# Commutative unital rings elementarily equivalent to prescribed product rings

Paola D’Aquino  Dipartimento di Matematica e Fisica, Università della Campania
L.Vanvitelli, Viale Lincoln 5, 81100 Caserta, Italy. E-mail:
<EMAIL_ADDRESS>and  Angus J. Macintyre  School of Mathematics,
University of Edinburgh, EH9 3FD Edinburgh, U.K. E-mail:
<EMAIL_ADDRESS>
###### Abstract.

The classical work of Feferman-Vaught [9] gives a powerful, constructive
analysis of definability in (generalized) product structures, and certain
associated enriched Boolean structures. Here, by closely related methods, but
in the special setting of commutative unital rings, we obtain a kind of
converse allowing us to determine in interesting cases, when a commutative
unital R is elementarily equivalent to a “nontrivial” product of a family of
commutative unital rings $R_{i}$. We use this in the model-theoretic analysis
of residue rings of models of Peano Arithmetic.

###### Key words and phrases:

Product of structures, models of Peano Arithmetic, idempotents, connected
rings

###### 1991 Mathematics Subject Classification:

03H15, 03C20, 03C40

##  1\. Introduction

In several recent investigations ([8], [4]) the following question was
considered:

When is a ring $R$ elementarily equivalent (in the standard first-order
language $\mathcal{L}_{rings}$ for unital rings) to some interesting product
$\prod_{i\in I}R_{i}$, where $R_{i}$ are commutative unital rings?

We do not define interesting here, but it is clearly not interesting to take
$I=\\{0\\}$ and $R_{0}=R$.

The question makes sense too for restricted products such as adèle rings, for
example $\mathbb{A}_{\mathbb{Q}}$, a restricted product of the rings
$\mathbb{Q}_{p}$ ($p$ prime) and $\mathbb{R}$.

The two cases that stimulated our work are

  1. i)

rings $\mathcal{M}/n\mathcal{M}$, where $\mathcal{M}$ is the ring associated
to a model of first-order Peano Arithmetic $PA$, in [4]. The hardest case is
when $n$ is nonstandard and divisible by infinitely many primes;

  2. ii)

classical adèle rings $\mathbb{A}_{K}$, $K$ a number field, in [8].

The ring $\mathcal{M}/n\mathcal{M}$ is not a classical product of the rings
$\mathcal{M}/q\mathcal{M}$, where $q$ runs through the maximal prime powers
dividing $n$, but is an internal product (in the sense of nonstandard
analysis). It turns out that $\mathcal{M}/n\mathcal{M}$ is elementarily
equivalent to the full external product of the $\mathcal{M}/q\mathcal{M}$’s.
This is rather surprising, and depends on refinements of the great work of
Feferman and Vaught from 1959, see [9].

Our work on case ii) contributes greatly to the understanding of the model
theory of rings $\mathcal{M}/n\mathcal{M}$, a topic previously neglected
(except when $n$ is prime, in [12]), but revived after Zilber’s [15].

### 1.1.

In this paper we refine a specially important case of the classical Feferman-
Vaught Theorem ([9, Theorem 3.1]) on products of first-order structures (some
results of [9] were anticipated by Mostowski in [13]). The general case
concerned a first-order language $\mathcal{L}$, and a family
$\\{R_{i}\\}_{i\in I}$ of $\mathcal{L}$-structures, out of which we form the
classical $\mathcal{L}$-structure $R$ which is the full product of the
$R_{i}$, and the objective was to analyze definability (and decidability) in
$R$ in terms of definability in the “stalks” $R_{i}$, and in terms of the
atomic Boolean algebra $\mathcal{P}(I)$. The method is amazingly general and
uniform, but has the defect that in general in [9] one has to appeal to
various auxiliary Boolean structures whose definability theory may be
complicated. But even in the general setting Weispfenning used the method (and
extensions to generalized products) to obtain a wide variety of important
algebraic applications ([14]). In this paper we restrict to the case when
$\mathcal{L}_{rings}$ is the usual first-order language for rings, and the
$R_{i}$ are unital, commutative rings, which are connected in the sense that
$0$ and $1$ are the only idempotents and $0\not=1$. We want a criterion (and
in fact we get a necessary and sufficient condition) for when a commutative
unital ring $R$ is elementarily equivalent to some product of connected rings
$R_{i}$ as above. What we get is a weak converse to Theorem 3.1 of [9]. That
theorem is of much greater generality than anything we need here, with many
notational complications. We work entirely in the language
$\mathcal{L}_{rings}$ of ring theory, coding the auxiliary Boolean formalism
of [9] in terms of ring idempotents, and in particular minimal idempotents
(the coding is always done in the ring $R$ where we are working). The analysis
of Theorem 3.1 of [9] in our ring theoretic setting leads to the isolation of
a set AXIOMS in the ring language, and a proof that products of connected
rings satisfy the axioms we isolate. Our main result Theorem 2.10 shows that
the models of AXIOMS are exactly the rings which are elementarily equivalent
to a product of connected unital rings. We have found no such “converse” in
[9].

Examples of connected rings 1\. Every integral domain is connected.

2\. Any commutative local ring is connected, as we now show. If $S$ is local
and $x\in S$ satisfies $x^{2}=x$ then either $x\in\mu$ or $x-1\in\mu$, where
$\mu$ is the unique maximal ideal of $S$. If $x\not\in\mu$ then $x$ is a unit,
and so $0=ux(x-1)=x-1$ where $u$ is the inverse of $x$. If $x-1\not\in\mu$
then $x-1$ is a unit, and hence $0=x(x-1)u$ where $u$ is the inverse of $x-1$.
So, either $x=1$ or $x=0$. It cannot happen that $x,x-1\in\mu$ otherwise
$1\in\mu$, which is a contradiction.

### 1.2.

The Feferman-Vaught analysis of products and generalized products of first
order $\mathcal{L}$-structures in [9] is done in terms of $\mathcal{L}$ and
also the language of Boolean algebras and expansions of it.

In the case of products of connected commutative unital rings $R=\prod_{i\in
I}R_{i}$ the hypothesis that the $R_{i}$ are connected is used to make the
following identification of the Boolean algebra of the power set of $I$ in
ring theoretic terms inside $R$. The elements of $R$ are functions $f$ defined
on $I$ such that $f(i)\in R_{i}$, and the Boolean algebra $\mathbb{B}$
associated to $R$ is the set of functions in $R$ which take values in
$\\{0,1\\}$, i.e. the set of idempotents of $R$ (here the hypothesis of
connectness of $R_{i}$ is crucial), with a natural Boolean structure defined
on it (see the next section). The individual element $i\in I$ is identified
with the function $\chi_{i}$ with $\chi_{i}(i)=1$, and $\chi_{i}(j)=0$ for
$j\in I$ and $j\not=i$, and these correspond to the atoms of $\mathbb{B}$.

In the case of a general commutative unital ring $R$ (not a product of
connected commutative unital rings) we will use the same notation $\mathbb{B}$
to denote the Boolean algebra of idempotent elements of $R$, and $e$ will be
an atom of $\mathbb{B}$. Note that for $R=\prod_{i\in I}R_{i}$ ($R_{i}$
connected commutative unital rings) the Boolean algebra $\mathbb{B}$ is
complete while in the general case $\mathbb{B}$ may not be complete. Inspired
by the Feferman-Vaught arguments in the proof of Theorem 3.1 in [9] we will
identify axioms, mainly in terms of $\mathbb{B}$, for $R$ in order to have $R$
elementarily equivalent to a product of connected commutative unital rings
which will be determined in terms of the atoms of $\mathbb{B}$.

We will find these properties, and prove a theorem applicable to both cases
(i) and (ii) of page 1.

Note that when $I$ is infinite and each $R_{i}$ is nontrivial then
$\prod_{i\in I}R_{i}$ is uncountable.

##  2\. Rings with connected stalks: a Feferman-Vaught theory

###  2.1. Basic idea

We will work in the languages of unital rings $\mathcal{L}_{rings}$ which
contains $+,-,\cdot,0,1$, and in the language of Boolean algebras
$\mathcal{L}_{B}$ which contains
$\boldsymbol{\wedge},\boldsymbol{\vee},\boldsymbol{\neg},0,1$. Let $R$ be any
commutative unital ring, $\mathbb{B}$ the Boolean algebra of idempotents of
$R$, interpretable in $R$ via

$\begin{array}[]{ccc}e\boldsymbol{\wedge}f&=&ef\\\
e\boldsymbol{\vee}f&=&1-(1-e)(1-f)\\\ &=&e+f-ef\\\ \boldsymbol{\neg}e&=&1-e\\\
0&=&0\\\ 1&=&1\end{array}$

One may note that on the set $\mathbb{B}$ we have the equivalent Boolean ring
structure $\mathbb{B}_{ring}$ (generally, not a subring of $R$) with

$\begin{array}[]{ccc}x\oplus
y&=&(x{\boldsymbol{\wedge}}\boldsymbol{\neg}y)\boldsymbol{\vee}(\boldsymbol{\neg}x\boldsymbol{\wedge}y)\\\
x\odot y&=&x\boldsymbol{\wedge}y\ (=x\cdot y)\\\ 0&=&0\\\ 1&=&1\end{array}$

On $\mathbb{B}$ we have the usual definable partial order $\leq$. Of special
importance to us are the atoms of $\mathbb{B}$ (equivalently, the minimal
idempotents $\not=0,1$).

######  Remark 2.1.

The set of idempotents of a ring $R$ is quantifier-free definable in the ring
language and via the above interpretation of
$\boldsymbol{\wedge},\boldsymbol{\vee},\boldsymbol{\neg}$ the Boolean
structure on $\mathbb{B}$ is quantifier-free in terms of $+,-,\cdot,0,1$ in
$R$. In this way any formula in $\mathcal{L}_{B}$ can be effectively
translated into a formula of $\mathcal{L}_{rings}$. We will refer to these
formulas as B-formulas. In this way we are going to work in a one sorted
language as opposed to the language in [9] which is a two sorted language, one
for the family of structures indexed by $I$, and the other of the Boolean
algebra.

######  Lemma 2.2.

For any idempotent element $e$ in $\mathbb{B}$ we have natural isomorphisms

$R/(1-e)R\cong eR\cong R_{e},$

where $R_{e}$ is the localization of $R$ at $\\{e^{n}:n\geq 0\\}$
$(=\\{1,e\\})$.

###### Proof.

The ring $R$ is the direct sum of the ideals $eR$ and $(1-e)R$, giving the
first isomorphism. The kernel of the map $R\rightarrow R_{e}$ consists of the
$x$ so that $xe=0$, and this is the same as $x=xe+x(1-e)=x(1-e)$, thus the
same as $x\in(1-e)R$. The ring $eR$ is unital with unit $e$. Using the
universal property of localization we can explicitly define the isomorphism
between $R_{e}$ and the ring $eR$ as $\frac{r}{s}\mapsto er$, where $s=1,e$. ∎

Notation.  An element of $R_{e}$ will be denoted by $f_{e}$ for some $f\in R$,
and by Lemma 2.2 it corresponds to $ef$ in $eR$.

By an atom in $R$ we will always mean an atom in the Boolean algebra
$\mathbb{B}$ of idempotents of the ring $R$, and the order $<$ is the order on
$\mathbb{B}$.

######  Corollary 2.3.

A nonzero idempotent $e$ of $R$ is an atom if and only if $R_{e}$ is
connected.

###### Proof.

($\Rightarrow$) If $e$ is an atom in $R$ then $R_{e}$ is the ring with two
elements which is clearly connected.

($\Leftarrow$) Let $e$ be a nonzero idempotent of $R$, and assume $R_{e}$ is
connected. We show that $e$ is an atom in $R$. Let $f\in R$ be an idempotent,
and $0<f<e$. Then $fe=f$ (by the Boolean operations on $\mathbb{B}$), and so
$f$ is not in the kernel of the localizing map at $e$ from $R$ to $R_{e}$
since $f\not=0$. By connectedness of $R_{e}$, $f=e$ in $R_{e}$, and so
$e(f-e)=0$ in $R$, which implies $ef=e$. Hence, $f=e$ in $R$, a contradiction.
∎

We have defined the localizations of $R$ at any idempotent, but the
interesting case is when $e$ is an atom. Only in this case we have that
$R_{e}$ is connected, and this is a fundamental property of the stalk $R_{e}$
of $R$. We will identify axioms which characterize commutative unital rings
which are elementarily equivalent to the product of their stalks at atoms.

Axiom 1. The Boolean algebra $\mathbb{B}$ is atomic.

This clearly holds in any product of commutative unital rings, and so in
particular it holds in products of commutative connected unital rings. Notice
that there are commutative rings where Axiom 1 fails, in particular take $R$
the Boolean ring associated to an atomless Boolean algebra.

The following property is well known, see e.g. [10]. We will often use it in
the proofs of our results.

######  Lemma 2.4.

If $f\in\mathbb{B}$, $f\not=0$ then

$f=\bigvee_{\begin{subarray}{c}e\leq f\\\ e\mbox{ {\tiny
atom}}\end{subarray}}e.$

The proof of the lemma follows easily from the following equivalences that are
proved on page 118 of [10]

  1. (1)

a Boolean algebra $B$ is atomic;

  2. (2)

every element $b\in B$ is the supremum of the set of atoms $a\leq b$;

  3. (3)

$1\in B$ is the supremum of all atoms.

###  2.2. Completeness

In a product $R=\prod_{i\in I}R_{i}$ where $R_{i}$’s are connected commutative
unital rings, the Boolean algebra $\mathbb{B}$ is complete. This means that
every non empty subset of $\mathbb{B}$ has a supremum (and hence also an
infimum) with respect to the usual order on the Boolean algebra.

This is not a first order condition, and we need to find a replacement. We use
a variant, adapted to ring theory.

For any $f_{0},\ldots,f_{n}\in R$, and any atom $e\in\mathbb{B}$ we will use
the notation $\overline{f}$ for the tuple $f_{0},\ldots,f_{n}$, and
$(\overline{f})_{e}$ for the tuple $(f_{0})_{e},\ldots,(f_{n})_{e}$ of
elements of the stalk $R_{e}$.

Axiom 2. For all formulas $\theta(x_{0},\ldots,x_{n})$ in
$\mathcal{L}_{rings}$, and for all $f_{0},\ldots,f_{n}\in R$ there is a unique
element $\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$ in $\mathbb{B}$ such
that

$R_{e}\models\theta((f_{0})_{e},\ldots,(f_{n})_{e})\mbox{ \hskip 7.22743pt iff
\hskip 7.22743pt}e\leq\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$

for all atoms $e$ in $\mathbb{B}$.

The elements $\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$ occur in [9]
with a different notation (see Definition 2.1 on page 63 of [9]). The
$\llbracket\ \cdot\ \rrbracket$ notation comes from Boolean-valued model
theory [2], not yet prominent when [9] was written.

Notice that Axiom 2 is a scheme of axioms, one for each formula
$\theta(x_{0},\ldots,x_{n})$. This is our substitute for completeness.

Axiom 2 is clearly true in products of unital connected commutative rings,
since $\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$ is
$\bigvee\\{i:R_{i}\models\theta(f_{0}(i),\ldots,f_{n}(i))\\}$.

###  2.3. Boolean values

We need to check preservation properties of
$\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$. We will use Axiom 2 in the
proofs.

######  Lemma 2.5.

Let $\theta_{1},\theta_{2}$ formulas, and $f_{0},\ldots,f_{n}\in R$. Then

$\llbracket(\theta_{1}\wedge\theta_{2})(\overline{f})\rrbracket=\llbracket\theta_{1}(\overline{f})\rrbracket\wedge\llbracket\theta_{2}(\overline{f})\rrbracket,$

where $\overline{f}=f_{0},\ldots,f_{n}$.

###### Proof.

Let $e$ be an atom. The following equivalences hold:

$e\leq\llbracket(\theta_{1}\wedge\theta_{2})(\overline{f})\rrbracket$ iff
$R_{e}\models(\theta_{1}\wedge\theta_{2})((\overline{f})_{e})$ iff
$R_{e}\models\theta_{1}((\overline{f})_{e})$ and
$R_{e}~{}\models~{}\theta_{2}~{}((\overline{f})_{e}~{})$ iff
$e\leq\llbracket\theta_{1}((\overline{f})_{e})\rrbracket$ and
$e\leq\llbracket\theta_{2}((\overline{f})_{e})\rrbracket$ iff
$e\leq\llbracket\theta_{1}((\overline{f})_{e})\rrbracket\wedge\llbracket\theta_{2}((\overline{f})_{e})\rrbracket$.

The result now follows from Lemma 2.4. ∎

######  Lemma 2.6.

Let $\theta$ be a formula, and $f_{0},\ldots,f_{n}\in R$. Then

$\llbracket\neg\theta(\overline{f})\rrbracket=\neg\llbracket\theta(\overline{f})\rrbracket.$

###### Proof.

Recall that for any atom $e$ and any $\alpha$ in a Boolean algebra either
$e\leq\alpha$ or $e\leq\neg\alpha$, and not both inequalities hold. Let $e$ be
an atom. The following equivalences hold:

$e\leq\llbracket\neg\theta(\overline{f})\rrbracket$ iff
$R_{e}\models\neg\theta((\overline{f})_{e})$ iff
$R_{e}\not\models\theta((\overline{f})_{e})$ iff
$e\not\leq\llbracket\neg\theta(\overline{f})\rrbracket$ iff

$e\leq\neg\llbracket\theta(\overline{f})\rrbracket$.

The result now follows from Lemma 2.4. ∎

######  Lemma 2.7.

Let $\theta_{1},\theta_{2}$ be formulas, and $f_{0},\ldots,f_{n}\in R$. Then

$\llbracket(\theta_{1}\vee\theta_{2})(\overline{f})\rrbracket=\llbracket\theta_{1}(\overline{f})\rrbracket\vee\llbracket\theta_{2}(\overline{f})\rrbracket,$

where $\overline{f}=f_{0},\ldots,f_{n}$.

###### Proof.

This follows from the two preceding lemmas. ∎

Axiom 3. For all formulas $\theta(v_{0},\ldots,v_{n},w)$ in
$\mathcal{L}_{rings}$, and for all $f_{0},\ldots,f_{n}\in R$ there is $g\in R$
such that

$\llbracket\exists
w\theta(\overline{f},w)\rrbracket\leq\llbracket\theta(\overline{f},g)\rrbracket$

where $\overline{f}=f_{0},\ldots,f_{n}$.

######  Lemma 2.8.

Axiom 3 is true in $\prod_{i\in I}R_{i}$ where $R_{i}$’s are connected
commutative unital rings.

###### Proof.

Let $h=\llbracket\exists w\theta(\overline{f},w)\rrbracket\in\mathbb{B}$ be
defined as

$h(i)=\left\\{\begin{array}[]{ll}1&\mbox{ if }R_{i}\models\exists
w\theta(f_{0}(i),\ldots,f_{n}(i),w)\\\ 0&\mbox{ if }R_{i}\not\models\exists
w\theta(f_{0}(i),\ldots,f_{n}(i),w)\end{array}\right.$

Now we define $g\in\prod R_{i}$ as follows

$g(i)=\left\\{\begin{array}[]{ll}g_{i}&\mbox{ if
}R_{i}\models\theta(f_{0}(i),\ldots,f_{n}(i),g_{i})\\\ 0&\mbox{ if
}R_{i}\not\models\exists
w\theta(f_{0}(i),\ldots,f_{n}(i),w)\end{array}\right.$

We use the axiom of choice in order to pick a $g_{i}\in R_{i}$ such that
$h(i)=1$ and $R_{i}\models\theta(f_{0}(i),\ldots,f_{n}(i),g_{i})$. If
$R_{i}\not\models\exists w\theta(f_{0}(i),\ldots,f_{n}(i),w)$ then $g(i)$ can
be chosen arbitrarily, we opted to define it $0$. Let
$t=\llbracket\theta(\overline{f},g)\rrbracket\in\mathbb{B}$ be defined as
follows

$t(i)=\left\\{\begin{array}[]{ll}1&\mbox{ if
}R_{i}\models\theta(f_{0}(i),\ldots,f_{n}(i),g(i))\\\ 0&\mbox{ if
}R_{i}\not\models\theta(f_{0}(i),\ldots,f_{n}(i),g(i))\end{array}\right.$

It is immediate to show that $h\leq t$. ∎

###  2.4. An axiom about atomic formulas

We add a further axiom, clearly true in products, and henceforward assumed.

Axiom 4. For any atomic formula $\theta(v_{0},\ldots,v_{n})$ in
$\mathcal{L}_{rings}$,

$R\models\theta(\overline{f})\mbox{\hskip 7.22743pt iff \hskip 7.22743pt
}\mathbb{B}\models\llbracket\theta(\overline{f})\rrbracket=1,$

where $\overline{f}=f_{0},\ldots,f_{n}\in R.$

Axiom 4 is clearly true in product of connected rings since the ring structure
on $\prod_{i\in I}R_{i}$ is defined coordinatewise.

###  2.5. Partitions and a Patching Axiom

In order to sketch a proof of a useful generalization of [9] to our more
restricted setting of rings $R$ satisfying the axioms listed above, we need to
review several notions of partitions used in [9].

In a Boolean algebra $B$ a partition is a finite sequence $Y_{0},\ldots,Y_{m}$
of elements of $B$ such that $1=Y_{0}\vee\ldots\vee Y_{m}$, and $Y_{i}\wedge
Y_{j}=0$ if $i\not=j$. (We do not insist that each $Y_{i}\not=0$, but we do
insist that the sequence is finite.) This notion can be easily formalized in
$\mathcal{L}_{B}$ (and in our context also in $\mathcal{L}_{rings}$) by the
formula

$Part_{m+1}(y_{0},\ldots,y_{m}):=(\bigvee_{0\leq j\leq
m}y_{j})=1\wedge\bigwedge_{0\leq i<j\leq m}(y_{i}\wedge y_{j}=0).$

In our context via the interpretation of the Boolean algebra of idempotents in
the ring $R$ we can give a meaning to the notation
$R\models\Phi(\llbracket\theta_{0}\rrbracket,\ldots,\llbracket\theta_{m}\rrbracket)$,
where $\Phi$ is the $B$-formula of $\mathcal{L}_{rings}$ corresponding to a
formula $\mathcal{L}_{B}$ (via the interpretation), and
$\theta_{0},\ldots,\theta_{m}$ are in $\mathcal{L}_{rings}$. Feferman and
Vaught work in a two sorted language, one for the structure and one for the
Boolean algebra, while we have only a one sorted language without the Boolean
sorts (see Remark 2.1). They introduce the following notions which occur in
Theorem 3.1 in [9]. We recall them and adapt them if necessary to our context.

1\. An acceptable sequence is a
$\zeta=\langle\Phi,\theta_{0},\ldots,\theta_{m}\rangle$ where $\Phi$ is a
$B$-formula in $\mathcal{L}_{rings}$ and $\theta_{0},\ldots,\theta_{m}$ are
$\mathcal{L}_{rings}$-formulas. From this one gets (after imposing trivial
syntactic constraints) an $\mathcal{L}_{rings}$-formula
$\Phi(\llbracket\theta_{0}(\overline{v})\rrbracket,\ldots,\llbracket\theta_{m}(\overline{v})\rrbracket)$
which defines in any ring $R$ satisfying the axioms we have given so far a
subset of $R^{|\overline{v}|}$. Note that $\zeta$ can be considered as a code
(i.e. a natural number) for
$\Phi(\llbracket\theta_{0}(\overline{v})\rrbracket,\ldots,\llbracket\theta_{m}(\overline{v})\rrbracket)$
by associating a natural number to each symbol of the language page 64 in [9].

2\. By a free variable in a sequence $\zeta$ we mean a variable $v$ that
appears free in at least one $\theta_{0},\ldots,\theta_{m}$. A standard
acceptable sequence is an acceptable sequence
$\zeta=\langle\Phi,\theta_{0},\ldots,\theta_{m}\rangle$ where the free
variables are exactly $v_{0},\ldots,v_{k}$ for some $k\geq 0$.

3\. A sequence of $\mathcal{L}_{rings}$-formulas
$\theta_{0}(\overline{v}),\ldots,\theta_{m}(\overline{v})$ is called a
partition sequence  if for any ring $R$ satisfying the axioms introduced so
far and for any $\overline{f}$ elements of $R$ the sequence

$\llbracket\theta_{0}(\overline{f})\rrbracket,\ldots,\llbracket\theta_{m}(\overline{f})\rrbracket$

forms a partition of the Boolean algebra $\mathbb{B}$ of idempotents of $R$.

Feferman and Vaught remark that in the language obtained by adding all the
$\zeta$ to the one sorted language of the structures $\mathcal{M}_{i}$ for
$i\in I$ the product $\prod_{i\in I}\mathcal{M}_{i}$ has an elimination of
quantifiers.

We now give an idea on how to produce partition sequences following page 67 of
[9]. This will be used in the inductive step of Theorem 2.10 when we have to
consider existential formulas.

Given an acceptable sequence
$\zeta^{\prime}=\langle\Phi^{\prime},\theta_{0}^{\prime},\ldots,\theta_{m^{\prime}}^{\prime}\rangle$
we can determine in an effective way another acceptable sequence
$\zeta=\langle\Phi,\theta_{0},\ldots,\theta_{m}\rangle$ with the same free
variables such that $\theta_{0},\ldots,\theta_{m}$ form a partition sequence
and for all models $R$ of our axioms

(1)
$R\models\Phi(\llbracket\theta_{0}(\overline{f})\rrbracket,\ldots,\llbracket\theta_{m}(\overline{f})\rrbracket)\mbox{
iff
}R\models\Phi^{\prime}(\llbracket\theta_{0}^{\prime}(\overline{f})\rrbracket,\ldots,\llbracket\theta_{m^{\prime}}^{\prime}(\overline{f})\rrbracket)$

for all $\overline{f}\in R$. The acceptable sequence $\zeta$ is constructed as
follows:

1\. let $m=2^{m^{\prime}+1}$ and $r_{0},\ldots r_{m}$ be the list of subsets
of $m^{\prime}+1=\\{0,\ldots,m^{\prime}\\}$.

For each $k\leq m$ define

$\theta_{k}:=\bigwedge_{k\in
r_{k}}\theta_{k}^{\prime}\wedge\bigwedge_{k\not\in
r_{k}}\neg\theta_{k}^{\prime};$

2\. for each $l\leq m^{\prime}$ let $s_{l}:=\\{k:k\leq m\mbox{ and }l\in
r_{k}\\}$ and

$\Phi(x_{0},\ldots,x_{m}):=\Phi^{\prime}(\bigvee_{k\in
s_{0}}x_{k},\ldots,\bigvee_{k\in s_{m^{\prime}}}x_{k}).$

It is easy to check that the $\theta_{k}$ form a partition, and that (1)
holds. Note that this argument is nothing more than the usual procedure of
constructing the disjunctive normal form of a propositional logic formula,
where the $\theta_{k}$ play the role of propositional variables.

We next have to introduce a new axiom which will use in the inductive step of
the proof of our main result, Theorem 2.10, when dealing with the existential
case. This axiom is necessary since we are working in a more general setting
than product of structures, and in our context we cannot prove the properties
in equation (15) on page 68 of [9]. This will be the content of our fifth and
final axiom.

Axiom 5, gives sufficient conditions for a ubiquitous “Patching Argument”, and
it would surely be worthwhile to give a more sheaf-theoretic, or topos-
theoretic, formulation.

Let $\phi(v_{0},….v_{m})$ be any formula in the language $\mathcal{L}_{B}$ of
Boolean algebras with free variables $v_{0},\ldots,v_{m}$. To
$\phi(v_{0},….v_{m})$ we associate a formula $\phi^{*}$ with free variables
$v_{0},\ldots,v_{m}$ in $\mathcal{L}_{B}$ expressing:

there exists a partition $W_{0},\ldots,W_{m}$ of $B$ such that $W_{j}\leq
v_{j}$ for each $j=0,\ldots,m$, and $\phi(W_{0},\ldots,W_{m})$ holds in $B$.

Axiom 5. Let $\phi(v_{0},….v_{m})$ be a formula in the language
$\mathcal{L}_{B}$ of Boolean algebras with free variables
$v_{0},\ldots,v_{m}$, and $\phi^{*}(v_{0},\ldots,v_{m})$ as above.

Let
$\theta_{0}(x_{0},\ldots,x_{k},x_{k+1}),\ldots,\theta_{m}(x_{0},\ldots,x_{k},x_{k+1})$
be formulas in the ring language and assume that

$(\theta_{0}(x_{0},\ldots,x_{k},x_{k+1}),\ldots,\theta_{m}(x_{0},\ldots,x_{k},x_{k+1}))$

is a partition. Then the following are equivalent in $\mathbb{B}$ for
$k$-tuples $\bar{f}\in R$

(1) there exists $g\in R$ such that
$\phi^{*}(\llbracket\theta_{0}(\bar{f},g)\rrbracket,\ldots,\llbracket\theta_{m}(\bar{f},g)\rrbracket)$
holds in $\mathbb{B}$;

(2) there is a partition $Y_{0},\ldots,Y_{m}$ of $\mathbb{B}$ such that for
each $j~{}=~{}0~{},~{}\ldots~{},m$, $Y_{j}\leq\llbracket\exists
x_{k+1}\theta_{j}(\bar{f},x_{k+1})\rrbracket$ and $\phi(Y_{0},\ldots,Y_{m})$
holds in $\mathbb{B}$.

######  Lemma 2.9.

Let $R=\prod_{i\in I}R_{i}$, where $R_{i}$ are connected unital rings and
$\mathbb{B}$ the boolean algebra of idempotents of $R$. Then Axiom 5 holds in
$R$.

###### Proof.

We use the same notation as in Axiom 5. Recall that in this case $\mathbb{B}$
can be identified with $\mathcal{P}(I)$, and a partition of $\mathbb{B}$ is a
finite collection of disjoint subsets of $I$ whose union is $I$. Moreover,

$\llbracket\xi(\bar{f})\rrbracket=\bigvee\\{i\in
I:R_{i}\models\xi(f_{0}(i),\ldots,f_{k}(i)\\}$

for any $\mathcal{L}_{rings}$-formula $\xi(\bar{x})$ and any
$\bar{f}\in\prod_{i\in I}R_{i}$.

2) $\Rightarrow$ 1) Let $Y_{0},\ldots,Y_{m}\in\mathbb{B}$ be a partition of
$\mathbb{B}$ such that for each $j~{}=~{}0~{},~{}\ldots~{},m$,
$Y_{j}\leq\llbracket\exists x_{k+1}\theta_{j}(\bar{f},x_{k+1})\rrbracket$ and
$\mathbb{B}\models\phi(Y_{0},\ldots,Y_{m})$. Notice that for each atom
$i\in\mathbb{B}$ there is a unique $j\in\\{1,\ldots,m\\}$ such that $i\leq
Y_{j}$ (this follows from the hypothesis that $Y_{0},\ldots,Y_{m}$ is a
partition of $\mathbb{B}$). Fix $j$, for all atoms $i$ such that $i\leq
Y_{j}\leq\llbracket\exists x_{k+1}\theta_{j}(\bar{f},x_{k+1})\rrbracket$, we
have $R_{i}\models\exists
x_{k+1}\theta_{j}(f_{0}(i),\ldots,f_{n}(i),x_{k+1})$. Using the axiom of
choice we are now going to define an element $g\in\prod_{i\in I}R_{e}$ as
follows:

for all atoms $i\leq Y_{j}$ we set $g(i)=g_{i}\in R_{i}$ for some $g_{i}\in
R_{i}$ such that $R_{i}\models\theta_{j}(f_{0}(i),\ldots,f_{n}(i),g_{i})$. We
do this for all $j=1,\ldots,m$, and so $g$ is well defined on all $i\in I$.

Clearly, $Y_{j}\leq\llbracket\theta_{j}(f_{0},\ldots,f_{k},g)\rrbracket$ for
$j~{}=~{}0~{},~{}\ldots~{},m$. Indeed,
$Y_{j}=\llbracket\theta_{j}(f_{0},\ldots,f_{k},g)\rrbracket$. If
$i\leq\llbracket\theta_{j}(f_{0},\ldots,f_{k},g)\rrbracket$ and $i\not\leq
Y_{j}$ then $i\leq Y_{h}$ for a unique $h\not=j$, and $i\leq
Y_{h}\leq\llbracket\theta_{h}(f_{0},\ldots,f_{k},g)\rrbracket$. Then
$\llbracket\theta_{j}(f_{0},\ldots,f_{k},g)\wedge\theta_{h}(f_{0},\ldots,f_{k},g)\rrbracket=\llbracket\theta_{j}(f_{0},\ldots,f_{k},g)\rrbracket\wedge\llbracket\theta_{h}(f_{0},\ldots,f_{k},g)\rrbracket\not=0$
which is a contradiction since
$(\theta_{0}(x_{0},\ldots,x_{k},x_{k+1}),\ldots,\theta_{m}(x_{0},\ldots,x_{k},x_{k+1}))$
is a partition. Hence,
$\mathbb{B}\models\phi^{*}(\llbracket\theta_{0}(\bar{f},g)\rrbracket,\ldots,\llbracket\theta_{m}(\bar{f},g)\rrbracket)$.

1) $\Rightarrow$ 2) Suppose there is $g\in R$ such that

$\mathbb{B}\models\phi^{*}(\llbracket\theta_{0}(\bar{f},g)\rrbracket,\ldots,\llbracket\theta_{m}(\bar{f},g)\rrbracket).$

So, there is a partition $w_{0},\ldots,w_{m}$ of $\mathbb{B}$ and
$w_{j}\leq\llbracket\theta_{j}(\bar{f},g)\rrbracket$ for all
$j~{}=~{}0~{},~{}\ldots~{},m$, and
$\mathbb{B}\models\phi(w_{0},\ldots,w_{k})$. Recall that

$\llbracket\theta_{j}(\bar{f},g)\rrbracket\leq\llbracket\exists
x_{k+1}\theta_{j}(\bar{f},x_{k+1})\rrbracket$

for all $j=1,\ldots,m$. Then 2) easily follows. ∎

The set of the five axioms will be denoted by $AXIOMS$. From $AXIOMS$ we will
prove our main result, Theorem 2.10 below. It is inspired by the fundamental
Theorem 3.1 of [9], but it is considerably less general, and uses quite
different notation.

######  Theorem 2.10.

For every $\mathcal{L}_{rings}$-formula $\theta(x_{0},\ldots,x_{k})$ there is
a partition
$(\theta_{0}(x_{0},\ldots,x_{k}),\ldots,\theta_{m}(x_{0},\ldots,x_{k}))$ of
ring formulas, and a Boolean algebra formula $\psi(y_{0},\ldots,y_{m})$ so
that for all $f_{0},\ldots,f_{k}\in R$, where $R$ is a ring satisfying
$AXIOMS$, and $\mathbb{B}$ is the Boolean algebra of idempotents of $R$

(2) $R\models\theta(\overline{f})\mbox{ \hskip 14.45377pt iff \hskip
14.45377pt}\mathbb{B}\models\psi(\llbracket\theta_{0}(\overline{f})\rrbracket,\ldots,\llbracket\theta_{m}(\overline{f})\rrbracket)$

where $\overline{f}=f_{0},\ldots,f_{k}.$

###### Proof.

As in [9] the proof proceeds by induction on the complexity of the formula
$\theta(x_{0},\ldots,x_{k})$. Using the five axioms we have identified we can
easily prove the cases of atomic formulas (use Axiom 4) and the Boolean
connectives (use Axiom 2 and Lemmas 2.5 and 2.6). When
$\theta(x_{0},\ldots,x_{k})=\exists x_{k+1}\xi(x_{0},\ldots,x_{k},x_{k+1})$
Axiom 5 guarantees that there are no contradictions in the patching argument
in the inductive step. Let

$(\xi_{0}(x_{0},\ldots,x_{k},x_{k+1}),\ldots,\xi_{m}(x_{0},\ldots,x_{k},x_{k+1}))$

be the partition in $\mathcal{L}_{rings}$-formulas and
$\psi(v_{0},\ldots,v_{m})$ the formula in $\mathcal{L}_{B}$ associated by the
inductive hypothesis to $\xi(x_{0},\ldots,x_{k},x_{k+1})$. Using Axiom 5 and
the inductive hypothesis the following are equivalent

  1. (1)

$R\models\exists x_{k+1}\xi(f_{0},\ldots,f_{k},x_{k+1})$

  2. (2)

there exists $g\in R$ such that $R\models\xi(f_{0},\ldots,f_{k},g)$

  3. (3)

$\mathbb{B}\models\psi(\llbracket\xi_{0}(\overline{f},g)\rrbracket,\ldots,\llbracket\xi_{m}(\overline{f},g)\rrbracket)$

  4. (4)

$\mathbb{B}\models\psi^{\prime}(\llbracket\exists
x_{k+1}\xi_{0}(\overline{f},x_{k+1})\rrbracket,\ldots,\llbracket\exists
x_{k+1}\xi_{m}(\overline{f},x_{k+1})\rrbracket)$ where
$\psi^{\prime}(v_{0},\ldots,v_{m})$ is the $\mathcal{L}_{B}$-formula

$\exists w_{0},\ldots,w_{m}(\bigwedge_{i=0}^{m}w_{i}\leq
v_{i}\wedge(w_{0},\ldots,w_{m}\mbox{ is partition of
$\mathbb{B}$})\wedge\psi(w_{0},\ldots w_{m})).$

Notice that $\exists x_{k+1}\xi_{0}(x_{0},\ldots,x_{k},x_{k+1}),\ldots,\exists
x_{k+1}\xi_{m}(x_{0},\ldots,x_{k},x_{k+1})$ may not be a partition. As
outlined on page 67 of [9] (and remarked also in this paper before introducing
axiom 5) there is an effective procedure which associates a partition to
$\exists x_{k+1}\xi_{0}(\overline{x},x_{k+1}),\ldots,\exists
x_{k+1}\xi_{m}(\overline{x},x_{k+1})$ mantaining the same free variables.

∎

The statement of Theorem 2.10 is uniform for all models of our axioms, and
this uniformity is fundamental. The partition

$(\theta_{0}(x_{0},\ldots,x_{k}),\ldots,\theta_{m}(x_{0},\ldots,x_{k}))$

of ring formulas, and the Boolean algebra formula $\psi(y_{0},\ldots,y_{m})$
depend only on $\theta(x_{0},\ldots,x_{k})$. Notice also that Theorem 2.10 is
effective.

Notice that equation (2) in Theorem 2.10 is a statement only about the ring
$R$. The left hand condition can be expressed as a condition on $R$ in terms
of $B$-formulas.

It is worthwhile to compare a model $R$ of our axioms, and the full product
$\prod_{e}R_{e}$ over all atoms $e$ of the Boolean algebra $\mathbb{B}$ of the
idempotents of $R$, which is also a model of the axioms. The obvious
difference between $R$ and $\prod_{e}R_{e}$ is that $R$ may be countable, with
$\mathbb{B}$ incomplete, whereas $\prod_{e}R_{e}$ will be uncountable when
$\mathbb{B}$ is infinite and each $R_{e}$ has cardinality $\geq 2$.

However, though both their Boolean algebras are atomic, and have the same
atoms (i.e. there is a natural bijection between their atoms, and we pretend
it is the identity), they will be different in general. The full product has
Boolean algebra $\mathbb{B}_{\prod}$ the power set of the set of atoms, and it
is complete. The Boolean algebra $\mathbb{B}$ associated to $R$ need not be
complete.

We construct a monomorphism from $\mathbb{B}$ to $\mathbb{B}_{\prod}$ as
follows. Let $r$ be a idempotent of $R$. Then in either of the atomic Boolean
algebras any nonzero element is the supremum of the atoms $e\leq r$. So we
naturally match $r$ to the supremum in $B_{\prod}$ of the atoms $e\leq r$
(identifying the atoms of $\mathbb{B}$ with those of $\mathbb{B}_{\prod}$).
This defines a map $F$ which is clearly an injective homomorphism of Boolean
rings, and therefore an injective morphism of Boolean algebras (which we also
call $F$.) It is a well known fact (see e.g. [10]) that for atomic Boolean
algebras $B_{1}$ and $B_{2}$, say with infinitely many atoms, any embedding is
elementary if it preserves the notions of atom and union of at most
$n\in\mathbb{N}$ atoms. This is certainly true in our situation, and so $F$ is
elementary as a map from $\mathbb{B}$ to the complete Boolean algebra
$\mathbb{B}_{\prod}$.

Also the Boolean value of sentences is the same in both algebras. (Note that
we have not tried yet to embed $R$ into the full product, this will be
analyzed in future work). At any rate, using Theorem 2.10, this is enough to
show that sentences have the same truth value in the two rings, $R$ and
$\prod_{e\mbox{ {\tiny atom}}}R_{e}$, which is stated in the next corollary.

######  Corollary 2.11.

Let $R$ as above. Then

$R\equiv\prod_{e\mbox{ {\tiny atom}}}R_{e}.$

###### Proof.

Notice that $R$ and the product of the $R_{e}$, for $e$ a minimum idempotent,
have the same Boolean algebra of idempotents, and the same stalks. It is
enough to apply the above theorem to sentences, and we use that $\mathbb{B}$
is an elementary substructure of $\mathbb{B}_{\prod}$. ∎

##  3\. Exotic examples from Arithmetic

### 3.1.

Since the 1930’s one has studied nonstandard models of arithmetic, in several
closely related formalisms. One formalism involves $+,\cdot,0,1$ and $<$. The
models for this formalism satisfy first-order Peano axioms, $PA$. See [11] for
a clear presentation. The formalism classically applies to the semiring
$\mathbb{N}$, and the model theory is pervaded by Gödel Incompleteness. The
other main formalism is inspired by the ring $\mathbb{Z}$, with the primitives
above, as well as $-$. There is little difference logically (in terms of
interpretability) between the two settings. In both the order $<$ is easily
definable from the other relations via some simple axioms and Lagrange’s
Theorem. We have proved, using Ax’s fundamental work [1] of 1968 that although
both set-ups are pervaded by Gödel Incompleteness, they seem rather immune to
any Incompletenss concerning the algebra of residue rings
$\mathcal{M}/n\mathcal{M}$, where $n$ is a (possibly) nonstandard power of a
prime in $\mathcal{M}$, for $\mathcal{M}$ a model of $PA$ (see [4]). The
analysis of the case $n$ composite is much harder, but can be handled by the
earlier analysis in the present paper. The key is to formulate and prove a
factorization theorem for $\mathcal{M}/n\mathcal{M}$ giving a meaning to

$\mathcal{M}/n\mathcal{M}=\prod\mathcal{M}/q\mathcal{M}$

where $q$ varies among the maximal prime powers dividing $n$. Using the
refined analysis of the rings $\mathcal{M}/q\mathcal{M}$ in [4] and the
previous results in Section 2 we are able to understand the first order theory
of the rings $\mathcal{M}/n\mathcal{M}$ for any $n\in\mathcal{M}$.

### 3.2.

We will work with the formalism for the semiring. If $\mathcal{M}$ is a
semiring then it is easy to interpret the associated ring
$\mathcal{M}^{\bullet}$ inside $\mathcal{M}$. Note that if $j$ is in the
semiring $\mathcal{M}$ then $\mathcal{M}/j\mathcal{M}$ is naturally isomorphic
to $\mathcal{M}^{\bullet}/j\mathcal{M}^{\bullet}$. Thus, for intelligibility
we conflate $\mathcal{M}$ and $\mathcal{M}^{\bullet}$ in what follows.

Gödel primitive recursive coding of syntax extends simply and uniformly to
$\mathcal{M}$, giving a meaning to (not necessarily finite) formulas, and
allowing certain inductive arguments about semantics. Details of such
arguments and results are to be found in [11]. For us the crucial notions
coded uniformly allowing us to prove properties about them in $PA$ are

  1. (1)

formulas, and specifically $\Delta_{0}$-formulas;

  2. (2)

satisfaction in $\mathcal{M}/n\mathcal{M}$, i.e. for any formula
$\psi(v_{0},\ldots,v_{k})$ in $\mathcal{L}_{ring}$ and any $k+1$-tuple
$c_{0},\ldots,c_{k}$ in $\mathcal{M}$ we can code

$\mathcal{M}/n\mathcal{M}\models\psi(c_{0}+n\mathcal{M},\ldots,c_{k}+n\mathcal{M}).$

Here $\mathcal{M}/n\mathcal{M}$ is represented as in elementary number theory
as $[0,n-1]$ with $+$ and $\cdot$ defined via remainders modulo $n$.

The essential point is (2). The indespensable uniformity is that there is a
single formula (not $\Delta_{0}$) $Sat(x,y,z)$, so that
$Sat(<\psi>,n,<\overline{c}>)$ expresses for all formulas $\psi$, and
$n,\overline{c}\in\mathcal{M}$ that

$\mathcal{M}/n\mathcal{M}\models\psi(c_{0}+n\mathcal{M},\ldots,c_{k}+n\mathcal{M}),$

where $<\psi>$ is a code for the formula $\psi$, $n\in\mathcal{M}$, and
$<\overline{c}>$ is a code for the $k+1$-tuple
$\overline{c}=c_{0},\ldots,c_{k}$ of elements of $\mathcal{M}$. This implies
that $\mathcal{M}/n\mathcal{M}$ is recursively saturated. Note that not all
rings elementarily equivalent to $\mathcal{M}/n\mathcal{M}$ are recursively
saturated.

Making this precise seems unecessary, as such codings and the notion of
recursive saturation are ubiquitous in the subject.

We may interpret $\mathcal{M}/n\mathcal{M}$ inside $\mathcal{M}$ in the usual
way using the $\Delta_{0}$-relation of congruence mod $n$, picking by
$\Delta_{0}$-induction the smallest positive representative in each coset
modulo the ideal $n\mathcal{M}$. We may also define on $[0,n-1]$ a
$\Delta_{0}$-ring structure using Euclidean division, and this is what we then
take as $\mathcal{M}/n\mathcal{M}$. Thus we have a bounded $\Delta_{0}$-set
and $\Delta_{0}$-relations $\oplus$, $\odot$, $\ominus$, with usual $0,1$
(notice that these functions are not related to those defined on the Boolean
ring in Section 2.1). The aim is to show that $\mathcal{M}/n\mathcal{M}$ is a
model of $AXIOMS$.

Now we recall the factoriality properties of $\mathcal{M}$. For fixed $n$ and
prime $p$ in $\mathcal{M}$ we can define in a $\Delta_{0}$-way the maximal
power of $p$ dividing $n$, and we can prove that it is unique. The set $Q$ of
maximal prime powers dividing $n$ is a bounded $\Delta_{0}$-set, and a
suitable version of the fundamental theorem of arithmetic can be proved in
$PA$, giving a meaning to $n$ is the product of prime powers in any non
standard model of $PA$. Moreover, there is a clear $\Delta_{0}$-version of the
Chinese Remainder Theorem ($\Delta_{0}$-CRT) which we state now for
convenience since we will appeal to it in many proofs (see [5] and [6] for
details and applications).

######  Theorem 3.1.

Let $\mathcal{M}$ be a model of $PA$ and $A$ a bounded $\Delta_{0}$-definable
set in $\mathcal{M}$. Let $f$ and $r$ be $\Delta_{0}$-functions such that
$f(a_{1}),f(a_{2})$ are pairwise coprime for all $a_{1},a_{2}\in A$ and
$a_{1}\not=a_{2}$, and $r(a)<f(a)$ for all $a\in A$. Suppose there exists
$w\in\mathcal{M}$ divisible by all elements of $f(A)$. Then there exists
$u<\prod_{a\in A}f(a)$ such that $u\equiv r(a)(\mbox{mod }f(a))$ for all $a\in
A$.

A much more general result holds giving a $\Delta_{0}$-definition of the
product relation $y=\prod_{i\leq n}F(n)$, for $F$ a $\Delta_{0}$-definable
function, see [3]. This has been also used by the authors in the analysis of
the multiplicative group of the residue field $\mathcal{M}/p\mathcal{M}$ where
$\mathcal{M}$ is a model of a weak fragment of Peano Arithmetic, and $p$ is a
prime in $\mathcal{M}$, see [7].

An alternative way of considering $\mathcal{M}/n\mathcal{M}$, which turns out
to be convenient in our setting, is to identify $\mathcal{M}/n\mathcal{M}$ as
a set of $\Delta_{0}$-definable functions. We now specify this.

To each $x\in\mathcal{M}/n\mathcal{M}$ we assign the $\Delta_{0}$-function
$\overline{x}$ defined on $Q$, the set of maximal prime powers $q$ dividing
$n$, as follows

$\begin{array}[]{ccccc}\overline{x}&:&Q&\rightarrow&\mathcal{M}/n\mathcal{M}\\\
&&q&\mapsto&\overline{x}(q)=rem(x,q)\end{array}$

where $rem(x,q)$ stands for the reminder of the division of $x$ by $q$ in
$\mathcal{M}$. We now identify the idempotents of $\mathcal{M}/n\mathcal{M}$,
i.e. the elements of the Boolean algebra $\mathbb{B}_{n}$ associated to
$\mathcal{M}/n\mathcal{M}$. Suppose $r\in\mathcal{M}/n\mathcal{M}$ is an
idempotent then $r^{2}-r\equiv 0(\mbox{mod }n)$ in $\mathcal{M}$. Therefore,
for each $q$ maximal prime power dividing $n$, $q$ divides $r^{2}-r$, so $q$
divides one, and only one, of $r$ or $r-1$. Conversely, if every $q$ in $Q$
divides either $r$ or $r-1$ then clearly $r$ is an idempotent in
$\mathcal{M}/n\mathcal{M}$. So, unless $n$ is a power of a prime (in which
case it is a local ring, and so connected) $\mathcal{M}/n\mathcal{M}$ is not
connected.

The minimal idempotents, i.e. the atoms in $\mathbb{B}_{n}$, are exactly the
$r$ such that for a unique $q\in Q$, $r\equiv 1($mod $q)$, and $r\equiv 0($mod
$q^{\prime})$ for all other $q^{\prime}\in Q$. By $\Delta_{0}$-CRT there are
such $r$ for each $q$. Note that $\mathcal{M}/n\mathcal{M}$ and the full
product $\prod_{q\in Q}\mathcal{M}/q\mathcal{M}$ have the same idempotents.

We have then identified each idempotent in $\mathbb{B}_{n}$ with a
$\Delta_{0}$-definable subset of $Q$, the set of maximal power of primes
dividing $n$. The atoms of $\mathbb{B}_{n}$ correspond to the subsets
$Q-\\{q\\}$, as $q$ varies in $Q$. To each atom $e$ in $\mathbb{B}_{n}$ we
associate the unique prime power $q_{e}$ such that $q_{e}\not|e$. We are aware
that $q_{e}$ is not the best notation for expressing the 1-1- correspondence
between atoms of $\mathbb{B}_{n}$ and prime powers dividing $n$. We hope that
no confusion will arise.

It is easy to show now that $(\mathcal{M}/n\mathcal{M})_{e}$ is isomorphic
$\mathcal{M}/q_{e}\mathcal{M}$. So, the elements of the localized ring at $e$
can be identified with elements of $\mathcal{M}$ which are $<q_{e}\leq n$.

Now we can verify that the ring $\mathcal{M}/n\mathcal{M}$ satisfies our
axioms.

Axiom 1. $\mathbb{B}_{n}$ is atomic.

Let $r\in\mathbb{B}_{n}$, $r\not=0$. There exists $q\in Q$, such that $r\equiv
1($mod $q)$, if not then $r=0$. Let $e$ be the atom such that $e\equiv 1($mod
$q)$ and $e\equiv 0($mod $q^{\prime})$ for any $q^{\prime}\in Q$ and
$q^{\prime}\not=q$. So, $re\equiv e(\mod n)$ which implies $e\leq r$. $\Box$

From now on an atom will always be an atom of $\mathbb{B}_{n}$. Let $e$ be an
atom. By Lemma 2.2 the rings $(\mathcal{M}/n\mathcal{M})_{e}$ and
$e(\mathcal{M}/n\mathcal{M})$ are isomorphic via the map $\frac{r}{s}\mapsto
er$ for $s=1,e$. Using this we can easily interpret
$(\mathcal{M}/n\mathcal{M})_{e}$ inside $\mathcal{M}/n\mathcal{M}$ as the ring
$e(\mathcal{M}/n\mathcal{M})$ which is definable (uniformly in $e$) inside
$\mathcal{M}/n\mathcal{M}$. We will also identify the ring
$e(\mathcal{M}/n\mathcal{M})$ with $\mathcal{M}/q_{e}\mathcal{M}$, where
$q_{e}$ is the unique prime power dividing $n$ such that $q_{e}\not|e$.

Axiom 2. Let $\theta(v_{0},\ldots,v_{k})$ be any
$\mathcal{L}_{rings}$-formula, and
$f_{0},\ldots,f_{k}\in\mathcal{M}/n\mathcal{M}$. We have to show that there is
a unique element $\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$ in
$\mathbb{B}_{n}$ such that

$e(\mathcal{M}/n\mathcal{M})\models\theta(ef_{0},\ldots,ef_{k})\mbox{ \hskip
7.22743pt iff \hskip
7.22743pt}e\leq\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$

for all atoms $e$ in $\mathbb{B}_{n}$. Let

$X=\\{e:e\mbox{ atom and
}e(\mathcal{M}/n\mathcal{M})\models\theta(ef_{0},\ldots,ef_{k})\\}.$

Clearly, $X$ is a bounded $\Delta_{0}$-set of $\mathcal{M}$ since
$\mathcal{M}/n\mathcal{M}$ lives in a $\Delta_{0}$-way on $[0,n-1]$, and so
the relativization of $\theta(f_{0},\ldots,f_{k})$ gets coded in a
$\Delta_{0}$-way. Now, $\Delta_{0}$-CRT implies the existence of the
idempotent $\gamma$ in $\mathcal{M}/n\mathcal{M}$ such that $\gamma\equiv
1($mod $q_{e})$ for all $e\in X$, and $\gamma\equiv 0($mod $q)$ for all other
prime powers $q$ in $Q$. So, we define $\bigvee
X=\gamma=\llbracket\theta(f_{0},\ldots,f_{n})\rrbracket$. $\Box$

As we have just shown in our setting the join of the set

$\\{e:e\mbox{ is an atom and
}e(\mathcal{M}/n\mathcal{M})\models\theta(ef_{0},\ldots,ef_{k})\\}$

always exists, and

$\bigvee\\{e:e\mbox{ is an atom and
}e(\mathcal{M}/n\mathcal{M})\models\theta(ef_{0},\ldots,ef_{k})\\}=\llbracket\theta(f_{0},\ldots,f_{k})\rrbracket.$

Axiom 3. Let $\exists v_{k+1}\theta(v_{0},\ldots,v_{k},v_{k+1})$ be an
$\mathcal{L}_{rings}$-formula, and
$f_{0},\ldots,f_{k}\in\mathcal{M}/n\mathcal{M}$. We need to show that there is
$g\in\mathcal{M}/n\mathcal{M}$ such that

$\llbracket\exists
w\theta(\overline{f},w)\rrbracket\leq\llbracket\theta(\overline{f},g)\rrbracket$

where $\overline{f}=f_{0},\ldots,f_{n}$.

Let

$X=\\{e:e\mbox{ is an atom and }e(\mathcal{M}/n\mathcal{M})\models\exists
v_{k+1}\theta(ef_{0},\ldots,ef_{k},v_{k+1})\\},$

and $\bigvee X=\llbracket\exists w\theta(\overline{f},w)\rrbracket$ in
$\mathcal{M}/n\mathcal{M}$. Then by $\Delta_{0}$-induction for each $e$ in $X$
we take the minimum $h\in\mathcal{M}/n\mathcal{M}$ such that
$e(\mathcal{M}/n\mathcal{M})\models\theta(ef_{0},\ldots,ef_{k},eh)$. We will
denote it by $h(e)$. Now we use again the $\Delta_{0}$-CRT in order to
determine $g\in\mathcal{M}/n\mathcal{M}$ such that $g\equiv h(e)(\mbox{mod
}q_{e})$ for all atoms $e\in X$, and $g\equiv 0(\mbox{mod }q)$ for all other
prime powers $q$ in $Q$.

The set

$Y=\\{e:e\mbox{ is an atom and
}e(\mathcal{M}/n\mathcal{M})\models\theta(ef_{0},\ldots,ef_{k},eg)\\}$

is $\Delta_{0}$-definable and bounded, and arguing as before we can show that
in $\mathcal{M}/n\mathcal{M}$ there exists the supremum of $Y$ which we denote
by $\llbracket\theta(\overline{f},g)\rrbracket$. From $X\subseteq Y$ it
follows

$\llbracket\exists
v_{k+1}\theta(\overline{f},v_{k+1})\rrbracket\leq\llbracket\theta(\overline{f},g)\rrbracket.$

. $\Box$

Axiom 4. Let $\theta(v_{0},\ldots,v_{k})$ be an atomic
$\mathcal{L}_{rings}$-formula, and
$\overline{f}=f_{1},\ldots,f_{k}\in\mathcal{M}/n\mathcal{M}$. We have to show
that $\mathbb{B}_{n}\models\llbracket\theta(\overline{f})\rrbracket=1$ iff
$\mathcal{M}/n\mathcal{M}\models\theta(\overline{f})$.

Now, $\mathbb{B}_{n}\models\llbracket\theta(\overline{f})\rrbracket=1$ iff
$\\{e:e\mbox{ atom and
}e(\mathcal{M}/n\mathcal{M})\models\theta(e\overline{f})\\}$ coincides with
the set of atoms of $\mathcal{M}/n\mathcal{M}$ iff
$\mathcal{M}/q_{e}\mathcal{M}\models\theta(\overline{f}_{q_{e}})$ for all
atoms $e$ iff $\mathcal{M}/n\mathcal{M}\models\theta(\overline{f})$. Notice
that the last equivalence follows from the fact that for a tuple to satisfy an
atomic formula corresponds to be a zero of a polynomial.

Axiom 5. The proof of this axiom in $\mathcal{M}/n\mathcal{M}$ follows the
line of the proof of Lemma 2.9. The only significant modification is in the
proof of 2) $\Rightarrow$ 1) where the element $g$ is obtained using
$\Delta_{0}$-CRT without appealing to the axiom of choice.

Now, arguing as in the proof of Corollary 2.11 we have

######  Theorem 3.2.

$\mathcal{M}/n\mathcal{M}\equiv\prod_{q\in Q}\mathcal{M}/q\mathcal{M}$.

In a forthcoming paper we will use Theorem 3.2 to describe thoroughly the
model theory of the rings $\mathcal{M}/n\mathcal{M}$ for any
$n\in\mathcal{M}$.

Derakhshan and Macintyre in [8] adapt our methods to the adèlic situation, at
the cost of modifying the preceeding axioms. The main difference is that the
Boolean algebras of idempotents are enriched with extra predicates identifying
those elements which are join of a finite sets of atoms.

Acknowledgements. The authors are very grateful to MSRI in Berkeley where this
paper was completed during the program Definability, Decidability and
Computability in Number Theory in July-August 2022. The first author was
partially supported by PRIN 2017-Mathematical Logic: Models, Sets, and
Computability. The second author was supported by a Leverhulme Emeritus
Fellowship.

We are grateful to Jamshid Derakhshan for many stimulating and helpful
discussions.

## References

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# Distance Antimagic Labeling of Zero-Divisor Graphs

V. Sivakumaran1,, K. Sankar2, S. Prabhu1
1Department of Mathematics, Rajalakshmi Engineering College, Thandalam,
Chennai 602105, India
2Department of Mathematics, Anna University, Chennai 600025, India
Corresponding author<EMAIL_ADDRESS>
###### Abstract

In this paper, we prove that for all $m\geq 1$ and $n=1$, the graph
$m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$, for all $n\geq 1$, and
$m=1$, the graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$, for all
$m\geq 1$,
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$,
for all prime $m\geq 3$,
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$ and
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ are all admit
distance antimagic labeling.

Keywords: Distance antimagic labeling, Zero-divisor graph, Join of graphs,
Cartesian product of graphs.

Mathematics Subject Classification: 05C78.

##  1 Introduction

In the year 1990, Beck [3] introduced Zero-divisor graph of a commutative
ring. Anderson and Livingston studied and developed the relation between
theoretical properties of the ring and graph theoretical properties of zero-
divisor graph $\Gamma(R)$ in [1].

Consider a commutative ring $R$ with unity and assume set of all zero-divisor
of $R$ be ${\mathbb{Z}}(R)$ and
${\mathbb{Z}}^{*}(R)={\mathbb{Z}}(R)\setminus\\{0\\}$ be the vertex set such
that any two distinct vertices $u,v\in{\mathbb{Z}}^{*}(R)$ are said to be
adjacent if and only if $uv=0$. This condition provides zero-divisor graph and
is denoted by $\Gamma(R)$. A graph $\Gamma({\mathbb{Z}}_{10})$ with
$V[\Gamma({\mathbb{Z}}_{10})]=\\{2,4,6,8,5\\}$ and edge set
$E[\Gamma({\mathbb{Z}}_{10})]=\\{(2,5),(4,5),(6,5),(8,5)\\}$ is given in
Figure 1. A graph $\Gamma({\mathbb{Z}}_{15})$ with
$V[\Gamma({\mathbb{Z}}_{15})]=\\{3,6,9,12,5,10\\}$ and edge set
$E[\Gamma({\mathbb{Z}}_{15})]=\\{(3,5)$, $(6,5)$, $(9,5)$, $(12,5)$, $(3,10)$,
$(6,10)$, $(9,10)$, $(12,10)\\}$ is given in Figure 2.

Figure 1: $\Gamma({\mathbb{Z}}_{10})$

Figure 2: $\Gamma({\mathbb{Z}}_{15})$

######  Definition 1.1.

[9] Let $H_{1},H_{2},\dots,H_{m}$ be the given $m$ copies of disjoint graphs.
Then a set of $k$ disjoint graphs. Then $H_{1}+H_{2}+\dots+H_{m}$ is obtained
from $H_{1},H_{2},\dots,H_{m}$ by joining every vertex of $H_{i}$ with every
vertex $H_{j}$, whenever $i\neq j$. join graph $G_{1}+G_{2}+\ldots+G_{k}$ is
obtained from $G_{1},G_{2},\dots,G_{k}$ by joining every vertex of $G_{i}$
with every vertex of $G_{j}$, whenever $i\neq j$.

######  Definition 1.2.

[9] The Cartesian product of the graphs $G$ and $H$, denoted $G\times H$, is
the graph with $V(G\times H)=\\{(u,v):u\in V(G)\text{ and }v\in V(H)\\}$ and
$E(G\times H)=\\{<(u,v),(u^{\prime},v^{\prime})>:u=u^{\prime}\text{ and
}vv^{\prime}\in E(H)\text{ or }v=v^{\prime}\text{ and }uu^{\prime}\in E(G)\\}$

If a graph $G$ is said to follow a DAML if for any two distinct vertices
$u_{1},u_{2}\in V(G)$ such that $N(u_{1})\neq N(u_{2})$.

##  2 Main Results

In the main result, we show that some join of zero-divisor this paper, we
prove that some join of zero-divisor graphs and some Cartesian product of
zero-divisor graphs are distance antimagic graphs.

######  Theorem 2.1.

The join graph $\Gamma({\mathbb{Z}}_{2m})+\Gamma({\mathbb{Z}}_{4})$ does not
have DAML for all prime $m\geq 3$.

###### Proof.

Consider the graphs $\Gamma({\mathbb{Z}}_{2m})$ and $\Gamma({\mathbb{Z}}_{4})$
for all prime numbers $m\geq 3$.
Let $G=\Gamma({\mathbb{Z}}_{2m})+\Gamma({\mathbb{Z}}_{4})$ be the join of
graphs as shown in Figure 3.

There exist $u,v\in V(G)$ with $N(u)=N(v)$. By the necessary condition for
distance antimagic graphs, the graph
$\Gamma({\mathbb{Z}}_{2m})+\Gamma({\mathbb{Z}}_{4})$ does not have DAML for
all prime $m\geq 3$.

Figure 3: The graph $\Gamma({\mathbb{Z}}_{2m})+\Gamma({\mathbb{Z}}_{4})$

∎

We state the following few theorems without proof. The proof of those theorems
can be done as the proof of Theorem 2.1.

######  Theorem 2.2.

For every prime $m\geq 3$, $\Gamma({\mathbb{Z}}_{2m})+G$,
$G\in\\{\Gamma({\mathbb{Z}}_{6}),\Gamma({\mathbb{Z}}_{9}),\overline{\Gamma({\mathbb{Z}}_{6})},\overline{\Gamma({\mathbb{Z}}_{9})}\\}$
does not admit DAML.

######  Theorem 2.3.

For every prime $m\geq 5$, $\Gamma({\mathbb{Z}}_{3m})+G$,
$G\in\\{\Gamma({\mathbb{Z}}_{4}),\Gamma({\mathbb{Z}}_{6}),\Gamma({\mathbb{Z}}_{9}),\overline{\Gamma({\mathbb{Z}}_{6})},\overline{\Gamma({\mathbb{Z}}_{9})}\\}$
does not admit DAML.

######  Theorem 2.4.

For all $m\geq 1$ and $n=1$ if and only if the graph
$m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$ admits DAML.

###### Proof.

Consider the zero-divisor graphs $\Gamma({\mathbb{Z}}_{9})$ and
$\Gamma({\mathbb{Z}}_{4})$.
Let $G=m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$ with $|V(G)|=2m+n$
for all $m\geq 1$, and $n\geq 1$.

Case (i): For $n=1$ and $m\geq 1$.

Consider the graph $m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})$ with
$|V(m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4}))|=2m+1$.
We label the vertices of $m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})$
as shown in Figure 4.

Figure 4: The graph $m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})$

Define
$f:V(m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4}))\to\\{1,2,\dots,2m+1\\}$
by
$f(v_{i})=i$, whenever $i\in\\{1,2,\ldots,2m+1\\}$ .

The weight of each vertex is determined by:
$w(v_{i})=\begin{cases}2m+2+i,&\text{if }i\text{ is odd}\\\ 2m+i,&\text{if
}i\text{ is even}\end{cases}$
and $w(v_{2m+1})=m(2m+1)$
It follows that $w(v_{i})\neq w(v_{j})$ for all $i\neq j$.

Hence, the graph $m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})$ admits
DAML for all $m\geq 1$.

Case (ii): For $n>1$ and $m\geq 1$.

Consider the graph $m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$ with
$|V(m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4}))|=2m+n$. We label the
vertices of $m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$ as shown in
Figure 5.

Figure 5: $m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$

Clearly, we have $N(v_{1})=N(v_{2})=\dots=N(v_{m})$.
Hence, the graph $m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$ does
not have DAML for all $m\geq 1$ and $n>1$.

Therefore, the graph $m\Gamma({\mathbb{Z}}_{9})+n\Gamma({\mathbb{Z}}_{4})$
admits DAML if and only if for $m\geq 1$ and $n=1$. ∎

######  Theorem 2.5.

For all $m\geq 1$, and $n\geq 1$, the graph
$m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{4})$ does not admit DAML.

###### Proof.

Consider the zero-divisor graphs $\Gamma({\mathbb{Z}}_{6})$ and
$\Gamma({\mathbb{Z}}_{4})$.
Let $G=m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{4})$ with
$|V(G)|=3m+n$, for $m\geq 1$ and $n\geq 1$.
We label the vertices of $G$ as in Figure 6.

Figure 6: $m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{4})$

We observe that $N(u_{1})=N(u_{3})$,
$N(u_{4})=N(u_{6}),\ldots,N(u_{3m-2})=N(u_{3m})$.
Hence, the graph $m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{4})$ does
not admit DAML for all $m\geq 1$ and $n\geq 1$. ∎

######  Theorem 2.6.

For all $m\geq 1$ and $n\geq 1$,
$m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{9})$ does not admit DAML.

Figure 7:  $m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{9})$

###### Proof.

Consider the zero-divisor graphs $\Gamma({\mathbb{Z}}_{6})$ and
$\Gamma({\mathbb{Z}}_{9})$. Let
$G=m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{9})$ with $|V(G)|=3m+2n$,
for $m,n\geq 1$. The vertices of $G$ labeled as in Figure 7. We have
$N(u_{1})=N(u_{3})$, $\dots$, $N(u_{3m-2})=N(u_{3m})$. Hence, the graph
$m\Gamma({\mathbb{Z}}_{6})+n\Gamma({\mathbb{Z}}_{9})$ does not admit DAML for
all $m\geq 1$ and $n\geq 1$. ∎

######  Corollary 2.1.

For all $m\geq 1$,
$m\Gamma({\mathbb{Z}}_{6})+\overline{\Gamma({\mathbb{Z}}_{9})}$ does not admit
DAML.

###### Proof.

Let $G=m\Gamma({\mathbb{Z}}_{6})+\overline{\Gamma({\mathbb{Z}}_{9})}$ be a
join graph and $|V(G)|=3m+2$. There exist
$u,v\in\overline{\Gamma({\mathbb{Z}}_{9})}(G)$, we have $N(u)=N(v)=3m$. Hence,
the graph $m\Gamma({\mathbb{Z}}_{6})+\overline{\Gamma({\mathbb{Z}}_{9})}$ does
not admit distance antimagic labeling. ∎

######  Theorem 2.7.

The join graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ admits DAML
if and only if $m=1$ and $n\geq 1$.

###### Proof.

Consider the zero-divisor graphs $\Gamma({\mathbb{Z}}_{6})$ and
$\Gamma({\mathbb{Z}}_{9})$.
Let $G=m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ with
$|V(G)|=3m+2n$ for all $m\geq 1$ and $n\geq 1$.

For $m>1$ and $n\geq 1$, Consider the graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ with
$|V(m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9}))|=3m+2n$.
We label the vertices of the graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ as in Figure
8.

Figure 8: The graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$

It is observed that $N(u_{3})=N(u_{6})=\dots=N(u_{3m})$. Hence, the graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ does not
admit DAML, for all $m>1$ & $n\geq 1$.

For $m=1$ and $n\geq 1$, consider the graph
$\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ with
$|V(\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9}))|=2n+3$. Let
$N=2n$ and the vertices of $G$ is labeled as shown in Figure 9.

Figure 9: The graph
$\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$

Let $f:V(G)\to\\{1,2,\dots,N+3\\}$ be a labeling function and is defined by
$f(v_{i})=i$, if $1\leq i\leq N+3$. The weight of each vertex is determined by
$w(v_{i})=\begin{cases}3N+i+7,&\text{if }i\text{ odd and }1\leq i\leq N-1\\\
3N+i+5,&\text{if }i\text{ even and }2\leq i\leq N\end{cases}$
$w(v_{N+1})=(N^{2}+3N+4)/2$,
$w(v_{N+2})=(N^{2}+3N+2)/2$ and
$w(v_{N+3})=(N^{2}+N)/2$.
It follows that $w(v_{i})\neq w(v_{j})$ for all $i\neq j$.
Then the graph $\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$
admits DAML for all $n\geq 1$.
Hence, the graph
$m\overline{\Gamma({\mathbb{Z}}_{6})}+n\Gamma({\mathbb{Z}}_{9})$ admits DAML
if and only if $m=1$ and $n\geq 1$. ∎

######  Theorem 2.8.

The graph
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
has distance antimagic labeling for all $m\geq 1$.

###### Proof.

Let
$G=[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
be a cartesian product graph with $|V(G)|=4m+2$ for $m\geq 1$. We prove this
theorem in two cases:

Case 1. Let $m=1$, then the graph
$[\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
be a Cartesian product graph of order 6. We label the vertices of
$[\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
as given in Figure 10.

Figure 10: The Cartesian product graph
$[\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$

Let
$f:V([\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9}))\to\\{1,2,3,4,5,6\\}$
be a labeling function and defined by
$f(v_{1})=1$, $f(v_{2})=2$, $f(v_{3})=3$, $f(v_{4})=4$, $f(v_{5})=5$,
$f(v_{6})=6$
The weight of each vertex is determined by
$w(v_{1})=12$, $w(v_{2})=10$, $w(v_{3})=11$, $w(v_{4})=9$, $w(v_{5})=13$,
$w(v_{6})=8$

It follows that the graph
$[\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
has distance antimagic labeling. Also we observe that the graph
$[\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
is a (8,1)-distance antimagic graph.

Case 2. Let $m\geq 2$, then the graph
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
be a Cartesian product graph of order $4m+2$.
We label the vertices of
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
as depicted in Figure 11.

Figure 11: The Cartesian product graph
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$

Let
$f:V([m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9}))\to\\{1,2,\dots,4m+2\\}$
be a labeling function and defined by
$f(v_{i})=i$, if $1\leq i\leq 4m+2$.
The weight of each vertex is determined by
$w(v_{i})=10m+3-2i$, if $1\leq i\leq m$
$w(v_{i})=10m+4-2i$, if $m+1\leq i\leq 2m$
$w(v_{i})=14m+4-2i$, if $2m+1\leq i\leq 3m$
$w(v_{i})=14m+3-2i$, if $3m+1\leq i\leq 4m$
$w(v_{4m+1})=4m^{2}+5m+2$
$w(v_{4m+2})=4m^{2}+5m+1$, for $m\geq 2$
Observe that $w(v_{i})\neq w(v_{j})$, for all $i\neq j$.
Therefore, the graph
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
has distance antimagic labeling for all $m\geq 2$. Hence, from Case 1 and 2,
the graph
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
has distance antimagic labeling for all $m\geq 1$. ∎

######  Theorem 2.9.

The graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$ admits
DAML for all prime $m\neq 2$.

###### Proof.

Consider the zero-divisor graphs $\Gamma({\mathbb{Z}}_{6})$ and
$\Gamma({\mathbb{Z}}_{2m})$.
Let $G=\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$ with
$|V(G)|=3m$, for all prime $m\neq 2$.

Case (i): For $m=3$,
$|V(\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{6}))|=9$.

We label the vertices of
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{6})$ as shown in Figure
12.

Figure 12: The graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{6})$

Let $f:V(G)\to\\{1,2,\dots,9\\}$ be a labeling function and is defined by
$f(v_{1})=1$, $f(v_{2})=2$, $f(v_{3})=3$, $f(v_{4})=4$, $f(v_{5})=5$,
$f(v_{6})=6$, $f(v_{7})=7$, $f(v_{8})=8$, $f(v_{9})=9$.
The weight of each vertex is determined by:
$w(v_{1})=19$, $w(v_{2})=24$, $w(v_{3})=23$,
$w(v_{4})=6$, $w(v_{5})=18$, $w(v_{6})=8$,
$w(v_{7})=3$, $w(v_{8})=11$, $w(v_{9})=5$.
It follows that the graph
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{6})$ admits DAML.

Case (ii): Let $m>3$ and prime.

We assume the graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$
with $|V(\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m}))|=3m$.
We label the vertices of
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$ as shown in Figure
13.

Figure 13: The graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$

$f:V(G)\to\\{1,2,\dots,3m\\}$ be a labeling function defined by
$f(v_{i}^{(j)})=\begin{cases}i,&\text{where }\ j=1\ \text{and}\ 1\leq i\leq
m\\\ m+i,&\text{where}\ j=2\ \text{and}\ 1\leq i\leq m\\\ 2m+i,&\text{where}\
j=3\ \text{and}\ 1\leq i\leq m\end{cases}$
The weight of each vertex is given by:
$w(v_{1}^{(1)})=(m^{2}+7m-2)/2$,
$w(v_{i}^{(1)})=3m-i+2$, if $2\leq i\leq m$,
$w(v_{1}^{(2)})=(3m^{2}+5m-2)/2$,
$w(v_{i}^{(2)})=4m-i+2$, if $2\leq i\leq m$,
$w(v_{i}^{(3)})=6m-2i+2$, if $1\leq i\leq m-1$ and
$w(v_{m}^{(3)})=(5m^{2}-3m+4)/2$.
Therefore, the graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$
admits DAML for all prime $m>3$.
Hence, the graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$
admits DAML for all prime $m\neq 2$. ∎

######  Theorem 2.10.

The graph $\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{m^{2}})$ admits
DAML for all prime $m>3$.

###### Proof.

It is observed that the graph
$\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{m^{2}})\cong P_{2}\times
K_{m-1}$ for all prime $m\geq 3$.

Already Sankar et al. [8] proved that, for all $n\geq 3$, $P_{2}\times K_{n}$
admits DAML.

Therefore, the graph
$\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{m^{2}})$ admits DAML for
all prime $m>3$. ∎

######  Theorem 2.11.

The graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ admits
DAML for all prime $m\geq 3$.

###### Proof.

Consider the zero-divisor graphs $\Gamma({\mathbb{Z}}_{6})$ and
$\Gamma({\mathbb{Z}}_{m^{2}})$ for all prime $m\geq 3$. Let
$G=\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ with
$|V(G)|=3m-3$.

Case (i): Let $m=3$. The graph
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{9})$ admits DAML with the
labels as in Figure 14. It is clear that
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{9})$ admits DAML.

Figure 14: The graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{9})$

Case (ii): Let $m>3$ and prime. Consider the graph
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ with
$|V(\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}}))|=3m-3$. We
label the vertices of
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ as shown in
Figure 15.

Figure 15: The graph
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$

Define a labeling function $f:V(G)\to\\{1,2,...,3m-3\\}$ by
$f(v_{i})=i$, if $1\leq i\leq 3m-3$.

Then the weight of each vertex is given by:

$w(v_{i})=\begin{cases}\frac{1}{2}[m^{2}+5m-4]-2i,&\text{if }1\leq i\leq
m-1\\\ \frac{1}{2}[3m^{2}+3m-4]-2i,&\text{if }m\leq i\leq 2m-2\\\
\frac{1}{2}[5m^{2}-9m+4]-i+[3m-(2j-1)],&\text{if }2m-1\leq i\leq 3m-3\text{
and }j=3+i-2m.\end{cases}$

It follows that $w(v_{i})\neq w(v_{j})$, for all $i\neq j$.
Therefore, the graph
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ admits DAML for
all $m>3$.

Hence, the graph $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$
admits DAML for all prime $m\geq 3$. ∎

######  Theorem 2.12.

The graph $\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{3m})$ admits
DAML for all odd $m$ and $m\not\equiv 0\ (\text{mod }3)$.

###### Proof.

Let $G=\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{3m})$ with
$|V(G)|=2m+2$ for all odd $m$ and $m\not\equiv 0\ (\text{mod }3)$.

We label the vertices of $G$ as given in Figure 16.

Figure 16: The graph $\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{3m})$

Let $f:V(G)\to\\{1,2,...,2m+2\\}$ be a labeling function defined by
$f(v_{i})=i$, for $1\leq i\leq 2m+2$.

Then the weight of each vertex is given by:

$w(v_{i})=\begin{cases}\frac{1}{2}(m^{2}+5m),&\text{for }i=1\\\
\frac{1}{2}(m^{2}+5m+2),&\text{for }i=2\\\ m+i+4,&\text{for }3\leq i\leq
m+1\\\ \frac{1}{2}(3m^{2}+3m-4),&\text{for }i=m+2\\\
\frac{1}{2}(3m^{2}+3m-2),&\text{for }i=m+3\\\ m+i+4,&\text{for }m+4\leq i\leq
2m+2\end{cases}$

It follows that $w(v_{i})\neq w(v_{j})$, for all $i\neq j$.
Hence, the graph $\Gamma({\mathbb{Z}}_{9})\times\Gamma({\mathbb{Z}}_{3m})$
admits DAML for all odd $m$ and
$m\not\equiv 0\ (\text{mod }3)$. ∎

##  3 Conclusion

In this paper, we found the distance antimagic labeling of
$[m\Gamma({\mathbb{Z}}_{9})+\Gamma({\mathbb{Z}}_{4})]\times\Gamma({\mathbb{Z}}_{9})$
for all $m\geq 1$, $\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{2m})$
for all primes $m\geq 3$ and
$\Gamma({\mathbb{Z}}_{6})\times\Gamma({\mathbb{Z}}_{m^{2}})$ for all primes
$m\geq 3$. The distance antimagic labeling of
$\Gamma({\mathbb{Z}}_{m})\times\Gamma({\mathbb{Z}}_{n})$ for all values of $m$
and $n$ and
$\Gamma({\mathbb{Z}}_{p})\times[\Gamma({\mathbb{Z}}_{m})\times\Gamma({\mathbb{Z}}_{n})]$
for all values of $p,m$ and $n$ are left for future research.

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