# Tensor products of $d$-fold matrix factorizations

Richie Sheng  Department of Mathematics, University of Utah, Salt Lake City,
UT, USA<EMAIL_ADDRESS>and  Tim Tribone  Department of Mathematics,
University of Utah, Salt Lake City, UT, USA<EMAIL_ADDRESS>timtribone.com

###### Abstract.

Consider a pair of elements $f$ and $g$ in a commutative ring $Q$. Given a
matrix factorization of $f$ and another of $g$, the tensor product of matrix
factorizations, which was first introduced by Knörrer and later generalized by
Yoshino, produces a matrix factorization of the sum $f+g$. We will study the
tensor product of $d$-fold matrix factorizations, with a particular emphasis
on understanding when the construction has a non-trivial direct sum
decomposition. As an application of our results, we construct indecomposable
maximal Cohen-Macaulay and Ulrich modules over hypersurface domains of a
certain form.

###### Key words and phrases:

matrix factorizations, tensor product, Ulrich modules, maximal Cohen-Macaulay
modules

###### 1991 Mathematics Subject Classification:

13C14, 13C13, 13H10, 13F25

##  1\. Introduction

Let $(Q,\mathfrak{n})$ be a regular local ring, $f$ a non-zero element of
$\mathfrak{n}^{2}$, and $z$ an indeterminate. In [Knö87], Knörrer introduced a
construction which allowed for the transfer of maximal Cohen-Macaulay (MCM)
modules between the hypersurface rings $Q/(f)$ and $Q\llbracket
z\rrbracket/(f+z^{2})$. The construction, which is most easily stated in terms
of matrix factorizations and is now commonly referred to as the tensor product
of matrix factorizations, has had far-reaching applications. In particular,
the careful analysis of the functor described by Knörrer is at the heart of
their famous periodicity theorem.

Knörrer’s functor is defined as follows: given a matrix factorization
$(\varphi,\psi)$ of $f$, that is, a pair of $n\times n$ square matrices with
entries in $Q$ satisfying

$\varphi\psi=f\cdot I_{n}=\psi\varphi,$

Knörrer defines a new matrix factorization of $f+z^{2}$ given by the pair of
$2n\times 2n$ block matrices

$(\varphi,\psi)\otimes(z,z)\coloneqq\left(\begin{pmatrix}zI_{n}&\varphi\\\
\psi&-zI_{n}\end{pmatrix},\begin{pmatrix}zI_{n}&\varphi\\\
\psi&-zI_{n}\end{pmatrix}\right).$

Yoshino later gave a generalization in [Yos98] which produces a matrix
factorization of any sum $f+g$ starting from a factorization of $f$ and a
factorization of $g$ using an analogous formula. In Yoshino’s language, the
above matrix factorization can be viewed as the tensor product of the matrix
factorization $(\varphi,\psi)$ of $f$ and of the $(1\times 1)$ matrix
factorization $(z,z)$ of $z^{2}$. Yoshino’s generalization has found
applications in homological mirror symmetry [Sei10], knot theory [KR08], and
generally in the theory of MCM modules over hypersurface rings and complete
intersection rings.

Our goal is to study a further generalization of Yoshino’s and, by extension,
Knörrer’s construction. Namely, we study the tensor product of $d$-fold matrix
factorizations. The techniques and ideas introduced in [Yos98] serve as a
framework for this article. We lift Yoshino’s formulas, and some of their
proofs, to the case of $d$-fold matrix factorizations for arbitrary $d\geq 2$.
The overall theme remains the same: given a matrix factorization of an element
$f$ and another of $g$, the tensor product produces a matrix factorization
(with the same number of factors) of the sum $f+g$.

It is worth noting that a $d$-fold version of Yoshino’s tensor product has
appeared in various other forms (several of which actually predate Yoshino’s
paper); see [Chi78], [BHS86], [BH89], [HUB91], [BES17], and [Hop21]. The
version we describe most closely resembles the tensor product of complexes
(really of “$d$-complexes”) and therefore we feel it serves as a good
candidate for a unified definition. We study some of the basic properties of
the tensor product, focusing mostly on those that are relevant to our main
goal: understanding its decomposability. As it turns out, the tensor product
of two indecomposable matrix factorizations often has many non-trivial direct
summands, so, understanding when the operation preserves indecomposability is
the main theme of Section 5. We also give one predictable situation in which
the tensor product always decomposes in Section 4.

Our main application is to the theory of Ulrich modules over hypersurface
domains. The authors of [HUB91] showed that any hypersurface ring (and more
generally, any strict complete intersection) has an Ulrich module using a
construction essentially equivalent to the $d$-fold tensor product. In Section
6, we reproduce their result using elementary properties of the tensor product
in the case of a hypersurface domain, and because of the results of Section 5,
we are able to give conditions (and examples) for when the Ulrich modules
produced are indecomposable.

##  2\. Background and notation

Let $Q$ be a commutative ring (always assumed to be Noetherian), $d\geq 2$ an
integer, and fix an element $f\in Q$. In this section, we provide a brief
background on $d$-fold matrix factorizations and set up the notations that
will be needed throughout. We also record several facts about the structure of
the category of $d$-fold matrix factorizations which will be needed in Section
5.

######  Definition 2.1.

A matrix factorization of $f$ (with $d$ factors) is a sequence of
$Q$-homomorphisms between finitely generated projective $Q$-modules

${X=(X_{0}}$${X_{d-1}}$${\cdots}$${X_{1}}$${X_{0})}$$\scriptstyle{\varphi_{0}}$$\scriptstyle{\varphi_{d-1}}$$\scriptstyle{\varphi_{2}}$$\scriptstyle{\varphi_{1}}$

satisfying

(2.1)
$\varphi_{i}\varphi_{i+1}\cdots\varphi_{0}\varphi_{1}\cdots\varphi_{i-2}\varphi_{i-1}=f\cdot
1_{X_{i-1}}$

for all $i\in\mathbb{Z}_{d}\coloneqq\mathbb{Z}/d\mathbb{Z}$ (see (2.1)). When
convenient, we will suppress the full sequence and refer to a matrix
factorization as an ordered $d$-tuple,
$X=(\varphi_{1},\varphi_{2},\dots,\varphi_{d-1},\varphi_{0})$. Alternatively,
we will view a matrix factorization $X=(X,\varphi)$ as a
$\mathbb{Z}_{d}$-graded projective $Q$-module,
$X=\bigoplus_{k\in\mathbb{Z}_{d}}X_{k}$, equipped with a degree $-1$ map
$\varphi:X\to X$ satisfying $\varphi^{d}=f\cdot 1_{X}$.

The collection of all $d$-fold matrix factorizations of $f$ forms an additive
category which we will denote by $\operatorname{MF}_{Q}^{d}(f)$. Namely, if
$(X,\varphi),(X^{\prime},\varphi^{\prime})\in\operatorname{MF}_{Q}^{d}(f)$,
then the direct sum is given by
$(X,\varphi)\oplus(X^{\prime},\varphi^{\prime})\coloneqq(X\oplus
X^{\prime},\varphi\oplus\varphi^{\prime})$. We will call a non-zero object in
$\operatorname{MF}_{Q}^{d}(f)$ indecomposable if it cannot be decomposed into
a direct sum of non-zero matrix factorizations. A morphism
$\alpha\in\textup{Hom}(X,X^{\prime})$ in $\operatorname{MF}_{Q}^{d}(f)$ is a
$d$-tuple of $Q$-homomorphisms,
$\alpha=(\alpha_{0},\alpha_{1},\dots,\alpha_{d-1})$, such that
$\alpha_{k-1}\varphi_{k}=\varphi_{k}^{\prime}\alpha_{k}$ for all
$k\in\mathbb{Z}_{d}$. Equivalently, $\alpha:X\to X^{\prime}$ is a degree zero
$Q$-linear map satisfying $\alpha\varphi=\varphi^{\prime}\alpha$. More details
on the category $\operatorname{MF}_{Q}^{d}(f)$ can be found in [Hop21] and
[Tri22].

If $f$ is a nonzerodivisor and each $X_{k}$, $k\in\mathbb{Z}_{d}$, is a free
$Q$-module, then (2.1) forces each of the ranks of the free $Q$-modules
$X_{k}$ to be equal [Eis80, Corollary 5.4]. In this case, we say that the
matrix factorization $X$ has rank $n$ where $n=\operatorname{rank}_{Q}X_{k}$,
$k\in\mathbb{Z}_{d}$.

The shift of a matrix factorization
$(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ will refer to the factorization
obtained by cyclically permuting the maps $\varphi_{k}$. We will denote this
factorization by

$TX=T(\varphi_{1},\varphi_{2},\dots,\varphi_{d-1},\varphi_{0})\coloneqq(\varphi_{2},\varphi_{3},\dots,\varphi_{0},\varphi_{1}).$

In particular, for $k\in\mathbb{Z}_{d}$, the degree $k$ piece of $TX$ is
$(TX)_{k}=X_{k+1}$. Similarly, the $i$th shift of $X$ cyclically permutes $X$
$i$ times:

$T^{i}X=(\varphi_{i+1},\varphi_{i+2},\dots,\varphi_{0},\varphi_{1},\dots,\varphi_{i-1},\varphi_{i}).$

We will refrain from using typical suspension notation as the shift operation
$T(-)$ does not induce the suspension functor in the triangulated stable
category associated to $\operatorname{MF}_{Q}^{d}(f)$ unless $d=2$ (see
[Tri22, Section 2]).

###  2.1. Notation and conventions

The main construction we will study is a $\mathbb{Z}_{d}$-graded analogue of
the tensor product of complexes (technically of $d$-complexes; see [HS21,
Section 3]). In this setting, the usual sign convention is replaced by
appropriate roots of unity. In order to give the construction in our desired
generality, we will assume throughout that the ring $Q$ is a
$\mathbb{Z}[\zeta]$-algebra where $\zeta$ is a primitive $d^{\textit{th}}$
root of unity in $\mathbb{C}$. That is, we will assume that there is a
(unital) ring homomorphism $\mathbb{Z}[\zeta]\to Q$. For instance, if $Q$ is a
domain contaning a primitive $d^{\textit{th}}$ roof of unity (in the usual
sense), then such an algebra structure exists. We will abuse notation and
write $\zeta\in Q$ to denote the image of $\zeta$ under this homomorphism.

Due to the cyclic nature of $d$-fold matrix factorization, it will be
convenient to take indices modulo $d$. Throughout, the reader should assume
that subscripts and other indexing variables are to be interpreted modulo $d$
unless otherwise indicated. In particular, components of a morphism will
(almost always) be taken modulo $d$ which we explain below.

If $\alpha:\bigoplus_{j=1}^{t}M_{j}\to\bigoplus_{i=1}^{s}N_{i}$ is a morphism
in an additive category, we use the notation $\alpha(i,j)$ to denote the
composition

$M_{j}\hookrightarrow\bigoplus_{j=1}^{t}M_{j}\xrightarrow{\,\alpha\,}\bigoplus_{i=1}^{s}N_{i}\twoheadrightarrow
N_{i}.$

We refer to $\alpha(i,j)$ as the $(i,j)$-component of $\alpha$ with respect to
the given direct sum decompositions. In this case, $\alpha$ can be represented
by a matrix $\alpha=(\alpha(i,j))_{1\leq i\leq s,1\leq j\leq t}$. Furthermore,
if both of the direct sum decompositions are indexed over $\mathbb{Z}_{d}$
with respect to the ordering $1\prec 2\prec\cdots\prec d-1\prec d=0$, then the
components of $\alpha$ are naturally indexed modulo $d$ as well, that is, for
any $1\leq i,j\leq d$ and $s,t\in\mathbb{Z}$, $\alpha(i+sd,j+td)=\alpha(i,j).$
The choice of ordering is simply to index the rows and columns of matrices in
the usual way. In other words, as a matrix, $\alpha$ is written as

$(\alpha(i,j))_{i,j\in\mathbb{Z}_{d}}=\begin{pmatrix}\alpha(1,1)&\alpha(1,2)&\dots&\alpha(1,d)\\\
\alpha(2,1)&\alpha(2,2)&\dots&\alpha(2,d)\\\ \vdots&\vdots&\ddots&\vdots\\\
\alpha(d,1)&\alpha(d,2)&\dots&\alpha(d,d)\end{pmatrix}.$

###  2.2. Idempotents in the category of matrix factorizations

Before proceeding, we record two important facts about morphisms of matrix
factorizations that will be needed in Section 5. They both hold in greater
generality than the setting of Section 5 so we include them here.

Let $Q$ be a commutative ring, $d\geq 2$, and $f\in Q$. Suppose
$(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ is of rank $n$. An idempotent
$e\in\operatorname{End}(X)$ defines a matrix factorization
$e(X)\coloneqq(\operatorname{Im}e,\varphi|_{\operatorname{Im}E})\in\operatorname{MF}_{Q}^{d}(f).$
A key aspect to the proofs in Section 5 will be to track the rank of $e(X)$.
Here we will show that, when $Q$ is local, idempotents in
$\operatorname{MF}_{Q}^{d}(f)$ can be put into a normal form which identifies
the rank of the matrix factorization $e(X)$.

We will say that a pair of morphisms $\alpha\in\textup{Hom}(X,X^{\prime})$ and
$\beta\in\textup{Hom}(Y,Y^{\prime})$ are equivalent if there exists
isomorphisms $\gamma$ and $\delta$ such that $\delta\alpha=\beta\gamma$. In
this case, we will write $\alpha\sim\beta$. Notice that equivalent idempotents
induce isomorphic summands, that is, if $e\sim e^{\prime}$, then $e(X)\cong
e^{\prime}(X^{\prime})$.

######  Lemma 2.2.

Let $X,X^{\prime}\in\operatorname{MF}_{Q}^{d}(f)$ and suppose
$\alpha=(\alpha_{0},\dots,\alpha_{d-1})\in\textup{Hom}(X,X^{\prime})$. For any
$k\in\mathbb{Z}_{d}$, replacing $\alpha_{k}$ with $A\alpha_{k}B$, for
isomorphisms $A$ and $B$, results in a morphism equivalent to $\alpha$.

###### Proof.

Let $k\in\mathbb{Z}_{d}$. For any isomorphisms of $Q$-modules
$A:X_{k}^{\prime}\to X_{k}^{\prime}$ and $B:X_{k}\to X_{k}$, we have a
commutative diagram

${X_{k+1}}$${X_{k}}$${X_{k-1}}$${X_{k+1}^{\prime}}$${X_{k}^{\prime}}$${X_{k-1}^{\prime}.}$$\scriptstyle{B^{-1}\varphi_{k+1}}$$\scriptstyle{\alpha_{k+1}}$$\scriptstyle{\varphi_{k}B}$$\scriptstyle{A\alpha_{k}B}$$\scriptstyle{\alpha_{k-1}}$$\scriptstyle{A\varphi_{k+1}^{\prime}}$$\scriptstyle{\varphi_{k}^{\prime}A^{-1}}$

Set $Y=(\varphi_{1},\dots,\varphi_{k}B,B^{-1}\varphi_{k+1},\dots,\varphi_{0})$
and
$Y^{\prime}=(\varphi_{1}^{\prime},\dots,\varphi_{k}^{\prime}A^{-1},A\varphi_{k+1}^{\prime},\dots,\varphi_{0})$.
It follows that $Y,Y^{\prime}\in\operatorname{MF}_{Q}^{d}(f)$ and the
commutative diagram above shows that

$\beta=(\alpha_{0},\alpha_{1},\dots,A\alpha_{k}B,\dots,\alpha_{d-1}):Y\to
Y^{\prime}$

is a morphism of matrix factorizations. Now, if we let
$\gamma=(1_{X_{0}},\dots,B^{-1},\dots,1_{X_{d-1}})$ and
$\delta=(1_{X_{0}^{\prime}},\dots,A,\dots,1_{X_{d-1}^{\prime}})$ we have that
$\delta\alpha=\beta\gamma$ implying that $\alpha$ and $\beta$ are equivalent.
∎

The next lemma provides a normal form for idempotents in
$\operatorname{MF}_{Q}^{d}(f)$. In particular, the lemma shows that
idempotents split in the additive category $\operatorname{MF}_{Q}^{d}(f)$.

######  Lemma 2.3.

Assume that $(Q,\mathfrak{n})$ is local and $f\in\mathfrak{n}$ is a
nonzerodivisor in $Q$. If $X\in\operatorname{MF}_{Q}^{d}(f)$ is of rank $n$
and $e=(e_{0},\dots,e_{d-1})\in\operatorname{End}(X)$ is an idempotent, then
there exists an integer $0\leq r\leq n$ such that $e$ is equivalent to a
morphism $(\epsilon,\epsilon,\dots,\epsilon)$ where

$\epsilon=\begin{pmatrix}I_{r}&0\\\ 0&0\end{pmatrix}:Q^{r}\oplus Q^{n-r}\to
Q^{r}\oplus Q^{n-r}.$

In particular, $e(X)$ is of rank $r$, $(1-e)(X)$ is of rank $n-r$, and $X\cong
e(X)\oplus(1-e)(X)$.

###### Proof.

If $e=0$ or $e=1$, there is nothing to prove so assume $e\neq 0,1$. Since $e$
is an idempotent of matrix factorizations, each component satisfies
$e_{k}^{2}=e_{k}$, that is, the $Q$-homomorphism $e_{k}:X_{k}\to X_{k}$ is an
idempotent for each $k\in\mathbb{Z}_{d}$. Therefore, there exists isomorphisms
$A_{k},B_{k}$ such that

$A_{k}e_{k}B_{k}=\begin{pmatrix}I_{r_{k}}&0\\\
0&0\end{pmatrix}:Q^{r_{k}}\oplus Q^{n-r_{k}}\to Q^{r_{k}}\oplus Q^{n-r_{k}}$

for some $0<r_{k}<n$. Applying Lemma 2.2 to each $e_{k}$,
$k\in\mathbb{Z}_{d}$, we may assume that

$e=\left(\begin{pmatrix}I_{r_{0}}&0\\\
0&0\end{pmatrix},\begin{pmatrix}I_{r_{1}}&0\\\
0&0\end{pmatrix},\dots,\begin{pmatrix}I_{r_{d-1}}&0\\\
0&0\end{pmatrix}\right)$

for integers $0<r_{0},r_{1},\dots,r_{d-1}<n$, and that
$\varphi_{k}:Q^{r_{k}}\oplus Q^{n-r_{k}}\to Q^{r_{k-1}}\oplus Q^{n-r_{k-1}}$
for each $k\in\mathbb{Z}_{d}$.

To finish the proof, it suffices to show that $r_{i-1}=r_{i}$ for each
$i\in\mathbb{Z}_{d}$. Let $i\in\mathbb{Z}_{d}$ and consider the commutative
diagram

${Q^{r_{i-1}}\oplus Q^{n-r_{i-1}}}$${Q^{r_{i}}\oplus
Q^{n-r_{i}}}$${Q^{r_{i-1}}\oplus Q^{n-r_{i-1}}}$${Q^{r_{i-1}}\oplus
Q^{n-r_{i-1}}}$${Q^{r_{i}}\oplus Q^{n-r_{i}}}$${Q^{r_{i-1}}\oplus
Q^{n-r_{i-1}}.}$$\scriptstyle{\varphi_{i+1}\varphi_{i+2}\cdots\varphi_{i-1}}$$\scriptstyle{e_{i-1}}$$\scriptstyle{e_{i}}$$\scriptstyle{\varphi_{i}}$$\scriptstyle{e_{i-1}}$$\scriptstyle{\varphi_{i+1}\varphi_{i+2}\cdots\varphi_{i-1}}$$\scriptstyle{\varphi_{i}}$

Decompose $\varphi_{i}$ and $\varphi_{i+1}\varphi_{i+2}\cdots\varphi_{i-1}$
with respect to these direct sum decompositions into

$\varphi_{i}=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\quad\text{ and
}\quad\varphi_{i+1}\varphi_{i+2}\cdots\varphi_{i-1}=\begin{pmatrix}A^{\prime}&B^{\prime}\\\
C^{\prime}&D^{\prime}\end{pmatrix}.$

Since
$(\varphi_{i},\varphi_{i+1}\varphi_{i+2}\cdots\varphi_{i-1})\in\operatorname{MF}_{Q}^{2}(f)$,
we have that

$\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\begin{pmatrix}A^{\prime}&B^{\prime}\\\
C^{\prime}&D^{\prime}\end{pmatrix}=\begin{pmatrix}f\cdot I_{r_{i-1}}&0\\\
0&f\cdot I_{n-r_{i-1}}\end{pmatrix}$

and

$\begin{pmatrix}A^{\prime}&B^{\prime}\\\
C^{\prime}&D^{\prime}\end{pmatrix}\begin{pmatrix}A&B\\\
C&D\end{pmatrix}=\begin{pmatrix}f\cdot I_{r_{i}}&0\\\ 0&f\cdot
I_{n-r_{i}}\end{pmatrix}.$

Using the fact that $e_{i-1}\varphi_{i}=\varphi_{i}e_{i}$, we can conclude
that $B=0$ and $C=0$. Similarly, the commutativity of the left square implies
that $B^{\prime}=0$ and $C^{\prime}=0$. Thus, $AA^{\prime}=f\cdot I_{r_{i-1}}$
and $A^{\prime}A=f\cdot I_{r_{i}}$ which is only possible if $r_{i-1}=r_{i}$
(see [Eis80, Corollary 5.4]).

The final statement follows by decomposing each
$\varphi_{i},i\in\mathbb{Z}_{d}$, along the direct sum decomposition
$\varphi_{i}:Q^{r}\oplus Q^{n-r}\to Q^{r}\oplus Q^{n-r}$ where $r$ is the
common value $r=r_{0}=r_{1}=\cdots=r_{d-1}$. ∎

We finish this section with a straightforward lemma that will also be useful
in Section 5.

######  Lemma 2.4.

Let $c_{1},c_{2},\dots,c_{d-1},c_{0}$ be elements of $Q$ such that
$c_{1}c_{2}\dots c_{d-1}c_{0}=1$. Then, for any
$(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$, $X$ is isomorphic to the
factorization

$(c_{1}\varphi_{1},c_{2}\varphi_{2},\dots,c_{d-1}\varphi_{d-1},c_{0}\varphi_{0})\in\operatorname{MF}_{Q}^{d}(f).$

In particular, if $c\in Q$ satisfies $c^{d}=1$, then
$cX\coloneqq(c\varphi_{1},\dots,c\varphi_{0})\cong X$. ∎

##  3\. The tensor product

In this section, we define the tensor product of $d$-fold matrix
factorizations and prove some of the basic properties that will be needed
going forward. Our definition closely follows the presentation given in
[Hop21] with minor modifications. Throughout this section, let $Q$ be
commutative ring, $d\geq 2$ an integer, and assume that $Q$ is a
$\mathbb{Z}[\zeta]$-algebra where $\zeta\in\mathbb{C}$ is a primitive
$d^{\textit{th}}$ root of unity. We also fix elements $f$ and $g$ in $Q$.

######  Definition 3.1.

For matrix factorizations $X=(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ and
$Y=(Y,\psi)\in\operatorname{MF}_{Q}^{d}(g)$ we define the tensor product
$X\otimes^{\zeta}Y=(X\otimes^{\zeta}Y,\Phi)\in\operatorname{MF}_{Q}^{d}(f+g)$
as follows. The underlying $\mathbb{Z}_{d}$-graded projective $Q$-module is
given by
$X\otimes^{\zeta}Y=\bigoplus_{k\in\mathbb{Z}_{d}}(X\otimes^{\zeta}Y)_{k}$
where

$(X\otimes^{\zeta}Y)_{k}=\bigoplus_{i+j\equiv k\mod d}X_{i}\otimes_{Q}Y_{j}.$

For homogeneous elements $x\in X$ and $y\in Y$, the map $\Phi$ is given by

$\Phi(x\otimes y)=\varphi(x)\otimes y+\zeta^{|x|}x\otimes\psi(y)$

where $|x|$ denotes the degree of the element $x\in X$, i.e. $x\in X_{i}$ if
and only if $|x|=i$.

In general, the isomorphism class of $X\otimes^{\zeta}Y$ depends on the choice
of root of unity, $\zeta\in Q$ (for instance, see Example 3.6). However, when
the root of unity is fixed, we will omit $\zeta$ from the notation.

This construction has appeared in various similar forms in [Chi78], [BHS86],
[BH89], [HUB91], [BES17], and [Hop21]. Each version has the same goal: to
produce a matrix factorization of $f+g$ from a factorization of $f$ and a
factorization of $g$. The techniques in each of the articles listed above are
quite distinct, though some of the constructions can be identified with each
other up to a choice of roots of unity and organization of the matrices (for
example, see Proposition 3.4). Our definition closely follows [Hop21, Section
2.5] except for a slightly different approach to the needed roots of unity.
The proof given by Hopkins explicitly uses the fact that $\zeta$ is a
distinguished primitive $d^{\textit{th}}$ root of unity so we are unable to
transfer their proof to our setting. However, with slight modification, the
proof of [BES17, Proposition 2.2] applies in our generality.

######  Lemma 3.2.

[BES17, Hop21] Let $X\in\operatorname{MF}_{Q}^{d}(f)$ and
$Y\in\operatorname{MF}_{Q}^{d}(g)$. Then $X\otimes Y$ defines a matrix
factorization of $f+g$ in $\operatorname{MF}_{Q}^{d}(f+g)$.

###### Proof.

Let $\rho:\mathbb{Z}[\zeta]\to Q$ denote the ring homomorphism defining the
$\mathbb{Z}[\zeta]$-algebra structure on $Q$. Notice that
$\prod_{i=0}^{d-1}(x-\rho(\zeta)^{i})=x^{d}-1$ in $Q[x]$. Hence the
coefficient of $x^{k}$ on the left is zero for each $0<k<d$. By picking
$z_{i}=\rho(\zeta)^{i}$, $0\leq i\leq d-1$, one can now follow the proof of
[BES17, Proposition 2.2] with minor adjustments to conclude that $X\otimes
Y\in\operatorname{MF}_{Q}^{d}(f+g)$. ∎

The operation
$(-)\otimes(-):\operatorname{MF}_{Q}^{d}(f)\times\operatorname{MF}_{Q}^{d}(g)\to\operatorname{MF}_{Q}^{d}(f+g)$
is functorial in both arguments. In particular, if $\alpha:X\to X^{\prime}$ is
a morphism in $\operatorname{MF}_{Q}^{d}(f)$, then, for homogeneous $x\in X$
and $y\in Y$,

$(\alpha\otimes 1_{Y})(x\otimes y)=\alpha(x)\otimes y$

forms a morphism $\alpha\otimes 1_{Y}:X\otimes Y\to X^{\prime}\otimes Y$ in
$\operatorname{MF}_{Q}^{d}(f+g)$. Similarly, if $\beta:Y\to Y^{\prime}$ is a
morphism in $\operatorname{MF}_{Q}^{d}(g)$, then $1_{X}\otimes\beta:X\otimes
Y\to X\otimes Y^{\prime}$ defines a morphism in
$\operatorname{MF}_{Q}^{d}(f+g)$.

It will often be useful to consider the components of the tensor product with
respect to a given direct sum decomposition. For each $k\in\mathbb{Z}_{d}$, we
will view $\Phi_{k}$ with respect to the direct sum decompositions
$\Phi_{k}:\bigoplus_{j=1}^{d}X_{j-1}\otimes
Y_{k+1-j}\to\bigoplus_{i=1}^{d}X_{i-1}\otimes Y_{k-i}$. For $1\leq i,j\leq d$,
the $(i,j)$-component of $\Phi_{k}$ is given by

(3.1)
$\Phi_{k}(i,j)=\begin{cases}\zeta^{i-1}1_{X_{i-1}}\otimes\psi_{k+1-i}&\text{
if }j\equiv i\mod d\\\ \varphi_{i}\otimes 1_{Y_{k-i}}&\text{ if }j\equiv
i+1\mod d\\\ 0&\text{ otherwise }\end{cases}.$

######  Example 3.3.

Let $Q=\Bbbk\llbracket x_{i},y_{i},z_{i}:i\in\mathbb{Z}_{3}\rrbracket$ where
$\Bbbk$ is a field containing a primitive third root of unity,
$\zeta\in\Bbbk$. Consider the rank one matrix factorizations
$X=(x_{1},x_{2},x_{0})\in\operatorname{MF}_{Q}^{3}(x_{1}x_{2}x_{0}),$
$Y=(y_{1},y_{2},y_{0})\in\operatorname{MF}_{Q}^{3}(y_{1}y_{2}y_{0})$, and
$Z=(z_{1},z_{2},z_{0})\in\operatorname{MF}_{Q}^{3}(z_{1}z_{2}z_{0})$. Then
$X\otimes^{\zeta}Y\in\operatorname{MF}_{Q}^{3}(x_{1}x_{2}x_{0}+y_{1}y_{2}y_{0})$
is the triple of $3\times 3$ matrices

$(A_{1},A_{2},A_{0})=\left(\begin{pmatrix}y_{1}&x_{1}&0\\\ 0&\zeta
y_{0}&x_{2}\\\
x_{0}&0&\zeta^{2}y_{2}\end{pmatrix},\begin{pmatrix}y_{2}&x_{1}&0\\\ 0&\zeta
y_{1}&x_{2}\\\
x_{0}&0&\zeta^{2}y_{0}\end{pmatrix},\begin{pmatrix}y_{0}&x_{1}&0\\\ 0&\zeta
y_{2}&x_{2}\\\ x_{0}&0&\zeta^{2}y_{1}\end{pmatrix}\right).$

If we tensor this with $Z$, we get a rank $9$ factorization of the sum
$f=x_{1}x_{2}x_{0}+y_{1}y_{2}y_{3}+z_{1}z_{2}z_{3}$, that is,
$(X\otimes^{\zeta}Y)\otimes^{\zeta}Z\in\operatorname{MF}_{Q}^{3}(f)$ is given
by triple of matrices

$(B_{1},B_{2},B_{0})=\left(\begin{pmatrix}z_{1}I_{3}&A_{1}&0\\\ 0&\zeta
z_{0}I_{3}&A_{2}\\\
A_{0}&0&\zeta^{2}z_{2}I_{3}\end{pmatrix},\begin{pmatrix}z_{2}I_{3}&A_{1}&0\\\
0&\zeta z_{1}I_{3}&A_{2}\\\
A_{0}&0&\zeta^{2}z_{0}I_{3}\end{pmatrix},\begin{pmatrix}z_{0}I_{3}&A_{1}&0\\\
0&\zeta z_{2}I_{3}&A_{2}\\\ A_{0}&0&\zeta^{2}z_{1}I_{3}\end{pmatrix}\right)$

where $I_{3}$ is a $3\times 3$ identity matrix. In Sections 5 and 6 we show
that both of these matrix factorizations are indecomposable and that each of
the corresponding cokernel modules are indecomposable Ulrich modules over the
appropriate hypersurface ring.

###  3.1. Properties of the tensor product

As above, we let $d\geq 2$ be an integer and $Q$ be a commutative
$\mathbb{Z}[\zeta]$-algebra where $\zeta\in\mathbb{C}$ is a primitive
$d^{\textit{th}}$ root of unity. We also fix elements $f,g\in Q$. Our first
observation is on the symmetry of the tensor product operation. Note that
since $\mathbb{Z}[\zeta]=\mathbb{Z}[\zeta^{-1}]$, we can also take the tensor
product with respect to $\zeta^{-1}$.

######  Proposition 3.4.

Let $X\in\operatorname{MF}_{Q}^{d}(f)$ and $Y\in\operatorname{MF}_{Q}^{d}(g)$.
Then there is a natural isomorphism of matrix factorizations
$X\otimes^{\zeta}Y\cong Y\otimes^{\zeta^{-1}}X$ in
$\operatorname{MF}_{Q}^{d}(f+g)$.

###### Proof.

Define a map
$\alpha:(X\otimes^{\zeta}Y,\Phi)\to(Y\otimes^{\zeta^{-1}}X,\Phi^{\prime})$
given by $\alpha(x\otimes y)=\zeta^{|x||y|}y\otimes x$ for homogeneous
elements $x\in X$ and $y\in Y$. Then $\alpha$ is invertible and satisfies

$\displaystyle\alpha\Phi(x\otimes y)$
$\displaystyle=\zeta^{|\varphi(x)||y|}y\otimes\varphi(x)+\zeta^{|x|}(\zeta^{|x||\psi(y)|}\psi(y)\otimes
x)$
$\displaystyle=\zeta^{(|x|-1)|y|}y\otimes\varphi(x)+\zeta^{|x|(1+|\psi(y)|)}\psi(y)\otimes
x$
$\displaystyle=\zeta^{-|y|}(\zeta^{|x||y|}y\otimes\varphi(x))+\zeta^{|x||y|}\psi(y)\otimes
x$ $\displaystyle=\Phi^{\prime}\alpha(x\otimes y)$

for any homogeneous $x\in X$ and $y\in Y$. This implies that
$X\otimes^{\zeta}Y\cong Y\otimes^{\zeta^{-1}}X$ in
$\operatorname{MF}_{Q}^{d}(f+g)$. The fact that $\alpha$ induces a natural
isomorphism (in both components) can be checked directly from the formula. ∎

######  Remark 3.5.

In the case $d=2$, the conclusion of Proposition 3.4 amounts to the
isomorphism $X\otimes Y\cong Y\otimes X$ [Yos98, Lemma 2.1]. However, this is
not the conclusion being made in the case that $d>2$; in general, the tensor
product construction with respect to a fixed root of unity is not symmetric
when $d>2$.

######  Example 3.6.

Let $Q=\Bbbk[x,y,z,u,v,w]$ for a field $\Bbbk$ containing a primitive $3$rd
root of unity, $\zeta\in\Bbbk$. Consider the rank one factorizations
$X=(x,y,z)\in\operatorname{MF}_{Q}^{3}(xyz)$ and
$Y=(u,v,w)\in\operatorname{MF}_{Q}^{3}(uvw)$. We claim that
$X\otimes^{\zeta}Y$ and $Y\otimes^{\zeta}X$ are not isomorphic in
$\operatorname{MF}_{Q}^{3}(xyz+uvw)$.

###### Proof.

Assume that $(X\otimes^{\zeta}Y,\Phi)\cong(Y\otimes^{\zeta}X,\Phi^{\prime})$,
that is, assume that there exist invertible matrices
$(\alpha_{0},\alpha_{1},\alpha_{2})$ with entries in $Q$ such that
$\alpha_{k-1}\Phi_{k}=\Phi_{k}^{\prime}\alpha_{k}$ for each
$k\in\mathbb{Z}_{3}$. Using the direct sum decomposition given in (3.1), we
will consider corresponding entries on either side of the equation
$\alpha_{0}\Phi_{1}=\Phi_{1}^{\prime}\alpha_{1}$. We will also write
$\text{c}(\alpha_{k}(i,j))$ to denote the constant term of the $(i,j)$th entry
of $\alpha_{k}$.

Comparing the (1,1) entries of the above equation gives us

$\alpha_{0}(1,1)u+\alpha_{0}(1,3)z=\alpha_{1}(1,1)x+\alpha_{1}(2,1)u.$

Equating coefficients of $u$ shows that
$\text{c}(\alpha_{0}(1,1))=\text{c}(\alpha_{1}(2,1))$, while the coefficients
of $x$ and $z$ give us that
$\text{c}(\alpha_{0}(1,3))=\text{c}(\alpha_{1}(1,1))=0$. Similarly, analyzing
the other entries, we find that the constant term of each entry of
$\alpha_{1}$ must be zero except possibly the $(1,2),(2,1),$ and $(3,3)$
entries which must satisfy
$\text{c}(\alpha_{1}(1,2))=\text{c}(\alpha_{1}(2,1))=\text{c}(\alpha_{1}(3,3)).$
Repeating this analysis with the second commutative square,
$\alpha_{1}\Phi_{2}=\Phi_{2}^{\prime}\alpha_{2}$, we get
$\text{c}(\alpha_{1}(1,2))=\zeta\text{c}(\alpha_{1}(2,1))=\zeta^{2}\text{c}(\alpha_{1}(3,3)).$
The two conclusions are only possible if each constant term is zero. Hence,
all entries of $\alpha_{1}$ have constant term zero contradicting that
$\alpha_{1}$ is invertible. ∎

Although the tensor product is not symmetric in general, Proposition 3.4
implies that the functorial properties given below, which we state only for
the left component, hold for both components. In particular, the tensor
product is additive in both components and associative. We omit the proofs of
additivity and associativity as they are straightforward computations
utilizing the corresponding properties of $Q$-modules.

######  Lemma 3.7.

  1. (i)

For any $X,X^{\prime}\in\operatorname{MF}_{Q}^{d}(f)$ and
$Y\in\operatorname{MF}_{Q}^{d}(g)$, there is a natural isomorphism $(X\oplus
X^{\prime})\otimes Y\cong(X\otimes Y)\oplus(X^{\prime}\otimes Y)$.

  2. (ii)

For any $f,g,h\in Q$ and matrix factorizations
$X\in\operatorname{MF}_{Q}^{d}(f),Y\in\operatorname{MF}_{Q}^{d}(g),$ and
$Z\in\operatorname{MF}_{Q}^{d}(h)$, there is an isomorphism $(X\otimes
Y)\otimes Z\cong X\otimes(Y\otimes Z)$.

∎

######  Lemma 3.8.

Let $X\in\operatorname{MF}_{Q}^{d}(f)$ and $Y\in\operatorname{MF}_{Q}^{d}(g)$.
Then

  1. (i)

$TX\otimes Y\cong T(X\otimes Y)=X\otimes TY$;

  2. (ii)

$T^{i}X\otimes T^{-i}Y\cong X\otimes Y$ for any $i\in\mathbb{Z}_{d}$.

###### Proof.

First notice that the equality in (i) follows from the definition and (ii)
follows from (i). So, it suffices to prove the isomorphism in (i).

Define a map $\alpha:(TX\otimes Y,\Phi)\to(T(X\otimes Y),\Phi^{\prime})$ given
by $\alpha(x\otimes y)=\zeta^{-|y|}x\otimes y$ for homogeneous elements $x\in
X$ and $y\in Y$. For homogeneous elements $x\in TX$ and $y\in Y$, $\Phi$ is
given by

$\Phi(x\otimes y)=\varphi(x)\otimes y+\zeta^{|x|-1}x\otimes\psi(y).$

Therefore,

$\displaystyle\alpha\Phi(x\otimes y)$
$\displaystyle=\zeta^{-|y|}\varphi(x)\otimes
y+\zeta^{|x|-1}\zeta^{-|\psi(y)|}x\otimes\psi(y)$
$\displaystyle=\zeta^{-|y|}(\varphi(x)\otimes y+\zeta^{|x|}x\otimes\psi(y))$
$\displaystyle=\zeta^{-|y|}\Phi^{\prime}(x\otimes y)$
$\displaystyle=\Phi^{\prime}(\alpha(x\otimes y)).$

Hence $\alpha$ is an invertible morphism $TX\otimes Y\to T(X\otimes Y)$. ∎

If $(Q,\mathfrak{n})$ is local and $f\in\mathfrak{n}$ is a nonzerodivisor, the
category $\operatorname{MF}_{Q}^{d}(f)$ is Frobenius exact [Tri22, Section 2].
The ring $Q$ is assumed to be regular in [Tri22, Section 2] but this
assumption is not needed for the Frobenius exact structure. The factorizations
of the form $\mathcal{P}_{i}\coloneqq
T^{i}(f,1,1,\dots,1)\in\operatorname{MF}_{Q}^{d}(f)$, for
$i\in\mathbb{Z}_{d}$, give a complete list of the indecomposable projective
objects in $\operatorname{MF}_{Q}^{d}(f)$. We will call a matrix factorization
$X\in\operatorname{MF}_{Q}^{d}(f)$ projective if it is isomorphic to a direct
sum of copies of $\mathcal{P}_{i}$ for $i\in\mathbb{Z}_{d}$, that is, if
$X\cong\bigoplus_{i\in\mathbb{Z}_{d}}\mathcal{P}_{i}^{s_{i}}$, $s_{i}\geq 0$.

######  Lemma 3.9.

Assume that $(Q,\mathfrak{n})$ is local, $f,g\in\mathfrak{n}$ are
nonzerodivisors, and that $\zeta\in Q$ is a distinguished111The proof of
[Hop21, Proposition 2.5.3] needs the fact that $\zeta$ is a distinguished
primitive $d^{\textit{th}}$ root of unity. In other words, the sum
$\sum_{i=0}^{n-1}\zeta^{i}$ must be invertible for all $1\leq n\leq d-1$ and
zero for $n=d$. primitive $d^{\textit{th}}$ root of unity. If $P$ is a
projective matrix factorization in $\operatorname{MF}_{Q}^{d}(f)$, then
$P\otimes Y$ is a projective matrix factorization in
$\operatorname{MF}_{Q}^{d}(f+g)$ for any $Y\in\operatorname{MF}_{Q}^{d}(g)$.

###### Proof.

The tensor product $\otimes^{\zeta}$ preserves homotopy equivalences; this is
proven in [Hop21, Proposition 2.5.3]. The conclusion of the lemma follows from
this and the fact that an object $P\in\operatorname{MF}_{Q}^{d}(h)$ is
homotopy equivalent to zero if and only if it is projective (see [Hop21,
Theorem 2.8.3] or [Tri22, Proposition 2.2.15] for more details). ∎

Recall that if $(X,\varphi)$ is a matrix factorization of a nonzerodivisor $f$
such that $\operatorname{rank}X=n$, then each $X_{k}$ is a free $Q$-module of
rank $n$. In particular, we may view $\varphi_{k}$ as a square matrix with
entries in $Q$. Note that if $\operatorname{rank}X=n$ and
$\operatorname{rank}Y=m$, then $\operatorname{rank}X\otimes Y=dnm$.

######  Proposition 3.10.

Let $f,g\in Q$ be nonzerodivisors and consider matrix factorizations
$X\in\operatorname{MF}_{Q}^{d}(f)$ of rank $n$ and
$Y\in\operatorname{MF}_{Q}^{d}(g)$ of rank $m$. Set $X\otimes
Y=(\Phi_{1},\Phi_{2},\dots,\Phi_{0})$. Then, for any $k\in\mathbb{Z}_{d}$,
$\det\Phi_{k}=(-1)^{nm(d+1)}(f+g)^{nm}$.

###### Proof.

First, note that it suffices to compute the determinant of $\Phi_{1}$. Indeed,
if $\det\Phi_{1}$ can be computed for arbitrary $X$ and $Y$, then one can
compute $\det\Phi_{k}$ for all $k\in\mathbb{Z}_{d}$ by applying Lemma 3.8.

Our strategy is to convert $\Phi_{1}$ into a lower triangular form using (not
necessarily invertible) row operations. We describe these row operations with
respect to the direct sum decompostion (3.1). We will also utilize the
following notation: If $(X,\varphi)$ is a matrix factorization and $r\geq 0$
is an integer, we let
$\varphi_{k}^{\\{r\\}}=\varphi_{k}\varphi_{k+1}\cdots\varphi_{k+r-1}$ for
$r>0$ and $\varphi_{k}^{\\{0\\}}=1_{X_{k-1}}$. Note that these superscripts
are not taken modulo $d$. In particular, $\varphi_{k}^{\\{d\\}}=f\cdot
1_{X_{k-1}}$ for any $k\in\mathbb{Z}_{d}$.

First we scale each row of $\Phi_{1}$ using the diagonal map

$\alpha(i,j)=\begin{cases}1_{X_{i-1}}\otimes\psi_{2}^{\\{d-i\\}}&\text{ if
}i=j\\\ 0&\text{ else}\end{cases}.$

Note that $\alpha$ is not an isomorphism (in general). The resulting product
is

$\alpha\Phi_{1}=\begin{pmatrix}1_{X_{0}}\otimes\psi_{2}^{\\{d\\}}&\varphi_{1}\otimes\psi_{2}^{\\{d-1\\}}&0&\cdots&0&0\\\
0&\zeta
1_{X_{1}}\otimes\psi_{2}^{\\{d-1\\}}&\varphi_{2}\otimes\psi_{2}^{\\{d-2\\}}&\cdots&0&0\\\
0&0&\zeta^{2}1_{X_{2}}\otimes\psi_{2}^{\\{d-2\\}}&\ddots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\ddots&\vdots\\\
0&0&0&&\zeta^{d-2}1_{X_{d-2}}\otimes\psi_{2}^{\\{2\\}}&\varphi_{d-1}\otimes\psi_{2}^{\\{1\\}}\\\
\varphi_{0}\otimes\psi_{2}^{\\{0\\}}&0&0&\cdots&0&\zeta^{d-1}1_{X_{d-1}}\otimes\psi_{2}^{\\{1\\}}\\\
\end{pmatrix}.$

Next, using invertible row operations on $\alpha\Phi_{1}$, we clear above the
main diagonal from right to left. We obtain the following lower triangular
matrix

$\beta=\begin{pmatrix}c_{d-1}(\varphi_{1}^{\\{d\\}}\otimes
1_{Y_{1}}+1_{X_{0}}\otimes\psi_{2}^{\\{d\\}})&0&0&\cdots&0&0\\\
c_{d-2}\varphi_{2}^{\\{d-1\\}}\otimes\psi_{2}^{\\{0\\}}&\zeta
1_{X_{1}}\otimes\psi_{2}^{\\{d-1\\}}&0&\cdots&0&0\\\
c_{d-3}\varphi_{3}^{\\{d-2\\}}\otimes\psi_{2}^{\\{0\\}}&0&\zeta^{2}1_{X_{2}}\otimes\psi_{2}^{\\{d-2\\}}&&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
c_{1}\varphi_{d-1}^{\\{2\\}}\otimes\psi_{2}^{\\{0\\}}&0&0&&\zeta^{d-2}1_{X_{d-2}}\otimes\psi_{2}^{\\{2\\}}&0\\\
\varphi_{0}\otimes\psi_{2}^{\\{0\\}}&0&0&\cdots&0&\zeta^{d-1}1_{X_{d-1}}\otimes\psi_{2}^{\\{1\\}}\\\
\end{pmatrix}$

where $c_{j}=(-1)^{j}\prod_{k=0}^{j}\zeta^{k}$, $1\leq j\leq d-1$. In
particular, we have that
$c_{d-1}=(-1)^{d-1}\prod_{k=0}^{d-1}\zeta^{k}=(-1)^{d-1}(-1)^{d+1}=1$. It
follows that the $\beta(1,1)=(f+g)\cdot 1_{X_{0}}\otimes 1_{Y_{1}}$ and
therefore

$\det\beta=(f+g)^{nm}\prod_{k=0}^{d-1}\zeta^{nmk}\cdot\det(\alpha)=(-1)^{nm(d+1)}(f+g)^{nm}\cdot\det(\alpha).$

Since $\alpha\Phi_{1}$ and $\beta$ differ by an isomorphism (with determinant
1), we have

$\det(\alpha)\det(\Phi_{1})=\det(\beta)=(-1)^{nm(d+1)}(f+g)^{nm}\cdot\det(\alpha).$

Since $g$ is a nonzerodivisor in $Q$, so is $\det(\alpha)$. The desired
conclusion follows by cancelling $\det(\alpha)$. ∎

##  4\. Knörrer’s direct sum decomposition

An important ingredient to the classification of simple hypersurface
singularities [BGS87, Theorem A] and Knörrer’s periodicity theorem [Knö87,
Theorem 3.1] is a simple direct sum decomposition result which says that if
$(\varphi,\varphi)$ is a matrix factorization of an element $f$ and $z$ is an
indeterminate, then the matrix factorization $(\varphi,\varphi)\otimes(z,z)$
of $f+z^{2}$ decomposes into a direct sum of the form

$(\varphi,\varphi)\otimes(z,z)\cong\left(\begin{pmatrix}zI-i\varphi&0\\\
0&zI+i\varphi\end{pmatrix},\begin{pmatrix}zI+i\varphi&0\\\
0&zI-i\varphi\end{pmatrix}\right)$

where $i$ is a square root of $-1$. The key here is that both
$(\varphi,\varphi)\in\operatorname{MF}^{2}(f)$ and
$(z,z)\in\operatorname{MF}^{2}(z^{2})$ satisfy
$T(\varphi,\varphi)=(\varphi,\varphi)$ and $T(z,z)=(z,z)$. This was
generalized by Yoshino [Yos98, Lemma 3.2] in the case $d=2$. We give a further
extension of these results to the case of $d$-fold matrix factorizations for
arbitrary $d\geq 2$.

Let $Q$ be a commutative ring and $d\geq 2$ an integer. Fix a primitive
$2d^{\textit{th}}$ root of unity, $\omega\in\mathbb{C}$. In this section we
will assume that $Q$ is a $\mathbb{Z}[\omega]$-algebra so that $Q$ contains
both a $2d^{\textit{th}}$ root of unity and a $d^{\textit{th}}$ root of unity
(the images of $\omega$ and $\omega^{2}$ respectively). We will abuse notation
and denote the image of $\omega$ in $Q$ by the same symbol.

######  Lemma 4.1.

Let $\omega\in\mathbb{C}$ be a primitive $2d^{\textit{th}}$ root of unity, and
$t$ be an integer such that $t+d$ is even. Then
$\sum_{j\in\mathbb{Z}_{d}}\omega^{-j^{2}+tj}\neq 0$.

###### Proof.

We first claim that the expression $\omega^{-j^{2}+tj}$ does not depend on the
representative chosen for $j$ modulo $d$ (this fact relies on $t+d$ being
even). If $j\equiv j^{\prime}\mod d$, then either $j\equiv j^{\prime}\mod 2d$,
in which case we are done, or $j\equiv j^{\prime}+d\mod 2d$. In the latter
case,

$-j^{2}\equiv-(j^{\prime})^{2}+d^{2}\mod 2d$

and $tj\equiv tj^{\prime}+td\mod 2d$. Thus,

$-j^{2}+tj\equiv-(j^{\prime})^{2}+d^{2}+tj^{\prime}+td\equiv-(j^{\prime})^{2}+tj^{\prime}+d(d+t)\equiv-(j^{\prime})^{2}+tj^{\prime}\mod
2d.$

Now, we claim that
$(\sum_{j\in\mathbb{Z}_{d}}\omega^{-j^{2}+tj})(\sum_{l\in\mathbb{Z}_{d}}\omega^{l^{2}-tl})=d$.
Performing the change of variable $s=l-j$, we have

$\displaystyle\left(\sum_{j\in\mathbb{Z}_{d}}\omega^{-j^{2}+tj}\right)\left(\sum_{l\in\mathbb{Z}_{d}}\omega^{l^{2}-tl}\right)$
$\displaystyle=\left(\sum_{j\in\mathbb{Z}_{d}}\omega^{-j^{2}+tj}\right)\left(\sum_{s\in\mathbb{Z}_{d}}\omega^{(s+j)^{2}-t(s+j)}\right)$
$\displaystyle=\sum_{s\in\mathbb{Z}_{d}}\left(\sum_{j\in\mathbb{Z}_{d}}\omega^{(s^{2}-st)+2sj}\right)$

The sum
$\sum_{j\in\mathbb{Z}_{d}}\omega^{(s^{2}-st)+2sj}=\omega^{s^{2}-st}\sum_{j\in\mathbb{Z}_{d}}\omega^{2sj}$
is equal to $d$ when $s=0$ in $\mathbb{Z}_{d}$ and $0$ otherwise. This proves
the claim and the desired conclusion follows. ∎

Let $\mathcal{C}$ be a $Q$-linear category, that is, an additive category such
that the abelian group $\textup{Hom}_{\mathcal{C}}(X,Y)$ has a $Q$-module
structure for all $X,Y\in\mathcal{C}$, and such that composition of morphisms
is $Q$-bilinear. In the proposition below, we will consider a pair of
commuting morphisms in $\mathcal{C}$. By this we mean a pair of morphisms
$A,B\in\textup{Hom}_{\mathcal{C}}(X,X)$, for some $X\in\mathcal{C}$, such that
$AB=BA$. The main result of this section is an application of the following
identity among matrices in $\mathcal{C}$.

######  Proposition 4.2.

Assume that $Q$ is a $\mathbb{Z}[\omega,1/d]$-algebra. Let $A$ and $B$ be
commuting morphisms in a $Q$-linear category $\mathcal{C}$ and set $\Phi$ to
be the $d\times d$ block morphism

$\Phi=\begin{pmatrix}B&A&0&\cdots&0\\\ 0&\omega^{2}B&A&\cdots&0\\\
\vdots&\vdots&\ddots&\ddots&\vdots\\\ 0&0&0&\ddots&A\\\
A&0&0&\cdots&\omega^{2(d-1)}B\end{pmatrix}.$

Then there exists isomorphisms $\alpha_{k},k\in\mathbb{Z}_{d}$, such that

$\alpha_{k-1}\Phi\alpha_{k}^{-1}=\begin{pmatrix}A-\omega^{2k-1}B&0&\dots&0\\\
0&A-\omega^{2k+1}B&\dots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&\dots&A-\omega^{2k-3}B\end{pmatrix}$

for each $k\in\mathbb{Z}_{d}$.

###### Proof.

To simplify the notation, we introduce the polynomial
$\mathfrak{p}(m)=-m^{2}+dm$, $m\in\mathbb{Z}$. For each $k\in\mathbb{Z}_{d}$,
we define a map $\alpha_{k}:X^{d}\to X^{d}$, where $X$ is the common
domain/codomain of $A$ and $B$, by the following rule: for any
$i,j\in\mathbb{Z}_{d}$, the $(i,j)$-component of $\alpha_{k}$ is defined to be
$\alpha_{k}(i,j)=\omega^{\mathfrak{p}(j-i-k)}1_{X}$. As in the proof of Lemma
4.1, this formula does not depend on the representatives chosen for $i,j,k$.
For each $k\in\mathbb{Z}_{d}$, let $\Phi_{k}^{\prime}$ denote the $d\times d$
block diagonal morphism in the statement of the proposition. We will show that
$\alpha_{k-1}\Phi_{k}=\Phi_{k}^{\prime}\alpha_{k}$ by computing the
$(i,j)$-component of each composition.

Fix $k\in\mathbb{Z}_{d}$. The only non-zero entries of the $j$th column of
$\Phi$ are $\Phi(j-1,j)=A$ and $\Phi(j,j)=\omega^{2(j-1)}B$. Therefore,

$\displaystyle(\alpha_{k-1}\Phi)(i,j)$
$\displaystyle=\alpha_{k-1}(i,j-1)A+\alpha_{k-1}(i,j)\omega^{2(j-1)}B$
$\displaystyle=\omega^{\mathfrak{p}(j-i-k)}A+\omega^{\mathfrak{p}(j-i-k+1)+2(j-1)}B$
$\displaystyle=\omega^{\mathfrak{p}(j-i-k)}A+\omega^{\mathfrak{p}(j-i-k)+d+2k+2(i-1)-1}B$

Similarly, the only non-zero entry in the $i$th row of $\Phi_{k}^{\prime}$ is
$A-\omega^{2k+2(i-1)-1}B$, so

$(\Phi_{k}^{\prime}\alpha_{k})(i,j)=\omega^{\mathfrak{p}(j-i-k)}A-\omega^{\mathfrak{p}(j-i-k)+2k+2(i-1)-1}B.$

Since $\omega^{d}=-1$, it follows that
$(\alpha_{k-1}\Phi)(i,j)=(\Phi_{k}^{\prime}\alpha_{k})(i,j)$ for all
$i,j,k\in\mathbb{Z}_{d}$.

It remains to show that $\alpha_{k}$ is an isomorphism for each
$k\in\mathbb{Z}_{d}$. It suffices to prove this in the case that $\alpha_{k}$
is the $d\times d$ matrix with entries in $\mathbb{Z}[\omega,1/d]$ given by
$\alpha_{k}(i,j)=\omega^{\mathfrak{p}(j-i-k)}$, $i,j\in\mathbb{Z}_{d}$. Fixing
$k\in\mathbb{Z}_{d}$, we observe that $\alpha_{k}$ is a circulant matrix, as
$\alpha_{k}(i,j)$ is a function of $j-i$, i.e. if $j-i=j^{\prime}-i^{\prime}$
so that $\alpha_{k}(i,j)$ and $\alpha_{k}(i^{\prime},j^{\prime})$ lie on the
same diagonal (modulo $d$) of $\alpha_{k}$, then
$\alpha_{k}(i,j)=\alpha_{k}(i^{\prime},j^{\prime})$. Since $\alpha_{k}$ is
circulant, the determinant is given by

$\displaystyle\det\alpha_{k}$
$\displaystyle=\prod_{s=1}^{d}(\alpha_{k}(1,1)+\alpha_{k}(1,2)\omega^{2s}+\alpha_{k}(1,3)\omega^{4s}+\dots+\alpha_{k}(1,d)\omega^{2(d-1)s})$
$\displaystyle=\prod_{s=1}^{d}\left(\sum_{j=1}^{d}\omega^{\mathfrak{p}(j-1-k)+2s(j-1)}\right).$

By Lemma 4.1, each of these sums is a unit in $\mathbb{Z}[\omega,1/d]$; take
$t=2+2k+2s+d$. Hence $\det\alpha_{k}$ is a unit implying that $\alpha_{k}$ is
invertible. ∎

######  Remark 4.3.

Proposition 4.2 can be viewed as a matrix analogy to the polynomial
factorization of $x^{d}+y^{d}$ into linear factors (see Example 4.6). Indeed,
this explains the presence of the $2d^{\textit{th}}$ roots of unity which
allow us to unify the even and odd case into one statement. In particular,
when $d$ is odd, the condition that $Q$ is a $\mathbb{Z}[\omega]$-algebra is
automatically satisfied by our usual assumption: if $\zeta\in\mathbb{C}$ is a
primitive $d^{\textit{th}}$ root of unity, then $\omega=-\zeta$ is a primitive
$2d^{\textit{th}}$ root of unity. Thus, $\mathbb{Z}[\zeta]=\mathbb{Z}[\omega]$
and therefore $Q$ is a $\mathbb{Z}[\zeta]$-algebra if and only if it is a
$\mathbb{Z}[\omega]$-algebra. This is, however, not true when $d$ is even. In
this case, the condition that $Q$ is a $\mathbb{Z}[\omega]$-algebra is
strictly stronger.

The main theorem of this section is a direct application of Proposition 4.2.

######  Theorem 4.4.

Let $f,g\in Q$ and assume that $Q$ is a $\mathbb{Z}[\omega,1/d]$-algebra.
Suppose $X\in\operatorname{MF}_{Q}^{d}(f)$ and
$Y\in\operatorname{MF}_{Q}^{d}(g)$ satisfy $TX=X$ and $TY=Y$. Then there
exists $Z\in\operatorname{MF}_{Q}^{d}(f+g)$ such that

$X\otimes^{\zeta}Y\cong\bigoplus_{i\in\mathbb{Z}_{d}}T^{i}Z$

where the tensor product is taken with respect to the $d^{\textit{th}}$ root
of unity $\zeta=\omega^{2}$.

###### Proof.

If $X=(\varphi,\varphi,\dots,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ and
$Y=(\psi,\psi,\dots,\psi)\in\operatorname{MF}_{Q}^{d}(g)$, then set
$B=1\otimes\psi$ and $A=\varphi\otimes 1$ and apply Proposition 4.2. The
factorization $Z$ is given by

$Z=\left(\varphi\otimes 1-\omega(1\otimes\psi),\varphi\otimes
1-\omega^{3}(1\otimes\psi),\dots,\varphi\otimes
1-\omega^{2d-1}(1\otimes\psi)\right)\in\operatorname{MF}_{Q}^{d}(f+g).$

∎

If the factorizations $X$ and $Y$ are only isomorphic to their shifts (not
necessarily equal), then a similar result is possible with additional
assumptions.

######  Corollary 4.5.

Assume $(Q,\mathfrak{n},\Bbbk)$ is a complete regular local ring with $\Bbbk$
algebraically closed of characteristic not dividing $d$ and fix non-zero
elements $f,g\in\mathfrak{n}^{2}$. Suppose $X\in\operatorname{MF}_{Q}^{d}(f)$
and $Y\in\operatorname{MF}_{Q}^{d}(g)$ satisfy $TX\cong X$ and $TY\cong Y$.
Then there exists $Z\in\operatorname{MF}_{Q}^{d}(f+g)$ such that

$X\otimes^{\zeta}Y\cong\bigoplus_{i\in\mathbb{Z}_{d}}T^{i}Z$

where the tensor product is taken with respect to the $d^{\textit{th}}$ root
of unity $\zeta=\omega^{2}$.

###### Proof.

The additional assumptions allow us to apply [LT23, Corollary 4.3], that is,
we may assume that $X=TX$ and $Y=TY$. The corollary now follows from Theorem
4.4. ∎

In the case $d=2$, Theorem 4.4 and Corollary 4.5 recover [Yos98, Lemma 3.2]
and by extension, [Knö87, Proposition 2.7].

######  Example 4.6.

Let $Q=\Bbbk[x,y]$ where $\Bbbk$ is a domain.

  1. (i)

Assume that $\Bbbk$ contains a primitive fourth root of unity, $i\in\Bbbk$.
Consider the rank one factorizations
$X=(x,x)\in\operatorname{MF}_{Q}^{2}(x^{2})$,
$Y=(y,y)\in\operatorname{MF}_{Q}^{2}(y^{2})$, and
$Z=(x-iy,x-i^{3}y)=(x-iy,x+iy)\in\operatorname{MF}_{Q}^{2}(x^{2}+y^{2})$. Then
Theorem 4.4 implies that $X\otimes Y\cong Z\oplus TZ$, that is,

$\left(\begin{pmatrix}y&x\\\ x&-y\end{pmatrix},\begin{pmatrix}y&x\\\
x&-y\end{pmatrix}\right)\cong\left(\begin{pmatrix}x-iy&0\\\
0&x+iy\end{pmatrix},\begin{pmatrix}x+iy&0\\\ 0&x-iy\end{pmatrix}\right).$

  2. (ii)

Assume that $\Bbbk$ contains a primitive third root of unity, $\zeta\in\Bbbk$.
Then $\omega=-\zeta$ is a primitive $6^{\textit{th}}$ root of unity. Consider
the rank one factorizations $X=(x,x,x)\in\operatorname{MF}_{Q}^{3}(x^{3})$,
$Y=(y,y,y)\in\operatorname{MF}_{Q}^{3}(y^{3})$, and

$Z=(x-\omega y,x-\omega^{3}y,x-\omega^{5}y)=(x+\zeta
y,x+y,x+\zeta^{2}y)\in\operatorname{MF}_{Q}^{3}(x^{3}+y^{3}).$

In this case, Theorem 4.4 implies that $X\otimes^{\zeta}Y\cong Z\oplus
TZ\oplus T^{2}Z.$

##  5\. Direct sum decompositions of tensor products

Following the work of [Yos98], we will consider tensor products of matrix
factorizations of polynomials $f$ and $g$ which are written in disjoint sets
of variables. In this setting, we are able to prove several decomposition
results, most of which are direct generalizations of the results shown in
[Yos98]. We begin by defining the key functor used in this section.

######  Definition 5.1.

Let $Q$ be a commutative ring, $d\geq 2$, and $f\in Q$. If
$(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$, and $I$ is an ideal of $Q$, then
we let $X_{I}$ denote the matrix factorization

$X_{I}\coloneqq(X\otimes_{Q}Q/I,\varphi\otimes_{Q}1_{Q/I})\in\operatorname{MF}_{Q/I}^{d}(f+I).$

Notice that $\operatorname{rank}X=\operatorname{rank}X_{I}$, if both ranks are
well-defined. Furthermore, reduction modulo $I$ forms a functor

$(-)_{I}:\operatorname{MF}_{Q}^{d}(f)\to\operatorname{MF}_{Q/I}^{d}(f+I).$

We will adopt the following notation throughout the rest of this section.

######  Notation 5.2.

Let $d\geq 2$ be an integer and let $\Bbbk$ be a field containing a primitive
$d^{\textit{th}}$ root of unity, $\zeta\in\Bbbk$. Fix the power series rings

  1. (i)

$S_{1}=\Bbbk\llbracket x_{1},x_{2},\dots,x_{a}\rrbracket$,

  2. (ii)

$S_{2}=\Bbbk\llbracket y_{1},y_{2},\dots,y_{b}\rrbracket$, and

  3. (iii)

$S=S_{1}\otimes_{\Bbbk}S_{2}\cong\Bbbk\llbracket
x_{1},\dots,x_{a},y_{1},\dots,y_{b}\rrbracket$,

and fix non-invertible non-zero elements $f\in S_{1}$ and $g\in S_{2}$. By
tensoring with $S$ we obtain faithful embeddings
$\operatorname{MF}_{S_{1}}^{d}(f)\hookrightarrow\operatorname{MF}_{S}^{d}(f)$
and
$\operatorname{MF}_{S_{2}}^{d}(g)\hookrightarrow\operatorname{MF}_{S}^{d}(g)$.
We will identify a factorization $X\in\operatorname{MF}_{S_{1}}^{d}(f)$ with
its image in $\operatorname{MF}_{S}^{d}(f)$, and similarly for
$Y\in\operatorname{MF}_{S_{2}}^{d}(g)$, in order to form the tensor product
$X\otimes Y\in\operatorname{MF}_{S}^{d}(f+g)$. Together with reduction modulo
$(x)\coloneqq S(x_{1},x_{2},\dots,x_{a})$ and reduction modulo $(y)\coloneqq
S(y_{1},y_{2},\dots,y_{b})$, we get the following diagram of functors:

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The interaction between these functors will be key going forward. Recall that
a matrix factorization $(X,\varphi)$, over a local ring $(Q,\mathfrak{n})$, is
said to be reduced if
$\operatorname{Im}\varphi_{k}\subseteq\mathfrak{n}X_{k-1}$ for each
$k\in\mathbb{Z}_{d}$. In other words, after choosing bases, the entries of the
matrix representing $\varphi_{k},k\in\mathbb{Z}_{d}$, live in the maximal
ideal of $Q$.

######  Proposition 5.3.

Let $(X,\varphi)\in\operatorname{MF}_{S_{1}}^{d}(f)$ and
$(Y,\psi)\in\operatorname{MF}_{S_{2}}^{d}(g)$.

  1. (i)

Suppose $X\in\operatorname{MF}_{S_{1}}^{d}(f)$ is reduced of rank $n$. Then
there is an isomorphism in $\operatorname{MF}_{S_{2}}^{d}(g)$

$(X\otimes Y)_{x}\cong\bigoplus_{i=1}^{d}(\zeta^{i-1}T^{1-i}Y)^{n}.$

  2. (ii)

Suppose $Y\in\operatorname{MF}_{S_{2}}^{d}(g)$ is reduced of rank $m$. Then
there is an isomorphism in $\operatorname{MF}_{S_{1}}^{d}(f)$

$(X\otimes Y)_{y}\cong\bigoplus_{i=1}^{d}(\zeta^{1-i}T^{1-i}X)^{m}.$

###### Proof.

Since $(X,\varphi)$ is reduced, the entries of $\varphi$ live in the ideal
$(x)$. This implies that $\varphi\otimes 1_{S/(x)}=0$. On the other hand,
since $\psi$ has entries in $S_{2}$, we may identify $\psi_{k}\otimes
1_{S/(x)}$ with $\psi_{k}$ and so, for any $j,k\in\mathbb{Z}_{d}$, we have

$(1_{X_{j}}\otimes\psi_{k})\otimes_{S}1_{S/(x)}\cong\psi_{k}^{\oplus n}.$

Considering (3.1), we obtain an isomorphism

$(X\otimes Y)_{x}\cong\bigoplus_{i=1}^{d}(\zeta^{i-1}T^{1-i}Y)^{n}.$

The second isomorphism follows similarly by considering the isomorphism
$X\otimes^{\zeta}Y\cong Y\otimes^{\zeta^{-1}}X$ given in Proposition 3.4. ∎

The powers of $\zeta$ can also be omitted from these isomorphisms by Lemma
2.4. In particular, this lemma is the $d$-fold analogue of [Yos98, Lemma 2.9].

In order to investigate the number of indecomposable direct summands of
$X\otimes Y$, we will work directly with idempotents in the endomorphism ring
of $X\otimes Y$. The next lemma shows how the functors $(-)_{x}$ and $(-)_{y}$
interact with morphisms of this form and more generally how they interact with
morphisms between tensor products.

######  Lemma 5.4.

  1. (i)

Assume $X\in\operatorname{MF}_{S_{1}}^{d}(f)$ is reduced of rank $n$,
$Y,Y^{\prime}\in\operatorname{MF}_{S_{2}}^{d}(g)$, and
$\alpha\in\textup{Hom}(X\otimes Y,X\otimes Y^{\prime})$. Then, for
$i,j\in\mathbb{Z}_{d}$, the $(i,j)$-component of the induced morphism

$\alpha_{x}:\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y)^{n}\to\bigoplus_{i=1}^{d}(\zeta^{i-1}T^{1-i}Y^{\prime})^{n}$

is given by the morphism

$\alpha_{x}(i,j)=(\alpha_{0}(i,j)_{x},\alpha_{1}(i,j)_{x},\dots,\alpha_{d-1}(i,j)_{x})=(\alpha_{k}(i,j)_{x})_{k=0}^{d-1}$

in $\operatorname{MF}_{S_{2}}^{d}(g)$.

  2. (ii)

Assume $Y\in\operatorname{MF}_{S_{2}}^{d}(g)$ is reduced of rank $m$,
$X,X^{\prime}\in\operatorname{MF}_{S_{1}}^{d}(f)$, and
$\beta\in\textup{Hom}(X\otimes Y,X^{\prime}\otimes Y)$. Then, for
$i,j\in\mathbb{Z}_{d}$, the $(i,j)$-component of the induced morphism

$\beta_{y}:\bigoplus_{j=1}^{d}(T^{1-j}X)^{m}\to\bigoplus_{i=1}^{d}(T^{1-i}X^{\prime})^{m}$

is given by the morphism

$\beta_{y}(i,j)=(\beta_{k}(2-i+k,2-j+k)_{y})_{k=0}^{d-1}$

in $\operatorname{MF}_{S_{1}}^{d}(f)$.

###### Proof.

The first statement follows directly from (3.1) and Proposition 5.3. To see
that (ii) holds, we introduce notation. Let $C$ be a $d\times d$ permutation
matrix corresponding to a permutation $\sigma$ of $\\{1,2,\dots,d\\}$. We
allow $C$ to act on direct sums of modules in the following way: if
$H=\bigoplus_{i=1}^{d}H_{i}$, then we set
$CH=\bigoplus_{i=1}^{d}H_{\sigma(i)}$ and view $C$ as a homomorphism $C:H\to
CH$ given by
$(h_{1},h_{2},\dots,h_{d})\mapsto(h_{\sigma(1)},h_{\sigma(2)},\dots,h_{\sigma(d)})$.

We will consider two such permutation matrices to justify the formula for
$\beta_{y}$:

$C=\scalebox{0.8}{$\begin{pmatrix}0&0&\cdots&0&1\\\ 1&0&\ddots&0&0\\\
0&1&\ddots&0&0\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&0\end{pmatrix}$}\quad\text{ and }\quad
B=\scalebox{0.8}{$\begin{pmatrix}1&0&\cdots&0&0\\\ 0&0&\ddots&0&1\\\
0&0&\ddots&1&0\\\ \vdots&\vdots&\ddots&\ddots\\\
0&1&\cdots&0&0\end{pmatrix}$}.$

If $h:\bigoplus_{j=1}^{d}H_{j}\to\bigoplus_{i=1}^{d}H_{i}^{\prime}$ is a
homomorphism, then conjugation by $C^{k}$, $k\in\mathbb{Z}_{d}$, cycles the
components of $h$; for any $i,j\in\mathbb{Z}_{d}$,

$(C^{-k}hC^{k})(i,j)=h(i+k,j+k):H_{j+k}\to H_{i+k}^{\prime}.$

On the other hand, noting that $B=B^{-1}$, conjugation by $B$ gives

$(BhB)(i,j)=h(2-i,2-j):H_{2-j}\to H_{2-i}^{\prime}.$

With this set up, we write $\beta_{y}$ with respect to the direct sum
decomposition (3.1), and apply these permutations to the components of
$\beta_{y}$. One can then check directly that $\beta_{y}$ induces the morphism

$\left(B(\beta_{0})_{y}B,BC^{-1}(\beta_{1})_{y}CB,\dots,BC^{-(d-1)}(\beta_{d-1})_{y}C^{d-1}B\right)$

from
$\bigoplus_{j=1}^{d}(T^{1-j}X)^{m}\to\bigoplus_{i=1}^{d}(T^{1-i}X^{\prime})^{m}$.
For any $i,j\in\mathbb{Z}_{d}$, the $(i,j)$-component of this morphism is

$\displaystyle\left((BC^{-k}(\beta_{k})_{y}C^{k}B)(i,j)\right)_{k=0}^{d-1}$
$\displaystyle=\left((C^{-k}(\beta_{k})_{y}C^{k})(2-i,2-j)\right)_{k=0}^{d-1}$
$\displaystyle=\left((\beta_{k})_{y}(2-i+k,2-j+k)\right)_{k=0}^{d-1}$
$\displaystyle=\left(\beta_{k}(2-i+k,2-j+k)_{y}\right)_{k=0}^{d-1}$

which is the formula claimed in (ii).

∎

###  5.1. Reduced matrix factorizations

In this section, we will restrict our attention to tensor products of reduced
matrix factorizations only.

######  Notation 5.5.

In addition to the notation given in (5.2), we will assume that
$X\in\operatorname{MF}_{S_{1}}^{d}(f)$ and
$Y\in\operatorname{MF}_{S_{2}}^{d}(g)$ are indecomposable reduced matrix
factorizations of $f$ and $g$ respectively. We denote the number of
indecomposable summands (counted with multiplicity) in the direct sum
decomposition of $X\otimes Y$ by $\\#(X\otimes Y)$. This quantity is well
defined since the Krull-Schmidt Theorem holds in
$\operatorname{MF}_{S}^{d}(f+g)$ when $S$ is complete as it is in this section
(see [Tri22, Section 3]). We also set $r=\gcd(n,m)$ where
$n=\operatorname{rank}X$ and $m=\operatorname{rank}Y$.

Our first result is a direct extension of [Yos98, Theorem 3.3].

######  Theorem 5.6.

The tensor product $X\otimes Y$ has at most $dr$ indecomposable summands.

###### Proof.

Let $Z\in\operatorname{MF}_{S}^{d}(f+g)$ be a summand of $X\otimes Y$. By
Lemma 5.3, $Z_{y}$ is isomorphic to a summand of
$\bigoplus_{i\in\mathbb{Z}_{d}}(T^{i}X)^{m}$ (we omit the powers of $\zeta$
using Lemma 2.4). Since $T^{i}X$ is indecomposable for each
$i\in\mathbb{Z}_{d}$, the Krull-Schmidt property of
$\operatorname{MF}_{S_{1}}^{d}(f)$ implies that
$Z_{y}\cong\bigoplus_{i\in\mathbb{Z}_{d}}(T^{i}X)^{t_{i}}$ for some integers
$0\leq t_{i}\leq m$. Hence, the rank of $Z_{y}$, which is the same as the rank
of $Z$, is $(t_{0}+\dots+t_{d-1})n$. Similarly, reduction modulo $(x)$ gives
us that the rank of $Z$ is equal to $(s_{0}+\dots+s_{d-1})m$ for some integers
$0\leq s_{i}\leq n$. Together, these imply that
$(t_{0}+\dots+t_{d-1})n=(s_{0}+\dots+s_{d-1})m$ which must be at least
$\operatorname{lcm}(n,m)=nm/r$. Since $X\otimes Y$ is of rank $dnm$ and we
have just shown that any summand must be of rank at least $nm/r$, it follows
that $\\#(X\otimes Y)\leq dr$. ∎

This upper bound can be improved if the factorizations are highly non-
symmetric (cf. [Yos98, Theorem 3.4]).

######  Theorem 5.7.

Suppose $X\not\cong T^{i}X$ and $Y\not\cong T^{j}Y$ for all $i,j\neq 0$. Then
$\\#(X\otimes Y)\leq r$.

The proof of Theorem 5.7 will require some preparation.

####  5.1.1. The radical of an additive category

Let $\mathcal{C}$ be a Krull-Schmidt category, that is, an additive category
such that every object decomposes into a finite sum of objects which have
local endomorphism rings. We refer the reader to [Kra15] for details. For
objects $X,Y\in\mathcal{C}$, the radical of $X$ and $Y$ is defined to be

$\operatorname{rad}_{\mathcal{C}}(X,Y)=\big{\\{}h\in\textup{Hom}_{\mathcal{C}}(X,Y):1_{Z}-ghk\text{
is invertible for all
}g\in\textup{Hom}_{\mathcal{C}}(Y,Z),k\in\textup{Hom}_{\mathcal{C}}(Z,X)\big{\\}}.$

In particular, the Jacobson radical of the ring
$\operatorname{End}_{\mathcal{C}}(X)$ coincides with
$\operatorname{rad}_{\mathcal{C}}(X,X)$. We recall several useful facts about
$\operatorname{rad}=\operatorname{rad}_{\mathcal{C}}$. Note that, while
working with objects in the abstract category $\mathcal{C}$, our indices are
not taken modulo $d$.

######  Lemma 5.8 ([Kra15]).

Let $X$ and $Y$ be objects in $\mathcal{C}$.

  1. (i)

Suppose $X=X_{1}^{n_{1}}\oplus\cdots\oplus X_{t}^{n_{t}}$ and
$Y=Y_{1}^{m_{1}}\oplus\cdots\oplus Y_{s}^{m_{s}}$ for indecomposable objects
$X_{i},Y_{j}$, and positive integers $n_{i},m_{j}$, for $1\leq i\leq t,1\leq
j\leq s$. Then
$\operatorname{rad}(X,Y)\cong\bigoplus_{i,j}\operatorname{rad}(X_{i}^{n_{i}},Y_{j}^{m_{j}})$.

  2. (ii)

Suppose that $X$ and $Y$ are indecomposable objects such that $X\not\cong Y$.
Then $\operatorname{rad}(X^{n},Y^{m})=\textup{Hom}(X^{n},Y^{m})$ for any
$n,m\geq 1$. ∎

######  Lemma 5.9.

Let $X_{1},\dots,X_{m}$ be indecomposable objects in $\mathcal{C}$ and set
$X=X_{1}^{n_{1}}\oplus\cdots\oplus X_{m}^{n_{m}}$ for positive integers
$n_{1},\dots,n_{m}$. Let $e=(e(i,j))_{i,j}$ be an idempotent in
$\operatorname{End}(X)$, where $e(i,j):X_{j}^{n_{j}}\to X_{i}^{n_{i}}$. If
$e(i,k)e(k,j)\in\operatorname{rad}(X_{j}^{n_{j}},X_{i}^{n_{i}})$ for all
$i,j,k$ where either $i\neq k$ or $j\neq k$, then there exist idempotents
$e_{i}\in\operatorname{End}(X_{i}^{n_{i}})$, $1\leq i\leq m$, such that
$e(i,i)-e_{i}\in\operatorname{rad}\operatorname{End}(X_{i}^{n_{i}})$ and
$e(X)\cong\bigoplus_{i=1}^{m}e_{i}(X_{i}^{n_{i}})$, where $e(X)$ denotes the
direct summand of $X$ given by the idempotent $e$.

###### Proof.

Since $e=e^{2}$, for each $1\leq i\leq m$, we have that

$e(i,i)=e^{2}(i,i)=\sum_{k=1}^{m}e(i,k)e(k,i).$

Since $e(i,k)e(k,i)\in\operatorname{rad}\operatorname{End}(X_{i}^{n_{i}})$ for
all $i\neq k$ by assumption, we have that

$e(i,i)\equiv
e(i,i)^{2}\mod\operatorname{rad}\operatorname{End}(X_{i}^{n_{i}}).$

Since $\mathcal{C}$ is a Krull-Schmidt category, we may lift $e(i,i)$ to an
idempotent in $\operatorname{End}(X_{i}^{n_{i}})$, that is, there exists an
idempotent $e_{i}\in\operatorname{End}(X_{i}^{n_{i}})$ such that
$e(i,i)-e_{i}\in\operatorname{rad}(X_{i}^{n_{i}},X_{i}^{n_{i}})$. Let
$\epsilon=\bigoplus_{i=1}^{m}e_{i}\in\operatorname{End}(X)$ and set
$\gamma=e-\epsilon$. We claim that
$\gamma^{2}\in\operatorname{rad}\operatorname{End}(X)$. To see this, first
notice that for each $i$,

$\gamma^{2}(i,i)=\sum_{k\neq
i}e(i,k)e(k,i)+(e(i,i)-e_{i})^{2}\in\operatorname{rad}(X_{i}^{n_{i}},X_{i}^{n_{i}}).$

If $i\neq j$, then

$\gamma^{2}(i,j)=(e(i,i)-e_{i})e(i,j)+e(i,j)(e(j,j)-e_{j})+\sum_{k\neq i,k\neq
j}e(i,k)e(k,j).$

Since $\operatorname{rad}_{\mathcal{C}}$ is an ideal in the category
$\mathcal{C}$, the fact that
$e(i,i)-e_{i}\in\operatorname{rad}(X_{i}^{n_{i}},X_{i}^{n_{i}})$ implies that
$(e(i,i)-e_{i})e(i,j)\in\operatorname{rad}(X_{j}^{n_{j}},X_{i}^{n_{i}})$.
Similarly, we have that
$e(i,j)(e(j,j)-e_{j})\in\operatorname{rad}(X_{j}^{n_{j}},X_{i}^{n_{i}})$.
Since the rest of the terms are in
$\operatorname{rad}(X_{j}^{n_{j}},X_{i}^{n_{i}})$ by assumption, we have that
$\gamma^{2}(i,j)\in\operatorname{rad}(X_{j}^{n_{j}},X_{i}^{n_{i}})$ for all
$i,j$. Lemma 5.8(i) then implies that
$\gamma^{2}\in\operatorname{rad}\operatorname{End}(X)$ as claimed.

We can now finish the proof of the lemma. Since
$\gamma^{2}\in\operatorname{rad}\operatorname{End}(X)$, we have that
$1_{X}-\gamma^{2}=(1_{X}+\gamma)(1_{X}-\gamma)$ is a unit in
$\operatorname{End}(X)$. Hence, both $(1_{X}-\gamma)$ and $(1_{X}+\gamma)$ are
units in $\operatorname{End}(X)$. Note that, since $e$ and $\epsilon$ are
idempotents and $\gamma=e-\epsilon$, we have

$(1_{X}+\gamma)\epsilon=\epsilon^{2}+\gamma\epsilon=(\epsilon+\gamma)\epsilon=e\epsilon=e(e-\gamma)=e-e\gamma=e(1_{X}-\gamma).$

That is, $e$ and $\epsilon$ are equivalent idempotents. It follows that
$e(X)\cong\epsilon(X)=\bigoplus_{i=1}^{m}e_{i}(X_{i}^{n_{i}})$. ∎

If, in addition, the indecomposable objects $X_{1},X_{2},\dots,X_{m}$ are
pairwise non-isomorphic, then Lemma 5.8(ii) implies that the assumptions of
the lemma are automatically satisfied. With this, we are able to prove the
Theorem 5.7. Our approach is a direct generalization of [Yos98, Theorem 3.4].

###### Proof of Theorem 5.7.

Suppose $\epsilon\in\operatorname{End}(X\otimes Y)$ is an idempotent. Applying
Lemma 5.3(i), we obtain an idempotent
$\epsilon_{x}\in\textup{Hom}(\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y))$. By
assumption, the indecomposable matrix factorizations $\zeta^{j-1}T^{1-j}Y\cong
T^{1-j}Y,j\in\mathbb{Z}_{d}$, are pairwise non-isomorphic. From Lemma 5.9, we
obtain idempotents $e_{j}\in\operatorname{End}(\zeta^{j-1}T^{1-j}Y)^{n}$,
$j\in\mathbb{Z}_{d}$, such that

$\epsilon_{x}(j,j)-e_{j}\in\operatorname{rad}\operatorname{End}(\zeta^{j-1}T^{1-j}Y)^{n})$

and $\epsilon_{x}((X\otimes
Y)_{x})\cong\bigoplus_{j=1}^{d}e_{j}(\zeta^{j-1}T^{1-j}Y)^{n})$. Since
$T^{1-j}Y$ is indecomposable for each $j\in\mathbb{Z}_{d}$, the Krull-Schmidt
theorem implies

(5.1)
$\bigoplus_{j=1}^{d}e_{j}(\zeta^{j-1}T^{1-j}Y)^{n})\cong\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y)^{t_{j}}$

for some integers $0\leq t_{j}\leq n$, $j\in\mathbb{Z}_{d}$.

Fix $j\in\mathbb{Z}_{d}$. We may apply Lemma 2.3 to obtain an equivalent
idempotent

$e_{j}\sim\begin{pmatrix}1_{t_{j}}&0\\\ 0&0\end{pmatrix}$

where $1_{t_{j}}$ denotes the identity morphism on the first summand of

$(\zeta^{j-1}T^{1-j}Y)^{n}=(\zeta^{j-1}T^{1-j}Y)^{t_{j}}\oplus(\zeta^{j-1}T^{1-j}Y)^{n-t_{j}}.$

Then, since
$\epsilon_{x}(j,j)-e_{j}\in\operatorname{rad}\operatorname{End}(\zeta^{j-1}T^{1-j}Y)^{n})$,
we use Lemma 2.2 to clear columns and rows and obtain

$\epsilon_{x}(j,j)\sim\begin{pmatrix}1_{t_{j}}&0\\\ 0&A_{j}\end{pmatrix}$

for some
$A_{j}\in\operatorname{rad}\operatorname{End}((\zeta^{j-1}T^{1-j}Y)^{n-t_{j}})$.
Putting this together for all $j\in\mathbb{Z}_{d}$, and applying Lemma 2.2
again, we obtain

$\epsilon_{x}\sim\begin{pmatrix}1&0\\\ 0&B\end{pmatrix}$

where $1$ is the identity morphism on
$\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y)^{t_{j}}$ and $B$ is a morphism in

$\operatorname{End}(\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y)^{n-t_{j}})$

satisfying $B(j,j)=A_{j}$ for each $j\in\mathbb{Z}_{d}$. This implies that
there is an isomorphism of free $Q$-modules,
$\operatorname{Im}\epsilon_{x}\cong\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y)^{t_{j}}\oplus\operatorname{Im}B$.
However, we know that $\epsilon_{x}((X\otimes
Y)_{x})\cong\bigoplus_{j=1}^{d}(\zeta^{j-1}T^{1-j}Y)^{t_{j}}$ and therefore,
counting ranks, $B$ must be zero. Hence $B(j,j)=A_{j}=0$ for each
$j\in\mathbb{Z}_{d}$. In other words,

$\epsilon_{x}(j,j)\sim\begin{pmatrix}1_{t_{j}}&0\\\ 0&0\end{pmatrix}\sim
e_{j}$

for each $j\in\mathbb{Z}_{d}$. Now, recall that the $(i,j)$-component of
$\epsilon_{x}$ is described by Lemma 5.4(i). In particular, for
$j\in\mathbb{Z}_{d}$,

$\epsilon_{x}(j,j)=(\epsilon_{0}(j,j)_{x},\epsilon_{1}(j,j)_{x},\dots,\epsilon_{d-1}(j,j)_{x}).$

It follows that $\epsilon_{i}(j,j)_{x}\sim\begin{pmatrix}I_{nt_{j}}&0\\\
0&0\end{pmatrix}$, for each $i\in\mathbb{Z}_{d}$, where $I_{nt_{j}}$ is an
identity matrix of rank $nt_{j}$. This allows us to make the following rank
count:

(5.2)
$\operatorname{rank}_{\Bbbk}(\epsilon_{i}(j,j)\otimes_{S}\Bbbk)=\operatorname{rank}_{\Bbbk}(\epsilon_{i}(j,j)_{x}\otimes_{S}\Bbbk)=nt_{j}\text{
for each }i,j\in\mathbb{Z}_{d}.$

Applying the same steps to $\epsilon_{y}$, this time using Lemma 5.4(ii), we
find that, for each $j\in\mathbb{Z}_{d}$, there exists an integer $0\leq
s_{j}\leq m$ such that

$\epsilon_{i}(2-j+i,2-j+i)_{y}\sim\begin{pmatrix}I_{ms_{j}}&0\\\
0&0\end{pmatrix}$

for all $i\in\mathbb{Z}_{d}$. In other words,

(5.3)
$\operatorname{rank}_{\Bbbk}(\epsilon_{i}(2-j+i,2-j+i)\otimes_{S}\Bbbk)=ms_{j}\text{
for each }i,j\in\mathbb{Z}_{d}.$

Now, let $j\in\mathbb{Z}_{d}$ and set $i=2j-2$. Then (5.2) and (5.3) imply
that

$nt_{j}=\operatorname{rank}_{\Bbbk}(\epsilon_{2j-2}(j,j)\otimes_{S}\Bbbk)=ms_{j}.$

On the other hand, if $i=2j-1$, then (5.2) and (5.3) imply that

$nt_{j+1}=\operatorname{rank}_{\Bbbk}(\epsilon_{2j-1}(j+1,j+1)\otimes_{S}\Bbbk)=ms_{j}.$

Together, these imply that $nt_{i}=ms_{j}$ for all $i,j\in\mathbb{Z}_{d}$, and
that $t\coloneqq t_{1}=t_{2}=\dots=t_{d}$ and $s\coloneqq
s_{1}=s_{2}=\dots=s_{d}$. This implies that the matrix factorization
$e(X\otimes Y)$ has rank $\sum_{j=1}^{d}nt_{j}=dnt$ which is also equal to
$\sum_{j=1}^{d}ms_{j}=dms$. Hence, the rank of the summand $e(X\otimes Y)$
must be at least $\operatorname{lcm}(dn,dm)=dnm/r$. Since
$\operatorname{rank}X\otimes Y=dnm$, and any summand must have rank at least
$dnm/r$, we have $\\#(X\otimes Y)\leq r$. ∎

In the following we will make repeated use of the following consequence of
Nakayama’s lemma: If $h:G\to F$ is homomorphism between finitely generated
free $S$-modules of the same rank, then $h$ is an isomorphism if and only if
$h\otimes 1_{S_{1}}$ is an isomorphism. The same holds for $h\otimes
1_{S_{2}}$, and in particular, $h,h\otimes 1_{S_{1}}$, and $h\otimes
1_{S_{2}}$ are isomorphisms simultaneously.

######  Theorem 5.10.

Let $\\{u_{1},u_{2},\dots,u_{d-1},u_{0}\\}$ be pairwise relatively prime
elements of the maximal ideal of $S_{1}$ and set
$X=(u_{1},u_{2},\dots,u_{d-1},u_{0})\in\operatorname{MF}_{S_{1}}^{d}(u_{1}u_{2}\cdots
u_{d-1}u_{0})$. If $Y\in\operatorname{MF}_{S_{2}}^{d}(g)$ satisfies $Y\cong
TY$, then $X\otimes Y$ is indecomposable.

###### Proof.

Notice that since $u_{i},u_{j}$ are relatively prime for $i\neq j$, if
$\alpha:(T^{j}X)^{m}\to(T^{i}X)^{m}$ is a morphism, then
$\alpha_{k}\otimes_{S_{1}}\Bbbk=0$ for all $k\in\mathbb{Z}_{d}$. Indeed, if
there exists $S_{1}$-homomorphisms satisfying $\beta u_{i}=u_{j}\gamma$, then
neither of $\beta$ nor $\gamma$ can have unit entries.

Set $Z=X\otimes Y$ and let $\epsilon\in\operatorname{End}(Z)$ be an
idempotent. It suffices to show that either $\epsilon=0$ or $\epsilon=1_{Z}$.
We first consider reduction modulo $(y)$ which gives us an idempotent
$\epsilon_{y}\in\operatorname{End}(Z_{y})\cong\operatorname{End}(\bigoplus_{i=1}^{d}(T^{1-i}X)^{m})$.
By the initial observation, for any $i\neq j$, the components of
$\epsilon_{y}(i,j)$ have entries only in the maximal ideal of $S_{1}$. More
specifically, Lemma 5.4(ii) gives us that

$\epsilon_{y}(i,j)=(\epsilon_{k}(2-i+k,2-j+k)_{y})_{k=0}^{d-1},$

and therefore, $\epsilon_{k}(2-i+k,2-j+k)_{y}\otimes\Bbbk=0$ for any $i\neq j$
and $k\in\mathbb{Z}_{d}$. Letting $i\neq j$ and $k$ vary over
$\mathbb{Z}_{d}$, and lifting back to $S$, we have

(5.4) $\epsilon_{k}(i,j)\otimes\Bbbk=0\text{ for all
}i,j,k\in\mathbb{Z}_{d}\text{ with }i\neq j.$

Next, we consider reduction modulo $(x)$. Note that since
$\operatorname{rank}X=1$, $\epsilon_{x}$ is an idempotent in
$\operatorname{End}(Z_{x})\cong\operatorname{End}(\bigoplus_{i=1}^{d}\zeta^{i-1}T^{1-i}Y)$.
Furthermore, since $TY\cong Y$, we have that $Z_{x}$ is isomorphic to a direct
sum of $d$ copies of $Y$. Let $\Lambda=\operatorname{End}(Y)$ which is a
(likely non-commutative) local ring in the sense that
$\Lambda/\operatorname{rad}\Lambda$ is a division ring.

By (5.4), $\epsilon_{k}(i,j)_{x}$ is not an isomorphism for all
$i,j,k\in\mathbb{Z}_{d}$ with $i\neq j$. It follows from Lemma 5.4(i) that
$\epsilon_{x}(i,j)$ is not an isomorphism of matrix factorizations for all
$i\neq j$. Since $\Lambda$ is local, this implies that
$\epsilon_{x}(i,j)\in\operatorname{rad}\Lambda$ for all $i\neq j$, and in
particular, $\epsilon_{x}(i,k)\epsilon_{x}(k,j)\in\operatorname{rad}\Lambda$
for all $i,j,k$ such that $i\neq k$ or $j\neq k$. Hence Lemma 5.9 applies; the
diagonal component $\epsilon_{x}(i,i)$ is an idempotent in the division ring
$\Lambda/\operatorname{rad}\Lambda$ for each $i\in\mathbb{Z}_{d}$. In other
words, for each $i\in\mathbb{Z}_{d}$, $\epsilon_{x}(i,i)$ is either an
automorphism of $Y$ or in $\operatorname{rad}\Lambda$.

First, consider the case when $\epsilon_{x}(i,i)\in\operatorname{rad}\Lambda$
for all $i\in\mathbb{Z}_{d}$. Since we showed that
$\epsilon_{x}(i,j)\in\operatorname{rad}\Lambda$ for all $i\neq j$ above, this
implies that $\epsilon_{x}\in\operatorname{rad}\operatorname{End}(Z_{x})$. An
idempotent in the radical must be zero and so we conclude that
$\epsilon_{x}=0$ in this case. It follows that
$\epsilon\in\operatorname{rad}\operatorname{End}(Z)$ and, using the same
reasoning, we have $\epsilon=0$.

Next, assume $\epsilon_{x}(q,q)$ is an automorphism for some
$q\in\mathbb{Z}_{d}$. We will prove that this implies that $\epsilon_{x}(i,i)$
is an automorphism for all $i\in\mathbb{Z}_{d}$. Since $\epsilon_{x}(q,q)$ is
an isomorphism, Lemma 5.4(i) implies that $\epsilon_{k}(q,q)_{x}$ is an
isomorphism for all $k\in\mathbb{Z}_{d}$ which in turn implies that
$\epsilon_{k}(q,q)$ is an isomorphism for all $k\in\mathbb{Z}_{d}$.

Let $i,j\in\mathbb{Z}_{d}$. To finish the proof, it suffices to show that
$\epsilon_{i}(j,j)$ is an isomorphism. In this case, combined with (5.4), this
would imply that the idempotent $\epsilon$ is an isomorphism (as it would be a
unit plus a radical element) which is only possible if it is the identity
morphism on $Z$.

To show the final claim, we return for a final time to reduction modulo $(y)$.
Since $X$ has rank one, any endomorphism
$\beta=(\beta_{0},\beta_{1},\dots,\beta_{d-1})\in\operatorname{End}((T^{1-p}X)^{m}),p\in\mathbb{Z}_{d}$,
has the property that $\beta_{0}=\beta_{1}=\cdots=\beta_{d-1}$. Thus, invoking
Lemma 5.4(ii) again, we have

$\epsilon_{i}(j,j)_{y}=\epsilon_{i+t}(2-j+t,2-j+t)_{y}\text{ for any
}t\in\mathbb{Z}_{d}.$

In particular, for $t=q+j-2$, we have
$\epsilon_{i}(j,j)_{y}=\epsilon_{i+t}(q,q)_{y}$, which is an isomorphism by
assumption. By Nakayama’s lemma, $\epsilon_{i}(j,j)$ is an isomorphism and
this completes the proof. ∎

Using similar techniques one can also show the following.

######  Theorem 5.11.

Let
$Y=(u_{1},u_{2},\dots,u_{d-1},u_{0})\in\operatorname{MF}_{S_{2}}^{d}(u_{1}u_{2}\cdots
u_{d-1}u_{0})$ be a rank one factorization. If
$X\in\operatorname{MF}_{S_{1}}^{d}(f)$ satisfies $T^{i}(X)\not\cong X$ for all
$i\neq 0$ in $\mathbb{Z}_{d}$, then $X\otimes Y$ is indecomposable. ∎

###  5.2. Strongly indecomposable matrix factorizations

In this section we obtain additional indecomposability results by introducing
the notion of strongly indecomposable222An object $X$ is sometimes called
strongly indecomposable if it’s endomorphism ring is local. Our definition
implies this but is stronger in general. matrix factorizations. We continue
using the assumptions and notations in (5.5).

######  Definition 5.12.

Let $(X,\varphi)\in\operatorname{MF}_{S}^{d}(f)$. The factorization $X$ is
said to be strongly indecomposable if it satisfies the following two
conditions.

  1. (i)

If $(\alpha,\beta)$ is a pair of $S$-homomorphisms satisfying
$\alpha\varphi_{j}=\varphi_{k}\beta$ for some $j\neq k$, then
$\alpha\otimes\Bbbk=\beta\otimes\Bbbk=0$.

  2. (ii)

If $(\alpha,\beta)$ is a pair of $S$-homomorphisms satisfying
$\alpha\varphi_{k}=\varphi_{k}\beta$ for some $k\in\mathbb{Z}_{d}$, then
$\alpha\otimes\Bbbk=\xi\cdot 1_{(X_{k-1}\otimes\Bbbk)}$ and
$\beta\otimes\Bbbk=\xi\cdot 1_{(X_{k}\otimes\Bbbk)}$ for some $\xi\in\Bbbk$.

In general, the assumption that a factorization is strongly indecomposable is
significantly more restrictive than indecomposability. The lemma below
illustrates this fact. However, in special cases of interest, strongly
indecomposable matrix factorizations are not hard to find. In particular, the
main goal of this section is to show that the tensor product construction
preserves strong indecomposability.

######  Lemma 5.13.

Suppose $(X,\varphi)\in\operatorname{MF}_{S}^{d}(f)$ is strongly
indecomposable and set $R=S/(f)$. Then the following hold.

  1. (i)

$X$ is an indecomposable matrix factorization.

  2. (ii)

$T^{i}X\not\cong X$ for all $i\neq 0$ in $\mathbb{Z}_{d}$.

  3. (iii)

$\operatorname{cok}\varphi_{i}$ is an indecomposable (maximal Cohen-Macaulay)
$R$-module for all $i\in\mathbb{Z}_{d}$.

  4. (iv)

$\operatorname{End}(X)/\operatorname{rad}\operatorname{End}(X)\cong\Bbbk$ and
$\operatorname{End}_{R}(\operatorname{cok}\varphi_{i})/\operatorname{rad}\operatorname{End}_{R}(\operatorname{cok}\varphi_{i})\cong\Bbbk$
for all $i\in\mathbb{Z}_{d}$.

∎

Items (iii) and (iv) of Lemma 5.13 are especially important consequences of
strong indecomposability since, in general, the modules associated to an
indecomposable $d$-fold matrix factorization (for $d>2$) need not be
indecomposable (see [LT23, Example 5.5]). We will utilize Lemma 5.13 along
with the following theorem to produce indecomposable maximal Cohen-Macaulay
and Ulrich modules in Section 6.

######  Theorem 5.14.

Let $X\in\operatorname{MF}_{S_{1}}^{d}(f)$ and
$Y\in\operatorname{MF}_{S_{2}}^{d}(g)$ be strongly indecomposable matrix
factorizations. Then $X\otimes Y$ is strongly indecomposable.

To prove Theorem 5.14 we need a lemma. Note that the faithful embeddings
$\operatorname{MF}_{S_{1}}^{d}(f)\hookrightarrow\operatorname{MF}_{S}^{d}(f)$
and
$\operatorname{MF}_{S_{2}}^{d}(g)\hookrightarrow\operatorname{MF}_{S}^{d}(g)$
preserve strong indecomposability.

######  Lemma 5.15.

Let $X\in\operatorname{MF}_{S_{1}}^{d}(f)$ and
$Y\in\operatorname{MF}_{S_{2}}^{d}(g)$ be strongly indecomposable of rank $n$
and $m$, respectively, and let $j,k\in\mathbb{Z}_{d}$.

  1. (i)

Assume $j\neq k$. If $\alpha$ and $\beta$ are $S$-homomorphisms satisfying
$\alpha(\varphi_{j}\otimes 1_{Y_{p}})=(\varphi_{k}\otimes
1_{Y_{p^{\prime}}})\beta$ for some $p,p^{\prime}\in\mathbb{Z}_{d}$, then
$\alpha\otimes\Bbbk=\beta\otimes\Bbbk=0$. Similarly, if
$\alpha(1_{X_{p}}\otimes\psi_{j})=(1_{X_{p^{\prime}}}\otimes\psi_{k})\beta$
for some $p,p^{\prime}\in\mathbb{Z}_{d}$, then
$\alpha\otimes\Bbbk=\beta\otimes\Bbbk=0$.

  2. (ii)

If $\alpha,\beta$, and $\gamma$ are $S$-homomorphisms satisfying
$\alpha(\varphi_{j}\otimes 1_{Y_{k-1}})=(\varphi_{j}\otimes 1_{Y_{k-1}})\beta$
and $\alpha(1_{X_{j-1}}\otimes\psi_{k})=(1_{X_{j-1}}\otimes\psi_{k})\gamma$,
then there exists $\xi\in\Bbbk$ such that
$\alpha\otimes\Bbbk,\beta\otimes\Bbbk$, and $\gamma\otimes\Bbbk$ are each
given by multiplication by $\xi$.

###### Proof.

By identifying $X_{j}\otimes Y_{p}$ and $X_{j-1}\otimes Y_{p}$ with
$X_{j}^{\oplus m}$ and $X_{j-1}^{\oplus m}$, respectively, we may view
$\varphi_{j}\otimes 1_{Y_{p}}$ as an $m\times m$ block diagonal map.
Similarly, we view $\varphi_{k}\otimes 1_{Y_{p^{\prime}}}:X_{k}^{\oplus m}\to
X_{k}^{\oplus m}$ as an $m\times m$ block diagonal map. If we view the
components of $\alpha$ and $\beta$ with respect to these direct sum
decompositions, we find that $\alpha(u,v)\varphi_{j}=\beta(u,v)\varphi_{k}$
for each $1\leq u,v\leq m$. Since $X$ is strongly indecomposable, the first
statement of (i) now follows. The second statement of (i) uses the same
argument.

For (ii), we first identify $X_{j-1}\otimes Y_{k}\cong Y_{k}^{\oplus n}$ and
$X_{j-1}\otimes Y_{k-1}\cong Y_{k-1}^{\oplus n}$. With respect to these bases,
$1_{X_{j-1}}\otimes\psi_{k}$ is an $n\times n$ block diagonal map. If we write
$\alpha$ and $\beta$ with respect to these decompositions, it follows that
$\alpha(u,v)\psi_{k}=\psi_{k}\gamma(u,v)$ for all $1\leq u,v\leq n$. Since $Y$
is strongly indecomposable, there exists $\xi_{u,v}\in\Bbbk$ such that, for
each pair $1\leq u,v\leq n$, $\alpha(u,v)\otimes\Bbbk$ and
$\gamma(u,v)\otimes\Bbbk$ are both given by multiplication by $\xi_{u,v}$.

Next, we consider $\varphi_{j}\otimes 1_{Y_{k-1}}$ with respect to the same
bases as in the previous paragraph, namely, $\varphi_{j}\otimes
1_{Y_{k-1}}:Y_{k-1}^{\oplus n}\to Y_{k-1}^{\oplus n}$. Note that with these
decompositions, $\varphi_{j}\otimes 1_{Y_{k-1}}$ is not diagonal. Instead, the
$(u,v)$-component is given by $\varphi_{j}(u,v)1_{Y_{k-1}}$ for any $1\leq
u,v\leq n$. With this setup, the equation $\alpha(\varphi_{j}\otimes
1_{Y_{k-1}})=(\varphi_{j}\otimes 1_{Y_{k-1}})\beta$ implies that

(5.5)
$\sum_{i=1}^{n}\alpha(u,i)\varphi_{j}(i,v)=\sum_{i=1}^{n}\varphi_{j}(u,i)\beta(i,v)$

for any $1\leq u,v\leq n$. Let $\alpha_{s,t}$ denote the submatrix of $\alpha$
formed by collecting the $(s,t)$ entry of each block $\alpha(u,v)$, $1\leq
u,v\leq n$. Similarly, we define $\beta_{s,t}$. By considering the $(s,t)$
entry of either side, equation (5.5) implies that
$\alpha_{s,t}\varphi_{j}=\varphi_{j}\beta_{s,t}$. Since $X$ is strongly
indecomposable, there exists $\xi_{s,t}\in\Bbbk$ such that $\alpha_{s,t}$ and
$\beta_{s,t}$ are both given by multiplication by $\xi_{s,t}$. It follows that
$\alpha(u,u)\otimes\Bbbk=\alpha(v,v)\otimes\Bbbk=\beta(u,u)\otimes\Bbbk=\beta(v,v)\otimes\Bbbk$
for any $1\leq u,v\leq n$, and
$\alpha(u,v)\otimes\Bbbk=\beta(u,v)\otimes\Bbbk=0$ for all $u\neq v$. The
desired conclusion of (ii) now follows by combining this with the analysis of
$\alpha$ and $\gamma$ above.

∎

###### Proof of Theorem 5.14.

Set $X\otimes Y=(X\otimes Y,\Phi)$ and let $j,k\in\mathbb{Z}_{d}$. Assume that
$\alpha$ and $\beta$ are $S$-homomorphisms satisfying

(5.6) $\alpha\Phi_{j}=\Phi_{k}\beta.$

Write $\alpha,\beta,\Phi_{j},$ and $\Phi_{k}$ with respect to the direct sum
decompositions given in (3.1). Then the $(s,t)$-component of either side of
(5.6) gives

$\displaystyle\alpha(s,t-1)(\varphi_{t-1}\otimes
1_{Y_{j-t+1}})+\alpha(s,t)(\zeta^{t-1}1_{X_{t-1}}\otimes\psi_{j-t+1})=$
$\displaystyle(\zeta^{s-1}1_{X_{s-1}}\otimes\psi_{k-s+1})\beta(s,t)+(\varphi_{s}\otimes
1_{Y_{k-s}})\beta(s+1,t)$

Reducing this modulo $y$, we obtain the equation

(5.7) $\alpha(s,t-1)_{y}(\varphi_{t-1}\otimes
1_{Y_{j-t+1}})_{y}=(\varphi_{s}\otimes 1_{Y_{k-s}})_{y}\beta(s+1,t)_{y},$

while reducing modulo $x$ gives

(5.8)
$\alpha(s,t)_{x}(\zeta^{t-1}1_{X_{t-1}}\otimes\psi_{j-t+1})_{x}=(\zeta^{s-1}1_{X_{s-1}}\otimes\psi_{k-s+1})_{x}\beta(s,t)_{x}.$

We consider three cases in order to show that the pair $(\alpha,\beta)$
satisfies both conditions of Definition 5.12.

  1. (1)

Assume that $s\neq t-1$ in $\mathbb{Z}_{d}$. Then, by applying Lemma 5.15(i)
to (5.7), we get that

$\alpha(s,t-1)_{y}\otimes\Bbbk=0=\beta(s+1,t)_{y}\otimes\Bbbk.$

This immediately implies that
$\alpha(s,t-1)\otimes\Bbbk=0=\beta(s+1,t)\otimes\Bbbk$. Thus, for any
$s^{\prime}\neq t^{\prime}$, we have that
$\alpha(s^{\prime},t^{\prime})\otimes\Bbbk=0=\beta(s^{\prime},t^{\prime})\otimes\Bbbk$.

  2. (2)

Assume that $j\neq k$ and set $s=t$ in (5.8). In this case, we have

$\alpha(s,s)_{x}(\zeta^{s-1}1_{X_{s-1}}\otimes\psi_{j-s+1})_{x}=(\zeta^{s-1}1_{X_{s-1}}\otimes\psi_{k-s+1})_{x}\beta(s,s)_{x}.$

Again, we apply Lemma 5.15(i). It follows that, when $j\neq k$,
$\alpha(s,s)\otimes\Bbbk=0=\beta(s,s)\otimes\Bbbk$ for all $s$. This, combined
with (1), establishes the first requirement of Definition 5.12.

  3. (3)

Assume that $j=k$. Setting $s=t$ in (5.8), we obtain the equation

$\alpha(s,s)_{x}(\zeta^{s-1}1_{X_{s-1}}\otimes\psi_{j-s+1})_{x}=(\zeta^{s-1}1_{X_{s-1}}\otimes\psi_{j-s+1})_{x}\beta(s,s)_{x}.$

Since $j=k$, we may apply Lemma 5.15(ii) to conclude that
$\alpha(s,s)\otimes\Bbbk$ and $\beta(s,s)\otimes\Bbbk$ are both given by
multiplication by a common element $\xi_{1}\in\Bbbk$.

On the other hand, if we set $s=t-1$ in (5.7), we get the equation

$\alpha(s,s)_{y}(\varphi_{s}\otimes 1_{Y_{j-s}})_{y}=(\varphi_{s}\otimes
1_{Y_{j-s}})_{y}\beta(s+1,s+1)_{y}$

which implies that $\alpha(s,s)\otimes\Bbbk$ and $\beta(s+1,s+1)\otimes\Bbbk$
are both given by multiplication by an element $\xi_{2}\in\Bbbk$. Letting $s$
vary over $\mathbb{Z}_{d}$, we find that $\xi_{1}=\xi_{2}$ and so the second
condition of Definition 5.12 is satisfied. This completes the proof.

∎

######  Example 5.16.

Suppose $\Bbbk$ is a field containing a primitive third root of unity,
$\zeta\in\Bbbk$. Let $\\{f_{1},f_{2},f_{3}\\}$ be pairwise relatively prime
elements of $S_{1}=\Bbbk\llbracket x_{1},\dots,x_{a}\rrbracket$ and
$\\{g_{1},g_{2},g_{3}\\}$ pairwise relatively prime elements of
$S_{2}=\Bbbk\llbracket y_{1},\dots,y_{b}\rrbracket$. Then
$X=(f_{1},f_{2},f_{3})$ is a rank one strongly indecomposable reduced matrix
factorization in $\operatorname{MF}_{S_{1}}^{3}(f_{1}f_{2}f_{3})$. Similarly,
$Y=(g_{1},g_{2},g_{3})$ has the same properties in
$\operatorname{MF}_{S_{2}}^{3}(g_{1}g_{2}g_{3})$. By Theorem 5.14, the tensor
product

$X\otimes Y=\left(\begin{pmatrix}g_{1}&f_{1}&0\\\ 0&\zeta g_{3}&f_{2}\\\
f_{3}&0&\zeta^{2}g_{2}\end{pmatrix},\begin{pmatrix}g_{2}&f_{1}&0\\\ 0&\zeta
g_{1}&f_{2}\\\
f_{3}&0&\zeta^{2}g_{3}\end{pmatrix},\begin{pmatrix}g_{3}&f_{1}&0\\\ 0&\zeta
g_{2}&f_{2}\\\ f_{3}&0&\zeta^{2}g_{1}\end{pmatrix}\right)$

is a strongly indecomposable reduced matrix factorization of
$f_{1}f_{2}f_{3}+g_{1}g_{2}g_{3}$ with rank three.

More generally, suppose $f$ can be written in the form

$f=\sum_{i=1}^{N}f_{i1}f_{i2}\cdots f_{id}$

where $\\{f_{i1},\dots,f_{id}\\}$ are pairwise relatively prime elements of a
power series ring over $\Bbbk$ such that each term of the sum comes from a
distinct set of variables. Then the tensor product
$\bigotimes_{i=1}^{N}(f_{i1},\dots,f_{id})$ is a strongly indecomposable
$d$-fold matrix factorization of $f$ of rank $d^{N-1}$.

##  6\. Constructing indecomposable Ulrich modules

The goal of this section is to construct Ulrich modules (and other maximal
Cohen-Macaulay modules) over a hypersurface domain using the tensor product
construction from Section 3. Several of the key results and methods of this
section can be found either explicitly or implicitly in [HUB91]. For instance,
they prove the existence of Ulrich modules in greater generality (over any
strict complete intersection). Our approach produces Ulrich modules over a
hypersurface domain using only elementary properties of the tensor product of
matrix factorizations and [Eis80, Proposition 5.6] (see also [LW12, Remark
8.8]). Using the results of Section 5, we are to prove that the Ulrich modules
produced are indecomposable over hypersurface rings of a certain form.

Let $(Q,\mathfrak{n})$ be a local ring and $f\in\mathfrak{n}$ a
nonzerodivisor. Assume further that $Q$ is a $\mathbb{Z}[\zeta]$-algebra where
$\zeta\in\mathbb{C}$ is a primitive $d^{\textit{th}}$ root of unity. Recall
that a matrix factorization $(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ is
reduced if $\operatorname{Im}\varphi_{k}\subseteq\mathfrak{n}X_{k-1}$ for each
$k\in\mathbb{Z}_{d}$. This means that, after choosing bases, the entries of
$\varphi_{k}$ lie in the maximal ideal of $Q$ for each $k\in\mathbb{Z}_{d}$.

######  Proposition 6.1.

Suppose that $I\subset Q$ is an ideal such that $f\in I^{d}$ for some $d\geq
2$. If $f$ can be written as $f=\sum_{i=1}^{N}f_{i1}f_{i2}\cdots f_{id}$ for
$f_{ij}\in I$ and some $N\geq 2$, then there exists a reduced matrix
factorization $(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ such that

  1. (i)

$\operatorname{rank}X=d^{N-1}$, and

  2. (ii)

$\det\varphi_{k}=\pm f^{d^{N-2}}$ for any $k\in\mathbb{Z}_{d}$.

###### Proof.

For each integer $1\leq i\leq N$, let $Y_{i}=(f_{i1},f_{i2},\dots,f_{id})$ be
the rank one matrix factorization of the product $f_{i}\coloneqq
f_{i1}f_{i2}\cdots f_{id}$ with $d$ factors. In other words,
$Y_{i}\in\operatorname{MF}_{Q}^{d}(f_{i})$ for each $1\leq i\leq N$. Then the
tensor product $X=\bigotimes_{i=1}^{N}Y_{i}$ is a reduced matrix factorization
of $f=\sum_{i=1}^{N}f_{i}$ of rank $d^{N-1}$. The determinant formula follows
from Proposition 3.10. ∎

A finitely generated module $M$ over a local ring $(A,\mathfrak{m})$ is called
maximal Cohen-Macaulay (MCM) if $0pt_{A}M=\dim A$. We will let $\mu_{A}(M)$
denote the number of elements in a minimal generating set of $M$ and the
Hilbert-Samuel multiplicity of an $A$-module will be denoted $e_{A}(M)$. If
$f\in\mathfrak{m}$, $\operatorname{ord}(f)$ will denote the order of $f$, that
is, largest integer $d$ such that $f\in m^{d}$. Recall that if $A$ is a
regular local ring and $f\in\mathfrak{m}$ has order $d$, then the multiplicity
of the ring $A/(f)$ is exactly $d=\operatorname{ord}(f)$ (see [HS06, Example
11.2.8]).

If $(Q,\mathfrak{n})$ is a Cohen-Macaulay local ring ($0pt_{Q}Q=\dim Q)$ and
$f\in\mathfrak{n}$ is a nonzerodivisor, then every module associated to a
matrix factorization $(X,\varphi)\in\operatorname{MF}_{Q}^{d}(f)$ defines an
MCM $R=Q/(f)$-module. More specifically,
$\operatorname{cok}(\varphi_{k}\varphi_{k+1}\cdots\varphi_{k+r-1})$ is an MCM
$R$-module for each $k\in\mathbb{Z}_{d}$ and $1\leq r\leq d$; this follows
directly from the Auslander–Buchsbaum formula.

######  Proposition 6.2.

Assume that $(Q,\mathfrak{n})$ is a regular local ring, $f\in\mathfrak{n}^{2}$
is irreducible, and set $R=Q/(f)$. Let $d=\operatorname{ord}(f)$ and write
$f=\sum_{i=1}^{N}f_{i1}f_{i2}\cdots f_{id}$ for $f_{ij}\in\mathfrak{n},N\geq
2$. Then, for each pair of integers $2\leq k\leq d,1\leq\ell\leq d-1$, there
exists a maximal Cohen-Macaulay $R$-module $M$ (with no free summands)
satisfying

  1. (i)

$\mu_{R}(M)=k^{N-1}$

  2. (ii)

$\operatorname{rank}_{R}(M)=\ell k^{N-2}$.

In particular, the multiplicity of $M$ is $e_{R}(M)=d\ell k^{N-2}$.

###### Proof.

Fix $2\leq k\leq d$ and $1\leq\ell\leq d-1$. For each integer $1\leq i\leq N$,
choose a partition of the product $f_{i}=f_{i1}f_{i2}\cdots f_{id}$ into $k$
products of the $f_{ij}$. Note that the index $i$ is not taken modulo $d$. The
choice of partition can be viewed as a rank one reduced matrix factorization
of $f_{i}$ with $k$ factors, call it
$Y_{i}\in\operatorname{MF}_{Q}^{k}(f_{i})$. Let
$(X,\varphi)\in\operatorname{MF}_{Q}^{k}(f)$ be the tensor product of
$Y_{1},Y_{2},\dots,Y_{N}$ as in the proof of Proposition 6.1 and set
$M=\operatorname{cok}\varphi_{1}\varphi_{2}\cdots\varphi_{\ell}$ (the choice
to end at $\varphi_{1}$ is not important; any composition of $\ell$ maps would
suffice). Then $X$ has rank $k^{N-1}$ by Proposition 6.1 and we have a minimal
free resolution of $M$ over $Q$

$0\to X_{\ell}\xrightarrow{\varphi_{1}\varphi_{2}\cdots\varphi_{\ell}}X_{0}\to
M\to 0.$

Hence, we may tensor this resolution with $R$ to obtain a minimal presentation
of $M$ over $R$. It follows that $M$ is an MCM $R$-module minimally generated
by $k^{N-1}$ elements.

Again by Proposition 6.1, we have $\det\varphi_{j}=\pm f^{k^{N-2}}$, for each
$j\in\mathbb{Z}_{d}$. Hence
$\det(\varphi_{1}\varphi_{2}\cdots\varphi_{\ell})=\pm f^{\ell k^{N-2}}$. Since
$f$ is irreducible, [Eis80, Proposition 5.6] implies that
$\operatorname{rank}_{R}(M)=\ell k^{N-2}$. Finally, since $R$ is a domain, the
multiplicity of $M$ may be computed as
$e_{R}(M)=e_{R}(R)\operatorname{rank}_{R}(M)=d\cdot\ell k^{N-2}$. ∎

The minimal number of generators of an MCM module $M$ over a local ring $A$ is
always bounded above by the modules multiplicity. In other words,
$\mu_{A}(M)\leq e_{A}(M)$. Maximal Cohen-Macaulay modules for which equality
is attained are called Ulrich modules. In [Ulr84], Ulrich asked if every
Cohen-Macaulay local ring has an Ulrich module. After remaining open for
nearly fourty years, Ulrich’s question was answered in the negative in
[IMWZ24]. They give examples of Gorenstein, and even complete intersection,
local domains which have no Ulrich modules. Earlier examples of non-Cohen-
Macaulay local rings which have no Ulrich modules were give by Yhee in
[Yhe23]. In light of these examples, the case of a hypersurface ring stands
out as unique.

######  Corollary 6.3.

[HUB91, Theorem 2.5] Assume that $(Q,\mathfrak{n})$ is a regular local ring,
$f\in\mathfrak{n}^{2}$ is irreducible, and set $R=Q/(f)$. Then there exists an
Ulrich $R$-module.

###### Proof.

This follows from Proposition 6.2 by picking $k=\operatorname{ord}(f)$ and
$\ell=1$. ∎

A subcategory $\mathcal{B}$ of an abelian category $\mathcal{A}$ is called
extension closed if it is closed under direct summands and if for each short
exact sequence $0\to L\to M\to N\to 0$, if $L,N\in\mathcal{B}$, then
$M\in\mathcal{B}$ as well. As it turns out, the full subcategory of Ulrich
modules over a Cohen-Macaulay local ring is rarely extension closed; in
dimension one this happens if and only if the ring is regular (see [DDD24,
Corollary 5.2.7]). The constructions of this section induce a similar result
for hypersurface domains.

######  Corollary 6.4.

Assume $(Q,\mathfrak{n})$ is a regular local ring, $f\in\mathfrak{n}$ is
irreducible, and set $R=Q/(f)$. Then the subcategory of Ulrich $R$-modules is
extension closed if and only if $R$ is regular.

###### Proof.

If $R$ is regular, then every MCM module is free. Hence every Ulrich module is
free and the subcategory of Ulrich modules is therefore extension closed.

Conversely, suppose that $R=Q/(f)$ is not regular, that is, suppose that
$f\in\mathfrak{n}^{2}$. Set $d=\operatorname{ord}(f)$. Since $d\geq 2$, we may
apply Proposition 6.2 to obtain a reduced matrix factorization
$X=(\varphi_{1},\varphi_{2},\dots,\varphi_{d-1},\varphi_{0})\in\operatorname{MF}_{Q}^{d}(f)$
satisfying

  1. (i)

$N=\operatorname{cok}\varphi_{1}$ and $L=\operatorname{cok}\varphi_{2}$ are
Ulrich $R$-modules, and

  2. (ii)

$M=\operatorname{cok}(\varphi_{1}\varphi_{2})$ satisfies
$\displaystyle\frac{\mu_{R}(M)}{e_{R}(M)}=\frac{1}{2}$.

In particular, $\operatorname{cok}(M)$ is not Ulrich. The result now follows
since the commutative diagram with exact rows

${0}$${X_{2}}$${X_{1}}$${L}$${0}$${0}$${X_{2}}$${X_{0}}$${M}$${0}$$\scriptstyle{\varphi_{2}}$$\scriptstyle{\varphi_{1}}$$\scriptstyle{\varphi_{1}\varphi_{2}}$

induces a short exact sequence of $R$-modules $0\to L\to M\to N\to 0$. Once
again, the choice of $\varphi_{1}$ and $\varphi_{2}$ was not important; any
pair of consecutive maps in $X$ induce a short exact sequence of this form. ∎

Combining the constructions of this section and the indecomposability results
of Section 5, we obtain the following.

######  Theorem 6.5.

Let $Q=\Bbbk\llbracket x_{1},\dots,x_{n}\rrbracket$ for a field $\Bbbk$, $f\in
Q$ be irreducible of order $e\geq 2$, and set $R=Q/(f)$. Assume further that
$\Bbbk$ contains a primitive $d^{\textit{th}}$ root of unity for all $2\leq
d\leq e$. Suppose that $f$ can be written as
$f=\sum_{i=1}^{N}f_{i1}f_{i2}\cdots f_{id}$, for
$f_{ij}\in\mathfrak{(}x_{1},\dots,x_{n})$ and $2\leq d\leq e$, such that

  1. (i)

for each $1\leq i,j\leq N$ with $i\neq j$, the power series
$\\{f_{i1},f_{i2},\dots,f_{id}\\}$ and $\\{f_{j1},f_{j2},\dots,f_{jd}\\}$ are
written in distinct sets of variables, and

  2. (ii)

the rank one factorization
$(f_{i1},f_{i2},\dots,f_{id})\in\operatorname{MF}_{Q}^{d}(f_{i1}f_{i2}\cdots
f_{id})$ is strongly indecomposable for each $1\leq i\leq N$.

Then there exists an indecomposable MCM $R$-module $M$ satisfying
$\operatorname{rank}_{R}(M)=d^{N-2}$, $\mu_{R}(M)=d^{N-1},$ and
$e_{R}(M)=ed^{N-2}$. Moreover, if $d=e$, then $M$ is an indecomposable Ulrich
$R$-module.

###### Proof.

The assumptions allow us to apply Theorem 5.14 to conclude that the tensor
product

$X=\bigotimes_{i=1}^{N}(f_{i1},f_{i2},\dots,f_{id})$

is a strongly indecomposable matrix factorization of $f$ in
$\operatorname{MF}_{Q}^{d}(f)$. If
$X=(\Phi_{1},\Phi_{2},\dots,\Phi_{d-1},\Phi_{0})$, then Lemma 5.13 implies
that $\operatorname{cok}\Phi_{i}$ is and indecomposable MCM $R$-module for
each $i\in\mathbb{Z}_{d}$. The rank, minimal number of generators, and
multiplicity counts all follow as in the proof of Proposition 6.2. ∎

Theorem 6.5 can be applied to any polynomial (or power series) of the form
given in Example 5.16. We give one specific instance of this below.

######  Example 6.6.

Let $Q=\Bbbk\llbracket x_{ij}:1\leq i\leq N,1\leq j\leq d\rrbracket$ where
$N,d\geq 2$, and $\Bbbk$ is a field containing a primitive $d^{\textit{th}}$
root of unity. Consider the polynomial

$f=\sum_{i=1}^{N}x_{i1}^{a_{i1}}x_{i2}^{a_{i2}}\cdots x_{id}^{a_{id}}\in Q,$

for integers $a_{ij}\geq 1$, and assume that $f$ is irreducible. Each of the
rank one matrix factorizations
$X_{i}=(x_{i1}^{a_{i1}},x_{i2}^{a_{i2}},\dots,x_{id}^{a_{id}})\in\operatorname{MF}_{Q}^{d}(x_{i1}^{a_{i1}}x_{i2}^{a_{i2}}\cdots
x_{id}^{a_{id}})$, $1\leq i\leq N$, is strongly indecomposable (see Example
5.16). Hence, Theorem 5.14 implies that
$X=\bigotimes_{i=1}^{N}X_{i}\in\operatorname{MF}_{Q}^{d}(f)$ is strongly
indecomposable of rank $d^{N-1}$. Set $X=(\Phi_{1},\Phi_{2},\dots,\Phi_{0})$
and $M_{i}=\operatorname{cok}\Phi_{i}$, $i\in\mathbb{Z}_{d}$.

If $a_{ij}=a\geq 1$ for all $i,j$, then $\operatorname{ord}(f)=da$. In this
case, each module $M_{i}$, $i\in\mathbb{Z}_{d}$, is an indecomposable MCM
$R$-module satisfying

$\frac{\mu_{R}(M_{i})}{e_{R}(M_{i})}=\frac{d^{N-1}}{(da)d^{N-2}}=\frac{1}{a}.$

If $a=1$, then each $M_{i}$, $i\in\mathbb{Z}_{d}$, is an indecomposable Ulrich
module minimally generated by $d^{N-1}$ elements and of rank $d^{N-2}$.

The notion of Ulrich complexity was introduced in [BES17] to quantify the
possible ranks of Ulrich modules over hypersurface rings. Indeed, the Ulrich
complexity of homogeneous polynomial of degree $d$ is the smallest integer $r$
such that there exists a matrix factorization
$(A,B)\in\operatorname{MF}^{2}(f)$ satisfying $\det(A)=f^{r}$ and such that
the entries of $A$ are linear forms. In this case, we will write
$\operatorname{uc}(f)=r$. The Ulrich complexity of several examples and
classes of polynomials was computed in [BES17] but, in general, Ulrich
complexity is difficult to determine. For an example of their results,
consider a generic trinomial
$f=x_{11}x_{12}x_{13}+x_{21}x_{22}x_{23}+x_{31}x_{32}x_{33}$ (i.e. $N=3,d=3$,
and $a=1$ from above). They show that $\operatorname{uc}(f)=3$ [BES17,
Proposition 4.10]. The matrix factorization of $f$ constructed in Example 6.6
produces Ulrich modules which attain this minimum and are (necessarily)
indecomposable.

If $f$ is a homogeneous polynomial which can be written as a sum of $N$
products of linear forms, then the tensor product construction gives an upper
bound for Ulrich complexity; $\operatorname{uc}(f)\leq d^{N-2}$ [BES17,
Theorem 2.1]. Interestingly, Example 6.6 produces indecomposable Ulrich
modules attaining this upper bound for any generic polynomial of the form
$f=\sum_{i=1}^{N}x_{i1}x_{{i2}}\cdots x_{id}$. If the Ulrich complexity is
strictly less than $d^{N-2}$ in this case, then it seems one would need a
construction more efficient than the tensor product to attain the minimum.

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