# Representation and Embedding of Pseudo MV-algebras with Square Roots II.
Closures

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, the main aim is to 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 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 present 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

## Introduction

In this paper, we continue with the study from [DvZa4], where we have studied
square roots on pseudo MV-algebras. Square roots on pseudo MV-algebra were
generalized from ones on MV-algebras in [DvZa3]. Motivation for studying
square roots and strict square roots and their basic properties can be found
in [DvZa3] together with the necessary notions and definitions. In this part,
sections, theorems, propositions, lemmas, examples, and equations are numbered
in continuation of [DvZa4].

##  5\. Embedding and strict square root closure

In this section, we study the possibility of embedding a pseudo MV-algebra $M$
into a pseudo MV-algebra with strict square root. Moreover, we introduce a
strict square root closure of a pseudo MV-algebra.

First, we recall some notes about divisible (pseudo) MV-algebras. Let
$M=\Gamma(G,u)$ be a pseudo MV-algebra, where $(G,u)$ is a unital
$\ell$-group.

###  5.1. Commutative case

Every MV-algebra $M=\Gamma(G,u)$ can be embedded into a divisible one: Let
$G^{d}$ be the group-theoretic divisible hull of the Abelian group $G$. Then
$G^{d}$ can be converted into an $\ell$-group assuming that an element $h\in
G^{d}$ is positive in $G^{d}$ if $nh\in G^{+}$ for some integer $n\geq 1$ (see
[AnFe, P. 23]). If $G^{d}$ is the divisible hull of the $\ell$-group $G$, then
$G^{d}$ is an $\ell$-group with strong unit $u$, and we use $M^{d}$ to denote
the MV-algebra $\Gamma(G^{d},u)$. If $M$ is a linearly ordered MV-algebra,
then $M^{d}$ is linearly ordered, too (see [DiLe, Cor 1.4.26], [Fuc, Lem
IV.5.A]). For more information about divisible MV-algebras, we suggest to see
[DiLe, DiSe, DvRi, LaLe].

######  Theorem 5.1.

[DvZa2, Prop 5.2] Every MV-algebra can be embedded in an MV-algebra with
square root.

###  5.2. Non-commutative case

For the non-commutative case, the situation is different. Holland, [Hol],
showed that every $\ell$-group could be embedded in a divisible $\ell$-group.
He proved that every $\ell$-group $G$ is an $\ell$-subgroup of $A(\Omega)$ for
some suitable totally ordered set $\Omega$, where $A(\Omega)$ is the set of
all order-preserving bijective maps on $\Omega$. But, if $G$ has a strong
unit, there is no guarantee that the divisible $\ell$-group $A(\Omega)$ has a
strong unit.

On the other side, let $M=\Gamma(G,u)$ be an arbitrary pseudo MV-algebra. Take
the divisible $\ell$-group $A(\Omega)$ in which $G$ can be embedded; for
simplicity, we can assume $G\subseteq A(\Omega)$. Set $G(u)=\\{g\in
A(\Omega)\mid g\leq nu$ for some integer $n\geq 1\\}$. Then $G(u)=\\{g\in
A(\Omega)\mid|g|\leq mu$ for some integer $m\geq 1\\}$ and $(G(u),u)$ is a
unital $\ell$-group. Let $g\in\Gamma(G(u),u)$. Given $n\geq 1$, there exists
$h\in A(\Omega)$ such that $nh=g$, so that $nh\geq 0$ which implies $h\geq 0$
(see [Dar, Prop 3.6]), and $0\leq h\leq nh\leq u$. Therefore,
$h\in\Gamma(G(u),u)$, proving that $M=\Gamma(G,u)$ can be embedded into a
divisible pseudo MV-algebra $\Gamma(G(u),u)$.

Using similar reasoning, it is possible to show that also $G(u)$ is divisible:
Let $g\in G(u)$, there is an integer $m\geq 1$ such that $|g|\leq mu$. Given
$n\geq 1$, there is an element $h\in A(\Omega)$ such that $nh=g$. Then
$|h|\leq n|h|=|nh|=|g|\leq mu$, implying $h\in G(u)$. Consequently, every
unital $\ell$-group $(G,u)$ can be embedded into a divisible unital
$\ell$-group. We note that there is no guarantee that $G(u)$ enjoys unique
extraction of roots.

In the sequel, we find some classes of unital $\ell$-groups that can be
embedded into divisible (two-divisible) unital $\ell$-groups.

The class of $\ell$-groups that can be embedded into a representable two-
divisible $\ell$-group will be denoted by $\mathcal{RDG}$. The class
$\mathcal{RDG}$ contains the class of MV-algebras. This class is closed under
direct products and subalgebras. In Example 5.4, we will see some
$\ell$-groups belonging to $\mathcal{RDG}$.

Recall that, in [DvZa3], we found some results about symmetric representing
pseudo MV-algebras with square root with the additional condition
$u/2\in\mathrm{C}(G)$. The following lemma shows that the assumption
$u/2\in\mathrm{C}(G)$ is superfluous. In each symmetric representing pseudo
MV-algebra $M=\Gamma(G,u)$ with square root, if $u/2$ exists, it always
belongs to the center of $G$.

######  Lemma 5.2.

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

  * (i)

If $M$ has a square root, then $M$ is symmetric. In addition, if $u/2$ is
defined, then $u/2\in\mathrm{C}(G)$.

  * (ii)

If $G$ is two-divisible and $u\in\mathrm{C}(G)$, then $r:M\to M$ defined by
$r(x)=(x-u)/2+u$ is a strict square root on $M$. Moreover,
$u/2\in\mathrm{C}(G)$.

###### Proof.

(i) Let $r:M\to M$ be a square root on $M$. Consider the subdirect embedding
$f:M\to\prod_{i\in I}M_{i}$, where each $M_{i}$ is linearly ordered. For each
$i\in I$, the onto homomorphism $\pi_{i}\circ f:M\to M_{i}$ and [DvZa3, Prop
3.9] imply that $M_{i}$ has a square root $r_{i}$ induced by $\pi_{i}\circ f$
and by $r$, where $\pi_{i}$ is the $i$-th natural projection map. If
$|M_{i}|=2$, then clearly $M_{i}$ is symmetric. Otherwise, the only Boolean
elements of $M_{i}$ are $0$ and $1$, so by Proposition 2.4(11)(b),
$r_{i}(0)^{-}\odot r_{i}(0)^{-}=0$ or $r_{i}(0)^{-}\odot r_{i}(0)^{-}=1$. The
equality $r_{i}(0)^{-}\odot r_{i}(0)^{-}=0$ implies that $r_{i}(0)^{-}\leq
r_{i}(0)$ consequently, $r_{i}(0)^{-}=r_{i}(0)$ ([DvZa3, (4.1)]) which means
$r_{i}$ is strict and so $M_{i}$ is symmetric by [DvZa3, Thm 5.6]. If
$r_{i}(0)^{-}\odot r_{i}(0)^{-}=1$, then $r_{i}(0)=0$. It follows from [DvZa3,
Thm 3.8] that $M_{i}$ is a Boolean algebra which is absurd (since $M_{i}$ is a
chain with more than $2$ elements). Hence, for each $i\in I$, the pseudo MV-
algebra $M_{i}$ is symmetric; consequently, $\prod_{i\in I}M_{i}$ is
symmetric, too. Therefore, $M$ is symmetric.

Let $u/2$ be defined. Assume that $M$ is a chain. We claim that
$u/2\in\mathrm{C}(G)$. Let $g\in G$. Without loss of generality, we assume
that $u/2+g\leq g+u/2$. Then $g+u=u+g=u/2+u/2+g\leq u/2+g+u/2$ and so
$g+u/2\leq u/2+g$. That is, $u/2+g=g+u/2$ which means $u/2\in\mathrm{C}(G)$.

Now, let $M$ be representable; then so is the unital $\ell$-group $(G,u)$. Let
$f:G\to\prod_{i\in I}G_{i}$ be a subdirect embedding where
$\\{(G_{i},u_{i})\mid i\in I\\}$ is a family of linearly ordered unital
$\ell$-groups. On the other hand, each $M_{i}=\Gamma(G_{i},u_{i})$ is
symmetric, where $u_{i}=\pi_{i}(f(u))$. For each $i\in I$, we have
$\pi_{i}(f(u/2))=(\pi_{i}(f(u)))/2=u_{i}/2$, since
$\pi_{i}(f(u/2))+\pi_{i}(f(u/2))=\pi_{i}(f(u))$. By the first step of the
proof, $\pi_{i}(f(u/2))\in\mathrm{C}(G_{i})$ for each $i\in I$. Let $g\in G$.
Then
$\pi_{i}(f(u/2+g))=\pi_{i}(f(u/2))+\pi_{i}(f(g))=\pi_{i}(f(g))+\pi_{i}(f(u/2))$
for all $i\in I$ and so $f(u/2+g)=f(g+u/2)$ which entails that $u/2+g=g+u/2$,
since $f$ is one-to-one. Therefore, $u/2\in\mathrm{C}(G)$.

(ii) Let $x,y\in M$. Then $r(x)\odot r(x)=((x-u)/2+u-u+(x-u)/2+u)\vee
0=(x-u+u)\vee 0=x$. Also, if $y\odot y\leq x$, then $y-u+y\leq(y-u+y)\vee
0\leq x$, so $2y-2u\leq x-u$, consequently $y-u\leq(x-u)/2$ and so
$y\leq(x-u)/2+u=r(x)$ (since $G$ is representable and $u\in\mathrm{C}(G)$). In
addition, $r(0)=(0-u)/2+u=u/2=u-r(0)=r(0)^{-}$. The mapping $r$ is a weak
strict square root and $r(0)=u/2$. Now, (i) implies that $u/2\in\mathrm{C}(G)$
and so $r(x)=(x-u)/2+u=(x+u)/2$. Therefore, $r$ is a strict square root; see
[DvZa3, Exm 3.7(iii)]. ∎

######  Theorem 5.3.

Let $(G;+,0)$ be an $\ell$-group belonging to $\mathcal{RDG}$ and $(H,u)$ be
an Abelian linearly ordered unital $\ell$-group. The pseudo MV-algebra
$M=\Gamma(H\,\overrightarrow{\times}\,G,(u,0))$ can be embedded in a pseudo
MV-algebra with strict square root.

###### Proof.

If $|M|=2$, then $M$ is a Boolean algebra and by [DvZa3, Thm 3.8], $\mbox{\rm
Id}_{M}$ is a square root on $M$. So, let $|M|\neq 2$. Consider the
$\ell$-group $H\,\overrightarrow{\times}\,G$. Since $H$ is a linearly ordered
Abelian $\ell$-group with strong unit $u$, there exists a linearly ordered
divisible $\ell$-group $H^{d}$ with the strong unit $w$ such that $(H,u)$ can
be embedded in $(H^{d},w)$ as a unital $\ell$-group. Let $f:H\to H^{d}$ be the
embedding with $f(u)=w$. By the assumptions, a representable divisible
$\ell$-group $D$ and an $\ell$-group embedding $g:G\to D$ exist. Consider the
lexicographic product $H^{d}\,\overrightarrow{\times}\,D$. The mapping
$\alpha:H\,\overrightarrow{\times}\,G\to H^{d}\,\overrightarrow{\times}\,D$
sending $(x,y)$ to $(f(x),g(y))$ is an $\ell$-group embedding. Clearly,
$H^{d}\,\overrightarrow{\times}\,D$ is a representable two-divisible
$\ell$-group with the strong unit $(w,0)$. For each $(x,y)\in
H^{d}\,\overrightarrow{\times}\,D$, $(x,y)/2=(x/2,y/2)$. Also,
$(w,0),(w,0)/2=(w/2,0)\in\mathrm{C}(H^{d}\,\overrightarrow{\times}\,D)$, so by
Lemma 5.2,
$r:\Gamma(H^{d}\,\overrightarrow{\times}\,D)\to\Gamma(H^{d}\,\overrightarrow{\times}\,D)$
defined by $r(x,y)=((x,y)-(w,0))/2+(w,0)$ is a strict square root on
$\Gamma(H^{d}\,\overrightarrow{\times}\,D)$.

On the other hand, a pseudo MV-algebra embedding
$\beta:\Gamma(H\,\overrightarrow{\times}\,G,(u,0))\to\Gamma(H^{d}\,\overrightarrow{\times}\,D,(w,0))$
can be induced by the unital $\ell$-group embedding
$\alpha:H\,\overrightarrow{\times}\,G\to H^{d}\,\overrightarrow{\times}\,D$.
Therefore, $M=\Gamma(H\,\overrightarrow{\times}\,G,(u,0))$ can be embedded in
a pseudo MV-algebra with square root. In addition, by Lemma 5.2,
$r:\Gamma(H^{d}\,\overrightarrow{\times}\,D,(w,0))\to\Gamma(H^{d}\,\overrightarrow{\times}\,D,(w,0))$
is strict, and so $M$ can be embedded in a pseudo MV-algebra with strict
square root. ∎

As a special case of Theorem 5.3, if $G$ is a representable divisible pseudo
MV-algebra, then clearly $M=\Gamma(H\,\overrightarrow{\times}\,G,(u,0))$ can
be embedded in a pseudo MV-algebra with strict square root.

######  Example 5.4.

(i) Take the non-Abelian $\ell$-group
$G_{1}:=\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}$
with a strong unit $u=(1,0,0)$, where $G_{1}$ is with the following operations
for all $(a,b,c),(x,y,z)\in G_{1}$:

$\displaystyle(a,b,c)+(x,y,z)=(a+x,b+y,c+z+ay),\ \ -(a,b,c,d)=(-a,-b,-c+ab).$

It is a linearly ordered and non-divisible $\ell$-group. It can be embedded
into linearly ordered divisible $\ell$-groups
$G_{2}=\mathbb{D}\,\overrightarrow{\times}\,\mathbb{D}\,\overrightarrow{\times}\,\mathbb{D}$,
$G_{3}=\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}$,
and
$G_{4}=\mathbb{R}\,\overrightarrow{\times}\,\mathbb{R}\,\overrightarrow{\times}\,\mathbb{R}$,
respectively, where the group operations are defined analogously as for
$G_{1}$, and $\mathbb{D}$ is the group of dyadic numbers in $\mathbb{R}$, i.e.
$\mathbb{D}=\\{m/2^{n}\mid m\in\mathbb{Z},n=0,1,\ldots\\}$.

Therefore, $G_{1}$ is an element of $\mathcal{RDG}$.

(ii) The non-Abelian $\ell$-group
$G_{5}:=\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}\,\overrightarrow{\times}\,\mathbb{Q}$
that was introduced in Example 4.11 is divisible and representable with a
strong unit. So, it belongs to $\mathcal{RDG}$.

(iii) By Theorem 5.3, if $H$ is a subgroup of $\mathbb{R}$, then
$H\,\overrightarrow{\times}\,G_{1}$ and $H\,\overrightarrow{\times}\,G_{2}$
are elements of $\mathcal{RDG}$.

(iv) A locally nilpotent group is a group whose finitely generated subgroups
are nilpotent. An $\ell$-group is locally nilpotent if it is locally nilpotent
as a group. According to Mal’cev (see [KoMe, Thm 7.3.2]), every locally
nilpotent linearly ordered group $G$ can be embedded in a divisible locally
nilpotent linearly ordered group. Therefore, every locally nilpotent linearly
ordered group $G$ belongs to $\mathcal{RDG}$. In addition, by Theorem 5.3,
$H\,\overrightarrow{\times}\,G\in\mathcal{RDG}$ for each Abelian unital
$\ell$-group $(H,1)$.

######  Remark 5.5.

Let $M=\Gamma(G,u)$ be an arbitrary linearly ordered pseudo MV-algebra. Then
$K:=\\{g\in G\mid n|g|<u,~{}\forall n\in\mathbb{N}\\}$ is the only maximal
convex $\ell$-subgroup of $G$; see [Dvu2, Prop 5.4, Thm 6.1]. Then
$I=K\cap[0,u]$ is a normal and maximal ideal of $M$. It follows that $M/I$ is
an MV-algebra which is isomorphic to a subalgebra of the MV-algebra of the
real unit interval $[0,1]$. We can show that $M$ is an $(H,1)$-perfect MV-
algebra, where $(H,1)=\Gamma^{-1}(M/I)\cong(G/H,u/H)$.

Now, using Theorem 5.3, we get that if $K\in\mathcal{RDG}$ and $M$ is a
strongly $(H,1)$-perfect MV-algebra, then $M$ can be embedded in a pseudo MV-
algebra with square root.

######  Theorem 5.6.

Let $M=\Gamma(G,u)$ be such that $G\in\mathcal{RDG}$ is a linearly ordered
$\ell$-group with strong unit $u$. The following statements hold:

  * (i)

$G$ can be embedded into a two-divisible linearly ordered group $A$ with a
strong unit. In addition, if $u\in\mathrm{C}(G)$, then $u\in\mathrm{C}(A)$.

  * (ii)

If $M$ is symmetric, then $M$ can be embedded in a linearly ordered pseudo MV-
algebra with strict square root.

###### Proof.

(i) Let $H$ be a representable two-divisible $\ell$-subgroup containing $G$ as
an $\ell$-subgroup. Then it enjoys unique extraction of roots. Set $G_{0}:=G$,
$G_{1}:=G_{0}/2+G_{0}/2$ and $G_{n+1}=G_{n}/2+G_{n}/2$, where $X/2:=\\{x/2\mid
x\in X\\}$ for each $X\subseteq H$. By the assumption, $G_{n}$’s are well-
defined. We have

$G=G_{0}\subseteq G_{1}\subseteq G_{2}\subseteq\cdots\subseteq G_{n}\subseteq
G_{n+1}\subseteq\cdots.$

Set $A:=\bigcup_{n\in\mathbb{N}}G_{n}$. For each $x,y\in A$ there exists
$n\in\mathbb{N}$ such that $x,y\in G_{n}$ and so $x+y\in G_{n+1}\subseteq A$
which means $A$ is a subgroup of $H$ containing $G$ (note that
$G=G_{0}\subseteq A$). Clearly, $A$ is two-divisible.

(1) Since $H$ is representable, $G/n$ is a chain, where $G/n=\\{x/n\mid x\in
G\\}$.

(2) We claim that for each $a\in A$, there exists $m\in\mathbb{N}$ such that
$|a|\leq mu$. Indeed, for $a\in G_{0}$, it is clear. If $a\in G_{1}$, then
$a=x/2+y/2$ for some $x,y\in G_{0}$ and so
$|a|=|x/2+y/2|\leq|x/2|+|y/2|+|x/2|+|y/2|$ (see [AnFe, Page 2, 1.1.3a]). By
(1), we can assume that $|x/2|\leq|y/2|$. It follows that $|a|\leq 4|y/2|$. If
$0\leq y/2$, then $|a|\leq 2y$. Choose $m\in\mathbb{N}$ such that $y\leq mu$,
then we have $|a|\leq 2mu$. If $y/2\leq 0$, then $|a|\leq 4(-y/2)=-2y$. Take
$m\in\mathbb{N}$ such that $-y\leq mu$, then we have $|a|\leq 2mu$. Let $1\leq
n$ and for each $a\in G_{n}$ there exists $m\in\mathbb{N}$ such that $|a|\leq
m$. Choose $a\in G_{n+1}$. Then $a=a_{1}+a_{2}$ where $a_{1},a_{2}\in G_{n}$.
By the assumption, there is $m\in\mathbb{N}$ such that $|a_{1}|,|a_{2}|\leq
mu$. It follows that $|a|=|a_{1}+a_{2}|\leq|a_{1}|+|a_{2}|+|a_{1}|\leq 3mu$.

(3) The group $A$ is linearly ordered. Since $H$ is representable, $A$ and $G$
are representable. Let $\varphi:A\to\prod_{i\in I}A_{i}$ be a subdirect
embedding of unital $\ell$-groups, where each $A_{i}$ is linearly ordered. For
each $i\in I$, set $B_{i}:=\pi_{i}\circ\varphi(G)$. Clearly,
$\varphi|_{G}:G\to\prod_{i\in I}B_{i}$ is a subdirect embedding. Since $G$ is
subdirectly irreducible (because $G$ is a chain), there exists $j\in I$ such
that $\pi_{j}\circ\varphi$ is an isomorphism, consequently,
$\pi_{j}\circ\varphi:G\to A_{j}$ is one-to-one. We claim that
$\pi_{j}\circ\varphi:A\to A_{j}$ is an isomorphism. Let $x\in A$ be such that
$\pi_{j}\circ\varphi(x)=0$. Then there exists $n\in\mathbb{N}$ such that $x\in
G_{n}$. We use induction on $n$. Clearly, if $x\in G_{0}=G$, then we get
$x=0$. If $x\in G_{1}$, then $x=x_{1}+x_{2}$ for some $x_{1},x_{2}\in G$.

$\displaystyle
0=\pi_{j}\circ\varphi(x)=\pi_{j}\circ\varphi(x_{1}+x_{2})=\pi_{j}\circ\varphi(x_{1})+\pi_{j}\circ\varphi(x_{2}),$

which implies $\pi_{j}\circ\varphi(x_{1})=-\pi_{j}\circ\varphi(x_{2})$ and so
$\pi_{j}\circ\varphi(2x_{1})=\pi_{j}\circ\varphi(-2x_{2})$. Since
$2x_{1},-2x_{2}\in G$, we have $2x_{1}=-2x_{2}$ and so $x_{1}=-x_{2}$ which
entails $x=0$. Now, let $1\leq n$ and each $\pi_{j}\circ\varphi$ be one-to-one
on $G_{n}$. As in the proof for $G_{1}$, we can show that
$\pi_{j}\circ\varphi$ is one-to-one on $G_{n+1}$. Hence
$\pi_{i}\circ\varphi:A\to A_{i}$ is an isomorphism, and consequently, $A_{i}$
is a chain. By (1)–(3), $G$ can be embedded in a linearly ordered two-
divisible $\ell$-group $A$ with strong unit.

Now, let $u\in\mathrm{C}(G)$. We will show that $u\in\mathrm{C}(A)$. First, we
show that for each $a\in G/2$, we have $a+u=u+a$. Otherwise, assume that
$a+u<u+a$ (because $A$ is a chain), then $a+a+u\leq a+u+a$. From
$u\in\mathrm{C}(G)$ and $2a\in G$, it follows that $u+2a=2a+u\leq a+u+a$ and
so $u+a\leq a+u$ which contradicts with the assumption. So, $u+a=a+u$ for all
$a\in G/2$ and consequently $a+u=u+a$ for all $a\in G_{1}$. We can easily
conclude that $a+u=u+a$ for all $a\in G_{1}/2$, which entails $u$ commutes
with all elements of $G_{2}$. By a similar way, $u$ commutes with all elements
of $A:=\bigcup_{n\in\mathbb{N}}G_{n}$.

(ii) By (i), there is an $\ell$-group embedding $f:G\to A$, where $A$ is a
linearly ordered two-divisible $\ell$-group with a strong unit $f(u)$. Then
$\Gamma(f):\Gamma(G,u)\to\Gamma(A,f(u))$ is an embedding of pseudo MV-
algebras. By Lemma 5.2, $r:\Gamma(A,f(u))\to\Gamma(A,f(u))$ defined by
$r(x)=(x-f(u))/2+f(u)$ is a strict square root. Therefore, the proof of part
(ii) is completed. ∎

######  Corollary 5.7.

Let $M=\Gamma(G,u)$ be a symmetric representable pseudo MV-algebra such that
$G\in\mathcal{RDG}$. Then $M$ can be embedded in a pseudo MV-algebra with
strict square root.

###### Proof.

Suppose that $(G,u)$ is a unital $\ell$-group and $f:G\to H$ is an
$\ell$-group embedding in a representable two-divisible $\ell$-group $H$. Let
$X:=X(M)$ be the set of all proper normal convex $\ell$-subgroup of $H$; we
can assume that $G\subseteq H$. Then $P\cap G$ ($P\cap M$) is a normal convex
$\ell$-subgroup of $G$ (normal ideal of the pseudo MV-algebra $M$) for every
$P\in X$. Clearly, $\bigcap_{P\in X}(P\cap M)=\\{0\\}$. Consider the
$\ell$-group embedding $f_{P}:M\to\prod_{P\in X}M/(P\cap M)$. For each $P\in
X$, $g_{P}:G/(P\cap G)\to H/P$ is an embedding, $H/P$ is two-divisible and
representable (it is a chain). In addition, for each $P\in X$, $G/(P\cap G)$
is a chain, hence by Theorem 5.6, $M/(P\cap M)$ can be embedded in a pseudo
MV-algebra $N_{P}$ with the strict square root $r_{P}$ (note that
$\Gamma(G/(P\cap G),u/(P\cap G))=M/(P\cap M)$). Let $g_{P}:M/(P\cap M)\to
N_{P}$ be the mentioned embedding for all $P\in X$. Consider the pseudo MV-
algebra $N:=\prod_{P\in X}N_{P}$ with the strict square root $r:N\to N$
defined by $r((x_{P})_{P\in X})=(r_{P}(x_{P}))_{P\in X}$. Then the mapping
$M\xrightarrow{\ \ \alpha\ \ }\prod_{P\in X}M/(P\cap
M)\xrightarrow{(g_{P})_{P\in X}}\prod_{P\in Y}N_{P}$ is an embedding of pseudo
MV-algebras. ∎

It was shown that if $M=\Gamma(G,u)$ is a symmetric pseudo MV-algebra with
$G\in\mathcal{RDG}$, then $M$ can be embedded into a pseudo MV-algebra with
strict square root. In other words, $M$ has a “cover” with strict square root.
Now, we find the least such “cover” for $M$.

######  Definition 5.8.

Let $M$ be a pseudo MV-algebra. A strict square root closure for $M$ is a
triple $(C,r,i)$, where $C$ is a pseudo MV-algebra, $r:C\to C$ is a strict
square root, and $i:M\to C$ is an embedding of pseudo MV-algebras such that if
$N$ is another pseudo MV-algebra with a strict square root $s$ and $j:M\to N$
is an injective homomorphism of pseudo MV-algebras, then there exists a unique
injective homomorphism $\alpha:C\to N$ satisfying the following conditions:

  * (C1)

$\alpha\circ i=j$;

  * (C2)

$\alpha(r(x))=s(\alpha(x))$ for all $x\in M$.

In the last definition, condition (C2) implies that $\alpha$ is a homomorphism
of pseudo MV-algebras with square root. In addition, if $(C,r,i)$ and
$(C^{\prime},r^{\prime},i^{\prime})$ are square root closures of $M$, then
$C\cong C^{\prime}$. Hence, $C$ will be denoted by $\mathbf{C}(M)$, which is
determined uniquely up to isomorphism. Sometimes, we need that $M,C,N$ belong
to a special class $\mathcal{V}$ of pseudo MV-algebras, e.g. the class of
linear pseudo MV-algebras, etc. If a pseudo MV-algebra $M$ has a strict square
root closure, $M$ has to be symmetric.

######  Lemma 5.9.

Let $(M_{1};\oplus,^{-},^{\sim},0,1)$ and $(M_{2};\oplus,^{-},^{\sim},0,1)$ be
pseudo MV-algebras such that $\mathbf{C}(M_{1})$ and $\mathbf{C}(M_{2})$
exist. The following statements hold:

  * (i)

$\mathbf{C}(M_{1})\cong\mathbf{C}(\mathbf{C}(M_{1}))$.

  * (ii)

If $f:M\to M_{2}$ is a one-to-one homomorphism of pseudo MV-algebras and
$\mathbf{C}(M_{1})$ exists, then $\mathbf{C}(M_{1})$ can be embedded into
$\mathbf{C}(M_{2})$.

###### Proof.

(i) Let $i_{1}:M\to\mathbf{C}(M_{1})$ and
$i_{2}:\mathbf{C}(M_{1})\to\mathbf{C}(\mathbf{C}(M_{1}))$ be the embeddings in
Definition 5.8 represented in the Figure 1.

$\mathbf{C}(M_{1})$$\mathbf{C}(\mathbf{C}(M_{1}))$$M_{1}$$\mathbf{C}(M_{1})$$\mathbf{C}(\mathbf{C}(M_{1}))$$i_{1}$$i_{2}$$i_{2}$$i_{1}$$\mbox{\rm
Id}_{\mathbf{C}(M_{1})}$$\alpha$$\beta$ Figure 1.  The diagram of strict
square root closures

There exist injective homomorphisms
$\alpha:\mathbf{C}(\mathbf{C}(M_{1}))\to\mathbf{C}(M_{1})$ and
$\beta:\mathbf{C}(M_{1})\to\mathbf{C}(\mathbf{C}(M_{1}))$ such that
$\alpha\circ i_{2}=\mbox{\rm Id}_{\mathbf{C}(M_{1})}$ and $\beta\circ
i_{1}=i_{2}\circ i_{1}$. It follows that
$\alpha\circ\beta:\mathbf{C}(M_{1})\to\mathbf{C}(M_{1})$ commutes the diagram,
that is, $\alpha\circ\beta\circ i_{1}=i_{1}$. On the other hand, $\mbox{\rm
Id}_{\mathbf{C}(M_{1})}\circ i_{1}=i_{1}$, so $\alpha\circ\beta=\mbox{\rm
Id}_{\mathbf{C}(M_{1})}$ which means $\alpha$ is onto. Therefore,
$\alpha:\mathbf{C}(\mathbf{C}(M_{1}))\to\mathbf{C}(M)$ is an isomorphism.

(ii) Let $i_{1}:M_{1}\to\mathbf{C}(M_{1})$ and
$i_{2}:M_{2}\to\mathbf{C}(M_{2})$ be the embeddings in Definition 5.8. Since
$i_{2}\circ f:M_{1}\to\mathbf{C}(M_{2})$ is an embedding, there exists a map
$\alpha:\mathbf{C}(M_{1})\to\mathbf{C}(M_{2})$ such that $\alpha\circ
i_{1}=i_{2}\circ f$. Since $\alpha$ is one-to-one, $\mathbf{C}(M_{1})$ can be
embedded into $\mathbf{C}(M_{2})$. ∎

######  Theorem 5.10.

Each MV-algebra has a strict square root closure.

###### Proof.

Let $M=\Gamma(G,u)$ be an MV-algebra and $i:G\to D$ be an $\ell$-group
embedding, where $D$ is an Abelian divisible unital $\ell$-group. We can
assume that $i$ is an inclusion map. Define $D_{0}=G$ and $D_{n}=D_{n-1}/2$
for all $n\in\mathbb{N}$. Note that since $D$ is representable, $D$ enjoys
unique extraction of roots. Assume that $H:=\bigcup_{i\in\mathbb{N}}D_{i}$.
Clearly, $G\subseteq H$ and $D_{n-1}\subseteq D_{n}$ for all $n\in\mathbb{N}$.
We assert that $\Gamma(H,u)$ is a strict square closure for $M$.

(1) Let $x,y\in G$. Then $x/2+y/2=(x+y)/2$ (because $D$ is commutative) and
$D_{1}$ is closed under $+$. Also, $2(x/2\wedge y/2)\leq x,y$, so $2(x/2\wedge
y/2)\leq x\wedge y$, that is $x/2\wedge y/2\leq(x\wedge y)/2$. On the other
hand, $x\wedge y\leq x,y$ implies that $(x\wedge y)/2\leq x/2,y/2$, so that
$(x\wedge y)/2\leq x/2\wedge y/2$. It follows that, $x/2\wedge y/2=(x\wedge
y)/2$. Similarly, $x/2\vee y/2=(x\vee y)/2$. Thus, $D_{1}$ is closed under
$\vee$ and $\wedge$. Similarly, we can prove that $D_{n}$ is closed under $+$,
$\vee$, and $\wedge$. Therefore, $H$ is an $\ell$-subgroup of $D$ which is
two-divisible. By Lemma 5.2, $r:\Gamma(H,u)\to\Gamma(H,u)$ defined by
$r(x)=(x+u)/2$ is a strict square root on $\Gamma(H,u)$ and
$i:\Gamma(G,u)\to\Gamma(H,u)$ is an embedding of MV-algebras.

(2) From (1), we conclude that for each $x\in H$, there exists
$n\in\mathbb{N}\cup\\{0\\}$ such that $2^{n}x\in G$. Let $n_{x}$ be the least
element $n$ of $\mathbb{N}\cup\\{0\\}$ such that $2^{n}x\in G$. For example,
if $x\in G$, then $n_{x}=0$, if $x\in D_{1}\setminus D_{0}$, then $n_{x}=1$,
and if $x\in D_{n}\setminus D_{n-1}$, then $n_{x}=n$.

(3) Let $N$ be an MV-algebra with a strict square root $s$, and $f:M\to N$ be
a homomorphism of MV-algebras. By Theorem 4.4, $(A,w):=\Gamma^{-1}(N)$ is a
two-divisible unital $\ell$-group with strong unit $w$. Consider the
$\ell$-group homomorphism $F:=\Gamma(f):G\to A$. Define $\alpha:H\to A$ by
$\alpha(x)=F(2^{m}x)/2^{m}$, where $2^{m}x\in G$. Let $x\in H$ and
$m\in\mathbb{N}$ such that $2^{m}x\in G$. Then $m=n_{x}p+q$ where
$p,q\in\mathbb{N}\cup\\{0\\}$ and $0\leq q<n_{x}$. By definition of $n_{x}$,
we have $q=0$ (otherwise, $2^{q}x\in G$ which is absurd). That is, $n_{x}|m$
and
$\alpha(x)=F(2^{m}x)/2^{m}=F(2^{n_{x}p}x)/2^{n_{x}p}=F(2^{n_{x}}x)/2^{n_{x}}$,
whence $F$ is well-defined. Also, $\alpha(0)=0$, $\alpha(u)=w$ and
$\alpha(i(g))=F(x)$ for all $g\in G$.

(4) Let $x,y\in H$ be given. Then there exist $m,n\in\mathbb{N}$ such that
$2^{m}x,2^{n}y\in G$ and so $2^{mn}x,2^{mn}y\in G$. Thus
$\alpha(x+y)=F(2^{mn}(x+y))/2^{mn}=F(2^{mn}x)/2^{mn}+F(2^{mn}y)/2^{mn}=F(2^{m}x)/2^{m}+F(2^{n}y)/2^{n}=\alpha(x)+\alpha(y)$
(since $F$ is a homomorphism and $G$, $H$ and $A$ enjoy unique extraction of
roots).

(5) If $x\leq y$, then $2^{mn}x\leq 2^{mn}y$ and so $F(2^{mn}x)\leq
F(2^{mn}y)$ consequently, $\alpha(x)=F(2^{mn}x)/2^{mn}\leq
F(2^{mn}y)/2^{mn}=\alpha(y)$. Hence, $\alpha:H\to A$ is ordered-preserving.
Moreover, if $M$, or equivalently $G$, is a chain, then $D$ and $H$ are also
chains. So, $\alpha$ is an $\ell$-group homomorphism.

(6) Similarly to (1), we can show that $x/2^{n}\wedge y/2^{n}=(x\wedge
y)/2^{n}$ for all $x,y\in G$, and for $2^{n}x,2^{n}y\in G$, we have $x\wedge
y=(2^{n}x\wedge 2^{n}y)/2^{n}$. Thus, $2^{n}x\wedge 2^{n}y=2^{n}(x\wedge y)$.
It follows that

$\displaystyle\alpha(x)\wedge\alpha(y)$ $\displaystyle=$
$\displaystyle\dfrac{F(2^{mn}x)}{2^{mn}}\wedge\dfrac{F(2^{mn}y)}{2^{mn}}=\dfrac{F(2^{mn}x)\wedge
F(2^{mn}y)}{2^{mn}}$ $\displaystyle=$ $\displaystyle\dfrac{F(2^{mn}x\wedge
2^{mn}y)}{2^{mn}}=\dfrac{F(2^{mn}(x\wedge y))}{2^{mn}}=\alpha(x\wedge y).$

(7) Analogously to (6), $\alpha$ preserves $\vee$.

(8) If $x\in D_{n}$, then
$\alpha(r(x))=\alpha((x+u)/2)=F(2^{n+1}(x+u)/2)/2^{n+1}=(F(2^{n}x)+F(2^{n}u))/2^{n+1}=F(2^{n}x)/2^{n+1}+w/2=(\alpha(x)+w)/2$.
Moreover, by Theorem 4.4, $A$ is two-divisible, so Lemma 5.2 implies that
$k:N\to N$, defined by $k(x)=(x+w)/2$, $x\in N$, is a square root on $N$,
consequently $k=s$ (because the square root on a pseudo MV-algebra is unique).
Hence, $s(\alpha(x))=(\alpha(x)+w)/2=\alpha(r(x))$.

Consequently, $\alpha$ is a one-to-one homomorphism of pseudo MV-algebras with
square roots. Summing up all the above cases, we conclude that $\Gamma(H,u)$
is a strict square closure for the MV-algebra $M$. ∎

The following notion will be used in the following statement. If $X$ is a
subset of an $\ell$-group $G$, then
$\sum_{i=1}^{n}X:=\\{x_{1}+\cdots+x_{n}\mid x_{1},\ldots,x_{n}\in X\\}$.

######  Corollary 5.11.

Let $M=\Gamma(G,u)$ be an MV-algebra, $\mathbf{C}(M)$ be its strict square
root closure, and assume $M\subseteq\mathbf{C}(M)$. For each
$x\in\mathbf{C}(M)$, there exists an integer $n\in\mathbb{N}$ such that
$2^{n}x\in\sum_{i=1}^{2^{n}}M$. That is,
$\mathbf{C}(M)=\bigcup_{n=0}^{\infty}(\sum_{i=1}^{2^{n}}M/2^{n})$.

###### Proof.

Consider the notations in the proof of Theorem 5.10. For each $n\in\mathbb{N}$
set $M_{n}:=\sum_{i=1}^{2^{n}}M/2^{n}$. According to Lemma 5.9,
$\mathbf{C}(M)$ is an MV-algebra with strict square root generated by $M$. By
Theorem 5.10, $\mathbf{C}(M)=\Gamma(H,u)$, where
$H=\bigcup_{i\in\mathbb{N}}D_{i}$. For each $y\in M$,
$y/2=u-(u-y+u)/2=r(y^{\prime})^{\prime}\in\mathbf{C}(M)$, where
$r(x)=(x+u)/2$, $x\in\mathbf{C}(M)$ is a (unique) strict square root on
$\mathbf{C}(M)$. It entails that $M/2\subseteq\mathbf{C}(M)$. Since $x,y\leq
u/2$ for all $x,y\in M/2$, we have $x+y=(x+y)\wedge u=x\oplus
y\in\mathbf{C}(M)$. That is, $M_{1}\subseteq\mathbf{C}(M)$. In a similar way,
we can show that $M_{n}\subseteq\mathbf{C}(M)$.

Now, let $x\in\mathbf{C}(M)$. If $x\in D_{0}=G$, then
$x\in\Gamma(G,u)=M/2^{0}$. If $x\in D_{1}$, then $x=g/2\leq u$ for some $g\in
D_{0}$, so $g\leq 2u$. Due to the Riesz Interpolation Property (RIP), there
exist $g_{1},g_{2}\in\Gamma(G,u)$ such that $g=g_{1}+g_{2}$, whence
$x=(g_{1}+g_{2})/2=g_{1}/2+g_{2}/2\in M/2+M/2$. On the other hand,
$g_{1}/2+g_{2}/2=g_{1}/2\oplus g_{2}/2\in M/2\oplus M/2$, where $\oplus$ is
the binary operation on the MV-algebra $\Gamma(H,u)$, since
$g_{1}/2,g_{2}/2\leq u/2$.

Let $0\leq n$. If $x\in D_{n}$, then $x=g/2^{n}$ for some $g\in G$, so
$2^{n}x=g\leq 2^{n}u$. Due to (RIP), there exist positive elements
$g_{1},\ldots,g_{2^{n}}\leq u$ of $G$ (that is, $g_{1},\ldots,g_{2^{n}}\in M$)
such that $g=\sum_{i=1}^{2^{n}}g_{i}$, whence
$x=g/2^{n}=(\sum_{i=1}^{2^{n}}g_{i})/2^{n}=\sum_{i=1}^{2^{n}}(g_{i}/2^{n})\in\sum_{i=1}^{2^{n}}M/2^{n}$.
Therefore,
$\mathbf{C}(M)\subseteq\bigcup_{n=0}^{\infty}(\sum_{i=1}^{2^{n}}M/2^{n})$, and
so, we have $\mathbf{C}(M)=\bigcup_{n=0}^{\infty}(\sum_{i=1}^{2^{n}}M/2^{n})$.
∎

The following example shows that there are non-commutative linearly ordered
symmetric pseudo MV-algebras without strict square roots that can be embedded
into a linearly ordered symmetric pseudo MV-algebra with strict square root
with some kind of minimality of the embedding.

######  Example 5.12.

Let
$G=\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}$,
$u=(1,0,0,0)$,
$H=\mathbb{D}\,\overrightarrow{\times}\,\mathbb{D}\,\overrightarrow{\times}\,\mathbb{D}\,\overrightarrow{\times}\,\mathbb{D}$,
where $\mathbb{D}$ is the group of dyadic numbers in $\mathbb{R}$, and let the
group operations on $G$ and $H$ be defined by (4.2). The groups $G$ and $H$
are linearly ordered, $u\in\mathrm{C}(G)$, $G$ is not two-divisible whereas
$H$ is two-divisible, and $G$ is a subgroup of $H$. The pseudo MV-algebras
$M=\Gamma(G,u)$ and $C=\Gamma(H,u)$ are symmetric and linearly ordered, $M$ is
not two-divisible, and $C$ is two-divisible. We assert that $\Gamma(H,u)$ is
the least two-divisible pseudo MV-algebra containing $\Gamma(G,u)$ and
contained in $\Gamma(H,u)$, moreover $\Gamma(H,u)$ is with strict square root.

Indeed, first, we show that $H$ is the least subgroup of $H$ that is two-
divisible and contains $G$. We have $\mathbb{D}=\\{m/2^{n}\mid
m\in\mathbb{Z},n\geq 0\\}$ and $\mathbb{D}$ is two-divisible. Let $H_{1}$ be a
two-divisible subgroup of $H$ containing $G$. Then
$\mathbb{D}\times\\{0\\}\times\\{0\\}\times\\{0\\},\\{0\\}\times\mathbb{D}\times\\{0\\}\times\\{0\\},\\{0\\}\times\\{0\\}\times\mathbb{D}\times\\{0\\},\\{0\\}\times\\{0\\}\times\\{0\\}\times\mathbb{D}\subseteq
H_{1}$. Take $(a,b,c,d)\in H$. Then
$(a,b,c,d)=(a,0,0,0)+(0,b,0,0)+(0,0,c,0)+(0,0,0,d-bc)\in H_{1}$, so that
$H_{1}=H$.

In Corollary 5.7, we showed that if $M=\Gamma(G,u)$ is a pseudo MV-algebra
such that $G\in\mathcal{RDG}$, then $M$ can be embedded into a pseudo MV-
algebra with strict square root. There is another way for embedding $M$ into a
pseudo MV-algebra with square root. We will prove that $M$ can be embedded in
a pseudo MV-algebra with square root, not necessarily strict.

######  Proposition 5.13.

Every symmetric pseudo MV-algebra $M=\Gamma(G,u)$, where $G\in\mathcal{RDG}$,
can be embedded into a symmetric pseudo MV-algebra with square root, not
necessarily strict.

###### Proof.

Let $M=\Gamma(G,u)$ be a symmetric pseudo MV-algebra such that
$G\in\mathcal{RDG}$. There exists an $\ell$-group embedding $h:G\to H$, where
$H$ is two-divisible and representable. Let $Y$ be the set of all normal prime
ideals of $H$. By [AnFe, Thm 4.1.1], the $\ell$-group $G$ is also
representable. Set $X=\\{h^{-1}(Q)\mid Q\in Y\\}$. Then each element of $X$ is
a normal convex $\ell$-subgroup of $G$ and $\bigcap_{P\in X}P=\\{0\\}$. It
follows that $f:G\to\prod_{P\in X}G/P$ defined by $f(g)=(g/P)_{P\in X}$ is a
one-to-one $\ell$-group homomorphism. Choose $f^{-1}(Q)=P\in X$. We know that
$\\{P\cap M\mid P\in X\\}$ is a set of normal prime ideals of $M$, and $M$ is
a subdirect product of the family $\\{M_{P}=M/(P\cap M)\mid P\in X\\}$ of
linearly ordered pseudo MV-algebras.

Every $M_{P}$ is one of the following three cases.

(i) If $|M_{P}|=1$, then we can remove $M_{P}$.

(ii) If $|M_{P}|=2$, then $M_{P}=\\{0/(P\cap M),1/(P\cap M)\\}$. We set
$r_{P}=\mbox{\rm Id}_{M_{P}}$ and $K_{P}=\Gamma(\mathbb{Z},1)$. Let
$g_{P}:M_{P}\to\Gamma(\mathbb{Z},1)$ be the only isomorphism between Boolean
algebras with two elements.

(iii) Suppose that $3\leq|M_{P}|$. We have $\Gamma(G/P,u/P)=M_{P}$. Since the
mapping $h_{P}:G/P\to H/Q$ sending $g/P$ to $h(g)/Q$ is an $\ell$-group
embedding and $H/Q$ is two-divisible, so $G/P$ is a linearly ordered group
belonging to $\mathcal{RDG}$. By the proof of Theorem 5.6(i), $A_{P}:=\bigcup
G_{P,n}$ is a two-divisible $\ell$-group, where $G_{P,0}=G_{P}$ and
$G_{P,n+1}=G_{P,n}/2+G_{P,n}/2$. Since $u\in\mathrm{C}(G)$, by that theorem
$(u/P)/2\in\mathrm{C}(A_{P})$. Lemma 5.2 implies that there exists a strict
square root $r_{P}$ on $K_{P}:=\Gamma(A_{P},u/P)$. Set $g_{P}=\Gamma(h_{P})$.

Now, define the pseudo MV-algebra $M_{0}:=\prod_{P\in X}K_{P}$, where $K_{P}$
is either $\Gamma(\mathbb{Z},1)$, if $|M_{P}|=2$ or $\Gamma(A_{P},u/P)$, if
$|M_{P}|>2$, and let $r:M_{0}\to M_{0}$ be the square root defined by
$r((x_{P})_{P\in X})=(r_{P}(x_{P}))_{P\in X}$. If all $|M_{P}|>2$, $r$ is a
strict square root on $M_{0}$, otherwise $r$ is not strict.

Consider the embedding $\varphi:M\xrightarrow{f_{M}}\prod_{P\in X}M/(P\cap
M)\xrightarrow{(g_{P})_{P\in X}}\prod_{P\in X}K_{P}=M_{0}$, where $f_{M}$ is
the restriction of $f$ onto $M$. Then $M$ can be embedded into a symmetric
pseudo MV-algebra $M_{0}$ with square root, not necessarily strict. ∎

A normal ideal $I$ of a pseudo MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$ is
called (i) a strict square ideal if $M/I$ is a pseudo MV-algebra with a strict
square root, and (ii) a Boolean ideal, if $M/I$ is a Boolean algebra
(equivalently, $x\wedge x^{-}\in I$ for all $x\in M$).

The following lemma will be useful for the proof of Proposition 5.15.

######  Lemma 5.14.

Let $I$ be a normal ideal of a pseudo MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$
with a square root $r$ and let $w=r(0)^{-}\odot r(0)^{-}$.

  * (i)

$I$ is a strict square ideal if and only if $w\in I$.

  * (ii)

If $J$ is a normal ideal of $M$ containing a strict square ideal $I$, then $J$
is a strict square ideal, too.

The intersection of each family of strict square ideals of $M$ is a strict
square ideal.

  * (iii)

The interval $[0,w]$ is the least square ideal of $M$.

###### Proof.

(i) Let $r$ be a square root on $M$. Due to [DvZa3, Prop 3.9], $r_{I}:M/I\to
M/I$ defined by $r_{I}(x/I)=r(x)/I$ is a square root on $M/I$.

$\displaystyle\mbox{ $I$ is strict }\Leftrightarrow r_{I}(0/I)^{-}\odot
r_{I}(0/I)^{-}=0/I\Leftrightarrow(r(0)^{-}\odot r(0)^{-})/I=0/I\Leftrightarrow
w\in I.$ (5.1)

(ii) The proof is a straight consequence of (i).

(iii) By [DvZa3, Prop 3.3(11)], $w$ is a Boolean element and by [DvZa3, Thm
4.3], $M\cong[0,w]\times[0,w^{-}]$ (the mapping $f:M\to[0,w]\times[0,w^{-}]$
defined by $f(x)=(x\wedge w,x\wedge w^{-})$ is the isomorphism). Therefore,
$[0,w]$ is a normal ideal of $M$, and $M/[0,w]$ is isomorphic to $[0,w^{-}]$
which is a strict pseudo MV-algebra. It entails that $[0,w]$ is a strict
square ideal. On the other hand, if $J$ is a strict square ideal of $M$, then
by (ii), $w\in J$ and so $[0,w]\subseteq J$. Therefore, $[0,w]$ is the least
square ideal of $M$. ∎

For each non-degenerate pseudo MV-algebra $M$, $X(M)$ denotes the set of all
proper normal prime ideals of $M$, $X(M)_{1}$ is the set of all Boolean normal
prime ideals of $M$ and $X(M)_{2}=X(M)\setminus X(M)_{1}$. In addition, we
define $I_{1}:=\bigcap X(M)_{1}$ and $I_{2}:=\bigcap X(M)_{2}$. If there is no
ambiguity, we will denote them simply by $X$, $X_{1}$, and $X_{2}$,
respectively.

######  Proposition 5.15.

Let $M=(M;\oplus,^{-},^{\sim},0,1)$ be a representable pseudo MV-algebra with
a square root $r$. The following statements hold:

  * (i)

If $r$ is strict, then $P$ is not a Boolean ideal, for each $P\in
X(M)\setminus\\{M\\}$.

  * (ii)

If $M\cong M_{1}\times M_{2}$ is the decomposition of $M$ that was mentioned
in [DvZa3, Thm 4.3], where $M_{1}$ is a Boolean algebra and $M_{2}$ is a
strict MV-algebra, then $M_{i}$ is isomorphic to a subalgebra of $\prod_{P\in
X_{i}}(M/P)$, for $i=1,2$.

  * (iii)

The ideal $M_{2}=\bigcap_{P\in X_{1}(M)}P$ is the least Boolean ideal and
$M_{1}=\bigcap_{P\in X_{2}(M)}P$ is the least strict square ideal of $M$. In
addition, $M$ is strict if and only if $\bigcap_{P\in X_{2}}P=\\{0\\}$.

###### Proof.

(i) Choose $P\in X(M)\setminus\\{M\\}$. Due to the onto homomorphism
$\pi_{P}:M\to M/P$ and [DvZa3, Prop 3.9], $r_{P}:M/P\to M/P$ defined by
$r_{P}(x/P)=r(x)/P$ is a square root, which is clearly strict, since $r$ is
strict. Hence $r_{P}(0/P)=r_{P}(0/P)^{-}$. If $M/P$ is a Boolean algebra, then
it has only two elements, that is, $M/P=\\{0/P,1/P\\}$ and by [DvZa3, Thm
3.8], $r_{p}(0/P)=0/P$, so $0/P=(0/P)^{-}=1/P$ which is absurd. Therefore,
$M/P$ is not a Boolean algebra.

(ii) According to [DvZa3, Thm 4.4], $M\cong M_{1}\times M_{2}$, where
$M_{1}=[0,w]$ is a Boolean algebra, $M_{2}=[0,w^{-}]$ is a strict MV-algebra,
and $w=r(0)^{-}\odot r(0)^{-}$. Set $E_{1}:=\\{(x_{P}/P)_{P\in
X(M)}\in\prod_{P\in X(M)}M/P\mid x_{P}=0,~{}\forall P\in X_{2}(M)\\}$ and
$E_{2}=\\{(x_{P}/P)_{P\in X(M)}\in\prod_{P\in X(M)}M/P\mid x_{P}=0,~{}\forall
P\in X_{1}(M)\\}$.

Let $a=(a_{P}/P)_{P\in X(M)}$, where $a_{P}=1$ if $P\in X_{1}(M)$ and
$a_{P}=0$ if $P\in X_{2}$. We define a mapping $g:\prod_{P\in X(M)}M/P\to
E_{1}\times E_{2}$ by $g(x)=(x\wedge a,x\wedge a^{\prime})$ for all
$x\in\prod_{P\in X(M)}M/P$. Then $g$ is an isomorphism. Let $f:M\to\prod_{P\in
X(M)}M/P$ be the natural subdirect embedding. Consider the embedding
$\varphi:=g\circ f$.

Case 1. If $P\in X_{2}(M)$, then $M/P$ is not a Boolean algebra, so
$r_{P}:M/P\to M/P$ defined by $r(x/P)=r(x)/P$ is strict, by [DvZa3, Cor 4.5],
and in other words, $P$ is a strict square ideal. We have $(r(0)^{-}\odot
r(0)^{-})/P=r_{P}(0/P)^{-}\odot r_{P}(0/P)^{-}=0/P$, that is, $w\in P$ and
$w^{-}/P=1/P$.

Case 2. If $P\in X_{1}(M)$, then $M/P$ is a Boolean algebra, so
$r_{P}(0/P)=0/P$. Thus $(r(0)^{-}\odot r(0)^{-})/P=r_{P}(0/P)^{\prime}\odot
r_{P}(0/P)=1/P$, so that $w/P=1/P$ and $w^{\prime}/P=0/P$.

Hence, $f(w)\in E_{1}$ and $f(w^{-})\in E_{2}$ yielding
$f(M_{1})=f([0,w])\subseteq E_{1}$ and $f(M_{2})=f([0,w^{-}])\subseteq E_{2}$.
The element $f(w)$ is a top element of $E_{1}$, and $f(w^{-})$ is a top
element of $E_{2}$. Therefore, $M_{1}$ is isomorphic to a subalgebra of
$E_{1}$, and $M_{2}$ is isomorphic to a subalgebra of $E_{2}$. In addition,
$M_{1}=\\{0\\}$ if and only if $w=0$ if and only if $E_{1}=\\{0\\}$.
Similarly, $M_{2}=\\{0\\}$ if and only if $w^{-}=0$ if and only if
$E_{2}=\\{0\\}$.

(iii) Clearly, $I=\bigcap_{P\in X_{1}(M)}P$ is a Boolean ideal of $M$ (it is
normal, too). If $J$ is a Boolean ideal of $M$, then $J=\bigcap\\{P\mid
J\subseteq P\\}$, since $M/J$ is representable. Each element $P$ of this set
is a Boolean ideal of $M$ (because it contains $J$). It follows that
$I\subseteq J$. Since $w^{-}\in\mathrm{B}(M)$, the set $M_{2}=[0,w^{-}]$ is an
ideal of $M$ which is normal (since $M\cong M_{1}\times M_{2}$ and
$\\{0\\}\times M_{2}$ is a normal ideal of $M_{1}\times M_{2}$). Also,
$M/M_{2}\cong M_{1}$ is a Boolean algebra, so $I\subseteq M_{2}$. On the other
hand, by the proof of part (ii), for each $P\in X_{1}(M)$, we have $w^{-}\in
P$, whence $M=[0,w^{\prime}]\subseteq I$. Therefore, $M_{2}=\bigcap_{P\in
X_{1}(M)}P$.

Let $K=\bigcap\\{P\mid P\in X_{2}(M)\\}$. According to (ii), each $P\in
X_{2}(M)$ is a strict square ideal, so by Lemma 5.14, $K$ is a strict square
ideal of $M$. Lemma 5.14 implies that $[0,w]\subseteq K$.

On the other hand, $[0,w]=\bigcap\\{P\in X(M)\mid[0,w]\subseteq
P\\}=\bigcap\\{P\in X(M)\mid w\in P\\}$, since $M$ is representable, so by
Lemma 5.14, each element of the set $\\{P\in X(M)\mid[0,w]\subseteq P\\}$ is a
strict square ideal. Consequently, $M/P$ is not a Boolean algebra. Hence
$\\{P\in X(M)\mid[0,w]\subseteq P\\}\subseteq X_{2}(M)$ and $K\subseteq[0,w]$.
Therefore, $K=[0,w]$ is the least strict square ideal of $M$; see Lemma
5.14(iii).

The proof of the other parts is straightforward. ∎

In the following theorem, we characterize representable pseudo MV-algebra with
square root using normal prime ideals.

######  Theorem 5.16.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a representable pseudo MV-algebra with a
square root $r$. Then $M$ is strict if and only if $M$ does not have any
proper Boolean normal prime ideal.

Consequently, if $M$ has a proper Boolean normal prime ideal, then $M$ is not
strict.

###### Proof.

Suppose that $M$ is strict. If $P$ is a Boolean normal prime ideal of $M$,
then $M/P$ is a Boolean algebra. By [DvZa3, Prop 3.9], $r_{P}(x)=r(x)/P$ is a
square root on $M/P$. Since $r$ is strict,
$r(0/P)=r(0)/P=r(0)^{-}/P=(r(0)/P)^{-}$ and the Booleanicity of $M/P$ entails
$r(0/P)=0/P$. It follows that $0/P=1/P$, which means $P=M$. Conversely, assume
that $M$ has no proper Boolean normal prime ideal. If $M$ is degenerate, then
clearly, the statement holds. Let $M$ be non-degenerate. The pseudo MV-algebra
$M$ is not a Boolean algebra. Otherwise, each prime ideal of $M$ is a Boolean
normal prime ideal. So, $X(M)=X(M)_{2}$. Set $w=r(0)^{-}\odot r(0)^{-}$. By
Lemma 5.14(iii), $w\in\bigcap\\{P\colon P\in X(M)_{2}\\}=\bigcap\\{P\colon
P\in X(M)\\}=\\{0\\}$. Therefore, $r(0)=r(0)^{-}$ and $r$ is a strict square
root on $M$. ∎

Finally, we present a criterion showing when an MV-algebra $N$ with strict
square root is a strict square root closure of an MV-algebra. It will be used
in Example 6.14.

######  Theorem 5.17.

Let $M=\Gamma(G,u)$ be an MV-algebra which is a subalgebra of an MV-algebra
$N=\Gamma(H,u)$ with strict square root. The following statements are
equivalent:

  * (i)

$N\cong\mathbf{C}(M)$.

  * (ii)

For each $h\in H$, there is an integer $n$ such that $2^{n}h\in G$.

  * (iii)

For each $x\in N$, there are an integer $n$ and elements
$x_{1},\ldots,x_{2^{n}}\in M$ such that
$x=x_{1}/2^{n}+\cdots+x_{2^{n}}/2^{n}$.

###### Proof.

Due to Mundici’s representation of MV-algebras, we can assume that $(G,u)$ is
a unital $\ell$-subgroup of the unital $\ell$-group $(H,u)$. Moreover, Theorem
4.4 says that $H$ is two-divisible, and $s(y)=(y+1)/2$, $y\in N$, is a unique
strict square root on $N$. Let $r$ be a unique strict square root on
$\mathbf{C}(M)$.

(i) $\Rightarrow$ (ii). If $\mathbf{C}(M)\cong N$, then statement (ii) follows
from the proof (2) of Theorem 5.10.

(ii) $\Rightarrow$ (iii). If $x\in N$, then by (ii), there is an integer
$n\geq 0$ such that $2^{n}x\in G$. Then $2^{n}x\leq 2^{n}u$, so by (RIP),
there are $x_{1},\ldots,x_{2^{n}}\in M$ such that
$2^{n}x=x_{1}+\cdots+x_{2^{n}}$, so that
$x=x_{1}/2^{n}+\cdots+x_{2^{n}}/2^{n}$.

(iii) $\Rightarrow$ (ii). If $h\in N$, there is $n$ and
$h_{1},\ldots,h_{2^{n}}\in M$ such that $2^{n}h=h_{1}+\cdots+h_{2^{n}}$ and
$2^{n}h\in G$. Let $h\in H^{+}$. Due to (RIP), there are elements
$h_{1},\ldots,h_{k}\in N$ such that $h=h_{1}+\cdots+h_{k}$. By (iii), every
$h_{i}$ is of the form
$h_{i}=x_{i1}/2^{n_{i}}+\cdots+x_{i2^{n_{i}}}/2^{n_{i}}$. We can assume
$n_{1}\leq\cdots\leq n_{k}$. Then
$x_{ij}/2^{n_{i}}=2^{n_{k}-n_{i}}x_{ij}/2^{n_{k}}$ and $2^{n_{k}-n_{i}}x_{ij}$
belongs to $G$. Therefore, for $n=n_{k}$, we have $2^{n}h\in G$. Finally, let
$h\in H$. It is of the form $h=h^{+}-h^{-}$, where $h^{+},h^{-}\in H^{+}$.
Now, it is easy to see that there is $n\in\mathbb{N}$ such that $2^{n}h\in G$.

(ii) $\Rightarrow$ (i). Let $i_{C}:M\to\mathbf{C}(M)$ be the embedding from
the definition of the strict square root closure and let $j_{N}:M\to N$ be the
inclusion embedding, $j_{N}(x)=x$, $x\in M$. There is a unique injective
homomorphism $\alpha:\mathbf{C}(M)\to N$ such that $\alpha\circ i_{C}=j_{N}$
and $\alpha\circ r=s\circ\alpha$.

Since (ii) and (iii) are equivalent, there are an integer $n$ and elements
$x_{1},\ldots,x_{2^{n}}\in M$ such that
$x=x_{1}/2^{n}+\cdots+x_{2^{n}}/2^{n}$. Since every $x_{i}\in M$, then
$i_{C}(x_{i}),i_{C}(x_{i})/2^{n}\in\mathbf{C}(M)$. Moreover,
$i_{C}(x_{1})/2^{n}+\cdots+i_{C}(x_{2^{n}})=i_{C}(x_{1})/2^{n}\oplus\cdots\oplus
i_{C}(x_{2^{n}})\in\mathbf{C}(M)$. Applying $\alpha$, we get

$\displaystyle\alpha(i_{C}(x_{1})/2^{n})+\cdots+\alpha(i_{C}(x_{2^{n}})/2^{n})$
$\displaystyle=\alpha(i_{C}(x_{1}))/2^{n}+\cdots+\alpha(i_{C}(x_{2^{n}}))/2^{n}$
$\displaystyle=j_{N}(x_{1})/2^{n}+\cdots+j_{N}(x_{2^{n}})/2^{n}$
$\displaystyle=x_{1}/2^{n}+\cdots+x_{2^{n}}/2^{n}=x,$

which proves that $\alpha(\mathbf{C}(M))=N$, and $\alpha$ is an injective and
surjective homomorphism. Consequently $N\cong\mathbf{C}(M)$. ∎

##  6\. Square root closure and strict square root closure

In this section, we introduce the notion of a square root closure of an MV-
algebra, and we compare it with the strict square root.

######  Definition 6.1.

A pseudo MV-algebra $M$ is called Boolean subdirectly irreducible if each
subdirect embedding $f:M\to\prod_{i\in J}M_{i}$ can be reduced to the
embedding $f_{2}:M\to\prod_{i\in J_{2}}M_{i}$, where $f_{2}(x)=(\pi_{i}\circ
f(x))_{i\in J_{2}}$, and $J_{2}:=\\{i\in J\mid M_{i}\mbox{ is not a Boolean
algebra}\\}$. Equivalently,

$\displaystyle f^{-1}(\\{(x_{i})_{i\in J}\in\prod_{i\in I}M_{i}\mid
x_{i}=0,~{}\forall i\in J_{2}\\})=\mbox{\rm Ker}f_{2}=\\{0\\}.$ (6.1)

######  Theorem 6.2.

Let $M$ be a pseudo MV-algebra. Then $M$ is Boolean subdirectly irreducible if
and only if each one-to-one homomorphism $f:M\to N_{1}\times N_{2}$ of pseudo
MV-algebras, where $N_{1}$ is a Boolean algebra and $N_{2}$ is a pseudo MV-
algebra, can be reduced to the embedding $\pi_{2}\circ f:M\to N_{2}$.

###### Proof.

First, assume that $M$ is Boolean subdirectly irreducible. Let $f:M\to
N_{1}\times N_{2}$ be a one-to-one homomorphism, where $N_{1}$ and $N_{2}$ are
a Boolean algebra and a pseudo MV-algebra, respectively. It suffices to show
that $\mbox{\rm Ker}(\pi_{2}\circ f)=\\{0\\}$. Clearly, $\mbox{\rm
Ker}(\pi_{2}\circ f)=f^{-1}(N_{1}\times\\{0\\})$. Set
$I:=f^{-1}(N_{1}\times\\{0\\})$ and $J=f^{-1}(\\{0\\}\times N_{2})$. We have
$I\cap J=\\{0\\}$, so $g:M\to M/I\times M/J$ defined by $g(x)=(x/I,x/J)$ is a
subdirect embedding, where $M/I$ is isomorphic to a subalgebra of $N_{2}$ and
$M/J$ is isomorphic to a subalgebra of $N_{1}$. Since $N_{1}$ is a Boolean
algebra, by the assumption $\pi_{2}\circ g:M\to M/I$ is an embedding, which
implies that $I=\\{0\\}$. Therefore, $\pi_{2}\circ f:M\to N_{2}$ is an
injective homomorphism.

Conversely, let $f:M\to\prod_{i\in I}M_{i}$ be a subdirect embedding. Set
$I_{1}=\\{i\in I\mid M_{i}\mbox{ is a Boolean algebra}\\}$ and
$I_{2}=I\setminus I_{1}$. Suppose that $E_{j}=\prod_{i\in I_{j}}M_{i}$ for
$j=1,2$. Consider the map $g:M\to E_{1}\times E_{2}$ defined by
$g(x)=((\pi_{i}\circ f(x))_{i\in I_{1}},(\pi_{i}\circ f(x))_{i\in I_{2}})$.
The map $g$ is a one-to-one homomorphism of pseudo MV-algebras. Since $E_{1}$
is a Boolean algebra, by the assumption, $\\{0\\}=\mbox{\rm Ker}(\pi_{2}\circ
g)=g^{-1}(E_{1}\times\\{0\\})=\\{x\in M\mid\pi_{i}\circ f(x)=0,~{}\forall i\in
I_{2}\\}$.

It follows that $f^{-1}(\\{(x_{i})_{i\in I}\in\prod_{i\in I}M_{i}\mid
x_{i}=0,~{}\forall i\in I_{2}\\})=\\{0\\}$. Hence, $M$ is Boolean subdirectly
irreducible. ∎

The following theorem shows a relation between Boolean subdirectly irreducible
pseudo MV-algebras and strict pseudo MV-algebras. Hence, it will provide us
with many examples of Boolean subdirectly irreducible pseudo MV-algebras.

######  Theorem 6.3.

Let $M$ be a pseudo MV-algebra with a square root $r$. Then $M$ is Boolean
subdirectly irreducible if and only if $r$ is strict.

###### Proof.

Let $r$ be a strict square root. If there exists a subdirect embedding $f:M\to
B\times A$, where $B$ is a Boolean algebra and $A$ is a pseudo MV-algebra,
then the onto homomorphism $\pi_{1}\circ f:M\to B$ induces a square root
$r_{1}:B\to B$ defined by $r_{1}(\pi_{1}\circ f(x))=\pi_{1}\circ f(r(x))$ for
all $x\in M$. It follows that $r_{1}(0)=\pi_{1}\circ f(r(0))=\pi_{1}\circ
f(r(0)^{-})=(\pi_{1}\circ f(r(0)))^{-}=r_{1}(0)^{-}$. On the other hand,
$r_{1}(0)=0$, since $B$ is a Boolean algebra (see [DvZa3, Thm 3.8]). Hence
$0=1$ on $B$, and so $B$ has only one element. It follows that $\mbox{\rm
Ker}(\pi_{2}\circ f)=f^{-1}(B\times\\{0\\})=\\{0\\}$. Therefore, by Theorem
6.2, $M$ is Boolean subdirectly irreducible.

Conversely, let $M$ be Boolean subdirectly irreducible. By [DvZa3, Thm 4.3],
$M\cong[0,v]\times[0,v^{-}]$ where $v=r(0)^{-}\odot r(0)^{-}$; the mapping
$f:M\to[0,v]\times[0,v^{\prime}]$ defined by $f(x)=(x\wedge v,x\wedge
v^{\prime})$ is an isomorphism

In addition, $[0,v]$ is a Boolean algebra and $[0,v^{-}]$ is a strict pseudo
MV-algebra. By the assumption, $f_{2}:M\to[0,v^{-}]$ defined by
$f_{2}(x)=x\wedge v^{-}$ is an embedding, so $v=0$. It follows that $r$ is
strict. ∎

######  Example 6.4.

(i) Let $S$ be a non-empty subset and for each $i\in S$, let $A_{i}$ be the
finite MV-algebra $M_{n}=\\{0,1/n,\ldots,n/n\\}$, where $n\in\mathbb{N}$. By
Theorem 4.4, the MV-algebra $\prod_{i\in S}A_{i}$ is Boolean subdirectly
irreducible, but it has no square root.

(ii) Each linearly ordered pseudo MV-algebra with more than two elements is
Boolean subdirectly irreducible. In addition, a direct product of any family
of such pseudo MV-algebras is Boolean subdirectly irreducible.

######  Proposition 6.5.

A representable pseudo MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$ is Boolean
subdirectly irreducible if and only if $\bigcap X(M)_{2}=\\{0\\}$.

###### Proof.

Let $M$ be a representable pseudo MV-algebra. If $M$ is Boolean subdirectly
irreducible, then the subdirect embedding $f:M\to\prod_{P\in X(M)}M/P$ implies
that $f_{2}:M\to\prod_{P\in X(M)_{2}}M/P$ defined by $f_{2}(x)=(x/P)_{P\in
X(M)_{2}}$ is an embedding, so $\bigcap X(M)_{2}=\bigcap_{P\in
X(M)_{2}}\mbox{\rm Ker}(\pi_{P}\circ f)=\mbox{\rm Ker}(f_{2})=\\{0\\}$.

Conversely, assume that $\bigcap X(M)_{2}=\\{0\\}$. Let $g:M\to\prod_{i\in
I}M_{i}$ be a subdirect embedding. Without loss of generality, we can assume
that each $M_{i}$ is linearly ordered. Let $I_{1}=\\{i\in I\mid M_{i}\mbox{ is
a Boolean algebra}\\}$ and $I_{2}=I\setminus I_{1}$. Consider the map
$g_{2}:M\to\prod_{i\in I_{2}}M_{i}$ defined by $g_{2}(x)=(\pi_{i}\circ
g(x))_{i\in I_{2}}$. Clearly, $\mbox{\rm Ker}(g_{2})=\bigcap_{i\in
I_{2}}\mbox{\rm Ker}(\pi_{i}\circ g)$. For each $i\in I_{2}$, $M/\mbox{\rm
Ker}(\pi_{i}\circ g)\cong M_{i}$, so $\mbox{\rm Ker}(\pi_{i}\circ g)\in
X(M)_{2}$, consequently, $\bigcap X(M)_{2}\subseteq\bigcap_{i\in
I_{2}}\mbox{\rm Ker}(\pi_{i}\circ g)$. Similarly, $\mbox{\rm Ker}(\pi_{i}\circ
g)\in X(M)_{1}$ for all $i\in I_{1}$ which implies that $\bigcap
X(M)_{1}\subseteq\bigcap_{i\in I_{1}}\mbox{\rm Ker}(\pi_{i}\circ g)$. Since
$(\bigcap_{i\in I_{2}}\mbox{\rm Ker}(\pi_{i}\circ g))\cap(\bigcap_{i\in
I_{1}}\mbox{\rm Ker}(\pi_{i}\circ g))=\\{0\\}$, we have $(\bigcap_{i\in
I_{2}}\mbox{\rm Ker}(\pi_{i}\circ g))\cap\bigcap X(M)_{1}=\\{0\\}$. We claim
that $\bigcap X(M)_{2}=\bigcap_{i\in I_{2}}\mbox{\rm Ker}(\pi_{i}\circ g)$. If
there exists $a\in(\bigcap_{i\in I_{2}}\mbox{\rm Ker}(\pi_{i}\circ
g))\setminus\bigcap X(M)_{2}$, then there exists $P\in X(M)_{2}$ such that
$a\notin P$. For each $b\in\bigcap X(M)_{1}$, $a\wedge b\in(\bigcap_{i\in
I_{2}}\mbox{\rm Ker}(\pi_{i}\circ g))\cap\bigcap X(M)_{1}=\\{0\\}\subseteq P$,
so $b\in P$, that is $\bigcap X(M)_{1}\subseteq P$. By definition,
$X(M)_{1}=\\{Q\in X(M)\mid M/Q\mbox{ is a Boolean algebra}\\}$, so each
element of $X(M)_{1}$ is a Boolean ideal of $M$, whence $\bigcap X(M)_{1}$ is
a Boolean ideal. Hence, $P$ is a Boolean ideal of $M$ (since $\bigcap
X(M)_{1}\subseteq P$) which contradicts with $P\in X(M)_{2}$. Therefore,
$\\{0\\}=\bigcap X(M)_{2}=\bigcap_{i\in I_{2}}\mbox{\rm Ker}(\pi_{i}\circ
g)=\mbox{\rm Ker}(g_{2})$ implying $g_{2}$ is one-to-one and $M$ is Boolean
subdirectly irreducible. ∎

######  Corollary 6.6.

If $(M;\oplus,^{-},^{\sim},0,1)$ is a representable pseudo MV-algebra, then
$M/I_{2}$ is Boolean subdirectly irreducible.

###### Proof.

Clearly, $M/I_{2}$ is representable. Let $Q\in X(M/I_{2})_{2}$. Then
$Q=P/I_{2}$ for some prime ideal $P$ of $M$ containing $I_{2}$. Since
$\frac{M/I_{2}}{P/I_{2}}\cong M/P$, we get $M/P$ is not a Boolean algebra,
which entails that $P\in X(M)_{2}$. Similarly, the converse also holds. Whence
$X(M/I_{2})_{2}=\\{P/I_{2}\mid P\in X(M)_{2}\\}$ and so $\bigcap
X(M/I_{2})_{2}=I_{2}/I_{2}=\\{0/I_{2}\\}$. By Proposition 6.5, $M/I_{2}$ is
Boolean subdirectly irreducible. ∎

######  Theorem 6.7.

Let $(M;\oplus,^{-},^{\sim},0,1)$ and $(N;\oplus,^{-},^{\sim},0,1)$ be pseudo
MV-algebras with square roots $r$ and $s$, respectively, and let $f:M\to N$ be
a homomorphism of pseudo MV-algebras. If $v=r(0)^{-}\odot r(0)^{-}$,
$w=s(0)^{-}\odot s(0)^{-}$, $M_{1}$, $M_{2}$, $N_{1}$ and $N_{2}$ be the
pseudo MV-algebras defined in [DvZa3, Thm 4.3], then the following statements
hold:

  * (i)

$f(M_{2})\subseteq N_{2}$;

  * (ii)

$f(M_{1})$ is isomorphic to a subalgebra of $N_{1}\times A$, where
$A=[0,f(v)\wedge w^{\prime}]$ is a Boolean subdirectly irreducible subalgebra
of $N_{2}$.

###### Proof.

According to [DvZa3, Thm 4.3], assume $M_{1}=[0,v]$, $M_{2}=[0,v^{-}]$,
$N_{1}=[0,w]$ and $N_{2}=[0,w^{-}]$. From $f(r(0))\odot f(r(0))=f(0)=0$ it
follows that $f(r(0))\leq s(0)$, so $s(0)^{-}\leq f(r(0)^{-})$ which entails
that $w=s(0)^{-}\odot s(0)^{-}\leq f(r(0)^{-})\odot f(r(0)^{-})=f(v)$. Thus,
$f(v^{-})\leq w^{-}$. Hence, $[0,f(v^{-})]\subseteq[0,w^{-}]=N_{2}$ and
$[0,w]\subseteq[0,f(v)]$.

Let $Q\in X(N)$. Since $Q$ is a proper prime ideal of $N$ and $f(v)\wedge
f(v^{-})=0\in Q$, only one of the statements $f(v)\in Q$ or $f(v^{-})\in Q$
holds. So, $X(N)$ is union of the disjoint sets $\\{Q\in X(N)\mid f(v)\in
Q\\}$ and $\\{Q\in X(N)\mid f(v^{-})\in Q\\}$. On the other hand,
$f(v)=(f(v)\wedge w)\vee(f(v)\wedge w^{-})=w\vee(f(v)\wedge w^{-})$, so
$[0,f(v)]\cong[0,w]\times[0,f(v)\wedge w^{-}]$. It follows that
$N\cong[0,w]\times[0,f(v)\wedge w^{-}]\times[0,f(v^{-})]$. Each set $[0,w]$,
$[0,f(v)\wedge w^{-}]$ and $[0,f(v^{-})]$ is a normal ideal of $N$ and so:

(i) If $x\in M_{2}$, then $f(x)\leq f(v^{-})\leq w^{\prime}$, so
$f(x)\in[0,w^{-}]=N_{2}$. That is, $f(M_{2})\subseteq N_{2}$.

(ii) If $x\in M_{1}$, then $f(x)\leq f(v)$ and so $f(x)=(f(x)\wedge
w)\vee(f(x)\wedge w^{-})=(f(x)\wedge w)\vee(f(x)\wedge f(v)\wedge w^{-})$.
Consider the injective homomorphism $g:M_{1}\to N_{1}\times[0,f(v)\wedge
w^{-}]$ defined by $g(x)=(f(x)\wedge w,f(x)\wedge f(v)\wedge w^{-})$. We show
that $[0,f(v)\wedge w^{-}]$ is a Boolean subdirectly irreducible subalgebra of
$N_{2}$. Since $f(v)\wedge w^{-}$ is a Boolean element of $N$, then clearly
$[0,f(v)\wedge w^{-}]$ is a subalgebra of $[0,w^{-}]$. Let $g:[0,f(v)\wedge
w^{-}]\to B\times D$ be a subdirect embedding of pseudo MV-algebras, where $B$
is a Boolean algebra and $D$ is a pseudo MV-algebra.

We claim that $\pi_{2}\circ g:[0,f(v)\wedge w^{-}]\to D$ is one-to-one,
equivalently, $g^{-1}(B\times\\{0\\})=\\{0\\}$.

Since $w^{-}=(f(v)\wedge w^{-})\vee(f(v^{-})\wedge w^{-})=(f(v)\wedge
w^{-})\vee f(v^{-})$ and $f(v),f(v^{-}),w^{-}\in\mathrm{B}(N)$, we get that
$[0,w^{-}]\cong[0,f(v)\wedge w^{-}]\times[0,f(v^{-})]$ and so
$\varphi:[0,w^{-}]\to(B\times D)\times[0,f(v^{-})]$ defined by
$\varphi(x)=(g(x\wedge f(v)\wedge w^{-}),x\wedge f(v^{-}))$ is a subdirect
embedding. Due to Theorem 6.3, $[0,w^{-}]$ is Boolean subdirectly irreducible,
so by (6.1), $\varphi^{-1}(B\times\\{0\\}\times\\{0\\})=\\{0\\}$,
consequently, by definition of $\varphi$, $g^{-1}(B\times\\{0\\})=\\{0\\}$.
So, the claim is true, and $[0,f(v)\wedge w^{-}]$ is a Boolean subdirectly
irreducible subalgebra of $N_{2}$. ∎

Now, we are ready to propose the main theorems of this part that will provide
a square root closure (not necessarily strict) for MV-algebras. First, we
state the following remark.

######  Remark 6.8.

Let $f:M\to N$ be an injective homomorphism of MV-algebras and $s:N\to N$ be a
square root of $N$. Let $I_{1}=\bigcap X(M)_{1}$ and $I_{2}=\bigcap X(M)_{2}$.
If $w=s(0)^{-}\odot s(0)^{-}$ and $N_{1}$ and $N_{2}$ be the MV-algebras that
were introduced in [DvZa3, Thm 4.3], then:

(i) The set $\\{f^{-1}(Q)\mid Q\in X(N)\\}$ represents each proper ideal $I$
of $M$, that is, for each proper ideal $I$ of $M$, we have
$I=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)\mbox{ and }f(I)\subseteq Q\\}$. Choose a
proper ideal $I$ of $M$. Clearly, $\downarrow f(I)=\\{y\in N\mid y\leq
f(x),~{}\exists x\in I\\}$ is the least ideal of $N$ containing $f(I)$. It
follows that $\downarrow f(I)=\bigcap\\{Q\in X(N)\mid\downarrow f(I)\subseteq
Q\\}=\bigcap\\{Q\in X(N)\mid f(I)\subseteq Q\\}$ and so $f^{-1}(\downarrow
f(I))=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)\mbox{ and }f(I)\subseteq Q\\}$. On the
other hand, if $x\in f^{-1}(\downarrow f(I))$, then $f(x)\leq f(a)$ for some
$a\in I$, whence $x\leq a$, since $f(x)=f(x\wedge a)$ and $f$ is one-to-one.
Thus, $x\in I$ and so $f^{-1}(\downarrow f(I))\subseteq I$. Clearly,
$I\subseteq f^{-1}(\downarrow f(I))$. That is, $I=f^{-1}(\downarrow f(I))$.
Therefore, $I=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)\mbox{ and }f(I)\subseteq
Q\\}$.

(ii) We divide $X(N)$ into three disjoint sets as follows:

$\displaystyle X(N)_{1},\quad Y_{1}=\\{Q\in X(N)_{2}\mid f^{-1}(Q)\in
X(M)_{1}\\},\quad Y_{2}=\\{Q\in X(N)_{2}\mid f^{-1}(Q)\in X(M)_{2}\\}.$

If $Q\in X(N)_{1}$, then $f^{-1}(Q)\in X(M)_{1}$, since $M/f^{-1}(Q)$ is
embedded in $N/Q$ and $N/Q$ is a Boolean algebra. So, $I_{1}=\bigcap
X(M)_{1}\subseteq\bigcap\\{f^{-1}(Q)\mid Q\in X(N)_{1}\cup Y_{1}\\}$. On the
other hand, according to (i), $I_{1}=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)\mbox{
and }f(I_{1})\subseteq Q\\}=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)\mbox{ and
}I_{1}\subseteq f^{-1}(Q)\\}$. For each $Q\in X(N)$, if $I_{1}\subseteq
f^{-1}(Q)$, then $f^{-1}(Q)$ is a Boolean ideal of $M$ and so $Q\in Y_{1}\cup
X(N)_{1}$, whence $\bigcap X(M)_{1}\subseteq\bigcap\\{f^{-1}(Q)\mid Q\in
X(N)_{1}\cup Y_{1}\\}\subseteq\bigcap\\{f^{-1}(Q)\mid Q\in X(N)_{1}\cup
Y_{1}\mbox{ and }I_{1}\subseteq f^{-1}(Q)\\}=I_{1}$. Hence

$\displaystyle I_{1}=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)_{1}\cup
Y_{1}\\}=f^{-1}\big{(}(\bigcap X(N)_{1})\cap(\bigcap Y_{1})\big{)}.$

Furthermore, clearly, $I_{2}\subseteq\bigcap\\{f^{-1}(Q)\mid Q\in Y_{2}\\}$.
By (i), $I_{2}=\bigcap\\{f^{-1}(Q)\mid Q\in X(N)\mbox{ and }f(I_{2})\subseteq
Q\\}$. So,

$\displaystyle I_{2}=(\bigcap\\{f^{-1}(Q)\mid f(I_{2})\subseteq Q\in
Y_{2}\\})\cap(\bigcap\\{f^{-1}(Q)\mid f(I_{2})\subseteq Q\in X(N)_{1}\cup
Y_{1}\\}).$

It follows that $M/I_{2}$ can be embedded into $M/A_{1}\times M/A_{2}$ by the
mapping $\alpha(x)=(x/A_{1},x/A_{2})$, where $A_{2}=\bigcap\\{f^{-1}(Q)\mid
f(I_{2})\subseteq Q\in Y_{2}\\}$ and $A_{1}=\bigcap\\{f^{-1}(Q)\mid
f(I_{2})\subseteq Q\in X(N)_{1}\cup Y_{1}\\}$. Since $M/A_{1}$ is a Boolean
algebra and by Corollary 6.6, $M/I_{2}$ is Boolean subdirectly irreducible, so
$\pi_{2}\circ\alpha$ is one-to-one. It follows that
$I_{2}=A_{2}=\bigcap\\{f^{-1}(Q)\mid f(I_{2})\subseteq Q\in Y_{2}\\}$, whence
$\bigcap\\{f^{-1}(Q)\mid Q\in Y_{2}\\}\subseteq I_{2}$. Therefore,

$\displaystyle I_{2}=\bigcap\\{f^{-1}(Q)\mid Q\in Y_{2}\\}.$ (6.2)

In Definition 5.8, we introduced the concept of a strict square root closure,
and in Theorem 5.10, we showed that each MV-algebra has a strict square root
closure. Now, we want to define also the concept of square root closure (not
necessarily a strict one) for MV-algebras.

######  Definition 6.9.

Let $M$ be an MV-algebra. A square root closure for M is a triple $(D,r,i)$,
where $D$ is an MV-algebra, $r:D\to D$ is a square root, and $i:M\to D$ is an
embedding of MV-algebras such that if $N$ is an MV-algebra with square root
$s$ and $j:M\to N$ is an injective homomorphism of MV-algebras, there exists a
unique injective homomorphism $\beta:D\to N$ satisfying the following
conditions:

  * (D1)

$\beta\circ i=j$ ;

  * (D2)

$\beta(r(x))\leq s(\beta(x))$ for all $x\in M$.

######  Remark 6.10.

(i) Similarly to strict square root closures, we can show that if an MV-
algebra $M$ has a square root closure, then it is unique up to isomorphism.
So, if it exists, the square root closure of $M$ is denoted by
$\mathbf{D}(M)$. An MV-algebra is called closed with respect to square root
closure, if $\mathbf{D}(M)\cong M$.

(ii) If $M$ is an MV-algebra with a square root $r$, then $(M,r,\mbox{\rm
Id}_{M})$ is a square root closure of $M$. Indeed, if $N$ is an MV-algebra
with a square root $s$ and $j:M\to N$ is an injective homomorphism, then for
$\beta=j$ clearly (D1) holds. For establishing (D2), we have
$\beta(r(x))\odot\beta(r(x))=\beta(r(x)\odot r(x))=\beta(x)$, so
$\beta(r(x))\leq s(\beta(x))$ for all $x\in M$. Therefore, $(M,r,\mbox{\rm
Id}_{M})$ is a square root closure of $M$. It follows that each Boolean
algebra $B$ has a square root closure, $(B,\mbox{\rm Id}_{B},\mbox{\rm
Id}_{B})$, since the identity map is a square root on $B$. Hence, if $M$ has a
square root, then $M$ is closed with respect to square root closure. It shows
an important difference between the “strict square root closure” and the
“square root closure”. According to Theorem 5.10, the strict square closure of
$B$, $\mathbf{C}(B)$, is two-divisible, so $\mathbf{C}(B)$ is not isomorphic
to $B$. But $\mathbf{D}(B)\cong B$.

######  Theorem 6.11.

Let $M=\Gamma(G,u)$ be an MV-algebra such that $u/2\in M$ and let
$\mathbf{D}(M)=(\Gamma(K,w),r,i)$ be its square root closure. Then
$\mathbf{D}(M)$ is the strict square root closure of $M$.

###### Proof.

Assume that $N=\Gamma(H,v)$ is an MV-algebra with a square root $s$ and
$j:M\to N$ is an embedding. We have $w=i(u)=i(u/2)\oplus
i(u/2)=i(u/2)+i(u/2)$, that is $w/2=i(u/2)$. If $y\in\mathbf{D}(M)$ is such
that $y\odot y\leq 0$, then $(y-w+y)\vee 0=0$, which entails that $y-w+y\leq
0$, equivalently, $y\leq w/2$. Now, from $w/2\odot w/2=0$ it follows that
$w/2=r(0)$, and so $r$ is strict. By Theorem 4.4, $\mathbf{D}(M)$ is a two-
divisible $\ell$-group and so $r(x)=(x+w)/2=x/2+w/2$ for all
$x\in\mathbf{D}(M)$; see [DvZa3, Thm 5.2].

On the other hand, $s(y)=(y+v)/2=y/2+v/2$ for all $y\in N$, by [DvZa3, Thm
5.2]. Due to Definition 6.9, there exists an injective homomorphism
$\beta:\mathbf{D}(M)\to N$ such that $\beta\circ i=j$. That is, (C1) holds.
Also, we have $\beta(r(x))=\beta(x/2+w/2)=\beta(x/2\oplus
w/2)=\beta(x/2)\oplus\beta(w/2)=\beta(x)/2\oplus\beta(w/2)=\beta(x)/2+\beta(w)/2=\beta(x)/2+v/2=s(\beta(x))$.
Thus, (C2) holds. Therefore, $\mathbf{D}(M)$ is a strict square root closure
for $M$. ∎

Let $(M;\oplus,^{\prime},0,1)$ be an MV-algebra, $I_{1}=\bigcap X(M)_{1}$,
$I_{2}=\bigcap X(M)_{2}$ and $f:M\to M/I_{1}\times M/I_{2}$ be defined by
$f(x)=(x/I_{1},x/I_{2})$, $x\in M$, satisfying the condition

$\displaystyle\exists a\in M\mbox{ such that }f(a)=(1/I_{1},0/I_{2}).$ (6.3)

Then $a\in\mathrm{B}(M)$. Since $(a/I_{1},a/I_{2})=f(a)=(1/I_{1},0/I_{2})$, we
have $a\in I_{2}$ and so $[0,a]\subseteq I_{2}$. Moreover, if $x\in I_{2}$,
then $f(x)=(x/I_{1},x/I_{2})=(x/I_{1},0/I_{2})\leq f(a)$, so $x\leq a$. Hence
$I_{2}=[0,a]$. A similar calculus shows that $I_{1}=[0,a^{\prime}]$.
Therefore, the element $a$ satisfying condition (6.3) is unique. Due to Remark
2.2, $[0,a]$ and $[0,a^{\prime}]$ are MV-algebras such that
$M/I_{1}=M/[0,a^{\prime}]\cong[0,a]=I_{2}$ and
$M/I_{2}=M/[0,a]\cong[0,a^{\prime}]=I_{1}$. It follows that $I_{2}$ is a
Boolean algebra and $I_{1}$ is a Boolean subdirectly irreducible MV-algebra
(see Corollary 6.6). Considering these notations and conditions, we present
the following theorem.

######  Theorem 6.12.

Let an MV-algebra $(M;\oplus,^{\prime},0,1)$ satisfy one of the following
conditions

  * (i)

$I_{1}=\\{0\\}$.

  * (ii)

$I_{2}=\\{0\\}$.

  * (iii)

$I_{1}\neq\\{0\\}\neq I_{2}$ and let (6.3) hold.

Then $M$ has a square root closure. Moreover, in the first case, $M$ is a
Boolean algebra and $\mathbf{D}(M)\cong M$. In the second case,
$\mathbf{D}(M)\cong\mathbf{C}(M)$, and in the third case,
$\mathbf{D}(M)\cong[0,a]\times\mathbf{C}([0,a^{\prime}])$ for a unique Boolean
element $a\neq 0,1$ satisfying (6.3).

###### Proof.

(i) If $I_{1}=\\{0\\}$, then according to the definition of $X(M)_{1}$,
$g:M\to\prod_{P\in X(M)_{1}}M/P$ is an embedding into a Boolean algebra, so
$M$ is a Boolean algebra. Hence by Remark 6.10(ii), $(M,\mbox{\rm
Id}_{M},\mbox{\rm Id}_{M})$ is the square root closure of $M$.

(ii) If $I_{2}=\\{0\\}$, by Corollary 6.6, $M$ is Boolean subdirectly
irreducible. Let $(\textbf{C}(M),i,r)$ be the strict square root closure of
$M$. We also claim that it is the square root closure of $M$. Assume that $N$
is an MV-algebra with a square root $s$ and $g:M\to N$ is an injective
homomorphism of MV-algebras. By [Höl, Thm 2.21],
$N\cong[0,v]\times[0,v^{\prime}]$, where $N_{1}=[0,v]$ is a Boolean algebra
and $N_{2}=[0,v^{\prime}]$ is an MV-algebra with strict square root, and
$v=s(0)^{\prime}\odot s(0)^{\prime}$ (the isomorphism is $x\longmapsto(x\wedge
v,x\wedge v^{\prime})$). Theorem 6.2 together with Proposition 6.5 imply that
$\pi_{2}\circ g:M\to[0,v^{\prime}]$ is one-to-one. Clearly, $\pi_{2}\circ
g(x)=g(x)\wedge v^{\prime}$ for all $x\in M$. Since $[0,v^{\prime}]$ is
strict, according to the definition of $\textbf{C}(M)$, there exists a unique
one-to-one homomorphism $\alpha:\textbf{C}(M)\to[0,v^{\prime}]$ such that
$\alpha\circ i=\pi_{2}\circ g$.

Let $f:M\to N_{1}\times N_{2}$ be the injective homomorphism defined by
$f(x)=(g(x)\wedge v,g(x)\wedge v^{\prime})$ for all $x\in M$. Assume that
$\textbf{C}(N_{1})$ is the strict square root closure of $N_{1}$ and consider
$N_{1}$ as a subalgebra of $\textbf{C}(N_{1})$. Clearly,
$\textbf{C}(N_{1})\times N_{2}$ is a strict MV-algebra. It follows that there
exists an embedding $\beta:\textbf{C}(M)\to\textbf{C}(N_{1})\times N_{2}$ such
that $\beta(x)=f(x)=(g(x)\wedge v,g(x)\wedge v^{\prime})$ for all $x\in M$.

$\textbf{C}(N_{1})\times N_{2}$$M$$\textbf{C}(M)$$i$$f$$\beta$ Figure 2.
Homomorphisms between $M$, $\textbf{C}(M)$, and $\textbf{C}(N_{1})\times
N_{2}$

Choose $x\in\textbf{C}(M)$. By Theorem 5.17(iii),
$x=\sum_{i=1}^{2^{n}}x_{i}/2^{n}$ for some $n\in\mathbb{N}$ and
$x_{1},\ldots,x_{2^{n}}\in M$. We have

$\displaystyle\beta(x)$ $\displaystyle=$
$\displaystyle\beta(\sum_{i=1}^{2^{n}}x_{i}/2^{n})=(\pi_{1}\circ\beta(\sum_{i=1}^{2^{n}}x_{i}/2^{n}),\pi_{2}\circ\beta(\sum_{i=1}^{2^{n}}x_{i}/2^{n}))$
$\displaystyle=$
$\displaystyle(\sum_{i=1}^{2^{n}}\pi_{1}\circ\beta(x_{i}/2^{n}),\sum_{i=1}^{2^{n}}\pi_{2}\circ\beta(x_{i}/2^{n}))=(\sum_{i=1}^{2^{n}}(\pi_{1}\circ\beta(x_{i}))/2^{n},\sum_{i=1}^{2^{n}}(\pi_{2}\circ\beta(x_{i}))/2^{n})$
$\displaystyle=$ $\displaystyle(\sum_{i=1}^{2^{n}}(g(x_{i})\wedge
v)/2^{n},\sum_{i=1}^{2^{n}}(g(x_{i})\wedge
v^{\prime})/2^{n})=(\sum_{i=1}^{2^{n}}(g(x_{i})\wedge v)/2^{n},\alpha(x)).$

For simplicity, we denote $\pi_{i}\circ\beta$ by $\beta_{i}$ for $i=1,2$.
Define $\varphi:\textbf{C}(M)\to N_{1}\times N_{2}$ by
$\varphi(x)=(\bigvee_{i=1}^{2^{n}}(g(x_{i})\wedge v),\beta_{2}(x))$. We will
show that $\varphi$ is an injective homomorphism. It suffices to show that
$\pi_{1}\circ\varphi:\textbf{C}(M)\to[0,v]$ is a homomorphism. By definition
of $\beta_{1}$, we have
$2^{n}\beta_{1}(x)=2^{n}(\sum_{i=1}^{2^{n}}(g(x_{i})\wedge
v)/2^{n})=\sum_{i=1}^{2^{n}}(g(x_{i})\wedge v)$, whence
$(2^{n}\beta_{1}(x))\wedge v=\bigoplus_{i=1}^{2^{n}}(g(x_{i})\wedge
v)=\bigvee_{i=1}^{2^{n}}(g(x_{i})\wedge v)=\pi_{1}\circ\varphi(x)$; note that
$g(x_{i})\wedge v\in N_{1}\subseteq\mathrm{B}(N)$ for all integers $1\leq
i\leq 2^{n}$. Hence, for all $x\in\textbf{C}(M)$ we have

$\displaystyle\pi_{1}\circ\varphi(x)=(2^{n}\beta_{1}(x))\wedge
v=2^{n}.\beta_{1}(x).$ (6.4)

Thus, $\pi_{1}\circ\varphi$ is well-defined. Moreover, $\varphi$ is one-to-
one, since $\alpha$ is one-to-one. In addition, $\varphi(0)=(0,0)$ and
$\varphi(1)=(g(1)\wedge v,g(1)\wedge v^{\prime})=(v,v^{\prime})$.

Now, we show that it preserves $\oplus$ and ′.

(a) From
$\pi_{1}\circ\varphi(x^{\prime})=\pi_{1}\circ\varphi(u-x)=\pi_{1}\circ\varphi(u-\sum_{i=1}^{2^{n}}x_{i}/2^{n})=\pi_{1}\circ\varphi(\sum_{i=1}^{2^{n}}(u-x_{i})/2^{n})=\sum_{i=1}^{2^{n}}(g(u-x_{i})/2^{n})=v-\sum_{i=1}^{2^{n}}g(x_{i})/2^{n}=v-\pi_{1}\circ\varphi(x)=(\pi_{1}\circ\varphi(x))^{\prime}$
it follows that $\pi_{1}\circ\varphi$ preserves the unary operation ′.

Let $x=\sum_{i=1}^{2^{n}}x_{i}/2^{n}$ and $y=\sum_{j=1}^{2^{m}}y_{j}/2^{n}$
for some $n,m\in\mathbb{N}$, where
$x_{1},\ldots,x_{2^{n}},y_{1},\ldots,y_{2^{m}}\in M$; by Corollary 5.11, such
representations exist. Without loss of generality, we can assume that $n=m$.
Otherwise, if e.g. $n<m$, then $x_{i}/2^{n}=2^{m-n}x_{i}/2^{m}$, and every
element $x_{i}$ will be now $2^{m-n}$-times. By (6.4), we get

$\displaystyle\pi_{1}\circ\varphi(x\oplus y)$ $\displaystyle=$ $\displaystyle
2^{n}.\beta_{1}(x\oplus y)=2^{n}.\beta(x)\oplus
2^{n}.\beta_{1}(y)=\pi_{1}\circ\varphi(x)\oplus\pi_{1}\circ\varphi(y).$

Therefore, $\varphi:\textbf{C}(M)\to N_{1}\times N_{2}$ is an injective
homomorphism. By definition of $\varphi$, if $x\in M$, then
$\varphi(x)=(g(x)\wedge v,g(x)\wedge v^{\prime})=f(x)$. Hence, $\textbf{C}(M)$
is a square root closure for $M$.

(iii) Let $I_{1}\neq\\{0\\}$ and $I_{2}\neq\\{0\\}$ and let (6.3) hold. By the
note right before this theorem, $I_{1}=[0,a^{\prime}]$, $I_{2}=[0,a]$, and
$a\in\mathrm{B}(M)\setminus\\{0,1\\}$. We claim that
$[0,a]\times\mathbf{C}([0,a^{\prime}])$ is the square root closure of $M$,
where $(\mathbf{C}([0,a^{\prime}]),i,r)$ is the strict square root of the MV-
algebra $[0,a^{\prime}]$. Let $r_{1}:[0,a]\to[0,a]$ be the identity map and
$r_{2}$ be the strict square root on $(\mathbf{C}([0,a^{\prime}]),i,r)$. Then
$(r_{1},r_{2})$ is a square root on $[0,a]\times\mathbf{C}([0,a^{\prime}])$.
Also, the map $f:M\to[0,a]\times\mathbf{C}([0,a^{\prime}])$ defined by
$f(x)=(x\wedge a,x\wedge a^{\prime})$ is an injective homomorphism. Assume
that $N$ is an MV-algebra with a square root $s$, $g:M\to N$ is an injective
homomorphism of MV-algebras, and $v$ is the element introduced in (ii). By
Remark 6.8(6.2), $I_{2}=\bigcap\\{g^{-1}(Q)\mid Q\in Y_{2}\\}$, so by
Proposition 5.15(iii), $[0,v]=\bigcap X_{2}^{N}\subseteq\bigcap
Y_{2}=\downarrow g(I_{2})=\downarrow g([0,a])$. That is, $v\leq g(a)$
consequently, $g(a^{\prime})\leq v^{\prime}$. We have

$\displaystyle g(a)$ $\displaystyle=$ $\displaystyle(g(a)\wedge
v)\vee(g(a)\wedge v^{\prime})=v\vee(g(a)\wedge v^{\prime}),$ $\displaystyle
v^{\prime}$ $\displaystyle=$ $\displaystyle(g(a^{\prime})\wedge
v^{\prime})\vee(g(a)\wedge v^{\prime})=g(a^{\prime})\vee(g(a)\wedge
v^{\prime}).$

(1) By (i), for the Boolean algebra $[0,a]$, we know that
$([0,a],f_{1},r_{1})$ is its square root closure as it is shown in the
commutative diagram in Figure 3, where $f_{1}=r_{1}$ is the identity map on
$[0,a]$ and $\alpha_{1}=g$.

$[0,g(a)]$$[0,a]$$[0,a]$$f_{1}$$g$$\alpha_{1}$ Figure 3.  The diagram of
square root closure of $[0,a]$

(2) Furthermore, the MV-algebra $[0,a^{\prime}]$ is Boolean subdirectly
irreducible. Also, $[0,v^{\prime}]\cong[0,g(a^{\prime})]\times[0,g(a)\wedge
v^{\prime}]$ and $\varphi:[0,v^{\prime}]\to[0,g(a^{\prime})]$ is an onto
homomorphism, so by [DvZa3, Prop 3.9],
$t:[0,g(a^{\prime})]\to[0,g(a^{\prime})]$ defined by
$t(\varphi(x))=\varphi(s(x))=s(x)\wedge g(a^{\prime})$ for all
$x\in[0,v^{\prime}]$ is a square root on $[0,g(a^{\prime})]$.

On the other hand, by Theorem 6.3, $[0,v^{\prime}]$ is Boolean subdirectly
irreducible, so $[0,g(a^{\prime})]$ is Boolean subdirectly irreducible, too,
because $[0,v^{\prime}]=[0,g(a^{\prime})]\times[0,g(a)\wedge v^{\prime}]$. By
Theorem 6.3, $t$ is a strict square root on $[0,g(a^{\prime})]$. According to
the strict square root closure definition, we have the commutative diagram in
Figure 4.

$[0,g(a^{\prime})]$$[0,a^{\prime}]$$\mathbf{C}([0,a^{\prime}])$$f_{2}$$g$$\alpha_{2}$
Figure 4.  The diagram of strict square root closures of $[0,a^{\prime}]$

From (1) and (2), it follows that
$\alpha:[0,a]\times\mathbf{C}([0,g(a^{\prime})])\to N$ defined by
$\alpha(x,y)=\alpha_{1}(x)\vee\alpha_{2}(y)$ is a homomorphism of MV-algebras
such that $\alpha\circ f=g$. Indeed, let $x\in M$. Then
$\alpha(f(x))=\alpha(x\wedge a,x\wedge a^{\prime})=\alpha_{1}(x\wedge
a)\vee\alpha_{2}(x\wedge a^{\prime})=g(x\wedge a)\vee g(x\wedge
a^{\prime})=g(x)$. That is, (D1) holds. Moreover, for each
$(x,y)\in[0,a]\times\mathbf{C}([0,g(a^{\prime})])$ we have
$\alpha((x,y))=\alpha((r_{1},r_{2})(x,y)\odot(r_{1},r_{2})(x,y))=\alpha((r_{1},r_{2})(x,y))\odot\alpha((r_{1},r_{2})(x,y))$
implies that $\alpha((r_{1},r_{2})(x,y))\leq s(\alpha((r_{1},r_{2})(x,y)))$,
that is (D2) holds. Therefore,
$([0,a]\times\mathbf{C}([0,g(a^{\prime})]),(r_{1},r_{2}),f)$ is the square
root closure of $M$. ∎

######  Remark 6.13.

(i) We note that due to Example 6.4(ii), every at least three-element linearly
ordered MV-algebra is Boolean subdirectly irreducible. Due to Proposition 6.5,
$I_{2}=\bigcap X(M)_{2}=\\{0\\}$. Theorem 6.12(ii) yields that it has a square
root closure, and $\mathbf{D}(M)=\mathbf{C}(M)$.

(ii) Every direct product of at least three-element linearly ordered MV-
algebras is Boolean subdirectly, i.e., $I_{2}=\\{0\\}$ and it has a square
root closure, and $\mathbf{D}(M)=\mathbf{C}(M)$.

(iii) Let $M_{n}=\\{1/n,2/n,\ldots,n/n\\}$, $n\geq 2$. Then
$X(M_{n})_{1}=\emptyset$, $X(M_{n})_{2}=\\{\\{0\\}\\}$, so $I_{1}=M$,
$I_{2}=\\{0\\}$ and (6.3) holds with the element $a=0$. The same is true for
$M=\Gamma(\mathbb{R},1)$, and $\mathbf{D}(M)=\mathbf{C}(M)$

(iv) Let $M=\\{0,1\\}$. Then $X(M)_{1}=\\{0\\}$ and $X(M)_{2}=M$, and the
element $a=1$ satisfies (6.3). Moreover, $\mathbf{D}(M)=\mathbf{C}(M)$.

(v) Let $M=\Gamma(\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z},(1,0))$.
Then $X(M)_{1}=\\{\\{(0,0)\\}\times\mathbb{Z}^{+}\\}$, $X(M)_{2}=\\{(0,0)\\}$,
$I_{1}=\\{0\\}\times\mathbb{Z}^{+}$, $I_{2}=\\{(0,0)\\}$, (6.3) fails. In
addition, $\mathbf{D}(M)=\mathbf{C}(M)$.

(vi) Let $A=[0,1]$ be the standard MV-algebra of the real unit interval and
$B$ an arbitrary Boolean algebra. For the MV-algebra $M:=B\times A$, we have

$\displaystyle X(M)=\\{P\times A\colon\mbox{ $P$ is a prime ideal of
$B$}\\}\cup\\{B\times\\{0\\}\\}.$

An ideal $I$ of $M$ belongs to $X(M)_{1}$ if and only if $I=P\times A$ for
some $P\in X(B)$. Therefore, $I_{1}=\bigcap X(M)_{1}=\\{0\\}\times A$ and
$X(M)_{2}=\\{B\times\\{0\\}\\}$, so that $I_{2}=B\times\\{0\\}$. The element
$a=(1,0)$ satisfies (6.3).

Now, using (i) of the latter remark and the criterion Theorem 5.17, we present
some examples of the square root closures and their relation with the strict
square root closures for $M_{n}=\\{0,1/n,2/n,\ldots,n/n\\}$ $(n\geq 1)$,
$\Gamma(\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z},(1,0))$, and
$M(\alpha)$, $\alpha$ rational in $[0,1]$. We define for each prime number
$p\geq 2$,

$\displaystyle\mathfrak{}\mathbb{D}(p)$
$\displaystyle=\big{\\{}\frac{j}{p^{m}}\mid j\in\mathbb{Z},\,m\geq
0\big{\\}},$ $\displaystyle\mathbb{D}_{2}(p)$
$\displaystyle=\big{\\{}\frac{1}{p}\frac{i}{2^{k}}\mid i\in\mathbb{Z},\,k\geq
0\big{\\}}.$

Then $(\mathbb{D}(p),1)$ is a unital $p$-divisible subgroup of
$(\mathbb{R},1)$ that is not two-divisible for any prime $p>2$, and
$(\mathbb{D}_{2}(p),1)$ is a unital two-divisible subgroup of $\mathbb{R}$
containing a copy of $\frac{1}{p}\mathbb{Z}$. In particular,
$\mathbb{D}:=\mathbb{D}_{2}(1)$ is the group of dyadic numbers with strong
unit $1$, and $\mathbb{D}(p)=\frac{1}{p}\mathbb{D}$.

######  Example 6.14.

Let $M_{n}=\Gamma(\frac{1}{n}\mathbb{Z},1)$ for $n\geq 1$.

(1) If $n=1$, $\mathbf{D}(M_{1})=M_{1}$,
$\mathbf{C}(M_{1})=\Gamma(\mathbb{D},1)$.

(2) If $n=2$,
$M_{2}\varsubsetneq\mathbf{C}(M_{2})=\Gamma(\mathbb{D},1)=\mathbf{D}(M_{2})$.

(3) If $n=p$ is a prime number, then $\mathbf{D}_{p}$ is not two-divisible,
see [DvZa2, Ex 4.15(2)], and
$\mathbf{D}(M_{p})=\Gamma(\mathbf{D}_{2}(p),1)=\mathbf{C}(M_{p})$.

(4) Let $n=p_{1}^{j_{1}}\cdots p_{k}^{j_{k}}$ be a compound number, where each
$p_{j}>1$ and $p_{j}$ is prime. Define

$\mathbf{D}_{2}(p_{1},\ldots,p_{k})=\frac{1}{p_{1}}\mathbb{D}+\cdots+\frac{1}{p_{k}}\mathbb{D}=\big{\\{}\frac{1}{p^{j_{1}}_{1}}\frac{i_{1}}{2^{m_{1}}}+\cdots+\frac{1}{p^{j_{k}}_{k}}\frac{i_{k}}{2^{m_{k}}}\mid
i_{1},\ldots,i_{k}\in\mathbb{Z},\,m_{1},\ldots,m_{k}\geq 0\big{\\}}.$

Then $(\mathbf{D}_{2}(p_{1},\ldots,p_{k}),1)$ is a unital subgroup of
$(\mathbb{R},1)$ that is two-divisible and containing a copy of
$\frac{1}{p_{i}}\mathbb{Z}$ for each $i=1,\ldots,k$. So
$M_{n}\varsubsetneq\mathbf{D}(M_{n})=\Gamma(\mathbf{D}_{2}(p_{1},\ldots,p_{k}),1)=\mathbf{C}(M_{n})$.

(5) If $M=\Gamma(\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z},(1,0))$, then
$M\varsubsetneq\mathbf{D}(M)=\Gamma(\mathbb{D}\,\overrightarrow{\times}\,\mathbb{D},(1,0))=\mathbf{C}(M)$.

(6) For each irrational $\alpha$, $0<\alpha<1/2$, let
$M(\alpha)=\\{m+n\alpha\mid m,n\in\mathbb{Z},0\leq m+n\alpha\leq 1\\}$,
$G(\alpha)=\\{m+n\alpha\mid m,n\in\mathbb{Z}\\}$, and
$\mathbb{D}(\alpha)=\\{d+m\alpha/2^{n}\mid d\in\mathbb{D},n\geq
0,m\in\mathbb{Z}\\}$. Then $M(\alpha)$ is an MV-subalgebra of $[0,1]$
generated by $\alpha$, and
$M(\alpha)=\Gamma(G(\alpha),1)\varsubsetneq\mathbf{D}(M(\alpha))=\mathbf{C}(M(\alpha))=\Gamma(\mathbb{D}(\alpha),1)$.

If $\beta$ is an irrational in $[0,1/2]$, $M(\alpha)\cong M(\beta)$ iff
$M(\alpha)=M(\beta)$ iff $\alpha=\beta$; see [CDM, Cor 7.2.6]. Whence, we have
uncountably many non-isomorphic MV-algebras $M(\alpha)$’s without square root;
see [DvZa2, Ex 4.15(1)].

(7) Let $M$ be the Boolean algebra $M=\\{0,1\\}^{n}$, $n\geq 1$, then
$M\cong\Gamma(\mathbb{Z}^{n},(1,\ldots,1))$ and
$M=\mathbf{D}(M)\varsubsetneq\Gamma(\mathbb{D}^{n},(1,\ldots,1))=\mathbf{C}(M)$.

(8) Let $M=B$ be a Boolean algebra. Due to the Stone theorem, there is an
algebra $\mathcal{S}(B)$ of subsets of some set $\Omega\neq\emptyset$ such
that $B\cong\mathcal{S}(B)$. Define $\mathcal{F}(B)=\\{\chi_{A}\mid
A\in\mathcal{S}(B)\\}$ and let $G(B)=\\{\sum_{i=1}^{j}n_{i}\chi_{A_{i}}\mid
n_{i}\in\mathbb{Z},\chi_{A_{i}}\in\mathcal{F}(B),i=1,\ldots,j,\,j\geq 1\\}$.
Then $(G(B),1)$ is a unital $\ell$-group of integer-valued functions such that
$B\cong\mathcal{F}(B)=\Gamma(G(B),1)$. If $\mathbb{D}(B)=\\{f/2^{n}\mid f\in
G(B),n\geq 0\\}$, then $(\mathbb{D}(B),1)$ is a two-divisible unital
$\ell$-group such that
$B\cong\mathcal{F}(B)=\mathbf{D}(B)\varsubsetneq\mathbf{C}(B)=\Gamma(\mathbb{D}(B),1)$.

(9) Let $M_{i}=\Gamma(G_{i},u_{i})$ be an MV-algebra and let
$\mathbf{C}(M_{i})=\Gamma(D_{i},u_{i})$, where $(D_{i},u_{i})$ is a two-
divisible unital $\ell$-group, $G_{i}\subseteq D_{i}$ for each $i=1,\ldots,n$.
Then $M:=\prod_{i=1}^{n}M_{i}=\Gamma(G,u)$, where $u=(u_{1},\ldots,u_{n})\in
G:=\prod_{i=1}^{n}G_{i}$. If $D:=D_{1}\times\cdots\times D_{n}$, then $D$ is a
two-divisible $\ell$-group. Let $(h_{1},\ldots,h_{n})\in D$. Due to the
criterion Theorem 5.17, for each $h_{i}\in D_{i}$, there is an integer $m_{i}$
such that $2^{m_{i}}h_{i}\in G_{i}$. Let $m=m_{1}+\cdots+m_{n}$. Then
$2^{m}(h_{1},\ldots,h_{n})=(2^{m}h_{1},\ldots,2^{m}h_{n})\in G$, which by the
mentioned criterion gives
$\mathbf{C}(\prod_{i=1}^{n}M_{i})\cong\Gamma(D,u)=\prod_{i=1}^{n}\Gamma(D_{i},u_{i})=\prod_{i=1}^{n}\mathbf{C}(M_{i})$.

At the end of the section, we present the following problem.

######  Problem 6.15.

Let $(M;\oplus,^{\prime},0,1)$ be an MV-algebra. Does it have the square root
closure? If yes, is $M/I_{1}\times\mathbf{C}(M/I_{2})$ the square closure of
$M$?

##  7\. Greatest Subalgebras with Square Root

In the last section, we embed special pseudo MV-algebras in pseudo MV-algebras
with square roots. Now, we define a square root of an individual element of a
pseudo MV-algebra and find the greatest subalgebra of a special pseudo MV-
algebra with weak square root.

######  Definition 7.1.

An element $a$ of a pseudo MV-algebra $(M;\oplus,^{-},^{\sim},0,1)$ is called
a square root of an element $x\in M$ if $a$ satisfies conditions (Sq1) and
(Sq2). That is, $a\odot a=x$ and if $y\odot y\leq x$, then $y\leq a$ for all
$y\in M$.

Clearly, (Sq2) implies that if $x\in M$ has a square root $a\in M$, then $a$
is unique, so we denote the element $a$ by $\sqrt{x}$. Moreover, in each
pseudo MV-algebra, the top element $1$ has a square root and $\sqrt{1}=1$. Let
$x\leq y$ be elements of $M$ such that $\sqrt{x}$ and $\sqrt{y}$ exist. Then
$\sqrt{x}\odot\sqrt{x}\leq y$ and (Sq2) implies that $\sqrt{x}\leq\sqrt{y}$.
Therefore,

$\displaystyle x\leq y\mbox{ implies that }\sqrt{x}\leq\sqrt{y}.$ (7.1)

If $\sqrt{x}$ exists for each $x\in M$, then the mapping $r:M\to M$ defined by
$r(x)=\sqrt{x}$, $x\in M$, is a weak square root mapping, introduced in
[DvZa3, Def 3.2]. If $M$ is an MV-algebra, then $r$ is a square root mapping
introduced in [Höl]; see also [DvZa3]. Moreover, not every weak square root
mapping on a pseudo MV-algebra is automatically a square root mapping (see
[DvZa3, Ex 6.1-6.2]); on MV-algebras, they are equivalent. In those examples,
$\sqrt{x}$ exists for each $x\in M$, but the mapping $x\mapsto\sqrt{x}$, $x\in
M$, is not a square root; it is only a weak square root.

######  Example 7.2.

(i) Consider the subalgebra $M_{3}=\\{0,1/3,2/3,1\\}$ of the MV-algebra of the
real unit interval $[0,1]$. Then $\sqrt{0}=1/3$, $\sqrt{1/3}=2/3$,
$\sqrt{1}=1$, but $2/3$ does not have any square root. In the MV-algebra
$M_{2n-1}=\Gamma(\frac{1}{2n-1}\mathbb{Z},1)$, the square root of $1/(2n-1)$
exists and is equal to $n/(2n-1)$, whereas, the square root of $2(n-1)/(2n-1)$
fails. Similarly, in $M_{2n}=\Gamma(\frac{1}{2n}\mathbb{Z},1)$,
$\sqrt{0}=n/(2n)$ and $1/(2n)$ does not have a square root.

(ii) Consider the $\ell$-group
$G:=\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}$
defined as follows: For $(a,b,c),(x,y,z)\in G$, let

$\displaystyle(a,b,c)+(x,y,z)=(a+x,b+y,c+z+ay),\quad-(a,b,c,d)=(-a,-b,-c+ab).$

By [AnFe, P 138, E41], $(G;+,(0,0,0))$ is a non-Abelian linearly ordered
group. The element $u=(1,0,0)$ is a strong unit of $G$, and $M=\Gamma(G,u)$ is
a pseudo MV-algebra.

The pseudo MV-algebra $\Gamma(G,u)$ has no square root. Let
$m,n\in\mathbb{N}$. Consider the element $(1,-2n,2m)\in M$. We have
$(1,-n,m)\odot(1,-n,m)=((1,-n,m)+(-1,0,0)+(1,-n,m))\vee(0,0,0)=((0,-n,m)+(1,-n,m))\vee(0,0,0)=(1,-2n,2m)\vee(0,0,0)=(1,-2n,2m)$.
On the other hand, assume that $(x,y,z)\in M$ such that
$(x,y,z)\odot(x,y,z)\leq(1,-2n,2m)$. If $x=0$, then clearly
$(x,y,x)\leq(1,-n,m)$. If $x=1$, then $y\leq 0$ and
$(1,2y,2z)=(x,y,z)\odot(x,y,z)\leq(1,-2n,2m)$, so $2y<-2n$ or $2y=-2n$ and
$2z\leq 2m$. In both cases, $(x,y,z)\leq(1,-n,m)$. That is,
$\sqrt{(1,-2n,2m)}=(1,-n,m)$.

######  Proposition 7.3.

Let $M=\Gamma(G,u)$ be a pseudo MV-algebra.

  * (i)

If $\sqrt{0}$ exists, then $\sqrt{0}\leq\sqrt{0}^{-}\wedge\sqrt{0}^{\sim}$ and
$\sqrt{0}=\max\\{x\wedge x^{-}\mid x\in M\\}$.

  * (ii)

The element $\sqrt{0}$ exists if and only if the set $\\{y\in M\mid 2y\leq
u\\}$ has a top element. In addition, if $M$ is representable and $u/2$
exists, then $\sqrt{0}$ exists and $\sqrt{0}=u/2$.

  * (iii)

If $\sqrt{x}$ exists, then $\sqrt{x}\leq(x\oplus\sqrt{0})\wedge(\sqrt{0}\oplus
x)$.

###### Proof.

(i) We have $\sqrt{0}\odot\sqrt{0}=0$ and [GeIo, Prop 1.9] imply that
$\sqrt{0}\leq\sqrt{0}^{-},\sqrt{0}^{\sim}$. Then
$\sqrt{0}=\sqrt{0}\wedge\sqrt{0}^{-}\in\\{x\wedge x^{-}\mid x\in M\\}$. For
each $x\in M$, $(x\wedge x^{-})\odot(x\wedge x^{-})\leq x\odot x^{-}=0$, so
$x\wedge x^{-}\leq\sqrt{0}$. Therefore, $\sqrt{0}=\max\\{x\wedge x^{-}\mid
x\in M\\}$.

(ii) Since $\sqrt{0}-u+\sqrt{0}\leq(\sqrt{0}-u+\sqrt{0})\vee
0=\sqrt{0}\odot\sqrt{0}=0$, we have $\sqrt{0}-u\leq-\sqrt{0}$ and so
$2\sqrt{0}\leq u$, that is $\sqrt{0}\in\\{y\in M\mid 2y\leq u\\}$. Now, let
$y\in M$ be such that $2y\leq u$, then $2y-u\leq 0$ and so $y-u\leq-y$ which
implies that $y-u+y\leq 0$. Hence $y\odot y=(y-u+y)\vee 0=0$, consequently,
$y\leq\sqrt{0}$. Therefore, $\sqrt{0}=\max\\{y\in M\mid 2y\leq u\\}$. Now,
assume that $M$ is representable and $u/2$ exists. For each $y\in\\{y\in M\mid
2y\leq u\\}$, we have $2y\leq u=2(u/2)$, so $y\leq u/2$. Hence
$u/2=\max\\{y\in M\mid 2y\leq u\\}=\sqrt{0}$.

(iii) From $(x^{\sim}\odot\sqrt{x})\odot(x^{\sim}\odot\sqrt{x})\leq
x^{\sim}\odot(\sqrt{x}\odot\sqrt{x})\leq x^{\sim}\odot x=0$ and (Sq2), it
follows that $x^{\sim}\odot\sqrt{x}\leq\sqrt{0}$, whence
$\sqrt{x}=x\vee\sqrt{x}=x\oplus(x^{\sim}\odot\sqrt{x})\leq x\oplus\sqrt{0}$.
In a similar way, using $(\sqrt{x}\odot x^{-})\odot(\sqrt{x}\odot
x^{-})\leq(\sqrt{x}\odot\sqrt{x})\odot x^{-}\leq x\odot x^{-}$, we can prove
that $\sqrt{x}\leq\sqrt{0}\oplus x$. ∎

######  Lemma 7.4.

Let $M=\Gamma(G,u)$ be a symmetric representable pseudo MV-algebra, $u/2\in
M$, $N$ be a linearly ordered pseudo MV-algebra and $f:M\to N$ be a surjective
homomorphism. If $\sqrt{x}$ exists for some $x\in M$, then $f(\sqrt{x})$
exists and

$f(\sqrt{x})=\left\\{\begin{array}[]{ll}(f(x)+f(u))/2=\sqrt{f(x)}&\mbox{ if
}f(x)\neq 0\\\ f(u)/2=\sqrt{0}&\ \mbox{if }f(x)=0.\end{array}\right.$

###### Proof.

Since $u/2$ is defined in $M$, clearly, $f(u)/2$ is defined in $N$. Let $x\in
M$ be such that $\sqrt{x}$ exists. If $f(x)\neq 0$, then
$0<f(x)=f(\sqrt{x}\odot\sqrt{x})=f(\sqrt{x})\odot
f(\sqrt{x})=(f(\sqrt{x})-f(u)+f(\sqrt{x}))\vee 0$, so
$f(x)=f(\sqrt{x})-f(u)+f(\sqrt{x})$ (since $N$ is a chain and symmetric). It
follows that $f(x)+f(u)=2f(\sqrt{x})$, that is $f(\sqrt{x})=(f(x)+f(u))/2$.
Clearly, $f(\sqrt{x})\odot f(\sqrt{x})=f(x)$. Let $y\in N$ be such that
$y\odot y\leq f(x)$. Then $2y\leq f(x)+f(u)=2f(\sqrt{x})$, consequently,
$y\leq f(\sqrt{x})$. Thus, $\sqrt{f(x)}=f(\sqrt{x})$.

If $f(x)=0$, by Proposition 7.3(ii), in $N$ we have, $\sqrt{0}=f(u)/2$ and so
$f(u/2)=f(u)/2=\sqrt{0}=\sqrt{f(x)}$. ∎

######  Proposition 7.5.

Let $M=\Gamma(G,u)$ be a symmetric representable pseudo MV-algebra such that
$u/2\in G$ and let $x\in M$. Then

  * (i)

$\sqrt{0}$ exists and $\sqrt{0}=u/2\in\mathrm{C}(G)$.

  * (ii)

$\sqrt{x}$ exists if and only if $(x+u)/2\in M$, in either case,
$\sqrt{x}=(x+u)/2$. In either case, $x/2\in M$. Moreover, if $M$ is a
subdirect product of {$\Gamma(G_{i},u_{i})\mid i\in I\\}$, and
$x=(x_{i})_{i}\in M$, then $\sqrt{x}$ exists if and only if $\sqrt{x_{i}}$
exists in $\Gamma(G_{i},u_{i})$ for each $i\in I$. In either case,
$\sqrt{x}=(\sqrt{x_{i}})_{i}$.

  * (iii)

If $\sqrt{x}$ exists, then $\sqrt{x^{-}}$ exists in $M$, and
$\sqrt{x^{-}}=\sqrt{x}\rightarrow\sqrt{0}=\sqrt{x^{\sim}}$.

  * (iv)

If $\sqrt{x}$ and $\sqrt{y}$ exist, then $\sqrt{x\vee y}$ and $\sqrt{x\wedge
y}$ exists in $M$, $\sqrt{x\vee y}=\sqrt{x}\vee\sqrt{y}$, and $\sqrt{x\wedge
y}=\sqrt{x}\wedge\sqrt{y}$.

  * (v)

Let $\sqrt{x}$ and $\sqrt{y}$ exist. If $x\oplus y=y\oplus x$, then
$\sqrt{x\oplus y}=(\sqrt{x}\odot\sqrt{0}^{-})\oplus\sqrt{y}$.

  * (vi)

Let $\sqrt{x}$ and $\sqrt{y}$ exist. If $x+y=y+x$, then $\sqrt{x\odot
y}=(\sqrt{x}\odot\sqrt{y})\vee\sqrt{0}$.

  * (vii)

$\sqrt{z\odot z}$ and $\sqrt{z\oplus z}$ exist for all $z\in M$.

###### Proof.

Let $(G,u)$ be a representable unital $\ell$-group, $\\{(G_{i},u_{i})\mid i\in
I\\}$ be a family of linearly ordered unital $\ell$-groups, and
$f:G\to\prod_{i\in I}G_{i}$ be a subdirect embedding. By Proposition 7.3(ii),
$\sqrt{0}=u/2$. Let $\pi_{i}:\prod_{i\in I}G\to G_{i}$ be the $i$-th natural
projection map, $i\in I$. We denote $\pi_{i}\circ f$ simply by $f_{i}$.

(i) Choose $g\in G$. For each $i\in I$, from
$u_{i}=f_{i}(u)=f_{i}(u/2+u/2)=f_{i}(u/2)+f_{i}(u/2)$, it follows that
$f_{i}(u)/2$ exists in $G_{i}$ and $f_{i}(u)/2=f_{i}(u/2)$. By Proposition
7.3(ii), $\sqrt{0_{i}}$ exists in the symmetric pseudo MV-algebra
$\Gamma(G_{i},f_{i}(u))$ and $\sqrt{0_{i}}=f_{i}(u/2)$. Since $G_{i}$ is
linearly ordered, we have: If $f_{i}(g)+f_{i}(u/2)\leq f_{i}(u/2)+f_{i}(g)$,
then $f_{i}(u/2)+f_{i}(g)+f_{i}(u/2)\leq f_{i}(u)+f_{i}(g)=f_{i}(g)+f_{i}(u)$,
so $f_{i}(u/2)+f_{i}(g)\leq f_{i}(g)+f_{i}(u/2)$. Analogously, we proceed if
$f_{i}(u/2)+f_{i}(g)\leq f_{i}(g)+f_{i}(u/2)$. Thus, $u/2\in\mathrm{C}(G)$.
Therefore, $\sqrt{0}$ exists, $\sqrt{0}=u/2$, and belongs to $\mathrm{C}(G)$.

(ii) Let $x=(x_{i})_{i}\in M$ be such that $(x+u)/2$ exists. Then
$(x+u)/2\odot(x+u)/2=x$ and if $t\odot t\leq x$, then $2t-u\leq(t-u+t)\vee
0=t\odot t\leq x$ implies that $2t\leq x+u$ and so $t\leq(x+u)/2$. That is,
$\sqrt{x}=(x+u)/2=((x_{i}+u_{i})/2)_{i}$.

Conversely, let $\sqrt{x}$ exist. Choose $i\in I$. Due to Lemma 7.4, we have

$f_{i}(\sqrt{x})=(f_{i}(x)+f_{i}(u))/2=\sqrt{f_{i}(x)}.$

So, $f_{i}(2\sqrt{x})=(f_{i}(x)+f_{i}(u))=f_{i}(x+u)$ for all $i\in I$,
consequently $2\sqrt{x}=(x+u)$ and so $\sqrt{x}=(x+u)/2$.

In addition, $x/2=(x+u)/2-u/2=\sqrt{x}-u/2\leq u-u/2\leq u$, so that $x/2\in
M$.

From these ideas, we have also $\sqrt{x}$ exists in $M$ iff $\sqrt{x_{i}}$
exists in $\Gamma(G_{i},u_{i})$, and then $\sqrt{x}=(\sqrt{x_{i}})_{i}$.

(iii) $\sqrt{x}\rightarrow\sqrt{0}=\big{(}(u-(x+u)/2)+u/2\big{)}\wedge
u=(u-x/2)\wedge u=u-x/2=(x^{-}+u)/2$. Hence by (ii), $\sqrt{x^{-}}$ exists and
is equal to $\sqrt{x}\rightarrow\sqrt{0}$. Since $M$ is symmetric,
$x^{-}=x^{\sim}$ and $\sqrt{x^{-}}=\sqrt{x^{\sim}}$.

(iv) Let $\sqrt{x}$ and $\sqrt{y}$ be defined. For the first part, it suffices
to show that $((x\vee y)+u)/2$ exists and is equal to $\sqrt{x}\vee\sqrt{y}$.
According to [AnFe, 1.1.10], we have $(x\vee y)+u=(x+u)\vee(y+u)$. Since
$(x+u)/2$ and $(y+u)/2$ exist, so $((x\vee y)+u)/2=(x+u)/2\vee(y+u)/2$ exists.
Therefore, by (ii), $\sqrt{x\vee y}=\sqrt{x}\vee\sqrt{y}$.

For the second part, we can use analogous statements as in the first one, or
we can use (iii) and the first part.

(v) Due to (iii), it is enough to assume that $M$ is linearly ordered. First,
we note that if $x+y=y+x$, then $x/2+y/2=x/2+x/2$. Indeed, since $M$ is a
chain assume that $x+y/2\leq y/2+x$, then $y+x=x+y\leq y/2+x+y/2$ and so
$y/2+x\leq x+y/2$. That is $x+y/2=y/2+x$. Similarly, using $x+y/2=y/2+x$, we
can prove that $x/2+y/2=y/2+x/2$.

Check $(\sqrt{x}\odot\sqrt{0}^{-})\oplus\sqrt{y}=((x+u)/2\odot
u/2)\oplus(y+u)/2=((x+u)/2-u+u/2)\oplus(y+u)/2=(x/2+(y+u)/2)\wedge u$. If
$u\leq x+y$, then $\sqrt{x\oplus y}=\sqrt{u}=u$. On the other hand, $u-y\leq
x$ and so $u/2-y/2=(u-y)/2\leq x/2$ which implies that $u/2\leq x/2+y/2$.
Thus, $u\leq x/2+y/2+u/2=x/2+(y+u)/2$, consequently
$(\sqrt{x}\odot\sqrt{0}^{-})\oplus\sqrt{y}=(x/2+(y+u)/2)\wedge u=u$, that is
$\sqrt{x\oplus y}=u=(\sqrt{x}\odot\sqrt{0}^{-})\oplus\sqrt{y}$. If $x\oplus
y<u$, then $y\oplus x<u$, so by the assumption we have $x+y=x\oplus y=y\oplus
x=y+x$. Hence $(\sqrt{x}\odot\sqrt{0}^{-})\oplus\sqrt{y}=(x/2+(y+u)/2)\wedge
u=((x+y+u)/2)\wedge u=((x+y)+u)/2=((x\oplus y)+u)/2=\sqrt{x\oplus y}$, since
$x+y=y+x$.

(vi) Again, we can assume $M$ is linearly ordered. By (i), we have

$\displaystyle(\sqrt{x}\odot\sqrt{y})\vee\sqrt{0}$ $\displaystyle=$
$\displaystyle\big{(}((x+u)/2-u+(y+u)/2)\vee 0\big{)}\vee u/2=(x+y)/2\vee u/2$
$\displaystyle=$ $\displaystyle((x+y-u)/2\vee 0)+u/2=\big{(}((x+y-u)\vee
0)/2\big{)}+u/2$ $\displaystyle=$ $\displaystyle((x\odot y)+u)/2.$

Hence by (iii), $\sqrt{x\odot y}=(\sqrt{x}\odot\sqrt{y})\vee\sqrt{0}$. Note
that since $G$ is linearly ordered, for each $x\in G$ with $x/2\in G$, we have
$x/2\vee 0=(x\vee 0)/2$.

(vii) By (iii), we assume $M$ is linearly ordered. Let $z\in M$. We have
$(z\vee\sqrt{0})\odot(z\vee\sqrt{0})=(z\odot z)\vee(z\odot\sqrt{0})$. If
$z\leq\sqrt{0}$, then $z\odot z=0=(z\odot z)\vee(z\odot\sqrt{0})$. If
$\sqrt{0}<z$, then $z\odot\sqrt{0}\leq z\odot z$ and so $(z\odot
z)\vee(z\odot\sqrt{0})=z\odot z$. Now, let $y\in M$ such that $y\odot y\leq
z\odot z$. If $y\odot y=0$, then $y\leq\sqrt{0}$ and so $y\leq z\vee\sqrt{0}$.
Otherwise, $0<y\odot y$ implies that $y\odot y=y-u+y=2y-u$ and $z\odot
z=z-u+z=2z-u$. Hence $2y-u\leq 2z-u$, so $2y\leq 2z$ consequently $y\leq z\leq
z\vee\sqrt{0}$. Therefore, $\sqrt{z\odot z}=z\vee\sqrt{0}$.

Since $\sqrt{z^{-}\odot z^{-}}$ exists, then by (iv), $\sqrt{z\oplus
z}=\sqrt{(z^{-}\odot z^{-})^{\sim}}$ exists. ∎

Let $X$ be a subalgebra of a pseudo MV-algebra $M$. We say that $X$ has the
weak square root property if $\sqrt{x}$ exists in $M$ for each $x\in X$,
$\sqrt{x}\in X$. In this case, the mapping $r_{X}(x)=\sqrt{x}$, $x\in X$, is a
weak square root on $X$.

We note that the Boolean skeleton $\mathrm{B}(M)$ of a pseudo MV-algebra has
the weak square root property iff $\sqrt{0}$ exists in $M$ and $\sqrt{0}=0$.
Indeed, let $\sqrt{0}=0$. The Boolean algebra $\mathrm{B}(M)$ has the unique
square root $r_{\mathrm{B}(M)}=\mbox{\rm Id}_{B(M)}$. Let $b\in\mathrm{B}(M)$
and let $y\in M$ be such that $y\odot y\leq b$. Then $(y\wedge
b^{-})\odot(y\wedge b^{-})\leq 0=\sqrt{0}$, so that $y\wedge b^{-}\leq 0$ and
$y\leq b$. This implies $\sqrt{b}$ exists in $M$, $\sqrt{b}=b$, and
$\mathrm{B}(M)$ has the weak square root property.

The converse statement is clear because the only square root mapping on a
Boolean algebra is the identity map.

######  Theorem 7.6.

Let $M=\Gamma(G,u)$ be a representable symmetric pseudo MV-algebra such that
$u/2\in G$ and define $X_{1}=\\{x\in M\mid\sqrt{x}\mbox{ exists}\\}$.

  * (i)

If $M$ is an MV-algebra, then $X_{1}$ is a subalgebra of $M$ and $X_{1}$ has
the weak square root property.

  * (ii)

If $M$ is an MV-algebra and $u/2^{n}$ exists for all $n\in\mathbb{N}$, then
the greatest subalgebra of $M$ with weak square root exits.

  * (iii)

We have $x\odot x\in X_{1}$ for all $x\in M$.

###### Proof.

(i) By Proposition 7.5(i), $0\in X_{1}$, so $0,1\in X_{1}$ (recall that
$\sqrt{1}=1$). Also, by Proposition 7.5(iii) and (iv), $X_{1}$ is closed under
the operations $\vee$, $\wedge$ and ′. In addition, since $\oplus$ is
commutative, Proposition 7.5(v) implies that $X_{1}$ is closed under $\oplus$.
Therefore, $X_{1}$ is a subalgebra of $M$.

(ii) Let $u/2^{n}$ exist for all $n\in\mathbb{N}$. Set $X_{n+1}:=\\{x\in
X_{n}\mid\sqrt{x}\mbox{ exists}\\}$ for all $n\in\mathbb{N}$. Using induction
and reasoning similar to (i), $X_{n+1}$ is a subalgebra of $X_{n}$ and so
$X:=\bigcap_{n\in\mathbb{N}}X_{n}$ is a subalgebra of $M$. Clearly, $X$ is a
pseudo MV-algebra with square root. It is the greatest subalgebra of $M$ with
square root.

(iii) Let $x\in M$. Use a standard subdirect decomposition to linear factors.
By Proposition 7.5(vi), $\pi_{i}(f(x))\vee\sqrt{0_{i}}$ is a square root of
$\pi_{i}(f(x))\odot\pi_{i}(f(x))=\pi_{i}(f(x\odot x))$. It follows that for
each $i\in I$, we have

$\displaystyle\pi_{i}(f(x\odot x))$ $\displaystyle=$
$\displaystyle\big{(}\pi_{i}(f(x))\vee\sqrt{0_{i}}\big{)}\odot\big{(}\pi_{i}(f(x))\vee\sqrt{0_{i}}\big{)}$
$\displaystyle=$
$\displaystyle\big{(}\pi_{i}(f(x))\vee\pi_{i}(f(\sqrt{0}))\big{)}\odot\big{(}\pi_{i}(f(x))\vee\pi_{i}(f(\sqrt{0}))\big{)}\quad\mbox{
by Lemma \ref{9.3}}$ $\displaystyle=$ $\displaystyle\pi_{i}\circ
f\big{(}(x\vee\sqrt{0})\odot(x\vee\sqrt{0})\big{)}.$

Hence, $f(x\odot x)=f\big{(}(x\vee\sqrt{0})\odot(x\vee\sqrt{0})\big{)}$, and
so $x\odot x=(x\vee\sqrt{0})\odot(x\vee\sqrt{0})$. On the other hand, if $y\in
M$ such that $y\odot y\leq x\odot x$, then
$\pi_{i}(f(y))\odot\pi_{i}(f(y))\leq\pi_{i}(f(x))\odot\pi_{i}(f(x))$.
Proposition 7.5(vii) and Lemma 7.4 entail
$\pi_{i}(f(y))\leq\pi_{i}(f(x))\vee\pi_{i}(f(\sqrt{0}))=\pi_{i}(f(x\vee\sqrt{0}))$
for all $i\in I$ which implies that $f(y)\leq f(x\vee\sqrt{0})$ and so $y\leq
x\vee\sqrt{0}$. Therefore, $\sqrt{x\odot x}$ exists and is equal to
$x\vee\sqrt{0}$. That is, $x\odot x\in X_{1}$. ∎

To illustrate the latter result, we have that if $M=M_{3}$, then $u/2\notin
M$, and $X_{1}=\\{0,1/3,1\\}$ is not a subalgebra of $M$. The same is true for
$M=M_{2n-1}$, where $u/2\notin M_{2n-1}$, $\sqrt{1/(2n-1)}=n/(2n-1)$, so
$1/(2n-1)\in X_{1}=\\{0,1/(2n-1),3/(2n-1),\ldots,(2n-1)/(2n-1)\\}$, but
$(1/(2n-1))^{\prime}=2(n-1)/(2n-1)\notin X_{1}$, and $X_{1}$ is not a
subalgebra. On the other side, if $M=M_{2n}$, then
$X_{1}=\\{0,2/2n,4/2n,\ldots,2n/2n\\}$ which is a subalgebra on $M$.

A close connection exists between the square root of $0$ and the square root
of idempotent elements of a pseudo MV-algebra. Indeed, in a pseudo MV-algebra
$M$, $\sqrt{0}$ exists if and only if $\sqrt{b}$ exists for all
$b\in\mathrm{B}(M)$:

######  Proposition 7.7.

Let $(M;\oplus,^{-},^{\sim},0,1)$ be a pseudo MV-algebra such that $\sqrt{0}$
exists. Then:

  * (i)

If $\sqrt{x}$ exists, then $\sqrt{x}\leq(x\oplus\sqrt{0})\wedge(\sqrt{0}\oplus
x)$.

  * (ii)

For all $b\in\mathrm{B}(M)$, $\sqrt{b}$ exists and $\sqrt{b}=b\vee\sqrt{0}$.

  * (iii)

If $\sqrt{0}=0$, then $\sqrt{b}=b$ for all $b\in\mathrm{B}(M)$. In addition,
if $x\in M$ is such that $\sqrt{x}$ exists, then $x\in\mathrm{B}(M)$.

###### Proof.

(i) Let $x\in M$ and $\sqrt{x}$ exist. Then $\sqrt{x}\odot\sqrt{x}=x$ implies
that $(\sqrt{x}\odot\sqrt{x})\odot x^{-}=0$ and
$x^{\sim}\odot(\sqrt{x}\odot\sqrt{x})=0$. Hence, $(\sqrt{x}\odot
x^{-})\odot(\sqrt{x}\odot x^{-})=0$ so $\sqrt{x}\odot x^{-}\leq\sqrt{0}$ and
$x^{\sim}\odot\sqrt{x}\leq\sqrt{0}$. Thus
$\sqrt{x}=x\vee\sqrt{x}=x\oplus(x^{\sim}\odot\sqrt{x})\leq x\oplus\sqrt{0}$.
Similarly, we can show that $\sqrt{x}\leq\sqrt{0}\oplus x$.

(ii) Let $b\in\mathrm{B}(M)$. The condition (Sq1) holds, because
$(b\vee\sqrt{0})\odot(b\vee\sqrt{0})=b\vee(b\odot\sqrt{0})=b$. From (i), we
conclude that $\sqrt{b}=b\vee\sqrt{0}$.

(iii) It is a straight consequence of (i) and (ii). ∎

######  Corollary 7.8.

Consider the assumptions of Theorem 7.6. Then $\mathrm{B}(M)\subseteq X$,
where $X:=\bigcap_{n\in\mathbb{N}}X_{n}$ is the greatest subalgebra of $M$
with square root.

###### Proof.

It follows from Theorem 7.6 and Proposition 7.7. ∎

##  8\. Conclusion

Square roots on MV-algebras were generalized in [DvZa3] to square roots on
pseudo MV-algebras. In the present study, which is divided into two parts, we
extended results on square roots and weak square roots.

Part I. We are interested in new characterizations of square roots with accent
to strict square roots; see Theorem 3.3. We showed that a strict square root
implies that the corresponding $\ell$-group is two-divisible, Theorem 4.1,
Theorem 4.4, and Theorem 4.9. We characterized semisimple MV-algebras with
strict square root; see Theorem 4.17.

Part II. We exhibited situations when a pseudo MV-algebra can be embedded into
a pseudo MV-algebra with square root, see, e.g., Theorem 5.3. We defined a
strict square root closure of a pseudo MV-algebra, showing that every MV-
algebra has a strict square root closure, Theorem 5.10. We also introduced a
square root closure of an MV-algebra, Theorem 6.12, and we compared both
notions of closures. Finally, we have investigated the existence of square
roots not for all elements but only for some. We showed that in some symmetric
representable pseudo MV-algebra such that one-half of the top element exists,
there is a maximal subalgebra with the weak square root property; see Theorem
7.6.

The research brought some interesting questions like (1) the existence of a
strict square root closure for some pseudo MV-algebras that are not
commutative, for example, for linearly ordered ones, and (2) the existence of
a maximal subalgebra with square root that is not weak.

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