# Representation and Embedding of Pseudo MV-algebras with Square Roots I.
Strict Square Roots

Anatolij Dvurečenskij${}^{{}^{1,2,3}}$, Omid Zahiri${}^{{}^{1,*}}$
1Mathematical Institute, Slovak Academy of Sciences, Štefánikova 49, SK-814 73
Bratislava, Slovakia  2Palacký University Olomouc, Faculty of Sciences, tř.
17. listopadu 12, CZ-771 46 Olomouc, Czech Republic  3Depart. Math.,
Constantine the Philosopher University in Nitra, Tr. A. Hlinku 1, SK-949 01
Nitra, Slovakia<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In [DvZa3], we started the investigation of pseudo MV-algebras with square
roots. In the present paper, we continue to study the structure of pseudo MV-
algebras with square roots focusing on their new characterizations. The paper
is divided into two parts. In the present first part, we investigate the
relationship between a pseudo MV-algebra with square root and its
corresponding unital $\ell$-group in the scene of two-divisibility.

In the second part, we find some conditions under which a particular class of
pseudo MV-algebras can be embedded into pseudo MV-algebras with square roots.
We introduce and investigate the concepts of a strict square root of a pseudo
MV-algebra and a square root closure, and we compare both notions. We show
that each MV-algebra has a square root closure. Finally, using the square root
of individual elements of a pseudo MV-algebra, we find the greatest subalgebra
of a special pseudo MV-algebra with weak square root.

###### Key words and phrases:

Pseudo MV-algebra, unital $\ell$-group, symmetric pseudo MV-algebra, square
root, strict square root, square root closure, two-divisibility, divisibility,
embedding

###### 2020 Mathematics Subject Classification:

06C15, 06D35

The paper acknowledges the support by the grant of the Slovak Research and
Development Agency under contract APVV-20-0069 and the grant VEGA No.
2/0142/20 SAV, A.D

The project was also funded by the European Union’s Horizon 2020 Research and
Innovation Programme on the basis of the Grant Agreement under the Marie
Skłodowska-Curie funding scheme No. 945478 - SASPRO 2, project 1048/01/01, O.Z

* Corresponding Author: Omid Zahiri

##  1\. Introduction

Chang [Cha1, Cha2] introduced MV-algebras as the algebraic semantics of the
$[0,1]$-valued Łukasiewicz logic. In the same sense, Boolean algebras do
algebraic semantics of the classical two-valued logic. Since then, the theory
of MV-algebras has been deeply investigated and studied in the different
aspects of mathematics and logic. MV-algebras form a category that is
categorically equivalent to the category of Abelian unital $\ell$-groups. This
principle result of the theory was presented by Mundici, [Mun].

More than twenty years ago, Georgescu and Iorgulescu, [GeIo], introduced a
non-commutative generalization of MV-algebras called pseudo MV-algebras. They
form an algebraic counterpart of the non-commutative Łukasiewicz logic; see,
for example, [Haj]. This concept also was independently defined by Rachůnek
[Rac] as generalized MV-algebras. Pseudo MV-algebras have a disjunction that
is not necessarily commutative and two negations, the left and right ones,
that could coincide even in non-commutative pseudo MV-algebras. We note that
other generalizations of pseudo MV-algebras are, for example, pseudo EMV-
algebras introduced in [DvZa0, DvZa1] and generalized MV-algebras from [GaTs].

Dvurečenskij, [Dvu1], using Bosbach’s notion of a semiclan, presented a
categorical equivalence between the category of pseudo MV-algebras and the
category of unital $\ell$-groups (not necessarily Abelian ones) that
generalizes a similar result by Mundici for MV-algebras. He showed that every
pseudo MV-algebra is always an interval in a unital $\ell$-group. Due to
Komori [Kom], it is well-known that the lattice of subvarieties of MV-algebras
is countable. In contrast, the lattice of subvarieties of pseudo MV-algebras
is uncountable; see [DvHo]. Hence, the structure of pseudo MV-algebras is much
richer than that of MV-algebras. Important varieties of pseudo MV-algebras are
the variety of symmetric pseudo MV-algebras, where the left and right
negations coincide, and the variety of representable pseudo MV-algebras; they
will be the playground for us where we will mainly work on the present paper.

Root operators are useful tools in studying algebraic structures with non-
idempotent binary operations. The study of the existence and uniqueness of
roots (in a particular case, square root) of an element or all elements of an
algebraic structure with binary operations has a long history. Mal’cev’s well-
known result, [KoMe, Thm 7.3.2], showed that the extraction of roots is unique
in a torsion-free locally nilpotent group. So, the equation $x^{n}=y^{n}$ has
at most one solution. Kontorovič, [Kon1, Kon2], also studied groups with
unique extraction of roots called R-groups. Baumslag in [Bau] developed the
theory of groups with unique roots.

In [Höl], Höhle studied square roots on integral, commutative, residuated
$\ell$-monoids, and especially for MV-algebras. He also introduced a strict
square root and proposed a classification of MV-algebras with square roots. In
particular, he showed that each MV-algebra with square root is exactly a
Boolean algebra (the square root is an identity map), or a strict MV-algebra
(the square root is strict), or isomorphic to the direct product of the first
two cases. He also investigated $\sigma$-complete, complete, divisible,
locally finite, and injective MV-algebras with square roots and provided their
relations. Bělohlávek, [Bel], continued in the study of the square root of
residuated lattices, and he proved that the class of all residuated lattices
with square roots is a variety. In [Amb], Ambrosio provided a study about
2-atomless MV-algebras and strict MV-algebras and proved that on this
structure, the concepts of strict and 2-atomless are equivalent. She also
proved that each strict MV-algebra contains a copy of the MV-algebra of all
rational dyadic numbers. We refer to [Höl, Amb, NPM] for more details about
square roots on MV-algebras.

In [DvZa2], we recently studied square roots on EMV-algebras which generalize
MV-algebras. We used square roots to characterize EMV-algebras and provided a
representation of EMV-algebras with square roots. Then we investigated
divisible and locally complete EMV-algebras with square roots. It was shown
that each strict EMV-algebra has a top element, so it is an MV-algebra.

In the next step, [DvZa3], we initiated a deed study of square roots on a
class of pseudo MV-algebras. We introduced and investigated the notion of a
square root and a weak square root on a pseudo MV-algebra. The class of pseudo
MV-algebras with square roots is equational, so it is a subvariety of pseudo
MV-algebras. We found that for each square root $r$ on a pseudo MV-algebra
$M$, the element $r(0)$ plays a significant role. It helped us classify the
pseudo MV-algebras class with square roots and proposed several examples. We
found a relationship between two-divisible $\ell$-groups and representable
symmetric pseudo MV-algebras with strict square roots.

The present work focuses on investigating square roots, and we extend our
research on pseudo MV-algebras, which was initiated in [DvZa3]. The main aims
of the present paper, which is divided into two parts, are:

Part I.

  * •

Present new characterizations and presentations of pseudo MV-algebras with
strict and non-strict square roots.

  * •

Show how two-divisibility is connected with the existence of square roots.

  * •

Investigate when the two-divisibility of a pseudo MV-algebra $M=\Gamma(G,u)$
entails the two-divisibility of $G$.

  * •

Study the possibility of embedding a pseudo MV-algebra into a pseudo MV-
algebra with square root.

  * •

Characterize square square roots on strongly $(H,1)$-perfect pseudo MV-
algebras.

Part II, [DvZa5].

  * •

Define and study the strict square root closure of a pseudo MV-algebra.

  * •

Define and study the square root closure of a pseudo MV-algebra and compare it
with the strict square root closure.

  * •

Investigate the square root of not all elements of a pseudo MV-algebra.

  * •

Find conditions when a maximal subalgebra exists in a pseudo MV-algebra with
weak square root.

The paper is organized as follows. Part I. Section 2 gathers basic
definitions, properties, and results about pseudo MV-algebras and square roots
that will be used in the next sections. Section 3 presents sufficient and
necessary conditions under which a pseudo MV-algebra has a strict or non-
strict square root. In Section 4, the relation between a pseudo MV-algebra
$M=\Gamma(G,u)$ with a square root and the two-divisibility of the unital
$\ell$-group $G$ is investigated. If $M$ is linearly ordered or an MV-algebra
with square root, then the $\ell$-group $G$ is two-divisible. We find a
sufficient and necessary condition under which a square root on a pseudo MV-
algebra $\Gamma(G,u)$ implies the two-divisibility of the $\ell$-group $G$. We
also characterize the lexicographic product of MV-algebras with square roots.

Part II. In Section 5, we find an answer to the question, “Is it possible to
embed a pseudo MV-algebra in a pseudo-MV-algebra with square root?” To answer
the problem, we study a class of pseudo MV-algebras that can be embedded into
a pseudo MV-algebra with strict square root. We define the concept of the
strict square root closure and prove that each MV-algebra has a strict square
root closure. Section 6 introduces a square root closure of an MV-algebra and
compares it with the strict square root. Section 7 describes a square root of
an individual element of a pseudo MV-algebra and finds the greatest subalgebra
of special pseudo MV-algebras with the weak square root property. The paper
contains interesting examples illustrating our results and some questions are
formulated.

##  2\. Preliminaries

In the section, we gather basic elements of $\ell$-groups, pseudo MV-algebras,
and square roots.

We will use groups $(G;+,0)$ written additively. A group $G$ is partially
ordered if there is a partial order $\leq$ on $G$ such that $f\leq g$ implies
$h_{1}+f+h_{2}\leq h_{1}+g+h_{2}$ for all $h_{1},h_{2}\in G$. If $\leq$ is a
lattice order, $G$ is said to be a lattice ordered or an $\ell$-group. An
element $u\geq 0$ is said to be a strong unit of $G$ if given $g\in G$, there
is an integer $n\geq 1$ such that $g\leq nu$. A couple $(G,u)$, where $u$ is a
fixed strong unit of $G$, is said to be a unital $\ell$-group. If $G$ is an
$\ell$-group, $\mathrm{C}(G)=\\{g\in G\colon g+h=h+g,\ \forall h\in G\\}$ is
the commutative center of $G$.

For more information about$\ell$-groups, we recommend to consult with e.g.,
[Dar, Gla, AnFe].

######  Definition 2.1.

[GeIo] A pseudo MV-algebra is an algebra $(M;\oplus,^{-},^{\sim},0,1)$ of type
$(2,1,1,0,0)$ such that the following axioms hold for all $x,y,z\in M$,

  * (A1)

$x\oplus(y\oplus z)=(x\oplus y)\oplus z$,

  * (A2)

$x\oplus 0=0\oplus x=x$,

  * (A3)

$x\oplus 1=1\oplus x=1$,

  * (A4)

$1^{-}=1^{\sim}=0$,

  * (A5)

$(x^{-}\oplus y^{-})^{\sim}=(x^{\sim}\oplus y^{\sim})^{-}$,

  * (A6)

$x\oplus(x^{\sim}\odot y)=y\oplus(y^{\sim}\odot x)=(x\odot y^{-})\oplus
y=(y\odot x^{-})\oplus x$,

  * (A7)

$x\odot(x^{-}\oplus y)=(x\oplus y^{\sim})\odot y$,

  * (A8)

$(x^{-})^{\sim}=x$,

where $x\odot y=(x^{-}\oplus y^{-})^{\sim}$. If the operation $\oplus$ is
commutative, equivalently $\odot$ is commutative, then $M$ is an MV-algebra.
(A6) defines $x\vee y$ and (A7) $x\wedge y$. We note that if $x^{\sim}=x^{-}$
for each $x\in M$, then $\oplus$ is not necessarily commutative. If
$x^{-}=x^{\sim}$ for each $x\in M$, $M$ is said to be symmetric.

We note that it can happen that $0=1$; in this case, $M$ is said to be
degenerate.

For example, if $(G,u)$ is a unital $\ell$-group, then
$\Gamma(G,u)=([0,u];\oplus,^{-},^{\sim},0,u)$ is a pseudo MV-algebra, where
$x\oplus y:=(x+y)\wedge u$, $x^{-}:=u-x$, and $x^{\sim}:=-x+u$. Moreover, due
to a basic representation of pseudo MV-algebras, see [Dvu1], every pseudo MV-
algebra is isomorphic to some $\Gamma(G,u)$ for a unique (up to isomorphism)
unital $\ell$-group $(G,u)$. In addition, the functor
$\Gamma:(G,u)\mapsto\Gamma(G,u)$ defines a categorical equivalence between the
category of unital $\ell$-groups and the category of pseudo MV-algebras. For
more information about the functor $\Gamma$ and its inverse $\Psi$, see
[Dvu1].

According to [Dvu1], we introduce a partial operation $+$ on any pseudo MV-
algebra $M$: Given $x,y\in M$, $x+y$ is defined if and only if $y\odot x=0$,
and in such a case, we set $x+y:=x\oplus y$. The operation $+$ is associative,
and using the $\ell$-group representation, it corresponds to the group
addition in the representing unital $\ell$-group.

For any integer $n\geq 0$ and any $x\in M$, we can define

$\displaystyle 0.x$ $\displaystyle=0,\quad\text{and}\quad n.x=(n-1).x\oplus
x,\quad n\geq 1,$ $\displaystyle 0x$ $\displaystyle=0,\quad\text{and}\quad
nx=(n-1)x+x,\quad n\geq 1,$

assuming $(n-1)x$ and $(n-1)x+x$ are defined in $M$. An element $x\in M$ is a
Boolean element if $x\oplus x=x$. The set $\mathrm{B}(M)$ denotes the set of
Boolean elements of $M$, which is a subalgebra of $M$ and a Boolean algebra;
it is a so-called Boolean skeleton of $M$.

In each pseudo MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$, we can define two
additional binary operations $\to$ and $\rightsquigarrow$ by

$x\to y:=x^{-}\oplus y,\quad x\rightsquigarrow y:=y\oplus x^{\sim}.$

The properties of the operations $\to$ and $\rightsquigarrow$ can be found
e.g. in [DvZa3, Prop 2.3].

Let $n\geq 2$ be an integer. A pseudo MV-algebra $M$ is said to be
$n$-divisible if, given an element $y\in M$, there is an element $x\in M$ such
that $nx$ exists in $M$ and $nx=y$. If $M$ is $n$-divisible for each $n\geq
1$, we say that $M$ is divisible. We note that $M$ is $n$-divisible iff given
$x\in M$, there is $y\in M$ such that $n.y=x$ and $(n-1).y\odot y^{-}=0$.
Analogously, an $\ell$-group $G$ is $n$-divisible if given $g\in G$, there is
$h\in G$ such that $nh=g$, and $G$ is divisible if it is $n$-divisible for
each $n\geq 2$. If $M=\Gamma(G,u)$ is an MV-algebra, then $M$ is $n$-divisible
iff $G$ is $n$-divisible. For pseudo MV-algebras $\Gamma(G,u)$,
$n$-divisibility of $G$ trivially implies $n$-divisibility of $M=\Gamma(G,u)$.
The converse also holds if, e.g., $G$ is linearly ordered and
$u/2\in\mathrm{C}(G)$; see Corollary 4.8.

An $\ell$-group $G$ enjoys unique extraction of roots if for every integer
$n\geq 1$ and $g,h\in G$, $ng=nh$ implies $g=h$. In the same way, we say that
a pseudo MV-algebra enjoys unique extraction of roots. We note that every
linearly ordered group, [Gla, Lem 2.1.4], (linearly ordered pseudo MV-algebra)
enjoys unique extraction of roots. The same applies to each representable
$\ell$-group, see [AnFe, Page 26].

######  Remark 2.2.

If $(M;\oplus,^{-},^{\sim},0,1)$ is a pseudo MV-algebra and $a\in M$, then
$([0,a];\oplus_{a},^{-_{a}},^{\sim_{a}},0,a)$ is a pseudo MV-algebra, where
$x\oplus_{a}y=(x\oplus y)\wedge a$, $x^{-a}=a\odot x^{-}$ and $x^{\sim
a}=x^{\sim}\odot a$. Indeed, by [Dvu1], we can assume that $M=\Gamma(G,u)$,
where $(G,u)$ is a unital $\ell$-group. For each $a\in M$ and each
$x,y\in[0,a]$, $a\odot x^{-}=a-u+(u-x)\odot a=a-x=x^{-a}$, $x^{\sim}\odot
a=-x+u-u+a=-x+a=x^{\sim a}$ and $(x\oplus y)\wedge a=(x+y)\wedge u\wedge
a=(x\oplus_{a}y)$. Therefore, by [GeIo, Exm 1.3],
$([0,a];\oplus_{a},^{-a},^{\sim a},0,a)$ is a pseudo MV-algebra.

A non-empty subset $I$ of a pseudo MV-algebra $M$ is called an ideal of $M$ if
(1) for each $y\in M$, $y\leq x\in I$ implies that $y\in I$; (2) $I$ is closed
under $\oplus$. An ideal $I$ of $M$ is said to be (i) prime if $x\wedge y\in
I$ implies $x\in I$ or $y\in I$; (ii) normal if $x\oplus I=I\oplus x$ for any
$x\in M$, where $x\oplus I:=\\{x\oplus i\mid i\in I\\}$ and $I\oplus
x=\\{i\oplus x\mid i\in I\\}$, (iii) proper if $I\neq M$, (iv) maximal if $I$
is a proper ideal of $M$ and it is not properly contained in any proper ideal
of $M$. We denote by $\text{MaxI}(M)$ and $\text{NormI}(M)$ the set of maximal
ideals and normal ideals, respectively, of $M$. Of course,
$\text{MaxI}(M)\neq\emptyset$ and $\text{NormI}(M)\neq\emptyset$, but their
intersection may be empty, see, e.g., [DvHo], what for MV-algebras is
impossible.

We recall that an ideal $I$ is normal if and only if given $x,y\in M$, $x\odot
y^{-}\in I$ if and only if $y^{\sim}\odot x\in I$, [GeIo, Lem 3.2]. For each
subset $I$ of $M$, we denote $I^{-}=\\{x^{-}\mid x\in I\\}$ and
$I^{\sim}=\\{x^{\sim}\mid x\in I\\}$.

The set of the prime ideals of $M$ is denoted by $\mathrm{Spec}(M)$.
Equivalent conditions, [GeIo, Thm 2.17], for an ideal $I$ to be prime are as
follows:

  * (P1)

$x\odot y^{-}\in I$ or $y\odot x^{-}$ for all $x,y\in M$.

  * (P2)

$x\odot y^{\sim}\in I$ or $y\odot x^{\sim}$ for all $x,y\in M$.

A one-to-one relationship exists between congruences and normal ideals of a
pseudo MV-algebra, [GeIo, Cor. 3.10]: If $I$ is a normal ideal of a pseudo MV-
algebra, then the relation $\sim_{I}$, defined by $x\sim_{I}y$ if and only if
$x\odot y^{-},y\odot x^{-}\in I$, is a congruence, and $M/I$ with the
following operations induced from $M$ is a pseudo MV-algebra, where
$M/I=\\{x/I\mid x\in M\\}$ and $x/I$ is the equivalence class containing $x$.

$\displaystyle x/I\oplus y/I=(x\oplus
y)/I,\quad(x/I)^{-}=x^{-}/I,\quad(x/I)^{\sim}=x^{\sim}/I,\quad 0/I,\quad 1/I.$

Conversely, if $\theta$ is a congruence on $M$, then $I_{\theta}=\\{x\in M\mid
x\sim 0\\}$ is a normal ideal such that $\sim_{I_{\theta}}=\sim$.

A pseudo MV-algebra $M$ is representable if $M$ is a subdirect product of a
linearly ordered pseudo MV-algebras system. By [Dvu2, Prop. 6.9], $M$ is
representable if and only if $a^{\bot}=\\{x\in M\mid x\wedge a=0\\}$ is a
normal ideal of $M$ for each $a\in M$. Moreover, the class of representable
pseudo MV-algebras forms a variety [Dvu2, Thm 6.11].

We will also use pseudo MV-algebras of the form
$\Gamma(H\,\overrightarrow{\times}\,G,(u,0))$, where $(H,u)$ is a linearly
ordered unital $\ell$-group, $G$ is an $\ell$-group, and
$\,\overrightarrow{\times}\,$ denotes the lexicographic product of $H$ and
$G$. We note that
$\Gamma(H,u)\cong\Gamma(H\,\overrightarrow{\times}\,O,(u,0))$, where $O$ is
the one-element zero group, i.e. $O=\\{0\\}$.

An important class of pseudo MV-algebras is the class of $(H,1)$-perfect
pseudo MV-algebras, where $(H,1)$ is a unital $\ell$-subgroup of the unital
group of real numbers $(\mathbb{R},1)$. They are roughly speaking isomorphic
to $\Gamma(H\,\overrightarrow{\times}\,G,(1,0))$, where $G$ is an
$\ell$-group, see [Dvu4, Thm 4.3], see also [Dvu3]. We recall that if $A,B$
are two subsets of $M$, then $A\leq B$ means $a\leq b$ for each $a\in A$ and
each $b\in B$. We say a pseudo MV-algebra $M$ is $H$-perfect, if there is a
decomposition $(M_{h}\mid h\in H)$ of $M$ such that

  * (i)

$M_{s}\neq\emptyset$ for each $s\in H$,

  * (ii)

$M_{s}\leq M_{t}$ if $s\leq t$, $s,t\in H$,

  * (iii)

$M_{s}^{-}=M_{1-s}=M^{\sim}_{s}$ for each $s\in H$,

  * (iv)

if $x\in M_{s}$ and $y\in M_{t}$, then $x\oplus y\in M_{s\oplus t}$, where
$s\oplus t=\min\\{s+y,1\\}$, $s,t\in H$.

An $(H,1)$-perfect pseudo MV-algebra $M=\Gamma(K,v)$ is strongly
$(H,1)$-perfect if there is a system of elements $(c_{t}\mid t\in H)$ such
that

  * (i)

$c_{t}\in H\cap\mathrm{C}(K)$, $t\in H$,

  * (ii)

if $s,t\in H$ and $s+t\in H$, then $c_{s}+c_{t}=c_{s+t}$,

  * (iii)

$c_{1}=1$.

According to [Dvu4, Thm 4.3], a pseudo MV-algebra $M$ is strongly
$(H,1)$-perfect iff there is an $\ell$-group $G$ such that
$M\cong\Gamma(H\,\overrightarrow{\times}\,G,(1,0))$.

A notion of a square root on MV-algebras was introduced in [Höl]. For pseudo
MV-algebras, it was introduced and studied in [DvZa3].

######  Definition 2.3.

A mapping $r:M\to M$ is said to be (i) a square root if it satisfies the
following conditions:

  * (Sq1)

for all $x\in M$, $r(x)\odot r(x)=x$,

  * (Sq2)

for each $x,y\in M$, $y\odot y\leq x$ implies $y\leq r(x)$,

  * (Sq3)

for each $x\in M$, $r(x^{-})=r(x)\to r(0)$ and
$r(x^{\sim})=r(x)\rightsquigarrow r(0)$,

and (ii) a weak square root if it satisfies only (Sq1) and (Sq2). A pseudo MV-
algebra $(M;\oplus,^{-},^{\sim},0,1)$ has square roots (weak square roots) if
there exists a square root (weak square root) $r$ on $M$. If $M$ is an MV-
algebras, then both notions of a square root coincide. A square root $r$ is
strict if $r(0)=r(0)^{-}$, equivalently, $r(0)=r(0)^{\sim}$. If $M$ has a
square root (weak square root), it is unique.

We note that it can happen that a pseudo MV-algebra has a weak square root but
no square root; see [DvZa3, Sec 6] for such examples.

The basic properties of square roots on pseudo MV-algebras can be found in the
following result:

######  Proposition 2.4.

[DvZa3] Let $r$ be a square root on a pseudo MV-algebra
$(M;\oplus,^{-},^{\sim},0,1)$. For each $x,y\in M$, we have:

  * (1)

$x\leq x\vee r(0)\leq r(x)$, $r(1)=1$, $(r(x)\odot r(0))\vee(r(0)\odot
r(x))\leq x$ and $r(x)\odot x=x\odot r(x)$.

  * (2)

$x\leq y$ implies that $r(x)\leq r(y)$.

  * (3)

$x\wedge y\leq r(x)\odot r(y),r(y)\odot r(x)$ and if $a\in\mathrm{B}(M)$ such
that $a\leq r(0)$, then $a=0$.

  * (4)

$x\leq r(x\odot x)$ and $r(x\odot x)\odot r(x\odot x)=r(x)\odot r(x)\odot
r(x)\odot r(x)=x\odot x$.

  * (5)

$(x\wedge x^{-})\vee(x\wedge x^{\sim})\leq r(0)$.

  * (6)

$r(x)\in\mathrm{B}(M)$ if and only if $r(x)=x$.

  * (7)

$r(x)\wedge r(y)=r(x\wedge y)$.

  * (8)

$r(x)\rightarrow r(y)\leq r(x\rightarrow y)$ and $r(x)\rightsquigarrow
r(y)\leq r(x\rightsquigarrow y)$. Moreover, $r(x)\odot r(y)\leq r(x\odot y)$
for all $x,y\in M$ if and only if $r(x)\rightarrow r(y)=r(x\rightarrow y)$ and
$r(x)\rightsquigarrow r(y)=r(x\rightsquigarrow y)$.

  * (9)

$r(x\vee y)=r(x)\vee r(y)$.

  * (10)

$r(x\odot y)\leq(r(x)\odot r(y))\vee r(0)$ and $r(x\odot x)=(r(x)\odot
r(x))\vee r(0)$. Consequently, if $r(0)\leq x$, then $r(x\odot x)=x$.

  * (11)

(a) $x\in\mathrm{B}(M)$ if and only if $r(x)=x\oplus r(0)$ if and only if
$r(x)=r(0)\oplus x$.

(b) $(r(0)\rightarrow 0)\odot(r(0)\rightarrow 0)=(r(0)\rightsquigarrow
0)\odot(r(0)\rightsquigarrow 0)\in\mathrm{B}(M)$.

(c) If $a,b\in M$, $a\leq b$, then $r([a,b])=[r(a),r(b)]$. In particular,
$r(M)=[r(0),1]$.

Properties (1)–(8) hold also for each weak square root on $M$.

##  3\. Characterizations of pseudo MV-algebras with square roots

Recently, in [DvZa3], we presented some characterizations of pseudo MV-
algebras with square roots using unital $\ell$-groups. In this section, we
study pseudo MV-algebras and find new conditions under which they have square
roots (strict or non-strict).

First, we gather some useful properties that will be used in the sequel.

######  Definition 3.1.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a pseudo MV-algebra and $a$ be an element
of $M$. We say that $M$ is

(i) $f$-isomorphic to $[0,a]$, if there is a bijective and order-preserving
map $f:M\to[0,a]$ such that $[0,a]$ with, $\oplus_{f}$ (the binary operation
is inherited from $M$ by $f$, that is, $f(x)\oplus_{f}f(y):=f(x\oplus y)$), -a
and ∼a form a pseudo MV-algebra such that $M\cong[0,a]$ and for all $x\in M$,
$f(x)\oplus_{f}f(x)=(f(x)\oplus f(x))\wedge a$.

(ii) isomorphic to $[0,a]$ if $(M;\oplus,^{-},^{\sim},0,1)$ is isomorphic to
$([0,a];\oplus_{a},^{-a},^{\sim a},0,a)$ (see Remark 2.2).

Note that in (ii), $\oplus_{a}$ is induced from $\oplus$, i.e.
$x\oplus_{a}y=(x\oplus y)\wedge a$. Clearly, if $M$ is isomorphic to $[0,b]$
for some $b\in M$, then there is an isomorphism $g:M\to[0,b]$ of pseudo MV-
algebras and so $M$ is $g$-isomorphic to $[0,b]$.

######  Proposition 3.2.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a pseudo MV-algebra with a square root
$r:M\to M$.

  * (i)

$[r(0),r(0)^{-}]\subseteq r(\mathrm{B}(M))$.

  * (ii)

$M$ is a Boolean algebra if $[r(0),r(0)^{-}]=M$, and $r$ is strict if
$[r(0),r(0)^{-}]=\\{r(0)\\}$.

  * (iii)

For each $x\in M$, there exists a unique element $R(x)\leq r(0)^{-}$ such that
$R(x)\oplus r(0)=r(x)$.

  * (iv)

$M$ is $f$-isomorphic to $[0,r(0)^{-}]$ for some bijection
$f:M\to[0,r(0)^{-}]$.

  * (v)

For each $x\in M$, there exists a unique element $R(x)\leq r(0)^{-}$ such that
$R(x)\oplus R(x)=x$.

  * (vi)

If $r(x)\odot r(y)\leq r(x\odot y)$ for all $x,y\in M$, then $M$ is isomorphic
to $[0,r(0)^{-}]$.

  * (vii)

For each $x\in M$, $r(x)=\max\\{y\wedge(y\rightarrow x)\mid y\in
M\\}=\max\\{y\wedge(y\rightsquigarrow x)\mid y\in M\\}$.

  * (viii)

$x\odot r(0)\leq x\odot x$ for all $x\in M$.

###### Proof.

Set $v:=r(0)^{-}\odot r(0)^{-}$. By [DvZa3, Prop 3.3] or Proposition 2.4(11),
$v\in\mathrm{B}(M)$. Then $r(0)^{\sim}=r(0)^{-}$, so that $v=r(0)^{\sim}\odot
r(0)^{\sim}$.

(i) Let $x\in[r(0),r(0)^{-}]$. Then $r(0)\leq x$ and [DvZa3, Prop 3.5(i)]
implies $x=x\vee r(0)=r(x\odot x)$, and so $r(0)^{-}=r(r(0)^{-}\odot
r(0)^{-})=r(v)$. Also, $x\leq r(0)^{-}$ implies $x\odot x\leq r(0)^{-}\odot
r(0)^{-}=v$. Consider the subset $[0,v]\subseteq M$ which is a subset of
$\mathrm{B}(M)$ ([DvZa3, Thm 4.3]). By Proposition 2.4(11)(c),
$r([0,v])=[r(0),r(v)]=[r(0),r(0)^{-}]$. In addition, $[0,v]=\\{x\odot x\mid
x\in[r(0),r(0)^{-}]\\}$.

(ii) It follows from (i) and [DvZa3, Thm 4.3].

(iii) Without loss of generality, we can assume that $M=\Gamma(G,u)$, where
$(G,u)$ is a unital $\ell$-group. For each $x\in M$ set
$R(x):=r(x^{\sim})^{-}$. By Proposition 2.4(2), $r(0)\leq r(x^{\sim})$ and so
$R(x)\leq r(0)^{-}$. By [DvZa3, Prop 3.5(v)], $R(x)\oplus r(0)=r(x)$. Now, let
$r(x)=y\oplus r(0)$ for some $y\leq r(0)^{-}$. Then $y+r(0)\leq
r(0)^{-}+r(0)=u-r(0)+r(0)=u$ and so $y+r(0)=(y+r(0))\wedge u=y\oplus r(0)$.
Similarly, $R(x)+r(0)=R(x)\oplus r(0)$. It follows that $R(x)+r(0)=R(x)\oplus
r(0)=y\oplus r(0)=y+r(0)$ entails that $R(x)=y$. In addition, from [DvZa3,
Prop 3.5(v)], we get that $r(x^{-})^{\sim}=R(x)=r(x^{\sim})^{-}$ since
$r(x^{-})^{\sim}\leq r(0)^{\sim}=r(0)^{-}$ and $r(x^{-})^{\sim}\oplus
r(0)=r(0)\oplus r(x^{-})^{\sim}=r(x)$.

(iv) Consider the mapping $f:M\to[0,r(0)^{-}]$ defined by
$f(x)=r(x^{\sim})^{-}$. From part (iii), we conclude that $f$ is well-defined,
one-to-one, and $f(x)=r(x^{-})^{\sim}$ for all $x\in M$. Also, if $y\leq
r(0)^{-}$, then $r(0)\leq y^{\sim}$ whence by [DvZa3, Prop 3.5(i)],
$y^{\sim}=y^{\sim}\vee r(0)=r(y^{\sim}\odot y^{\sim})$. It follows that
$y=r(y^{\sim}\odot y^{\sim})^{-}=r(z^{\sim})^{-}\in\mathrm{Im}(f)$, where
$z=y\oplus y$. So, $f$ is a bijection map. Note that, for each $x\in M$, by
[DvZa3, Prop 3.5(v)], $r(0)\oplus x=x\oplus r(0)$, consequently

$\displaystyle r(0)^{-}\odot x=x\odot r(0)^{-},\quad\forall x\in M.$ (3.1)

Consider the following operations of $[0,r(0)^{-}]$. If $x,y\in[0,r(0)^{-}]$,
there exist $a,b\in M$ such that $x=f(a)$ and $y=f(b)$. We set

$\displaystyle x\oplus_{r}y:=f(a\oplus b)=r((a\oplus b)^{\sim})^{-},\quad\
x^{-r}=x^{-}\odot r(0)^{-},\quad\ x^{\sim r}=x^{\sim}\odot r(0)^{-}.$

Let $x,y,z\in M$.

(1) $f(0)=r(0^{\sim})^{-}=r(1)^{-}=1^{-}=0$ and
$f(1)=r(1^{\sim})^{-}=r(0)^{-}$. For each $x,y\in M$,
$f(x)\oplus_{r}f(y)=r((x\oplus y)^{\sim})^{-}=f(x\oplus y)$.

(2) By (Sq3), (3.1), and (iii), we have (recall that in (iii), we have proved
that $r(x^{-})^{\sim}=r(x^{\sim})^{-}$ for all $x\in M$)

$\displaystyle f(x^{-})$
$\displaystyle=r(x^{-\sim})^{-}=r(x)^{-}=r(x)^{-}\wedge
r(0)^{-}=(r(x)^{-}\oplus r(0))\odot r(0)^{-}=(r(x)\to r(0))\odot r(0)^{-}$
$\displaystyle=r(x^{-})\odot r(0)^{-}=r(x^{-})^{\sim-}\odot
r(0)^{-}=r(x^{\sim})^{--}\odot r(0)^{-}=f(x)^{-}\odot r(0)^{-}=f(x)^{-r},$
$\displaystyle f(x^{\sim})$ $\displaystyle=r(x)^{\sim}=r(x)^{\sim}\wedge
r(0)^{\sim}=r(0)^{\sim}\odot(r(0)\oplus
r(x)^{\sim})=r(0)^{\sim}\odot(r(x)\rightsquigarrow r(0))=r(0)^{\sim}\odot
r(x\rightsquigarrow 0)$ $\displaystyle=r(0)^{\sim}\odot
r(x^{\sim})=r(0)^{\sim}\odot r(x^{\sim})^{-\sim}=r(0)^{\sim}\odot
f(x)^{\sim}=f(x)^{\sim}\odot r(0)^{-}=f(x)^{\sim r}.$

(3) $f(0)\oplus_{r}f(x)=r((0\oplus x)^{\sim})^{-}=r(x^{\sim})^{-}=f(x)$.
Similarly, $f(x)\oplus_{r}f(0)=f(x)$. We have $f(x)\oplus_{r}f(1)=r((1\oplus
x)^{\sim})^{-}=r(1^{\sim})^{-}=r(0)^{-}=r((x\oplus
1)^{\sim})^{-}=f(1)\oplus_{r}f(x)$ and $(r(0)^{-})^{\sim
r}=(r(0)^{-})^{\sim}\odot r(0)^{-}=r(0)\odot r(0)^{-}=0$ and
$(r(0)^{-})^{-r}=(r(0)^{\sim})^{-}\odot r(0)^{-}=r(0)\odot r(0)^{-}=0$ (by
[DvZa3, Lem 3.15]).

(4) By the definition of $\oplus_{r}$, we have
$f(x)\oplus_{r}(f(y)\oplus_{r}f(z))=f(x)\oplus_{r}(r((y\oplus
z)^{\sim})^{-})=f(x)\oplus_{r}f(y\oplus z)=f(x\oplus(y\oplus z))=f((x\oplus
y)\oplus z)$. Similarly, $(f(x)\oplus_{r}f(y))\oplus_{r}f(z)=f((x\oplus
y)\oplus z)$. Hence, $\oplus_{r}$ is associative.

(5) By [DvZa3, Prop 3.5(v)], $(f(x)^{-r})^{\sim r}=(f(x)^{-}\odot
r(0)^{-})^{\sim}\odot r(0)^{-}=(r(0)\oplus f(x))\odot r(0)^{-}=(f(x)\oplus
r(0))\odot r(0)^{-}=f(x)\wedge r(0)^{-}=f(x)$.

(6) By (1) and (2), $(f(x)^{-r}\oplus_{r}f(y)^{-r})^{\sim
r}=(f(x^{-})\oplus_{r}f(y^{-}))^{\sim r}=f((x^{-}\oplus y^{-})^{\sim})$ and
$(f(x)^{\sim r}\oplus_{r}f(y)^{\sim
r})^{-r}=(f(x^{\sim})\oplus_{r}f(y^{\sim}))^{-r}=f((x^{\sim}\oplus
y^{\sim})^{-})$.

(7) In a similar way, using (1) and (2), we can show that identities (A7) and
(A8) hold.

(8) By Proposition 2.4(10), $f(x\oplus x)=r((x\oplus
x)^{\sim})^{-}=r(x^{\sim}\odot x^{\sim})^{-}=((r(x^{\sim})\odot
r(x^{\sim}))\vee r(0))^{-}=(r(x^{\sim})^{-}\oplus r(x^{\sim})^{-})\wedge
r(0)^{-}=(f(x)\oplus f(x))\wedge r(0)^{-}$.

Therefore, $([0,r(0)^{-}];\oplus_{r},^{-r},^{\sim r},0,r(0)^{-})$ is a pseudo
MV-algebras, $f$ is an isomorphism, and $M$ is isomorphic to $[0,r(0)^{-}]$.

(v) By [DvZa3, Prop 3.5(i)], $r(x^{\sim})^{-}\oplus r(x^{\sim})^{-}=x$ and by
(iii), $r(x^{\sim})^{-}\leq r(0)^{-}$. If $y\leq r(0)^{-}$ is an element of
$M$ such that $y\oplus y=x$, then due to (iv), there exists $z\in M$ such that
$r(z^{\sim})^{-}=f(z)=y$. We have $x=y\oplus y=r(z^{\sim})^{-}\oplus
r(z^{\sim})^{-}=z$, and so $y=r(x^{\sim})^{-}$.

(vi) Set $b:=r(0)^{-}$. Consider the bijection map $f$ defined in the proof of
(iv). By part (iv) and the definition, it suffices to show that
$f(x)\oplus_{b}f(y)=f(x\oplus y)$ for all $x,y\in M$. Since $r(x)\odot
r(y)\leq r(x\odot y)$ and $r(0)\leq r(x\odot y)$, by Proposition 2.4(10), we
get $(r(x)\odot r(y))\vee r(0)=r(x\odot y)$ for all $x,y\in M$. It follows
that

$\displaystyle f(x\oplus y)$ $\displaystyle=$ $\displaystyle r((x\oplus
y)^{\sim})^{-}=r((y^{\sim}\odot x^{\sim}))^{-}=\left(\left(r(y^{\sim})\odot
r(x^{\sim})\right)\vee r(0)\right)^{-}$ $\displaystyle=$
$\displaystyle\left(r(x^{\sim})^{-}\oplus r(y^{\sim})^{-}\right)\wedge
r(0)^{-}=(f(x)\oplus f(y))\wedge r(0)^{-}.$

Similarly, we can show that if the mapping $f$ is a homomorphism, then
$r(x)\odot r(y)\leq r(x\odot y)$ for all $x,y\in M$.

(vii) Let $x\in M$ and $\Omega_{x}:=\\{y\wedge(y\rightarrow x)\mid y\in M\\}$.
Then $r(x)\odot r(x)=x$ implies that $r(x)\leq r(x)\rightarrow x$, so
$r(x)\wedge(r(x)\rightarrow x)=r(x)$ and $r(x)\in\Omega_{x}$.

Now, for each $y\in M$, by [DvZa3, Prop 2.3(i)], we have

$\displaystyle\big{(}y\wedge(y\rightarrow
x)\big{)}\odot\big{(}y\wedge(y\rightarrow x)\big{)}\leq y\odot(y\rightarrow
x)=y\wedge x\leq x,$

which means $y\wedge(y\rightarrow x)\leq r(x)$. Therefore,
$r(x)=\max\Omega_{x}=\max\\{y\wedge(y\rightarrow x)\mid y\in M\\}$.

In a similar way, we can show that $r(x)=\max\\{y\wedge(y\rightsquigarrow
x)\mid y\in M\\}$.

(viii) For each $x\in M$, we have $r(x\odot x)=x\vee r(0)$, by [DvZa3, Prop
3.3(10)]. It follows that $(x\odot x)\vee(x\odot r(0))\vee(r(0)\odot x)=(x\vee
r(0))\odot(x\vee r(0))=x\odot x$, so $x\odot r(0)\leq x\odot x$. ∎

According to Proposition 3.2(iv), if $M$ is a pseudo MV-algebra with a square
root $r$, then $M$ is $f$-isomorphic to $[0,r(0)^{-}]$, where
$f:M\to[0,r(0)^{-}]$ is defined by $f(x)=r(x^{\sim})^{-}$ for all $x\in M$.

######  Theorem 3.3.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a pseudo MV-algebra. Then $M$ has a
square root if and only if there exists $b\in M$ satisfying the following
conditions:

  * (i)

$b^{-}\leq b$ and $b\odot x=x\odot b$ for all $x\in M$,

  * (ii)

$f(x)\oplus f(x)=x$,

  * (iii)

$M$ is $f$-isomorphic to $[0,b]$ for some bijection $f:M\to[0,b]$.

In addition, if in condition (iii), $M$ is isomorphic to the $[0,b]$, then
$r(x)\odot r(y)\leq r(x\odot y)$ for all $x,y\in M$.

###### Proof.

Let $M=\Gamma(G,u)$, where $(G,u)$ is a unital $\ell$-group.

First, assume that there exists $b\in M$ satisfying the conditions (i)–(iii).
Define $r:M\to M$ by

$\displaystyle r(x)=b^{-}+f(x),\quad x\in M.$

Since for each $x\in M$, $b^{-}+f(x)\leq b^{-}+b=u$, we have
$b^{-}+f(x)=b^{-}\oplus f(x)\in M$. In a similar way, $f(x)+b^{-}=f(x)\oplus
b^{-}$. Thus, $r$ is one-to-one. Also, for all $x\in M$,
$b^{-}+f(x)=b^{-}\oplus f(x)=(f(x)^{\sim}\odot b)^{-}=(b\odot
f(x)^{\sim})^{-}=f(x)\oplus b^{-}=f(x)+b^{-}$. We claim that $r$ is a square
root.

(1) From (i), it follows that $b^{\sim}=b^{-}$. Hence, $r(0)=b^{-}\oplus
f(0)=b^{-}=b^{\sim}$.

(2) Since $M$ is $f$-isomorphic to $[0,b]$, $f(x^{-})^{\sim}=(f(x)^{-}\odot
b)^{\sim}=b^{\sim}\oplus f(x)=b^{-}\oplus f(x)=r(x)$ for each $x\in M$.

(3) By (ii) and (2), we get that $r(x)\odot r(x)=f(x^{-})^{\sim}\odot
f(x^{-})^{\sim}=(f(x^{-})\oplus f(x^{-}))^{\sim}=(x^{-})^{\sim}=x$.

(4) Since $M$ is $f$-isomorphic to $[0,b]$, property (2) implies that
$r(x\rightarrow 0)=r(x^{-})=f(x^{--})^{\sim}=(f(x^{-})^{-}\odot
b)^{\sim}=b^{\sim}\oplus f(x^{-})=b^{-}\oplus
f(x^{-})^{\sim-}=b^{-}\oplus(f(x^{-})^{\sim})^{-}=r(x)^{-}\oplus
b^{-}=r(x)\rightarrow r(0)$.

(5) Let $y,x\in M$ such that $y\odot y\leq x$. Then $f(y\odot y)\leq f(x)$.

$\displaystyle f(y\odot y)$ $\displaystyle=$ $\displaystyle f((y^{-}\oplus
y^{-})^{\sim})=f(y^{-}\oplus y^{-})^{\sim}\odot b=\big{(}(f(y^{-})\oplus
f(y^{-}))\wedge b\big{)}^{\sim}\odot b$ $\displaystyle=$
$\displaystyle\left(\left(f(y)^{-}\odot b\right)\oplus\left(f(y)^{-}\odot
b\right)\right)^{\sim}\odot b=\left(f(y)^{-}\odot
b\right)^{\sim}\odot\left(f(y)^{-}\odot b\right)^{\sim}\odot b$
$\displaystyle=$ $\displaystyle(r(y)\odot r(y))\odot b=y\odot b\mbox{ by (1),
(2), and (3)}.$

From (2), we conclude that $y\leq y\vee b^{-}=(y\odot b)\oplus b^{-}=f(y\odot
y)\oplus b^{-}\leq f(x)\oplus b^{-}=r(x)$.

(3)–(5) imply $r$ is a square root on $M$.

Conversely, if $M$ has a square root $r$, then it suffices to set
$b:=r(0)^{-}$ and $f(x)=r(x^{\sim})^{-}$ for all $x\in M$. By Proposition
3.3(iv), conditions (i)–(iii) hold.

Now, let $M$ be isomorphic to $[0,b]$. Consider the isomorphism $g:M\to[0,b]$.
By the first part and the note right after Definition 3.1,
$r(x)=g(x^{-})^{\sim}=g(x)\oplus b^{-}$ is a square root on $M$. Let $x,y\in
M$.

$\displaystyle r(x\odot y)$ $\displaystyle=$ $\displaystyle f(y^{-}\oplus
x^{-})^{\sim}=f(y^{-}\oplus x^{-})^{\sim}=\big{(}(f(y^{-})\oplus
f(x^{-}))\wedge b\big{)}^{\sim}\mbox{ by the assumption}$ $\displaystyle=$
$\displaystyle\big{(}(f(y)^{-}\odot b)\oplus(f(x)^{-}\odot
b)\big{)}^{\sim}\vee b^{\sim}=\left(f(x)^{-}\odot
b\right)^{\sim}\odot\left(f(y)^{-}\odot b\right)^{\sim}\vee b^{\sim}$
$\displaystyle=$ $\displaystyle(r(y)\odot r(y))\vee b\mbox{ by (2) and (3)}.$

Therefore, $M$ satisfies the condition $r(x)\odot r(y)\leq r(x\odot y)$ for
all $x,y\in M$. ∎

######  Corollary 3.4.

Let $M$ be an MV-algebra. Then $M$ has a square root if and only if there
exists $b\in M$ such that $b^{-}\leq b$, $M$ is $f$-isomorphic to the MV-
algebra $[0,b]$ for some isomorphism $f:M\to[0,b]$, and for all $x\in M$,
$f(x)\oplus f(x)=x$.

###### Proof.

It follows from Theorem 3.3. Note that if $r$ is a square root on $M$, then
for each $x,y\in M$, $r(x)\odot r(y)\odot r(x)\odot r(y)=r(x)\odot r(x)\odot
r(y)\odot r(y)=x\odot y$ and so by (Sq2), $r(x)\odot r(y)\leq r(x\odot y)$. ∎

##  4\. Square roots and two-divisibility

In [DvZa3], we studied the existence of a square root on a pseudo MV-algebra
$M=\Gamma(G,u)$, where the unital $\ell$-group $(G,u)$ is two-divisible. In
this section, we study this question in more detail, in particular, we
characterize square roots on strongly $(H,1)$-perfect pseudo MV-algebras.

Note that if $(G;+,0)$ is an $\ell$-group and $x,y\in G$ such that
$(x+y)/2=x/2+y/2$, then $(x/2+y/2)+(x/2+y/2)=x+y=(x/2+x/2)+(y/2+y/2)$ and so
$y/2+x/2=x/2+y/2$. It yields that $x+y=y+x$. Hence, if $G$ is two-divisible
that enjoys unique extraction of roots, then $G$ satisfies the identity
$(x+y)/2=x/2+y/2$ if and only if $G$ is Abelian.

More generally, if $G$ is a two-divisible $\ell$-group that enjoys unique
extraction of roots and satisfies the following inequality

$x/2+y/2\leq(x+y)/2,\quad x,y\in G,$

then $x/2=((x-y)+y)/2\geq(x-y)/2+y/2$ and so $x/2-y/2\geq(x-y)/2$ for all
$x,y\in G$. Substituting $y$ by $-y$ in the last inequality, we get
$x/2+y/2=x/2-(-y)/2\geq(x+y)/2$ for all $x,y\in M$. That is,

$x/2+y/2=(x+y)/2,\quad x,y\in G.$

In a similar way, $x/2+y/2\geq(x+y)/2$ for all $x,y\in G$ implies that
$x/2+y/2=(x+y)/2$ for all $x,y\in G$.

######  Theorem 4.1.

Let a pseudo MV-algebra $M=\Gamma(G,u)$ be the direct product of linearly
ordered pseudo MV-algebras and let $M$ be with strict square root. Then $G$ is
two-divisible.

###### Proof.

Let $r$ be a strict root on $M$. We note that by [DvZa3, Thm 5.6], $M$ is
symmetric, $u/2$ exists and belongs to the center of $\mathrm{C}(G)$ of $G$,
and by [DvZa3, Thm 5.2], $r(x)=(x+u)/2$ for each $x\in M$. Since $M$ is
representable, so is $G$, and by [AnFe, Page 26], $G$ enjoys unique extraction
of roots.

For each $x\in G$ if $x/2$ exists, then
$(x/2+u/2)+(x/2+u/2)=x/2+x/2+u/2+u/2=x+u$ which implies that
$(x+u)/2=x/2+u/2$. Similarly, $(x+nu)/2=x/2+nu/2$. Also, $-(x/2)-(x/2)=-x$
implies that $(-x)/2$ exists and is equal to $-x/2$. Consequently,
$(x-u)/2=x/2-u/2$.

(I) Let us first assume $M$ is linearly ordered; then so is $G$. We show that
for every $g\in G$, the element $g/2$ is defined in $G$.

(1) If $g\in[0,u]$, then $g\in M$, $r(g)=(g+u)/2$ is defined in $[0,u]$, and
$(g+u)/2-u/2=(g+u-u)/2=g/2\in G$.

(2) Assume $g\in G^{+}$ is such that $nu\leq g\leq(n+1)u$ for some integer
$n\geq 0$. From $0\leq g-nu\leq u$ and (1), it follows that $(g-nu)/2$ exists.
Hence, $(g-nu)/2+n(u/2)=(g-nu)/2+nu/2=g/2$.

(3) If $g\in G^{-}$, then $-g\in G^{+}$ and $(-g)/2=-(g/2)$ is defined in $G$,
consequently $g/2$ exists in $G$ and is unique.

Summarizing (1)–(3), $G$ is two-divisible.

(II) Now, let $M=\prod_{i\in I}M_{i}$, where each $M_{i}$ is a linearly
ordered pseudo MV-algebra. Then, for each $i\in I$,
$M_{i}\cong\Gamma(G_{i},u_{i})$ with a linearly ordered unital $\ell$-group
$(G_{i},u_{i})$. Without loss of generality, we can assume
$M_{i}=\Gamma(G_{i},u_{i})$. Define $r_{i}:M_{i}\to M_{i}$ by
$r_{i}(x_{i})=\pi_{i}(r(x))$, $x_{i}\in M_{i}$ and $x=(x_{j})_{j}$, where
$\pi_{i}$ is the $i$-th projection from $M$ onto $M_{i}$. By [DvZa3, Prop
3.9], $r_{i}$ is a strict square root on $M_{i}$. Due to [DvZa3, Thm 5.2],
$u_{i}/2,u_{i}\in\mathrm{C}(G_{i})$ for each $i\in I$. By part (I), every
$G_{i}$ is two-divisible.

We describe the unital $\ell$-group $(G,u)$: Put $u=(u_{i})_{i}$, then

$G=\\{g=(g_{i})_{i}\in\prod_{i\in I}G_{i}\mid\exists n\in\mathbb{N}:-nu\leq
g\leq nu\\}.$

Let $g\in G^{+}$, then $g=(g_{i})_{i}$, where each $g_{i}\geq 0$. Therefore,
$g_{i}/2\geq 0$ exists in $G_{i}$ for each $i\in I$, and $g/2=(g_{i}/2)_{i}$
exists in $G$ while $0\leq g/2\leq g\leq nu$. Now take an arbitrary $g\in G$.
There is an integer $n\in\mathbb{N}$ such that $-nu\leq g\leq nu$. Then
$g+nu\geq 0$ so that $(g+nu)/2$ is defined. Therefore, $(g+nu)/2-n(u/2)=(g+nu-
nu)/2=g/2$ is defined in $G$. ∎

######  Corollary 4.2.

Let $M=\Gamma(G,u)$ be a representable pseudo MV-algebra with strict square
root. Then $(G,u)$ can be embedded into a two-divisible unital $\ell$-group.

###### Proof.

Let $r$ be a strict square root on $M$. Let $M_{i}=\Gamma(G_{i},u_{i})$ for
each $i\in I$ be a linearly ordered pseudo MV-algebra, and $f:M\to
M_{0}:=\prod_{i\in I}M_{i}$ be a subdirect embedding. Without loss of
generality, we can assume that each $M_{i}$ is non-degenerate. By [DvZa3, Prop
3.9], $r_{i}:M_{i}\to M_{i}$, defined by $r_{i}(\pi_{i}\circ
f(x))=\pi_{i}\circ f(r(x))$ for all $x\in M$, is a square root on $M_{i}$.
Since $M_{i}$ is a chain, by [DvZa3, Cor 4.5 and 5.7(3)], $r_{i}$ is strict or
$|M_{i}|=2$. If $|M_{i}|=2$, then $M_{i}=\\{0,1\\}$, so by [DvZa3, Thm 3.8],
$r_{i}(0_{i})=0_{i}$. On the other hand, since $r$ is strict we have
$r_{i}(0_{i})=r_{i}(\pi_{i}\circ f(0))=\pi_{i}\circ f(r(0))=\pi_{i}\circ
f(r(0)^{-})=(\pi_{i}\circ f(r(0)))^{-}=r_{i}(0_{i})^{-}=1_{i}$, which means
$M_{i}$ is degenerate, that is absurd. So, $|M_{i}|\neq 2$ for all $i\in I$.
It follows from Theorem 4.1, that for each $i\in I$, $G_{i}$ is two-divisible.
If we define $G_{0}$ by

$G_{0}=\\{g=(g_{i})_{i}\in\prod_{i\in I}G_{i}\mid\exists
n\in\mathbb{N}:-nu_{0}\leq g\leq nu_{0}\\},$ (4.1)

where $u_{0}=(u_{i})_{i}$, then $(G_{0},u_{0})$ is a two-divisible unital
$\ell$-group in which $(G,u)$ can be embedded. ∎

Theorem 4.1 entails the following question:

######  Problem 4.3.

Does Theorem 4.1 hold if $M=\Gamma(G,u)$ is a subdirect product of linearly
ordered pseudo MV-algebras?

Now, we show that Theorem 4.1 holds for every MV-algebra with strict square
root.

######  Theorem 4.4.

Let $M=\Gamma(G,u)$ be an MV-algebra with a strict square root $r$, where
$(G,u)$ is a unital $\ell$-group. Then $G$ is two-divisible.

###### Proof.

By [DvZa3, Thm 5.2], $r(x)=(x+u)/2$ for each $x\in M$. So,
$r(u-x)=(u-x+u)/2=x/2$ which means $x/2$ exists for all $x\in M$. Now, let
$g\in G^{+}$. Then there exists $n\in\mathbb{N}$ such that $g\leq nu$. The
Riesz interpolation property (see [Dar, Thm 1.3.11]) implies that
$g=\sum_{i=1}^{n}x_{i}$ where $0\leq x_{i}\leq u$. By the assumption,
$x_{i}/2\in G$ for all $i\in\\{1,2,\ldots,n\\}$. It follows that
$g=\sum_{i=1}^{n}x_{i}=2(\sum_{i=1}^{n}(x_{i}/2))$, that is
$g/2=\sum_{i=1}^{n}(x_{i}/2)\in G$. Now, choose an arbitrary element $g\in G$.
Then $g=g^{+}-g^{-}$. Since $g^{+},g^{-}\in G^{+}$, the elements
$(g^{+})/2,(g^{-})/2$ exist. It follows that
$g=g^{+}-g^{-}=(g^{+})/2-(g^{-})/2+(g^{+})/2-(g^{-})/2$. Hence, $g/2$ exists,
and $G$ is two-divisible. ∎

######  Corollary 4.5.

Let $M=\Gamma(G,u)$ be an MV-algebra with a square root $r$, where $(G,u)$ is
a unital $\ell$-group. Then $G$ is isomorphic to the direct product of unital
$\ell$-groups $(G_{1},u_{1})$ and $(G_{2},u_{2})$, where $G_{1}$ is a
subdirect product of copies of $(\mathbb{Z},1)$, and $G_{2}$ is a two-
divisible $\ell$-group.

###### Proof.

By [Höl, Thm 2.21], $M\cong M_{1}\times M_{2}$, where $M_{1}$ is a Boolean
algebra and $M_{2}$ is an MV-algebra with a strict square root $s:M_{2}\to
M_{2}$. Let $M_{i}=\Gamma(G_{i},u_{i})$, for $i=1,2$. Since $M_{2}$ is strict,
by Theorem 4.4, $G_{2}$ is two-divisible. If $X$ is the set of all prime ideal
of $M_{1}$, then $f:M_{1}\to\prod_{P\in X}M_{1}/P$, defined by
$f(x)=(x/P)_{P\in X}$, is a subdirect embedding. Clearly, $\prod_{P\in
X}M_{1}/P\cong\prod_{P\in X}\\{0,1\\}=\prod_{P\in X}\Gamma(\mathbb{Z},1)$.
Therefore, $G_{1}$ can be embedded in $\prod_{P\in X}\mathbb{Z}$. ∎

In the following example, we show that Theorem 4.1 does not hold for MV-
algebras with non-strict square root.

######  Example 4.6.

Let $B=\\{0,1\\}$ with $0<1$ be a two-element Boolean algebra. By [Höl,
DvZa3], $r:=\mbox{\rm Id}_{B}$ is a square root on $B$ that is not strict.
Then $M=\Gamma(\mathbb{Z},1)$, where $(\mathbb{Z},1)$ is the unital
$\ell$-group of integers with the strong unit $1$. Clearly, $\mathbb{Z}$ is
not two-divisible.

######  Proposition 4.7.

Let $M=\Gamma(G,u)$ be a representable pseudo MV-algebra with a square root
$r$. If $G$ is two-divisible, then $r$ is strict.

Conversely, if $M$ is the direct product of linearly ordered pseudo MV-
algebras, then $G$ is two-divisible.

###### Proof.

Let $G$ be two-divisible. We claim that $w=r(0)^{-}\odot r(0)^{-}=0$. By
Proposition 2.4(11), $w\in\mathrm{B}(M)$ and by the proof of Case 3 in [DvZa3,
Thm 4.3], $([0,w];\oplus,^{-_{w}},^{\sim_{w}},0,w)$ is a Boolean algebra, so
$[0,w]\subseteq\mathrm{B}(M)$. Choose $b\in[0,w]$. Since $G$ is two-divisible,
$b/2\in G$ exists, and $G$ is a representable unital $\ell$-group (since $M$
is representable), so $G$ enjoys unique extraction of roots and $0\leq b/2\leq
b\leq u$. It follows that $b/2\in[0,w]$ and so $b/2=b/2\oplus
b/2=(b/2+b/2)\wedge u=b\wedge u=b$. Consequently, $b=b/2=0$. Therefore, $w=0$.
Applying the proof of Case 2 in [DvZa3, Thm 4.3], we have $r$ is strict.

The converse follows from Theorem 4.1. ∎

######  Corollary 4.8.

Let a pseudo MV-algebra $M=\Gamma(G,u)$ be a direct product of linearly
ordered pseudo MV-algebras. The following statements are equivalent:

  * (i)

The pseudo MV-algebra $M$ has a strict square root.

  * (ii)

The pseudo MV-algebra $M$ is two-divisible and $u/2\in\mathrm{C}(G)$.

  * (iii)

The $\ell$-group $G$ is two-divisible and $u/2\in\mathrm{C}(G)$.

In either case, $(x+u)/2$ is defined in $M$ for each $x\in M$, and
$r(x)=(x+u)/2$, $x\in M$, is a strict square root on $M$.

###### Proof.

(i) $\Rightarrow$ (ii). It was established in [DvZa3, Thm 5.6].

(ii) $\Rightarrow$ (i). If $x\in M$, then $x/2$ and $u/2$ are defined in $M$,
therefore, $(x+u)/2=(x/2)+(u/2)$ exists in $M$, and the mapping
$r(x)=(x+u)/2$, $x\in M$, is, in fact, a strict square root on $M$.

(i) $\Rightarrow$ (iii). By the equivalence (i) and (ii), we have
$u/2\in\mathrm{C}(G)$. Theorem 4.1 entails implication (i) $\Rightarrow$
(iii).

(iii) $\Rightarrow$ (i). The mapping $r(x)=(x+u)/2$, $x\in M$, is a strict
square root on $M$; see [DvZa3, Ex 3.7(iii)]. ∎

The following result is a partial answer to the question posed in Problem 4.3
above.

We note that a pseudo MV-algebra $M=\Gamma(G,u)$ is said to be dense in $G$,
if for each $g\in G^{+}$, there exists $x\in M$ and $n\in\mathbb{N}$ such that
$g=nx$.

######  Theorem 4.9.

Let $M=\Gamma(G,u)$ be a representable pseudo MV-algebra, where $(G,u)$ is a
unital $\ell$-group. The following statements are equivalent:

  * (i)

The $\ell$-group $G$ is two-divisible and $u/2\in\mathrm{C}(G)$.

  * (ii)

The pseudo MV-algebra $M$ is dense in $G$ and $M$ has a strict square root.

###### Proof.

Let $g\in G$ and take the positive and negative parts of $g$: $g^{+}:=g\vee 0$
and $g^{-}:=-(g\wedge 0)=-g\vee 0$. Then $g^{+}\wedge g^{-}=0$ so that
$g^{+}+g^{-}=g^{-}+g^{+}$, and $g+(-g\vee 0)=0\vee g=(-g\vee 0)+g$. That is,
$g=g^{+}-g^{-}=-g^{-}+g^{+}$.

(i) $\Rightarrow$ (ii). Suppose that $G$ is two-divisible and
$u/2\in\mathrm{C}(G)$. For each $g\in G^{+}$, there exists $n\in\mathbb{N}$
such that $g\leq nu\leq 2^{n}u$. It follows that $0\leq g/2^{n}\leq u$. Hence
$M$ is dense in $G$. Consider the mapping $r:M\to M$ defined by
$r(x)=(x+u)/2$. By [DvZa3, Exm 3.7(iii)], $r$ is a square root on $M$. In
addition, $r(0)=u/2=u-r(0)=r(0)^{-}$, whence $r$ is strict.

(ii) $\Rightarrow$ (i). Assume that $M$ is dense in $G$ and $r:M\to M$ is a
strict square root on $M$. There exist $m,n\in\mathbb{N}$ such that $g^{+}=nx$
and $g^{-}=my$ for some $x,y\in M$. Since $r$ is strict, by the first step in
the proof of Theorem 4.1, $x/2,y/2\in M$. We have
$g^{+}=nx=n(x/2+x/2)=n(x/2)+n(x/2)$ and $g^{-}=my=m(y/2+y/2)=m(y/2)+m(y/2)$.
The elements $c=g^{+}/2$ and $d=g^{-}/2$ exist in $G$.

On the other hand, since $M$ is representable, $G$ also is representable. Let
$h:M\to\prod_{i\in I}M_{i}$ be a subdirect embedding, where $\\{M_{i}\mid i\in
I\\}$ is a family of linearly ordered pseudo MV-algebras. Suppose that
$M_{i}=\Gamma(G_{i},u_{i})$ for all $i\in I$, where $G_{i}$ an $\ell$-group
with the strong unit $u_{i}$. Then $\\{(G_{i},u_{i})\mid i\in I\\}$ is a
family of linearly ordered unital $\ell$-groups, and let $f:G\to\prod_{i\in
I}G_{i}$ be a subdirect embedding.

We have $f(g^{+})=f(g)\vee f(0)=f(g)\vee(0)_{i\in I}$ and $f(g^{-})=-f(g)\vee
f(0)=-f(g)\vee(0)_{i\in I}$. Suppose that $f(g)=(g_{i})_{i\in I}$,
$f(g^{+})=(g^{+}_{i})_{i\in I}$, $f(g^{-})=(g^{-}_{i})_{i\in I}$,
$f(c)=(c_{i})_{i\in I}$ and $f(d)=(d_{i})_{i\in I}$. The linearity of $G_{i}$
gives $g^{+}_{i}\neq 0\Leftrightarrow g_{i}>0\Leftrightarrow g^{-}_{i}=0$ and
$g^{-}_{i}\neq 0\Leftrightarrow g_{i}<0\Leftrightarrow g_{i}^{+}=0$. Since
$2c_{i}=g^{+}_{i}$, $2d_{i}=g^{-}_{i}$ and $G_{i}$ is a chain, $c_{i}\neq
0\Leftrightarrow g^{+}_{i}\neq 0$ and $d_{i}\neq 0\Leftrightarrow
g^{-}_{i}\neq 0$ for all $i\in I$. Hence, $c_{i}\neq 0\Leftrightarrow
d_{i}=0$, which yields $c_{i}+d_{i}=d_{i}+c_{i}$ and therefore, we have
$c_{i}-d_{i}=-d_{i}+c_{i}$. It follows that $f(c-d)=f(c)-f(d)=(c_{i})_{i\in
I}-(d_{i})_{i\in I}=-(d_{i})_{i\in I}+(c_{i})_{i\in I}=-f(d)+f(c)=f(-d+c)$
consequently, $c-d=-d+c$. Whence $2(c-d)=2c-2d=g^{+}-g^{-}=g$. That is, $g/2$
exists in $G$. Therefore, $G$ is two-divisible. ∎

In [DvZa3, Thm 5.6], we showed that if $M=\Gamma(G,u)$ is a representable
pseudo MV-algebra with a strict square root $r$, then $u/2$ exists, belongs to
$\mathrm{C}(G)$, and $r(0)=u/2$. A similar result for the general case is as
follows:

######  Proposition 4.10.

Let $r$ be an arbitrary square root on a representable pseudo MV-algebra
$M=\Gamma(G,u)$ and $w=r(0)^{-}\odot r(0)^{-}$. Then $(u-w)/2$ exists, is
equal to $r(0)$, and for all $x\in M$, $x+(u-w)/2=(u-w)/2+x$. In particular,
$(u-w)/2=r(0)=(-w+u)/2\in\mathrm{C}(G)$.

###### Proof.

Since $w\in\mathrm{B}(M)$, $w\vee r(0)=w\oplus r(0)=(r(0)^{-}\odot
r(0)^{-})\oplus r(0)=r(0)^{-}\vee r(0)=r(0)^{-}$. Also, by Proposition
2.4(11), $w\odot r(0)=r(0)\odot w=0$. Thus, $w+r(0)=w\oplus r(0)=r(0)^{-}$
entails that $r(0)^{-}-r(0)=w$. Similarly, $r(0)+w=r(0)\oplus w=r(0)^{-}$.
From $u=r(0)^{-}+r(0)=w+r(0)+r(0)$ we get that $-w+u=2r(0)$ and so
$r(0)=(-w+u)/2$ exists. On the other hand, by [DvZa3, Lem 3.15],
$u=r(0)+r(0)^{\sim}=r(0)+r(0)^{-}=r(0)+r(0)+w$. Thus, $u-w=2r(0)$ and
$r(0)=(u-w)/2$. Consequently, $(u-w)/2=r(0)=(-w+u)/2$.

(I) First, assume that $M$ is a chain. Choose $x\in M$.

(1) If $x+(u-w)/2<u$, by [DvZa3, Prop 3.5(vi)], we have
$x+(u-w)/2=x\oplus(u-w)/2=x\oplus r(0)=r(0)\oplus x=(r(0)+x)\wedge u$. Since
$M$ is a chain, we have $r(0)+x\leq u$, and whence $x+(u-w)/2=(r(0)+x)\wedge
u=r(0)+x=(u-w)/2+x$.

(2) If $x+(u-w)/2=u$, then $x+r(0)=u$ and so
$x=u-r(0)=r(0)^{-}=r(0)^{\sim}=-r(0)+u=-((u-w)/2)+u$. It follows that
$(u-w)/2+x=u$.

(3) If $u<x+(u-w)/2$, then for $y=x-((u-w)/2)$, we have $y+(u-w)/2=x$, so
$0\leq y+(u-w)/2\leq u$. Hence, by the first parts of the proof,
$x=x-((u-w)/2)+(u-w)/2=y+(u-w)/2=(u-w)/2+y=(u-w)/2+x-((u-w)/2)$ which implies
that $x+(u-w)/2=((u-w)/2)+x$.

(II) Now, let $X:=X(M)$ be the set of all proper normal prime ideals of $M$.
Then $r_{P}:M/P\to M/P$ defined by $r_{P}(x/P)=r(x)/P$ ($x\in M$) is a square
root on $M/P$; see [DvZa3, Prop 3.9]. Given $P\in X$, let $\hat{P}$ be the
$\ell$-ideal of $G$ generated by $P$. Then, due to the representation theorem
of pseudo MV-algebras by unital $\ell$-groups, [Dvu1],
$M/P=\Gamma(G/\hat{P},u/P)$, and since $M/P$ is a chain, then $G/\hat{P}$ is a
linearly ordered group. The map $f:M\to\prod_{P\in X}M/P$, $f(x)=(x/P)_{P\in
X}$, is a subdirect embedding, and $\hat{f}:(G,u)\to\prod_{P\in
X}(G/\hat{P},u/P)$ is also a subdirect embedding; it is an extension of $f$.
Then $f(r(x))=(r_{P}(x/P))_{P\in X}$, so that $f(r(0))=(r(0)/P)_{P\in X}$ and
$f(w)=(w/P)_{P\in X}=(w_{P})_{P\in X}$, where $w_{P}=r_{P}(0/P)^{-}\odot
r_{P}(0/P)^{-}$. Due to part (I), we have
$(x/P)+_{P}(u/P-_{P}w_{P})/2=(u_{P}-_{P}w/P)/2+_{P}(x/P)$ for each $x\in M$,
where $+_{P}$ and $-_{P}$ are the addition and subtraction, respectively, in
$G/\hat{P}$. Therefore,

$\displaystyle\hat{f}(x+((u-w)/2))$ $\displaystyle=$
$\displaystyle\hat{f}(x)+\hat{f}((u-w)/2)$ $\displaystyle=$
$\displaystyle\big{(}x/P\big{)}_{P\in
X}+_{P}\big{(}(u_{P}-_{P}w/P)/2\big{)}_{P\in X}$ $\displaystyle=$
$\displaystyle\big{(}x/P+_{P}(u_{P}-_{P}w/P)/2\big{)}_{P\in X}$
$\displaystyle=$ $\displaystyle\big{(}(u/P-_{P}w/P)/2+_{P}x/P\big{)}_{P\in X}$
$\displaystyle=$
$\displaystyle\hat{f}((u-w)/2)+\hat{f}(x)=\hat{f}((u-w)/2)+x),$

giving $x+(u-w)/2=(u-w)/2+x$ for each $x\in M$ as stated.

Since, every $x\in G^{+}$ is of the form $x=x_{1}+\cdots+x_{n}$ for some
$x_{1},\ldots,x_{n}\in M$, we have $x+(u-w)/2=(u-w)/2+x$. The equality holds
for each $x\in G^{-}$. If $x=g\in G$, then the equality holds for both $g^{+}$
and $g^{-}$, and finally, for each $g\in G$. ∎

Consequently, the latter proposition says that if $r$ is a square root on a
representable pseudo MV-algebra $M=\Gamma(G,u)$, the element
$r(0)=(u-w)/2=(-w+u)/2\in\mathrm{C}(G)$. If $r$ is strict, equivalently,
$w=0$, then $r(0)=u/2\in\mathrm{C}(G)$ as it was established in [DvZa3, Thm
5.6], and Proposition 4.10 generalizes the situation known only for strict
square roots.

In [DvZa3, Prob 4.7], we proposed a question whether the class of pseudo MV-
algebras with square roots satisfying $r(x)\odot r(x)\leq r(x\odot y)$ for all
$x,y\in M$ is a proper subvariety of the variety of pseudo MV-algebras with
square roots. In the sequel, we give a partial answer to it.

######  Proposition 4.11.

Let $\mathcal{V}$ be the variety of pseudo MV-algebras with square roots
satisfying the inequality

$\displaystyle r(x)\odot r(y)\leq r(x\odot y).$ (4.2)

Then $\mathcal{V}$ properly contains the variety $\mathcal{W}$ of MV-algebras
with square roots. In addition, each representable symmetric pseudo MV-algebra
with square root is contained in $\mathcal{V}$.

###### Proof.

According to Proposition 2.4(8), we know that
$\mathcal{W}\subseteq\mathcal{V}$.

First, assume that $M$ is a linearly ordered pseudo MV-algebra with a square
root $r$. If $r(0)=0$, then $M$ is a Boolean algebra and $r$ is an identity
map, which means inequality (4.2) holds. Otherwise, by [DvZa3, Thms 5.1, 5.6],
$M$ is two-divisible and symmetric with a square root $r(x)=(x+u)/2$ for all
$x\in M$. We have

$\displaystyle r(x)\odot r(y)$ $\displaystyle=$
$\displaystyle((x+u)/2-u+(y+u)/2)\vee 0=(x+y)/2,$ $\displaystyle r(x\odot y)$
$\displaystyle=$ $\displaystyle(((x-u+y)\vee 0)+u)/2=((x+y)\vee
u)/2=(x+y)/2\vee u/2,\mbox{ since $M$ is a chain}$ $\displaystyle=$
$\displaystyle(r(x)\odot r(y))\vee r(0).$

Therefore, $M\in\mathcal{V}$ implies $\mathcal{W}$ is a proper subvariety of
$\mathcal{V}$.

Let $M$ be a symmetric representable pseudo MV-algebra with a square root $r$
and $X$ be the set of all normal prime ideals of $M$. Consider the subdirect
embedding $f:M\to\prod_{P\in X}M/P$ defined by $f(x)=(x/P)_{P\in X}$. By
[DvZa3, Prop 3.9], the onto homomorphism $\pi_{P}\circ f:M\to M/P$ induces a
square root $r_{P}:M/P\to M/P$ defined by $r_{P}(x/P)=r(x)/P$ for all $x\in
M$. By the first part, $r_{P}(x/P\odot y/P)=(r_{P}(x/P)\odot r_{P}(y/P))\vee
r_{P}(0/P)$. It follows that

$\displaystyle f\big{(}(r(x)\odot r(y))\vee r(0)\big{)}$ $\displaystyle=$
$\displaystyle\big{(}f(r(x))\odot f(r(y))\big{)}\vee
f(r(0))=\big{(}(\frac{r(x)}{P})_{P\in X}\odot(\frac{r(y)}{P})_{P\in
X}\big{)}\vee(\frac{r(0)}{P})_{P\in X}$ $\displaystyle=$
$\displaystyle\big{(}(r_{P}(\frac{x}{P}))_{P\in
X}\odot(r_{P}(\frac{y}{P}))_{P\in X}\big{)}\vee(r_{P}(\frac{0}{P}))_{P\in X}$
$\displaystyle=$ $\displaystyle\big{(}(r_{P}(x/P)\odot r_{P}(y/P))\vee
r_{P}(0/P)\big{)}_{P\in X}=\big{(}r_{P}(x/P\odot y/P)\big{)}_{P\in X}$
$\displaystyle=$ $\displaystyle\big{(}r(x\odot y)/P\big{)}_{P\in X}=f(r(x\odot
y)).$

So, $(r(x)\odot r(y))\vee r(0)=r(x\odot y)$, which entails that
$M\in\mathcal{V}$. According to Proposition 2.4(8), we know that
$\mathcal{W}\subseteq\mathcal{V}$. ∎

We present an example of a linearly ordered pseudo MV-algebra
$M\in\mathcal{V}\setminus\mathcal{W}$.

######  Example 4.12.

Let $\mathbb{Q}$ be the set of all rational numbers and
$G=\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}$.
Consider the following binary operation on $G$:

$\displaystyle(a,b,c,d)+(x,y,z,w)=(a+x,b+y,c+z,d+w+bz),\quad\forall(a,b,c,d),(x,y,z,w)\in
G.$ (4.3)

Similarly to [AnFe, Page 138, E41], we can show that $(G;+,(0,0,0,0))$ is a
linearly ordered group, where $-(a,b,c,d)=(-a,-b,-c,-d+bc)$. Clearly, $G$ is
not Abelian. The element $u=(1,0,0,0)$ is a strong unit of $G$: Indeed, if
$(a,b,c,d)\in G^{+}$, then for each integer $n>1+\max\\{|a|,|b|,|c|,|d|\\}$ we
have $nu=(n,0,0,0)>(a,b,c,d)$. In addition, $G$ is two-divisible: For each
$(a,b,c,d)$ consider the element $(a/2,b/2,c/2,(4d-bc)/8)\in G$. Then
$(a/2,b/2,c/2,(4d-bc)/8)+(a/2,b/2,c/2,(4d-bc)/8)=(a,b,c,(4d-bc)/4+bc/4)=(a,b,c,d)$.
On the other hand, $u/2=(1/2,0,0,0)$ and
$u/2+(a,b,c,d)=(1/2,0,0,0)+(a,b,c,d)=(1/2+a,b,c,d)=(a,b,c,d)+(1/2,0,0,0)=(a,b,c,d)+u/2$
which means $u/2\in\mathrm{C}(G)$ (consequently, $u\in\mathrm{C}(G)$). By
[DvZa3, Exm 3.7], $M=\Gamma(G,u)$ is a pseudo MV-algebra with a square root
$r$ defined by $r(x)=(x+u)/2$ for all $x\in M$. By Proposition 4.11,
$M\in\mathcal{V}$, but $M\notin\mathcal{W}$.

######  Lemma 4.13.

Let $r$ be a square root on a pseudo MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$
and $S\subseteq M$. The following properties hold:

  * (i)

If $\bigwedge S$ exists, then $\bigwedge r(S)$ exists and is equal to
$r(\bigwedge S)$.

  * (ii)

If $\bigvee S$ exists, then $\bigvee r(S)$ exists and is equal to $r(\bigvee
S)$.

###### Proof.

(i) By Proposition 2.4(2), for each $s\in S$, we have $r(\bigwedge S)\leq
r(s)$. Let $x\in M$ be a lower bound for $r(S)$. Then $x\leq r(s)$ for all
$s\in S$ and so $x\odot x\leq r(s)\odot r(s)=s$. It follows that $x\odot
x\leq\bigwedge S$, which implies that $x\leq r(\bigwedge S)$ (by (Sq2)).
Hence, $r(\bigwedge S)$ is the greatest lower bound of $r(S)$, that is,
$\bigwedge r(S)=r(\bigwedge S)$.

(ii) Clearly, $r(\bigvee S)$ is an upper bound for the set $r(S)$. Let $b\in
M$ be such that $r(s)\leq b$ for all $s\in S$. Then $b\in[r(0),1]$. According
to Proposition 2.4(11), we know that $r(M)=[r(0),1]$, so there exists $a\in M$
such that $r(a)=b$ which implies that $r(s)\leq r(a)$ for all $s\in S$. From
Proposition 2.4(7), we get that $r(a\wedge x)=r(a)\wedge r(x)=r(x)$, whence
$a\wedge x=x$ and so $s\leq a$ for all $s\in S$ (since $r$ is a one-to-one
ordered-preserving map). Hence, $\bigvee S\leq a$ and $r(\bigvee S)\leq
r(a)=b$. Therefore, $r(\bigvee S)$ is the least upper bound of the set $r(S)$.
∎

Recall that, for each Boolean element $a$ of a pseudo MV-algebra
$(M;\oplus,^{-},^{\sim},0,1)$, the set $[a,1]$ is closed under $\oplus$ and
$([a,1];\oplus_{a},^{-a},^{\sim a},a,1)$ is a pseudo MV-algebra, where
$x^{-a}=x^{-}\vee a$, $x^{\sim a}=x^{\sim}\vee a$, and $x\oplus_{a}y=x\oplus
y$, $x,y\in[a,1]$. In addition, for all $x,y\in[a,1]$, $a=a\odot a\leq x\odot
y$, so $x\odot_{a}y:=(x\odot y)\vee a=x\odot y$.

######  Proposition 4.14.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a pseudo MV-algebra with a square root
$r$ such that the element $a=\bigvee_{n\in\mathbb{N}}r^{n}(0)$ exists. Then
the pseudo MV-algebra $([a,1];\oplus_{a},^{-a},^{\sim a},a,1)$ is a Boolean
algebra.

###### Proof.

Set $a=\bigvee_{n\in\mathbb{N}}r^{n}(0)$. Since $r^{n}(0)\leq r^{n+1}(0)$ for
all $n\in\mathbb{N}$, by Lemma 4.13, we have
$r(a)=r(\bigvee_{n\in\mathbb{N}}r^{n}(0))=\bigvee_{n\in\mathbb{N}}r^{n+1}(0)=\bigvee_{n=2}^{\infty}r^{n}(0)=\bigvee_{n\in\mathbb{N}}r^{n}(0)=a$.
Thus $a=r(a)\odot r(a)=a\odot a$, that is $a\in\mathrm{B}(M)$, consequently by
the remark mentioned just before this proposition,
$([a,1];\oplus_{a},^{-a},^{\sim a},a,1)$ is a pseudo MV-algebra. Clearly,
$r([a,1])\subseteq[a,1]$, so by the note just before this proposition, $r$ is
a square root on the pseudo MV-algebra $[a,1]$. Now, $r(a)=a$, and [DvZa3, Thm
3.8] imply that $r=\mbox{\rm Id}_{[a,1]}$, so that
$([a,1];\oplus_{a},^{-a},^{\sim a},a,1)$ is a Boolean algebra. ∎

From Proposition 4.14, we get that if $(M;\oplus,^{-},^{\sim},0,1)$ is a
$\sigma$-complete pseudo MV-algebra with a square root $r$, then
$[\bigvee_{n\in\mathbb{N}}r^{n}(0),1]\subseteq\mathrm{B}(M)$.

For example, let $M_{1}=M_{2}=M_{3}=[0,1]$ be the MV-algebra of the unit real
interval and $M_{2}=M_{4}=\\{0,1\\}$. Consider the MV-algebra
$M=\prod_{i=1}^{5}M_{i}$. Define $r_{i}:M_{i}\to M_{i}$ by $r_{i}(x)=(x+1)/2$,
$i=1,2,3$ and $r_{2}=r_{4}=\mbox{\rm Id}_{\\{0,1\\}}$. The mapping
$r=(r_{1},r_{2},r_{3},r_{4},r_{5})$ is a square root on $M$. We have
$a:=\bigvee_{n\in\mathbb{N}}r^{n}(0)=(1,0,1,0,1)$ and in the MV-algebra $M$,
we have $[a,1]=\\{a,(1,1,1,0,1),(1,0,1,1,1),(1,1,1,1,1)\\}$, which is a
Boolean algebra.

Moreover, in a pseudo MV-algebra $M$ with a square root $r$, we have
$\bigvee_{n\in\mathbb{N}}r^{n}(0)=1$ if and only if $|U_{r}|=1$ where
$U_{r}:=\\{x\in M\mid r^{n}(0)\leq x,~{}\forall n\in\mathbb{N}\\}$. We note if
$I$ is a normal ideal of $M$ with a square root $r$, then $r/I:M/I\to M/I$,
defined by $r/I(x/I)=r(x)/I$, $x/I\in M/I$, is a square root on $M/I$, [DvZa3,
Cor 3.10].

In [DvZa3, Prop 5.5(vi)], we proved that if $r$ is a square root on a pseudo
MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$, then $r(0)\oplus x=x\oplus r(0)$ for
all $x\in M$. We show that $r(0)\odot x=x\odot r(0)$ for all $x\in M$.

######  Proposition 4.15.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a pseudo MV-algebra with a square root
$r$. Then $r(0)\odot x=x\odot r(0)$ for all $x\in M$.

###### Proof.

If $M$ is a Boolean algebra, then the proof is evident. Choose $x\in M$.

(i) If $M$ is strict, then by [DvZa3, Prop 3.5(vi)], $x\odot
r(0)=(r(0)^{-}\oplus x^{-})^{\sim}=(r(0)\oplus x^{-})^{\sim}=(x^{-}\oplus
r(0))^{\sim}=(x^{-}\oplus r(0)^{-})^{\sim}=r(0)\odot x$.

(ii) If $M$ is neither strict nor a Boolean algebra, then $v=r(0)^{-}\odot
r(0)^{-}\neq 0,1$. According to [DvZa3, Thm 4.3],
$M\cong[0,v]\times[0,v^{-}]$, where $[0,v]$ is a Boolean algebra and
$[0,v^{-}]$ is a strict pseudo MV-algebra. Let $r_{1}:[0,v]\to[0,v]$ and
$r_{2}:[0,v^{-}]\to[0,v^{-}]$ be square roots, where $r_{1}$ is the identity
map on $[0,v]$ and $r_{2}$ is strict. Define $s:M\to M$ by $s(x)=r_{1}(x\wedge
v)\vee r_{2}(x\wedge v^{-})$. Then $s(x)\odot s(x)=(r_{1}(x\wedge v)\vee
r_{2}(x\wedge v^{-}))\odot(r_{1}(x\wedge v)\vee r_{2}(x\wedge v^{-}))=(x\wedge
v)\vee\big{(}r_{1}(x\wedge v)\odot r_{2}(x\wedge v^{-})\big{)}\vee(x\wedge
v^{-})=(x\wedge v)\vee 0\vee(x\wedge v^{-})=x$.

In addition, if $y\in M$ such that $y\odot y\leq x$, then $((y\wedge
v)^{2}\vee(y\wedge v^{-})^{2})=((y\wedge v)\vee(y\wedge v^{-}))\odot((y\wedge
v)\vee(y\wedge v^{-}))\leq(x\wedge v)\vee(x\vee v^{-})$, it follows that
$(y\wedge v)^{2}\leq(x\wedge v)$ and $(y\wedge v^{-})^{2}\leq(x\wedge v^{-})$,
since $v\wedge v^{-}=0$. So, $y\wedge v\leq r_{1}(x\wedge v)$ and $z\wedge
v^{-}\leq r_{2}(x\wedge v^{-})$, consequently, $y=(y\wedge v)\vee(y\wedge
v^{-})\leq r_{1}(x\wedge v)\vee r_{2}(x\wedge v^{-})=s(x)$. Hence, $s$ is a
square root on $M$, so $s=r$. We have $x\odot r(0)=x\odot s(0)=((x\wedge
v)\vee(x\wedge v^{-}))\odot(r_{1}(0)\vee r_{2}(0))=(x\wedge v^{-})\odot
r_{2}(0)$. By (i), $(x\wedge v^{-})\odot r_{2}(0)=r_{2}(0)\odot(x\wedge
v^{-})$. In a similar way, $r_{2}(0)\odot(x\wedge v^{-})=r(0)\odot x$.
Therefore, $x\odot r(0)=(x\wedge v^{-})\odot r_{2}(0)=r_{2}(0)\odot(x\wedge
v^{-})=r(0)\odot x$. ∎

######  Corollary 4.16.

Let $x$ be an element of a pseudo MV-algebra $M$ with a square root $r$. The
following statements hold:

  * (i)

$r(x^{n})=r(x)^{n}\vee r(0)$ for all $n\in\mathbb{N}$.

  * (ii)

$\bigvee_{n\in\mathbb{N}}r(x^{n})$ exists if and only if
$\bigvee_{n\in\mathbb{N}}r(x)^{n}$ exists.

###### Proof.

(i) For $n=1$, the proof is clear, since $r(0)\leq r(x)$. For $n=2$, the
statement was proved in [DvZa3, Prop 3.3(10)]. Let for $2\leq n$ we have
$r(x^{n})=r(x)^{n}\vee r(0)$. By [DvZa3, Prop 3.3(10)],
$r(x^{n+1})\leq(r(x)\odot r(x^{n}))\vee r(0)$. Then,

$\displaystyle(r(x)\odot r(x^{n}))\odot(r(x)\odot r(x^{n}))$
$\displaystyle=r(x)\odot\big{(}(r(x)^{n}\vee r(0))\odot r(x)\big{)}\odot
r(x^{n})$ $\displaystyle=r(x)\odot\big{(}(r(x)^{n}\odot r(x))\vee(r(0)\odot
r(x))\big{)}\odot r(x^{n})$ $\displaystyle=r(x)\odot\big{(}(r(x)\odot
r(x)^{n})\vee(r(x)\odot r(0))\big{)}\odot r(x^{n})\mbox{, by Proposition
\ref{ns1}}$ $\displaystyle=r(x)\odot r(x)\odot(r(x)^{n}\vee r(0))\odot
r(x^{n})$ $\displaystyle=x\odot r(x^{n})\odot r(x^{n})=x^{n+1},$

so, by (Sq2), $r(x)\odot r(x^{n})\leq r(x^{n+1})$. Clearly, $r(0)\leq
r(x^{n+1})$, whence $(r(x)\odot r(x^{n}))\vee r(0)\leq r(x^{n+1})$. Therefore,

$\displaystyle r(x^{n+1})$ $\displaystyle=$ $\displaystyle(r(x)\odot
r(x^{n}))\vee r(0)=\Big{(}r(x)\odot\big{(}r(x)^{n}\vee r(0)\big{)}\Big{)}\vee
r(0)$ $\displaystyle=$ $\displaystyle r(x)^{n+1}\vee(r(x)\odot r(0))\vee r(0)$
$\displaystyle=$ $\displaystyle r(x)^{n+1}\vee r(0).$

Therefore, $r(0)\vee r(x)^{n}=r(x^{n})$ for all $n\in\mathbb{N}$

(ii) It follows from part (i). ∎

######  Theorem 4.17.

Let $M=\Gamma(G,u)$ be a semisimple MV-algebra with a square root $r$. Then
$r$ is strict if and only if $\bigvee_{n\in\mathbb{N}}r^{n}(0)=u$.

###### Proof.

The proof is clear for $|M|=1$, so we assume that $M$ is non-degenerate.
Suppose that $r$ is a strict square root on $M$ and $Y=\text{MaxI}(M)$ is the
set of all maximal ideals of $M$. Then $r(0)=u-r(0)$, and so $r(0)=u/2$. Since
$M$ is semisimple, $\bigcap Y=\\{0\\}$. Consider the natural embedding
$f:M\to\prod_{I\in Y}M/I$. For each $I\in Y$, the MV-algebra $M/I$ is
isomorphic to a subalgebra of the unit real interval $[0,1]$. In addition,
$|M/I|>2$ and $r_{I}:M/I\to M/I$ defined by $r_{I}(x/I)=r(x)/I$ is a strict
square root on the linearly ordered MV-algebra $M/I$. By [DvZa3, Thm 5.1], for
any $n\in\mathbb{N}$, $r_{I}^{n}(0/I)=(2^{n}-1)u/2^{n}$. Since $M/I$ is
isomorphic to a subalgebra of $[0,1]$, we have
$\bigvee_{n\in\mathbb{N}}r_{I}^{n}(0/I)=\bigvee_{n\in\mathbb{N}}(2^{n}-1)u/2^{n}=1$.
We show that $\bigvee_{n\in\mathbb{N}}r^{n}(0)=u$. Let $a\in M$ be an upper
bound for the set $\\{r^{n}(0)\mid n\in\mathbb{N}\\}$. We have

$\displaystyle(a/I)_{I\in Y}=f(a)$ $\displaystyle\geq$ $\displaystyle
f(r^{n}(0))=f((2^{n}-1)u/2^{n})=(((2^{n}-1)u/I)/2^{n})_{I\in Y}.$

Hence, $a/I=u/I$ for all $I\in Y$ and so $a=u$.

Conversely, let $\bigvee_{n\in\mathbb{N}}r^{n}(0)=u$. We claim that $r$ is
strict. If $B$ is a Boolean algebra, by [DvZa3, Thm 3.8], $r(0)=0$, so that
$u=\bigvee_{n\in\mathbb{N}}r^{n}(0)=0$, which is absurd. Otherwise, by [Höl,
Thm 2.21], $M\cong[0,v]\times[0,v^{\prime}]$, where $v=r(0)^{\prime}\odot
r(0)^{\prime}$, $[0,v]$ is a Boolean algebra and $[0,v^{\prime}]$ is a strict
MV-algebra. According [DvZa3, Thm 5.3], $r(x)=(x\wedge v)\vee((x\wedge
v^{\prime})+v^{\prime})/2$ for all $x\in M$ and so $r(0)\leq v^{\prime}$. In a
similar way, $r^{2}(0)=r(r(0))=(r(0)\wedge v)\vee(r(0)+v^{\prime})/2\leq
v^{\prime}$ and $r^{n}(0)\leq v^{\prime}$ for all $n\in\mathbb{N}$. From
$\bigvee_{n\in\mathbb{N}}r^{n}(0)=u$ we get $u\leq v^{\prime}$ which means
$u=v^{\prime}$. Therefore, $M$ is strict. ∎

######  Proposition 4.18.

Let $r$ be a square root on a representable pseudo MV-algebra
$(M;\oplus,^{-},^{\sim},0,1)$. Then

  * (i)

$I:=\\{x\in M\mid x\leq r^{n}(0)^{-},~{}\forall n\in\mathbb{N}\\}$ is an ideal
of $M$.

  * (ii)

If $M$ is linearly ordered, then $I$ is a normal and maximal ideal.

  * (iii)

If $I$ is a normal ideal of $M$, then $|U_{r/I}|=1$.

###### Proof.

(i) Since $r(0)\leq r^{2}(0)\leq\cdots\leq r^{n}(0)\leq\cdots$, we have
$r(0)^{-}\geq r^{2}(0)^{-}\geq\cdots\geq r^{n}(0)^{-}\geq\cdots$. Clearly,
$x\leq y\in I$ implies that $x\in I$. Let $x,y\in I$. Then $x,y\leq
r^{n}(0)^{-}$ for all $n\in\mathbb{N}$. Choose $m\in\mathbb{N}$. $x\oplus
y\leq r^{m+1}(0)^{-}\oplus r^{m+1}(0)^{-}=(r^{m+1}(0)\odot
r^{m+1}(0))^{-}=r^{m}(0)^{-}$. Thus, $x\oplus y\in I$. Therefore, $I$ is an
ideal of $M$.

(ii) If $|M|=2$, then $M=\\{0,1\\}$, $r=\mbox{\rm Id}_{M}$, and clearly, $I$
is normal. Let $3\leq|M|$ and $M=\Gamma(G,u)$, where $(G,u)$ is a unital
$\ell$-group. By [DvZa3, Thm 4.3], $r$ is strict. [DvZa3, Thm 5.6] implies
that $u/2$ exists, $u/2,u\in\mathrm{C}(G)$, and $M$ is symmetric. Moreover,
$r(x)=(x+u)/2$ for each $x\in M$ (by [DvZa3, Thm 4.3]). Since for each $x\in
M$, $r(x^{-})^{-}=u-(u+(u-x))/2=x/2$, then $r(0)^{-}=u/2$,
$r^{2}(0)^{-}=r(r(0)^{-})^{-}=u/4,\ldots,r^{n+1}(0)^{-}=r((u/2^{n})^{-})^{-}=u/2^{n+1}$.
Set $H:=I\cup-I$. Then $H$ is an $\ell$-ideal (a normal convex
$\ell$-subgroup) of $G$.

(1) If $0\geq x\geq y\in-I$, then by (i), $0\leq-x\leq-y\in I$ and so
$x\in-I$. Now, since $G$ is linearly ordered, by (i) $H=I\cup-I$ is convex.

(2) Let $x,y\in I$. We can assume $x\leq y$. Then $x+y\leq y+y$. If $u\leq
y+y$, then $r(0)^{-}=u/2\leq y$, which contradicts with $y\in I$. Thus $y+y<u$
and so $y+y=y\oplus y\in I\subseteq H$. If $x,y\in-I$, then $-x,-y\in I$, so
$-(y+x)=-x+-y\in I$ and consequently $y+x\in-I\subseteq H$.

(3) Let $x\in I$ and $y\in-I$. If $0\leq x+y$, then $0\leq x+y\leq x$ and (1)
imply that $x+y\in I\subseteq H$. Otherwise, $y+x\leq 0$, then similarly by
(1), $y\leq y+x\leq 0$ implies that $y+x\in-I\subseteq H$.

(4) Let $g\in G$ and $x\in I$. We claim that $g+x-g\in I$. If $g+x-g>u$, then
$x\geq-g+u+g=u-g+g=u$ implies that $x\notin G^{+}\setminus I$. Clearly, $0\leq
g+x-g$ (since $g+x-g<0$ implies that $x\leq 0$ and so $x=0$). Consequently,
$g+x-g=0$. Thus $0\leq g+x-g\leq u$. If $g+x-g\notin H$, there exists
$n\in\mathbb{N}$ such that $r^{n}(0)^{-}\leq g+x-g$. It follows that
$g+2^{n}x-g=2^{n}(g+x-g)\geq 2^{n}r^{n}(0)^{-}=2^{n}(u/2^{n})=u$ and so
$2^{n}x\geq-g+u+g=u$, which contradicts with $x\in I$. Similarly, we can show
that for each $g\in G$ and $x\in-I$, $g+x-g\in-I$. Hence, $H$ is a normal
ideal (convex $\ell$-subgroup) of $G$, and so is its corresponding ideal in
the pseudo MV-algebra $M$, that is, $I=H\cap M$ is a normal ideal of $M$.

On the other hand, if $J$ is an ideal of $M$ properly containing $I$, then for
each $x\in J\setminus I$, there exists $n\in\mathbb{N}$ such that
$r^{n}(0)^{-}<x$. It follows from the first step of this part that
$u/2^{n}\leq x$ and so $u\in J$, which means $J=M$. Therefore, $I$ is a
maximal ideal of $M$.

(iii) Let $I$ be normal. Consider the square root $r/I:M/I\to M/I$ defined by
$r/I(x/I)=r(x)/I$. Choose $x/I\in U_{r/I}$. Then $r_{r/I}^{n}(0/I)\leq x/I$
for all $n\in\mathbb{N}$ and so $r^{n}(0)/I\leq x/I$ for all $n\in\mathbb{N}$.
It follows that $r^{n}(0)\odot x^{-}\in I$. For each $n\in\mathbb{N}$ we have
$r^{n}(0)^{-}\vee x^{-}=r^{n}(0)^{-}\oplus(r^{n}(0)\odot x^{-})\in I$, which
means $x^{-}\in I$. Hence $x/I=1/I$, that is $|U_{r/I}|=1$. ∎

From the note right before Proposition 4.18, it follows that if $M$ is a
linearly pseudo MV-algebra with a square root $r$, then
$\bigvee\\{r^{n}(0)/I\mid n\in\mathbb{N}\\}=1/I$. In addition, Proposition
4.14 implies that if $\bigvee\\{r^{n}(0)\mid n\in\mathbb{N}\\}$ exists, then
it is equal to $1$.

In the sequel, we characterize strongly $(H,1)$-perfect MV-algebras with
square roots.

######  Proposition 4.19.

Let $M=\Gamma(H\,\overrightarrow{\times}\,G,(1,0))$ be a strongly
$(H,1)$-perfect pseudo MV-algebra with a square root $r$. Then $M$ satisfies
precisely one of the following statements:

  * (i)

If $H\cong\mathbb{Z}$, then $|G|=1$ and $r(0)=0$. The converse holds, too.

  * (ii)

If $r$ is strict, then $H$ is two-divisible. The converse holds, too.

###### Proof.

Suppose that $M$ has a square root $r$. By [Dvu4, Thm 3.4], $I:=\\{0\\}\times
G^{+}$ is a normal and maximal ideal of $M$, and $M/I\cong\Gamma(H,1)$. Due to
[DvZa3, Prop 3.9], there is a square root $t:M/I\to M/I$ defined by
$t(x/I)=r(x)/I$ for all $x\in M$. Consequently, $\Gamma(H,1)$ has a square
root $s$. Since $\Gamma(H,1)$ is linearly ordered and symmetric, [DvZa3, Thm
4.5] implies $|\Gamma(H,1)|=2$ or $t$ is strict.

(i) If $|\Gamma(H,1)|=2$ or equivalently $H\cong\mathbb{Z}$, then
$M=(\\{0\\}\times G^{+})\cup(\\{1\\}\times G^{-})$. Let $r((0,0))=(a,b)$, then
$(0,0)=(a,b)\odot(a,b)=((2a,2b)-(1,0))\vee(0,0)=(2a-1,2b)\vee(0,0)$ and so
$a=0$. Clearly, for each $b\leq c\in G^{+}$ we have $(0,c)\odot(0,c)=(0,0)$,
whence by (Sq2), $(0,c)\leq r((0,0))=(0,b)$. Thus, $b$ is the top element of
$G$. In a similar way, we can show that $r((0,g))=(0,b)$, where $g\in G^{+}$.
Injectivity of $r$ implies that $|G|=1$. The proof of the converse is
straightforward because $|G|=1$ implies that $|M|=2$. Hence, $M$ is a Boolean
algebra, and so $\mbox{\rm Id}_{M}$ is the only square root on $M$.

(ii) Let $r$ be strict. By (i), $H$ is not isomorphic to $\mathbb{Z}$, so
$2<|\Gamma(H,1)|$, and $s$ is strict. Due to Theorem 4.1, $H$ is two-
divisible. Conversely, let $H$ be two-divisible. Then $1/2\in H$. Thus
$(1/2,0)\in M$. Clearly, $(1/2,0)\odot(1/2,0)=(0,0)$. For each $(a,b)\in M$,
$(a,b)\odot(a,b)\leq(0,0)$ implies that $(a,b)-(1,0)+(a,b)\leq(0,0)$
consequently, $(2a,2b)=2(a,b)\leq(1,0)=2(1/2,0)$. If $2a<1$, then clearly
$(a,b)\leq(1/2,0)$. If $2a=1$, then $a=1/2$ and $2b\leq 0$. From [Dar, Prop
3.6], it follows that $b\leq 0$ and so $(a,b)\leq(1/2,0)$. That is,
$r((0,0))=(1/2,0)$. Therefore, $r$ is strict. ∎

######  Theorem 4.20.

Let $M=\Gamma(H\,\overrightarrow{\times}\,G,(1,0))$ be a strongly
$(H,1)$-perfect pseudo MV-algebra such that $G$ enjoys unique extraction of
roots and $2<|M|$. If $M$ has a square root, then $H$ and $G$ are two-
divisible. In general, if $G$ does not enjoy unique extraction of roots, then
for each $g\in G$, the set $\\{x\in G\mid 2x=g\\}$ has a top element in $G$.

###### Proof.

Suppose that $r$ is a square root on $M$. Since $2<|M|$, by Proposition 4.19,
$r$ is strict and $H$ is two-divisible, so $1/2\in H$. Choose $g\in G$. Then
$(1/2,g)\in M$. Let $r((1/2,g))=(a,b)\in M$. Then

$\displaystyle(1/2,g)=(a,b)\odot(a,b)=\big{(}(a,b)-(1,0)+(a,b)\big{)}\vee(0,0)=(2a-1,2b)\vee(0,0).$

If $2a-1<0$, then $(2a-1,2b)\vee(0,0)=(0,0)\neq(1/2,g)$ and if $2a-1=0$, then
$(2a-1,2b)\vee(0,0)=(0,2b)\neq(1/2,g)$. It follows that $0<2a-1$ and so
$(2a-1,2b)\vee(0,0)=(2a-1,2b)$ that implies $2a-1=1/2$ and $2b=g$. That is
$a=3/4$, $2b=g$, and $G$ is two-divisible. Since $G$ enjoys unique extraction
of roots, we have $b=g/2$ and so $r((1/2,g))=(3/4,g/2)$.

Now, let $G$ not enjoy the unique extraction of roots. Choose $g\in G$. By the
first step of the proof, we know that $\\{x\in G\mid 2x=g\\}\neq\emptyset$ and
$r(1/2,g)=(3/4,d)$ for some $d\in G$ with $2d=g$. If $x\in G$ such that
$2x=g$, then $(3/4,x)\odot(3/4,x)=(1/2,2x)=(1/2,g)$, so $(3/4,x)\leq(3/4,d)$
which implies that $x\leq d$. Therefore, $d=\max\\{x\in G\mid 2x=g\\}$. ∎

######  Corollary 4.21.

Let $M=\Gamma(H\,\overrightarrow{\times}\,G,(1,0))$ be a strongly
$(H,1)$-perfect pseudo MV-algebra with a square root $r$, and let $G$ enjoy
unique extraction of roots. Then

$\displaystyle
r((x,y))=\big{(}\frac{x+1}{2},\frac{y}{2}\big{)},\quad\forall(x,y)\in
M\setminus\\{(a,b)\in M\mid a\neq 0\\}.$ (4.4)

###### Proof.

Choose $(x,y)\in M$. Let $r((x,y))=(a,b)$. Then $(2a-1,2b)\vee(0,0)=(x,y)$.

(i) If $x=0=y$ then there are two cases, $2a-1<0$ or $2a-1=0=2b$. By
Proposition 4.19, $r((x,y))=(1/2,0)=((x+1)/2,y/2)$.

(ii) If $x=0$ and $0<y$. Then $(x,y)=(2a-1,2b)\vee(0,0)=(0,2b\vee 0)$. So,
$2a-1=x=0$ and $2b\vee 0=y$, which means $r((x,y))=((x+1)/2,y/2)=(1/2,y/2)$.

$2a-1=0$, $2b\geq 0$ and $(2a-1,2b)\vee(0,0)=(2a-1,2b)$. It follows that
$a=(x+1)/2$ and $b=y/2$. That is, $r(x,y)=((x+1)/2,y/2)$.

(iii) If $x>0$, then $(2a-1,2b)=(2a-1,2b)\vee(0,0)=(x,y)$ and so $a=(x+1)/2$
and $b=y/2$. That is, $r(x,y)=((x+1)/2,y/2)$. ∎

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