Symmetric Functions and Rings of Multinumbers Associated with Finite Groups

Abstract

In this paper, we introduce ${\omega }_n$-symmetric polynomials associated with the finite group ${\omega }_n,$ which consists of roots of unity, and groups of permutations acting on the Cartesian product of Banach spaces ${\ell }_1.$ These polynomials extend the classical notions of symmetric and supersymmetric polynomials on ${\ell }_1.$ We explore algebraic bases in the algebra of ${\omega }_n$-symmetric polynomials and derive corresponding generating functions. Building on this foundation, we construct rings of multisets (multinumbers), defined as equivalence classes on the underlying space under the action of ${\omega }_n$-symmetric polynomials, and investigate their fundamental properties. Furthermore, we examine the ring of integer multinumbers associated with the group ${\omega }_n,$ proving that it forms an integral domain when n is prime or $n=4.$

1. Introduction

Contemporary trends in mathematics require the use of various generalizations of fundamental concepts. For example, the concept of fuzzy sets supposes that any element x of a fuzzy set M may have a fractional membership in $M.$ In other words, the membership function takes values in the interval $[0,1]$ (see, e.g., [1]). The concept of multisets gives us another generalization of the set. For multisets, the membership function takes values in the set of positive integers ${\mathbb{Z}}_{+}$ [2]. From this point of view, the idea of multisets is similar and, in some sense, complementary to the fuzzy set theory. For example, for a multiset of numbers $\underset{n_1}{\underbrace{x_1,\dots x_1}},\underset{n_2}{\underbrace{x_2,\dots x_2}},\dots$, we can say that $x_i$ has a greater membership than $x_j$ if $n_i>n_j.$ As multisets are unordered collections of elements with finite multiplicities, they appear as natural domains of symmetric functions, that is, functions of several or infinitely many variables, invariant with respect to permutations of the variables. In particular, multisets of numbers (or multinumbers), such that ${\displaystyle \sum _{i=1}^{\infty }\left|x_i\right|<\infty }$ can be considered as a natural domain of symmetric functions on the space ${\ell }_1$ of absolutely summing number sequences. This set of multinumbers can be identified with the quotient set ${\ell }_1/\sim$ with respect to an appropriated relation of equivalence. Such an approach was fruitful for the investigation of spectra of algebras of symmetric analytic functions on complex ${\ell }_1$ (see [3,4,5]). For the classical multiset theory, we refer the reader to [2,6] and the references therein. The set of multisets has a natural algebraic structure of unital commutative semiring. This semiring can be embedded into a ring of multinumbers that consists of classes of equivalence of two-side sequences in ${\ell }_1\times {\ell }_1.$ The constructed ring corresponds to so-called supersymmetric functions on ${\ell }_1\times {\ell }_1$ [7].

In this paper, we consider rings of multinumbers associated with the Banach space ${\ell }_1^n={\ell }_1\times \cdots \times {\ell }_1,$ and the finite group ${\omega }_n$ of roots of $1.$ Note that symmetric functions appear, in particular, in neural networks, blockchain technology [8,9,10], cryptography [11], and quantum physics [12,13,14,15]. Applications of symmetric functions in neural networks and blockchain technology are based on the fact that usually, data have some symmetry and, so, all functions over data should be symmetric in some sense. In particular, symmetric polynomials can be used as hash functions, and if we have a basis of symmetric polynomials, then the corresponding hash functions have no collisions. The computation of the multiplicative inverse to a given multinumber is a complicated computational problem, and so it can be used for creating cryptographic systems with public keys [11]. Generating functions of sequences of algebraic bases of symmetric polynomials appear as grand canonical partition functions in quantum systems. The presented investigation works with a new type of symmetry on the Cartesian products of ${\ell }_1,$ corresponding bases of symmetric polynomials, their generating functions, and rings of multisets. Thus, the obtained results may have applications in the above-mentioned branches of informatics and physics.

In Section 2, we construct ${\omega }_n$-symmetric polynomials on the nth Cartesian power of ${\ell }_1,$ and some basic properties of the polynomials. In Section 3, we consider algebraic bases in algebras of ${\omega }_n$-symmetric polynomials and generating functions associated with these bases. In Section 4, we introduce rings of multinumbers, associated with algebras of ${\omega }_n$-symmetric polynomials, and consider some properties of such rings and their subrings. An example for $n=4$ is considered in Section 5. In Section 6, we investigate subrings of integer multinumbers associated with ${\omega }_n$-symmetric polynomials and show that if n is a prime number or $n=4,$ then the corresponding subrings of integer multinumbers is an integral domain. In Section 7, we discuss the main results of the paper and consider some directions for further investigations.

The general theory of polynomials on Banach spaces can be found in [16,17].

2. Main Definitions

Let ${\omega }_n=\{{\alpha }_0,{\alpha }_1,\cdots ,{\alpha }_{n-1}\}$ be the group of nth roots of unity. We suppose that ${\alpha }_j$ are naturally ordered, that is ${\alpha }_0=1$ and ${\alpha }_1^j={\alpha }_j=e^{i2\pi j/n}$, $0\le j<n.$ Then, in particular, the inverse element of ${\alpha }_j$ is ${\alpha }_{n-j}$, $j>0.$

Let us denote by ${\mathbb{N}}_{{\omega }_n}$ a matrix of indexes consisting of n rows with entries $m{\alpha }_j$ for $0\le j\le n-1$ and $m\in \mathbb{N}.$ Also, we denote by ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ the Banach space of all absolutely summing complex sequences indexed by numbers in ${\mathbb{N}}_{{\omega }_n}.$ The symbol ${\ell }_1={\ell }_1\left({\mathbb{N}}_{{\omega }_1}\right)$ means the classical Banach space of absolutely summing complex sequences. Every element x in ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ has the representation

$$x={\left(x_{m{\alpha }_j}\right)}_{m\in \mathbb{N},0\le j<n}=\left(\begin{array}{c}{x}_{{\alpha }_0}\\ x_{{\alpha }_1}\\ ⋮\\ x_{{\alpha }_{n-1}}\end{array}\right)=\left(\begin{array}{ccccc}{x}_{1{\alpha }_0},& x_{2{\alpha }_0},& \cdots ,& x_{m{\alpha }_0},& \cdots \\ x_{1{\alpha }_1},& x_{2{\alpha }_1},& \cdots & x_{m{\alpha }_1},& \cdots \\ ⋮& ⋮& \ddots & ⋮& ⋮\\ x_{1{\alpha }_{n-1}},& x_{2{\alpha }_{n-1}},& \cdots ,& x_{m{\alpha }_{n-1}},& \cdots \end{array}\right),$$

where $x_{{\alpha }_j}=(x_{1{\alpha }_j},x_{2{\alpha }_j},\cdots ,x_{m{\alpha }_j},\cdots )$ is in ${\ell }_1$ for all $j\in \{0,1,2,\cdots ,n-1\},$ and

$$\parallel x\parallel =\sum _{i=1}^{\infty }\sum _{j=0}^{n-1}|x_{i{\alpha }_j}|.$$

Let us consider the following polynomials

$${\Omega }_k^{\left(n\right)}\left(x\right)=\sum _{j=0}^{n-1}{\alpha }_j{F}_k\left(x_{{\alpha }_j}\right)=\sum _{j=0}^{n-1}{\alpha }_j\sum _{i=1}^{\infty }{x}_{i{\alpha }_j}^k,k\in \mathbb{N}$$

on ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$

Definition 1.

A polynomial P on ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ is said to be ${\omega }_n$-symmetric if it can be represented as an algebraic combination of polynomials ${\left\{{\Omega }_k^{\left(n\right)}\right\}}_{k=1}^{\infty }.$

An ${\omega }_1$-symmetric polynomial on ${\ell }_1\left({\mathbb{N}}_{{\omega }_1}\right)$ is called symmetric, and an ${\omega }_2$-symmetric polynomial on ${\ell }_1\left({\mathbb{N}}_{{\omega }_2}\right)$ is called supersymmetric. Symmetric polynomials and analytic functions on Banach spaces have been considered by many authors (see, e.g., [3,18,19,20,21,22,23,24]). Supersymmetric polynomials on ${\ell }_1\left({\mathbb{N}}_{{\omega }_2}\right)$ were considered in [7,11,13] (for supersymmetric polynomials in combinatorics, see [25,26,27]). The group ${\omega }_2$ consists of two elements ${\alpha }_0=1$, ${\alpha }_1=-1$ and will use more convenient notation

$$\left(y\right|x)=(\dots ,y_2,y_1|x_1,x_2,\dots )$$

instead of

$$\left(\begin{array}{c}{x}_{{\alpha }_0}\\ x_{{\alpha }_1}\end{array}\right)=\left(\begin{array}{ccc}{x}_{1{\alpha }_0},& x_{2{\alpha }_0},& \cdots \\ x_{1{\alpha }_1},& x_{2{\alpha }_1},& \cdots \end{array}\right),$$

where $x=x_{{\alpha }_0},$ and $y=x_{{\alpha }_1}.$ Also, ${\omega }_4=\{{\alpha }_0=1,{\alpha }_1=i,{\alpha }_2=-1,{\alpha }_3=-i\},$ and we will use notation

$$\left(\begin{array}{ccc}y& |& x\\ r& |& s\end{array}\right)=\left(\begin{array}{c}{x}_{{\alpha }_0}\\ x_{{\alpha }_1}\\ x_{{\alpha }_2}\\ x_{{\alpha }_3}\end{array}\right)$$

(1)

for $x=x_{{\alpha }_0}$, $y=x_{{\alpha }_2}$, $s=x_{{\alpha }_1},$ and $r=x_{{\alpha }_3}.$

Similarly, as in the case of symmetric and supersymmetric polynomials [4,7,11], we define the operation “symmetric addition”, “•” on ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$:

$$x•y=\left(\begin{array}{cccccccc}{x}_{1{\alpha }_0},& y_{1{\alpha }_0},& x_{2{\alpha }_0},& y_{2{\alpha }_0},& \cdots ,& x_{m{\alpha }_0},& y_{m{\alpha }_0},& \cdots \\ x_{1{\alpha }_1},& y_{1{\alpha }_1},& x_{2{\alpha }_1},& y_{2{\alpha }_1},& \cdots ,& x_{m{\alpha }_1},& y_{m{\alpha }_1},& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ x_{1{\alpha }_{n-1}},& y_{1{\alpha }_{n-1}},& x_{2{\alpha }_{n-1}},& y_{2{\alpha }_{n-1}},& \cdots ,& x_{m{\alpha }_{n-1}},& y_{m{\alpha }_{n-1}},& \cdots \end{array}\right).$$

We will use the brief notation

$$•\sum _{k=1}^m{x}_k=(\cdots (x_1•x_2)•x_3)\cdots )•x_m,x_k\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$$

Let us consider the case of ${\ell }_1\left({\mathbb{N}}_{{\omega }_1}\right)={\ell }_1.$ The functions

$${\Omega }_k^{\left(1\right)}\left(x\right)=F_k\left(x\right)=\sum _{i=1}^{\infty }{x}_i^k,k\in \mathbb{N}$$

form the algebraic basis of power symmetric polynomials in ${\mathcal{P}}_s\left({\ell }_1\right).$ For given x and y in ${\ell }_1$, the symmetric product $x\diamond y$ is the resulting sequence of ordering the set $\{x_i{y}_j:i,j\in \mathbb{N}\}$ with one single index in some fixed order (see, e.g., [11]). Note that for every $k\in \mathbb{N}$, $F_k(x\diamond y)=F_k\left(x\right)F_k\left(y\right)$ and for every symmetric polynomial $P,$ $P\left((x•y)\diamond z\right)=P\left((x\diamond z)•(y\diamond z)\right).$

The symmetric product can be defined for the general case ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ in the following way

$$x\diamond y=\left(\begin{array}{c}{x}_{{\alpha }_0}\\ x_{{\alpha }_1}\\ ⋮\\ x_{{\alpha }_{n-1}}\end{array}\right)\diamond \left(\begin{array}{c}{y}_{{\alpha }_0}\\ y_{{\alpha }_1}\\ ⋮\\ y_{{\alpha }_{n-1}}\end{array}\right)=\left(\begin{array}{c}{z}_{{\alpha }_0}\\ z_{{\alpha }_1}\\ ⋮\\ z_{{\alpha }_{n-1}}\end{array}\right),$$

where

$$z_{{\alpha }_j}=•\sum _{{\alpha }_j={\alpha }_i{\alpha }_k}{x}_{{\alpha }_j}\diamond y_{{\alpha }_k}.$$

Here, we consider $x_{{\alpha }_j}$, $y_{{\alpha }_k},$ and $z_{{\alpha }_j}$ as elements in ${\ell }_1\left({\mathbb{N}}_{{\omega }_1}\right),$ where the operation ⋄ is already defined, and we suppose that ${\ell }_1\left({\mathbb{N}}_{{\omega }_1}\right)$ is included to ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ by

$$z_{{\alpha }_j}⟼\left(\begin{array}{c}0\\ ⋮\\ 0\\ z_{{\alpha }_j}\\ 0\\ ⋮\\ 0\end{array}\right).$$

Proposition 1.

For every $x$, $y\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ and $k\in \mathbb{N},$

$${\Omega }_k^{\left(n\right)}(x\diamond y)={\Omega }_k^{\left(n\right)}\left(x\right){\Omega }_k^{\left(n\right)}\left(y\right).$$

Proof. 

Using the straightforward computations and the definition of “⋄”, we have

$$\begin{array}{cc}\hfill {\Omega }_k^{\left(n\right)}(x\diamond y)& ={\Omega }_k^{\left(n\right)}\left(•\sum _{m=0}^{n-1}•\sum _{{\alpha }_m={\alpha }_i{\alpha }_j}{x}_{{\alpha }_i}\diamond y_{{\alpha }_j}\right)=\sum _{m=0}^{n-1}{\alpha }_m\sum _{{\alpha }_m={\alpha }_i{\alpha }_j}{x}_{{\alpha }_i}^k{y}_{{\alpha }_j}^k\hfill \\ & =\sum _{i=0}^{n-1}{\alpha }_i\sum _{l=1}^{\infty }{x}_{l{\alpha }_i}^k\sum _{j=0}^{n-1}{\alpha }_j\sum _{l=1}^{\infty }{y}_{l{\alpha }_j}^k={\Omega }_k^{\left(n\right)}\left(x\right){\Omega }_k^{\left(n\right)}\left(y\right).\hfill \end{array}$$

□

3. Bases in Algebras of ${\mathsf{\omega }}_{\mathit{n}}$-Symmetric Polynomials

Let ${\mathcal{P}}_s\left({\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)\right)$ be the algebra of all ${\omega }_n$-symmetric polynomials on ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$ In the case $n=1$, we denote it just as ${\mathcal{P}}_s\left({\ell }_1\right).$ It is well-known [3,19,28] that polynomials $F_k={\Omega }_k^{\left(1\right)}$, $k\in \mathbb{N}$ form an algebraic basis in ${\mathcal{P}}_s\left({\ell }_1\right)$ (the basis of power symmetric polynomials). As the restriction of ${\Omega }_k^{\left(n\right)}$ to the subspace ${\ell }_1\subset {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ spanned on vectors $x_{{\alpha }_0}$ is equal to $F_k$ for every $k,$ polynomials ${\Omega }_k^{\left(n\right)}$, $k\in \mathbb{N}$ are algebraically independent. On the other hand, by the definition of ${\omega }_n$-symmetric polynomials, every polynomial in ${\mathcal{P}}_s\left({\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)\right)$ is a finite algebraic combination of ${\Omega }_k^{\left(n\right)}$, $k\in \mathbb{N}.$ Thus, we have proven the following proposition.

Proposition 2.

Polynomials ${\Omega }_k^{\left(n\right)}$, $k\in \mathbb{N}$ form an algebraic basis in ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ for every $n\in \mathbb{N}.$

Note that ${\mathcal{P}}_s\left({\ell }_1\right)$ admits other algebraic bases. Let

$$G_n\left(x\right)=\sum _{1\le i_1<\dots <i_n}{x}_{i_1}\dots x_{i_n}$$

be the basis of elementary symmetric polynomials and

$$H_n\left(x\right)=\sum _{1\le i_1\le \dots \le i_n}{x}_{i_1}\dots x_{i_n}$$

be the basis of complete symmetric polynomials.

Well-known Newton’s formulas connect the classical bases of symmetric polynomials. In particular,

$$m{G}_m=\sum _{k=1}^m{(-1)}^{k-1}{G}_{m-k}{F}_k,m\in \mathbb{N},$$

(2)

and

$$m{H}_m=\sum _{k=1}^m{H}_{m-k}{F}_k,m\in \mathbb{N}.$$

(3)

Let $\mathcal{F}\left(x\right)\left(t\right)$, $\mathcal{G}\left(x\right)\left(t\right)$, and $\mathcal{H}\left(x\right)\left(t\right)$ be generating functions of the sequences $F_n\left(x\right)$, $G_n\left(x\right)$, and $H_n\left(x\right)$, respectively. In other words the generating functions are formal series

$$\mathcal{F}\left(x\right)\left(t\right)=\sum _{n=1}^{\infty }{t}^{n-1}{F}_n\left(x\right),$$

(4)

$$\mathcal{G}\left(x\right)\left(t\right)=\sum _{n=0}^{\infty }{t}^n{G}_n\left(x\right),G_0=1,$$

(5)

$$\mathcal{H}\left(x\right)\left(t\right)=\sum _{n=0}^{\infty }{t}^n{H}_n\left(x\right),H_0=1,$$

where $x\in {\ell }_1$, $t\in \mathbb{C}.$ The following formulas for the generating functions are well-known in combinatorics ([29], pp. 4–5) and they are true if $x\in {\ell }_1$ and t belong to the corresponding domain of convergence.

$$\mathcal{G}\left(x\right)\left(t\right)={\displaystyle \frac{1}{\mathcal{H}(-x)\left(t\right)}},$$

(6)

and

$$\mathcal{G}\left(x\right)\left(t\right)=exp\left(-\sum _{n=1}^{\infty }{t}^n{\displaystyle \frac{F_n(-x)}{n}}\right)=exp\left(-{\int }_0^t\mathcal{F}(-x)\left(\zeta \right)d\zeta \right).$$

(7)

Let us consider the domain of $\mathcal{G}\left(x\right)\left(t\right)$ in more detail. According to [4], $\parallel G_n\parallel = 1/n!$ and (5) converges for every $x\in {\ell }_1$ and $t\in \mathbb{C},$ but $\parallel F_n\parallel = 1$ and (5) converges if $\left|t\right|\parallel x\parallel < 1.$ In particular, the function

$$\sum _{n=1}^{\infty }{t}^n{\displaystyle \frac{F_n(-x)}{n}}=\sum _{n=1}^{\infty }{(-1)}^{n+1}{t}^n{\displaystyle \frac{F_n\left(x\right)}{n}}$$

in (7) converges only if $\left|t\right|\parallel x\parallel < 1.$ In other words,

$$exp\left(-\sum _{n=1}^{\infty }{t}^n{\displaystyle \frac{F_n(-x)}{n}}\right)$$

is formally defined if $\left|t\right|\parallel x\parallel < 1,$ but can be extended to the whole space $\mathbb{C}\times {\ell }_1$ because

$$\begin{array}{cc}\hfill exp\left(-\sum _{n=1}^{\infty }{t}^n{\displaystyle \frac{F_n(-x)}{n}}\right)& =exp\left(\sum _{n=1}^{\infty }{(-1)}^{n+1}{t}^n\sum _{k=1}^{\infty }{\displaystyle \frac{x_k^n}{n}}\right)\hfill \\ & =exp\left(\sum _{k=1}^{\infty }log(1+t{x}_k)\right)=\prod _{k=1}^{\infty }(1+t{x}_k)\hfill \end{array}$$

and

$$\prod _{k=1}^{\infty }(1+t{x}_k)=\mathcal{G}\left(x\right)\left(t\right)$$

converges for every $x\in {\ell }_1$ and $t\in \mathbb{C}.$

Let us denote by ${\Lambda }_{{\omega }_n}$ the algebraic isomorphism from ${\mathcal{P}}_s\left({\ell }_1\right)$ to ${\mathcal{P}}_s\left({\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)\right)$, such that

$${\Lambda }_{{\omega }_n}:F_k⟼{\Omega }_k^{\left(n\right)}.$$

In [7], it was computed that if ${\Lambda }_{{\omega }_2}\left(G_k\right)=W_k,$ then

$$W_k\left(y\right|x)=\sum _{m=0}^k{G}_m\left(x\right)H_{k-m}(-y),k\in \mathbb{N}$$

(8)

and

$$\mathcal{W}\left(y\right|x)\left(t\right):=\sum _{n=0}^{\infty }{t}^n{W}_n\left(y\right|x)={\displaystyle \frac{\mathcal{G}\left(x\right)\left(t\right)}{\mathcal{G}\left(y\right)\left(t\right)}}.$$

Note that the function ${\displaystyle \frac{\mathcal{G}\left(x\right)\left(t\right)}{\mathcal{G}\left(y\right)\left(t\right)}}$ is well defined if the denominator is not equal to zero. Thus, for a fixed $\left(y\right|x),$ the radius of convergence of (8) is equal to $1/max|y_j|.$

In the general case, applying ${\Lambda }_{{\omega }_n}$ to (2), we have

$${\Lambda }_{{\omega }_n}\left(G_m\right)=\sum _{k=1}^m{(-1)}^{k-1}{\Lambda }_{{\omega }_n}\left(G_{m-k}\right){\Lambda }_{{\omega }_n}\left(F_k\right)=\sum _{k=1}^m{(-1)}^{k-1}{\Lambda }_{{\omega }_n}\left(G_{m-k}\right){\Omega }_k^{\left(n\right)}.$$

Using (7), we can formally write

$$\begin{array}{cc}\hfill {\Lambda }_{{\omega }_n}\left(\mathcal{G}\right)\left(x\right)\left(t\right)& =\sum _{m=1}^{\infty }{t}^m{\Lambda }_{{\omega }_n}\left(G_m\right)\left(x\right)\hfill \\ \hfill & =exp\left(-\sum _{m=1}^{\infty }{t}^m{\displaystyle \frac{{\Omega }_m^{\left(n\right)}(-x)}{m}}\right)\hfill \\ \hfill & =\prod _{k=0}^{n-1}exp\left({\alpha }_k\sum _{m=1}^{\infty }{(-1)}^{m+1}{t}^m{\displaystyle \frac{F_m\left(x_{{\alpha }_k}\right)}{m}}\right)\hfill \\ \hfill & =\prod _{k=0}^{n-1}exp\left({\alpha }_k\sum _{j=1}^{\infty }log(1+t{x}_{j{\alpha }_k})\right)\hfill \\ \hfill & =\prod _{k=0}^{n-1}\prod _{j=1}^{\infty }{(1+t{x}_{j{\alpha }_k})}^{{\alpha }_k}.\hfill \end{array}$$

(9)

As the function

$$\sum _{m=1}^{\infty }{(-1)}^{m+1}{t}^m{\displaystyle \frac{F_m\left(x_{{\alpha }_k}\right)}{m}}$$

is well defined if $\left|t\right|\parallel x_{\alpha }\parallel < 1$, $x_{\alpha }\in {\ell }_1\subset {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right),$ Formula (9) defines an analytic function of t in the disk $\left|t\right|<1/\parallel x\parallel$ for every fixed $x\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$ Therefore, we have the following theorem.

Theorem 1.

The generating functions ${\Lambda }_{{\omega }_n}\left(\mathcal{G}\right)\left(x\right)\left(t\right)$ and ${\Lambda }_{{\omega }_n}\left(\mathcal{H}\right)\left(x\right)\left(t\right)$ of the sequences ${\Lambda }_{{\omega }_n}\left(G_m\right)\left(x\right)$ and ${\Lambda }_{{\omega }_n}\left(H_m\right)\left(x\right)$, $m\in \mathbb{N}$ are of the form

$${\Lambda }_{{\omega }_n}\left(\mathcal{G}\right)\left(x\right)\left(t\right)=\sum _{m=1}^{\infty }{t}^m{\Lambda }_{{\omega }_n}\left(G_m\right)\left(x\right)=\prod _{k=0}^{n-1}\prod _{j=1}^{\infty }{(1+t{x}_{j{\alpha }_k})}^{{\alpha }_k},$$

$${\Lambda }_{{\omega }_n}\left(\mathcal{H}\right)\left(x\right)\left(t\right)=\sum _{m=1}^{\infty }{t}^m{\Lambda }_{{\omega }_n}\left(H_m\right)\left(x\right)=\prod _{k=0}^{n-1}\prod _{j=1}^{\infty }{(1-t{x}_{j{\alpha }_k})}^{-{\alpha }_k},$$

and for every $x\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$, they are analytic function on the open disk $\left|t\right|<1/\parallel x\parallel .$

Proof. 

The first formula is proven in (9) and the second can be obtained from the first using (6). □

Let us notice that polynomials ${\Lambda }_{{\omega }_n}\left(G_m\right)\left(x\right)$ and ${\Lambda }_{{\omega }_n}\left(H_m\right)\left(x\right)$ can be computed by

$${\Lambda }_{{\omega }_n}\left(G_m\right)\left(x\right)={\displaystyle \frac{d^m}{d{t}^m}}{|}_{t=0}{\Lambda }_{{\omega }_n}\left(\mathcal{G}\right)\left(x\right)\left(t\right)$$

and

$${\Lambda }_{{\omega }_n}\left(H_m\right)\left(x\right)={\displaystyle \frac{d^m}{d{t}^m}}{|}_{t=0}{\Lambda }_{{\omega }_n}\left(\mathcal{H}\right)\left(x\right)\left(t\right).$$

Let us consider the case ${\ell }_1\left({\mathbb{N}}_{{\omega }_4}\right).$ If

$$u=\left(\begin{array}{ccc}y& |& x\\ r& |& s\end{array}\right)\in {\ell }_1\left({\mathbb{N}}_{{\omega }_4}\right)$$

is defined as in (1), then

$${\Omega }_k^{\left(4\right)}\left(u\right)=F_k\left(x\right)-F_k\left(y\right)+i(F_k\left(s\right)-F_k\left(r\right)),k\in \mathbb{N}.$$

Thus,

$$\begin{array}{cc}\hfill {\Lambda }_{{\omega }_4}\left(\mathcal{G}\right)\left(u\right)\left(t\right)& =exp\left(\sum _{m=1}^{\infty }{t}^m{\displaystyle \frac{F_m(-x)}{m}}-\sum _{m=1}^{\infty }{t}^m{\displaystyle \frac{F_m(-y)}{m}}+i\sum _{m=1}^{\infty }{t}^m{\displaystyle \frac{F_m(-s)}{m}}-i\sum _{m=1}^{\infty }{t}^m{\displaystyle \frac{F_m(-r)}{m}}\right)\hfill \\ \hfill & ={\displaystyle \frac{\mathcal{G}\left(x\right)\left(t\right)}{\mathcal{G}\left(y\right)\left(t\right)}}{\left({\displaystyle \frac{\mathcal{G}\left(s\right)\left(t\right)}{\mathcal{G}\left(r\right)\left(t\right)}}\right)}^i.\hfill \end{array}$$

(10)

As in the general case, ${\Lambda }_{{\omega }_4}\left(\mathcal{G}\right)\left(u\right)\left(t\right)$ is a well-defined analytic function of t in the domain $\left|t\right|<1/\parallel u\parallel$ for any fixed $u\in {\ell }_1\left({\mathbb{N}}_{{\omega }_4}\right).$

Proposition 3.

Let u be of the form

$$u=\left(\begin{array}{ccc}0& |& 0\\ r& |& s\end{array}\right)\in {\ell }_1\left({\mathbb{N}}_{{\omega }_4}\right),$$

and suppose that s and r are such that $F_k\left(s\right)$ and $F_k\left(r\right)$ are real numbers for every $k\in \mathbb{N}.$ Then $\left|{\Lambda }_{{\omega }_4}\left(\mathcal{G}\right)\left(u\right)\left(t\right)\right|=1$ for every $u\ne 0$ and real $\left|t\right|<1/\parallel u\parallel .$

Proof. 

From (10), we have that for this special case,

$$\begin{array}{cc}\hfill {\Lambda }_{{\omega }_4}\left(\mathcal{G}\right)\left(u\right)\left(t\right)& =expi\left(\sum _{m=1}^{\infty }{t}^m\left({\displaystyle \frac{F_m(-s)}{m}}-{\displaystyle \frac{F_m(-r)}{m}}\right)\right)\hfill \\ & =cos\left(\sum _{m=1}^{\infty }{t}^m\left({\displaystyle \frac{F_m(-s)}{m}}-{\displaystyle \frac{F_m(-r)}{m}}\right)\right)+isin\left(\sum _{m=1}^{\infty }{t}^m\left({\displaystyle \frac{F_m(-s)}{m}}-{\displaystyle \frac{F_m(-r)}{m}}\right)\right).\hfill \end{array}$$

As ${\sum }_{m=1}^{\infty }{t}^m\left({\displaystyle \frac{F_m(-s)}{m}}-{\displaystyle \frac{F_m(-r)}{m}}\right)$ is real, $\left|{\Lambda }_{{\omega }_4}\left(\mathcal{G}\right)\left(u\right)\left(t\right)\right|=1.$ □

4. Rings of Multinumbers Associated with ${\mathcal{P}}_{\mathit{s}}\left({\ell }_{\mathbf{1}}\left({\mathbb{N}}_{{\mathsf{\omega }}_{\mathit{n}}}\right)\right)$

Let us introduce the following relation of equivalence on ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$ We say that $u\sim v$, $u,v\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$ if and only if ${\Omega }_k^{\left(n\right)}\left(u\right)={\Omega }_k^{\left(n\right)}\left(v\right)$ for every $k\in \mathbb{N}$ or (that is the same), $u\sim v$ if and only if $P\left(u\right)=P\left(v\right)$ for every ${\omega }_n$-symmetric polynomial $P.$ The set of classes of the equivalence will be denotde by ${\mathcal{M}}^{{\omega }_n}={\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)/\sim .$ For a given $u\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)$, we denote by $\left[u\right]$ the class in ${\mathcal{M}}^{{\omega }_n}$ containing $u.$ Note that the operations “•” and “⋄” can be extended to the following operations of addition and multiplication on ${\mathcal{M}}^{{\omega }_n}$:

$$\left[u\right]+\left[v\right]=[u•v]\mathrm{and}\left[u\right]\left[v\right]=[u\diamond v].$$

Let us show that this definition does not depend on representatives. Indeed, if $u^{\prime }\sim u$ and $v^{\prime }\sim v,$ then for every $k\in \mathbb{N},$

$${\Omega }_k^{\left(n\right)}(u^{\prime }•v^{\prime })={\Omega }_k^{\left(n\right)}\left(u^{\prime }\right)+{\Omega }_k^{\left(n\right)}\left(v^{\prime }\right)={\Omega }_k^{\left(n\right)}\left(u\right)+{\Omega }_k^{\left(n\right)}\left(v\right)={\Omega }_k^{\left(n\right)}(u•v),$$

and

$${\Omega }_k^{\left(n\right)}(u^{\prime }\diamond v^{\prime })={\Omega }_k^{\left(n\right)}\left(u^{\prime }\right){\Omega }_k^{\left(n\right)}\left(v^{\prime }\right)={\Omega }_k^{\left(n\right)}\left(u\right){\Omega }_k^{\left(n\right)}\left(v\right)={\Omega }_k^{\left(n\right)}(u\diamond v).$$

Thus, $[u^{\prime }•v^{\prime }]=[u•v]$ and $[u^{\prime }\diamond v^{\prime }]=[u\diamond v].$

Let

$$u=\left(\begin{array}{c}{u}_{{\alpha }_0}\\ u_{{\alpha }_1}\\ ⋮\\ u_{{\alpha }_{n-1}}\end{array}\right)\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$$

We denote by $u^{-}$ the following element in $\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right),$

$$u^{-}=\left(\begin{array}{c}{u}_{{\alpha }_{n-1}}\\ u_{{\alpha }_0}\\ u_{{\alpha }_1}\\ ⋮\\ u_{{\alpha }_{n-3}}\\ u_{{\alpha }_{n-2}}\end{array}\right)•\left(\begin{array}{c}{u}_{{\alpha }_{n-2}}\\ u_{{\alpha }_{n-1}}\\ u_{{\alpha }_0}\\ ⋮\\ u_{{\alpha }_{n-4}}\\ u_{{\alpha }_{n-3}}\end{array}\right)•\cdots •\left(\begin{array}{c}{u}_{{\alpha }_1}\\ u_{{\alpha }_2}\\ u_{{\alpha }_3}\\ ⋮\\ u_{{\alpha }_{n-1}}\\ u_{{\alpha }_0}\end{array}\right).$$

Proposition 4.

Let $n\ge 2.$ Then, for every $u\in {\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right),$ we have $u•u^{-}\sim 0,$ where 0 is the zero-vector in ${\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right).$

Proof. 

For every $k\in \mathbb{N},$

$$\begin{array}{cc}\hfill {\Omega }_k^{\left(n\right)}(u•u^{-})& ={\Omega }_k^{\left(n\right)}\left(\begin{array}{c}{u}_{{\alpha }_0}•u_{{\alpha }_1}•\cdots •u_{{\alpha }_{n-1}}\\ u_{{\alpha }_0}•u_{{\alpha }_1}•\cdots •u_{{\alpha }_{n-1}}\\ ⋮\\ u_{{\alpha }_0}•u_{{\alpha }_1}•\cdots •u_{{\alpha }_{n-1}}\end{array}\right)\hfill \\ & =\left({\alpha }_0+{\alpha }_1+\cdots +{\alpha }_{n-1}\right)F_n\left(u_{{\alpha }_0}•u_{{\alpha }_1}•\cdots •u_{{\alpha }_{n-1}}\right)=0\hfill \end{array}$$

because ${\alpha }_0+{\alpha }_1+\cdots +{\alpha }_{n-1}=0.$ Hence, $u•u^{-}\sim 0.$ □

We will use notations $-\left[u\right]=\left[u^{-}\right]$ and $\left[u\right]-\left[v\right]=\left[u\right]+(-[v\left]\right).$

Let P be an ${\omega }_n$-symmetric polynomial and $u\in {\mathcal{M}}^{{\omega }_n}.$ We can define $P\left(\right[u\left]\right)=P\left(u\right)$ and the value does not depend on the representative by the definition of the equivalence “∼”.

Theorem 2.

The set ${\mathcal{M}}^{{\omega }_n}$ with the introduced operations of addition, multiplication, and the operation of taking the inverse $\left[u\right]\mapsto -\left[u\right]$ is a commutative unital ring if $n\ge 2$ and a commutative unital semiring if $n=1.$ Polynomials ${\Omega }_k^{\left(n\right)}$ generate ring homomorphisms $\left[u\right]\mapsto {\Omega }_k^{\left(n\right)}\left(u\right)$ from ${\mathcal{M}}^{{\omega }_n}$ to $\mathbb{C}.$

Proof. 

For $n=1$, the statement of this theorem was proven in [30] and for $n=2$ in [7]. We consider the case $n>2.$ Clearly, both the addition and multiplication are commutative and $-\left[u\right]+\left[u\right]=\left[0\right]$, $\left[u\right]\in {\mathcal{M}}^{{\omega }_n}$, $n\ge 2.$ Let us check the distributivity low. For every $k\in \mathbb{N},$

$$\begin{array}{cc}\hfill {\Omega }_k^{\left(n\right)}\left(\left[u\right]\left(\right[v]+[w\left]\right)\right)& ={\Omega }_k^{\left(n\right)}\left(u\diamond (v•w)\right)\hfill \\ & ={\Omega }_k^{\left(n\right)}\left(u\right)\left({\Omega }_k^{\left(n\right)}\left(v\right)+{\Omega }_k^{\left(n\right)}\left(w\right)\right)={\Omega }_k^{\left(n\right)}\left(\left[u\right]\left[v\right]+\left[v\right]\left[w\right]\right).\hfill \end{array}$$

Thus, $\left[u\right]\left(\right[v]+[w\left]\right)=\left[u\right]\left[v\right]+\left[v\right]\left[w\right].$ Similarly, we can check the associativity of addition and multiplication. From the definition of multiplication, we can see that

$$\mathbf{1}:=\left[\begin{array}{c}\left(\begin{array}{cccc}1,& 0,& 0,& \cdots \\ 0,& 0,& 0,& \cdots \\ ⋮& ⋮& \ddots & ⋮\\ 0,& 0,& 0,& \cdots \end{array}\right)\end{array}\right]$$

is the multiplicative unity of ${\mathcal{M}}^{{\omega }_n}.$ Also, we know that ${\Omega }_k^{\left(n\right)}(\left[u\right]+\left[v\right])={\Omega }_k^{\left(n\right)}\left(\left[u\right]\right)+{\Omega }_k^{\left(n\right)}\left(\left[v\right]\right)$ and ${\Omega }_k^{\left(n\right)}\left(\left[u\right]\left[v\right]\right)={\Omega }_k^{\left(n\right)}\left(\left[u\right]\right){\Omega }_k^{\left(n\right)}\left(\left[v\right]\right).$ Thus, ${\Omega }_k^{\left(n\right)}$ are ring homomorphisms for all $k.$ □

Let us consider some obvious subrings of ${\mathcal{M}}^{{\omega }_n}.$ The next proposition easily follows from definitions.

Proposition 5.

(i) 

If m divides $n,$ then ${\omega }_m$ is a subgroup of ${\omega }_n$ and so ${\mathcal{M}}^{{\omega }_m}$ is a subring of ${\mathcal{M}}^{{\omega }_n}.$

(ii) 

Let $U\subset \mathbb{C}$ be a multiplicative semigroup containing 1 and 0. Then,

$${\mathcal{M}}_U^{{\omega }_n}:=\{\left[u\right]\in {\mathcal{M}}^{{\omega }_n}:u_{j{\alpha }_k}\in U,j\in \mathbb{N},0\le k\le n-1\}$$

is a subring of ${\mathcal{M}}^{{\omega }_n}.$

Example 1.

The following semirings ${\mathcal{M}}_U^{{\omega }_n}$ can be interesting:

• ${\mathcal{M}}_{\mathbb{R}}^{{\omega }_n},$ where $\mathbb{R}$ is the set of real numbers;
• ${\mathcal{M}}_{{\mathbb{R}}_{+}}^{{\omega }_n},$ where ${\mathbb{R}}_{+}$ is the set of nonnegative real numbers;
• ${\mathcal{M}}_{{\mathbb{Z}}_{+}}^{{\omega }_n},$ where ${\mathbb{Z}}_{+}$ is the set of nonnegative integer numbers;
• ${\mathcal{M}}_{\mathbb{D}}^{{\omega }_n},$ where $\mathbb{D}$ is the closed unit disk in $\mathbb{C}$;
• ${\mathcal{M}}_{S\cup \left\{0\right\}}^{{\omega }_n},$ where S is the unit circle in $\mathbb{C}.$

5. Example ${\ell }_{\mathbf{1}}\left({\mathbb{N}}_{{\mathsf{\omega }}_{\mathbf{4}}}\right)$

As we observed, any element in ${\mathcal{M}}^{{\omega }_4}$ can be written as

$$\left[z\right]=\left[\begin{array}{c}\left(\begin{array}{ccc}y& |& x\\ t& |& s\end{array}\right)\end{array}\right]$$

(11)

for some elements $\left[\right(y\left|x\right)]$, $\left[\right(t\left|s\right)]$ in ${\mathcal{M}}^{{\omega }_2}.$ Let us denote by $\mathcal{I}$ the “outer” imaginary unit,

$$\mathcal{I}=\left[\begin{array}{c}\left(\begin{array}{ccc}0& |& 0\\ 0& |& 1\end{array}\right)\end{array}\right].$$

Note that

$${\mathcal{I}}^2=\left[\begin{array}{c}\left(\begin{array}{ccc}1& |& 0\\ 0& |& 0\end{array}\right)\end{array}\right]$$

which is the “outer” negative unit. Clearly,

$$\left[z\right]=\left[\begin{array}{c}\left(\begin{array}{ccc}y& |& x\\ 0& |& 0\end{array}\right)\end{array}\right]+\mathcal{I}\left[\begin{array}{c}\left(\begin{array}{ccc}t& |& s\\ 0& |& 0\end{array}\right)\end{array}\right].$$

Taking into account that the subring of ${\mathcal{M}}^{{\omega }_4}$ consisting of elements of the form $\left[\begin{array}{c}\left(\begin{array}{ccc}y& |& x\\ 0& |& 0\end{array}\right)\end{array}\right]$ is isomorphic to ${\mathcal{M}}^{{\omega }_2}$ under the natural map

$${\mathcal{M}}^{{\omega }_4}\ni \left[\begin{array}{c}\left(\begin{array}{ccc}y& |& x\\ 0& |& 0\end{array}\right)\end{array}\right]⟼\left[\left(y\right|x)\right]\in {\mathcal{M}}^{{\omega }_2},$$

we can write $\left[z\right]=\left[\right(y\left|x\right)]+\mathcal{I}[\left(t\right|s\left)\right].$ Let us denote $\Re \left[z\right]:=\left[\right(y\left|x\right)],$ and $\Im \left[z\right]=\left[\right(t\left|s\right)].$

Proposition 6.

Let $\left[z\right]\in {\mathcal{M}}^{{\omega }_4}.$ Then, $\left[z\right]=\left[z^{\prime }\right]$ in ${\mathcal{M}}^{{\omega }_4}$ if and only if $\Re \left[z\right]=\Re \left[z^{\prime }\right]$ and $\Im \left[z\right]=\Im \left[z^{\prime }\right].$

Proof. 

Let $\left[z\right]=\left[z^{\prime }\right],$ where $\left[z\right]=\left[\right(y\left|x\right)]+\mathcal{I}[\left(t\right|s\left)\right]$ and $\left[z^{\prime }\right]=\left[\left(y^{\prime }\right|x^{\prime })\right]+\mathcal{I}\left[\left(t^{\prime }\right|s^{\prime })\right].$ Then, for every $k\in \mathbb{N},$

$$\begin{array}{cc}\hfill {\Omega }_k^{\left(4\right)}\left(z\right)& =F_k\left(x\right)-F_k\left(y\right)+i\left(F_k\left(s\right)-F_k\left(t\right)\right)\hfill \\ & =F_k\left(x^{\prime }\right)-F_k\left(y^{\prime }\right)+i\left(F_k\left(s^{\prime }\right)-F_k\left(t^{\prime }\right)\right)\hfill \\ & ={\Omega }_k^{\left(4\right)}\left(z^{\prime }\right).\hfill \end{array}$$

Thus, $F_k\left(x\right)-F_k\left(y\right)=F_k\left(x^{\prime }\right)-F_k\left(y^{\prime }\right)$ and $F_k\left(s\right)-F_k\left(t\right)=F_k\left(s^{\prime }\right)-F_k\left(t^{\prime }\right)$ for every $k.$ Hence, $\left[\left(y\right|x)\right]=\left[\left(y^{\prime }\right|x^{\prime })\right]$ and $\left[\left(t\right|s)\right]=\left[\left(t^{\prime }\right|s^{\prime })\right].$ □

Let $\left[z_1\right]=\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_1& |& x_1\\ t_1& |& s_1\end{array}\right)\end{array}\right],$ and $\left[z_2\right]=\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_2& |& x_2\\ t_2& |& s_2\end{array}\right)\end{array}\right]$ be in ${\mathcal{M}}^{{\omega }_4}.$ Then,

$$\begin{array}{cc}\hfill \left[z_1\right]+\left[z_2\right]& =\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_1& |& x_1\\ t_1& |& s_1\end{array}\right)\end{array}\right]+\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_2& |& x_2\\ t_2& |& s_2\end{array}\right)\end{array}\right]\hfill \\ & =\left[\left(y_1\right|x_1)\right]+\mathcal{I}\left[\left(t_1\right|s_1)\right]+\left[\left(y_2\right|x_2)\right]+\mathcal{I}\left[\left(t_2\right|s_2)\right]\hfill \\ & =\left(\left[\left(y_1\right|x_1)\right]+\left[\left(y_2\right|x_2)\right]\right)+\mathcal{I}\left(\left[\left(t_1\right|s_1)\right]+\left[\left(t_2\right|s_2)\right]\right)\hfill \\ & =[\left(y_1\right|x_1)•\left(y_2\right|x_2)]+\mathcal{I}[\left(t_1\right|s_1)•\left(t_2\right|s_2)]\hfill \\ & =\left[(y_1•y_2|x_1•x_2)\right]+\mathcal{I}\left[(t_1•t_2|s_1•s_2)\right]=\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_1•y_2& |& x_1•x_2\\ t_1•t_2& |& s_1•s_2\end{array}\right)\end{array}\right].\hfill \end{array}$$

Also, for the multiplication, we have

$$\begin{array}{cc}\hfill \left[z_1\right]\left[z_2\right]& =\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_1& |& x_1\\ t_1& |& s_1\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{ccc}{y}_2& |& x_2\\ t_2& |& s_2\end{array}\right)\end{array}\right]=\left(\left[\left(y_1\right|x_1)\right]+\mathcal{I}\left[\left(t_1\right|s_1)\right]\right)\left(\left[\left(y_2\right|x_2)\right]+\mathcal{I}\left[\left(t_2\right|s_2)\right]\right)\hfill \\ & =\left[\left(y_1\right|x_1)\right]\left[\left(y_2\right|x_2)\right]-\left[\left(t_1\right|s_1)\right]\left[\left(t_2\right|s_2)\right]+\mathcal{I}\left[\left(t_1\right|s_1)\right]\left[\left(y_2\right|x_2)\right]+\mathcal{I}\left[\left(t_2\right|s_2)\right]\left[\left(y_1\right|x_1)\right]\hfill \\ & =\left[((y_1\diamond x_2)•(x_1\diamond y_2)|(x_1\diamond x_2)•(y_1\diamond y_2))\right]\hfill \\ & -\left[((t_1\diamond s_2)•(s_1\diamond t_2)|(s_1\diamond s_2)•(t_1\diamond t_2))\right]\hfill \\ & +\mathcal{I}\left[((t_1\diamond x_2)•(s_1\diamond y_2)|(s_1\diamond x_2)•(t_1\diamond y_2))\right]\hfill \\ & +\mathcal{I}\left[((t_2\diamond x_1)•(s_2\diamond y_1)|(s_2\diamond x_1)•(t_2\diamond y_1))\right]\hfill \end{array}$$

$$=\left[\begin{array}{c}\left(\begin{array}{c}(y_1\diamond x_2)•(x_1\diamond y_2)•(s_1\diamond s_2)•(t_1\diamond t_2)|(x_1\diamond x_2)•(y_1\diamond y_2)•(t_1\diamond s_2)•(s_1\diamond t_2)\\ (t_1\diamond x_2)•(s_1\diamond y_2)•(t_2\diamond x_1)•(s_2\diamond y_1)|(s_1\diamond x_2)•(t_1\diamond y_2)•(s_2\diamond x_1)•(t_2\diamond y_1)\end{array}\right)\end{array}\right].$$

Thus, in particular,

$$-\mathcal{I}=\left[\begin{array}{c}\left(\begin{array}{ccc}1& |& 0\\ 0& |& 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{ccc}0& |& 0\\ 0& |& 1\end{array}\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{ccc}0& |& 0\\ 1& |& 0\end{array}\right)\end{array}\right].$$

For a given ring $\mathcal{R}$, we denote the matrix ring of $m\times m$ matrices with entries in $\mathcal{R}$ by $M_m\left(\mathcal{R}\right).$ It is well-known that there exists an isomorphism between the ring of complex numbers $\mathbb{C}=\{a+bi: a,b\in \mathbb{R}\}$ and the ring of quadratic matrices $M_2\left(\mathbb{R}\right)$ of a form

$$a+bi⟼\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right).$$

Clearly, the set of matrices

$$\left(\begin{array}{cc}\left[u\right]& \left[v\right]\\ -\left[v\right]& \left[u\right]\end{array}\right),\left[u\right],\left[v\right]\in {\mathcal{M}}^{{\omega }_2}$$

form a subring in $M_2\left({\mathcal{M}}^{{\omega }_2}\right)$, which we denote by $M_2^0\left({\mathcal{M}}^{{\omega }_2}\right).$

Let us consider the following map $\mathcal{A}:{\mathcal{M}}^{{\omega }_4}\to M_2\left({\mathcal{M}}^{{\omega }_2}\right)$

$$\mathcal{A}\left(\left[z\right]\right)=\left(\begin{array}{cc}\Re \left[z\right]& \Im \left[z\right]\\ -\Im \left[z\right]& \Re \left[z\right]\end{array}\right).$$

Theorem 3.

The map $\mathcal{A}:{\mathcal{M}}^{{\omega }_4}\to M_2^0\left({\mathcal{M}}^{{\omega }_2}\right)$ is a ring isomorphism.

Proof. 

For every quadratic $2\times 2$ matrix $M=\left(\begin{array}{cc}\left[u\right]& \left[v\right]\\ -\left[v\right]& \left[u\right]\end{array}\right)$ with entries in $\left[u\right]=\left[\right(y\left|x\right)]$ and $\left[v\right]=\left[\right(t\left|s\right)]$ in ${\mathcal{M}}^{{\omega }_2}$, there exists $\left[z\right]\in {\mathcal{M}}^{{\omega }_4}$, defined as follows:

$$\left[z\right]=\left[u\right]+\mathcal{I}\left[v\right]=\left(\begin{array}{cc}\left[u\right]& \left[v\right]\\ -\left[v\right]& \left[u\right]\end{array}\right)=\left(\begin{array}{cc}\left[\right(y\left|x\right)]& \left[\right(t\left|s\right)]\\ \left[\right(s\left|t\right)]& \left[\right(y\left|x\right)]\end{array}\right)=\left[\begin{array}{c}\left(\begin{array}{ccc}y& |& x\\ t& |& s\end{array}\right)\end{array}\right]$$

and such that

$$\mathcal{A}\left(\left[z\right]\right)=\left(\begin{array}{cc}\Re \left[z\right]& \Im \left[z\right]\\ -\Im \left[z\right]& \Re \left[z\right]\end{array}\right)=M.$$

Thus, $\mathcal{A}$ is surjective.

If $\mathcal{A}\left(\left[z_1\right]\right)=\mathcal{A}\left(\left[z_2\right]\right)$, then $\Re \left[z_1\right]=\Re \left[z_2\right]$ and $\Im \left[z_1\right]=\Im \left[z_2\right],$ that is, $\left[z_1\right]=\left[z_2\right]$, and so $\mathcal{A}$ is injective. Also,

$$\begin{array}{cc}\hfill \mathcal{A}(\left[z_1\right]+\left[z_2\right])& =\mathcal{A}[z_1•z_2]\hfill \\ & =\left(\begin{array}{cc}\Re \left([z_1•z_2]\right)& \Im \left([z_1•z_2\right)\\ -\Im \left([z_1•z_2]\right)& \Re \left([z_1•z_2]\right)\end{array}\right)=\left(\begin{array}{cc}\Re \left[z_1\right]+\Re \left[z_2\right]& \Im \left[z_1\right]+\Im \left[z_2\right]\\ -\Im \left[z_1\right]-\Im \left[z_2\right]& \Re \left[z_1\right]+\Re \left[z_2\right]\end{array}\right)\hfill \\ & =\left(\begin{array}{cc}\Re \left[z_1\right]& \Im \left[z_1\right]\\ -\Im \left[z_1\right]& \Re \left[z_1\right]\end{array}\right)+\left(\begin{array}{cc}\Re \left[z_2\right]& \Im \left[z_2\right]\\ -\Im \left[z_2\right]& \Re \left[z_2\right]\end{array}\right)=\mathcal{A}\left(\left[z_1\right]\right)+\mathcal{A}\left(\left[z_2\right]\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathcal{A}\left(\left[z_1\right]\left[z_2\right]\right)& =\mathcal{A}\left(\left(\Re \left[z_1\right]+\mathcal{I}\Im \left[z_1\right]\right)\left(\Re \left[z_2\right]+\mathcal{I}\Im \left[z_2\right]\right)\right)\hfill \\ & =\mathcal{A}\left(\Re \left[z_1\right]\Re \left[z_2\right]-\Im \left[z_1\right]\Im \left[z_2\right]+\mathcal{I}\left(\Re \left[z_1\right]\Im \left[z_2\right]+\Re \left[z_2\right]\Im \left[z_1\right]\right)\right)\hfill \\ & =\left(\begin{array}{cc}\Re \left[z_1\right]\Re \left[z_2\right]-\Im \left[z_1\right]\Im \left[z_2\right]& \Re \left[z_1\right]\Im \left[z_2\right]+\Re \left[z_2\right]\Im \left[z_1\right]\\ -\left(\Re \left[z_1\right]\Im \left[z_2\right]+\Re \left[z_2\right]\Im \left[z_1\right]\right)& \Re \left[z_1\right]\Re \left[z_2\right]-\Im \left[z_1\right]\Im \left[z_2\right]\end{array}\right)\hfill \\ & =\left(\begin{array}{cc}\Re \left[z_1\right]& \Im \left[z_1\right]\\ -\Im \left[z_1\right]& \Re \left[z_1\right]\end{array}\right)\left(\begin{array}{cc}\Re \left[z_2\right]& \Im \left[z_2\right]\\ -\Im \left[z_2\right]& \Re \left[z_2\right]\end{array}\right)=\mathcal{A}\left(\left[z_1\right]\right)\mathcal{A}\left(\left[z_2\right]\right).\hfill \end{array}$$

Hence, the map $\mathcal{A}$ is an isomorphism. □

Clearly that the matrix $\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$, $a,b\in \mathbb{C}$ is always invertible if its determinant, $a^2+b^2\ne 0.$ However, it is not so for multinumbers.

Example 2.

By direct computations, we have

$$\left(\begin{array}{cc}\left[\right(-1\left|1\right)]& 0\\ 0& \left[\right(-1\left|1\right)]\end{array}\right)\left(\begin{array}{cc}\left[\right(0|1,-1)]& 0\\ 0& \left[\right(0|1,-1)]\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).$$

Thus, both $\left[\right(-1\left|1\right)]+\mathcal{I}[0]$ and $\left[\right(0|1,-1)]+\mathcal{I}[0]$ are devisors of zero and cannot to be invertible. On the other hand, ${\left[(-1|1)\right]}^2=\left[(-1,-1|1,1)\right]\ne 0,$ and ${\left[\left(0\right|1,-1)\right]}^2=\left[\left(0\right|1,1,-1,-1)\right]\ne \left[0\right].$

In [7], it is proven that if $\parallel x\parallel +\parallel y\parallel < 1$ for some $x$, $y\in {\ell }_1,$ then

$$\left[\left(0\right|1)\right]-\left[\left(y\right|x)\right]=\left[\left(x\right|1•y)\right]=\left[(\dots ,x_2,x_1|1,y_1,y_2,\dots )\right]$$

is invertible, and

$${\left(\left[\right(0\left|1\right)]-[\left(y\right|x\left)\right]\right)}^{-1}=\sum _{m=0}^{\infty }{\left[\left(y\right|x)\right]}^m\in {\mathcal{M}}^{{\omega }_2}.$$

Proposition 7.

Let

$$\left[z\right]=\left[u\right]+\mathcal{I}\left[v\right]=\left(\begin{array}{cc}\left[u\right]& \left[v\right]\\ -\left[v\right]& \left[u\right]\end{array}\right)\in {\mathcal{M}}^{{\omega }_4}$$

such that ${\left[u\right]}^2+{\left[v\right]}^2$ is invertible in ${\mathcal{M}}^{{\omega }_2}.$ Then, $\left[z\right]$ is invertible, and

$${\left[z\right]}^{-1}=\left(\left[u\right]-\mathcal{I}\left[v\right]\right){\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}=\left(\begin{array}{cc}\left[u\right]{\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}& -\left[v\right]{\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}\\ {\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}& \left[u\right]{\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}\end{array}\right).$$

Proof. 

By direct calculations,

$$\left[z\right]{\left[z\right]}^{-1}=\left(\left[u\right]+\mathcal{I}\left[v\right]\right)\left(\left[u\right]-\mathcal{I}\left[v\right]\right){\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}=\left({\left[u\right]}^2+{\left[v\right]}^2\right){\left({\left[u\right]}^2+{\left[v\right]}^2\right)}^{-1}=\mathbf{1}.$$

□

6. The Ring of Integer Multinumbers ${\mathcal{Z}}^{{\mathsf{\omega }}_{\mathit{n}}}$

The following example shows that, in the general case, ${\mathcal{M}}^{{\omega }_n}$ contains devisors of zero if $n>1.$

Example 3.

Let $\left[u\right]$ and $\left[v\right]$ in ${\mathcal{M}}^{{\omega }_n}$ be such that

$$u=\left(\begin{array}{c}{\alpha }_0\\ {\alpha }_1\\ {\alpha }_2\\ \cdots \\ {\alpha }_{n-1}\end{array}\right)andv=\left(\begin{array}{ccccc}{\alpha }_0,& {\alpha }_1,& {\alpha }_2,& \dots ,& {\alpha }_{n-1}\\ 0,& 0,& 0,& \dots ,& 0\\ 0,& 0,& 0,& \dots ,& 0\\ 0,& 0,& 0,& \dots ,& 0\\ 0,& 0,& 0,& \dots ,& 0\end{array}\right).$$

Then

$$\left[u\right]\left[v\right]=\left[\begin{array}{c}\left(\begin{array}{ccccc}{\alpha }_0,& {\alpha }_1,& {\alpha }_2,& \dots ,& {\alpha }_{n-1}\\ {\alpha }_0,& {\alpha }_1,& {\alpha }_2,& \dots ,& {\alpha }_{n-1}\\ {\alpha }_0,& {\alpha }_1,& {\alpha }_2,& \dots ,& {\alpha }_{n-1}\\ {\alpha }_0,& {\alpha }_1,& {\alpha }_2,& \dots ,& {\alpha }_{n-1}\\ {\alpha }_0,& {\alpha }_1,& {\alpha }_2,& \dots ,& {\alpha }_{n-1}\end{array}\right)\end{array}\right]=\left[0\right].$$

Let us denote ${\mathcal{Z}}^{{\omega }_n}:={\mathcal{M}}_{{\mathbb{Z}}_{+}}^{{\omega }_n}.$ This ring can be considered as a subring of integer multinumbers in ${\mathcal{M}}_{{\mathbb{Z}}_{+}}^{{\omega }_n}.$ If $u={\left(u_{m{\alpha }_j}\right)}_{m\in \mathbb{N},0\le j<n}\in {\mathcal{Z}}^{{\omega }_n},$ then only a finite number of components $u_{m{\alpha }_j}$ are nonzero and all of them are positive integers.

We need the following technical result, which probably is well-known.

Lemma 1.

Let n be a prime number and $m_0,\dots ,m_{n-1}\in \mathbb{Z}.$ Then

$$m_0+{\alpha }_1{m}_1+\cdots +{\alpha }_{n-1}{m}_{n-1}=0$$

implies that $m_0=m_1=\cdots =m_{n-1}.$

Proof. 

As $m_0(1+{\alpha }_1+\cdots +{\alpha }_{n-1})=0,$ we have that

$${\alpha }_1(m_1-m_0)+\cdots +{\alpha }_{n-1}(m_{n-1}-m_0)=0.$$

On the other hand, it is known that any $n-1$ of the elements $1,{\alpha }_1,\dots ,{\alpha }_{n-1}$ form a basis for $\mathbb{Z}\left[{\alpha }_1\right]$ over $\mathbb{Z}$ [31] (p. 23). Thus, $m_j-m_0=0$ for all $j=0,\cdots ,n-1.$ □

In [11], an isomorphism of ${\mathcal{Z}}^{{\omega }_2}$ was constructed to a ring of polynomials with integer coefficients and of infinite many variables. From there, it was deduced that ${\mathcal{Z}}^{{\omega }_2}$ is an integral domain. This approach can be extended to the general case. Let ${\mathbb{C}}^{\infty }$ be the linear space of all complex sequences $t=(t_1,t_2,\dots ).$ We denote by $\mathbb{C}\left[{\mathbb{C}}^{\infty }\right]$ the ring of all polynomials on ${\mathbb{C}}^{\infty }.$ Any polynomial Q in $\mathbb{C}\left[{\mathbb{C}}^{\infty }\right]$ is a finite linear combination over $\mathbb{C}$ of polynomials $t_{j_1}^{k_1}\cdots t_{j_m}^{k_m},$ that is

$$Q(t_1,t_2,\dots )=\sum a_{j_1\dots j_m}^{k_1\dots k_m}{t}_{j_1}^{k_1}\cdots t_{j_m}^{k_m},$$

(12)

for some complex $a_{j_1\dots j_m}^{k_1\dots k_m},$ where we sum all indexes $j_1,\dots .j_m,k_1,\dots ,k_m$ that go over a finite subset of ${\mathbb{Z}}_{+}.$ Let us consider a subring ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right]$ of $\mathbb{C}\left[{\mathbb{C}}^{\infty }\right]$ consisting of polynomials (12), such that coefficients $a_{j_1\dots j_m}^{k_1\dots k_m}$ are of the form ${\sum }_{i=0}^{n-1}{\alpha }_i{R}_{i;j_1\left(i\right)\dots j_m\left(i\right)}^{k_1\left(i\right)\dots k_m\left(i\right)}$ for ${\alpha }_i\in {\omega }_n$ and $R_{i;j_1\left(i\right)\dots j_m\left(i\right)}^{k_1\left(i\right)\dots k_m\left(i\right)}\in {\mathbb{Z}}_{+}.$

Theorem 4.

Let n be a prime positive integer number or $n=4.$

(i) 

There exists a ring isomorphism from ${\mathcal{Z}}^{{\omega }_n}$ to ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right].$

(ii) 

${\mathcal{Z}}^{{\omega }_n}$ is an integral domain.

(iii) 

For every $t=(t_1,t_2,\dots )\in {\mathbb{C}}^{\infty }$ the mapping

$$u\mapsto \nu \left(\right[u\left]\right)\left(t\right)$$

is a complex-valued ring homomorphism.

Proof. 

As in [11], for every positive integer $a=p_1^{k_1}\cdots p_m^{k_m},$ where $\left(p_j\right)$ is the sequence of all prime numbers $(2,3,5,\dots ,)$ we define ${\nu }_0\left(a\right):=t_1^{k_1}\cdots t_m^{k_m}$ and extend it to ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right]$ by

$$\nu \left[u\right]=\nu \left[\begin{array}{c}\left(\begin{array}{ccccc}{u}_{1{\alpha }_0},& u_{2{\alpha }_0},& \cdots ,& u_{m{\alpha }_0},& \cdots \\ u_{1{\alpha }_1},& u_{2{\alpha }_1},& \cdots & u_{m{\alpha }_1},& \cdots \\ ⋮& ⋮& \ddots & ⋮& ⋮\\ u_{1{\alpha }_{n-1}},& u_{2{\alpha }_{n-1}},& \cdots ,& u_{m{\alpha }_{n-1}},& \cdots \end{array}\right)\end{array}\right]=\sum _{k=0}^{n-1}{\alpha }_k\sum _{j=1}^{\infty }{\nu }_0\left(u_{j{\alpha }_k}\right).$$

As every element $\left[u\right]\in {\mathcal{Z}}^{{\omega }_n}$ contains only a finite number of nonzero components $u_{j{\alpha }_k},$ the range

$$\nu \left(\left[u\right]\right)=\sum _{k=0}^{n-1}{\alpha }_k\sum _{j=1}^{\infty }{\nu }_0\left(u_{j{\alpha }_k}\right)$$

has a finite number of addends and so belongs to ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right].$ Let $\left[u\right]=\left[0\right]$ and n be prime numbers. Then, for every $m\in \mathbb{N},$

$${\Omega }_m^{\left(n\right)}\left(\left[u\right]\right)=\sum _{k=0}^{n-1}{\alpha }_k{F}_m\left(u_{{\alpha }_k}\right).$$

Thus, as n is prime, by Lemma 1, $F_m\left(u_{{\alpha }_k}\right)=F_m\left(u_{{\alpha }_j}\right)$ for all k and $j,$ and so $u_{{\alpha }_k}\sim u_{{\alpha }_j}$ as elements in ${\ell }_1.$ In other words, there is a permutation $\sigma$, such that

$$u_{{\alpha }_k}=(u_{1{\alpha }_k},u_{2{\alpha }_k},\dots )=(u_{\sigma \left(1\right){\alpha }_j},u_{\sigma \left(2\right){\alpha }_j},\dots ).$$

In particular, $\nu \left(u_{{\alpha }_k}\right)=\nu \left(u_{{\alpha }_j}\right)$ for all k and j in $\{0,1,\dots ,n-1\}.$ Let $\nu \left(u_{{\alpha }_k}\right)={\alpha }_{k}cP(t_1,\dots ,t_m)$ for some $c\in {\mathbb{Z}}_{+}$ and $P\in \mathbb{C}\left[{\mathbb{C}}^{\infty }\right].$ Then,

$$\nu \left(\left[u\right]\right)=cP(t_1,\dots ,t_m)\sum _{k=0}^{n-1}{\alpha }_k=0.$$

Hence, $\nu \left(\right[u\left]\right)$ does not depend on the representative.

Now, let $\left[u\right]=\left[0\right]$ and $n=4.$ Then, by Proposition 6, both the real part of $\left[u\right]$ and the imaginary part of $\left[u\right]$ are equal to zero. That is, $\Re \left[u\right]=\left[\right(y\left|x\right)]=0,$ and $\Im \left[u\right]=\left[\right(r\left|s\right)]=0.$ Then, $y\sim x$ and $r\sim s$ are elements in ${\ell }_1.$ Let ${\nu }_0\left(x\right)=P\left(t\right)$ and ${\nu }_0\left(s\right)=Q\left(t\right)$ for some $P,Q\in \mathbb{C}\left[{\mathbb{C}}^{\infty }\right].$ Then,

$$\nu \left(\right[u\left]\right)=P\left(t\right)-P\left(t\right)+i\left(Q\right(t)-Q(t\left)\right)=0$$

and so, $\nu \left(\right[u\left]\right)$ does not depend on the representative.

We can see that

$$\nu (-\left[u\right])=\sum _{k=0}^{n-1}{\alpha }_{n-k}\sum _{j=1}^{\infty }\nu \left(u_{j{\alpha }_k}\right)=\sum _{k=0}^{n-1} -{\alpha }_k\sum _{j=1}^{\infty }\nu \left(u_{j{\alpha }_k}\right)=-\nu \left[u\right],$$

$$\nu (\left[u\right]+\left[v\right])=\sum _{k=0}^{n-1}{\alpha }_k\sum _{j=1}^{\infty }\left(\nu \left(u_{j{\alpha }_k}\right)+\nu \left(v_{j{\alpha }_k}\right)\right)=\nu \left[u\right]+\nu \left[v\right],$$

and

$$\nu \left(\left[u\right]\left[v\right]\right)=\sum _{k=0}^{n-1}\sum _{m+l=k}{\alpha }_k\sum _{i,j=1}^{\infty }\nu \left(u_{i{\alpha }_m}\right)\nu \left(v_{j{\alpha }_l}\right)=\nu \left[u\right]\nu \left[v\right].$$

Hence, $\nu$ is a homomorphism. If $a\ne b,$ then ${\nu }_0\left(a\right)\ne {\nu }_0\left(b\right)$, because of the uniqueness of the representation of any integer as a product of prime numbers. From here, we can deduce that $\nu$ is injective. On the other hand, for every polynomial $Q(t_1,t_2,\dots )\in {\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right],$

$$Q(t_1,t_2,\dots ,)=\sum _{i=0}^{n-1}{\alpha }_i\sum R_{i;j_1\left(i\right)\dots j_m\left(i\right)}^{k_1\left(i\right)\dots k_m\left(i\right)}{t}_{j_1\left(i\right)}^{k_1\left(i\right)}\cdots t_{j_m\left(i\right)}^{k_m\left(i\right)},$$

${\alpha }_i\in {\omega }_n$, $R_{i;j_1\dots j_m}^{k_1\dots k_m}\in {\mathbb{Z}}_{+},$ we have a unique preimage $\left[u\right]\in {\mathcal{Z}}^{{\omega }_n}$, such that $u=u_{{\alpha }_0}•\cdots •u_{{\alpha }_{n-1}},$ and

$$u_{{\alpha }_i}=•\sum {\left(p_{j_1\left(i\right)}^{k_1\left(i\right)}\cdots p_{j_m\left(i\right)}^{k_m\left(i\right)}\right)}^{•R_{i;j_1\left(i\right)\dots j_m\left(i\right)}^{k_1\left(i\right)\dots k_m\left(i\right)}}=•\sum \underset{R_{i;j_1\left(i\right)\dots j_m\left(i\right)}^{k_1\left(i\right)\dots k_m\left(i\right)}}{\underbrace{\left(p_{j_1\left(i\right)}^{k_1\left(i\right)},\cdots ,p_{j_m\left(i\right)}^{k_m\left(i\right)},,,\dots ,,,p_{j_1\left(i\right)}^{k_1\left(i\right)},\cdots ,p_{j_m\left(i\right)}^{k_m\left(i\right)}\right)}}.$$

Hence, $\nu$ is surjective. Therefore, $\nu$ is a ring isomorphism.

The ring ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right]$ is a subring of all polynomials on ${\mathbb{C}}^{\infty }$ that evidently has no devisors of zero. Thus, ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right]$ is an integral domain and ${\mathcal{Z}}^{{\omega }_n}$ is an integral domain as well.

For every fixed $t\in {\mathbb{C}}^{\infty },$ the mapping $\left[u\right]\mapsto \nu \left(\right[u\left]\right)\left(t\right)$ is a complex-valued ring homomorphism as a composition of two homomorphisms. □

Let n be a prime number or $n=4.$ We say that a multinumber $\left[u\right]$ is prime if $\nu \left(\right[u\left]\right)$ is an irreducible polynomial in ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right].$ Let us recall that a polynomial $P\in \mathbb{C}\left[{\mathbb{C}}^{\infty }\right]$ is irreducible if P is not equal to a product $P_1{P}_2$ of two polynomials in $\mathbb{C}\left[{\mathbb{C}}^{\infty }\right]$, such that $deg{P}_1>1,$ and $deg{P}_2>1.$ Thus, if P is irreducible, then $cP$ is irreducible for any constant $c\in \mathbb{C}.$ It is well-known that every polynomial on a linear space (in particular on ${\mathbb{C}}^{\infty }$) can be represented as a product of irreducible polynomials, and this representation is unique up to permutations and multiplicative constants. Applying these observations to polynomials in ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right],$ we can assert that every polynomial in ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right]$ can be represented as a product of irreducible polynomials in ${\omega }_n{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right]$, and if $P=P_1\cdots P_k=Q_1\cdots Q_m,$ then $k=m,$ there is a permutation $\sigma$ on $\{1,\dots ,k\}$ and there are some constants $c_1,\dots ,c_k\in {\omega }_n{\mathbb{Z}}_{+}$, such that $c_1\cdots c_k=1,$ and $P_i=c_i{Q}_{\sigma \left(i\right)}$, $i=1,\dots ,k.$

According to the definition of $\nu ,$

$${\nu }^{-1}\left({\alpha }_k\right)=\left(\begin{array}{c}0\\ ⋮\\ 0\\ 1\\ 0\\ ⋮\\ 0\end{array}\right)⤎k\mathrm{th}\mathrm{position},$$

${\alpha }_k\in {\omega }_n.$ As every constant in ${\omega }_n{\mathbb{Z}}_{+}$ is a finite additive combination of elements in ${\omega }_n,$ ${\nu }^{-1}\left({\omega }_n{\mathbb{Z}}_{+}\right)$ is the additive span of elements ${\nu }^{-1}\left({\alpha }_k\right),$ ${\alpha }_k\in {\omega }_n.$

Corollary 1.

Let n be a prime number or $n=4.$ Any element $\left[u\right]$ in ${\mathcal{Z}}^{{\omega }_n}$ can be represented as a product of prime elements and if $\left[u\right]=\left[u_1\right]\cdots \left[u_k\right]=\left[v_1\right]\cdots \left[v_m\right],$ then $k=m,$ there is a permutation σ on $\{1,\dots ,k\}$ and there are some multinumbers $\left[b_1\right],\dots ,\left[b_k\right]\in {\nu }^{-1}\left({\omega }_n{\mathbb{Z}}_{+}\right)$, such that $\left[b_1\right]\cdots \left[b_k\right]=\left[1\right],$ and $\left[u_i\right]=\left[b_i\right]\left[v_{\sigma \left(i\right)}\right]$, $i=1,\dots ,k.$

Example 4.

Consider the following Problem: factor

$$\left[u\right]=\left[\begin{array}{c}\left(\begin{array}{cccc}24,& 24,& 3,& 3\\ 0,& 0,& 0,& 0\\ 0,& 0,& 0,& 0\end{array}\right)\end{array}\right]$$

into prime multipliers in ${\mathcal{Z}}^{{\omega }_3}.$

Solution. Note that $24=2^3·3.$ Thus, we have that $\nu \left(u\right)\left(t\right)=2t_1^3{t}_2+2t_2=2t_2(t_1+{\alpha }_0)(t_1+{\alpha }_1)(t_1+{\alpha }_2)$ is the representation of $\nu \left(u\right)\left(t\right)$ as a product of irreducible polynomials in ${\omega }_3{\mathbb{Z}}_{+}\left[{\mathbb{C}}^{\infty }\right],$ where ${\alpha }_0=1$, ${\alpha }_1=-{\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{i}{2}},$ and ${\alpha }_2=-{\displaystyle \frac{\sqrt{3}}{2}}-{\displaystyle \frac{i}{2}}.$ From the definition of $\nu ,$

$${\nu }^{-1}\left(2t_2\right)=\left[\begin{array}{c}\left(\begin{array}{cc}3,& 3\\ 0,& 0\\ 0,& 0\end{array}\right)\end{array}\right],{\nu }^{-1}(t_1+1)=\left[\begin{array}{c}\left(\begin{array}{cc}2,& 1\\ 0,& 0\\ 0,& 0\end{array}\right)\end{array}\right],$$

$${\nu }^{-1}(t_1+{\alpha }_1)=\left[\begin{array}{c}\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)\end{array}\right],{\nu }^{-1}(t_1+{\alpha }_2)=\left[\begin{array}{c}\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)\end{array}\right]$$

are prime factors. By straightforward computations, we can check that

$$\left[\begin{array}{c}\left(\begin{array}{cc}3,& 3\\ 0,& 0\\ 0,& 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{cc}2,& 1\\ 0,& 0\\ 0,& 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{cccc}24,& 24,& 3,& 3\\ 0,& 0,& 0,& 0\\ 0,& 0,& 0,& 0\end{array}\right)\end{array}\right].$$

As ${\alpha }_1{\alpha }_2=1,$

$${\alpha }_1\left[\begin{array}{c}\left(\begin{array}{cc}3,& 3\\ 0,& 0\\ 0,& 0\end{array}\right)\end{array}\right]{\alpha }_2\left[\begin{array}{c}\left(\begin{array}{cc}2,& 1\\ 0,& 0\\ 0,& 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{cc}0,& 0\\ 3,& 3\\ 0,& 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{cc}0,& 0\\ 0,& 0\\ 2,& 1\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)\end{array}\right]\left[\begin{array}{c}\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)\end{array}\right]$$

is another representation of $\left[u\right].$

7. Discussion

Algebras of ${\omega }_n$-symmetric polynomials ${\mathcal{P}}_s\left({\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)\right)$ are natural generalizations of algebras symmetric and supersymmetric polynomials on ${\ell }_1.$ Other generalizations can be found in [22,23,32,33]. Such types of algebras are generated by countable sets of basis polynomials. It is important for applications to know explicit representations of different algebraic bases. We know that polynomials ${\Omega }_k^{\left(n\right)}$, $k\in \mathbb{N},$ by definition, form an algebraic basis in ${\mathcal{P}}_s\left({\ell }_1\left({\mathbb{N}}_{{\omega }_n}\right)\right),$ and we have explicit representations only for ${\Lambda }_{{\omega }_2}\left(G_m\right)$ and ${\Lambda }_{{\omega }_2}\left(H_m\right)$ (see [13]). The general case of ${\Lambda }_{{\omega }_n}\left(G_m\right)$ and ${\Lambda }_{{\omega }_n}\left(H_m\right)$ can be obtained using generating functions and the mth derivative. In further investigations, we aim to derive closed formulas for ${\Lambda }_{{\omega }_n}\left(G_m\right)$ and ${\Lambda }_{{\omega }_n}\left(H_m\right)$ if $n>2.$

The construction of multinumbers associated with ${\omega }_n$ can extend to other, potentially noncommutative groups. In this case, we will obtain a ring of multisets consisting of elements in a given group. Another way to obtain a noncommutative ring is by considering matrixes of multinumbers. In Section 5, we observed that ${\mathcal{M}}^{{\omega }_4}$ is isomorphic to a ring of matrix of elements in ${\mathcal{M}}^{{\omega }_2}.$ Using similar construction and the known matrix representation of quaternions, we can suppose that the ring of elements of the form

$$\left(\begin{array}{cc}\left[u\right]& \left[v\right]\\ \\ -\left[v\right]& \overline{\left[u\right]}\end{array}\right),\left[u\right],\left[v\right]\in {\mathcal{M}}^{{\omega }_4}$$

provides a representation of a ring of quaternionic multinumbers. Here, $\overline{\left[u\right]}=\Re \left[u\right]-\Im \left[u\right].$

Let us notice that the additive group $\left({\mathcal{M}}^{{\omega }_2},+\right)$ is the Grothendieck group (or group of differences) of the commutative monoid $\left({\mathcal{M}}^{{\omega }_1},+\right)$ (see [34] for details on Grothendieck groups). However, the ring of multinumbers ${\mathcal{M}}^{{\omega }_n}$ for $n>2$ is more complicated. It is easy to check that Lemma 1 is true only for a prime number n, but Theorem 4 is proven also for $n=4.$ We do not know if the theorem is true for other composite numbers. For this purpose, we need a description of symmetries of ${\mathcal{M}}^{{\omega }_n}$ in the general case. Such a description would be useful, also, for topologization of ${\mathcal{M}}^{{\omega }_4}.$