# The Coven-Meyerowitz tiling conditions for 3 odd prime factors

Izabella Łaba and Itay Londner

###### Abstract.

It is well known that if a finite set $A\subset\mathbb{Z}$ tiles the integers
by translations, then the translation set must be periodic, so that the tiling
is equivalent to a factorization $A\oplus B=\mathbb{Z}_{M}$ of a finite cyclic
group. We are interested in characterizing all finite sets
$A\subset\mathbb{Z}$ that have this property. Coven and Meyerowitz [3]
proposed conditions (T1), (T2) that are sufficient for $A$ to tile, and
necessary when the cardinality of $A$ has at most two distinct prime factors.
They also proved that (T1) holds for all finite tiles, regardless of size. It
is not known whether (T2) must hold for all tilings with no restrictions on
the number of prime factors of $|A|$.

We prove that the Coven-Meyerowitz tiling condition (T2) holds for all integer
tilings of period $M=(p_{i}p_{j}p_{k})^{2}$, where $p_{i},p_{j},p_{k}$ are
distinct odd primes. The proof also provides a classification of all such
tilings.

###### Contents

  1. 1 Introduction
  2. 2 Notation and preliminaries
    1. 2.1 Multisets and mask polynomials
    2. 2.2 Array coordinates and geometric representation
    3. 2.3 Divisor set and divisor exclusion
    4. 2.4 Standard tiling complements
    5. 2.5 Box product
    6. 2.6 Cuboids
    7. 2.7 Tiling reductions
    8. 2.8 Saturating sets
    9. 2.9 Fibers and cofibered structures
  3. 3 Classification results
    1. 3.1 Classification results
    2. 3.2 Outline of the proof
  4. 4 Toolbox
    1. 4.1 Divisors
    2. 4.2 Cyclotomic divisibility
    3. 4.3 Saturating sets
    4. 4.4 Fibering lemmas
  5. 5 Structure on unfibered grids
    1. 5.1 Diagonal boxes
    2. 5.2 Extended corners
  6. 6 Unfibered grids with missing top differences
    1. 6.1 A structure result
    2. 6.2 Special unfibered structures: odd $M$
    3. 6.3 Special unfibered structures: even $M$
  7. 7 Resolving diagonal boxes
    1. 7.1 Preliminary results
    2. 7.2 Case (DB1)
    3. 7.3 Case (DB2)
  8. 8 Corners
    1. 8.1 Cases (C1) and (C2)
    2. 8.2 Case (C3)
  9. 9 Fibered grids
    1. 9.1 Intersections of $\mathcal{I},\mathcal{J},\mathcal{K}$
    2. 9.2 Toolbox for fibered grids
    3. 9.3 Fibering on lower scales
    4. 9.4 Proof of Proposition 9.2
    5. 9.5 Proof of Proposition 9.3
    6. 9.6 Proof of Proposition 9.4
  10. 10 Acknowledgement

##  1\. Introduction

We say that a set $A\subset\mathbb{Z}$ tiles the integers by translations if
there is a set $T\subset\mathbb{Z}$ such that every integer $n$ can be
represented uniquely as $n=a+t$ with $a\in A$ and $t\in T$. Throughout this
article, we assume that $A$ is finite. It is well known (see [33]) that any
tiling of $\mathbb{Z}$ by a finite set $A$ must be periodic, i.e. $T=B\oplus
M\mathbb{Z}$ for some finite set $B\subset\mathbb{Z}$ such that $|A|\,|B|=M$.
Equivalently, $A\oplus B$ is a factorization of the cyclic group
$\mathbb{Z}_{M}$.

We are interested in determining which finite sets $A\subset\mathbb{Z}$ have
this property, and, in particular, in a characterization proposed by Coven and
Meyerowitz [3]. In order to state their conditions, we need to introduce some
notation. By translational invariance, we may assume that
$A,B\subset\\{0,1,\dots\\}$ and that $0\in A\cap B$. The characteristic
polynomials (also known as mask polynomials) of $A$ and $B$ are

$A(X)=\sum_{a\in A}X^{a},\ B(x)=\sum_{b\in B}X^{b}.$

Then the tiling condition $A\oplus B=\mathbb{Z}_{M}$ is equivalent to

(1.1) $A(X)B(X)=1+X+\dots+X^{M-1}\ \mod(X^{M}-1).$

Let $\Phi_{s}(X)$ be the $s$-th cyclotomic polynomial, i.e., the unique monic,
irreducible polynomial whose roots are the primitive $s$-th roots of unity.
Alternatively, $\Phi_{s}$ can be defined inductively via the identity

(1.2) $X^{n}-1=\prod_{s|n}\Phi_{s}(X).$

In particular, (1.1) is equivalent to

(1.3) $|A||B|=M\hbox{ and }\Phi_{s}(X)\ |\ A(X)B(X)\hbox{ for all }s|M,\ s\neq
1.$

Since $\Phi_{s}$ are irreducible, each $\Phi_{s}(X)$ with $s|M$ must divide at
least one of $A(X)$ and $B(X)$.

The following result is due to Coven and Meyerowitz [3].

######  Theorem 1.1.

[3] Let $S_{A}$ be the set of prime powers $p^{\alpha}$ such that
$\Phi_{p^{\alpha}}(X)$ divides $A(X)$. Consider the following conditions.

(T1) $A(1)=\prod_{s\in S_{A}}\Phi_{s}(1)$,

(T2) if $s_{1},\dots,s_{k}\in S_{A}$ are powers of different primes, then
$\Phi_{s_{1}\dots s_{k}}(X)$ divides $A(X)$.

Then:

  * •

if $A$ satisfies (T1), (T2), then $A$ tiles $\mathbb{Z}$;

  * •

if $A$ tiles $\mathbb{Z}$ then (T1) holds;

  * •

if $A$ tiles $\mathbb{Z}$ and $|A|$ has at most two distinct prime factors,
then (T2) holds.

While (T1) is relatively easy to prove, (T2) turns out to be much deeper and
more difficult. Coven and Meyerowitz [3] proved that if (T2) holds, then the
tiling complement $B$ in the tiling $A\oplus B=\mathbb{Z}_{M}$ may be replaced
by an explicit, highly structured “standard” tiling complement (defined here
in Section 2.4). We prove in [23] that having a tiling complement of this type
is in fact equivalent to (T2). In this formulation, (T2) bears some
resemblance to questions on replacement of factors in theory of factorizations
of abelian groups (see [43] for an overview of the latter).

The proof of Theorem 1.1 in [3] is based on an inductive argument. Coven and
Meyerowitz use a theorem of Tijdeman [46] to prove that if $A$ tiles the
integers, then it also tiles $\mathbb{Z}_{M}$ for some $M$ which has the same
prime factors as $|A|$. Hence, if $|A|$ has at most two distinct prime
factors, then we may assume that so does $M$. The authors then use Sands’s
theorem [36], which states that, in any tiling $A\oplus B=\mathbb{Z}_{M}$ with
$M$ divisible by at most 2 primes, at least one of $A$ and $B$ must be
contained in $p\mathbb{Z}_{M}$ for some prime $p|M$. Coven and Meyerowitz use
this to decompose the given tiling into tilings of smaller groups while
keeping track of the (T2) property. We also note that if $|A|$ is a prime
power, then the Coven-Meyerowitz characterization simplifies further since
(T2) is vacuous; in this case, the result had been proved earlier by Newman
[33].

The Coven-Meyerowitz proof does not extend to the general case. Sands’s factor
replacement theorem is false if $M$ has three or more prime factors, with
counterexamples in [42], [24]. On the other hand, we prove in [23, Corollary
6.2] (using a relatively minor modification of the argument in [3]) that if
$A\oplus B=\mathbb{Z}_{M}$, and if $|A|$ and $|B|$ share at most two distinct
prime factors, then both $A$ and $B$ satisfy (T2). (See also [45], [38].) Thus
the simplest case that is not covered by these methods is when
$|A|=|B|=p_{i}p_{j}p_{k}$, where $p_{i},p_{j},p_{k}$ are distinct primes.

Our main result is the following theorem.

######  Theorem 1.2.

Let $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$, where $p_{i},p_{j},p_{k}$ are distinct
odd primes. Assume that $A\oplus B=\mathbb{Z}_{M}$, with
$|A|=|B|=p_{i}p_{j}p_{k}$. Then both $A$ and $B$ satisfy (T2).

We also obtain a classification of all tilings $A\oplus B=\mathbb{Z}_{M}$,
where $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$. Our main results in that regard are
Theorems 3.1 and 3.2. Since those theorems require some notation and
definitions, we postpone their statements until Section 3.

The proof of Theorem 1.2 relies on the methods and concepts introduced in
[23]. In order to keep this article reasonably self-contained modulo results
that can be used as black boxes, we provide a summary of the concepts and
results that we will need here, specialized to the 3-prime setting, in Section
2. We then state our classification results in Section 3.1. In Section 3.2, we
discuss our strategy and some of the main new ideas of the proof. The rest of
the paper is devoted to the proof of Theorems 1.2, 3.1, and 3.2.

Since [3], there has been essentially no progress on proving (T2), except for
a few special cases that either assume particular structure of the tiling (see
e.g. [21]) or are covered by the methods of [3] as in [23, Corollary 6.2] (see
e.g. [38]). However, there has been recent work on other questions related to
tiling. For instance, Bhattacharya [1] has established the periodic tiling
conjecture in $\mathbb{Z}^{2}$, with a quantitative version due to Greenfeld
and Tao [12]. There has also been interesting work on tilings of the real line
by a function (see [18] for a survey and some open questions).

The Coven-Meyerowitz tiling conditions have implications for the ongoing work
on Fuglede’s spectral set conjecture [9]. Fuglede conjectured that a set
$\Omega\subset\mathbb{R}^{n}$ of positive $n$-dimensional Lebesgue measure
tiles $\mathbb{R}^{n}$ by translations if and only if it is spectral, in the
sense that the space $L^{2}(\Omega)$ admits an orthogonal basis of exponential
functions. While the conjecture has been disproved in its full generality in
dimensions 3 and higher [44], [19], [20], [7], [32], [8], significant
connections between tiling and spectrality do exist (see [4] for an overview
of the problem in dimension 1), and there is a large body of work
investigating such connections from many points of view. In higher dimensions,
the conjecture has been proved for convex sets in $\mathbb{R}^{n}$, by
Iosevich, Katz and Tao [13] for $n=2$, Greenfeld and Lev [11] for $n=3$, and
by Lev and Matolsci [28] for general $n$. There have been many recent results
on special cases of the finite abelian group analogue of the conjecture [14],
[30], [5], [15], [16], [17], [29], [38], [39], [40], [6].

If (T2) could be proved for all finite integer tiles, this would imply the
“tiling implies spectrum” part of Fuglede’s spectral set conjecture for all
compact tiles in dimension 1, as well as for all cyclic groups
$\mathbb{Z}_{M}$. This follows from the results of [25], [26], [22]. Proving
(T2) for specific tiling problems does not resolve the full conjecture, but it
does imply that the conjecture holds in those settings. In that regard, our
Theorem 1.2 combined with [22, Theorem 1.5] and [23, Corollary 6.2] has the
following immediate corollary.

######  Corollary 1.3.

Let $M={p_{i}^{2}p_{j}^{2}p_{k}^{2}}$.

(i) The “tiling implies spectrum” part of Fuglede’s spectral set conjecture
holds for the cyclic group $\mathbb{Z}_{M}$. In other words, if
$A\subset\mathbb{Z}_{M}$ tiles $\mathbb{Z}_{M}$ by translations, then it is
spectral.

(ii) Let $A\subset\mathbb{Z}$ be a finite set such that $A$ tiles
$\mathbb{Z}_{M}$, and let ${F=\bigcup_{a\in A}[a,a+1]}$, so that $F$ tiles
$\mathbb{R}$ by translations. Then $F$ is spectral.

Indeed, let $M$ and $A$ be as in Corollary 1.3. If $|A|\neq p_{i}p_{j}p_{k}$,
then both $A$ and $B$ satisfy (T2) by [23, Corollary 6.2]. If on the other
hand $|A|=p_{i}p_{j}p_{k}$, then both $A$ and $B$ satisfy (T2) by Theorem 1.2.
In both cases, spectrality follows from [22, Theorem 1.5]. (While Theorem 1.5
in [22] is stated for unions of finite intervals as in (ii), the same argument
applies in the finite group setting. See e.g. [4].)

##  2\. Notation and preliminaries

This section summarizes the relevant definitions and results of [23],
specialized to the 3-prime case. All material due to other authors is
indicated explicitly as such.

###  2.1. Multisets and mask polynomials

Throughout this paper, we will assume that
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$, where $p_{i},p_{j},p_{k}$ are
distinct primes and $n_{i},n_{j},n_{k}\in\mathbb{N}$. The indices
$\\{i,j,k\\}$ can be thought of as a permutation of $\\{1,2,3\\}$; however, we
will always use $i,j,k$ for this purpose, freeing up numerical subscripts for
other purposes. While the full proof of Theorem 1.2 requires that
$n_{i}=n_{j}=n_{k}=2$ and that $p_{i},p_{j},p_{k}\neq 2$, many of our
intermediate results are valid under weaker assumptions as indicated.

We will always work in either $\mathbb{Z}_{M}$ or in $\mathbb{Z}_{N}$ for some
$N|M$. We use $A(X)$, $B(X)$, etc. to denote polynomials modulo $X^{M}-1$ with
integer coefficients. Each such polynomial
$A(X)=\sum_{a\in\mathbb{Z}_{M}}w_{A}(a)X^{a}$ is associated with a weighted
multiset in $\mathbb{Z}_{M}$, with weights $w_{A}(x)$ assigned to each
$x\in\mathbb{Z}_{M}$, which we will also denote by $A$. (If the coefficient of
$X^{x}$ in $A(X)$ is 0, we set $w_{A}(x)=0$.) In particular, if $A$ has 0-1
coefficients, then $w_{A}$ is the characteristic function of a set
$A\subset\mathbb{Z}_{M}$. We will use $\mathcal{M}(\mathbb{Z}_{M})$ to denote
the family of all weighted multisets in $\mathbb{Z}_{M}$, and reserve the
notation $A\subset\mathbb{Z}_{M}$ for sets.

If $N|M$, then any $A\in\mathcal{M}(\mathbb{Z}_{M})$ induces a weighted
multiset $A$ mod $N$ in $\mathbb{Z}_{N}$, with the corresponding mask
polynomial $A(X)$ mod $(X^{N}-1)$ and induced weights

(2.1) $w_{A}^{N}(x)=\sum_{x^{\prime}\in\mathbb{Z}_{M}:x^{\prime}\equiv x\,{\rm
mod}\,N}w_{A}(x^{\prime}),\ \ x\in\mathbb{Z}_{N}.$

We will continue to write $A$ and $A(X)$ for $A$ mod $N$ and $A(X)$ mod
$X^{N}-1$, respectively, while working in $\mathbb{Z}_{N}$.

If $A,B\in\mathcal{M}(\mathbb{Z}_{M})$, we will use $A+B$ to indicate the
weighted multiset corresponding to the mask polynomial $(A+B)(X)=A(X)+B(X)$,
with the weight function $w_{A+B}(x)=w_{A}(x)+w_{B}(x)$. We use the
convolution notation $A*B$ to denote the weighted sumset of $A$ and $B$, so
that $(A*B)(X)=A(X)B(X)$ and

$w_{A*B}(x)=(w_{A}*w_{B})(x)=\sum_{y\in\mathbb{Z}_{M}}w_{A}(x-y)w_{B}(y).$

If one of the sets is a singleton, say $A=\\{x\\}$, we will simplify the
notation and write $x*B=\\{x\\}*B$. The direct sum notation $A\oplus B$ is
reserved for tilings, i.e., $A\oplus B=\mathbb{Z}_{M}$ means that
$A,B\subset\mathbb{Z}_{M}$ are both sets and $A(X)B(X)=\frac{X^{M}-1}{X-1}$
mod $X^{M}-1$. We will not use derivatives of polynomials in this paper, hence
notation such as $A^{\prime}$, $A^{\prime\prime}$, etc., will be used to
denote auxiliary multisets and polynomials rather than derivatives.

###  2.2. Array coordinates and geometric representation

For $\nu\in\\{i,j,k\\}$, define
$M_{\nu}:=M/p_{\nu}^{n_{\nu}}=\prod_{\kappa\neq\nu}p_{\kappa}^{n_{\kappa}}$.
Then each $x\in\mathbb{Z}_{M}$ can be written uniquely as

$x=\sum_{\nu\in\\{i,j,k\\}}\pi_{\nu}(x)M_{\nu},\ \
\pi_{\nu}(x)\in\mathbb{Z}_{p_{\nu}^{n_{\nu}}}.$

This sets up an isomorphism
$\mathbb{Z}_{M}=\mathbb{Z}_{p_{i}^{n_{i}}}\oplus\mathbb{Z}_{p_{j}^{n_{j}}}\oplus\mathbb{Z}_{p_{k}^{n_{k}}}$,
and identifies each element $x\in\mathbb{Z}_{M}$ with an element of a
3-dimensional lattice with coordinates $(\pi_{i}(x),\pi_{j}(x),\pi_{k}(x))$.
The tiling $A\oplus B=\mathbb{Z}_{M}$ can then be interpreted as a tiling of
that lattice.

For $D|M$, a $D$-grid in $\mathbb{Z}_{M}$ is a set of the form

$\Lambda(x,D):=x+D\mathbb{Z}_{M}=\\{x^{\prime}\in\mathbb{Z}_{M}:\
D|(x-x^{\prime})\\}$

for some $x\in\mathbb{Z}_{M}$. A few special cases have a geometric
interpretation of interest. A line through $x\in\mathbb{Z}_{M}$ in the
$p_{\nu}$ direction is the set $\ell_{\nu}(x):=\Lambda(x,M_{\nu})$, and a
plane through $x\in\mathbb{Z}_{M}$ perpendicular to the $p_{\nu}$ direction,
on the scale $M_{\nu}p_{\nu}^{\alpha_{\nu}}$, is the set
$\Pi(x,p_{\nu}^{\alpha_{\nu}}):=\Lambda(x,p_{\nu}^{\alpha_{\nu}}).$

An $M$-fiber in the $p_{\nu}$ direction is a set of the form $x*F_{\nu}$,
where $x\in\mathbb{Z}_{M}$ and

(2.2) $F_{\nu}=\\{0,M/p_{\nu},2M/p_{\nu},\dots,(p_{\nu}-1)M/p_{\nu}\\}.$

Thus $x*F_{\nu}=\Lambda(x,M/p_{\nu})$. A set $A\subset\mathbb{Z}_{M}$ is
$M$-fibered in the $p_{\nu}$ direction if there is a subset $A^{\prime}\subset
A$ such that $A=A^{\prime}*F_{\nu}$.

If $N=p_{i}^{\alpha_{1}}p_{j}^{\alpha_{j}}p_{k}^{\alpha_{k}}$ is a divisor of
$M$, with $0\leq\alpha_{\nu}\leq n_{\nu}$, we let

$D(N):=p_{i}^{\gamma_{i}}p_{j}^{\gamma_{j}}p_{k}^{\gamma_{k}},\ \hbox{ where
}\gamma_{\nu}=\max(0,\alpha_{\nu}-1)\hbox{ for }\nu\in\\{i,j,k\\}.$

We will also write $N_{\nu}=M/p_{\nu}$ for $\nu\in\\{i,j,k\\}$.

###  2.3. Divisor set and divisor exclusion

For $N|M$ and $A\subset\mathbb{Z}_{M}$, we define

(2.3) ${\mathrm{Div}}_{N}(A):=\\{(a-a^{\prime},N):\ a,a^{\prime}\in A\\}$

When $N=M$, we will omit the subscript and write
${\mathrm{Div}}(A)={\mathrm{Div}}_{M}(A)$. Informally, we will refer to the
elements of ${\mathrm{Div}}(A)$ as the divisors of $A$ or differences in $A$.
A theorem due to Sands [36] states that $A\oplus B=\mathbb{Z}_{M}$ if and only
if $A,B\subset\mathbb{Z}_{M}$ are sets such that $|A|\,|B|=M$ and

(2.4) ${\mathrm{Div}}(A)\cap{\mathrm{Div}}(B)=\\{M\\}.$

We will refer to this as divisor exclusion.

In cases when we need to indicate where a particular divisor of $A$ must
occur, we will use the following notation for localized divisor sets. If
$A,A_{1},A_{2}\subset\mathbb{Z}_{M}$ and $a_{0}\in\mathbb{Z}_{M}$, we will
write

(2.5) $\begin{split}{\mathrm{Div}}_{N}(A_{1},A_{2})&:=\\{(a_{1}-a_{2},N):\
a_{1}\in A_{1},a_{2}\in A_{2}\\},\\\
{\mathrm{Div}}_{N}(A,a_{0})={\mathrm{Div}}_{N}(a_{0},A)&:={\mathrm{Div}}_{N}(A,\\{a_{0}\\})=\\{(a-a_{0},N):\
a\in A\\}.\end{split}$

For example, if $A\oplus B=\mathbb{Z}_{M}$, we will often need to consider
${\mathrm{Div}}(A_{1},A_{2})$, where $A_{1}$ and $A_{2}$ are restrictions of
$A$ to geometric structures such as planes or lines.

###  2.4. Standard tiling complements

Suppose that $A\oplus B=\mathbb{Z}_{M}$, and let

$\mathfrak{A}_{\nu}(A)=\left\\{\alpha_{\nu}\in\\{1,2,\dots,n_{\nu}\\}:\
\Phi_{p_{\nu}^{\alpha_{\nu}}}(X)|A(X)\right\\}$

The standard tiling complement $A^{\flat}$ is defined via its mask polynomial

(2.6)
$A^{\flat}(X)=\prod_{\nu\in\\{i,j,k\\}}\prod_{\alpha_{\nu}\in\mathfrak{A}_{\nu}(A)}\left(1+X^{M_{\nu}p_{\nu}^{\alpha_{\nu}-1}}+\dots+X^{(p_{\nu}-1)M_{\nu}p_{\nu}^{\alpha_{\nu}-1}}\right).$

Figure 1. The standard sets
$A^{\flat},B^{\flat}\subset\mathbb{Z}_{p_{i}^{2}p_{j}^{2}}$
with $p_{i}=3,p_{j}=5$ and
$\Phi_{p_{i}^{2}}\Phi_{p_{j}^{2}}|A,\Phi_{p_{i}}\Phi_{p_{j}}|B$.

Then $A^{\flat}(X)$ satisfies (T2) and has the same prime power cyclotomic
divisors as $A(X)$. For each prime power $s|M$, $\Phi_{s}$ divides exactly one
of $A$ and $B$ [3], hence $A^{\flat}$ is also uniquely determined by $M$ and
$B$. Coven and Meyerowitz proved in [3] that if a finite tile satisfies (T2),
it has a standard tiling complement. We prove the converse in [23].

######  Proposition 2.1.

Let $A\oplus B=\mathbb{Z}_{M}$. Then $A^{\flat}\oplus B=\mathbb{Z}_{M}$ if and
only if $B$ satisfies (T2).

We say that the tilings $A\oplus B=\mathbb{Z}_{M}$ and $A^{\prime}\oplus
B=\mathbb{Z}_{M}$ are T2-equivalent if

(2.7) $A\hbox{ satisfies (T2) }\Leftrightarrow A^{\prime}\hbox{ satisfies
(T2).}$

Since $A$ and $A^{\prime}$ tile the same group $\mathbb{Z}_{M}$ with the same
tiling complement $B$, they must have the same cardinality and the same prime
power cyclotomic divisors. We will sometimes say simply that $A$ is
T2-equivalent to $A^{\prime}$ if both $M$ and $B$ are clear from context.
Usually, $A^{\prime}$ will derived from $A$ using certain permitted
manipulations such as fiber shifts (Lemma 2.9). In particular, if we can prove
that either $A$ or $B$ in a given tiling is T2-equivalent to a standard tiling
complement, this resolves the problem completely in that case.

######  Corollary 2.2.

Suppose that the tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to the
tiling $A^{\flat}\oplus B=\mathbb{Z}_{M}$. Then $A$ and $B$ satisfy (T2).

###  2.5. Box product

Let $A\subset\mathbb{Z}_{M}$ and $N|M$. For $x\in\mathbb{Z}_{M}$, define

$\displaystyle\mathbb{A}^{N}_{m}[x]$ $\displaystyle=\\#\\{a\in A:\
(x-a,N)=m\\}.$

We may think of $A^{N}_{m}[x]$, with $x$ fixed and $m$ ranging over the
divisors of $N$, as the entries of the $N$-box
$\mathbb{A}^{N}[x]=(\mathbb{A}^{N}_{m}[x])_{m|N}$ [23]. If $N=M$, we will
usually omit the superscript and write
$\mathbb{A}^{M}_{m}[x]=\mathbb{A}_{m}[x]$. If $C\subset\mathbb{Z}_{M}$, we
write $\mathbb{A}_{m}^{N}[C]:=\sum_{c\in C}\mathbb{A}^{N}_{m}[c]$.
Furthermore, if $X\subset\mathbb{Z}_{M}$ and $x\in\mathbb{Z}_{M}$, define

$\mathbb{A}^{N}_{m}[x|X]=\\#\\{a\in A\cap X:\ (x-a,N)=m\\}.$

If $A,B\subset\mathbb{Z}_{M}$, we define the box product of the associated
$M$-boxes as

(2.8)
$\langle\mathbb{A}[x],\mathbb{B}[y]\rangle=\sum_{m|M}\frac{1}{\phi(M/m)}\mathbb{A}_{m}[x]\mathbb{B}_{m}[y].$

Here $\phi$ is the Euler totient function: if
$n=\prod_{\iota=1}^{L}q_{\iota}^{r_{\iota}}$, where $q_{1},\dots,q_{L}$ are
distinct primes and $r_{\iota}\in\mathbb{N}$, then
$\phi(n)=\prod_{\iota=1}^{L}(q_{\iota}-1)q_{\iota}^{r_{\iota}-1}.$

######  Theorem 2.3.

([23]; following [10, Theorem 1]) If $A\oplus B=\mathbb{Z}_{M}$ is a tiling,
then

(2.9) $\langle\mathbb{A}^{M}[x],\mathbb{B}^{M}[y]\rangle=1\ \ \forall
x,y\in\mathbb{Z}_{M}.$

###  2.6. Cuboids

######  Definition 2.4.

(i) A cuboid type $\mathcal{T}$ on $\mathbb{Z}_{N}$ is an ordered triple
$\mathcal{T}=(N,\vec{\delta},T)$, where:

  * •

$N=\prod_{\nu\in\\{i,j,k\\}}p_{\nu}^{n_{\nu}-\alpha_{\nu}}$ is a divisor of
$M$, with $0\leq\alpha_{\nu}\leq n_{\nu}$ for each $\nu$,

  * •

$\vec{\delta}=(\delta_{i},\delta_{j},\delta_{k})$, with $0\leq\delta_{\nu}\leq
n_{\nu}-\alpha_{\nu}$,

  * •

the template $T$ is a nonempty subset of $\mathbb{Z}_{N}$.

(ii) A cuboid $\Delta$ of type $\mathcal{T}$ is a weighted multiset
corresponding to a mask polynomial of the form

(2.10) $\Delta(X)=X^{c}\prod_{\nu\in\mathfrak{J}}(1-X^{d_{\nu}}),$

where $\mathfrak{J}=\mathfrak{J}_{\vec{\delta}}:=\\{\nu:\delta_{\nu}\neq
0\\}$, and $c,d_{\nu}$ are elements of $\mathbb{Z}_{M}$ such that
$(d_{\nu},N)=N/p_{\nu}^{\delta_{\nu}}$ for $\nu\in\\{i,j,k\\}$. The vertices
of $\Delta$ are the points

(2.11) $x_{\vec{\epsilon}}=c+\sum_{\nu\in\mathfrak{J}}\epsilon_{\nu}d_{\nu}:\
\vec{\epsilon}=(\epsilon_{\nu})_{\nu\in\mathfrak{J}}\in\\{0,1\\}^{|\mathfrak{J}|},$

with weights
$w_{\Delta}(x_{\vec{\epsilon}})=(-1)^{\sum_{\nu\in\mathfrak{J}}\epsilon_{\nu}}$.

(iii) Let $A\in\mathcal{M}(\mathbb{Z}_{N})$, and let $\Delta$ be a cuboid of
type $\mathcal{T}$. Define

$\mathbb{A}^{\mathcal{T}}[\Delta]=\mathbb{A}^{N}_{N}[\Delta*T]=\sum_{\vec{\epsilon}\in\\{0,1\\}^{k}}w_{\Delta}(x_{\vec{\epsilon}})\mathbb{A}^{N}_{N}[x_{\vec{\epsilon}}*T],$

where we recall that $x*T=\\{x+t:\ t\in T\\}$, so that

$\mathbb{A}^{N}_{N}[x_{\vec{\epsilon}}*T]:=\sum_{t\in
T}\mathbb{A}^{N}_{N}[x_{\vec{\epsilon}}+t].$

For consistency, we will also write
$\mathbb{A}^{\mathcal{T}}[x]=\mathbb{A}^{N}_{N}[x*T]$ for
$x\in\mathbb{Z}_{M}.$

An important special case is as follows: for $N|M$, an $N$-cuboid is a cuboid
of type $\mathcal{T}=(N,\vec{\delta},T)$, where $N|M$, $T(X)=1$, and
$\delta_{\nu}=1$ for all $\nu$ such that $p_{\nu}|N$. Thus, $N$-cuboids have
the form

(2.12) $\Delta(X)=X^{c}\prod_{p_{\nu}|N}(1-X^{d_{\nu}}),$

with $(d_{\nu},N)=N/p_{\nu}$ for all $\nu$ such that $p_{\nu}|N$. We reserve
the term “$N$-cuboid”, without cuboid type explicitly indicated, to refer to
cuboids as in (2.12); for cuboids of any other type, we will always specify
$\mathcal{T}$.

Cuboids provide useful criteria to determine cyclotomic divisibility
properties of mask polynomials. We say that a multiset
$A\in\mathcal{M}(\mathbb{Z}_{M})$ is $\mathcal{T}$-null if for every cuboid
$\Delta$ of type $\mathcal{T}$,

(2.13) $\mathbb{A}^{\mathcal{T}}[\Delta]=0.$

For $A\in\mathcal{M}(\mathbb{Z}_{N})$, we have $\Phi_{N}(X)|A(X)$ if and only
if $\mathbb{A}^{N}_{N}[\Delta]=0$ for every $N$-cuboid $\Delta$. This has been
known and used previously in the literature, see e.g. [41, Section 3], or [15,
Section 3]. In particular, for any $N|M$, $\Phi_{N}$ divides $A$ if and only
if it divides the mask polynomial of $A\cap\Lambda(x,D(N))$ for every
$x\in\mathbb{Z}_{M}$.

More generally, if for every $m|N$ the polynomial $\Phi_{m}(X)$ divides at
least one of $A(X)$, $T(X)$, or $\Delta(X)$ for every $\Delta$ of type
$\mathcal{T}(N,\vec{\delta},T)$, then $A$ is $\mathcal{T}$-null. We use such
cuboid types to test for divisibility by combinations of cyclotomic
polynomials. For example, we have the following.

  * •

Assume that $n_{i}\geq 2$, and let $\mathcal{T}=(M,\vec{\delta},1)$, with
$\delta_{i}=2$ and $\delta_{j}=\delta_{k}=1$. Then

$\displaystyle\Phi_{M}\Phi_{M/p_{i}}|A$ $\displaystyle\Leftrightarrow A\hbox{
is }\mathcal{T}\hbox{-null}.$

  * •

Assume that $n_{i}=2$, and let $\mathcal{T}=(M,\vec{\delta},T)$, where
$\delta_{i}=0$, $\delta_{j}=\delta_{k}=1$, and

$T(X)=\frac{X^{M/p_{i}}-1}{X^{M/p_{i}^{2}}-1}=1+X^{M/p_{i}^{2}}+\dots+X^{(p_{i}-1)M/p_{i}^{2}}.$

If $\Phi_{M}\Phi_{M/p_{i}^{2}}|A$, then $A$ is $\mathcal{T}$-null.

###  2.7. Tiling reductions

The general formulations of the tiling reductions below are provided in
Theorems 6.1, 6.5, and Corollary 6.6 in [23]. The additional assumption that
$p_{\nu}\parallel|B|$ in Theorem 2.5 ensures that in any tiling
$A^{\prime}\oplus B^{\prime}=\mathbb{Z}_{N_{\nu}}$ with $|A^{\prime}|=|A|$ and
$|B^{\prime}|=|B|/p_{\nu}$, $|A^{\prime}|$ and $|B^{\prime}|$ have only two
common factors. Hence the assumption (ii) of [23, Theorem 6.1] is satisfied by
[23, Corollary 6.2], and we deduce that $A$ and $B$ both satisfy (T2). The
assumption that $p_{\nu}\parallel|A|$ in Corollary 2.7 serves the same
purpose.

######  Theorem 2.5.

(Subgroup reduction) [3, Lemma 2.5] Let
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$. Assume that $A\oplus
B=\mathbb{Z}_{M}$, and that $A\subset p_{\nu}\mathbb{Z}_{M}$ for some
$\nu\in\\{i,j,k\\}$ such that $p_{\nu}\parallel|B|$. Then $A$ and $B$ satisfy
(T2).

######  Theorem 2.6.

Assume that $A\oplus B=\mathbb{Z}_{M}$ and $\Phi_{p_{\nu}^{n_{\nu}}}|A$ for
some $\nu\in\\{i,j,k\\}$. Define

(2.14) $A_{p_{\nu}}=\\{a\in A:\ 0\leq\pi_{\nu}(a)\leq
p_{\nu}^{n_{\nu}-1}-1\\}.$

Then the following are equivalent:

(i) For any translate $A^{\prime}$ of $A$ we have $A^{\prime}_{p_{\nu}}\oplus
B=\mathbb{Z}_{M/p_{\nu}}$.

(ii) For every $d$ such that $p_{\nu}^{n_{\nu}}|d|M$, at least one of the
following holds:

(2.15) $\Phi_{d}|A,$ (2.16)
$\Phi_{d/p_{\nu}}\Phi_{d/p_{\nu}^{2}}\dots\Phi_{d/p_{\nu}^{n_{\nu}}}\mid B.$

######  Corollary 2.7.

(Slab reduction) Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$. Assume that
$A\oplus B=\mathbb{Z}_{M}$, and that there exists a $\nu\in\\{i,j,k\\}$ such
that $\Phi_{p_{\nu}^{2}}|A$, $p_{\nu}\parallel|A|$, and $A,B$ obey the
condition (ii) of Theorem 2.6. (In particular, this holds if $A$ is
$M$-fibered in one of the $p_{i},p_{j},p_{k}$ directions.) Then $A$ and $B$
satisfy (T2).

###  2.8. Saturating sets

Let $A\oplus B=\mathbb{Z}_{M}$, and $x,y\in\mathbb{Z}_{M}$. Define

$A_{x,y}:=\\{a\in A:\ (x-a,M)=(y-b,M)\hbox{ for some }b\in B\\},$
$A_{x}:=\\{a\in A:(x-a,M)\in{\mathrm{Div}}(B)\\}=\bigcup_{b\in B}A_{x,b}.$

We will refer to $A_{x}$ as the saturating set for $x$. The sets $B_{y,x}$ and
$B_{y}$ are defined similarly, with $A$ and $B$ interchanged.

By divisor exclusion, $A_{a}=\\{a\\}$ for all $a\in A$. For
$x\in\mathbb{Z}_{M}\setminus A$, saturating spaces are more robust, but are
still subject to geometric constraints based on divisor exclusion. For
$x,x^{\prime}\in\mathbb{Z}_{M}$ such that
$(x-x^{\prime},M)=p_{i}^{\alpha_{i}}p_{j}^{\alpha_{j}}p_{k}^{\alpha_{k}}$,
with $0\leq\alpha_{\nu}\leq n_{\nu}$, define

(2.17)
$\begin{split}{\mathrm{Span}}(x,x^{\prime})&=\bigcup_{\nu:\alpha_{\nu}<n_{\nu}}\Pi(x,p_{\nu}^{\alpha_{\nu}+1}),\\\
{\mathrm{Bispan}}(x,x^{\prime})&={\mathrm{Span}}(x,x^{\prime})\cup{\mathrm{Span}}(x^{\prime},x).\end{split}$

Then for any $x,x^{\prime},y\in\mathbb{Z}_{M}$, we have

(2.18) $A_{x^{\prime},y}\subset A_{x,y}\cup{\mathrm{Bispan}}(x,x^{\prime}),$

and in particular,

(2.19) $A_{x}\subset\bigcap_{a\in A}{\mathrm{Bispan}}(x,a).$

For example, if $x\in\mathbb{Z}_{M}\setminus A$ satisfies $(x-a,M)=M/p_{i}$
for some $a\in A$, then

$A_{x}\subset{\mathrm{Bispan}}(x,a)=\Pi(x,p_{i}^{n_{i}})\cup\Pi(a,p_{i}^{n_{i}}).$

In the sequel, whenever we evaluate saturating sets, we will always start with
geometric restrictions based on (2.19).

###  2.9. Fibers and cofibered structures

The following is a simplified version of the definitions and results of [23,
Section 8], restricted to $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$. Most of this
article can be read with just these definitions, if the reader is willing to
substitute $n_{i}=n_{j}=n_{k}=2$ in all arguments. On those few occasions when
the more general versions are necessary even in that case, we have to refer
the reader to [23, Section 8].

Let $N|M$, $c\in\mathbb{N}$, and $\nu\in\\{i,j,k\\}$. An $N$-fiber in the
$p_{\nu}$ direction with multiplicity $c$ is a set $F\subset\mathbb{Z}_{M}$
such that $F$ mod $N$ has the mask polynomial

$F(X)\equiv
cX^{a}(1+X^{N/p_{\nu}}+X^{2N/p_{\nu}}+\dots+X^{(p_{\nu}-1)N/p_{\nu}})\mod(X^{N}-1)$

for some $a\in\mathbb{Z}_{M}$. We will say sometimes that $F$ passes through
$a$, or is rooted at $a$. A set $A\subset\mathbb{Z}_{M}$ is $N$-fibered in the
$p_{\nu}$ direction if it can be written as a union of disjoint $N$-fibers in
the $p_{\nu}$ direction, all with the same multiplicity.

Fiber chains are convolutions of fibers in the same direction on different
scales. For $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$, the only fiber chains of interest
that are not fibers on some scale are multisets $\tilde{F}$ with mask
polynomials

$\tilde{F}(X)=cX^{a}(1+X^{M_{\nu}}+X^{2M_{\nu}}+\dots+X^{(p_{\nu}^{2}-1)M_{\nu}}),\
\ \nu\in\\{i,j,k\\},\ a\in\mathbb{Z}_{M},\ c\in\mathbb{N}.$

If $F\subset\mathbb{Z}_{M}$ is an $M$-fiber in the $p_{\nu}$ direction, we say
that an element $x\in\mathbb{Z}_{M}$ is at distance $m$ from $F$ if $m|M$ is
the maximal divisor such that $(z-x,M)=m$ for some $z\in F$. It is easy to see
that such $m$ exists.

Let $A\oplus B=\mathbb{Z}_{M}$ be a tiling. We will often be interested in
finding “complementary” fibers and fibered structures in $A$ and $B$, in the
following sense.

######  Definition 2.8 (Cofibers and cofibered structures).

Let $A,B\subset\mathbb{Z}_{M}$, with $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$, and let
$\nu\in\\{i,j,k\\}$.

(i) We say that $F\subset A,G\subset B$ are $(1,2)$-cofibers in the $p_{\nu}$
direction if $F$ is an $M$-fiber and $G$ is an $M/p_{\nu}$-fiber, both in the
$p_{\nu}$ direction.

(ii) We say that the pair $(A,B)$ has a (1,2)-cofibered structure in the
$p_{\nu}$ direction if

  * •

$B$ is $M/p_{\nu}$-fibered in the $p_{\nu}$ direction,

  * •

$A$ contains at least one “complementary” $M$-fiber $F\subset A$ in the
$p_{\nu}$ direction, which we will call a cofiber for this structure.

The advantage of cofibered structure is that it permits fiber shifts as
described below. In many cases, we will be able to use this to reduce the
given tiling to a simpler one.

######  Lemma 2.9 (Fiber-Shifting Lemma).

Let $A\oplus B=\mathbb{Z}_{M}$. Assume that the pair $(A,B)$ has a
$(1,2)$-cofibered structure, with a cofiber $F\subset A$. Let $A^{\prime}$ be
the set obtained from $A$ by shifting $F$ to a point $x\in\mathbb{Z}_{M}$ at a
distance $M/p_{i}^{2}$ from it. Then $A^{\prime}\oplus B=\mathbb{Z}_{M}$, and
$A$ is T2-equivalent to $A^{\prime}$.

In order to identify cofibered structures in $(A,B)$, we will use saturating
sets, via the following lemma.

######  Lemma 2.10.

Assume that $A\oplus B=\mathbb{Z}_{M}$ is a tiling, with
$M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$. Suppose that $x\in\mathbb{Z}_{M}\setminus A$,
$b\in B$, $M/p_{\nu}\in{\mathrm{Div}}(A)$, and $A_{x,b}\subset\ell_{\nu}(x)$
for some $\nu\in\\{i,j,k\\}$ and $b\in B$. Then

(2.20)
$\mathbb{A}^{M}_{M/p_{\nu}^{2}}[x]\mathbb{B}^{M}_{M/p_{\nu}^{2}}[b]=\phi(p_{\nu}^{2}).$

with the product saturated by a $(1,2)$-cofiber pair $(F,G)$ such that
$F\subset A$ is at distance $M/p_{i}^{2}$ from $x$ and $G\subset B$ is rooted
at $b$. In particular, if $A_{x}\subset\ell_{\nu}(x)$, then the pair $(A,B)$
has a $(1,2)$-cofibered structure.

##  3\. Classification results

###  3.1. Classification results

We are now ready to state our classification results and provide a more
detailed outline of the proof. Let $A\oplus B=\mathbb{Z}_{M}$, where
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$. By (1.3), we have $\Phi_{s}(X)\
|\ A(X)B(X)$ for all $s|M$ such that $s\neq 1$. In particular, $\Phi_{M}$
divides at least one of $A(X)$ and $B(X)$. Without loss of generality, we may
assume that $\Phi_{M}|A$.

We have $\Phi_{M}|A$ if and only if $\Phi_{M}$ divides $A\cap\Lambda(x,D(M))$
for every $x\in\mathbb{Z}_{M}$ (see Section 2.6). This implies structure
results for restrictions of $A$ to such grids. Let $\Lambda:=\Lambda(a,D(M))$
for some $a\in A$, so that $A\cap\Lambda$ is nonempty. It is easy to see that
$\Phi_{M}|F_{\nu}$ for each $\nu\in\\{i,j,k\\}$. By the classic results on
vanishing sums of roots of unity [2], [34], [35], [37], [31], [27], $\Phi_{M}$
divides $A\cap\Lambda$ if and only if

$(A\cap\Lambda)(X)=\sum_{\nu\in\\{i,j,k\\}}Q_{\nu}(X)F_{\nu}(X),$

where $Q_{i},Q_{j},Q_{k}$ are polynomials with integer coefficients depending
on both $A$ and $\Lambda$.

A particularly simple case occurs when $A(X)=Q_{\nu}(X)F_{\nu}(X)$ for a
single $\nu\in\\{i,j,k\\}$, so that $A$ is fibered on all $D(M)$ grids in the
same direction. However, much more complicated structures are also possible.
For instance, $A\cap\Lambda$ may be $M$-fibered in different directions on
different $D(M)$ grids $\Lambda$, or there may exist a $D(M)$ grid $\Lambda$
such that $A\cap\Lambda$ contains non-intersecting $M$-fibers in two or three
different directions. An additional issue is that the polynomials
$Q_{i},Q_{j},Q_{k}$ are not required to have nonnegative coefficients. In such
cases, there may be points $a\in A$ such that $a*F_{\nu}\not\subset A$ for any
$\nu$, due to cancellations between $M$-fibers in different directions.

Our classification results, and our proof of (T2), split into cases according
to the fibering properties of $A$. Theorems 3.1 and 3.2 summarize our main
findings in the unfibered and fibered case, respectively.

######  Theorem 3.1.

Let $A\oplus B=\mathbb{Z}_{M}$, where $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$ is odd.
Assume that $|A|=|B|=p_{i}p_{j}p_{k}$, $\Phi_{M}|A$, and there exists a $D(M)$
grid $\Lambda$ such that $A\cap\Lambda$ is nonempty and is not $M$-fibered in
any direction. Assume further, without loss of generality, that $0\in\Lambda$.
Then $A^{\flat}=\Lambda$, and the tiling $A\oplus B=\mathbb{Z}_{M}$ is
T2-equivalent to $\Lambda\oplus B=\mathbb{Z}_{M}$ via fiber shifts. By
Corollary 2.2, both $A$ and $B$ satisfy (T2).

######  Theorem 3.2.

Let $A\oplus B=\mathbb{Z}_{M}$, where $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$ is odd.
Assume that $|A|=|B|=p_{i}p_{j}p_{k}$, $\Phi_{M}|A$, and for every $a\in A$,
the set $A\cap\Lambda(a,D(M))$ is $M$-fibered in at least one direction
(possibly depending on $a$).

(I) Suppose that there exists an element $a_{0}\in A$ such that

(3.1) $a_{0}*F_{\nu}\subset A\ \ \forall\nu\in\\{i,j,k\\}.$

Then the tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to $\Lambda\oplus
B=\mathbb{Z}_{M}$ via fiber shifts, where $\Lambda:=\Lambda(a_{0},D(M))$. By
Corollary 2.2, both $A$ and $B$ satisfy (T2).

(II) Assume that (3.1) does not hold for any $a_{0}\in A$. Then at least one
of the following holds.

  * •

$A\subset\Pi(a,p_{\nu})$ for some $a\in A$ and $\nu\in\\{i,j,k\\}$. By Theorem
2.5, both $A$ and $B$ satisfy (T2).

  * •

There exists a $\nu\in\\{i,j,k\\}$ such that (possibly after interchanging $A$
and $B$) the conditions of Theorem 2.6 are satisfied in the $p_{\nu}$
direction. By Corollary 2.7, both $A$ and $B$ satisfy (T2).

A more detailed breakdown of the case (II) of Theorem 3.2 is provided in
Theorem 9.1.

###  3.2. Outline of the proof

In the rest of this section, we provide an outline of the proof of Theorems
3.1 and 3.2. We assume that $A\oplus B=\mathbb{Z}_{M}$, where
$M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$ is odd, $|A|=|B|=p_{i}p_{j}p_{k}$, and
$\Phi_{M}|A$. Some of our arguments apply to tilings with more general $M$. In
order to be able to sketch the main ideas without interruptions, we postpone
the discussion of such extensions to the end of the section.

We begin with general arguments that are needed in both fibered and unfibered
cases. In Section 4, we develop technical tools we will use throughout the
article. Lemma 4.1 is from [23]; several of the other results in that section
are specific to the 3-prime setting.

Assume first that $\Phi_{M}|A$ and that there exists a $D(M)$ grid $\Lambda$
such that $A\cap\Lambda$ is not $M$-fibered in any direction. In Sections 5
and 6, we prove that $A\cap\Lambda$ must then contain at least one of two
special structures, either diagonal boxes (Proposition 5.2) or an extended
corner (Proposition 5.5). Large parts of the argument are combinatorial and
apply to all $A\subset\mathbb{Z}_{M}$ such that $\Phi_{M}|A$; however, to get
the full strength of our results, we need to use saturating set techniques,
hence the tiling assumption is necessary.

Some of our technical tools work only when all the “top differences” are
divisors of $A$, i.e.,

(3.2) $\\{m:\ D(M)|m|M\\}\subset{\mathrm{Div}}(A).$

We therefore must pay special attention to the cases where (3.2) fails. A
classification of such structures is provided in Section 6. This analysis is
also needed in the fibered case (Theorem 3.2) when fibering, or lack thereof,
on lower scales must be considered.

We resolve diagonal boxes and extended corner structures in Sections 7 and 8,
respectively. In Theorem 7.1, we prove that if $A\cap\Lambda$ contains
diagonal boxes, then $A$ is T2-equivalent to either $\Lambda$ (in which case
we are done) or to another tile $A^{\prime}$ containing an extended corner. We
then prove in Theorem 8.1 that if $A\cap\Lambda$ contains an extended corner
structure, then $A$ is T2-equivalent to $\Lambda$. Theorem 3.1 follows by
combining Theorem 7.1 and Theorem 8.1.

The main idea of that part of the proof is that all such tilings can be
obtained via fiber shifts (Lemma 2.9) from the standard tiling

(3.3) $A^{\flat}\oplus B^{\flat}=\mathbb{Z}_{M},$

where $A^{\flat}=\Lambda(0,D(M))$, and $B^{\flat}$ is the standard tiling
complement with $\Phi_{p_{i}}\Phi_{p_{j}}\Phi_{p_{k}}|B^{\flat}$. Similar
tilings were constructed by Szabó ([42]; see also [24]). We prove that all
tilings satisfying the assumptions of Theorems 3.1 must in fact have this
form. Starting with an unfibered grid in the given tiling, and using
saturating set methods, we are able to locate the shifted fibers and shift
them back into place, returning to (3.3). This proves (T2) and provides full
information about the structure of the tiling.

In Section 9, we consider the fibered case. In the simple case when all of $A$
is $M$-fibered in the same direction, we can apply Corollary 2.7 and be done;
however, it is possible for $A$ to be fibered in different directions on
different $D(M)$-grids. The proof breaks down into cases, according to how the
fibers in different directions interact.

Suppose first that there exists an element $a_{0}\in A$ such that (3.1) holds.
This case turns out to be similar to that of unfibered grids and is resolved
by similar methods, ending in T2-equivalence to (3.3).

Assume now that no such element exists. Our main intermediate result in this
case is that, in fact, only two fibering directions are allowed (see Theorem
9.1 for more details). This breaks down further into cases according to
fibering properties and cyclotomic divisibility, with each case terminating in
either the subgroup reduction (Theorem 2.5) or slab reduction (Theorem 2.6 and
Corollary 2.7).

While our final result is restricted to the case when
$M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$ is odd, many of our methods and intermediate
results apply under weaker assumptions. Whenever a significant part of the
argument can be run in a more general case with little or no additional
effort, we do so, assuming that $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$
for more general $n_{i},n_{j},n_{k}\geq 2$ and $p_{i},p_{j},p_{k}\geq 2$. For
example, the classification of unfibered grids in Sections 5 and 6 allows all
$n_{\nu}$ to be arbitrary and $M$ to be either odd or even. The resolution of
the $p_{i}$ extended corner case in Section 8 works for both odd and even $M$,
with $n_{i}\geq 2$ arbitrary and with only a few additional lines needed to
accommodate the even case. On the other hand, the arguments in Section 9 are
limited to the odd $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$ case from the beginning.

Although we expect that the main conclusions should be the same in the even
case, many of our technical tools work differently when one of the primes is
equal to 2. We would like to draw the reader’s attention to the basic fibering
argument in Lemma 4.9. This argument does not work when $p_{i}=2$, and indeed,
in Section 6 we provide examples of unfibered grids in the even case where the
fibering conclusions of the lemma fail. In particular, the unfibered
structures in Lemma 6.5 do not have a counterpart in the odd case.
Additionally, with fewer geometric restrictions coming from (2.19), saturating
set arguments can be more difficult to run.

The constraint $n_{i}=n_{j}=n_{k}=2$ is often needed in arguments based on
divisor exclusion. For example, while (2.19) provides geometric restrictions
on saturating sets, we often need additional constraints based on availability
of divisors, and with $n_{i}=n_{j}=n_{k}=2$ there are fewer divisors available
to begin with. In the fibered case, several of our proofs terminate in an
essentially 2-dimensional (therefore easier) problem after we have run out of
scales in one direction. In order to allow arbitrary $n_{i},n_{j},n_{k}$
throughout the argument, we expect that a systematic way to induct on scales
may be necessary.

##  4\. Toolbox

###  4.1. Divisors

The first lemma is Lemma 8.9 of [23], specialized to the 3-prime case.

######  Lemma 4.1 (Enhanced divisor exclusion).

Let $A\oplus B=\mathbb{Z}_{M}$, with
$M=\prod_{\iota\in\\{i,j,k\\}}p_{\iota}^{n_{\iota}}$. Let
$m=\prod_{\iota\in\\{i,j,k\\}}p_{\iota}^{\alpha_{\iota}}$ and
$m^{\prime}=\prod_{\iota\in\\{i,j,k\\}}p_{\iota}^{\alpha^{\prime}_{\iota}}$,
with $0\leq\alpha_{\iota},\alpha^{\prime}_{\iota}\leq n_{\iota}$. Assume that
at least one of $m,m^{\prime}$ is different from $M$, and that for every
$\iota\in\\{i,j,k\\}$ we have

(4.1) $\hbox{ either }\alpha_{\iota}\neq\alpha^{\prime}_{\iota}\hbox{ or
}\alpha_{\iota}=\alpha^{\prime}_{\iota}=n_{\iota}.$

Then for all $x,y\in\mathbb{Z}_{M}$ we have

$\mathbb{A}^{M}_{m}[x]\,\mathbb{A}^{M}_{m^{\prime}}[x]\,\mathbb{B}^{M}_{m}[y]\,\mathbb{B}^{M}_{m^{\prime}}[y]=0.$

In other words, there are no configurations $(a,a^{\prime},b,b^{\prime})\in
A\times A\times B\times B$ such that

(4.2) $(a-x,M)=(b-y,M)=m,\ \ (a^{\prime}-x,M)=(b^{\prime}-y,M)=m^{\prime}.$

###### Proof.

If we did have a configuration as in (4.2), then, under the assumption (4.1)
for all $\iota$, we would have

$(a-a^{\prime},M)=(b-b^{\prime},M)=\prod_{\iota\in\\{i,j,k\\}}p_{\iota}^{\min(\alpha_{\iota},\alpha^{\prime}_{\iota})},$

with the right side different from $M$. But that is prohibited by divisor
exclusion. ∎

###  4.2. Cyclotomic divisibility

######  Lemma 4.2.

Let $A\in\mathcal{M}(\mathbb{Z}_{M})$, and let $m,s|M$ with $s\neq 1$. Suppose
that for every $a\in A$, $\Phi_{s}$ divides $A\cap\Lambda(a,m)$. Then
$\Phi_{s}|A$.

###### Proof.

Write $\mathbb{Z}_{M}=\bigcup_{\nu}\Lambda_{\nu}$, where $\Lambda_{\nu}$ are
pairwise disjoint $m$-grids. Accordingly, $A(X)=\sum_{\nu}A_{\nu}(X)$, where
$A_{\nu}=A\cap\Lambda_{\nu}$. If $A$ is disjoint from $\Lambda_{\nu}$, we have
$A_{\nu}(X)\equiv 0$. If on the other hand $A\cap\Lambda_{\nu}\neq\emptyset$,
then $\Phi_{s}|A_{\nu}$. Summing up in $\nu$, we get $\Phi_{s}|A$. ∎

The next two lemmas are based on a combinatorial interpretation of
divisibility by prime power cyclotomics. For $A\subset\mathbb{Z}_{M}$ and
$1\leq\alpha\leq n_{i}$, we have $\Phi_{p_{i}^{\alpha}}(X)|A(X)$ if and only
if

(4.3)
$|A\cap\Pi(x,p_{i}^{\alpha})|=\frac{1}{p_{i}}|A\cap\Pi(x,p_{i}^{\alpha-1})|\ \
\forall x\in\mathbb{Z}_{M},$

so that the elements of $A$ are uniformly distributed mod $p_{i}^{\alpha}$
within each residue class mod $p_{i}^{\alpha-1}$. This in particular limits
the number of elements that a tiling set may have in a plane on some scale.

######  Lemma 4.3 (Plane bound).

Let $A\oplus B=\mathbb{Z}_{M}$, where
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$ and
$|A|=p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}p_{k}^{\beta_{k}}$. Then for every
$x\in\mathbb{Z}_{M}$ and $0\leq\alpha_{i}\leq n_{i}$ we have

$|A\cap\Pi(x,p_{i}^{n_{i}-\alpha_{i}})|\leq
p_{i}^{\alpha_{i}}p_{j}^{\beta_{j}}p_{k}^{\beta_{k}}.$

######  Corollary 4.4.

Let $A\oplus B=\mathbb{Z}_{M}$, where
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$ and
$|A|=p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}p_{k}^{\beta_{k}}$ with $\beta_{i}>0$.
Suppose that for some $x\in\mathbb{Z}_{M}$ and $1\leq\alpha_{0}\leq n_{i}$

(4.4)
$|A\cap\Pi(x,p_{i}^{n_{i}-\alpha_{0}})|>p_{i}^{\beta_{i}-1}p_{j}^{\beta_{j}}p_{k}^{\beta_{k}},$

then $\Phi_{p_{i}^{n_{i}-\alpha}}|A$ for at least one
$\alpha\in\\{0,\ldots,\alpha_{0}-1\\}$.

###### Proof.

Suppose that $\Phi_{p_{i}^{n_{i}-\alpha}}\nmid A$ for all
$\alpha\in\\{0,\ldots,\alpha_{0}-1\\}$. It follows that there must exist a
$\gamma\geq\alpha_{0}$ such that $\Phi_{p_{i}^{n_{i}-\gamma}}|A$. The latter
implies, by (4.4), that $|A|>\prod_{\nu}p_{\nu}^{\beta_{\nu}}$, which is a
contradiction. ∎

###  4.3. Saturating sets

######  Lemma 4.5 (No missing joints).

Let $A\oplus B=\mathbb{Z}_{M}$, where
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$. Suppose that

(4.5) $\\{D(M)|m|M\\}\cap{\mathrm{Div}}(B)=\emptyset,$

and that for some $x\in\mathbb{Z}_{M}$ there exist $a_{i},a_{j},a_{k}\in A$
such that

(4.6) $(x-a_{i},M)=M/p_{i},\ (x-a_{j},M)=M/p_{j},\ (x-a_{k},M)=M/p_{k}.$

Then $x\in A$.

###### Proof.

Suppose that $x\not\in A$, and let $\Delta$ be the $M$-cuboid with vertices
$x,a_{i},a_{j},a_{k}$. By (4.6), the saturating set $A_{x}$ is contained in
the vertex set of $\Delta$. But that is impossible by (4.5). ∎

######  Lemma 4.6 (Flat corner).

Let $A\oplus
B=\mathbb{Z}_{M},M=p_{i}^{2}p_{j}^{n_{j}}p_{k}^{n_{k}},|A|=p_{i}p_{j}p_{k}$.
Suppose that (4.5) holds, and that $A$ contains the following $3$-point
configuration: for some $x\in\mathbb{Z}_{M}\setminus A$ there exist
$a,a_{j},a_{k}\in A$ such that

$(a-a_{j},M)=(a_{k}-x,M)=M/p_{j},\,(a-a_{k},M)=(a_{j}-x,M)=M/p_{k}$

Then $A_{x}\subset\ell_{i}(x)$, and the pair $(A,B)$ has a $(1,2)$-cofibered
structure, with an $M$-cofiber in $A$ at distance $M/p_{i}^{2}$ from $x$.

###### Proof.

Fix $b\in B$. Then

$A_{x,b}\subset\ell_{i}(x)\cup\ell_{i}(a)\cup\ell_{i}(a_{j})\cup\ell_{i}(a_{k}).$

By (4.5), we have the following.

  * •

$A_{x,b}\cap\ell_{i}(x)\neq\emptyset$ implies that
$\mathbb{A}_{M/p_{i}^{2}}[x]\mathbb{B}_{M/p_{i}^{2}}[b]>0$, hence
$M/p_{i}^{2}p_{j}$, $M/p_{i}^{2}p_{k}$,
$M/p_{i}^{2}p_{j}p_{k}\in{\mathrm{Div}}(A)$ and
$M/p_{i}^{2}\in{\mathrm{Div}}(B)$.

  * •

$A_{x,b}\cap\ell_{i}(a)\neq\emptyset$ implies that
$\mathbb{A}_{M/p_{i}^{2}p_{j}p_{k}}[x|\ell_{i}(a)]\mathbb{B}_{M/p_{i}^{2}p_{j}p_{k}}[b]>0$,
hence $M/p_{i}^{2}$, $M/p_{i}^{2}p_{j}$,
$M/p_{i}^{2}p_{k}\in{\mathrm{Div}}(A)$ and
$M/p_{i}^{2}p_{j}p_{k}\in{\mathrm{Div}}(B)$.

  * •

$A_{x,b}\cap\ell_{i}(a_{j})\neq\emptyset$ implies that
$\mathbb{A}_{M/p_{i}^{2}p_{k}}[x|\ell_{i}(a_{j})]\mathbb{B}_{M/p_{i}^{2}p_{k}}[b]>0$,
hence $M/p_{i}^{2}$, $M/p_{i}^{2}p_{j}$,
$M/p_{i}^{2}p_{j}p_{k}\in{\mathrm{Div}}(A)$ and
$M/p_{i}^{2}p_{k}\in{\mathrm{Div}}(B)$.

  * •

$A_{x,b}\cap\ell_{i}(a_{k})\neq\emptyset$ implies that
$\mathbb{A}_{M/p_{i}^{2}p_{j}}[x|\ell_{i}(a_{k})]\mathbb{B}_{M/p_{i}^{2}p_{j}}[b]>0$,
hence $M/p_{i}^{2}$, $M/p_{i}^{2}p_{k}$,
$M/p_{i}^{2}p_{j}p_{k}\in{\mathrm{Div}}(A)$ and
$M/p_{i}^{2}p_{j}\in{\mathrm{Div}}(B)$.

It follows that $A_{x,b}$ cannot intersect more than one of the above lines.
We now show it cannot intersect any line other than $\ell_{i}(x)$. Indeed,
assume that $A_{x,b}\subset\ell_{i}(a)$, then

$\displaystyle\phi(p_{i}^{2}p_{j}p_{k})$
$\displaystyle=\mathbb{A}_{M/p_{i}^{2}p_{j}p_{k}}[x|\ell_{i}(a)]\mathbb{B}_{M/p_{i}^{2}p_{j}p_{k}}[b]$
$\displaystyle=\mathbb{A}_{M/p_{i}^{2}}[a]\sum_{y:(y-b,M)=M/p_{j}p_{k}}\mathbb{B}_{M/p_{i}^{2}}[y].$

Observe that $\mathbb{A}_{M/p_{i}^{2}}[a]\leq\phi(p_{i}^{2})$, and since
$M/p_{i},M/p_{i}^{2}\notin{\mathrm{Div}}(B)$, we have
$\mathbb{B}_{M/p_{i}^{2}}[y]\leq 1$ for all $y\in\mathbb{Z}_{M}\setminus B$
with $(y-b,M)=M/p_{j}p_{k}$. It follows that both must hold with equality.
Now, if $p_{i}>p_{j}$ then $\mathbb{A}_{M/p_{i}^{2}}[a]=\phi(p_{i}^{2})$
implies that

$\displaystyle|A\cap\Pi(a,p_{k}^{n_{k}})|$
$\displaystyle\geq\mathbb{A}[a]+\mathbb{A}_{M/p_{i}^{2}}[a]$
$\displaystyle=1+\phi(p_{i}^{2})>p_{i}p_{j}$

which contradicts Lemma 4.3. If on the other hand $p_{i}>p_{j}$, then
$\mathbb{B}_{M/p_{i}^{2}}[y]=1$ for all $y$ as above implies
$\\{D(M)|m|M\\}\cap{\mathrm{Div}}(B)\neq\emptyset$, which contradicts (4.5).
Hence $A_{x,b}\cap\ell_{i}(a)=\emptyset$.

Next, assume that $A_{x,b}\subset\ell_{i}(a_{k})$, then

$\displaystyle\phi(p_{j}p_{i}^{2})$
$\displaystyle=\mathbb{A}_{M/p_{i}^{2}p_{j}}[x|\ell_{i}(a_{k})]\mathbb{B}_{M/p_{i}^{2}p_{j}}[b]$
$\displaystyle=\mathbb{A}_{M/p_{i}^{2}}[a_{k}]\sum_{y:(y-b,M)=M/p_{j}}\mathbb{B}_{M/p_{i}^{2}}[y]$

By the same argument as above, we deduce that
$\mathbb{A}_{M/p_{i}^{2}}[a_{k}]=\phi(p_{i}^{2})$ and
$\mathbb{B}_{M/p_{i}^{2}}[y]=1$ for all $y\in\mathbb{Z}_{M}\setminus B$ with
$(y-b,M)=M/p_{j}$. When $p_{i}>p_{j}$, we therefore get
$|A\cap\Pi(a_{k},p_{k}^{n_{k}})|>p_{i}p_{j}$, which is a contradiction. When
$p_{i}<p_{j}$, we must have
$\\{M/p_{j},M/p_{i}p_{j}\\}\cap{\mathrm{Div}}(B)\neq\emptyset$, contradicting
(4.5). Thus $A_{x,b}\cap\ell_{i}(a_{k})=\emptyset$. By symmetry,
$A_{x,b}\cap\ell_{i}(a_{j})=\emptyset$. ∎

###  4.4. Fibering lemmas

Lemma 4.7 below is a simple version of the de Bruijn-Rédei-Schoenberg theorem
for cyclic groups $\mathbb{Z}_{N}$, where $N$ has at most two distinct prime
factors. This was essentially proved in [2]; see also [27, Theorem 3.3].

######  Lemma 4.7.

(Cyclotomic divisibility for 2 prime factors) Let
$A\in\mathcal{M}(\mathbb{Z}_{N})$ for some $N\mid M$ such that $p_{i}\nmid N$.
Then:

(i) $\Phi_{N}|A$ if and only if $A$ is a linear combination of $N$-fibers in
the $p_{j}$ and $p_{k}$ direction with non-negative integer coefficients.

(ii) Assume that $\Phi_{N}|A$ and that (6.12) holds for some
$c_{0}\in\mathbb{N}$. Then for every $D(N)$-grid $\Lambda$, $A\cap\Lambda$ is
$N$-fibered in either the $p_{j}$ or the $p_{k}$ direction.

Lemma 4.8 is a localized version of the above.

######  Lemma 4.8.

(Flat cuboids) Let $A\subset\mathbb{Z}_{M}$. Assume that $\Phi_{M}|A$, and
that there is a plane $\Pi:=\Pi(z,p_{i}^{n_{i}})$ such that $A\cap\Pi$ is a
disjoint union of $M$-fibers in the $p_{j}$ and $p_{k}$ directions. Then for
every parallel plane $\Pi^{\prime}:=\Pi(z^{\prime},p_{i}^{n_{i}})$, where
$(z-z^{\prime},M)=M/p_{i}$, we have that $A\cap\Pi^{\prime}$ is a disjoint
union of $M$-fibers in the $p_{j}$ and $p_{k}$ directions.

###### Proof.

Consider a 2-dimensional cuboid $\Delta^{\prime}$ with vertices
$x^{\prime},x^{\prime}+d_{j},x^{\prime}+d_{k},x^{\prime}+d_{j}+d_{k}$, where
$x^{\prime}\in\Pi$, $(d_{j},M)=M/p_{j}$, $(d_{k},M)=M/p_{k}$. Let $x\in\Pi$ be
the point such that $(x-x^{\prime},M)=M/p_{i}$, and let $\Delta$ be the
2-dimensional cuboid with vertices $x,x+d_{j},x+d_{k},x+d_{j}+d_{k}$. By the
fibering property of $A\cap\Pi$, we have $\mathbb{A}_{M}[\Delta]=0$. Since
$\Phi_{M}|A$, we also have $\mathbb{A}_{M}[\Delta-\Delta^{\prime}]=0$, hence
$\mathbb{A}_{M}[\Delta^{\prime}]=0$. If we consider $A\cap\Pi^{\prime}$ (after
translation) as a subset of $\mathbb{Z}_{M/p_{i}^{n_{i}}}$, it follows that
$\Phi_{M/p_{i}^{n_{i}}}\mid(A\cap\Pi^{\prime})(X)$. By Lemma 4.7,
$A\cap\Pi^{\prime}$ is a union of fibers as claimed. ∎

######  Lemma 4.9.

(Missing top difference implies fibering) Let $N|M$ with $p_{i}p_{j}p_{k}|N$.
Assume that $A\in\mathcal{M}(\mathbb{Z}_{N})$ satisfies $\Phi_{N}|A$, and that
(6.12) holds for some positive integer $c_{0}$.

(i) Suppose that $p_{i}\neq 2$, and that there exists a grid
$\Lambda=\Lambda(x_{0},D(N))$ for some $x_{0}\in\mathbb{Z}_{N}$ such that

(4.7) $N/p_{i}\not\in{\mathrm{Div}}_{N}(A\cap\Lambda).$

Then $A\cap\Lambda$ is $N$-fibered in one of the $p_{j}$ and $p_{k}$
directions. In particular, if $N/p_{i}\not\in{\mathrm{Div}}_{N}(A)$, then
$A\cap\Lambda$ is $N$-fibered in one of the $p_{j}$ and $p_{k}$ directions for
every $D(N)$-grid $\Lambda$.

(ii) Suppose that $p_{i},p_{j}\neq 2$, and that
$N/p_{i},N/p_{j}\not\in{\mathrm{Div}}_{N}(A)$. Then $A$ is $N$-fibered in the
$p_{k}$ direction.

###### Proof.

Let $\Lambda:=\Lambda(x_{0},D(N))$ be a grid where (4.7) holds. We first prove
that each $a\in A\cap\Lambda$ belongs to an $N$-fiber in the $p_{\nu}$
direction for at least one $\nu\in\\{j,k\\}$. Suppose, for contradiction, that
there exists an $a\in A\cap\Lambda$ that does not have this property. Then
there are $x_{i},x_{j}\in\mathbb{Z}_{N}$ such that

$\mathbb{A}^{N}[x_{j}]=\mathbb{A}^{N}[x_{k}]=0\,\text{ and
}(x_{j}-a,N)=N/p_{j},\,(x_{k}-a,N)=N/p_{k}.$

Since $p_{i}>2$ and $\mathbb{A}^{N}_{N/p_{i}}[a]=0$, there exist at least two
distinct elements $x_{i},x_{i}^{\prime}\in\mathbb{Z}_{N}$ satisfying

$\mathbb{A}^{N}[x_{i}]=\mathbb{A}^{N}[x_{i}^{\prime}]=0,\
(a-x_{i},N)=(a-x_{i}^{\prime},N)=N/p_{i}.$

Consider two $N$-cuboids, each with vertices at $a,x_{i},x_{j}$, and with
another vertex at $x_{k}$ and $x_{k}^{\prime}$ respectively. By the cyclotomic
divisibility assumption, each of those cuboids must be balanced. This can only
happen if there are two elements $a_{ijk},a_{ijk}^{\prime}\in\mathbb{Z}_{N}$
at the opposite vertex of each cuboid from $a$ (that is, with
$(a-a_{ijk},N)=(a-a_{ijk}^{\prime},N)=D(N)$) satisfying
$\mathbb{A}^{N}[a_{ijk}]=\mathbb{A}^{N}[a_{ijk}^{\prime}]=c_{0}$. However,
this leads to a contradiction, since $(a_{ijk}-a^{\prime}_{ijk},N)=N/p_{i}$.
We therefore conclude that each $a\in A\cap\Lambda$ belongs to an $N$-fiber in
at least one direction as indicated.

Next, suppose that $a_{j},a_{k}\in A\cap\Lambda$ belong to $N$-fibers in,
respectively, the $p_{j}$ and $p_{k}$ direction. If $a_{j}*F_{j}$ and
$a_{k}*F_{k}$ do not intersect, then
$N/p_{i}\in{\mathrm{Div}}_{N}(a_{j}*F_{j},a_{k}*F_{k})$, contradicting the
assumption (4.7). We may therefore assume that $a_{j}=a_{k}=a$ and that
$a*F_{j},a*F_{k}\subset A$.

Consider any $N$-cuboid with one vertex at ${a}$. Then the vertices at
distance $N/p_{j}$ and $N/p_{k}$ from $a$ belong to $A$, and, by (4.7), the
vertices at distance $N/p_{i}p_{j}$ and $N/p_{i}p_{k}$ from $a$ cannot be in
$A$. The only way to balance the cuboid is for the vertex at distance
$N/p_{j}p_{k}$ from $a$ to be in $A$. Allowing such cuboids to vary, we see
that $A\cap\Pi\cap\Lambda$ must be $N$-fibered in both of the $p_{j}$ and
$p_{k}$ directions. This proves part (i) of the lemma.

Suppose now that $N/p_{i},N/p_{j}\not\in{\mathrm{Div}}_{N}(A)$. By part (i),
$A$ must be $N$-fibered in one of the $p_{j}$ and $p_{k}$ directions on each
$D(N)$-grid. However, it cannot be $N$-fibered in the $p_{j}$ direction on any
grid, since $N/p_{j}\not\in{\mathrm{Div}}(A)$. Part (ii) follows. ∎

##  5\. Structure on unfibered grids

###  5.1. Diagonal boxes

Throughout this section, we will use the following notation. Let
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$, and let $\Lambda$ be a fixed
$D(M)$ grid such that $A\cap\Lambda\neq\emptyset$. We identify $\Lambda$ with
$\mathbb{Z}_{p_{i}}\oplus\mathbb{Z}_{p_{j}}\oplus\mathbb{Z}_{p_{k}}$, and
represent each point $x\in\Lambda$ as
$(\lambda_{i}x,\lambda_{j}x,\lambda_{k}x)$ in the implied coordinate system.

######  Definition 5.1.

Let $A\subset\mathbb{Z}_{M}$. We say that $A\cap\Lambda$ contains diagonal
boxes if there are nonempty sets $I\subset\mathbb{Z}_{p_{i}}$,
$J\subset\mathbb{Z}_{p_{j}}$, $K\subset\mathbb{Z}_{p_{k}}$, such that

$I^{c}:=\mathbb{Z}_{p_{i}}\setminus I,\ J^{c}:=\mathbb{Z}_{p_{j}}\setminus J,\
K^{c}:=\mathbb{Z}_{p_{k}}\setminus K$

are also nonempty, and

(5.1) $(I\times J\times K)\cup(I^{c}\times J^{c}\times K^{c})\subset
A\cap\Lambda.$

Figure 2. A pair of diagonal boxes.

######  Proposition 5.2.

Let $A\oplus B=\mathbb{Z}_{M}$ be a tiling,
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$, and assume that $\Phi_{M}\mid A$.
Suppose that there is a $D(M)$-grid $\Lambda$ such that $A\cap\Lambda$ is not
a union of disjoint $M$-fibers (possibly in different directions). Then
$A\cap\Lambda$ contains diagonal boxes.

###### Proof.

Write $D=D(M)$ for short. We first construct a set $A_{0}\subset A\cap\Lambda$
as follows. If $A\cap\Lambda$ contains no $M$-fibers, let
$A_{0}:=A\cap\Lambda$. If $A\cap\Lambda$ does contain an $M$-fiber $F$,
consider the set $(A\cap\Lambda)\setminus F$ (if there is more than one such
fiber, just choose one arbitrarily and remove it). If this set contains no
$M$-fibers, we let $A_{0}$ be that set; otherwise continue by induction. The
procedure terminates when no more $M$-fibers can be found. The remaining set
$A_{0}$ is nonempty by our assumption on $A$, contains no $M$-fibers, and
$\Phi_{M}|A_{0}$; however, $A_{0}$ need not be a tiling complement.

We remark that $A_{0}$ is not necessarily uniquely determined by $A$, as the
fiber removal procedure may lead to different outcomes depending on the order
in which fibers are removed. In that event, we fix one such set $A_{0}$ and
keep it fixed throughout the proof.

For future reference, we record a lemma.

######  Lemma 5.3.

Suppose that $x\in A\setminus A_{0}$. Then there exists a $\nu\in\\{i,j,k\\}$
such that $x*F_{\nu}\subset(A\setminus A_{0})$.

###### Proof.

This follows directly from the construction, since any point $x\in A\setminus
A_{0}$ would have been removed from $A$ together with an $M$-fiber (in some
direction) containing $x$. ∎

For each $x\in\Lambda$, we define

$\displaystyle I(x)$ $\displaystyle=\\{l\in\mathbb{Z}_{p_{i}}:\ \
(l,\lambda_{j}x,\lambda_{k}x)\in A_{0}\\},$ $\displaystyle
I^{c}(x)=\mathbb{Z}_{p_{i}}\setminus I(x),$ $\displaystyle J(x)$
$\displaystyle=\\{l\in\mathbb{Z}_{p_{j}}:\ \ (\lambda_{i}x,l,\lambda_{k}x)\in
A_{0}\\},$ $\displaystyle J^{c}(x)=\mathbb{Z}_{p_{j}}\setminus J(x),$
$\displaystyle K(x)$ $\displaystyle=\\{l\in\mathbb{Z}_{p_{k}}:\ \
(\lambda_{i}x,\lambda_{j}x,l)\in A_{0}\\},$ $\displaystyle
K^{c}(x)=\mathbb{Z}_{p_{k}}\setminus K(x).$

Let $a\in A_{0}$ be an element such that $|K(a)|$ is maximal, in the sense
that

(5.2) $|K(a^{\prime})|\leq|K(a)|\ \ \forall a^{\prime}\in A_{0}.$

By translational invariance, we may assume that $a=(0,0,0)$. Observe that
$I(a),J(a),K(a)$ are all nonempty since $a\in A_{0}$, and
$I^{c}(a),J^{c}(a),K^{c}(a)$ are all nonempty since $a$ does not belong to an
$M$-fiber in $A_{0}$ in any direction.

Claim 1. For all $a^{\prime}=(0,0,l)\in A_{0}$, with $l\in K(a)$, we have
$I(a^{\prime})\times J(a^{\prime})\times\\{l\\}\subset A_{0}$.

###### Proof.

Since $K(a^{\prime})=K(a)$, it suffices to prove this with $a^{\prime}=a$. Let
$a_{i}=(l_{i},0,0)$ and $a_{j}=(0,l_{j},0)$ for some $l_{i}\in I(a)$,
$l_{j}\in J(a)$, and let $z=(l_{i},l_{j},0)\in I(a)\times J(a)\times\\{0\\}$.
We need to prove that $z_{0}\in A_{0}$. Suppose, for contradiction, that
$z\not\in A_{0}$, and consider the cuboid with vertices at $a,a_{i},a_{j}$,
and $u_{l_{k}}=(0,0,l_{k})$. For $l_{k}\in K(a)$, the cuboid can only be
balanced if both $(l_{i},0,l_{k})$ and $(0,l_{j},l_{k})$ are elements of
$A_{0}$. Furthermore, let $l_{k}\in K^{c}(a)$, then for the cuboid to be
balanced we still need at least one of $(l_{i},0,l_{k})$ and $(0,l_{j},l_{k})$
to be in $A_{0}$. This implies that $\max(|K(a_{i})|,|K(a_{j})|)>|K(a)|$,
contradicting the maximality assumption (5.2). Hence Claim 1 follows. ∎

Claim 2. For all $a^{\prime}=(0,0,l)\in A_{0}$ with $l\in K(a)$, we have
$I^{c}(a^{\prime})\times J^{c}(a^{\prime})\times K^{c}(a)\subset A_{0}$.

###### Proof.

Again, it suffices to prove this with $a^{\prime}=a$. Let $x\in
I^{c}(a^{\prime})\times J^{c}(a^{\prime})\times K^{c}(a)$. Considering the
$M$-cuboid with opposite vertices at $a$ and $x$, we see that it can only be
balanced if $x\in A_{0}$ and

(5.3)
$(\lambda_{i}x,\lambda_{j}x,0),(\lambda_{i}x,0,\lambda_{k}x),(0,\lambda_{j}x,\lambda_{k}x)\not\in
A_{0},$

as claimed. ∎

Claim 3. Suppose that for all $l^{\prime},l^{\prime\prime}\in K(a)$, we have
either $(I(a^{\prime})\times J(a^{\prime}))\subseteq(I(a^{\prime\prime})\times
J(a^{\prime\prime}))$ or $(I(a^{\prime\prime})\times
J(a^{\prime\prime}))\subseteq(I(a^{\prime})\times J(a^{\prime}))$, where
$a^{\prime}=(0,0,l^{\prime})$, $a^{\prime\prime}=(0,0,l^{\prime\prime})$. Then
$A_{0}$ contains diagonal boxes.

###### Proof.

Under the assumptions of the claim, the sets $I(a^{\prime})\times
J(a^{\prime})$ with $a^{\prime}\in A_{0}$ and $M/p_{k}\mid a-a^{\prime}$ have
a minimal element. Without loss of generality, we may assume that

$(I(a)\times J(a))\subseteq(I(a^{\prime})\times J(a^{\prime}))\ \ \forall
a^{\prime}\in A_{0}\hbox{ such that }M/p_{k}\mid a-a^{\prime}.$

Then Claims 1 and 2 imply that $A_{0}$ (and therefore $A$) contains the
diagonal boxes

$(I(a)\times J(a)\times K(a))\cup(I^{c}(a)\times J^{c}(a)\times K^{c}(a)),$

which proves the claim. ∎

It remains to consider the case when there exist
$l^{\prime},l^{\prime\prime}\in K(a)$ such that for
$a^{\prime}=(0,0,l^{\prime})$, $a^{\prime\prime}=(0,0,l^{\prime\prime})$, we
have

(5.4) $I(a^{\prime})\not\subset I(a^{\prime\prime})\hbox{ and
}J(a^{\prime\prime})\not\subset J(a^{\prime}).$

For this to be possible, $K(a)$ must have at least two distinct elements, and
each of $I(a^{\prime})$ and $J(a^{\prime\prime})$ must have at least one
element different from $0$. Since none of $a,a^{\prime},a^{\prime\prime}$
belongs to an $M$-fiber in $A_{0}$, we must have

(5.5) $p_{i},p_{j},p_{k}>2.$

Furthermore, this configuration implies that

(5.6) $\\{m:\ D|m|M\\}\subset{\mathrm{Div}}(A_{0})\subset{\mathrm{Div}}(A).$

Indeed, we have

$\\{m:\
D|m|M\\}\setminus\\{M/p_{i}p_{j}\\}\subset{\mathrm{Div}}((I(a^{\prime})\times
J(a^{\prime})\times\\{l^{\prime}\\})\cup(I(a^{\prime\prime})\times
J(a^{\prime\prime})\times\\{l^{\prime\prime}\\})),$
$M/p_{i}p_{j}\in{\mathrm{Div}}(I^{c}(a^{\prime})\times
J^{c}(a^{\prime})\times\\{l_{k}\\},I^{c}(a^{\prime\prime})\times
J^{c}(a^{\prime\prime})\times\\{l_{k}\\})\hbox{ for any }l_{k}\in K^{c}(a).$

Claim 4. Suppose that (5.4) holds for some $l^{\prime},l^{\prime\prime}\in
K(a)$, with $a^{\prime}=(0,0,l^{\prime})$,
$a^{\prime\prime}=(0,0,l^{\prime\prime})$. Then

$I^{c}(a^{\prime\prime})\times J^{c}(a^{\prime})\times K^{c}(a)\subset A_{0}.$

###### Proof.

We may assume that $a^{\prime\prime}=a$ and $l=0$. We will also write
$l_{k}=l^{\prime}$, so that $a^{\prime}=(0,0,l_{k})$ Let $l_{i}\in
I(a^{\prime})\setminus I(a),l_{j}\in J(a)\setminus J(a^{\prime}),l_{i}^{c}\in
I^{c}(a^{\prime})$ and $l_{j}^{c}\in J^{c}(a)$. We also fix $l^{c}_{k}\in
K^{c}(a)$. For the purpose of this proof, we will need to consider points with
coordinates $(\beta_{i},\beta_{j},\beta_{k})$ such that

$\beta_{\nu}\in\\{0,l_{\nu},l_{\nu}^{c}\\}\text{ for }\nu\in\\{i,j,k\\}.$

Let $x_{i}:=(l_{i},0,0)$, $x_{j}:=(0,l_{j},l_{k})$, and
$a_{i}:=(l_{i},0,l_{k}),a_{j}:=(0,l_{j},0)$. Then $x_{i},x_{j}\not\in A_{0}$,
and $a_{i},a_{j}\in A_{0}$. By Claim 2, we have

(5.7) $\begin{split}z_{i}&:=(l_{i}^{c},l_{j},l_{k}^{c})\in A_{0}\hbox{ since
}l_{j}\in J^{c}(a^{\prime}),\\\ z_{j}&:=(l_{i},l_{j}^{c},l_{k}^{c})\in
A_{0}\hbox{ since }l_{i}\in I^{c}(a).\\\ \end{split}$

Let $z=(l_{i},l_{j},l_{k}^{c})\in I^{c}(a)\times J^{c}(a^{\prime})\times
K^{c}(a)$ be the point such that

(5.8) $(z-z_{i},M)=M/p_{i},\ (z-z_{j},M)=M/p_{j}.$

We need to prove that $z\in A_{0}$.

We begin with the following reduction.

Claim 4’. Let $x_{ij}=(l_{i},l_{j},0)$ and $y_{ij}=(l_{i},l_{j},l_{k})$.
Suppose that at least one of the following holds: either

(5.9) $\\{x_{ij},y_{ij}\\}\cap A\neq\emptyset,$

or else there exists a set $A^{\prime}\subset\mathbb{Z}_{M}$, identical to $A$
except possibly along the line $\ell_{j}(x_{i})$, such that $A^{\prime}\oplus
B=\mathbb{Z}_{M}$ and

(5.10) $x_{i}*F_{j}\subset A^{\prime},\text{ hence }x_{ij}\in A^{\prime}.$

Then $z\in A_{0}$.

###### Proof.

For notational consistency, if (5.9) holds, we let $A^{\prime}=A$. We have

(5.11) $(x_{ij}-z,M)=(y_{ij}-z,M)=M/p_{k}.$

Suppose that at least one of $x_{ij}$, $y_{ij}$ belongs to $A^{\prime}$. By
(5.8) and (5.6), the assumptions of Lemma 4.5 hold for $z$ and $A^{\prime}$.
Hence $z\in A^{\prime}$. Since $A^{\prime}$ and $A$ may differ only along
$\ell_{j}(x_{i})$, we must in fact have $z\in A$.

Next, we claim that $z\in A_{0}$. Indeed, assume for contradiction that $z\in
A\setminus A_{0}$. By Lemma 5.3, there is a $\nu\in\\{i,j,k\\}$ such that
$z*F_{\nu}\subset(A\setminus A_{0})$. But by (5.7) and (5.8),
$z_{i}\in(z*F_{i})\cap A_{0}$ and $z_{j}\in(z*F_{j})\cap A_{0}$. It follows
that $\nu=k$, and $z*F_{k}\subset(A\setminus A_{0})$. By (5.11), we have
$x_{ij},y_{ij}\in z*F_{k}$, and in particular $x_{ij},y_{ij}\in A$.

Consider now the point $x_{i}$. We have

(5.12) $(x_{i}-a,M)=M/p_{i},\ \ (x_{i}-a_{i},M)=M/p_{k},\ \
(x_{i}-x_{ij},M)=M/p_{j},$

with $a,a^{\prime}\in A_{0},x_{ij}\in A$. Taking also (5.6) into account, we
see that $x_{i}$ satisfies the assumptions of Lemma 4.5 applied to $A$.
Therefore $x_{i}\in A$.

Since $x_{i}\not\in A_{0}$, we must have $x_{i}\in A\setminus A_{0}$. By Lemma
5.3 and (5.12), it follows that $x_{i}*F_{j}\subset A\setminus A_{0}$.
However, consider the point $x_{ij}\in x_{i}*F_{j}$. We have already seen that
$z$ must have been removed from $A$ together with the fiber $z*F_{k}$, which
also contains $x_{ij}$. This is a contradiction, since we are not allowed to
remove the same point twice. ∎

To prove Claim 4, it remains to prove that at least one of the conditions
(5.9), (5.10) in Claim 4’ must hold. Assume first that $x_{i}\in A\setminus
A_{0}$. By Lemma 5.3, $A\setminus A_{0}$ must contain an $M$-fiber
$x_{i}*F_{\nu}$ for some $\nu\in\\{i,j,k\\}$. Since $a\in(x_{i}*F_{i})\cap A$
and $a_{i}\in(x_{i}*F_{k})\cap A$, we must have $\nu=j$. This clearly implies
$x_{ij}\in A$. The same argument applies with $i$ and $j$ interchanged.

We are left with the case when $x_{i},x_{j},x_{ij},y_{ij}\notin A$. Assume
also for contradiction that $z\notin A_{0}$. Consider the $M$-cuboid with
vertices at $x_{ij},z,z_{i}$ and $z_{j}$, and note that $z_{i},z_{j}\in
A_{0}$. In order to balance this cuboid in $A_{0}$, at least one of the points
$y_{i}:=(l_{i}^{c},l_{j},0)$ and $y_{j}:=(l_{i},l_{j}^{c},0)$ must be in
$A_{0}$. However, if $y_{j}\in A_{0}$, then we can apply Lemma 4.5 again, this
time to $A$ and $x_{i}$, using (5.6), (5.12), and $(x_{i}-y_{j},M)=M/p_{j}$.
Hence $x_{i}\in A$. We assumed that $x_{i}\not\in A_{0}$, hence by Lemma 5.3
there must be a $\nu\in\\{i,j,k\\}$ such that $x_{i}*F_{\nu}\subset A\setminus
A_{0}$. However, the points $a,a_{i},y_{j}\in A_{0}$ block all three
directions, contradicting the assumption that $y_{j}\in A_{0}$.

We therefore have $y_{i}\in A_{0}$. Now, considering the saturating set
$A_{x_{i}}$, we have

$A_{x_{i}}\subset\bigcap_{\hat{a}\in\\{a,a_{i},z_{j},y_{i}\\}}{\mathrm{Bispan}}(x_{i},\hat{a})=\ell_{j}(x_{i})\cup\\{a,a_{j}\\}.$

By (5.6), we must in fact have $A_{x_{i}}\subset\ell_{j}(x_{i})$. It follows
that the pair $(A,B)$ has a $(\mathcal{P}_{A},\mathcal{P}_{B})$-cofibered
structure of depth $\gamma$ in the $p_{j}$ direction for some $\gamma\geq 2$,
with $1\in\mathcal{P}_{A}$ and an $M$-fiber in the $p_{j}$ direction in $A$ at
distance $M/p_{j}^{\gamma}$ from $x_{i}$. Applying Lemma 2.9 to shift that
fiber to $x_{i}$, we obtain a set $A^{\prime}$ with the desired properties,
thus concluding the proof of Claim 4. ∎

We can now complete the proof of the proposition. We claim that

(5.13) $(I\times J\times K)\cup(I^{c}\times J^{c}\times K^{c})\subset A_{0},$

where

$I=\bigcap_{l\in K(a)}I(a_{l}),\ \ J=\bigcap_{l\in K(a)}J(a_{l}),\ \ K=K(a),$
$I^{c}=\bigcup_{l\in K(a)}I^{c}(a_{l}),\ \ J^{c}=\bigcup_{l\in
K(a)}J^{c}(a_{l}),\ \ K^{c}=K^{c}(a),$

and $a_{l}=(0,0,l)$. It is clear that $I\times J\times K\subset A_{0}$.
Suppose now that $(l_{i},l_{j},l_{k})\in I^{c}\times J^{c}\times K^{c}$. Then
there are $l,l^{\prime}\in K(a)$ such that $l_{i}\in I^{c}(a_{l})$ and
$l_{j}\in J^{c}(a_{l^{\prime}})$. If $l=l^{\prime}$, or if $l_{i}\in
I^{c}(a_{l^{\prime}})$, then $(l_{i},l_{j},l_{k})\in A_{0}$ by Claim 2 applied
to $a_{l^{\prime}}$. The case $l_{j}\in J^{c}(a_{l})$ is similar. Assume
therefore that $l_{i}\in I(a_{l^{\prime}})\setminus I(a_{l})$ and $l_{j}\in
J(a_{l})\setminus J(a_{l^{\prime}})$. But then (5.4) holds with
$a^{\prime},a^{\prime\prime}$ replaced by $a_{l},a_{l^{\prime}}$. Applying
Claim 4, we see that $(l_{i},l_{j},l_{k})\in A_{0}$ in that case as well. This
ends the proof of the proposition. ∎

###  5.2. Extended corners

We continue to write $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$. Assume that
$A\oplus B=\mathbb{Z}_{M}$ is a tiling, $\Phi_{M}\mid A$, and there exists a
$D(M)$-grid $\Lambda$ such that $A\cap\Lambda\neq\emptyset$ and $A\cap\Lambda$
is not $M$-fibered in any direction. By Proposition 5.2, if $A\cap\Lambda$ is
not a union of disjoint $M$-fibers, then $A\cap\Lambda$ contains diagonal
boxes. It remains to consider the case when $A\cap\Lambda$ is a union of
disjoint $M$-fibers. In that case, we claim that $A\cap\Lambda$ contains the
following structure.

######  Definition 5.4.

Suppose that $A\subset\mathbb{Z}_{M}$, and let $\Lambda$ be a $D(M)$ grid.

(i) We say that $A\cap\Lambda$ contains a $p_{i}$ corner if there exist
$a,a_{i}\in A\cap\Lambda$ with $(a-a_{i},M)=M/p_{i}$ satisfying

(5.14) $A\cap(a*F_{j}*F_{k})=a*F_{j},\ \
A\cap(a_{i}*F_{j}*F_{k})=a_{i}*F_{k}.$

(ii) We say that $A\cap\Lambda$ contains a $p_{i}$ extended corner if there
exist $a,a_{i}\in A\cap\Lambda$ such that $(a-a_{i},M)=M/p_{i}$ and

  * •

$A\cap(a*F_{j}*F_{k})$ is $M$-fibered in the $p_{j}$ direction but not in the
$p_{k}$ direction,

  * •

$A\cap(a_{i}*F_{j}*F_{k})$ is $M$-fibered in the $p_{k}$ direction but not in
the $p_{j}$ direction.

We now prove that unfibered grids as described above must contain extended
corners.

######  Proposition 5.5.

Let $D=D(M)$, and let $\Lambda$ be a $D$-grid. Assume that $A\cap\Lambda$ is a
union of disjoint $M$-fibers, but is not fibered in any direction. Then
$A\cap\Lambda$ contains a $p_{\nu}$ extended corner for some
$\nu\in\\{i,j,k\\}$.

###### Proof.

Fix $A$ and $\Lambda$ as in the statement of the proposition. We will say that
$\kappa:A\cap\Lambda\to\\{i,j,k\\}$ is an assignment function if
$A\cap\Lambda$ can be written as

$A\cap\Lambda=\bigcup_{a\in A\cap\Lambda}(a*F_{\kappa(a)}),$

where for any $a,a^{\prime}\in A\cap\Lambda$, the fibers $a*F_{\kappa(a)}$ and
$a^{\prime}*F_{\kappa(a^{\prime})}$ are either identical or disjoint. Thus, if
$a^{\prime}\in a*F_{\kappa(a)}$, then $\kappa(a^{\prime})=\kappa(a)$.

We recall [23, Proposition 7.10]:

######  Proposition 5.6.

Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$, and let $D=D(M)$. Assume that
$A\oplus B=\mathbb{Z}_{M}$ is a tiling and that there exists a $D$-grid
$\Lambda$ such that $A\cap\Lambda$ is a nonempty union of disjoint $M$-fibers.
Then there is a subset $\\{\nu_{1},\nu_{2}\\}\subset\\{i,j,k\\}$ of
cardinality 2 such that $A\cap\Lambda$ is a union of disjoint $M$-fibers in
the $p_{\nu_{1}}$ and $p_{\nu_{2}}$ directions.

Hence there exists an assignment function $\kappa$ that takes at most two
distinct values. (In fact, the proof of [23, Proposition 7.10] shows that this
is true for any assignment function.) Without loss of generality, we may
assume that $\kappa(a)\in\\{j,k\\}$ for all $a\in A\cap\Lambda$. We claim that
this implies that $A\cap\Lambda$ contains a $p_{i}$ extended corner.

Split $\Lambda$ into 2-dimensional grids
$\Pi_{\iota}:={x_{\iota}}*F_{j}*F_{k}$, $\iota=0,1,\dots,p_{i}-1$. Then for
each $\iota$, the set $A\cap\Pi_{\iota}$ is a union of disjoint fibers in at
least one of the $p_{j}$ or $p_{k}$ direction. Moreover, we are assuming that
$A\cap\Lambda$ is not fibered in either the $p_{j}$ or $p_{k}$ direction.
Therefore for each $\nu\in\\{j,k\\}$, there must be at least one $\iota(\nu)$
such that $A\cap\Pi_{\iota(\nu)}$ is fibered only in the $p_{\nu}$ direction.
Choosing $a\in A\cap\Pi_{\iota(j)}$ and $a_{i}\in A\cap\Pi_{\iota(k)}$ with
$(a-a_{i})=M/p_{i}$, we see that the condition (ii) of Definition 5.4 is
satisfied. ∎

A $p_{i}$ corner is a special case of a $p_{i}$ extended corner, with only one
fiber in each of the planes through $a$ and $a_{i}$. This is one of the
special structures that occur when $\Phi_{M}\mid A$, $A\cap\Lambda$ is not
$M$-fibered in any direction, and $\\{D(M)|m|M\\}\not\subset{\mathrm{Div}}(A)$
(see Section 6).

In addition to the present purpose of classification of unfibered grids in on
scale $M$, we will also refer to Definition 5.4 (ii) in the fibered case, in
the proofs of Proposition 9.14 (Claim 1) and Lemma 9.30.

##  6\. Unfibered grids with missing top differences

Let $A\subset\mathbb{Z}_{M}$, and let $\Lambda$ be a $D(M)$-grid such that
$A\cap\Lambda\neq\emptyset$. The purpose of this section is to classify all
possible unfibered grids $A\cap\Lambda$ under the assumption that $\Phi_{M}|A$
and that ${\mathrm{Div}}(A)$ does not contain all $m$ such that $D(M)|m|M$. We
do not assume in this section that $A$ is a tiling complement.

###  6.1. A structure result

######  Proposition 6.1.

Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$. Assume that
$A\subset\mathbb{Z}_{M}$ satisfies $\Phi_{M}|A$. Let $\Lambda$ be a
$D(M)$-grid such that $A\cap\Lambda\neq\emptyset$. Suppose that $A\cap\Lambda$
is not $M$-fibered in any direction, and that (5.6) fails, i.e.

(6.1) $\\{m:\ D(M)|m|M\\}\not\subset{\mathrm{Div}}(A\cap\Lambda).$

Then $A\cap\Lambda$ is a union of at most one set of diagonal boxes

(6.2) $A_{1}=(I_{1}\times J_{1}\times K_{1})\cup(I_{1}^{c}\times
J_{1}^{c}\times K_{1}^{c}),$

where $I_{1}\subset\mathbb{Z}_{p_{i}}$, $J_{1}\subset\mathbb{Z}_{p_{j}}$,
$K_{1}\subset\mathbb{Z}_{p_{k}}$ are non-empty sets such that
$I_{1}^{c}:=\mathbb{Z}_{p_{i}}\setminus I_{1},\
J_{1}^{c}:=\mathbb{Z}_{p_{j}}\setminus J_{1},\
K_{1}^{c}:=\mathbb{Z}_{p_{k}}\setminus K_{1}$ are also non-empty, and possibly
additional $M$-fibers in one or more directions, disjoint from $A_{1}$ and
from each other. Furthermore, if $A\cap\Lambda$ does contain a set of diagonal
boxes (6.2), then at least one of the sets $I_{1},J_{1},K_{1}$ has cardinality
1, and at least one of the sets $I^{c}_{1},J^{c}_{1},K^{c}_{1}$ has
cardinality 1.

######  Remark 6.1.

For simplicity, we only state and prove Proposition 6.1 for sets
$A\subset\mathbb{Z}_{M}$. However, if we assume instead that
$A\in\mathcal{M}(\mathbb{Z}_{N})$ for some $N|M$, and that (6.12) holds (i.e.,
$A\cap\Lambda$ is a multiset of constant multiplicity $c_{0}$), the same
argument applies except that the diagonal boxes and fibers in the conclusion
also have multiplicity $c_{0}$.

###### Proof.

We begin as in the proof of Proposition 5.2. Let $A_{0}\subset A\cap\Lambda$
be a set constructed by removing $M$-fibers from $A\cap\Lambda$ until none are
left, so that $A\cap\Lambda$ is the union of $A_{0}$ and some number of
$M$-fibers in one or more directions, disjoint from $A_{0}$ and from each
other. If $A_{0}=\emptyset$, we are done. Otherwise, we proceed with Claims 1,
2, and 3 from the proof of Proposition 5.2, noting that this part does not
require the use of saturating sets (hence $A$ need not be a tile). At that
point, the only remaining case in the proof of Proposition 5.2 is when there
exist $l^{\prime},l^{\prime\prime}\in K(a)$ such that (5.4) holds. However, in
that case we have (5.6), which contradicts (6.1). Therefore, under the
assumptions of Proposition 6.1, $A_{0}$ contains diagonal boxes $A_{1}$ as in
(6.2). Moreover, the cardinality statement must hold, since otherwise we would
not have (6.1).

As in Proposition 5.2, $A_{0}$ need not be unique and may depend on the order
in which the fibers are removed, and $A_{1}$ may then depend on the choice of
$A_{0}$. We fix one such choice of $A_{0}$ and $A_{1}$, and keep it fixed for
the remainder of the proof.

We claim that $A_{0}=A_{1}$. To prove this, assume for contradiction that
$A_{0}\setminus A_{1}$ is nonempty. We clearly have $\Phi_{M}|A_{0}$ and
$\Phi_{M}|A_{1}$, therefore $\Phi_{M}|A_{0}(X)-A_{1}(X)$. Since the set
$A_{0}\setminus A_{1}$ is nonempty and contains no fibers, it must contain
another set of diagonal boxes

(6.3) $A_{2}=(I_{2}\times J_{2}\times K_{2})\cup(I_{2}^{c}\times
J_{2}^{c}\times K_{2}^{c}),$

with obvious notation. Furthermore, since $A_{2}\subset A_{0}\setminus A_{1}$,
we must have

(6.4) $A_{1}\cap A_{2}=\emptyset.$

We first claim that at least one of

$I_{1}=I_{2},\ I_{1}=I_{2}^{c},\ J_{1}=J_{2},\ J_{1}=J_{2}^{c},\ K_{1}=K_{2},\
K_{1}=K_{2}^{c},$

must hold. Indeed, by (6.4), we must have

$(I_{1}\times J_{1}\times K_{1})\cap(I_{2}\times J_{2}\times
K_{2})=\emptyset,$

so that at least one of $I_{1}\cap I_{2}$, $J_{1}\cap J_{2}$, $K_{1}\cap
K_{2}$ is empty. Without loss of generality, we may assume that $K_{1}\cap
K_{2}=\emptyset$, so that

(6.5) $K_{1}\subset K_{2}^{c},\ \ K_{2}\subset K_{1}^{c}.$

We also have

$(I_{1}\times J_{1}\times K_{1})\cap(I_{2}^{c}\times J_{2}^{c}\times
K_{2}^{c})=\emptyset,$

and since $K_{1}\cap K_{2}^{c}\neq\emptyset$, one of $I_{1}\cap I_{2}^{c}$ and
$J_{1}\cap J_{2}^{c}$ is empty. Without loss of generality, we may assume that
$J_{1}\cap J^{c}_{2}=\emptyset$, so that

(6.6) $J_{1}\subset J_{2},\ \ J_{2}^{c}\subset J_{1}^{c}.$

Next,

$(I_{1}^{c}\times J_{1}^{c}\times K_{1}^{c})\cap(I_{2}^{c}\times
J_{2}^{c}\times K_{2}^{c})=\emptyset.$

By (6.6), we have $J^{c}_{1}\cap J_{2}^{c}\neq\emptyset$, therefore one of
$I_{1}^{c}\cap I_{2}^{c}$ and $K^{c}_{1}\cap K_{2}^{c}$ is empty. If
$K^{c}_{1}\cap K_{2}^{c}=\emptyset$, then $K_{2}^{c}\subset K_{1}$, which
together with (6.5) shows that $K_{1}=K_{2}^{c}$ and proves the claim. If
$I_{1}^{c}\cap I_{2}^{c}=\emptyset$, we get that

(6.7) $I_{1}^{c}\subset I_{2},\ \ I_{2}^{c}\subset I_{1}.$

Using that

$(I_{1}^{c}\times J_{1}^{c}\times K_{1}^{c})\cap(I_{2}\times J_{2}\times
K_{2})=\emptyset.$

and taking (6.5) and (6.7) into account, we see that $J_{1}^{c}\cap
J_{2}=\emptyset$. But then $J_{2}\subset J_{1}$, which together with (6.6)
proves the claim.

We may assume without loss of generality that

(6.8) $K_{1}=K_{2}^{c}.$

This implies that

(6.9) $(I_{1}\times J_{1})\cap(I_{2}\times J_{2})=\emptyset\hbox{ and
}(I_{1}^{c}\times J_{1}^{c})\cap(I_{2}^{c}\times J_{2}^{c})=\emptyset,$

since otherwise we would have an $M$-fiber in the $p_{k}$ direction in
$A_{0}$. But also, considering the box layers with third coordinate $l\in
K_{1}$ and $l^{\prime}\in K_{2}$, we have

(6.10) $(I_{1}\times J_{1})\cap(I_{2}^{c}\times J_{2}^{c})=\emptyset\hbox{ and
}(I_{1}^{c}\times J_{1}^{c})\cap(I_{2}\times J_{2})=\emptyset,$

The first parts of (6.9) and (6.10) imply that either

(6.11) $I_{1}\cap I_{2}=J_{1}\cap J_{2}^{c}=\emptyset,$

or else the same holds with $I,J$ interchanged. Assume that (6.11) holds. Then
$I_{2}\subset I_{1}^{c}$ and $J_{2}^{c}\subset J_{1}^{c}$, so that in order
for the second parts of (6.9) and (6.10) to hold, we must have

$I_{1}^{c}\cap I_{2}^{c}=J_{1}^{c}\cap J_{2}=\emptyset.$

But this implies that $I_{1}^{c}\subset I_{2}$ and $J_{2}^{c}\subset
J_{1}^{c}$. Hence $I_{1}=I_{2}^{c}$ and $J_{1}=J_{2}$. But then $A_{0}$
contains $M$-fibers in the $p_{i}$ direction, a contradiction. ∎

###  6.2. Special unfibered structures: odd $M$

######  Lemma 6.2.

Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$ and
$N=M/p_{i}^{\alpha_{i}}p_{j}^{\alpha_{j}}p_{k}^{\alpha_{k}}$ with
$\alpha_{\iota}<n_{\iota}$ for all $\iota\in\\{i,j,k\\}$. Assume that $2\nmid
M$, and that $A\in\mathcal{M}(\mathbb{Z}_{N})$ satisfies $\Phi_{N}|A$. Let
$\Lambda$ be a $D(N)$-grid such that $A\cap\Lambda\neq\emptyset$. Assume
further that

  * •

there exists a $c_{0}\in\mathbb{N}$ such that

(6.12) $\mathbb{A}^{N}_{N}[x]\in\\{0,c_{0}\\}\,\text{ for all }x\in\Lambda,$

  * •

$A\cap\Lambda$ is not $M$-fibered in any direction.

Then there is a permutation of $\\{i,j,k\\}$ such that

(6.13)
$\\{D(N)|m|N\\}\setminus{\mathrm{Div}}(A\cap\Lambda)\subseteq\\{N/p_{i}p_{j},N/p_{i}p_{k}\\}.$

Moreover, (6.13) holds with equality if and only if there exists
$x\in\mathbb{Z}_{M}\setminus A$ such that

$\mathbb{A}^{N}_{N/p_{i}}[x]=c_{0}\phi(p_{i}),\,\mathbb{A}^{N}_{N/p_{j}p_{k}}[x]=c_{0}\phi(p_{j}p_{k}),$
$\mathbb{A}^{N}_{m}[x]=0\hbox{ for all
}m\in\\{D(N)|m|N\\}\setminus\\{N/p_{i},N/p_{j}p_{k}\\}.$

We will refer to this structure as a $p_{i}$-full plane.

Figure 3. Full plane structure on a $D(M)$ grid.

###### Proof.

We may assume that $M=N$ and $c_{0}=1$. We will also write $D=D(M)$ for short.
By Proposition 6.1 and the assumption that $A\cap\Lambda$ is not fibered,
$A\cap\Lambda$ must contain at least one of the following:

(a) two non-overlapping $M$-fibers in different directions, say $p_{i}$ and
$p_{j}$,

(b) diagonal boxes as in Definition 5.1, and possibly additional $M$-fibers in
one or more directions, disjoint from the diagonal boxes and from each other.

In the first case, we have
$\\{D|m|M\\}\setminus{\mathrm{Div}}(A\cap\Lambda)\subseteq\\{M/p_{i}p_{j}\\}$,
just based on these two fibers. It remains to consider the second case.
Suppose that $A\cap\Lambda$ contains diagonal boxes

$(I\times J\times K)\cup(I^{c}\times J^{c}\times K^{c})$

with $I,J,K,I^{c},J^{c},K^{c}$ all nonempty. In order for
$\\{D|m|M\\}\setminus{\mathrm{Div}}(A\cap\Lambda)$ to be nonempty, we must
have

(6.14) $\min(|I|,|J|,|K|)=\min(|I^{c}|,|J^{c}|,|K^{c}|)=1.$

We may assume without loss of generality that $|I^{c}|=|J|=1$. Since
$p_{\iota}\geq 3$ for all $\iota$, it follows that $|I|,|J^{c}|$, and at least
one of $|K|,|K^{c}|$ are greater than 1. Assume that $|K^{c}|>1$. Then

(6.15) $\begin{split}M/p_{i}&\in{\mathrm{Div}}(I\times J\times K),\\\
M/p_{j},M/p_{k},M/p_{j}p_{k}&\in{\mathrm{Div}}(I^{c}\times J^{c}\times
K^{c}),\\\ M/p_{i}p_{j}p_{k}&\in{\mathrm{Div}}(I\times J\times K,I^{c}\times
J^{c}\times K^{c}).\end{split}$

This implies (6.13). Furthermore, if (6.13) holds with equality, then $|K|=1$,
since otherwise we would also have $M/p_{i}p_{k}\in{\mathrm{Div}}(I\times
J\times K)$. This proves the second conclusion of the lemma, with $x$ equal to
the unique element of $I^{c}\times J\times K$. Note that if we add an
$M$-fiber in any direction to this structure, then equality in (6.13) can no
longer hold. ∎

######  Lemma 6.3.

Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$ and
$N=M/p_{i}^{\alpha_{i}}p_{j}^{\alpha_{j}}p_{k}^{\alpha_{k}}$ with
$\alpha_{\iota}<n_{\iota}$ for all $\iota\in\\{i,j,k\\}$. Assume that $2\nmid
M$, and that $A\in\mathcal{M}(\mathbb{Z}_{N})$ satisfies $\Phi_{N}|A$. Let
$\Lambda$ be a $D(N)$-grid such that $A\cap\Lambda\neq\emptyset$. Assume
further that

  * •

(6.12) holds for all $x\in\Lambda$,

  * •

$A\cap\Lambda$ is not fibered in any direction,

and that

(6.16) $\\{D|m|N\\}\setminus Div(A\cap\Lambda)=\\{N/p_{i}p_{j}\\}.$

Then $A\cap\Lambda$ has one of the following, mutually exclusive, possible
structures:

  1. (_i_)

($p_{k}$-corner, cf. Definition 5.4 (i).) For each $x\in\mathbb{Z}_{N}$, the
set $A\cap\Lambda\cap\Pi(x,p_{k}^{n_{k}-\alpha_{k}})$ is either empty or
consists of a single $N$-fiber in one of the $p_{i}$ or $p_{j}$ directions.
Since $A\cap\Lambda$ is not fibered, there has to be at least one of each.

Figure 4. Corner structure on a $D(M)$ grid.

  2. (_ii_)

($p_{k}$-almost corner) There exist
$x_{0},x_{1},\ldots,x_{\phi(p_{k})}\in\mathbb{Z}_{N}$ with
$(x_{l}-x_{l^{\prime}},N)=N/p_{k}$ for $l\neq l^{\prime}$, and two disjoint
sets $\mathcal{L}_{i},\mathcal{L}_{j}\subset\mathbb{Z}_{p_{k}}$ satisfying
$|\mathcal{L}_{i}|,|\mathcal{L}_{j}|>1$ and
$\mathcal{L}_{i}\cup\mathcal{L}_{j}=\\{0,1,\ldots,\phi(p_{k})\\}$, such that
for all $z\in\Lambda$ we have

$\mathbb{A}^{N}_{N}[z]=\begin{cases}c_{0}&\hbox{ if }(z-x_{l},N)=N/p_{i}\hbox{
for some }l\in\mathcal{L}_{i}\\\ &\ \hbox{ or }(z-x_{l},N)=N/p_{j}\hbox{ for
some }l\in\mathcal{L}_{j}\\\ 0&\hbox{ otherwise. }\end{cases}$

In particular, $\mathbb{A}^{N}_{N}[x_{l}]=0$ and
$\mathbb{A}^{N}_{N/p_{i}}[x_{l}]=c\phi(p_{i})$ for all $l\in\mathcal{L}_{i}$,
and similarly with $i$ and $j$ interchanged.

Figure 5. An almost corner structure on a $D(M)$ grid.

###### Proof.

We may assume that $M=N$ and $c_{0}=1$, and proceed as in the proof of Lemma
6.2. Suppose first that (a) holds (i.e., $A\cap\Lambda$ contains two non-
overlapping $M$-fibers in two different directions), but $A$ does not contain
diagonal boxes. Then the $M$-fibers must be in the $p_{i}$ and $p_{j}$
directions, or else (6.16) would be violated. Moreover, having any two such
fibers in the same plane $\Pi(x,p_{k}^{n_{k}-\alpha_{k}})$ would also violate
(6.16), hence (i) holds in this case.

In case (b), $A\cap\Lambda$ contains diagonal boxes. We proceed as in the
proof of Lemma 6.2 to get (6.14). Since
$M/p_{i}p_{j}\not\in{\mathrm{Div}}(A)$, we must have

$\min(|I|,|J|)=\min(|I^{c}|,|J^{c}|)=1.$

Therefore we may again assume without loss of generality that $|I^{c}|=|J|=1$
and that $|K^{c}|>1$, and get (6.15). We now consider two cases.

  * •

If $|K|=1$, the diagonal boxes are as in the conclusion of Lemma 6.2, and
instead of (6.16) we have (6.13) with equality.

  * •

If $|K|>1$, the diagonal boxes present the structure described in (ii), with
$\mathcal{L}_{i}=K$ and $\mathcal{L}_{j}=K^{c}$.

The only way we can add an $M$-fiber to either of these structures without
adding $M/p_{i}p_{j}$ to ${\mathrm{Div}}(A)$ is to add an $M$-fiber in the
$p_{k}$ direction rooted at $x$ (in the first case, with $x$ defined as in
Lemma 6.2), or at $x_{0}$ (in the second case, with $x_{0}$ specified in (ii))
That, however, puts us in the case (i) of the lemma. ∎

###  6.3. Special unfibered structures: even $M$

######  Lemma 6.4.

Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$ with $2|M$, and
$N=M/p_{i}^{\alpha_{i}}p_{j}^{\alpha_{j}}p_{k}^{\alpha_{k}}$ with
$\alpha_{\iota}<n_{\iota}$ for all $\iota\in\\{i,j,k\\}$. Assume that
$A\in\mathcal{M}(\mathbb{Z}_{N})$ satisfies $\Phi_{N}|A$. Let $\Lambda$ be a
$D(N)$-grid such that $A\cap\Lambda\neq\emptyset$. Assume further that

  * •

(6.12) holds for all $x\in\Lambda$,

  * •

$A\cap\Lambda$ is not fibered in any direction.

Assume that $N/p_{\iota}\in{\mathrm{Div}}(A\cap\Lambda)$ for all
$\iota\in\\{i,j,k\\}$, but
$\\{D(N)|m|N\\}\not\subset{\mathrm{Div}}(A\cap\Lambda)$. Then for some
permutation of $\\{i,j,k\\}$ we have

$\\{D(N)|m|N\\}\setminus{\mathrm{Div}}(A\cap\Lambda)=\\{N/p_{i}p_{j}\\},$

and $A\cap\Lambda$ has the $p_{k}$ corner structure in the sense of Definition
5.4 (i): for each $x\in\Lambda$, the set
$A\cap\Lambda\cap\Pi(x,p_{k}^{n_{k}-\alpha_{k}})$ is either empty or consists
of a single $N$-fiber in one of the $p_{i}$ or $p_{j}$ directions. Since
$A\cap\Lambda$ is not fibered, there has to be at least one of each.

###### Proof.

We may assume that $M=N$ and $c_{0}=1$. Assume without loss of generality that
$p_{k}=2$. In this case, in order for
$N/p_{k}\in{\mathrm{Div}}(A\cap\Lambda)$, we must have at least one $N$-fiber
$F$ in the $p_{k}$ direction in $A\cap\Lambda$. Let $A_{1}$ be the set
obtained from $A\cap\Lambda$ by removing all $N$-fibers in the $p_{k}$
direction. Since $A\cap\Lambda$ is not fibered, $A_{1}$ is nonempty and
satisfies (6.1). By Proposition 6.1, it must contain either at least one fiber
in another direction or a set of diagonal boxes, each disjoint from $F$.

  * •

Suppose that $A_{1}$ contains diagonal boxes as in Definition 5.1. Without
loss of generality, we may assume that $F$ is rooted at a point $a\in
I^{c}\times J\times K$. Then

$\displaystyle M/p_{i},M/p_{i}p_{k}$ $\displaystyle\in{\mathrm{Div}}(F,I\times
J\times K),$ $\displaystyle M/p_{j},M/p_{j}p_{k}$
$\displaystyle\in{\mathrm{Div}}(F,I^{c}\times J^{c}\times K^{c}),$
$\displaystyle M/p_{i}p_{j}p_{k}$ $\displaystyle\in{\mathrm{Div}}(I\times
J\times K,I^{c}\times J^{c}\times K^{c}).$

It follows that the only missing divisor can be $M/p_{i}p_{j}$. In order to
avoid that divisor within each box, we must have

$\min(|I|,|J|)=\min(|I^{c}|,|J^{c}|)=1.$

Taking into account the differences $(a^{\prime}-a^{\prime\prime},M)$, where
$a^{\prime}\in F$ and $a^{\prime\prime}$ belongs to one of the boxes, we see
that the only possible case is $|I^{c}|=|J|=1$. Then $A$ contains the
$M$-fiber in the $p_{i}$ direction rooted at the unique point $x\in
I^{c}\times J\times K$, and the $M$-fiber in the $p_{j}$ direction rooted at
the unique point $x^{\prime}\in I^{c}\times J\times K^{c}$. Any other points
in $A\cap\Lambda$ would add $M/p_{i}p_{j}$ to ${\mathrm{Div}}(A\cap\Lambda)$.
Thus the conclusion of the lemma holds.

  * •

If $A_{1}$ contains no diagonal boxes, then it must contain a fiber in at
least one other direction. This case is identical to the corresponding case of
Lemma 6.3 (i), for some permutation of $\\{i,j,k\\}$.

∎

######  Lemma 6.5.

Let $M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$ with $2|M$, and
$N=M/p_{i}^{\alpha_{i}}p_{j}^{\alpha_{j}}p_{k}^{\alpha_{k}}$ with
$\alpha_{\iota}<n_{\iota}$ for all $\iota\in\\{i,j,k\\}$. Assume that
$A\in\mathcal{M}(\mathbb{Z}_{N})$ satisfies $\Phi_{N}|A$. Let $\Lambda$ be a
$D(M)$-grid such that $A\cap\Lambda\neq\emptyset$. Assume further that
$p_{k}=2$, and that

  * •

(6.12) holds for all $x\in\Lambda$,

  * •

$A\cap\Lambda$ is not fibered in any direction,

  * •

$\\{N/p_{i},N/p_{j},N/p_{k}\\}\not\subset{\mathrm{Div}}(A\cap\Lambda)$.

Then

(6.17) $N/p_{k}\not\in{\mathrm{Div}}(A\cap\Lambda),$

and there is a pair of diagonal boxes

$A_{0}=(I\times J\times K)\cup(I^{c}\times J^{c}\times K^{c})\subset\Lambda,$

as in Definition 5.1, such that for all $z\in A\cap\Lambda$ we have

$\mathbb{A}^{N}_{N}[z]=\begin{cases}c_{0}&\hbox{ if }z\in A_{0},\\\ 0&\hbox{
otherwise. }\end{cases}$

Figure 6. An even almost corner structure on a $D(M)$ grid
with $M/p_{k}\notin{\mathrm{Div}}(A)$, $p_{k}=2$.

###### Proof.

We may assume that $M=N$ and $c_{0}=1$. If $M/p_{i}$ or $M/p_{j}$ is not in
${\mathrm{Div}}(A\cap\Lambda)$, then $A\cap\Lambda$ is fibered by Lemma 4.9,
contradicting the assumptions of the lemma. Therefore (6.17) holds. Invoking
Proposition 6.1 again, we see that $A\cap\Lambda$ must contain either diagonal
boxes or at least two non-overlapping $M$-fibers in different directions. The
second case cannot be reconciled with (6.17). Therefore $A\cap\Lambda$ must
contain a set $A_{0}$ of diagonal boxes. Notice that adding an $M$-fiber in
any direction to $A_{0}$ would introduce $M/p_{k}$ as a divisor of
$A\cap\Lambda$. Therefore $A\cap\Lambda=A_{0}$. ∎

##  7\. Resolving diagonal boxes

######  Theorem 7.1.

Let $A\oplus B=\mathbb{Z}_{M}$ be a tiling, $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$
with $p_{i},p_{j},p_{k}\geq 3$, $|A|=|B|=p_{i}p_{j}p_{k}$, and assume that
$\Phi_{M}\mid A$. Let $D=D(M)$, and let $\Lambda$ be a $D(M)$-grid such that
$A\cap\Lambda\neq\emptyset$. Assume further that $A\cap\Lambda$ is not fibered
in any direction, and that one of the following holds: either

(7.1) $\\{m:\ D(M)|m|M\\}\subset{\mathrm{Div}}(A\cap\Lambda)$

and $A\cap\Lambda$ contains diagonal boxes as in Definition 5.1, or else
$A\cap\Lambda$ has one of the structures described in Lemma 6.2 (full plane)
or Lemma 6.3 (ii) (almost corner). Then at least one of the following is true:

  * •

The tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to $\Lambda\oplus
B=\mathbb{Z}_{M}$ via fiber shifts. Thus $\Lambda$ is a translate of
$A^{\flat}$, and by Corollary 2.2, both $A$ and $B$ satisfy (T2).

  * •

The tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to a tiling
$A^{\prime}\oplus B$, where $A^{\prime}\cap\Lambda$ contains a $p_{\nu}$
corner structure as in Definition 5.4 (i) for some $\nu\in\\{i,j,k\\}$.

We remark that, in the case when (7.1) holds, $A\cap\Lambda$ might be larger
than just a pair of diagonal boxes. For example, it could contain diagonal
boxes and some number of $M$-fibers in various directions disjoint from the
boxes. However, any such additional structures can only make our task easier.

We split the proof into cases. Since $p_{i},p_{j},p_{k}\geq 3$, at least one
of $I$ and $I^{c}$ must have cardinality greater than 1, and similarly for
each of the pairs $J,J^{c}$ and $K,K^{c}$. We claim that it suffices to
consider the following two cases.

Case (DB1): The tiling $A\oplus B$ satisfies the assumptions of Theorem 7.1,
and additionally $\min(|I|,|J^{c}|,|K^{c}|)\geq 2$.

Case (DB2): The tiling $A\oplus B$ satisfies the assumptions of Theorem 7.1,
and additionally $|I^{c}|=|J^{c}|=|K^{c}|=1$.

Indeed, if either $|I|=|J|=|K|=1$ or $|I^{c}|=|J^{c}|=|K^{c}|=1$, then we are
in the case (DB2), possibly after relabelling $I,J,K$ as $I^{c},J^{c},K^{c}$
and vice versa. Suppose now that neither of these holds, say $|I|\geq 2$ and
$|J^{c}|\geq 2$. Since $p_{k}\geq 3$, at least one of $K$ and $K^{c}$ must
have cardinality at least 2. If $|K^{c}|\geq 2$, we are in the case (DB1). All
other cases are identical, possibly after a permutation of the indices $i,j,k$
and relabelling some or all of the sets $I,J,K$ as $I^{c},J^{c},K^{c}$ and
vice versa.

We will proceed to resolve the cases (DB1) and (DB2) in Sections 7.2 and 7.3,
respectively. Throughout this section, we continue to use the notation of
Section 5.1.

###  7.1. Preliminary results

######  Lemma 7.2.

Let $A\oplus B=\mathbb{Z}_{M}$ be a tiling, $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$,
$|A|=|B|=p_{i}p_{j}p_{k}$, and assume that $\Phi_{M}\mid A$. Let $D=D(M)$, and
let $\Lambda$ be a $D$-grid such that $A\cap\Lambda\neq\emptyset$. Assume that
(7.1) holds. Suppose that $A\cap\Lambda$ is not fibered, and that it contains
diagonal boxes

(7.2) $A_{0}:=(I\times J\times K)\cup(I^{c}\times J^{c}\times K^{c})\subset
A\cap\Lambda,$

with $I,J,K,I^{c},J^{c},K^{c}$ as in Definition 5.1. If one of the
“complementary boxes” is contained in $A$, say

(7.3) $(I^{c}\times J\times K)\subset A,$

then the tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to $\Lambda\oplus
B=\mathbb{Z}_{M}$. Consequently, $A$ and $B$ both satisfy T2.

###### Proof.

Throughout the proof, we assume that $A\oplus B=\mathbb{Z}_{M}$ is a tiling
satisfying the assumptions of the lemma.

Claim 1. Assume that (7.3) holds. Then either $A\oplus B=\mathbb{Z}_{M}$ is
T2-equivalent to $\Lambda\oplus B=\mathbb{Z}_{M}$, or

(7.4) $(I^{c}\times J\times K^{c})\cap A=\emptyset.$

###### Proof.

Suppose that there is an $a\in(I^{c}\times J\times K^{c})\cap A$, and let

$Z=\Pi(a,p_{i}^{2})\cap I^{c}\times J^{c}\times K.$

For any $z\in Z$, let $a_{j},a_{k}$ be points such that
$(z-a_{j},M)=(a_{k}-a,M)=M/p_{j}$ and $(z-a_{k},M)=(a_{j}-a,M)=M/p_{k}$. Then
$a_{j},a_{k}\in A$, since $a_{j}\in I^{c}\times J\times K$ and $a_{k}\in
I^{c}\times J^{c}\times K^{c}$. By Lemma 4.6, either $z\in A$, or else
$A_{z}\subset\ell_{i}(x)$, with a $(1,2)$ cofibered structure for $(A,B)$ in
the $p_{i}$ direction and an $M$-fiber in $A$ at distance $M/p_{i}^{2}$ from
$z$ as a cofiber. We can use Lemma 2.9 to shift that fiber to $z$. Repeating
this procedure for all $z\in Z\setminus A$, we get a new set
$A_{1}\subset\mathbb{Z}_{M}$ such that $A_{1}\oplus B=\mathbb{Z}_{M}$ and
$A_{1}$ is T2-equivalent to $A$. Moreover, $Z\subset A_{1}$.

Now, any $x\in(I^{c}\times J^{c}\times K)\setminus Z$ satisfies the
assumptions of Lemma 4.5 applied to $A_{1}$, with $(x-z,M)=M/p_{i}$ for some
$z\in Z$, $(x-a^{\prime}_{j},M)=M/p_{j}$ for some $a^{\prime}_{j}\in
I^{c}\times J\times K$, and $(x-a^{\prime}_{k},M)=M/p_{k}$ for some
$a^{\prime}_{k}\in I^{c}\times J^{c}\times K^{c}$. It follows that

(7.5) $I^{c}\times J^{c}\times K\subset A.$

We can then apply Lemma 4.6 and the fiber shifting argument again, first to
all points in $I^{c}\times J\times K^{c}$ with the flat corner configurations
in planes perpendicular to the $p_{i}$ direction, then to all points in
$I\times J^{c}\times K$ with the flat corner configurations in planes
perpendicular to the $p_{k}$ direction. Thus $A_{1}$ is T2-equivalent to a set
$A_{2}$ that satisfies $A_{2}\oplus B=\mathbb{Z}_{M}$, continues to obey
(7.3)–(7.5), and moreover has the property that

$(I^{c}\times J\times K^{c})\cup(I\times J^{c}\times K)\subset A_{2}.$

Finally, we cannot have any points in $I\times J^{c}\times K^{c}$ that are not
in $A_{2}$, since that would contradict Lemma 4.5. Thus $A_{2}=\Lambda$. ∎

Claim 1’. Assume that (7.3) holds. Then either $A\oplus B=\mathbb{Z}_{M}$ is
T2-equivalent to $\Lambda\oplus B=\mathbb{Z}_{M}$, or

(7.6) $(I^{c}\times J^{c}\times K)\cap A=\emptyset.$

###### Proof.

This is identical to the proof of Claim 1, with the $p_{j}$ and $p_{k}$
directions interchanged, and with $J$ and $K^{c}$ replaced by $J^{c}$ and $K$.
∎

Claim 2. Assume that (7.3), (7.4), and (7.6) hold. Then

(7.7) $I\times J^{c}\times K^{c}\subset A.$

###### Proof.

Let $x\in I\times J^{c}\times K^{c}$. Considering any $M$-cuboid with one
vertex at $x$ and another in $I^{c}\times J\times K$, we see that it can only
be balanced if $x\in A$. ∎

We can now finish the proof of the lemma. It suffices to consider the case
when (7.3) and (7.7) hold, so that

(7.8) $Y:=(I\times J\times\mathbb{Z}_{p_{i}})\cup(I^{c}\times
J^{c}\times\mathbb{Z}_{p_{i}})\subset A.$

If there are any elements $a\in(A\cap\Lambda)\setminus Y$, we can repeat the
argument in the proof of Claim 1 to prove that the tiling $A\oplus
B=\mathbb{Z}_{M}$ is T2-equivalent to $\Lambda\oplus B=\mathbb{Z}_{M}$. If on
the other hand $A\cap\Lambda=Y$, then this set is fibered in the $p_{i}$
direction, contradicting the assumptions of the lemma. ∎

######  Corollary 7.3.

Let $A\oplus B=\mathbb{Z}_{M}$ be a tiling, $M=p_{i}^{2}p_{j}^{2}p_{k}^{2}$,
$|A|=|B|=p_{i}p_{j}p_{k}$, and assume that $\Phi_{M}\mid A$. Let $D=D(M)$, and
let $\Lambda$ be a $D$-grid such that $A\cap\Lambda\neq\emptyset$. Assume that
(5.6) holds. Suppose that $A\cap\Lambda$ is not fibered, that it contains
diagonal boxes (7.2) with $I,J,K,I^{c},J^{c},K^{c}$ as in Definition 5.1, and
that it also contains at least one $M$-fiber disjoint from these boxes. Then
the tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to $\Lambda\oplus
B=\mathbb{Z}_{M}$. Consequently, $A$ and $B$ both satisfy T2.

###### Proof.

We may assume without loss of generality that the $M$-fiber $F\subset
A\cap\Lambda$ is in the $p_{j}$ direction, with $F\subset
I^{c}\times\mathbb{Z}_{p_{j}}\times K$. By translational invariance, we may
further assume that

$F=\\{0\\}\times\mathbb{Z}_{p_{j}}\times\\{0\\},$

with $0\in I^{c}$ and $0\in K$.

We first claim that there is a set $A_{1}\subset\mathbb{Z}_{M}$, either equal
to $A$ or T2-equivalent to it, such that $A_{1}\oplus B=\mathbb{Z}_{M}$ and

(7.9) $\\{0\\}\times J\times K\subset A_{1}.$

Indeed, if $K=\\{0\\}$, then $\\{0\\}\times J\times K\subset F\subset A$ and
there is nothing to prove. Otherwise, let $l_{j}\in J$ and $l_{k}\in
K\setminus\\{0\\}$. Then the point $x=(0,l_{j},l_{k})$ either belongs to $A$,
or else it satisfies the assumptions of Lemma 4.6, with $(0,l_{j},l_{k})$,
$(l_{i},l_{j},0)$, and $(l_{i},l_{j},l_{k})$ all in $A$ for any $l_{i}\in I$.
By Lemma 4.6, in the latter case we have $A_{x}\subset\ell_{j}(x)$, implying a
$(1,2)$ cofibered structure. We can then use Lemma 2.9 to shift the
$M$-cofiber in $A$ to $x$. After all such shifts have been performed, we
arrive at $A_{1}$.

By a similar argument but with $p_{i}$ and $p_{k}$ interchanged, we may
further replace $A_{1}$ by a T2-equivalent set $A_{2}$ such that $A_{2}\oplus
B=\mathbb{Z}_{M}$, (7.9) continues to hold for $A_{2}$, and

(7.10) $I^{c}\times J^{c}\times\\{0\\}\subset A_{2}.$

Next, we replace $A_{2}$ by a T2-equivalent set $A_{3}$ such that $A_{3}\oplus
B=\mathbb{Z}_{M}$, (7.9) and (7.10) both continue to hold for $A_{3}$, and

(7.11) $I^{c}\times J\times\\{0\\}\subset A_{3}.$

Indeed, consider a point $z=(l^{\prime}_{i},l_{j},0\\}\in I^{c}\times
J\times\\{0\\}$, with $l^{\prime}_{i}\neq 0$. (If no such $l^{\prime}_{i}$
exists, (7.11) holds with $A_{3}=A_{2}$.) If $z\not\in A_{2}$, then it
satisfies the assumptions of Lemma 4.6, with $(0,l_{j},0)$,
$(0,l^{\prime}_{j},0)$, and $(l_{i},l^{\prime}_{j},0)$ all in $A_{2}$ for any
$l^{\prime}_{j}\in J^{c}$.Then $A_{z}\subset\ell_{k}(z)$, and again our
conclusion follows by Lemma 2.9.

Finally, we may pass to another T2-equivalent set $A_{4}$ such that
$A_{4}\oplus B=\mathbb{Z}_{M}$ and

(7.12) $I^{c}\times J\times K\subset A_{4}.$

If $I^{c}=\\{0\\}$ or $K=\\{0\\}$, this follows from (7.9) or (7.11), with
$A_{4}=A_{3}$. Otherwise, we let $w=(l^{\prime}_{i},l_{j},l_{k}\\}\in
I^{c}\times J\times K$ with $l^{\prime}_{i}\neq 0$ and $l_{k}\neq 0$, and
repeat the argument from the proof of (7.9) with the first coordinate $0$
replaced by $l^{\prime}_{i}$.

With (7.12) in place, the corollary now follows from Lemma 7.2. ∎

###  7.2. Case (DB1)

Assume that $A\oplus B=\mathbb{Z}_{M}$ is a tiling satisfying the assumptions
of Theorem 7.1. Let $D=D(M)$, and let $\Lambda$ be the $D(M)$ grid provided by
the assumption of the theorem. Additionally, we assume that

(7.13) $\min(|I|,|J^{c}|,|K^{c}|)\geq 2.$

If $A\cap\Lambda$ has one of the structures described in Lemma 6.2 (full
plane) or Lemma 6.3 (ii) (almost corner), then in both cases we have
$A\cap\Lambda=(I\times J\times K)\cup(I^{c}\times J^{c}\times K^{c})$ with no
other points permitted, so that

(7.14) $(I^{c}\times J\times K)\cap A=\emptyset.$

(Note that (7.13) covers the cases of a $p_{i}$ full plane and a $p_{j}$
almost corner structure. See the end of this section for more details.)

If on the other hand (7.1) holds and $I^{c}\times J\times K\subset A$, the
conclusion of Theorem 7.1 follows by Lemma 7.2. We may therefore assume that
there exists a point

(7.15) $x\in(I^{c}\times J\times K)\setminus A.$

######  Lemma 7.4.

Assume (DB1), and let $x$ satisfy (7.15). Then for every $b\in B$ we have
exactly one of the following:

(7.16) $A_{x,b}\subset\ell_{j}(x),\hbox{ with
}\mathbb{A}_{M/p_{j}^{2}}[x]\mathbb{B}_{M/p_{j}^{2}}[b]=\phi(p_{j}^{2}),$
(7.17) $A_{x,b}\subset\ell_{k}(x),\hbox{ with
}\mathbb{A}_{M/p_{k}^{2}}[x]\mathbb{B}_{M/p_{k}^{2}}[b]=\phi(p_{k}^{2}).$

Furthermore:

  * •

$\mathbb{A}_{M/p_{j}}[x]\cdot\mathbb{A}_{M/p_{k}}[x]=0$,

  * •

if $\mathbb{A}_{M/p_{j}}[x]>0$, then (7.16) cannot hold for any $b\in B$, and
if $\mathbb{A}_{M/p_{k}}[x]>0$, then (7.17) cannot hold for any $b\in B$,

  * •

if (7.16) holds for some $b\in B$, then the product
$\langle\mathbb{A}[x],\mathbb{B}[b]\rangle$ is saturated by a $(1,2)$-cofiber
pair in the $p_{j}$ direction, with the $A$-cofiber at distance $M/p_{j}^{2}$
from $x$ and the $B$-fiber rooted at $b$. The same is true for (7.17), with
$j$ and $k$ interchanged.

###### Proof.

The assumptions (DB1) and (7.15) imply in particular that

(7.18) $\mathbb{A}_{M/p_{i}}[x]\geq 2,$

so that $A_{x}\subset\Pi(x,p_{i}^{2})$. Moreover,

$A_{x}\subset\bigcap_{a:(x-a,M)=M/p_{j}p_{k}}{\mathrm{Bispan}}(x,a)=\Pi(x,p_{j}^{2})\cup\Pi(x,p_{k}^{2}),$

which together with (7.18) proves that
$A_{x}\subset\ell_{j}(x)\cup\ell_{k}(x)$. By Lemma 4.1, for each $b\in B$ we
must in fact have either $A_{x,b}\subset\ell_{j}(x)$ or
$A_{x,b}\subset\ell_{k}(x)$. If $A_{x,b}\subset\ell_{j}(x)$, then the second
part of (7.16) follows since $M/p_{j}\in{\mathrm{Div}}(A)$, and Lemma 2.10
implies the existence of a cofiber pair as described in the last part of the
lemma. The same applies with $j$ and $k$ interchanged.

Next, suppose that $\mathbb{A}_{M/p_{j}}[x]>0$. Then
$A_{x}\subset{\mathrm{Bispan}}(x,a)$ for any $a\in A$ with $(x-a,M)=M/p_{j}$.
But this is clearly not compatible with (7.16) for any $b\in B$. The same
applies with $j$ and $k$ interchanged.

On the other hand, one of (7.16) and (7.17) must hold for each $b$, therefore
we cannot have both $\mathbb{A}_{M/p_{j}}[x]>0$ and
$\mathbb{A}_{M/p_{j}}[x]>0$. This ends the proof of the lemma. ∎

######  Lemma 7.5.

Assume (DB1). Suppose that there is a point $x\in\Lambda$ such that (7.15)
holds and

(7.19) $\max(\mathbb{A}_{M/p_{j}}[x],\mathbb{A}_{M/p_{k}}[x])>0.$

Then the conclusion of Theorem 7.1 holds.

###### Proof.

Assume without loss of generality that $\mathbb{A}_{M/p_{j}}[x]>0$. Then
$A_{x}\subset\ell_{j}(x)$ by Lemma 7.4. By Lemma 2.10, the pair $(A,B)$ has a
(1,2)-cofibered structure in the $p_{j}$ direction with a cofiber in $A$ at
distance $M/p_{j}^{2}$ from $x$. We apply Lemma 2.9 to shift the $M$-cofiber
in $A$ to $x$. Let $A_{1}$ be the set thus obtained.

We note that (7.19) does not hold for either the full plane structure or the
almost corner structure. Therefore (7.1) must hold, and since
$A\cap\Lambda\subset A_{1}\cap\Lambda$, the same holds for $A_{1}$.
Furthermore, $A_{1}$ contains the diagonal boxes inherited from $A$ as well as
the added $M$-fiber through $x$ in the $p_{j}$ direction, disjoint from the
boxes. By Corollary 7.3, the tiling $A_{1}\oplus B=\mathbb{Z}_{M}$ (therefore
also $A\oplus B=\mathbb{Z}_{M}$) is T2-equivalent to $\Lambda\oplus
B=\mathbb{Z}_{M}$. ∎

It remains to consider the complementary case when

(7.20) $\mathbb{A}_{M/p_{j}}[x]=\mathbb{A}_{M/p_{k}}[x]=0\ \ \forall
x\in(I^{c}\times J\times K)\setminus A.$

######  Lemma 7.6.

Assume that (DB1) holds, that $(I^{c}\times J\times K)\setminus
A\neq\emptyset$, and that (7.20) holds. Fix an index $l_{i}\in I^{c}$ such
that $(\\{l_{i}\\}\times J\times K)\setminus A$ is nonempty, and define

$\mathcal{X}=\mathcal{X}(l_{i}):=\\{l_{i}\\}\times J\times K.$

Then either $A$ is T2-equivalent to a set $A^{\prime}\subset\mathbb{Z}_{M}$
such that $A^{\prime}\oplus B=\mathbb{Z}_{M}$ and every point
$x^{\prime}\in\mathcal{X}$ belongs to a fiber in $A^{\prime}$ in the $p_{j}$
direction, or else the same holds with $j$ and $k$ interchanged.

###### Proof.

We fix $l_{i}$ as in the statement of the lemma, and keep it fixed for the
duration of the proof. Observe first that, by (7.20), we must have
$x^{\prime}\not\in A$ for all $x^{\prime}\in I^{c}\times J\times K$ with
$(x^{\prime}-x,M)\in\\{M/p_{j},M/p_{k}\\}$, Applying (7.20) to all such
$x^{\prime}$, we see that $\mathcal{X}\cap A=\emptyset$. Let

$\mathcal{X}_{j}=\\{x\in\\{l_{i}\\}\times J\times K:\ \exists b\in B\hbox{
such that }A_{x,b}\subset\ell_{j}(x)\\},$

and similarly for $\mathcal{X}_{k}$. Then $\mathcal{X}\subset\Pi$, where
$\Pi:=\Pi(x_{0},p_{i}^{2})$ for any $x_{0}\in\mathcal{X}$. By Lemma 7.4, we
have $\mathcal{X}=\mathcal{X}_{j}\cup\mathcal{X}_{k}$.

Suppose that $\mathcal{X}_{k}\neq\emptyset$, with $x_{0}\in\mathcal{X}_{k}$,
and let $x_{1},x_{2},\dots,x_{|J|-1}$, be the distinct points in $\mathcal{X}$
such that $(x_{0}-x_{\nu},M)=M/p_{j}$ for $\nu\neq 0$. By the definition of
$\mathcal{X}_{k}$, there exists $b\in B$ such that

(7.21) $A_{x_{\nu},b}\subset\ell_{k}(x_{\nu})$

for $\nu=0$. We claim that (7.21) also holds for $\nu=1,\dots,|J|-1$, with the
same $b\in B$. Indeed, suppose that $A_{x_{\nu},b}\subset\ell_{j}(x_{\nu})$
for some $\nu\geq 1$. By Lemma 2.10, the product
$\langle\mathbb{A}[x_{\nu}],\mathbb{B}[b]\rangle$ would be saturated by a
$(1,2)$ cofiber pair in the $p_{j}$ direction. But then we would also have
$A_{x_{0},b}\subset\ell_{j}(x_{0})$, with the product saturated by the same
cofiber pair, contradicting the assumption that (7.21) holds for $\nu=0$.

By Lemma 2.10 again, it follows that if $\mathcal{X}_{k}\neq\emptyset$, then
$A$ contains an $M$-fiber in the $p_{k}$ direction at distance $M/p_{k}^{2}$
from each of the points $x_{0},x_{1},x_{2},\dots,x_{|J|-1}$. Since
$\ell_{k}(x_{\nu})\subset\Pi$, all these fibers are contained in $\Pi$.

If we had both $\mathcal{X}_{j}\neq\emptyset$ and
$\mathcal{X}_{k}\neq\emptyset$, we would get such sets of fibers in both
directions, all contained in $\Pi$. But then $\Pi$ contains $|J^{c}|\,|K^{c}|$
points of $I^{c}\times J^{c}\times K^{c}$, at least $|J|$ fibers in $A$ in the
$p_{k}$ direction, and at least $|K|$ fibers in $A$ in the $p_{j}$ direction,
disjoint from $\Lambda\cap\Pi$ and from each other. Thus

$|A\cap\Pi|\geq(p_{j}-|J|)(p_{k}-|K|)+p_{k}|J|+p_{j}|K|=p_{j}p_{k}+|J||K|.$

This contradicts Lemma 4.3.

It follows that at least one of $\mathcal{X}_{j}$ and $\mathcal{X}_{k}$ must
be empty. Assume without loss of generality that $\mathcal{X}_{j}=\mathcal{X}$
and $\mathcal{X}_{k}=\emptyset$. It follows that $A_{x}\subset\ell_{j}(x)$ for
all $x\in\mathcal{X}$. By Lemma 2.10, the pair $(A,B)$ has a $(1,2)$-cofibered
structure in the $p_{k}$ direction, and for every $x\in\mathcal{X}$ there is a
cofiber in $A$ at distance $M/p_{j}^{2}$ from $x$. We can now use Lemma 2.9 to
shift all of the aforementioned cofibers in the $p_{j}$ direction, obtaining
the new set $A^{\prime}$ as indicated in the conclusion of the lemma. ∎

We can now conclude the proof of Theorem 7.1 under the assumption that (7.13),
(7.15), and (7.20) hold.

  * •

If (7.1) holds, then the set $A^{\prime}\subset\mathbb{Z}_{M}$ from Lemma 7.6
still satisfies (7.1), and $A^{\prime}\cap\Lambda$ contains the diagonal boxes
inherited from $A$ as well as the additional fibers added in Lemma 7.6. By
Corollary 7.3, $A^{\prime}$ is T2-equivalent to $\Lambda$, and the theorem
follows.

  * •

Suppose that $A\cap\Lambda$ has the $p_{i}$ full plane structure (Lemma 6.2),
with $|I^{c}|=|J|=|K|=1$. Then there is only one index $l_{i}\in I^{c}$, and
for that $l_{i}$, $\mathcal{X}=\\{x\\}$ is a single point. In this case, Lemma
7.6 identifies a direction $p_{\nu}$ for some $\nu\in\\{j,k\\}$, such that the
pair $(A,B)$ has a $(1,2)$-cofibered structure in the $p_{\nu}$ direction with
a cofiber in $A$ at distance $M/p_{\nu}^{2}$ from $x$. Assume without loss of
generality that $\nu=j$. After shifting the cofiber to $x$ as permitted by
Lemma 2.9, we arrive at a $p_{j}$ corner structure, where
$A^{\prime}\cap\Lambda$ is the union of one $M$-fiber in the $p_{i}$ direction
and $(p_{j}-1)$ $M$-fibers in the $p_{k}$ directions, each in a different
plane perpendicular to the $p_{j}$ direction.

  * •

Assume now that $A\cap\Lambda$ has the $p_{j}$ almost corner structure (Lemma
6.3 (ii)), with $|K|=|I^{c}|=1$ and $|J|,|J^{c}|>1$. Then there is again only
one index $l_{i}\in I^{c}$, and the set $\mathcal{X}=\\{l_{i}\\}\times J\times
K$ for that value of $l_{i}$ has dimensions $1\times|J|\times 1$. We again
apply Lemma 7.6. If the fiber identified and shifted in the lemma was in the
$p_{j}$ direction, then we are in the $p_{j}$ corner situation again, with
$A^{\prime}\cap\Lambda$ consisting of non-overlapping fibers in the $p_{i}$
and $p_{k}$ directions. If on the other hand the shifted fibers were in the
$p_{k}$ direction, then we have
$\\{D|m|M\\}\subset{\mathrm{Div}}(A^{\prime})$, and $A^{\prime}\cap\Lambda$
contains the diagonal boxes inherited from $A$ as well as the added fibers
disjoint from the boxes. By Corollary 7.3, the tiling $A^{\prime}\oplus
B=\mathbb{Z}_{M}$ (therefore also $A\oplus B=\mathbb{Z}_{M}$) is T2-equivalent
to $\Lambda\oplus B=\mathbb{Z}_{M}$.

###  7.3. Case (DB2)

We now assume that $A\oplus B=\mathbb{Z}_{M}$ is a tiling satisfying the
assumptions of Theorem 7.1 on a $D$-grid $\Lambda$. Additionally, we assume
that

(7.22) $|I^{c}|=|J^{c}|=|K^{c}|=1.$

Since $p_{i},p_{j},p_{k}$ are all odd, the assumption (7.22) implies that
$\min(|I|,|J|,|K|)\geq 2$, and in particular (7.1) holds. Let also

$I^{c}\times J^{c}\times K^{c}=\\{a_{0}\\}.$

######  Lemma 7.7.

Assume (DB2). Suppose that there exists a point $x\in\Lambda\setminus A$ such
that $A_{x}$ is contained in one of the lines $\ell_{i}(x)$, $\ell_{j}(x)$,
$\ell_{k}(x)$. Then the tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent to
$\Lambda\oplus B=\mathbb{Z}_{M}$. Hence the conclusion of Theorem 7.1 holds.

###### Proof.

Suppose that $x\in\Lambda\setminus A$ and $A_{x}\subset\ell_{k}(x)$. By Lemma
2.10, the pair $(A,B)$ has a (1,2)-cofibered structure in the $p_{k}$
direction. By Lemma 2.9, the cofiber in $A\setminus\Lambda$ can be shifted to
a fiber in $\Lambda$ rooted at $x$. This yields a new tiling $A^{\prime}\oplus
B=\mathbb{Z}_{M}$, T2-equivalent to $A\oplus B=\mathbb{Z}_{M}$, such that
$A^{\prime}\cap\Lambda$ contains both the diagonal boxes
$\\{a_{0}\\}\cup(I\times J\times K)$ and the added $M$-fiber. In light of
(7.1), we can apply Corollary 7.3 to conclude the proof. ∎

######  Lemma 7.8.

Assume (DB2). If $(A\cap\Lambda)\setminus(\\{a_{0}\\}\cup(I\times J\times
K))\neq\emptyset$, then the tiling $A\oplus B=\mathbb{Z}_{M}$ is T2-equivalent
to $\Lambda\oplus B=\mathbb{Z}_{M}$.

###### Proof.

Observe first that if $A\cap\Lambda$ contains a flat corner in the sense of
Lemma 4.6, then this is sufficient to prove our conclusion. Indeed, Lemma 4.6
implies then that the point $x\not\in A$ in the flat corner configuration
satisfies the assumptions of Lemma 7.7, which we then apply.

Suppose that there is an element $a\in A\cap(I^{c}\times J\times K)$. In order
to avoid a flat corner situation at points $x\in
I^{c}\times\\{\lambda_{j}a\\}\times K$, all such points must belong to $A$.
Similarly, we must have $I^{c}\times J\times\\{\lambda_{k}a\\}\subset A$, in
order to avoid a flat corner at those points. By Lemma 4.5 and (7.1), we must
in fact have $I^{c}\times J\times K\subset A$. But then the conclusion follows
by Lemma 7.2. Since the assumption (DB2) is symmetric with respect to all
permutations of the indices $\\{i,j,k\\}$, the same argument applies when
$A\cap(I\times J^{c}\times K)$ or $A\cap(I\times J\times K^{c})$ is nonempty.

Assume now that $a\in A\cap(I^{c}\times J^{c}\times K)$. Consider an
$M$-cuboid with one vertex at $a_{0}$, another at $a$, and a third one at
$a^{\prime}\in I\times J\times K$ with $(a-a^{\prime},M)=M/p_{i}p_{j}$ and
$(a_{0}-a,M)=D$. In order to balance this cuboid, at least one of the vertices
in $I^{c}\times J\times K$, $I\times J^{c}\times K$, or $I\times J\times
K^{c}$, must be in $A$. But then we are in the situation from the last
paragraph. Since this case is also symmetric with respect to all permutations
of $\\{i,j,k\\}$, we are done. ∎

######  Lemma 7.9.

Assume (DB2), and let $N_{k}=M/p_{k}$. If $M/p_{k}^{2}\in{\mathrm{Div}}(A)$,
then $\Phi_{N_{k}}|A$.

###### Proof.

We apply Lemma 4.9 to $B$, on scale $N=N_{k}$. Since
$N_{k},N_{k}/p_{k}\not\in{\mathrm{Div}}(B)$, it follows that $B$ is
$N_{k}$-fibered in one of the $p_{i}$ and $p_{j}$ directions. However, that is
impossible, given that ${\mathrm{Div}}(B)\cap\\{D|m|M\\}=\\{M\\}$ due to
(7.1). ∎

We now begin the proof of Theorem 7.1 under the assumption (DB2). By Lemma
7.8, it suffices to consider the case when

(7.23) $(A\cap\Lambda)\setminus(\\{a_{0}\\}\cup(I\times J\times
K))=\emptyset.$

Let $x\in I\times J^{c}\times K$, in particular $x\not\in A$. Fix $b\in B$.
Then

$A_{x,b}\subset\ell_{i}(x)\cup\ell_{i}(a_{0})\cup\ell_{k}(x)\cup\ell_{k}(a_{0}).$

We claim that $A_{x,b}$ cannot intersect more than one of the above lines.
Indeed, by (7.1) we have

  * •

$A_{x,b}\cap\ell_{i}(x)\neq\emptyset$ implies
$\mathbb{A}_{M/p_{i}^{2}}[x]\mathbb{B}_{M/p_{i}^{2}}[b]>0$, hence
$M/p_{i}^{2}p_{k}\in{\mathrm{Div}}(A)$ and $M/p_{i}^{2}\in{\mathrm{Div}}(B)$.

  * •

$A_{x,b}\cap\ell_{k}(x)\neq\emptyset$ implies
$\mathbb{A}_{M/p_{k}^{2}}[x|\ell_{k}(x)]\mathbb{B}_{M/p_{k}^{2}}[b]>0$, hence
$M/p_{i}p_{k}^{2}\in{\mathrm{Div}}(A)$ and $M/p_{k}^{2}\in{\mathrm{Div}}(B)$.

  * •

$A_{x,b}\cap\ell_{i}(a_{0})\neq\emptyset$ implies
$\mathbb{A}_{M/p_{i}^{2}p_{k}}[x|\ell_{i}(a_{0})]\mathbb{B}_{M/p_{i}^{2}p_{k}}[b]>0$,
hence $M/p_{i}^{2}\in{\mathrm{Div}}(A)$ and
$M/p_{i}^{2}p_{k}\in{\mathrm{Div}}(B)$.

  * •

$A_{x,b}\cap\ell_{k}(a_{0})\neq\emptyset$ implies
$\mathbb{A}_{M/p_{i}p_{k}^{2}}[x|\ell_{k}(a_{0})]\mathbb{B}_{M/p_{i}p_{k}^{2}}[b]>0$,
hence $M/p_{k}^{2}\in{\mathrm{Div}}(A)$ and
$M/p_{i}p_{k}^{2}\in{\mathrm{Div}}(B)$.

It is then easy to check that $A_{x,b}$ cannot have nonempty intersection with
any two of the above lines, either by Lemma 4.1 or due to direct divisor
conflict.

######  Lemma 7.10.

Assume (DB2) and (7.23), and let $x$ be as above. Then
$A_{x,b}\cap\ell_{k}(a_{0})=\emptyset$.

###### Proof.

Assume for contradiction that

(7.24) $A_{x,b}\subset\ell_{k}(a_{0}).$

Then $M/p_{k}^{2}\in{\mathrm{Div}}(A)$, hence by Lemma 7.9 we have
$\Phi_{M/p_{k}}|A$. We are also assuming as part of (DB2) that $\Phi_{M}|A$.
It follows that $A$ is null with respect to all cuboids of type
$(M,(1,1,2),1)$.

Consider any such cuboid with one vertex at $a_{0}$, a second one at a point
$x_{j}\in I^{c}\times J\times K^{c}$ so that $(x_{j}-a_{0},M)=M/p_{j}$, and a
third one at a point $z\in\ell_{k}(x)$ with $(x-z,M)=M/p_{k}^{2}$. By (7.22),
none of the cuboid vertices in the plane $\Pi(a_{0},p_{k}^{2})$ except for
$a_{0}$ are in $A$. In order to balance this cuboid, one of the vertices
$z_{i}$ and $z_{j}$ must be in $A$, where $(z_{i}-z,M)=M/p_{i}$ and
$(z_{j}-z,M)=M/p_{j}$. However, if $z_{j}\in A$, then
$M/p_{i}p_{k}^{2}\in{\mathrm{Div}}(\\{z_{j}\\},I\times J\times
K)\subset{\mathrm{Div}}(A)$, which contradicts (7.24) due to divisor conflict.
It follows that $z_{i}\in A$.

Repeating this argument for all $z\in\ell_{k}(x)$ with $(x-z,M)=M/p_{k}^{2}$,
we get that

(7.25) $\mathbb{A}_{M/p_{k}^{2}}[a_{0}]=\phi(p_{k}^{2}).$

If $p_{i}<p_{k}$, then we must in fact have $p_{i}\leq p_{k}-2$ (since both
are odd primes), so that $\phi(p_{k}^{2})=p_{k}(p_{k}-1)>p_{i}p_{k}$. In this
case, (7.25) contradicts Lemma 4.3.

Assume now that $p_{i}>p_{k}$. In this case, (7.24) and (7.25) imply that

$\displaystyle 1$
$\displaystyle=\frac{1}{\phi(p_{i}p_{k}^{2})}\mathbb{A}_{M/p_{i}p_{k}^{2}}[x|\ell_{k}(a_{0})]\mathbb{B}_{M/p_{i}p_{k}^{2}}[b]$
$\displaystyle=\frac{1}{\phi(p_{i}p_{k}^{2})}\mathbb{A}_{M/p_{k}^{2}}[a_{0}]\mathbb{B}_{M/p_{i}p_{k}^{2}}[b]$
$\displaystyle=\frac{1}{\phi(p_{i})}\mathbb{B}_{M/p_{i}p_{k}^{2}}[b].$

But when $p_{i}>p_{k}$, this is not possible without introducing $M/p_{i}$ or
$M/p_{i}p_{k}$ as divisors of $B$, which would contradict (7.1). ∎

Repeating Lemmas 7.9 and 7.10 with $k$ replaced by $i$, we get that for each
$b\in B$ we must have one of

$A_{x,b}\subset\ell_{i}(x)\hbox{ or }A_{x,b}\subset\ell_{k}(x).$

Now let $x^{\prime}\in I\times J\times K^{c}$. The same analysis, with the
$p_{j}$ and $p_{k}$ directions interchanged, shows that for each $b\in B$ we
must have one of

$A_{x^{\prime},b}\subset\ell_{i}(x^{\prime})\hbox{ or
}A_{x^{\prime},b}\subset\ell_{j}(x^{\prime}).$

If $A_{x,b}\subset\ell_{k}(x)$ and
$A_{x^{\prime},b^{\prime}}\subset\ell_{j}(x^{\prime})$ for some
$b,b^{\prime}\in B$, we use the same fiber crossing argument as in the case
(DB1). By Lemma 2.10, $A$ must contain an $M$-fiber in the $p_{k}$ direction
at distance $M/p_{k}^{2}$ from $x$, as well as an $M$-fiber in the $p_{j}$
direction at distance $M/p_{j}^{2}$ from $x^{\prime}$, both of them in the
plane $\Pi:=\Pi(x,p_{i}^{2})$ and disjoint from each other. The same plane
also contains $(p_{j}-1)(p_{k}-1)$ elements of $I\times J\times K$. Therefore

$|\Pi\cap A|\geq(p_{j}-1)(p_{k}-1)+p_{j}+p_{k}=p_{j}p_{k}+1>p_{j}p_{k},$

which contradicts Lemma 4.3.

It follows that either $A_{x,b}\subset\ell_{i}(x)$ for all $b\in B$, or else
$A_{x^{\prime},b}\subset\ell_{i}(x^{\prime})$ for all $b\in B$. An application
of Lemma 7.7 concludes the proof of the theorem.

##  8\. Corners

In this section we address the extended corner structure, defined in
Definition 5.4. We will assume the following.

Assumption (C): Assume that $A\oplus B=\mathbb{Z}_{M}$, where
$M=p_{i}^{n_{i}}p_{j}^{n_{j}}p_{k}^{n_{k}}$, $|A|=p_{i}p_{j}p_{k}$, and
$\Phi_{M}|A$. Moreover, assume that $A$ contains a $p_{i}$ extended corner in
the sense of Definition 5.4 on a $D(M)$ grid $\Lambda$, that is, there exist
$a,a_{i}\in A\cap\Lambda$ with $(a-a_{i},M)=M/p_{i}$ such that

  * •

$A\cap(a*F_{j}*F_{k})$ is $M$-fibered in the $p_{j}$ direction but not in the
$p_{k}$ direction,

  * •

$A\cap(a_{i}*F_{j}*F_{k})$ is $M$-fibered in the $p_{k}$ direction but not in
the $p_{j}$ direction.

In particular,

(8.1) $a*F_{j}\subset A,\ \ a_{i}*F_{k}\subset A,$ (8.2)
$a_{i}*F_{j}\not\subset A,\ \ a*F_{k}\not\subset A.$

We fix such $a,a_{i}$ for the rest of this section.

The following theorem is our main result for the extended corner structure.

######  Theorem 8.1.

Assume that (C) holds with $n_{j}=n_{k}=2$. Then the tiling $A\oplus
B=\mathbb{Z}_{M}$ is T2-equivalent to $\Lambda\oplus B=\mathbb{Z}_{M}$ via
fiber shifts. By Corollary 2.2, both $A$ and $B$ satisfy (T2).

We first note that when $A$ contains a $p_{i}$ corner,

(8.3) $\\{D(M)|m|M\\}\setminus{\mathrm{Div}}(A)\subseteq\\{M/p_{j}p_{k}\\}$

######  Lemma 8.2 (Size-Divisor Lemma).

Assume that $A\subset\mathbb{Z}_{M}$, and that there exist $a,a_{i}\in A$ with
$(a-a_{i},M)=M/p_{i}$ such that

(8.4) $a*F_{j}\subset A,\ \ a_{i}*F_{k}\subset A,\ \ a*F_{k}\not\subset A.$

(In particular, this is true if (C) holds.) Suppose that
$\Phi_{M}\Phi_{M/p_{j}}|A$. Then we have the following.

(i) For all $a_{k}\in A$ such that $(a_{k}-a_{i},M)=M/p_{k}$ and
$\mathbb{A}_{M/p_{i}}[a_{k}|\ell_{k}(a)]=0$, we have

$\mathbb{A}_{M/p_{j}^{2}}[a]+\mathbb{A}_{M/p_{j}^{2}}[a_{k}]\geq\phi(p_{j}^{2}),$

(ii) $M/p_{j}^{2},\,M/p_{i}p_{j}^{2}p_{k}\in{\mathrm{Div}}(A)$. Moreover, if
$\mathbb{A}_{M/p_{j}^{2}}[a]>0$, then $M/p_{i}p_{j}^{2}\in{\mathrm{Div}}(A)$,
and if $\mathbb{A}_{M/p_{j}^{2}}[a]<\phi(p_{j}^{2})$, then
$M/p_{j}^{2}p_{k}\in{\mathrm{Div}}(A)$.

###### Proof.

Under the assumptions of the lemma, $A$ is $\mathcal{T}$-null with respect to
the cuboid type $\mathcal{T}=(M,\vec{\delta},1)$, where
$\vec{\delta}=(1,2,1)$. Now the lemma follows by considering all possible
$\mathcal{T}$ cuboids with vertices at $a,a_{i},a_{k}$ and $x_{k}$, where
$x_{k}\not\in A$ satisfies $(x_{k}-a,M)=M/p_{k}$ and
$(x_{k}-a_{k},M)=M/p_{i}$. ∎

######  Corollary 8.3.

Assume (C). Then $\Phi_{N_{j}}\Phi_{N_{k}}\nmid A$.

###### Proof.

Assume that $\Phi_{N_{j}}\Phi_{N_{k}}|A$. By Lemma 8.2 ($i$), we have

$|A\cap(\Pi(a,p_{i}^{n_{i}})\cup\Pi(a_{i},p_{i}^{n_{i}}))|\geq
p_{j}+\phi(p_{j}^{2})+p_{k}+\phi(p_{k}^{2})=p_{j}^{2}+p_{k}^{2}>2p_{j}p_{k}$

hence
$\max\\{|A\cap(\Pi(a,p_{i}^{n_{i}}))|,|A\cap(\Pi(a_{i},p_{i}^{n_{i}}))||\\}>p_{j}p_{k}$,
which contradicts Lemma 4.3. ∎

By Corollary 8.3, it suffices to consider the following sets of assumptions.

Assumption (C1): Assume that (C) holds, $p_{k}<p_{j}$, and $\Phi_{N_{k}}|B$.

Assumption (C2): Assume that (C) holds, $p_{k}<p_{j}$, $\Phi_{N_{j}}|B$, and

(8.5) $M/p_{j}p_{k}\notin{\mathrm{Div}}(B).$

In particular, (8.5) holds if either $\mathbb{A}_{M/p_{j}p_{k}}[a]>0$ or
$\mathbb{A}_{M/p_{j}p_{k}}[a_{i}]>0$.

Assumption (C3): Assume that (C) holds, $p_{k}<p_{j}$, $\Phi_{N_{j}}|B$, and

$\mathbb{A}_{M/p_{j}p_{k}}[a]=\mathbb{A}_{M/p_{j}p_{k}}[a_{i}]=0.$

(In this case, $A\cap\Lambda$ contains a $p_{i}$ corner in the sense of
Definition 5.4 (i).)

We first prove in Section 8.1 that under either of the assumptions (C1) or
(C2), the conclusion of Theorem 8.1 holds. We then prove in Section 8.2 that
if (C3) holds, then we must also have $\Phi_{N_{k}}|B$, so that we may return
to (C1) to complete the proof.

###  8.1. Cases (C1) and (C2)

The first two cases are similar and will be considered together.

######  Lemma 8.4.

Assume that (C1) holds with $n_{k}=2$. Then:

(i) Every element of $B$ belongs to an $N_{k}$-fiber in $B$ in either the
$p_{j}$ or $p_{k}$ direction (not necessarily the same for all $b$).

(ii) Furthermore, suppose that there is a $b\in B$ that does not belong to an
$N_{k}$ fiber in the $p_{j}$ direction in $B$. Then for all
$x_{j}\in\mathbb{Z}_{M}\setminus A$ such that $(a_{i}-x_{j},M)=M/p_{j}$ we
have $A_{x_{j},b}\subset\ell_{k}(x_{j})$, with the product saturated by an
$N_{k}$-fiber in $B$ in the $p_{k}$ direction, rooted at $b$, and an
$M$-cofiber in $A$ at distance $M/p_{k}^{2}$ from $x_{j}$.

Moreover, this lemma does not require the assumption $p_{k}<p_{j}$, therefore
the same conclusions with $j$ and $k$ interchanged hold under the assumptions
(C2) and (C3).

###### Proof.

Observe that, by (8.3),

(8.6) $\mathbb{B}^{N_{k}}_{N_{k}/p_{i}}[b]=0\text{ for all }b\in B.$

Assume first that $p_{i}=2$, and suppose that there is a $b_{0}\in B$ that
does not belong to an $N$-fiber in either the $p_{k}$ or the $p_{j}$
direction. In this case, by (8.6) and Lemma 6.5 with $N=N_{k}$ we have

(8.7) $N_{k}/p_{k}=M/p_{k}^{2}\in{\mathrm{Div}}(B).$

Let $x_{j}\in\mathbb{Z}_{M}\setminus A$ with $(a_{i}-x_{j},M)=M/p_{j}$, and
consider the saturating set $A_{x_{j}}$. Let $a_{j}$ be the element of $A$
satisfying $(a-a_{j},M)=M/p_{j},(x_{j}-a_{j},M)=M/p_{i}$. We have

$A_{x_{j},b_{0}}\subset\ell_{k}(x_{j})\cup\ell_{k}(a_{i})\cup\ell_{k}(a)\cup\ell_{k}(a_{j}).$

By (8.3), no top level divisors except possibly $M/p_{j}p_{k}$ are available
to contribute to the product
$\langle\mathbb{A}[x_{j}],\mathbb{B}[b_{0}]\rangle$. Therefore, if
$A_{x_{j},b_{0}}\cap\ell_{k}(a_{j})\neq\emptyset$, we must have

$\mathbb{A}_{M/p_{i}p_{k}^{2}}[x_{j}|\ell_{k}(a_{j})]=\mathbb{A}_{M/p_{k}^{2}}[a_{j}]>0$

which contradicts (8.7). Similarly
$A_{x_{j},b_{0}}\cap\ell_{k}(a)\neq\emptyset$ implies

$\mathbb{A}_{M/p_{i}p_{j}p_{k}^{2}}[x_{j}|\ell_{k}(a)]=\mathbb{A}_{M/p_{k}^{2}}[a]>0$

which, again, contradicts (8.7). Next, if
$A_{x_{j},b_{0}}\cap\ell_{k}(a_{i})\neq\emptyset$, then by (8.7)

(8.8)
$\mathbb{A}^{N_{k}}_{N_{k}/p_{j}}[x_{j}|\ell_{k}(a_{i})]\mathbb{B}^{N_{k}}_{N_{k}/p_{j}}[b_{0}]=\mathbb{A}_{M/p_{j}p_{k}}[x_{j}|\ell_{k}(a_{i})]\mathbb{B}_{M/p_{j}p_{k}}[b_{0}]>0,$

and if $A_{x_{j},b}\cap\ell_{k}(x_{j})\neq\emptyset$, then

(8.9)
$\mathbb{A}^{N_{k}}_{N_{k}/p_{k}}[x_{j}]\mathbb{B}^{N_{k}}_{N_{k}/p_{k}}[b_{0}]=\mathbb{A}_{M/p_{k}^{2}}[x_{j}]\mathbb{B}_{M/p_{k}^{2}}[b_{0}]>0.$

By Lemma 4.1, for a fixed $b_{0}$ we cannot have both (8.9) and (8.8). Hence
either $A_{x_{j},b_{0}}\subset\ell_{k}(a_{i})$ or
$A_{x_{j},b_{0}}\subset\ell_{k}(x_{j})$. In the first case, we have

$\mathbb{A}_{M/p_{j}p_{k}}[x_{j}|\ell_{k}(a_{i})]\mathbb{B}_{M/p_{j}p_{k}}[b_{0}]=\mathbb{A}_{M/p_{k}}[a_{i}]\mathbb{B}_{M/p_{j}p_{k}}[b_{0}]=\phi(p_{j}p_{k});$

since $\mathbb{A}_{M/p_{k}}[a_{i}]=\phi(p_{k})$, it follows that
$\mathbb{B}_{M/p_{j}p_{k}}[b_{0}]=\mathbb{B}^{N_{k}}_{N_{k}/p_{j}}[b_{0}]=\phi(p_{j})$,
hence $b_{0}$ belongs to an $N_{k}$-fiber in the $p_{j}$ direction,
contradicting the choice of $b_{0}$. In the second case, we have

$\mathbb{A}_{M/p_{k}^{2}}[x_{j}]\mathbb{B}_{M/p_{k}^{2}}[b_{0}]=\phi(p_{k}^{2});$

since $M/p_{k}\in{\mathrm{Div}}(A)$, this is only possible if $A$ and $B$
contain a configuration as in part (ii) of the lemma. In this case, it follows
that $b_{0}$ belongs to an $N_{k}$-fiber in the $p_{k}$ direction,
contradicting our initial choice of $b_{0}$. This concludes the proof of (i).
Furthermore, if the conclusion of (ii) fails for any $x_{j}$ chosen as above,
then we must always be in the first case for that $x_{j}$, and therefore every
$b\in B$ must in fact belong to an $N_{k}$ fiber in the $p_{j}$ direction, a
contradiction.

Assume now that $p_{i}>2$. Then part (i) of the lemma follows directly from
Lemma 4.9, (8.6) and the fact that $B$ satisfies (6.12) with $N=N_{k}$ and
$c=1$. For (ii), suppose that there is a $b\in B$ that does not belong to an
$N_{k}$ fiber in $B$ in the $p_{j}$ direction. By (i), there is an $N_{k}$
fiber in $B$ in the $p_{k}$ direction containing $b$. In particular, this
implies that (8.7) holds. The rest of the proof of (ii) is the same as for the
$p_{i}=2$ case. ∎

######  Corollary 8.5.

(i) Assume (C1) with $n_{k}=2$. Then the pair $(A,B)$ has a $(1,2)$-cofibered
structure in the $p_{k}$ direction. Moreover, for each
$x_{j}\in\mathbb{Z}_{M}\setminus A$ such that $(a_{i}-x_{j},M)=M/p_{j}$, $A$
has a cofiber at distance $M/p_{k}^{2}$ from $x_{j}$.