yx\right)_{\infty}}=\sum_{j=0}^{\infty}\frac{y^{j}}{\left(q\right)_{j}}x^{j}\left(1/x\right)_{j}.$

Let us denote
$d_{n}(x|q)\allowbreak=\allowbreak\sum_{j=0}^{n}\QATOPD[]{n}{j}_{q}\left(x\right)_{n-j}x^{j}\left(1/x\right)_{j}$
and let us find its generating function. We have:

$\displaystyle\sum_{n=0}^{\infty}\frac{y^{n}}{\left(q\right)_{n}}d_{n}(x|q)=\sum_{n=0}^{\infty}y^{n}\sum_{j=0}^{n}\frac{\left(x\right)_{n-j}}{\left(q\right)_{n-j}}\frac{x^{j}\left(1/x\right)_{j}}{\left(q\right)_{j}}=$
$\displaystyle\sum_{j=0}^{\infty}\frac{y^{j}x^{j}\left(1/x\right)_{j}}{\left(q\right)_{j}}\sum_{n=j}^{\infty}\frac{y^{n-j}}{\left(q\right)_{n-j}}\left(x\right)_{n-j}=\frac{\left(y\right)_{\infty}}{\left(yx\right)_{\infty}}\frac{\left(yx\right)_{\infty}}{\left(y\right)_{\infty}}=1.$

Hence $d_{n}(x|q)\allowbreak=\allowbreak 0$ for $n\geq 1.$

We also have the following almost elementary observation:

Following (3.18) for $n>0$ and $a\allowbreak=\allowbreak 1$ we have he
following identity, true for $n\geq 1:$

$\sum_{j=0}^{n}\QATOPD[]{n}{j}_{q}(-1)^{j}q^{\binom{j}{2}}=0.$

As a corollary we have the following relationship:

######  Corollary 2.

1\. $\forall x,q\in\mathbb{C}$,$\left|q\right|\neq 1:$

$\left[x^{n-j}/(q)_{n-j}\right]^{-1}\allowbreak=\allowbreak\left[(-1)^{n-j}x^{n-j}q^{\binom{n-j}{2}}/(q)_{n-j}\right].$

In particular, taking $x\allowbreak=\allowbreak 1-q,$ we have:

$\left[1/\left[n-j\right]_{q}!\right]^{-1}=\left[(-1)^{n-j}q^{\binom{n-j+1}{2}}/\left[n-j\right]_{q}!\right].$

Further multiplying from the left-hand side by the matrix
$\left[\left\\{\left(q\right)_{n}\right\\}\right]$ and from the right-hand
side by the matrix $\left[\left\\{\left(q\right)_{j}^{-1}\right\\}\right],$ we
get the following relationship:

(3.25)
$\left[x^{n-j}\QATOPD[]{n}{j}_{q}\right]^{-1}=\left[\left(-x\right)^{n-j}q^{\binom{n-j}{2}}\QATOPD[]{n}{j}_{q}\right].$

2\. Following (3.24), we get:for all $x,y\in\mathbb{C}$:

$\left[y^{n-j}\left(x\right)_{n-j}/\left(q\right)_{n-j}\right]^{-1}=\left[y^{n-j}x^{n-j}\left(1/x\right)_{n-j}/\left(q\right)_{n-j}\right]$

and after applying similar trick with diagonal matrix multiplication we get:

$\left[y^{n-j}\QATOPD[]{n}{j}_{q}\left(x\right)_{n-j}\right]^{-1}=\left[y^{n-j}x^{n-j}\QATOPD[]{n}{j}_{q}\left(1/x\right)_{n-j}\right]$

###### Proof.

We start with the assertion of Proposition 3.25 and multiply both sides of
this identity from the left-hand side by
$\left[\left\\{\left(q\right)_{n}^{-1}\right\\}\right]$ and from the right-
hand side by $\left[\left\\{\left(q\right)_{j}\right\\}\right]$ and then apply
assertion $4.$ of Remark 1. On the way we utilize the definition of
$\QATOPD[]{n}{j}_{q}$ given by (3.21).

######  Remark 7.

Notice that when passing with $q$ to $1$ in the last assertion of the
Corollary 2 we get (3.5).

Recall, e.g., following [6], that the following polynomials

$R_{n}(x|q)\allowbreak=\allowbreak\sum_{j=0}^{n}\QATOPD[]{n}{j}_{q}x^{j},$

are called Rogers-Szegö polynomials and play an important rôle in the
$q-$series theory. Following , e.g., [6], we know that the generating function
of the polynomials $\left\\{R_{s}\right\\}$ are given by the following
formula:

$\sum_{s\geq
0}\frac{t^{s}}{\left(q\right)_{s}}R_{s}(x|q)=\frac{1}{\left(t\right)_{\infty}\left(tx\right)_{\infty}}.$

Hence, by simple operation of series multiplication and making use of the
formula (3.20) we deduce that the following family of polynomials:

$\hat{R}_{n}(x|q)=(-1)^{n}\sum_{s=0}^{n}\QATOPD[]{n}{s}_{q}x^{s}q^{\binom{s}{2}+\binom{n-s}{2}},$

defined for $n\geq 0$ satisfy the following identity:

$\sum_{j=0}^{n}\QATOPD[]{n}{j}_{q}R_{j}(x|q)\hat{R}_{n-j}(x|q)=\delta_{n,0}.$

Consequently, we have

$\left[R_{n-j}(x|q)/\left(q\right)_{n-j}\right]^{-1}=\left[\hat{R}_{n-j}(x|q)/\left(q\right)_{n-j}\right].$

Following [6], formulae (3.16), (4.10), and (5.15), and the proceeding each of
these formulae definitions, we have:

$\displaystyle\left[h_{n-j}(x|q)/\left[n-j\right]_{q}!\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[b_{n-j}(x|q)/\left[n-j\right]_{q}!\right],~{}\left[h_{n-j}(x|a,q)/\left[n-j\right]_{q}!\right]^{-1}=\left[\hat{h}_{n-j}(x|a,q)/\left[n-j\right]_{q}!\right],$
$\displaystyle\left[Q_{n-j}(x|a,b,q)/\left[n-j\right]_{q}!\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\hat{Q}_{n-j}(x|a,b,q)/\left[n-j\right]_{q}!\right],.$

where $\left\\{h_{n}(x|q\right\\},$ $\left\\{h_{n}(x|a,q)\right\\},$
$\left\\{Q_{n}(x|a,b,q)\right\\},$ are respectively the so-called $q-$Hermite,
big $q-$Hermite, Al-Salam–Chihara polynomials. They constitute a part of the
so-called Askey-Wilson scheme of orthogonal polynomials defined by their
three-term recurrences given by the formulae respectively (3.1), (4.1), (5.1)
in [6]. The families of polynomials $\left\\{b_{n}(x|q)\right\\},$
$\left\\{\hat{h}_{n}(x|a,q)\right\\},$ $\left\\{\hat{Q}_{n}(x|a,b,q)\right\\}$
are defined by their three-term recurrences given by the formulae respectively
(3.14), (4.9) and unnumbered formula proceeding (5.15). Although these
families of polynomials were defined for real $x,a,b,q$ all from the segment
$[-1,1]$, we can extend their ranges to all complex numbers since they are
polynomials.

###  3.4. Classical Orthogonal polynomials

In this subsection we will present only two, but important examples involving
classical polynomials

####  3.4.1. Laguerre polynomials

Let $L_{n}(x)$ denote $n-$th Laguerre polynomial. Let us define the following
family of orthogonal polynomials:

$\mathcal{L}_{n}(x)=\left\\{\begin{array}[]{ccc}1&\text{if}&n=0\\\
x-1&\text{if}&n=1\\\
\sum_{j=2}^{n}\binom{n-2}{j-2}x^{j}/j!&\text{if}&n>1\end{array}\right..$

######  Proposition 4.

For all $n>m\geq 0:\sum_{j=m}^{n}L_{n-j}(x)\mathcal{L}_{j-m}(x)=0,$
Consequently, we have:
$\left[L_{n-j}(x)\right]^{-1}=\left[\mathcal{L}_{n-j}(x)\right]$

###### Proof.

First, notice that the we can take $k\allowbreak=\allowbreak j-m$ and prove
the identity
$\sum_{k=0}^{n-m}L_{n-m-k}(x)\mathcal{L}_{k}(x)\allowbreak=\allowbreak 0.$
Now, we use the concept of the generating function and use assertion of Remark
4. Consequently, we realize that our identity is equivalent to fact that the
generating function of the polynomials $\left\\{\mathcal{L}_{k}\right\\}$ is
equal to the inverse of the generating function of the Laguerre polynomials.
But it is well-known that

$\sum_{n\geq 0}t^{n}L_{n}(x)=\frac{1}{1-t}\exp(-\frac{t}{1-t}x).$

Hence it remains to show that

$\sum_{n\geq 0}t^{n}\mathcal{L}_{n}(x)=(1-t)\exp(\frac{t}{1-t}x).$

We have

$\displaystyle\sum_{n\geq 0}t^{n}\mathcal{L}_{n}(x)=1+t(x-1)+\sum_{n\geq
2}t^{n}\sum_{j=2}^{n}\binom{n-2}{j-2}x^{j}/j!$
$\displaystyle=1+t(x-1)+t^{2}\sum_{m\geq
0}t^{m}\sum_{k=0}^{m}\binom{m}{k}x^{k+2}/(k+2)!$
$\displaystyle=1+t(x-1)+t^{2}\sum_{k\geq
0}t^{k}x^{k+2}/(k+2)!\sum_{k=m}^{\infty}\binom{m}{k}t^{k-m}$
$\displaystyle=1+t(x-1)+\sum_{k\geq
0}t^{k+2}x^{k+2}/(k+2)!\frac{1}{(1-t)^{k+1}}$
$\displaystyle=1+t(x-1)+(1-t)\sum_{k\geq
0}\left(\frac{tx}{1-t}\right)^{k+2}\frac{1}{(k+2)!}$
$\displaystyle=1+t(x-1)+(1-t)(\exp(\frac{tx}{1-t})-1-\frac{tx}{1-t}).$

####  3.4.2. Hermite polynomials

Following an identity presented in, say [6](unnumbered formula below (8.17)),
we have

$\left[He_{n-j}(x)/(n-j)!\right]^{-1}=\left[i^{n-j}He_{n-j}(ix)/(n-j)!\right],$

where $i$ denotes an imaginary unit, and $He_{n}(x)$ $n-$th so-called
probabilistic Hermite polynomial, i.e., the ones orthogonal with respect to
the measure with the density $\exp(-x^{2}/2)/\sqrt{2\pi}$.

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