On properties of square-free elements in commutative cancellative monoids

TL;DR

This paper investigates the properties of square-free elements in commutative cancellative monoids, exploring factorizations, atomicity, and gcd properties, and characterizes submonoids with specific square-free element behaviors.

Contribution

It provides a full characterization of submonoids of factorial monoids where all square-free elements are also square-free in the larger monoid, and explores conditions for atoms to be square-free.

Findings

Characterization of submonoids where square-free elements are preserved

Conditions under which atoms are square-free in submonoids

Relationships between factorial properties and square-free elements

Abstract

We discuss various square-free factorizations in monoids in the context of: atomicity, ascending chain condition for principal ideals, decomposition, and a greatest common divisor property. Moreover, we obtain a full characterization of submonoids of factorial monoids in which all square-free elements of a submonoid are square-free in a monoid. We also present factorial properties implying that all atoms of a submonoid are square-free in a monoid.

Equations

$$\begin{array}[]{ccccccc}&&{\rm(i)}&\Leftarrow&\text{\rm atm}&\Leftarrow&\text{\rm ACCP}\\ &&\mbox{\raisebox{-3.0pt}{$\Uparrow$}}&\mbox{\begin{psfrags}\rotatebox{-45.0}{$\Leftarrow$}\end{psfrags}}&&\mbox{\begin{psfrags}\rotatebox{-135.0}{$\Rightarrow$}\end{psfrags}}&\\ {\rm(ii)}&\Rightarrow&{\rm(iii)}&&{\rm(iv)}&&\\ \Downarrow&&&&\Downarrow&&\\ {\rm(v)}&&&&{\rm(vi)}&&\end{array}$$

$$\begin{array}[]{ccccc}{\rm(ii)}&\Leftrightarrow&{\rm(iii)}&\Rightarrow&{\rm(iv)}.\end{array}$$

$$\begin{array}[]{ccccc}{\rm(ii)}&\Leftrightarrow&{\rm(iii)}&\Leftrightarrow&{\rm(iv)}.\end{array}$$

$$\text{(i)}\wedge \neg \text{(v)},\text{(i)}\wedge \neg \text{(vi)},\text{(v)}\wedge \neg \text{(i)},\text{(v)}\wedge \neg \text{(vi)},\text{(vi)}\wedge \neg \text{(i)},\text{(vi)}\wedge \neg \text{(v)}.$$

$$a^{2}b\in M\Rightarrow a,ab\in M.$$

$$a^{2}b\in M\Rightarrow a,ab\in M,$$

$$s_0{s}_1^2{s}_2^{2^2}\dots s_n^{2^n}\in M\Rightarrow s_i{s}_{i+1}{s}_{i+2}^2{s}_{i+3}^{2^2}\dots s_n^{2^{n-i-1}}\in M,i=0,\dots ,n-1,\text{mbox}and{s}_n\in M,$$

$$s_1{s}_2^2{s}_3^3\dots s_n^n\in M\Rightarrow s_n,s_{n-1}{s}_n,s_{n-2}{s}_{n-1}{s}_n,\dots ,s_1{s}_2\dots s_n\in M,$$

$$s_1{s}_2\dots s_n\in M\Rightarrow s_1,s_2,\dots ,s_n\in M,$$

$$ab\in M\Rightarrow a,b\in M.$$

$$s_{1}s_{2}^{2}s_{3}^{2^{2}}\ldots s_{n}^{2^{n-1}},\;\;\big{(}s_{1}s_{2}^{2}s_{3}^{2^{2}}\ldots s_{n}^{2^{n-1}}\big{)}s_{0}\in M.$$

$$s_{2}s_{3}^{2}s_{4}^{2^{2}}\ldots s_{n}^{2^{n-2}},\;\;\big{(}s_{2}s_{3}^{2}s_{4}^{2^{2}}\ldots s_{n}^{2^{n-2}}\big{)}s_{1}\in M.$$

$$\big{(}s_{1}s_{2}^{2}s_{3}^{2^{2}}\ldots s_{n}^{2^{n-1}}\big{)}s_{0},\>\big{(}s_{2}s_{3}^{2}s_{4}^{2^{2}}\ldots s_{n}^{2^{n-2}}\big{)}s_{1},\ldots,\>s_{n-1}s_{n}^{2}s_{n-2},\>s_{n}s_{n-1},\>s_{n}\in M.$$

$$s_0{s}_1^2{s}_2^{2^2}\dots s_n^{2^n}=a^{2}b\in M.$$

$$s_0{s}_1{s}_2^2{s}_3^{2^2}\dots s_n^{2^{n-1}},s_1{s}_2{s}_3^2{s}_4^{2^2}\dots s_n^{2^{n-2}},\dots ,s_{n-2}{s}_{n-1}{s}_n^2,s_{n-1}{s}_n,s_n\in M.$$

$$a=s_{n}(s_{n-1}s_{n})\big{(}s_{n-2}s_{n-1}s_{n}^{2}\big{)}\big{(}s_{n-3}s_{n-2}s_{n-1}^{2}s_{n}^{2^{2}}\big{)}\ldots\big{(}s_{1}s_{2}s_{3}^{2}s_{4}^{2^{2}}\ldots s_{n}^{2^{n-2}}\big{)}\in M.$$

$$t_1{t}_2^2{t}_3^{2^2}\dots t_r^{2^{r-1}}=\left(\prod _{i=0}^{r-1}{t}_i{t}_{i+1}{t}_{i+2}^2{t}_{i+3}^{2^2}\dots t_r^{2^{r-i-1}}\right)t_r\in M.$$

$$s_1{s}_2{s}_3^2{s}_4^2\dots s_n^{\lceil \frac{n}{2}\rceil }=t_0{t}_1{t}_2^2\dots t_r^{2^{r-1}}\in M,$$

$$s_2{s}_3{s}_4^2{s}_5^2\dots s_n^{\lfloor \frac{n}{2}\rfloor }=t_1{t}_2^2{t}_3^{2^2}\dots t_r^{2^{r-1}}\in M.$$

$$s_1^{\lceil \frac{1}{2^l}\rceil }{s}_2^{\lceil \frac{2}{2^l}\rceil }\dots s_{n-1}^{\lceil \frac{n-1}{2^l}\rceil }{s}_n^{\lceil \frac{n}{2^l}\rceil },s_1^{1-\lceil \frac{1}{2^l}\rceil }{s}_2^{2-\lceil \frac{2}{2^l}\rceil }\dots s_{n-1}^{n-1-\lceil \frac{n-1}{2^l}\rceil }{s}_n^{n-\lceil \frac{n}{2^l}\rceil }\in M.$$

$$s_1^{\lceil \frac{1}{2^{l+1}}\rceil }{s}_2^{\lceil \frac{2}{2^{l+1}}\rceil }\dots s_{n-1}^{\lceil \frac{n-1}{2^{l+1}}\rceil }{s}_n^{\lceil \frac{n}{2^{l+1}}\rceil }=s_1^{\prime }{s}_2^{\prime }(s_3^{\prime }{)}^2(s_4^{\prime }{)}^2\dots (s_q^{\prime }{)}^{\lceil \frac{q}{2}\rceil }\in M$$

$$s_1^{\lceil \frac{1}{2^l}\rceil -\lceil \frac{1}{2^{l+1}}\rceil }{s}_2^{\lceil \frac{2}{2^l}\rceil -\lceil \frac{2}{2^{l+1}}\rceil }\dots s_n^{\lceil \frac{n}{2^l}\rceil -\lceil \frac{n}{2^{l+1}}\rceil }=s_2^{\prime }{s}_3^{\prime }(s_4^{\prime }{)}^2(s_5^{\prime }{)}^2\dots (s_q^{\prime }{)}^{\lfloor \frac{q}{2}\rfloor }\in M.$$

$$s_1^{1-\lceil \frac{1}{2^{l+1}}\rceil }{s}_2^{2-\lceil \frac{2}{2^{l+1}}\rceil }\dots s_n^{n-\lceil \frac{n}{2^{l+1}}\rceil }=$$

$$s_1^{1-\lceil \frac{1}{2^l}\rceil }{s}_2^{2-\lceil \frac{2}{2^l}\rceil }\dots s_n^{n-\lceil \frac{n}{2^l}\rceil }\cdot s_1^{\lceil \frac{1}{2^l}\rceil -\lceil \frac{1}{2^{l+1}}\rceil }{s}_2^{\lceil \frac{2}{2^l}\rceil -\lceil \frac{2}{2^{l+1}}\rceil }\dots s_n^{\lceil \frac{n}{2^l}\rceil -\lceil \frac{n}{2^{l+1}}\rceil }\in M.$$

$$s_1{s}_2{s}_3\dots s_n{s}_{n+1},s_2{s}_3^2{s}_4^3\dots s_n^{n-1}{s}_{n+1}^n\in M.$$

$$s_{n+1},s_n{s}_{n+1},s_{n-1}{s}_n{s}_{n+1},\dots ,s_2{s}_3\dots s_n{s}_{n+1}\in M.$$

$$a{\text{mbox}\mathrm{rpr}}_{M}b\Rightarrow a{\text{mbox}\mathrm{rpr}}_{H}b,$$

$$a{\nsim }_{M}b\Rightarrow a{\text{mbox}\mathrm{rpr}}_{H}b,$$

$$s_1{s}_2^2{s}_3^3\dots s_n^n\in M\Rightarrow s_1,s_2,\dots ,s_n\in M,$$

$$q_1^{k_1}\dots q_n^{k_n}\in M\Rightarrow q_1^{c_i^{(1)}}\dots q_n^{c_i^{(n)}}\in M\text{mbox}foreachi,$$

Full text

1 Introduction

Throughout this paper by a monoid we mean a commutative cancellative monoid. We adopt the notation from [12].

Let $H$ be a monoid. We denote by $H^{\times}$ the group of all invertible elements of $H$. Two elements $a,b\in H$ are called relatively prime if they have no common non-invertible divisors, what we denote by $a\mathbin{\mbox{\small\rm rpr}}b$. The set of all atoms in $H$ will be denoted by $\operatorname{{\cal A}}(H)$. Recall that an element $a\in H$ is called square-free if it cannot be presented in the form $a=b^{2}c$, where $b,c\in H$ and $b\not\in H^{\times}$. The set of all square-free elements in $H$ we will denote by $\operatorname{{\cal S}}(H)$.

The main motivation of this paper is connected with the following two properties concerning a submonoid $M\subset H$. The first one is that all atoms of $M$ are square-free in $H$:

• (1.1)

$\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(H)$.

The second one is that all square-free elements of $M$ are square-free in $H$:

• (1.2)

$\operatorname{{\cal S}}(M)\subset\operatorname{{\cal S}}(H)$.

These properties are related to the famous Jacobian conjecture (for details see Section 2).

If $H$ is a factorial monoid and a submonoid $M\subset H$ satisfies $M^{\times}=H^{\times}$ and $\operatorname{\text{\sf q}}(M)\cap H=M$, then condition (1.2) can be expressed in a factorial way (see [15], Theorem 3.4 – formulated in terms of rings, but in fact valid for monoids):

• (1.3)

for every $a\in H$, $b\in\operatorname{{\cal S}}(H)$, if $a^{2}b\in M$, then $a,b\in M$.

Recall also (see [15], Theorem 3.6) that under these assumptions a submonoid $M$ satisfying (1.2) is root closed in $H$. Recently Angermüller showed in [4], Proposition 32, that under the same assumptions a submonoid $M$ satisfying (1.1) is root closed in $H$. A submonoid $M\subset H$ is called root closed in $H$ if, for every $a\in H$ and $n\geq 1$, $a^{n}\in M$ implies $a\in M$.

Recall two questions concerning the conditions (1.1) and (1.2) in the case of a UFD, stated in [14]. We have asked if they are equivalent under some natural assumptions (like $M^{\times}=H^{\times}$), and if not, can the condition (1.1) be expressed in a form of factoriality, similarly to (1.3)?

In Section 4 we present a factorial property implying (1.1), weaker than (1.3), namely:

• (1.4)

for every $a\in H$, $b\in\operatorname{{\cal S}}(H)$, if $a^{2}b\in M$, then $a,ab\in M$.

In Theorem 4.3 we show that property (1.4) has natural equivalent forms with respect to various square-free factorizations.

In Theorem 5.1 we obtain full description of submonoids of a factorial monoid, satisfying (1.2), as factorial submonoids generated (up to irreducibles) by any set of pairwise relatively prime non-invertible square-free elements. We also obtain the answer to a question, when (1.1) and (1.2) are equivalent, expressing (1.2) as a conjunction of (1.1) and the property that any two non-associated atoms of $M$ are relatively prime in $H$. Moreover, we refer in Theorem 5.1 to various square-free factorizations, in particular equivalence between (1.2) and (1.3) holds without the assumption $\operatorname{\text{\sf q}}(M)\cap H=M$.

Section 6 is devoted to properties of radical elements. Reinhart in [21] introduced the notions of radical element and radical factoriality of a monoid. An element $a\in H$ is called radical if its principal ideal $aH$ is a radical ideal. A monoid $H$ is called radical factorial if every element is a product of radical elements. As we already observed in [16], Lemma 3.2 b), every radical element is square-free. So we have the following diagram of relations on elements of a monoid:

• (1.5)

\begin{array}[]{ccc}\text{prime}&\Rightarrow&\text{atom}\\ \Downarrow&&\Downarrow\\ \text{radical}&\Rightarrow&\text{square-free}\\ \end{array}

A radical element is an analog of a square-free one in the same way as a prime element is an analog of an atom. Moreover, a radical element is a generalization of a prime in the same way as a square-free element is a generalization of an atom.

How these analogies and generalizations work, we show in Section 6. In Propositions 6.5 – 6.7 we study the uniqueness of factorizations. In Proposition 6.4 we prove that in a decomposition monoid all square-free elements are radical. Recall that a monoid $H$ is called a decomposition monoid if every element $a\in H$ is primal, that is, for every $b,c\in H$ such that $a\mid bc$ there exist $a_{1},a_{2}\in H$ such that $a=a_{1}a_{2}$, $a_{1}\mid b$ and $a_{2}\mid c$. A domain $R$ is pre-Schreier if the multiplicative monoid $R\setminus\{0\}$ is a decomposition monoid. The notion of a pre-Schreier domain was introduced by Zafrullah in [23], see also [7] and the references given there.

In Sections 2 and 7 we discuss square-free factorizations in monoids in the context of the following properties: atomicity, ACCP, decomposition, GCD. We collect all relationships in Proposition 3.4. This is a generalization and extension of Proposition 1 from [17]. In Section 7 we consider possible classifications of monoids with respect to square-free factorizations and we state questions about existence of monoids. Some examples are presented in Section 8.

We refer to the following diagram of relations of monoids:

• (1.6)

\begin{array}[]{rcl}&&\text{BF}\;\;\Rightarrow\;\;\text{ACCP}\;\;\Rightarrow\;\;\text{atomic}\\ &\mbox{\begin{psfrags}\rotatebox{30.0}{\Rightarrow}\end{psfrags}}&\\ \text{factorial}&&\\ &\mbox{\begin{psfrags}\rotatebox{-30.0}{\Rightarrow}\end{psfrags}}&\\ &&\text{GCD}\;\;\Rightarrow\;\;\text{decomposition}\;\;\Rightarrow\;\;\text{atoms are primes}\end{array}

Remember that

• (1.7)

$\text{atomic}\;\wedge\;\text{atoms are primes}\;\;\Rightarrow\;\;\text{factorial}$

Finally, in Section 9 we concern a natural question about the possible number of square-free elements in a monoid.

2 Connections with the Jacobian conjecture

The Jacobian conjecture, stated by Keller ([18]) in 1939 is one of the most important open problems stimulating modern mathematical research (see [22]), with long lists of false proofs and equivalent formulations. For more information we refer the reader to van den Essen’s book [10].

Jacobian conjecture. Let $k$ be a field of characteristic [math]. For every polynomials $f_{1},\dots,f_{n}\in k[x_{1},\dots,x_{n}]$ with $n\geq 2$, if

• (2.1)

\left|\begin{array}[]{ccc}\dfrac{\partial f_{1}}{\partial x_{1}}&\cdots&\dfrac{\partial f_{1}}{\partial x_{n}}\\ \vdots&&\vdots\\ \dfrac{\partial f_{n}}{\partial x_{1}}&\cdots&\dfrac{\partial f_{n}}{\partial x_{n}}\end{array}\right|\in k\setminus\{0\},

then $k[f_{1},\dots,f_{n}]=k[x_{1},\dots,x_{n}]$.

Now, we will describe some topics of an approach to the conjecture in terms of irreducibility and square-freeness. For more details we refer the reader to our survey article [14].

Under the assumption that $f_{1},\dots,f_{n}$ are algebraically independent over $k$, the Jacobian condition (2.1) is equivalent to any of the following ones ([6], [13], [15]):

• (2.2)

every atom of $k[f_{1},\dots,f_{n}]$ is square-free in $k[x_{1},\dots,x_{n}]$,

• (2.3)

every square-free element of $k[f_{1},\dots,f_{n}]$ is square-free in $k[x_{1},\dots,x_{n}]$.

Under the same assumption, the assertion of the conjecture: $k[f_{1},\dots,f_{n}]=k[x_{1},\dots,x_{n}]$ is equivalent to the following one ([5], [1], [13]):

• (2.4)

every atom of $k[f_{1},\dots,f_{n}]$ is an atom of $k[x_{1},\dots,x_{n}]$.

Hence, in particular, the existence of a non-trivial example for (2.2), where by ”non-trivial” we mean ”not satisfying (2.4)”, is equivalent to the negation of the Jacobian conjecture.

Recall a generalization of the Jacobian conjecture formulated in [15].

Conjecture. Let $k$ be a field of characteristic [math]. For every polynomials $f_{1},\dots,f_{r}\in k[x_{1},\dots,x_{n}]$ with $n\geq 2$ and $r\in\{2,\dots,n\}$, if

• (2.5)

\gcd\big{(}\left|\begin{array}[]{ccc}\dfrac{\partial f_{1}}{\partial x_{j_{1}}}&\cdots&\dfrac{\partial f_{1}}{\partial x_{j_{r}}}\\ \vdots&&\vdots\\ \dfrac{\partial f_{r}}{\partial x_{j_{1}}}&\cdots&\dfrac{\partial f_{r}}{\partial x_{j_{r}}}\end{array}\right|,\;1\leq j_{1}<\ldots<j_{r}\leq n\big{)}\in k\setminus\{0\},

then $k[f_{1},\dots,f_{r}]$ is algebraically closed in $k[x_{1},\dots,x_{n}]$.

By Nowicki’s characterization ([20], Theorem 5.5, [19], Theorem 4.1.5, [8], 1.4) the assertion above is equivalent to: ”$R$ is a ring of constants for some $k$-derivation of $k[x_{1},\dots,x_{n}]$”.

Under the assumption that $f_{1},\dots,f_{r}$ are algebraically independent over $k$, the generalized Jacobian condition (2.5) is equivalent to any of the following ones ([15]):

• (2.6)

every atom of $k[f_{1},\dots,f_{r}]$ is square-free in $k[x_{1},\dots,x_{n}]$,

• (2.7)

every square-free element of $k[f_{1},\dots,f_{r}]$ is square-free in $k[x_{1},\dots,x_{n}]$.

3 Square-free factorizations in monoids

The aim of this section is to recall and extend some observations from [17]. The statements in that paper were formulated for rings, but the arguments are valid for monoids, since we were working only with the multiplicative structure of rings. In particular, Lemma 1 and Lemma 2 e) of [17] take the following form.

Lemma 3.1**.**

Let $H$ be a monoid. If $a\in\operatorname{{\cal S}}(H)$ and $a=b_{1}b_{2}\dots b_{n}$, then $b_{1}$, $b_{2}$, $\dots$, $b_{n}\in\operatorname{{\cal S}}(H)$ and $b_{i}\mathbin{\mbox{\small\rm rpr}}b_{j}$ for $i\neq j$.

Lemma 3.2**.**

Let $H$ be a decomposition monoid. If $a_{1},\dots,a_{n}\in\operatorname{{\cal S}}(H)$ and $a_{i}\mathbin{\mbox{\small\rm rpr}}a_{j}$ for all $i\neq j$, then $a_{1}\ldots a_{n}\in\operatorname{{\cal S}}(H)$.

As an immediate consequence we obtain.

Corollary 3.3**.**

If $H$ is a decomposition monoid and $a_{1},\dots,a_{n}\in\operatorname{{\cal A}}(H)$, $a_{i}\nsim a_{j}$ for $i\neq j$, then $a_{1}\ldots a_{n}\in\operatorname{{\cal S}}(H)$.

In [17], Proposition 1, we considered three types of square-free factorizations – (ii), (iii), (iv) in Proposition 3.4 below. In [17] we did not consider condition denoted (i) below as a separate one, as well as atomicity implying it. Moreover, we considered in [17], Proposition 1, only one type of square-free extraction – (vi) in Proposition 3.4 below. Here we add a second type of square-free extraction – (v) as easily following from (ii) for an arbitrary monoid. Finally, implications ${\rm(vi)}\Rightarrow{\rm(ii)}$ and ${\rm(vi)}\Rightarrow{\rm(iv)}$ in [17], Proposition 1 b) were formulated for GCD-domains, but the proofs were based only on [17], Lemma 2 e). This is why implications ${\rm(iii)}\Rightarrow{\rm(ii)}$ and ${\rm(iii)}\Rightarrow{\rm(iv)}$ below hold for arbitrary decomposition monoids.

Proposition 3.4**.**

Let $H$ be a monoid. Consider the following conditions:

(i)*  for every $a\in H$ there exist $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $a=s_{1}s_{2}\dots s_{n}$,*

(ii)*  for every $a\in H$ there exist $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{i}\mid s_{i+1}$ for $i=1,\dots,n-1$, and $a=s_{1}s_{2}\dots s_{n}$,*

(iii)*  for every $a\in H$ there exist $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{i}\mathbin{\mbox{\small\rm rpr}}s_{j}$ for $i\neq j$, and $a=s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}$,*

(iv)*  for every $a\in H$ there exist $n\geq 0$ and $s_{0},s_{1},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $a=s_{0}s_{1}^{2}s_{2}^{2^{2}}\ldots s_{n}^{2^{n}}$,*

(v)*  for every $a\in H$ there exist $b\in H$ and $c\in\operatorname{{\cal S}}(H)$ such that $a=bc$ and $a\mid c^{n}$ for some $n\geq 1$,*

(vi)*  for every $a\in H$ there exist $b\in H$ and $c\in\operatorname{{\cal S}}(H)$ such that $a=b^{2}c$.*

a)* The following implications hold:*

[TABLE]

b)* If $H$ is a decomposition monoid, then*

[TABLE]

c)* If $H$ is a GCD-monoid, then*

[TABLE]

Note that, according to (v), under the assumption $a=bc$ the condition ”$a\mid c^{n}$ for some $n\geq 1$” is equivalent to ”$b\mid c^{n}$ for some $n\geq 1$”.

Recall that every radical element is square-free ([16], Lemma 3.2 b), so radical factorial monoids studied by Reinhart in [21] satisfy condition (i).

Remark 3.5*.*

The statement that there are (in general) no other implications than the ones stated above is equivalent to the existence of the following counter-examples.

Non-factorial GCD-monoids satisfying:

[TABLE] 2. 2.

A decomposition non-GCD monoid satisfying $\text{\rm(iv)}\wedge\neg\text{\rm(v)}$. 3. 3.

Non-decomposition monoids satisfying:  $\text{\rm(ii)}\wedge\neg\text{\rm(vi)}$,  $\text{\rm(iii)}\wedge\neg\text{\rm(v)}$. 4. 4.

Non-factorial ACCP-monoids satisfying:  $\neg\text{\rm(iii)}$,  $\neg\text{\rm(v)}$. 5. 5.

An atomic non-ACCP monoid satisfying $\neg\text{\rm(vi)}$. 6. 6.

A non-atomic monoid satisfying (ii).

4 Sufficient conditions for ${\cal A}(M)\subset{\cal S}(H)$

In this section we study a factorial property (1.4) implying that all atoms of a submonoid are square-free in a monoid. We show that this property is, in general, not a necessary one. However, it is interesting by itself since it has natural equivalent forms with respect to several square-free factorizations, what we obtain in Theorem 4.3.

Proposition 4.1**.**

Let $H$ be a monoid satisfying condition (vi) of Proposition 3.4. Let $M$ be a submonoid of $H$ such that for every $a\in H$, $b\in\operatorname{{\cal S}}(H)$,

[TABLE]

Then $\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(H)$.

Proof.

Suppose that there exists some $c\in\operatorname{{\cal A}}(M)$ such that $c\notin\operatorname{{\cal S}}(H)$. Then $c=a^{2}b$ for some $a\in H$, $b\in\operatorname{{\cal S}}(H)$. Since $a^{2}b\in M$, then $a,ab\in M$. Note that $a\notin H^{\times}$, because $c\notin\operatorname{{\cal S}}(H)$, so $a,ab\notin M^{\times}$, a contradiction. ∎

The converse implication is not valid:

Example 4.2*.*

Consider a monoid $H={\mathbb{N}}^{3}$ and its submonoid $M=\langle(1,1,0),$ $(1,0,1)\rangle$. Then $\operatorname{{\cal A}}(M)=\{(1,1,0),(1,0,1)\}$, so $\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(H)$, but for $a=(1,0,0)\in H$, $b=(0,1,1)\in\operatorname{{\cal S}}(H)$ we have $2a+b\in M$ and $a,a+b\notin M$.

Observe that in the above example the monoid $M$ satisfies $\operatorname{\text{\sf q}}(M)\cap H=M$, and under this condition properties (1.3) and (1.4) are equivalent.

The most difficult part of Theorem’s 4.3 proof is the connection between ${\rm(i)}\Leftrightarrow{\rm(ii)}$ and ${\rm(iii)}\Leftrightarrow{\rm(iv)}\Leftrightarrow{\rm(v)}$, i.e. the equivalence of (ii) and (iii).

Theorem 4.3**.**

Let $H$ be a factorial monoid. Let $M\subset H$ be a submonoid such that $M^{\times}=H^{\times}$. The following conditions are equivalent:

(i)*  for every $a\in H$ and $b\in\operatorname{{\cal S}}(H)$,*

[TABLE]

(ii)*  for every $n\geq 0$ and $s_{0},s_{1},\dots,s_{n}\in\operatorname{{\cal S}}(H)$,*

[TABLE]

(iii)*  for every $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$,*

[TABLE]

(iv)*  for every $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{i}\mid s_{i+1}$ for $i=1,\ldots,n-1$,*

[TABLE]

(v)*  for every $a\in H$ and $b\in\operatorname{{\cal S}}(H)$ such that $a\mid b^{n}$ for some $n\geq 1$,*

[TABLE]

Proof.

$\text{\rm(i)}\Rightarrow\text{\rm(ii)}$

Assume (i). Consider elements $s_{0},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{0}s_{1}^{2}s_{2}^{2^{2}}\ldots s_{n}^{2^{n}}\in M$. Since \big{(}s_{1}s_{2}^{2}s_{3}^{2^{2}}\ldots s_{n}^{2^{n-1}}\big{)}^{2}s_{0}\in M, from (i) we obtain

[TABLE]

Then, since \big{(}s_{2}s_{3}^{2}s_{4}^{2^{2}}\ldots s_{n}^{2^{n-2}}\big{)}^{2}s_{1}\in M, from (i) we obtain

[TABLE]

Continuing, finally we receive:

[TABLE]

$\text{\rm(ii)}\Rightarrow\text{\rm(i)}$

Assume (ii). Consider $a\in H,b\in\operatorname{{\cal S}}(H)$ such that $a^{2}b\in M$. We can express $a$ in the form $a=s_{1}s_{2}^{2}s_{3}^{2^{2}}\ldots s_{n}^{2^{n-1}}$, where $s_{i}\in\operatorname{{\cal S}}(H)$ for $i=1,\ldots,n$. Put $s_{0}=b$. Thus we receive:

[TABLE]

Using the assumption we obtain:

[TABLE]

We see that $ab=s_{0}s_{1}s_{2}^{2}s_{3}^{2^{2}}\ldots s_{n}^{2^{n-1}}\in M$. Moreover:

[TABLE]

$\text{\rm(ii)}\Rightarrow\text{\rm(iii)}$

Assume (ii). We write ${\lceil x\rceil}$ and ${\lfloor x\rfloor}$ for respectively the ceiling and the floor of a real number $x$.

Step I. If $s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}\in M$, where $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$, then $s_{1}s_{2}s_{3}^{2}s_{4}^{2}\ldots s_{n}^{\lceil\frac{n}{2}\rceil},\;s_{2}s_{3}s_{4}^{2}s_{5}^{2}\ldots s_{n}^{\lfloor\frac{n}{2}\rfloor}\in M.$

Let $a=s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}\in M$, where $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$. Then the element $a$ can be presented in the form $a=t_{0}t_{1}^{2}t_{2}^{2^{2}}\ldots t_{r}^{2^{r}}$, where $t_{i}=s_{1}^{c^{(1)}_{i}}\ldots s_{n}^{c^{(n)}_{i}}\in\operatorname{{\cal S}}(H)$, $i=0,\ldots,r$ and $k=\sum_{i=0}^{r}c_{i}^{(k)}2^{i}$ with $c_{i}^{(k)}\in\{0,1\}$, $k=0,1,\ldots,n$ (see the proof of (vi)$\Rightarrow$(ii) in [17], Proposition 1). From (ii) we get $t_{i}t_{i+1}t_{i+2}^{2}t_{i+3}^{2^{2}}\ldots t_{r}^{2^{r-i-1}}\in M$, $i=0,\ldots,r-1$ and $t_{r}\in M$. In particular, $t_{0}t_{1}t_{2}^{2}\ldots t_{r}^{2^{r-1}}\in M$. Moreover:

[TABLE]

By the definition of $c^{(j)}_{i}$, we have

[TABLE]

[TABLE]

Step II. If $s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}\in M$, where $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$, then $s_{1}s_{2}s_{3}\ldots s_{n},\;s_{2}s_{3}^{2}s_{4}^{3}\ldots s_{n}^{n-1}\in M.$

Assume that $s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}\in M$, where $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$. We prove by induction on $l$ that

[TABLE]

Put $q={\lceil\frac{n}{2^{r}}\rceil}$. Then $(q-1)2^{r}<n\leq q2^{r}$. Put $s_{i}^{\prime}=s_{(i-1)2^{r}+1}s_{(i-1)2^{r}+2}\ldots s_{i2^{r}}$ for $i=1,\ldots,q-1$ and $s_{q}^{\prime}=s_{(q-1)2^{r}+1}s_{(q-1)2^{r}+2}\ldots s_{n}$. Note that $s_{1}^{\prime},s_{2}^{\prime},\ldots,s_{q}^{\prime}\in\operatorname{{\cal S}}(H)$ and $s_{i}^{\prime}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}^{\prime}$ for $i\neq j$, because $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$. We have $s_{1}^{\lceil\frac{1}{2^{l}}\rceil}s_{2}^{\lceil\frac{2}{2^{l}}\rceil}\ldots s_{n-1}^{\lceil\frac{n-1}{2^{l}}\rceil}s_{n}^{\lceil\frac{n}{2^{l}}\rceil}=s_{1}^{\prime}(s_{2}^{\prime})^{2}\ldots(s_{q}^{\prime})^{q}.$ If $s_{1}^{\lceil\frac{1}{2^{l}}\rceil}s_{2}^{\lceil\frac{2}{2^{l}}\rceil}\ldots s_{n-1}^{\lceil\frac{n-1}{2^{l}}\rceil}s_{n}^{\lceil\frac{n}{2^{l}}\rceil}\in M$, then by step I we obtain that

[TABLE]

and

[TABLE]

If moreover $s_{1}^{1-\lceil\frac{1}{2^{l}}\rceil}s_{2}^{2-\lceil\frac{2}{2^{l}}\rceil}\ldots s_{n}^{n-\lceil\frac{n}{2^{l}}\rceil}\in M$, then also

[TABLE]

[TABLE]

There exists $r\in\mathbb{N}$ such that $2^{r}>n$. Then for every $1\leq t\leq n$ we have $\lceil\frac{t}{2^{r}}\rceil=1$. Consequently, $s_{1}s_{2}s_{3}\ldots s_{n},\;s_{2}s_{3}^{2}s_{4}^{3}\ldots s_{n}^{n-1}\in M$.

Step III. We prove (iii) by induction on $n$. For $n=1$ it is clear. Assume the assertion for $n$ and consider $s_{1},s_{2},\dots,s_{n},s_{n+1}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$, such that $s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}s_{n+1}^{n+1}\in M$. By step II we have

[TABLE]

Then by the inductive assumption we have

[TABLE]

$\text{\rm(iii)}\Rightarrow\text{\rm(ii)}$

Assume (iii). We prove (ii) by induction on $n$. For $n=0$ it is clear.

We assume the assertion for $n$, that is, if $s_{0},s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(M)$, then $s_{0}s_{1}^{2}s_{2}^{2^{2}}\ldots s_{n}^{2^{n}}\in M$ implies $s_{n-l}\prod\limits_{j=0}^{l-1}s_{n-l+j+1}^{2^{j}}\in M$ for every $l\in\{0,1,\ldots,n\}$.

We prove the assertion for $n+1$. Let $a=s_{0}s_{1}^{2}s_{2}^{2^{2}}\ldots s_{n+1}^{2^{n+1}}\in M$, where $s_{0},s_{1},\ldots,s_{n+1}\in\operatorname{{\cal S}}(H)$. Then the element $a$ can be presented in the form $a=t_{1}t_{2}^{2}t_{3}^{3}\ldots t_{m}^{m}$, where $m=2^{n+2}-1$ and $t_{1},\ldots,t_{m}\in\operatorname{{\cal S}}(H)$, $t_{i}\mathbin{\mbox{\small\rm rpr}}_{H}t_{j}$, for $i\neq j$ (for details see the proof of ${\rm(ii)}\Rightarrow{\rm(vi)}$ in [17], Proposition 1b). From (iii) we have $t_{m},t_{m-1}t_{m},\ldots,t_{1}t_{2}\ldots t_{m}\in M$. Note that $m$ is odd. Multiplying the elements of the form $t_{r}t_{r+1}\ldots t_{m}$ for all odd $r$ we obtain $t_{1}t_{2}t_{3}^{2}t_{4}^{2}\ldots t_{m}^{\lceil\frac{m}{2}\rceil}\in M$. Multiplying the elements of that form for all even $r$ we obtain $t_{2}t_{3}t_{4}^{2}t_{5}^{2}\ldots t_{m}^{\lfloor\frac{m}{2}\rfloor}\in M$. Since $t_{2}t_{3}t_{4}^{2}t_{5}^{2}\ldots t_{m}^{\lfloor\frac{m}{2}\rfloor}=s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n+1}^{2^{n}}$, by the inductive assumption we have $s_{n+1-l}\prod\limits_{j=0}^{l-1}s_{(n+1)-l+j+1}^{2^{j}}\in M$ for $l\in\{0,1,\ldots,n\}$. Moreover, $t_{1}t_{2}t_{3}^{2}t_{4}^{2}\ldots t_{m}^{\lceil\frac{m}{2}\rceil}=s_{0}s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n+1}^{2^{n}}$, which gives the assertion for $l=n+1$.

$\text{\rm(iii)}\Leftrightarrow\text{\rm(iv)}$ follows from the equivalence of presentations (ii) and (iii) in Proposition 3.4 (for details see [17], the proofs of ${\rm(iv)}\Rightarrow{\rm(vi)}$ in Proposition 1a and ${\rm(vi)}\Rightarrow{\rm(iv)}$ in Proposition 1b).

$\text{\rm(iv)}\Rightarrow\text{\rm(v)}$

Assume (iv). Consider $a\in H$, $b\in\operatorname{{\cal S}}(H)$ such that $a\mid b^{n}$ for some $n\geq 1$, and $ab\in M$. Let $a=s_{1}s_{2}\ldots s_{m}$, where $s_{1},\ldots,s_{m}\in\operatorname{{\cal S}}(H)$, $s_{i}\mid s_{i+1}$ for $i=1,\ldots,m-1$. Then $s_{m}\mid b^{n}$, hence $s_{m}\mid b$, because $s_{m}\in\operatorname{{\cal S}}(H)$. We have $s_{1}s_{2}\ldots s_{m}b=ab\in M$. By (iv) we obtain $s_{1},s_{2},\ldots,s_{m},b\in M$, so $a,b\in M$.

$\text{\rm(v)}\Rightarrow\text{\rm(iv)}$

Assume (v). Let $s_{1}s_{2}\ldots s_{n}\in M$, where $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mid s_{i+1}$ for $i=1,\ldots,n-1$. Put $a=s_{1}s_{2}\ldots s_{n-1}$, $b=s_{n}$. Then $a\mid b^{n-1}$. By (v) we have $s_{1}s_{2}\ldots s_{n-1}\in M$ and $s_{n}\in M$, and the assertion follows by induction. ∎

5 Necessary and sufficient conditions for

${\cal S}(M)\subset{\cal S}(H)$

In this section we obtain a full characterization of submonoids of a factorial monoid for which all square-free elements of a submonoid are square-free in a monoid.

Let us note that the formulation and the proof of Proposition 4.1 from [16] involve only the multiplicative structure of a domain. Thus we have the equivalence of the conditions (vi) – (viii) of the following Theorem 5.1. For the same reason implication $\text{\rm(viii)}\Rightarrow\text{\rm(i)}$ of Theorem 5.1 follows from the proof of implication $\text{\rm(ii)}\Rightarrow\text{\rm(i)}$ of Theorem 3.4 from [15].

Theorem 5.1**.**

Let $H$ be a factorial monoid. Let $M\subset H$ be a submonoid such that $M^{\times}=H^{\times}$. The following conditions are equivalent:

(i)*  $\operatorname{{\cal S}}(M)\subset\operatorname{{\cal S}}(H)$,*

(ii)*  $\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(H)$ and, for every $a,b\in M$,*

[TABLE]

(iii)*  $\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(H)$ and, for every $a,b\in\operatorname{{\cal A}}(M)$,*

[TABLE]

(iv)*  $M=H^{\times}\times\operatorname{\text{\sf F}}(B)$, where $B$ is any set of pairwise relatively prime (in $H$) non-invertible square-free elements of $H$,*

(v)*  for every $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$,*

[TABLE]

(vi)*  for every $n\geq 1$, $k_{1},\dots,k_{n}\geq 0$ and $q_{1},\dots,q_{n}\in\operatorname{{\cal A}}(H)$ such that $q_{i}\not\sim_{H}q_{j}$ for $i\neq j$,*

[TABLE]

where $k_{j}=c^{(j)}_{r}2^{r}+\ldots+c^{(j)}_{0}2^{0}$ for $j=1,\dots,n$, with $c^{(j)}_{i}\in\{0,1\}$ for $i=0,\dots,r$.

(vii)*  for every $n\geq 0$ and $s_{0},\dots,s_{n}\in\operatorname{{\cal S}}(H)$,*

[TABLE]

(viii)*  for every $a\in H$ and $b\in\operatorname{{\cal S}}(H)$,*

[TABLE]

Proof.

First, observe that $H$ is a BF-monoid and the submonoid $M$ satisfies $M^{\times}=H^{\times}\cap M$, so $M$ is also a BF-monoid, by [12], Corollary 1.3.3, p. 17. In particular, $M$ is atomic.

$\text{\rm(i)}\Rightarrow\text{\rm(iii)}$  Assume $\operatorname{{\cal S}}(M)\subset\operatorname{{\cal S}}(H)$. Since $\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(M)$, we have $\operatorname{{\cal A}}(M)\subset\operatorname{{\cal S}}(H)$.

Suppose that there exist $a,b\in\operatorname{{\cal A}}(M)$ such that $a\nsim_{M}b$ and $a$, $b$ are not relatively prime in $H$. Then $t=\gcd_{H}(a,b)\in H\setminus H^{\times}$, so $a=tu$, $b=tv$ for some $u,v\in H$, $u\mathbin{\mbox{\small\rm rpr}}_{H}v$. Since $a,b\in\operatorname{{\cal A}}(M)$, we have $a,b\in\operatorname{{\cal S}}(H)$, but $u\mid_{H}a$, $v\mid_{H}b$, so $u,v\in\operatorname{{\cal S}}(H)$, and then $uv\in\operatorname{{\cal S}}(H)$, because $u\mathbin{\mbox{\small\rm rpr}}_{H}v$.

Now, we have $ab=t^{2}uv\not\in\operatorname{{\cal S}}(H)$, so $ab\not\in\operatorname{{\cal S}}(M)$, that is, $ab=c^{2}d$ for some $c\in M\setminus M^{\times}$, $d\in M$. We may assume that $c\in M\setminus M^{\times}$ is minimal (with respect to natural length function in $H$) satisfying the following property: ”there exist $a,b,d\in H$ such that $c\mid_{H}a,b$ and $ab=c^{2}d$”. We have $c^{2}d=t^{2}uv$, where $uv\in\operatorname{{\cal S}}(H)$, so $c\mid_{H}t$, because $H$ is factorial, and then $t=cw$ for some $w\in H$.

We obtain $a=tu=cwu$, so $uv\in\operatorname{{\cal S}}(H)$, since $a\in\operatorname{{\cal S}}(H)$. We have $ac=c^{2}wu\not\in\operatorname{{\cal S}}(H)$, so $ac\not\in\operatorname{{\cal S}}(M)$, hence $ac=e^{2}h$ for some $e\in M\setminus M^{\times}$, $h\in M$. Since $e^{2}h=c^{2}wu$, where $wu\in\operatorname{{\cal S}}(H)$, we infer $e\mid_{H}c$, and then also $e\mid_{H}a$. We have obtained $e\mid_{H}a,c$ and $ac=e^{2}h$, so $e\sim_{H}c$ by the minimality of $c$. Then $e\sim_{M}c$, because $M^{\times}=H^{\times}$. But $ac=e^{2}h$, so $a\sim_{M}eh\sim_{M}ch$. Then $a\sim_{M}c$, since $a\in\operatorname{{\cal A}}(M)$ and $c\in M\setminus M^{\times}$.

Analogously we show that $b\sim_{M}c$, so $a\sim_{M}b$, a contradiction.

$\text{\rm(ii)}\Rightarrow\text{\rm(iii)}$  It is enough to note that for every $a,b\in\operatorname{{\cal A}}(M)$,

[TABLE]

Namely, if $a,b\in\operatorname{{\cal A}}(M)$ are not relatively prime in $M$, then $a=cd$ and $b=ce$ for some $c\in M\setminus M^{\times}$, $d,e\in M$, so $d,e\in M^{\times}$ and $a\sim_{M}b$.

$\text{\rm(iii)}\Rightarrow\text{\rm(ii)}$  Assume (iii) and consider elements $a,b\in M$ such that $a\mathbin{\mbox{\small\rm rpr}}_{M}b$. We already know that $M$ is atomic. Let $a=a_{1}\dots a_{m}$ and $b=b_{1}\dots b_{n}$ be factorizations into atoms in $M$. Since $a\mathbin{\mbox{\small\rm rpr}}_{M}b$, for all $i,j$ we have $a_{i}\nsim_{M}b_{j}$, so $a_{i}\mathbin{\mbox{\small\rm rpr}}_{H}b_{j}$, but then $a\mathbin{\mbox{\small\rm rpr}}_{H}b$.

$\text{\rm(iii)}\Rightarrow\text{\rm(iv)}$  Assume (iii). Let $B$ be a maximal (with respect to inclusion) set of pairwise non-associated (in $M$) atoms of $M$. By (iii) the elements of $B$ are pairwise relatively prime in $H$. $H$ is a factorial monoid, so $B$ generates a free submonoid. Since $M$ is atomic and $M^{\times}=H^{\times}$, we obtain $M=H^{\times}\times\operatorname{\text{\sf F}}(B)$.

$\text{\rm(iv)}\Rightarrow\text{\rm(v)}$  Assume (iv). Let $a=s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}\in M$, where $s_{1},\ldots,s_{n}\in\operatorname{{\cal S}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$. By (iv), the element $a$ can be presented in the form $a=ct_{1}t_{2}^{2}t_{3}^{3}\ldots t_{m}^{m}$ with $c\in H^{\times}$, $t_{i}=\prod_{j=1}^{r_{i}}b_{j}^{(i)}\in M$, $r_{i}\geq 0$, $m\geq n$, and pairwise different all $b_{j}^{(i)}\in B$. Since $b_{j}^{(i)}$ are square-free and pairwise relatively prime in $H$, then $t_{1},\ldots,t_{m}$ are also square-free and pairwise relatively prime in $H$. Finally, for $i=1,\dots,n$ we have $s_{i}\sim_{H}t_{i}$, so $s_{i}\in M$.

$\text{\rm(v)}\Rightarrow\text{\rm(vi)}$  Assume (v). Let $a=q_{1}^{k_{1}}\dots q_{n}^{k_{n}}\in M$, where $q_{1},\dots,q_{n}\in\operatorname{{\cal A}}(H)$, $q_{i}\not\sim_{H}q_{j}$ for $i\neq j$, and $k_{1},\dots,k_{n}\geq 0$. Put $m=\max(k_{1},\ldots,k_{n})$. For $l=1,\dots,m$ denote $s_{l}=\prod_{j\colon k_{j}=l}q_{j}$. Then $s_{1},s_{2},\dots,s_{m}\in\operatorname{{\cal S}}(H)$ and $s_{i}\mathbin{\mbox{\small\rm rpr}}_{H}s_{j}$ for $i\neq j$. We have $a=s_{1}s_{2}^{2}\ldots s_{m}^{m}$, so $s_{1},s_{2},\dots,s_{m}\in M$, by (v).

Now, let $k_{j}=c^{(j)}_{r}2^{r}+\ldots+c^{(j)}_{0}2^{0}$ for $j=1,\dots,n$, with $c^{(j)}_{i}\in\{0,1\}$ for $i=0,\dots,r$. Note that if $k_{j_{1}}=k_{j_{2}}$, then $c^{(j_{1})}_{i}=c^{(j_{2})}_{i}$ for each $i$, so we may denote $d^{(l)}_{i}=c^{(j)}_{i}$ for each $j$ such that $k_{j}=l$, where $l=1,\dots,m$. Then $q_{1}^{c^{(1)}_{i}}\dots q_{n}^{c^{(n)}_{i}}=s_{1}^{d^{(1)}_{i}}\dots s_{m}^{d^{(m)}_{i}}\in M$. ∎

The only type of factorizations from Proposition 3.4 we haven’t considered in Theorem 4.3 nor Theorem 5.1 is (i). There is no surprise that in this case we obtain a divisor-closed submonoid.

Proposition 5.2**.**

Let $H$ be a monoid such that each element $a\in H$ can be presented in the form $a=s_{1}s_{2}\ldots s_{n}$, where $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$. Let $M\subset H$ be a submonoid. The following conditions are equivalent:

(i)*  for every $a,b\in H$,*

[TABLE]

(ii)*  for every $n\geq 1$ and $s_{1},s_{2},\dots,s_{n}\in\operatorname{{\cal S}}(H)$,*

[TABLE]

6 Radical elements and the uniqueness of factorizations

Let $H$ be a monoid. Recall from [21] that an element $a\in H$ is called radical if the principal ideal $aH$ is radical, equivalently, if for arbitrary $b\in H$ and $n\geq 1$,

[TABLE]

Denote by $\operatorname{{\cal R}}(H)$ the set of radical elements of $H$, and by $\operatorname{{\cal P}}(H)$ the set of prime elements.

Clearly, every prime element is radical:

[TABLE]

This is an analog of the fact that every atom is square-free.

Note also that every radical element is square-free, see [16], Lemma 3.2 b), what is an analog of the fact that a prime element is an atom.

Proposition 6.1**.**

Let $H$ be a monoid. Then

[TABLE]

The next lemma completes Lemma 3.1.

Lemma 6.2**.**

Let $H$ be a monoid and let $a\in\operatorname{{\cal R}}(H)$ and $b\in H$. If $b\mid a$, then $b\in\operatorname{{\cal R}}(H)$.

Proof.

Let $a\in\operatorname{{\cal R}}(H)$ and $b\mid a$. Let $c\in H$ and $b\mid c^{n}$ for some $n\geq 1$. By assumption we have $a=bd$, where $d\in H$. Then $a\mid c^{n}d^{n}$ and this implies $a\mid cd$, so $b\mid c$. ∎

In Lemma 6.3 a), b) below we recall Lemma 2 a), d) from [17] in terms of monoids.

Lemma 6.3**.**

Let $H$ be a decomposition monoid.

a)* Let $a,b,c\in H$. If $a\mid bc$ and $a\mathbin{\mbox{\small\rm rpr}}b$, then $a\mid c$.*

b)* Let $a_{1},\dots,a_{n},b\in H$. If $a_{i}\mathbin{\mbox{\small\rm rpr}}b$ for $i=1,\dots,n$, then $a_{1}\ldots a_{n}\mathbin{\mbox{\small\rm rpr}}b$.*

c)* Let $a,b_{1},\dots,b_{n}\in H$. If $a\mid b_{1}\ldots b_{n}$, then there exist $a_{1},\dots,a_{n}\in H$ such that $a=a_{1}\ldots a_{n}$ and $a_{i}\mid b_{i}$ for $i=1,\dots,n$.*

d)* Let $a_{1},\dots,a_{n}\in\operatorname{{\cal S}}(H)$, $b\in H$. If $a_{i}\mathbin{\mbox{\small\rm rpr}}a_{j}$ for $i\neq j$ and $a_{i}\mid b$ for $i=1,\dots,n$, then $a_{1}\ldots a_{n}\mid b$.*

Proof.

c) Simple induction.

d) Induction. Assume the assertion for $n$. Consider $a_{1},\dots,a_{n},a_{n+1}\in\operatorname{{\cal S}}(H)$, $a_{i}\mathbin{\mbox{\small\rm rpr}}a_{j}$ for $i\neq j$, and $b\in H$ such that $a_{i}\mid b$ for $i=1,\dots,n+1$. Put $a=a_{1}\ldots a_{n}$. Then, by the induction hypothesis, $a\mid b$, so $b=ac$ for some $c\in H$. Moreover, $a\mathbin{\mbox{\small\rm rpr}}a_{n+1}$ by b). Since $a_{n+1}\mid ac$, by a) we obtain $a_{n+1}\mid c$, and than $aa_{n+1}\mid ac$. ∎

Now we can prove that in a decomposition monoid every square-free element is radical. This is an analog of the fact that in a decomposition monoid atoms are primes.

Proposition 6.4**.**

Let $H$ be a decomposition monoid. Then

[TABLE]

Proof.

Let $a\in\operatorname{{\cal S}}(H)$. Assume that $a\mid b^{n}$ for some $b\in H$ and $n\geq 1$. Then, by Lemma 6.3 c), there exist $a_{1},\dots,a_{n}\in H$ such that $a=a_{1}\ldots a_{n}$ and $a_{i}\mid b$ for $i=1,\dots,n$. Observe that $a_{1},\dots,a_{n}\in\operatorname{{\cal S}}(H)$ and $a_{i}\mathbin{\mbox{\small\rm rpr}}a_{j}$ for $i\neq j$, by Lemma 3.1, so $a_{1}\ldots a_{n}\mid b$ by Lemma 6.3 d). ∎

In the rest of this section we concern uniqueness properties of factorizations (ii) – (iv) and extractions (v), (vi) from Proposition 3.4. In an arbitrary monoid we have the uniqueness of factorization (ii) and extraction (v) for radical elements.

Proposition 6.5**.**

Let $H$ be a monoid.

a)* For every $r_{1},\dots,r_{n},t_{1},\dots,t_{n}\in\operatorname{{\cal R}}(H)$ such that $r_{i}\mid r_{i+1}$ and $t_{i}\mid t_{i+1}$, $i=1,\dots,n-1$, if*

[TABLE]

then $r_{i}\sim t_{i}$ for $i=1,\dots,n$.

b)* For every $a,c\in H$, $b,d\in\operatorname{{\cal R}}(H)$ such that $a\mid b^{m}$ and $c\mid d^{n}$ for some $m,n\geq 1$, if*

[TABLE]

then $a\sim c$ and $b\sim d$.

Proof.

a)  Assume that $r_{1}r_{2}\dots r_{n}\sim t_{1}t_{2}\dots t_{n},$ where $r_{1},\dots,r_{n},t_{1},\dots,t_{n}\in\operatorname{{\cal R}}(H)$, $r_{i}\mid r_{i+1}$ and $t_{i}\mid t_{i+1}$ for $i=1,\dots,n-1$. We have $r_{n}\mid t_{1}\dots t_{n}$, so $r_{n}\mid t_{n}^{n}$. Since $r_{n}\in\operatorname{{\cal R}}(H)$ we obtain $r_{n}\mid t_{n}$. Analogously, we get $t_{n}\mid r_{n}$. Hence $r_{n}\sim t_{n}$ and $r_{1}\dots r_{n-1}\sim t_{1}\dots t_{n-1}$. Then we repeat the above reasoning for $r_{n-1}$ and $t_{n-1}$, etc.

b)  Assume that $ab\sim cd$, where $a,c\in H$, $b,d\in\operatorname{{\cal R}}(H)$, $a\mid b^{m}$ and $c\mid d^{n}$ for some $m,n\geq 1$. We see that $b\mid cd$, so $b\mid d^{n+1}$. Since $b\in\operatorname{{\cal R}}(H)$ we obtain $b\mid d$. Analogously, we get $d\mid b$, so $b\sim d$, and then $a\sim c$. ∎

In a decomposition monoid we have the uniqueness of factorization (iii) from Proposition 3.4.

Proposition 6.6**.**

Let $H$ be a decomposition monoid. For every $s_{1},\dots,s_{n}$, $t_{1}$, $\dots$, $t_{n}\in\operatorname{{\cal S}}(H)$ such that $s_{i}\mathbin{\mbox{\small\rm rpr}}s_{j}$ and $t_{i}\mathbin{\mbox{\small\rm rpr}}t_{j}$ for $i\neq j$, if

[TABLE]

then $s_{i}\sim t_{i}$ for $i=1,\dots,n$.

Proof.

Assume that $s_{1}s_{2}^{2}s_{3}^{3}\ldots s_{n}^{n}\sim t_{1}t_{2}^{2}t_{3}^{3}\ldots t_{n}^{n}$, where $s_{1},\dots,s_{n},t_{1},\dots,t_{n}\in\operatorname{{\cal R}}(H)$, $s_{i}\mathbin{\mbox{\small\rm rpr}}s_{j}$ and $t_{i}\mathbin{\mbox{\small\rm rpr}}t_{j}$ for $i\neq j$. Put $s^{\prime}_{i}=s_{i}\ldots s_{n}$, $t^{\prime}_{i}=t_{i}\ldots t_{n}$ for $i=1,\ldots,n$. Then

[TABLE]

Note that $s^{\prime}_{i},t^{\prime}_{i}\in\operatorname{{\cal S}}(H)$ for $i=1,\ldots,n$ by Lemma 3.2. Since $s^{\prime}_{i+1}\mid s^{\prime}_{i}$ and $t^{\prime}_{i+1}\mid t^{\prime}_{i}$ for $i=1,\ldots,n-1$, from Proposition 6.5 a) we obtain $s^{\prime}_{i}\sim t^{\prime}_{i}$ for $i=1,\ldots,n$. Then $s_{i}\sim t_{i}$ for $i=1,\dots,n$. ∎

Finally, recall from [17], Proposition 2 (i), (ii), the uniqueness of factorization (iv) and extraction (vi) for a GCD-monoid. It was formulated for a GCD-domain, but the proof is valid for a GCD-monoid.

Proposition 6.7**.**

Let $H$ be a GCD-monoid.

a)* For every $s_{0},s_{1},\dots,s_{n},t_{0},t_{1},\dots,t_{n}\in\operatorname{{\cal S}}(H)$, if*

[TABLE]

then $s_{i}\sim t_{i}$ for $i=0,\dots,n$.

b)* For every $a,c\in H$, $b,d\in\operatorname{{\cal S}}(H)$, if*

[TABLE]

then $a\sim c$ and $b\sim d$.

7 Classifications of monoids with respect to

square-free factorizations

In this section we show how to organize all the variety of cases when properties considered in Proposition 3.4 hold or do not. We would like to emphasize two advantages of this situations. First: it yields mostly non-trivial questions about existence of 7, 19, 24, or even 60 monoids, respectively. Second: it provides many ways of classifying monoids with respect to possesing or not different square-free factorizations or extractions, which may be more subtle than with respect to irreducible factorizations.

There are $7$ possible combinations of logical values for properties (i) – (iv).

[TABLE]

We would like to involve the following properties of monoids: ACCP, atomicity, GCD, decomposition. We introduce the value of ”ACCP/atm” as follows.

[TABLE]

Similarly, we introduce the value of ”GCD/decomp”.

[TABLE]

Now, we can collect all possibilities for conditions (i) – (vi) in Proposition 3.4, taking into account the properties mentioned above. By $1^{\ast}$ below we denote that $1$ as the value of ”ACCP/atm” is possible only when the value of ”GCD/decomp” is [math], and also $1$ as the value of ”GCD/decomp” is possible only when the value of ”ACCP/atm” is [math]. In the leftmost column we indicate the number of cases for ”ACCP/atm” and ”GCD/decomp” with respect to given values of (i) – (iv). In the rightmost column we indicate the number of cases for extractions (v) and (vi) also with respect to (i) – (iv).

[TABLE]

Let us extract possible combinations of (i) – (iv) for: atomic, ACCP, decomposition and GCD-monoids. We have:

• •

6 possible combinations for atomic monoids,

[TABLE]

• •

3 possible combinations for ACCP-monoids,

[TABLE]

• •

4 possible combinations for decomposition monoids,

[TABLE]

• •

3 possible combinations for GCD-monoids.

[TABLE]

There are 24 classes of monoids with respect to properties:

ACCP, atomicity, GCD, decomposition, (i) – (iv).

The question if all of them are non-empty is, in our opinion, of fundamental importance.

Extraction (vi) is a basic tool for exploring properties of subrings connected with square-free elements. This is why we think it is reasonable to consider whole set of properties (i) – (vi). There arises a question if all combinations of logical values are possible, i.e., a question about 19 examples.

There are 60 classes of monoids with respect to all properties:

ACCP, atomicity, GCD, decomposition, (i) – (vi).

We don’t think that all of them are non-empty. It may be true, e.g., that for ACCP-monoids there is ${\rm(ii)}\Leftrightarrow{\rm(v)}$. Hence, we state a question about 60 examples of monoids.

8 Some examples

Example 8.1*.*

Put

[TABLE]

where $p$, $q$ are positive integers.

Then $B_{p,q}$ is a non-factorial GCD-monoid for any $p$, $q$.

a) $B_{1,1}$ satisfies all conditions (i) – (vi), in particular, it is a non-atomic monoid satisfying (ii), mentioned in Remark 3.5.6.

b) if $q$ is even, then $B_{p,q}$ satisfies (vi) and no one of (i) – (v), in particular, it is a non-factorial GCD-monoid satisfying $\text{\rm(vi)}\wedge\neg\text{\rm(i)}$ as well as $\text{\rm(vi)}\wedge\neg\text{\rm(v)}$, mentioned in Remark 3.5.1.

c) if $q$ is odd and $(p,q)\neq(1,1)$, then $B_{p,q}$ satisfies no one of the conditions (i) – (vi).

Monoid $B_{1,1}$ gives an important argument in the discussion of how property (i) extends atomicity in the context of diagram (1.6):

[TABLE]

Namely, we loose connection with the lower line of the diagram since $B_{1,1}$ satisfies the strongest one – GCD – and is not factorial, so in general the conjunction of (i) and GCD does not imply factoriality.

Example 8.2*.*

Let ${\mathbb{Q}}_{\geq 0}$ denote the set of all non-negative rational numbers. $H=({\mathbb{Q}}_{\geq 0},+)$ is a GCD-monoid, because $\gcd(a,b)=\min\{a,b\}$ for all $a,b\in H$. It satisfies condition (vi), because for any $a\in H$ we have $a=\frac{a}{2}+\frac{a}{2}+0$ and $0\in\operatorname{{\cal S}}(H)$. However it do not satisfy any of conditions (i) – (iv), because $\operatorname{{\cal S}}(H)=\{0\}$ and $0+\ldots+0\neq a$ for $a\neq 0$. Neither condition (v), because if $c\in\operatorname{{\cal S}}(H)$, then $c=0$ and $0+\ldots+0$ is not divisible in $({\mathbb{Q}}_{\geq 0},+)$ by a non-zero $a$ (here $a\mid b$ iff $a\leq b$). Clearly $H$ is also non-factorial.

Example 8.3*.*

For a non-negative integer $k$ we denote by ${\mathbb{N}}_{\geq k}$ the set of integers greater or equal to $k$. Then $H=({\mathbb{N}}_{\geq 2}\cup\{0\},+)$ is not a decomposition monoid, since $\operatorname{{\cal A}}(H)=\{2,3\}$ and $\operatorname{{\cal P}}(H)=\emptyset$. See also section 8.

Example 8.4*.*

Let $L$ and $F$ be fields such that $L\subset F$. Consider $T=L+xF[x]$. Then the atoms of the ring $T$ are known:

Theorem 8.5**.**

([3], Theorems 2.9 and 5.3)*.

$T$ is half-factorial domain and $\operatorname{{\cal A}}(T)=\{ax,a\in F\}\cup\{a(1+xf(x)),a\in L,f\in F[x],1+xf(x)\in\operatorname{{\cal A}}(F[x])\}.$*

We can also determine all the square-free elements of $T$:

Proposition 8.6**.**

Let $T=L+xF[x]$, and $f\in F[x]$. Then $f\in\operatorname{{\cal S}}(T)$ iff $f\in\operatorname{{\cal S}}(F[x])\wedge f(0)\in L$.

Proof.

Suppose that $f\notin\operatorname{{\cal S}}(F[x])$ or $f(0)\notin L$. If $f(0)\notin L$, then $f\notin T$, so $f\notin\operatorname{{\cal S}}(T)$. Now, assume that $f\notin\operatorname{{\cal S}}(F[x])$. Then $f=g^{2}h$, where $g\in F[x]\setminus F,h\in F[x]$. Let $g=a_{n}x^{n}+\ldots+a_{1}x+a_{0},h=b_{m}x^{m}+\ldots+b_{1}x+b_{0}$. We have $f=(a_{n}x^{n}+\ldots+a_{1}x+a_{0})^{2}(b_{m}x^{m}+\ldots+b_{1}x+b_{0})$. Then f=\Big{(}\dfrac{a_{n}}{a_{0}}x^{n}+\ldots+\dfrac{a_{1}}{a_{0}}x+1\Big{)}^{2}(b_{m}a_{0}^{2}x^{m}+\ldots+b_{1}a_{0}^{2}x+b_{0}a_{0}^{2}), where $b_{0}a_{0}^{2}=f(0)\in L$.

Now, suppose that $f\notin\operatorname{{\cal S}}(T)$. If $f\notin T$, then $f(0)\notin L$. Now, assume that $f\in T$. Then we have $f=g^{2}h$, where $g\in T\setminus L,h\in T$. This implies $g\in F[x]\setminus F,h\in F[x]$. ∎

In particular, if $T=\mathbb{R}+x\mathbb{C}[x]$, then

$\operatorname{{\cal A}}(T)=\{ax,a\in\mathbb{C}\}\cup\{a(1+bx),a\in\mathbb{R},b\in\mathbb{C}\setminus\{0\}\}$

$\operatorname{{\cal S}}(T)=\{1\}\cup\{ax,a\in\mathbb{C}\}\cup\{a(1+bx),a\in\mathbb{R},b\in\mathbb{C}\setminus\{0\}\}\cup\{x(a+bx),a\in\mathbb{R},b\in\mathbb{C}\setminus\{0\}\}$.

Using Proposition 8.6 we easily verify that $L+xF[x]$ fulfills (i) – (vi).

If $F$ and $L$ are finite fields and it is a proper extension, then $L+xF[x]$ is a non-factorial ACCP domain (see [2], [9]).

9 The number of square-free elements of a reduced monoid

It is obvious that an arbitrary non-negative integer can be the number of atoms of a monoid. For example it can be the number of its free generators. In a group every element is square-free, since there is no non-invertible element. Hence, any positive integer can be the number of square-free elements of a monoid. It is not such obvious, but still true, that an arbitrary positive integer can be the number of square-free elements of a reduced monoid. It also remains valid if we assume that this reduced monoid is cancellative.

For integers $a,b$ such that $a\geq b$, we define $[a,b]=\{c\in{\mathbb{Z}};a\leq c\leq b\}$, that is, the set of all consecutive integers from $a$ to $b$.

Theorem 9.1**.**

Let $n$ be a positive integer. Then there exists a reduced cancellative monoid $H$ such that $\#\operatorname{{\cal S}}(H)=n$.

Proof.

Let $m$ be an integer $\geq 2$. Consider a monoid

[TABLE]

with the operation of addition.

Clearly $\operatorname{{\cal A}}(H)=\{m\}\cup[2m+1,3m-1]$ and $\#\operatorname{{\cal A}}(H)=m$. Then $\operatorname{{\cal S}}(H)=\{0,m\}\cup[2m+1,3m-1]\cup[3m+1,4m-1]$ and consequently $\#\operatorname{{\cal S}}(H)=2m$.

Now let $m$ be an integer $\geq 3$ and consider a monoid

[TABLE]

In this case $\operatorname{{\cal A}}(H)=\{m,2m-1\}\cup[2m+1,3m-2]$ and $\#\operatorname{{\cal A}}(H)=m$. Then $\operatorname{{\cal S}}(H)=\{0,m,2m-1\}\cup[2m+1,3m-1]\cup[3m+1,4m-3]$ and finally $\#\operatorname{{\cal S}}(H)=2m-1$.

So far we have proved the assertion for $n\geq 4$. If $n=1$ we can take $H=\{0\}$. If $n=2$ we may consider $H={\mathbb{N}}_{\geq 0}$. If $n=3$ we can take $H={\mathbb{N}}_{\geq 2}\cup\{0\}$. ∎

Note that the proof could not be based solely on the monoids of the form $H_{k}={\mathbb{N}}_{\geq k}\cup\{0\}$, because $\#\operatorname{{\cal S}}(H_{k})$ grows faster than $k$.