Derived and stable equivalences of centralizer matrix algebras

Xiaogang Li and Changchang Xi∗

†† ∗ Corresponding author. Email<EMAIL_ADDRESS>Fax: 0086 10
68903637.††2020 Mathematics Subject Classification: Primary 16E35, 18G80,
20C05, 15A27; Secondary 18G65, 16D90, 16S50, 05A05.††Keywords: Centralizer
matrix algebra; Elementary divisor; Morita equivalence; Derived equivalence;
Stable equivalence, Symmetric group.

###### Abstract

The centralizer of a matrix in a full matrix algebra is called a principal
centralizer matrix algebra. Characterizations are presented for principal
centralizer matrix algebras to be Morita equivalent, almost $\nu$-stable
derived equivalent, derived equivalent, and stably equivalent of Morita type,
respectively, in terms of new equivalence relations on square matrices. These
equivalence relations on matrices are introduced in a natural way by their
elementary divisors. Thus the categorical equivalences are reduced to
questions in linear algebra. Consequently, principal centralizer matrix
algebras of permutation matrices are Morita equivalent if and only if they are
derived equivalent. Moreover, two representation-finite, principal centralizer
matrix algebras over a perfect field are stably equivalent of Morita type if
and only if they are stably equivalent. Further, derived equivalences between
the principal centralizer matrix algebras of permutation matrices induce the
ones of their $p$-regular parts and $p$-singular parts of the given
permutations.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Derived equivalences of algebras
    2. 2.2 Modules over quotients of polynomial algebras
    3. 2.3 Centralizer matrix algebras of representation-finite type
  3. 3 Derived and stable equivalences of centralizer matrix algebras
    1. 3.1 Characterizations of Morita and derived equivalences: Proof of Theorem 1.2
    2. 3.2 Relations among derived, Morita and stable equivalences: Proof of Corollary 1.3
    3. 3.3 Restrictions of derived equivalences
  4. 4 Examples and questions

##  1 Introduction

Let $R$ be a field and $n$ a natural number. We denote by $[n]$ the set of
integers $\\{1,\cdots,n\\}$ and by $M_{n}(R)$ the full $n\times n$ matrix
algebra over $R$ with the identity matrix $I_{n}$. For a nonempty set $X$ of
$M_{n}(R)$, the centralizer matrix algebra $S_{n}(X,R)$ of $X$ in $M_{n}(R)$
is defined by

$S_{n}(X,R):=\\{a\in M_{n}(R)\mid ax=xa,\;\forall\;x\in X\\}.$

In case of $X=\\{c\\}$, we write $S_{n}(c,R)$ for $S_{n}(X,R)$. Clearly,
$S_{n}(X,R)=\cap_{c\in X}S_{n}(c,R)$. The $R$-algebra $S_{n}(c,R)$ is termed
as a _principal centralizer matrix algebra_.

Clearly, if $X$ consists of invertible matrices, then the centralizer matrix
algebras $S_{n}(X,R)$ are a special class of invariant algebras, which can be
dated back to the classical invariant theory (see [31]). If $X$ consists of
nilpotent matrices over an algebraically closed field $R$, then all nilpotent
matrices in $S_{n}(X,R)$ form a variety which is of significant interest in
semisimple Lie algebras (see [24, 25]). Further, the centralizer matrix
algebras $S_{n}(X,R)$ also appear in the representation theory of finite
groups. For instance, if $R$ is a field of characteristic $p>0$ and $G$ is a
finite group with a Sylow $p$-subgroup $P$, then $G$ acts on the set
$G\backslash P$ of left cosets of $P$ in $G$ by left multiplication, and the
set $R[G\backslash P]$ of all $R$-linear combination of the left cosets of $P$
in $G$ is a permutation module over the group algebra $R[G]$ of $G$. Moreover,
${\rm End}_{R[G]}(R[G\backslash P])$ is isomorphic to a centralizer matrix
algebra $S_{r}(G_{0},R)$, where $r=|G:P|$ and $G_{0}$ is a subgroup of
permutation matrices in $M_{r}(R)$. In the literature, ${\rm
End}_{R[G]}(R[G\backslash P])$ is also called a modular Hecke algebra (see [2,
7]). Note that $G_{0}$ is isomorphic to $G/O_{p}(G)$, where $O_{p}(G)$ stands
for the largest normal $p$-subgroup of $G$. Alperin suggested to study the
endomorphism ring ${{\rm End}}_{R[G]}(R[G\backslash P])$ for attacking the
famous Alperin’s weight conjecture (see [1]).

Centralizer matrix algebras have been studied in various aspects of
mathematics, such as invariant subspaces or orbits in [5], and maximal doubly
stochastic matrices in [9]. Recently, in a series of papers [34, 35, 36], a
lot of new structural and homological properties of $S_{n}(c,R)$ is revealed.
For instance, $S_{n}(c,R)$ is always a cellular $R$-algebra if $R$ is an
algebraically closed field, and the famous Auslander-Reiten (or Auslander-
Alperin) conjecture on stable equivalences holds true for $S_{n}(c,R)$ over an
arbitrary field $R$. The conjecture states that stably equivalent algebras
should have the same number of non-projective non-isomorphic simple modules.
Further, $S_{n}(c,R)$ is always a Gorenstein algebra and captures the
Auslander algebra of the truncated polynomial algebra $R[x]/(x^{n})$ for all
$n\in\mathbb{N}$, which has played an important role in the classification of
parabolic subgroups of classical groups with a finite number of orbits on the
unipotent radical (see [12]).

In this note, we study relations between principal centralizer matrix
algebras. Particularly, we are interested in when they are Morita, derived and
stably equivalent of Marita type. These equivalences are of great importance
in the representation theory of algebras and groups [30]. Our question here
can be formulated precisely as follows.

Question: Let $R$ be a field, $c\in M_{n}(R)$ and $d\in M_{m}(R)$. What are
the necessary and sufficient conditions for $S_{n}(c,R)$ and $S_{m}(d,R)$ to
be Morita, derived or stably equivalent (of Morita type)?

The answer to this question is closely related to the minimal polynomials of
$c$ and $d$, and our characterizations of Morita, derived and stable
equivalences are given surprisingly in a very elementary way, namely in terms
of elementary divisors of matrices $c$ and $d$.

To state our main result precisely, we first introduce several equivalence
relations on square matrices.

Let $R[x]$ be the polynomial algebra over a field $R$ in one variable $x$.
Given polynomials $f(x)$ and $g(x)$ of positive degree, we define $f(x)\leq
g(x)$ if $f(x)$ divides $g(x)$, that is, $g(x)=f(x)h(x)$ with $h(x)\in R[x]$.

For $c\in M_{n}(R)$, let

$\mathcal{E}_{c}\subset R[x]$ denote the set of distinct elementary divisors
of $c$. Here we understand that the elements of a set are pairwise different,
and the ones of a multiset are allowed to be duplicate.

$\mathcal{M}_{c}:=\\{f(x)\in\mathcal{E}_{c}\mid f(x)\mbox{ is maximal with
respect to the order }\leq\\},$ the set of maximal divisors of $c$.

For $f(x)\in\mathcal{M}_{c}$, we define

$P_{c}(f(x)):=\\{i\geq 1\mid\exists\mbox{ irreducible polynomial }p(x)\mbox{
such that }p(x)\leq f(x),p(x)^{i}\in\mathcal{E}_{c}\\},$ the set of power
indices of $f(x)$ in $\mathcal{E}_{c}$.

$\mathcal{R}_{c}:=\\{f(x)\in\mathcal{M}_{c}\mid f(x)\mbox{ is a reducible
polynomial}\\},$ the set of reducible maximal divisors of $c.$

The polynomials in $\mathcal{R}_{c}$ are exactly those
$f(x)\in\mathcal{M}_{c}$ with $P_{c}(f(x))\neq\\{1\\}$.

Let $\mathbb{Z}_{>0}$ be the set of all positive integers and
$s\in\mathbb{Z}_{>0}$. For a subset $T:=\\{m_{1},m_{2},\cdots,m_{s}\\}$ of
$\mathbb{Z}_{>0}$ with $m_{1}>m_{2}>\cdots>m_{s}$, we define a multiset
$\mathcal{H}_{T}:=\\{\\{m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s}\\}\\}$ and a
set $\mathcal{J}_{T}=:\\{m_{1},m_{1}-m_{2},\cdots,m_{1}-m_{s}\\}$. If
$T=\\{m_{1}\\}$, then $\mathcal{H}_{T}=\mathcal{J}_{T}=T$.

Now we introduce a few new equivalence relations on square matrices.

######  Definition 1.1.

Two matrices $c\in M_{n}(R)$ and $d\in M_{m}(R)$ are said to be

$(1)$ $M$-equivalent, written $c\stackrel{{\scriptstyle M}}{{\sim}}d$, if
there is a bijection $\pi$ between $\mathcal{M}_{c}$ and $\mathcal{M}_{d}$
such that $R[x]/(f(x))\simeq R[x]/((f(x))\pi)$ as algebras and
${P_{c}(f(x))}={P_{d}((f(x))\pi)}$ for all $f(x)\in\mathcal{M}_{c}$, where
$(f(x))\pi$ denotes the image of $f(x)$ under the map $\pi$.

$(2)$ $D$-equivalent, written $c\stackrel{{\scriptstyle D}}{{\sim}}d$, if
there is a bijection $\pi$ between $\mathcal{M}_{c}$ and $\mathcal{M}_{d}$
such that $R[x]/(f(x))\simeq R[x]/((f(x))\pi)$ as algebras and
$\mathcal{H}_{P_{c}(f(x))}=\mathcal{H}_{P_{d}((f(x))\pi)}$ for all
$f(x)\in\mathcal{M}_{c}$.

$(3)$ $AD$-equivalent, written $c\stackrel{{\scriptstyle AD}}{{\sim}}d$, if
there is a bijection $\pi$ between $\mathcal{M}_{c}$ and $\mathcal{M}_{d}$
such that $R[x]/(f(x))\simeq R[x]/((f(x))\pi)$ as algebras and
$P_{c}(f(x))={P_{d}((f(x))\pi)}$ or
$P_{c}(f(x))=\mathcal{J}_{P_{d}((f(x))\pi)}$ for all $f(x)\in\mathcal{M}_{c}$.

$(4)$ $SM$-equivalent, written $c\stackrel{{\scriptstyle SM}}{{\sim}}d$, if
there is a bijection $\pi$ between $\mathcal{R}_{c}$ and $\mathcal{R}_{d}$
such that $R[x]/(f(x))\simeq R[x]/((f(x))\pi)$ as algebras and
${P_{c}(f(x))}={P_{d}((f(x))\pi)}$ or
$P_{c}(f(x))=\mathcal{J}_{P_{d}((f(x))\pi)}$ for all $f(x)\in\mathcal{R}_{c}$.

Next, we characterize Morita, derived and stable equivalences between
principal centralizer matrix algebras in terms of these equivalence relations
on matrices. Our main result reads as follows.

######  Theorem 1.2.

Let $R$ be a field, $c\in M_{n}(R)$ and $d\in M_{m}(R)$. Then

$(1)$ $S_{n}(c,R)$ and $S_{m}(d,R)$ are Morita equivalent if and only if
$c\stackrel{{\scriptstyle M}}{{\sim}}d$.

$(2)$ $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent if and only if
$c\stackrel{{\scriptstyle D}}{{\sim}}d$.

$(3)$ $S_{n}(c,R)$ and $S_{m}(d,R)$ are almost $\nu$-stable derived equivalent
if and only if $c\stackrel{{\scriptstyle AD}}{{\sim}}d$.

$(4)$ Assume that either $R$ is perfect or both $c$ and $d$ are invertible
matrices of finite order. Then $S_{n}(c,R)$ and $S_{m}(d,R)$ are stably
equivalent of Morita type if and only if $c\stackrel{{\scriptstyle
SM}}{{\sim}}d$.

Thus the existence of a Morita equivalence, an almost $\nu$-stable derived
equivalence and a derived equivalence between principal centralizer matrix
algebras can be read off from the elementary divisors of given matrices
directly, and therefore is a problem in linear algebra.

As a consequence, we have the following corollary.

######  Corollary 1.3.

Let $R$ be a field, $c\in M_{n}(R)$ and $d\in M_{m}(R)$.

$(1)$ If $c$ and $d$ are permutation matrices, then $S_{n}(c,R)$ and
$S_{m}(d,R)$ are Morita equivalent if and only if they are derived equivalent.

$(2)$ Suppose that $R$ is perfect and that $S_{n}(c,R)$ and $S_{m}(d,R)$ are
representation-finite. Then $S_{n}(c,R)$ and $S_{m}(d,R)$ are stably
equivalent of Morita type if and only if they are stably equivalent.

$(3)$ If $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent or stably
equivalent, then they have the same dominant dimension.

Moreover, for the centralizer matrix algebras of permutation matrices, a
derived equivalence between them gives rise to derived equivalences of smaller
centralizer matrix algebras corresponding to $p$-regular and $p$-singular
parts of permutations. For details, we refer to Proposition 3.14.

The paper is organized as follows. In Section 2 we fix notation, recall basic
definitions and terminologies, and prove a few preliminary lemmas needed in
the later proofs. In Section 3 we prove the main result and its corollary. In
Section 4 we present examples to show that even for principal centralizer
matrix algebras over a field, the notions of Morita equivalences, almost
$\nu$-stable derived equivalences and derived equivalences are different,
though they may coincide in some cases. Finally, we propose some open
questions for further investigation of centralizer matrix algebras. For
example, how to characterize generally the stable equivalences of centralizer
matrix algebras?

##  2 Preliminaries

In this section we recall some basic definitions and terminologies, and
prepare a few lemmas for our proofs.

###  2.1 Derived equivalences of algebras

In this paper, $R$ is a field unless stated otherwise. By an algebra we mean a
finite-dimensional unitary associative algebra over $R$.

Let $A$ be an algebra. By ${\rm rad}(A)$ and $LL(A)$ we denote the Jacobson
radical and the Loewy length of $A$, respectively. Let $A^{{}^{\rm op}}$ and
$A^{e}$ stand for the opposite algebra and the enveloping algebra
$A\otimes_{R}A^{\rm op}$ of $A$, respectively. By a module we always mean a
left module unless stated otherwise. We write $A\mbox{{\rm-mod}}$ for the
category of all finitely generated left $A$-modules, $A\mbox{{\rm-
mod}}_{\mathscr{P}}$ for the full subcategory of $A\mbox{{\rm-mod}}$
consisting of modules without any nonzero projective summands, and $A$-proj
(respectively, $A$-inj) for the full subcategory of $A\mbox{{\rm-mod}}$
consisting of projective (respectively, injective) $A$-modules.

For an $A$-module $M$, we denote by ${{\rm add}}(M)$ the full subcategory of
$A\mbox{{\rm-mod}}$ consisting of all modules isomorphic to direct summands of
direct sums of finitely many copies of $M$. Let $\ell(M)$ stand for the
composition length of $M$, $\mathcal{B}(M)$ for the basic module of $M$, and
$M_{\mathscr{P}}$ for the submodule of $M$ without any nonzero projective
summand such that $M/M\\!_{\mathscr{P}}$ is projective.

The stable module category of $A$ is denoted by $A$-mod which has the same
objects as $A$-mod does, but the morphism set $\underline{{\rm Hom}}_{A}(X,Y)$
of objects $X$ and $Y$ is the quotient of ${\rm Hom}_{A}(X,Y)$ modulo
$\mathcal{P}(X,Y)$, the set of all homomorphisms that factorize through
projective $A$-modues. If $M$ is a non-projective indecomposable $A$-module,
then $\mathcal{P}(M,M)\subseteq{\rm rad}({\rm End}_{A}(M))$.

When we say the number of a specific class of $A$-modules such as projective,
injective, simple and indecomposable $A$-modules, we always refer to the
number of isomorphism classes of them.

For homomorphisms $f:X\to Y$ and $g:Y\to Z$ in $A\mbox{{\rm-mod}}$, we write
$fg$ for their composition. This implies that the image of an element $x\in X$
under $f$ is denoted by $(x)f$. Thus ${\rm Hom}_{A}(X,Y)$ is naturally a left
${\rm End}_{A}(X)$\- and right ${\rm End}_{A}(Y)$-bimodule. The composition of
functors between categories is written from right to left, that is, for two
functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and
$G:\mathcal{D}\rightarrow\Sigma$, we write $G\circ F$, or simply $GF$, for the
composition of $F$ with $G$. The image of an object $X\in\mathcal{C}$ under
$F$ is written as $F(X).$

Let $D:A\mbox{{\rm-mod}}\to A^{{}^{\rm op}}\mbox{{\rm-mod}}$ be the usual
duality of the algebra $A$. The Nakayama functor $\nu_{A}:=D{\rm
Hom}_{A}(-,A)\simeq D(A)\otimes_{A}-$ from $A\mbox{{\rm-mod}}$ to itself
restricts to an equivalence: $A\mbox{{\rm-
proj}}\stackrel{{\scriptstyle\sim}}{{\to}}A$-inj. An $A$-module $M$ is said to
be $\nu$-_stably projective_ if $\nu^{i}_{A}{M}$ is projective for all $i\geq
0$. For example, if $e^{2}=e\in A$ satisfies ${\rm add}(\nu_{A}{Ae})={\rm
add}(Ae)$, then $Ae$ is $\nu$-stably projective. In this case, $e\in A$ is
said to be _$\nu$ -stable_. Let $A$-stp denote the full subcategory of
$A\mbox{{\rm-mod}}$ consisting of all $\nu$-stably projective $A$-modules. The
_Frobenius part_ of $A$ is defined (up to Morita equivalence) to be the
algebra $eAe$ where $e$ is an idempotent such that ${\rm add}(Ae)=A$-stp (see
[15]).

Let ${\mathscr{D}}(A)$ (respectively, ${\mathscr{D}}^{b}(A)$) be the unbounded
(respectively, bounded) derived category of $A\mbox{{\rm-mod}}$. They are
$R$-linear, triangulated categories. Algebras $A$ and $B$ are said to be
_derived equivalent_ if their derived categories ${\mathscr{D}}^{b}(A)$ and
${\mathscr{D}}^{b}(B)$ are equivalent as $R$-linear triangulated categories.
An $R$-linear triangle equivalence
$F:{\mathscr{D}}^{b}(A)\rightarrow{\mathscr{D}}^{b}(B)$ is called a _derived
equivalence_ between $A$ and $B$.

A special class of derived equivalences, called almost $\nu$-stable derived
equivalences, was introduced in [13] to establish relations between derived
equivalences and stable equivalences of Morita type (see [13] for more
details). One of the interesting properties of almost $\nu$-stable derived
equivalences is that such a derived equivalence between finite-dimensional
algebras always induces a stable equivalence of Morita type (see [13, Theorem
1.1]), and thus preserves global and dominant dimensions of algebras. Recall
that finite-dimensional algebras $A$ and $B$ are _stably equivalent of Morita
type_ [6] if there exist left-right projective (projective as left and right
module) bimodules ${}_{A}M_{B}$ and ${}_{B}N_{A}$ such that
$M\otimes_{B}N\simeq A\oplus P$ as $A^{e}$-modules for some projective
$A^{e}$-module $P$ and $N\otimes_{A}M\simeq B\oplus Q$ as $B^{e}$-modules for
some projective $B^{e}$-module $Q$. Clearly, the exact functor
$N\otimes_{A}-:A\mbox{{\rm-mod}}{}\rightarrow B\mbox{{\rm-mod}}$ induces a
stable equivalence
$N\otimes_{A}-:A\mbox{{\rm-{\text@underline{mod}}}}\rightarrow B$-mod.

It is known that $R$ and any separable $R$-algebra are stably equivalent of
Morita type. Recall that an $R$-algebra $A$ is separable over $R$ if $A$ is a
projective module over $A^{e}$.

Algebras $A$ and $B$ are said to be Morita equivalent if $A\mbox{{\rm-mod}}$
and $B\mbox{{\rm-mod}}$ are equivalent as $R$-linear categories. Clearly,
Morita equivalences are almost $\nu$-stable derived equivalences.

To get derived equivalences, the following corollary of [15, Theorem 1.1]
provides a convenient way. For further information on constructions of derived
equivalences of algebras, we refer to [33].

Let $\mathcal{C}$ be an additive category and $\mathcal{D}$ a full subcategory
of $\mathcal{C}$. For $Y\in\mathcal{C},$ a morphism $f:M\rightarrow Y$ with
$M\in\mathcal{D}$ is called a _right $\mathcal{D}$-approximation_ of $Y$ if
each morphism $D\to Y$ with $D\in\mathcal{D}$ factorizes through $f$. Dually,
one defines a _left $\mathcal{D}$-approximation_ of an object $X$ in
$\mathcal{C}.$

A sequence $X\stackrel{{\scriptstyle
g}}{{\rightarrow}}M\stackrel{{\scriptstyle f}}{{\rightarrow}}Y$ in
$\mathcal{C}$ with $M\in\mathcal{D}$ is called a _$\mathcal{D}$ -split
sequence_ [15] if $g$ is both a kernel of $f$ and a left
$\mathcal{D}$-approximation of $X$, and if $f$ is both a cokernel of $g$ and a
right $\mathcal{D}$-approximation of $Y$.

######  Lemma 2.1.

[15] Let $A$ be an algebra, and let $\mathcal{C}$ be a full subcategory of
$A\mbox{{\rm-mod}}$ and $M$ an object in $\mathcal{C}$. Suppose $X\rightarrow
M^{\prime}\rightarrow Y$ is an ${{\rm add}}(M)$-split sequence in
$\mathcal{C}$. Then the endomorphism algebras ${{\rm
End}}_{\mathcal{C}}(M\oplus X)$ and ${{\rm End}}_{\mathcal{C}}(M\oplus Y)$ are
derived equivalent via a tilting module.

The next simple observation characterizes Morita equivalences.

######  Lemma 2.2.

Let $A$ be an algebra and $M,N\in A\mbox{{\rm-mod}}$. Then ${{\rm
End}}_{A}(M)$ and ${{\rm End}}_{A}(N)$ are Morita equivalent if and only if
${{\rm add}}(M)$ and ${{\rm add}}(N)$ are equivalent as $R$-linear categories.

######  Lemma 2.3.

[13, Section 3, Remark] Let $A$ be a self-injective algebra and $X\in
A\mbox{{\rm-mod}}$. Then the endomorphism algebras ${{\rm End}}_{A}(A\oplus
X)$ and ${{\rm End}}_{A}(A\oplus\Omega_{A}(X))$ are almost $\nu$-stable
derived equivalent, where $\Omega_{A}(X)$ stands for the syzygy of $X$.

######  Lemma 2.4.

[8, Theorem 4.4] Let $A$ and $B$ be symmetric algebras, and let $F$ be an
almost $\nu$-stable derived equivalence between two gendo-symmetric algebras
${\rm End}_{A}(A\oplus M)$ and ${\rm End}_{B}(B\oplus N)$, where $M$ and $N$
are basic non-zero modules without projective summands. Then $A$ and $B$ are
(almost $\nu$-stable) derived equivalent. Furthermore, $F$ induces a stable
equivalence $\overline{F}:A$-mod $\rightarrow B$-mod with $\overline{F}(M)=N$.

######  Lemma 2.5.

Let $A$ and $B$ be commutative self-injective algebras, ${}_{A}M$ and
${}_{B}N$ be faithful modules over $A$ and $B$, respectively. If the
endomorphism algebras ${{\rm End}}_{A}(M)$ and ${{\rm End}}_{B}(N)$ are
derived equivalent, then $A\simeq Z({\rm End}_{A}(M))\simeq Z\big{(}{{\rm
End}}_{B}(N))\simeq B$, where $Z(C)$ denotes the center of an algebra $C$.

Proof. For an algebra $C$ and a faithful $C$-module $X$, one always has an
embedding $Z(C)\hookrightarrow Z({\rm End}_{C}(X))$. Thus $A\hookrightarrow
Z({\rm End}_{A}(M))$ since $A$ is commutative. Note that a faithful module
over a self-injective algebra is clearly a generator-cogenerator. This implies
that $M_{{\rm End}_{A}(M)}$ is a right faithful module and the bimodule
${}_{A}M_{{\rm End}_{A}(M)}$ has the double centralizer property. Thus there
is an embedding $Z({\rm End}_{A}(M))\hookrightarrow{\rm End}_{{\rm
End}_{A}(M)^{\rm op}}(M)\simeq A.$ Hence $A\simeq Z({\rm End}_{A}(M))$. Now,
assume that ${{\rm End}}_{A}(M)$ and ${{\rm End}}_{B}(N)$ are derived
equivalent. Then $Z\big{(}{{\rm End}}_{A}(M)\big{)}\simeq Z\big{(}{{\rm
End}}_{B}(N)\big{)}$ by [27, Proposition 9.2], and therefore $A\simeq Z({\rm
End}_{A}(M))\simeq Z\big{(}{{\rm End}}_{B}(N))\simeq B$. $\square$

###  2.2 Modules over quotients of polynomial algebras

In this section we remind basic facts on modules over the polynomial algebra
$R[x]$. Throughout this section, $R$ is a field unless stated otherwise.

Let $f(x)$ be an irreducible polynomial in $R[x]$ and $A:=R[x]/(f(x)^{n})$ for
$n\in\mathbb{N}$. Then $A$ is a self-injective algebra by [29, Corollary
4.37]. Further, $A$ is a local, commutative, symmetric, Nakayama algebra (see,
for instance [4, Example, p.127]). Thus $A$ has $n$ indecomposable modules
$M_{f(x)}(i):=R[x]/(f(x)^{i})$ for $i\in[n].$ For simplicity, we often write
$M(i)$ for $M_{f(x)}(i)$ and understand $M(0)=0$. Clearly, ${\rm
Hom}_{A}(M(i),A)\simeq{\rm Hom}_{R}(M(i),R)\simeq M(i)$ as $A$-modules for all
$i\in[n].$

Let $B:=R[x]/(f(x)^{m})$ for $m<n$. Then there is a canonical surjective
homomorphism $\pi:A\rightarrow B$ of $R$-algebras, and each $B$-module can be
viewed as an $A$-module via $\pi$. Up to isomorphism, indecomposable
$A$-modules coming from $B$-modules are exactly those $M(i)$ with $i\in[m].$
For $M,N\in B\mbox{{\rm-mod}}$, ${{\rm Hom}}_{A}(M,N)={{\rm Hom}}_{B}(M,N)$.

For an irreducible polynomial $g(x)\in R[x]$ and a positive integers $m$, if
$A\simeq R[x]/(g(x)^{m})$ as $R$-algebras, then
$n=LL(R[x]/(f(x)^{n}))=LL(R[x]/(g(x)^{m}))=m$, and for $t\in[n]$, the
indecomposable $R[x]/(g(x)^{n})$-module $R[x]/(g(x)^{t})$ is isomorphic to the
$A$-module $R[x]/(f(x)^{t})$.

######  Lemma 2.6.

Let $a,b,c,d\in\\{0,1,\cdots,n\\}$ such that $b<a<c$, $b<d<c$ and $a+d=b+c$.
If ${}_{A}X\in A\mbox{{\rm-mod}}$ has no indecomposable direct summands $N$
with $b<\ell(N)<c$ and ${}_{A}Y:={}_{A}X\oplus M(b)\oplus M(c)$, then there is
an ${{\rm add}}(_{A}Y)$-split sequence $0\rightarrow M(a)\rightarrow
M(b)\oplus M(c)\rightarrow M(d)\rightarrow 0$.

Proof. Let $g:M(b)\to M(d)$ and $h:M(c)\to M(d)$ be the canonical injective
and surjective homomorphisms, respectively. We define
$v:=\left(\begin{smallmatrix}g\\\ h\\\ \end{smallmatrix}\right).$ Then
$v:M(b)\oplus M(c)\to M(d)$ is a surjective homomorphism. Similarly, let
$p:M(a)\to M(b)$ and $q:M(a)\to M(c)$ be the canonical surjective and
injective homomorphisms, respectively, and $u:=(-p,q)$. Then $u:M(a)\to
M(b)\oplus M(c)$ is an injective homomorphism. By the definition of $M(i)$, we
have $uv=0.$ It follows from $a+d=b+c$ that the sequence

$(\star)\quad 0\longrightarrow M(a)\stackrel{{\scriptstyle
u}}{{\longrightarrow}}M(b)\oplus M(c)\stackrel{{\scriptstyle
v}}{{\longrightarrow}}M(d)\longrightarrow 0$

of $A$-modules is exact. We shall show that $u$ and $v$ are left and right
${{\rm add}}(_{A}Y)$-approximations of $M(a)$ and $M(d)$, respectively. In
fact, we need only to show that $v$ is a right ${{\rm
add}}(_{A}Y)$-approximation of $M(d)$ because applying the dual functor ${{\rm
Hom}}_{R}(-,R)$ shows that $u$ is a left ${{\rm add}}(_{A}Y)$-approximation of
$M(a)$. To show that $v$ is a right ${{\rm add}}(_{A}Y)$-approximation of
$M(d)$, it suffices to prove that any homomorphism $h:Z\to M(d)$ with $Z$ an
indecomposable summand of $Y$ factorizes through $v$. By assumption, either
$\ell(Z)\leq b$ or $\ell(Z)\geq c.$ Suppose $\ell(Z)\leq b.$ Then
$\ell((Z)h)\leq b=\ell((M(b))g)$, and therefore $(Z)h\subset(M(b))g.$ Let
$s:Z\to M(b)\oplus M(c)$ be the homomorphism defined by
$(z)s:=(((z)h)g^{-1},0)$ for $z\in Z.$ Clearly, $h=sv$. Suppose $\ell(Z)\geq
c.$ Then max$\\{a,b,c,d\\}\leq\ell(Z)$. Let $B:=R[x]/(f(x)^{\ell(Z)})$. Then
$B$ is the quotient of $A$ by the ideal $(f(x))^{n-\ell(Z)}$, $Z\simeq
M(\ell(Z))=R[x]/(f(x)^{\ell(Z)})=B$ as $A$-modules, and the exact sequence
$(\star)$ can be viewed as the one of $B$-modules. So the exactness of ${{\rm
Hom}}_{B}(Z,-)$ implies that $h$ factorizes through $v$ in $B\mbox{{\rm-
mod}}$. Since ${{\rm Hom}}_{A}(M,N)={{\rm Hom}}_{B}(M,N)$ for $M,N\in
B\mbox{{\rm-mod}}$, $h$ factorizes through $v$ in $A\mbox{{\rm-mod}}$.
$\square$

######  Lemma 2.7.

Let $n=\sum^{s}_{i=1}\ell_{i}$ with $\ell_{i}\in\mathbb{Z}_{>0}$. For
$\sigma\in\Sigma_{s}$ and $j\in[s]$, define $M_{j}:=M(\sum^{j}_{i=1}\ell_{i})$
and $M^{\sigma}_{j}:=M(\sum^{j}_{i=1}\ell_{(i)\sigma})$. Then the endomorphism
algebras ${{\rm End}}_{A}(\bigoplus^{s}_{j=1}M_{j})$ and ${{\rm
End}}_{A}(\bigoplus^{s}_{j=1}M^{\sigma}_{j})$ are derived equivalent.

Proof. The symmetric group $\Sigma_{s}$ is generated by all transpositions
$(t,t+1),t\in[s-1]$. In particular, $\sigma\in\Sigma_{s}$ can be written as a
product of those transpositions, say $\sigma=\prod^{k}_{i=1}(t_{i},t_{i}+1)$
for $t_{i}\in[s-1].$ Set $\sigma_{k+1}:=1$ and
$\sigma_{r}:=\prod^{k}_{i=r}(t_{i},t_{i}+1)$ for all $r\in[k].$ Then
$(t_{r},t_{r}+1)\sigma_{r}=\sigma_{r+1}$.

It suffices to show that there is a derived equivalence between ${{\rm
End}}_{A}\big{(}\bigoplus^{s}_{j=1}M^{\sigma_{r}}_{j}\big{)}$ and ${{\rm
End}}_{A}\big{(}\bigoplus^{s}_{j=1}M^{\sigma_{r+1}}_{j}\big{)}$ for all
$r\in[k].$ For any $\tau\in\Sigma_{s}$, we define
$\sum^{t_{r}-1}_{i=1}\ell_{(i)\tau}=0$ if $t_{r}=1$. For $r\in[k]$, let
$a_{r}=\ell_{(t_{r}+1)\sigma_{r+1}}+\sum^{t_{r}-1}_{i=1}\ell_{(i)\sigma_{r+1}},b_{r}=\sum^{t_{r}-1}_{i=1}\ell_{(i)\sigma_{r+1}},c_{r}=\sum^{t_{r}+1}_{i=1}\ell_{(i)\sigma_{r+1}},d_{r}=\sum^{t_{r}}_{i=1}\ell_{(i)\sigma_{r+1}}$
and $Y_{r}:=\bigoplus_{j\neq t_{r}}M^{\sigma_{r+1}}_{j}.$ Then
$b_{r}<a_{r}<c_{r},b_{r}<d_{r}<c_{r},a_{r}+d_{r}=b_{r}+c_{r}$ and $Y_{r}$
contains $M\big{(}\sum^{t_{r}-1}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}\oplus
M\big{(}\sum^{t_{r}+1}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}$ as a direct
summand. Further, $Y_{r}$ has no indecomposable direct summand $Z$ with
$b_{r}<\ell(Z)<c_{r}.$ It then follows from Lemma 2.6 that there is an ${{\rm
add}}(Y_{r})$-split sequence

$0\longrightarrow
M\big{(}\ell_{(t_{r}+1)\sigma_{r+1}}+\sum^{t_{r}-1}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}\longrightarrow
M\big{(}\sum^{t_{r}-1}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}\oplus
M\big{(}\sum^{t_{r}+1}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}\longrightarrow
M\big{(}\sum^{t_{r}}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}\longrightarrow 0.$

Clearly, $\bigoplus^{s}_{j=1}M^{\sigma_{r}}_{j}=Y_{r}\oplus
M\big{(}\ell_{(t_{r}+1)\sigma_{r+1}}+\sum^{t_{r}-1}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}$
and $\bigoplus^{s}_{j=1}M^{\sigma_{r+1}}_{j}=Y_{r}\oplus
M\big{(}\sum^{t_{r}}_{i=1}\ell_{(i)\sigma_{r+1}}\big{)}.$ By Lemma 2.1, ${{\rm
End}}_{A}(\bigoplus^{s}_{j=1}M^{\sigma_{r}}_{j})$ and ${{\rm
End}}_{A}(\bigoplus^{s}_{j=1}M^{\sigma_{r+1}}_{j})$ are derived equivalent.
$\square$

######  Remark 2.8.

The sums “$\sum^{j}_{i=1}\ell_{i}$” and “$\sum^{j}_{i=1}\ell_{(i)\sigma}$”
appearing in Lemma 2.7 are related to the definition of $D$-equivalences of
matrices. For $s\geq 2$ and a series of integers
$m_{s}>m_{s-1}>\cdots>m_{1}\geq 1$, let $\ell_{1}:=m_{1}$ and
$\ell_{i}:=m_{i}-m_{i-1}$ for $2\leq i\leq s$. Then
$m_{j}=\sum^{j}_{i=1}\ell_{i}$ for $j\in[s].$ For another series of integers
$n_{s}>n_{s-1}>\cdots>n_{1}\geq 1$, if
$\\{\\{m_{s}-m_{s-1},\cdots,m_{1}\\}\\}=\\{\\{n_{s}-n_{s-1},\cdots,n_{1}\\}\\}$,
then there exists some $\sigma\in\Sigma_{s}$ such that
$n_{j}=\sum^{j}_{i=1}\ell_{(i)\sigma}$ for $j\in[s].$ Moreover, if
$\\{\\{m_{s}-m_{s-1},\cdots,m_{1}\\}\\}=\\{\\{n_{s}-n_{s-1},\cdots,n_{1}\\}\\}$
and if there are two irreducible polynomials $f(x)$ and $g(x)$ in $R[x]$ such
that $R[x]/(f(x)^{m_{s}})\simeq R[x]/(g(x)^{n_{s}})$ as algebras, then it
follows from Lemma 2.7 that ${\rm
End}_{R[x]/(f(x)^{m_{s}})}(\bigoplus_{k\in[s]}R[x]/(f(x)^{m_{k}}))$ and ${\rm
End}_{R[x]/(g(x)^{n_{s}})}(\bigoplus_{k\in[s]}R[x]/(g(x)^{n_{k}}))$ are
derived equivalent.

Two algebras $\Lambda$ and $\Gamma$ are said to be stably equivalent if their
stable module categories $\Lambda$-mod and $\Gamma$-mod are equivalent as
$R$-linear categories. Let $F$ be a stable equivalence between $\Lambda$ and
$\Gamma$. Then $F$ induces a one-to-one correspondence between
$\Lambda\mbox{{\rm-mod}}_{\mathscr{P}}$ and $\Gamma\mbox{{\rm-
mod}}_{\mathscr{P}}.$

Now, suppose that $G:A\mbox{{\rm-{\text@underline{mod}}}}\to A$-mod is a
stable equivalence. For $n\geq 2$, $\Gamma_{n-1}:=\\{M(i)\mid
i\in[n-1]\\}\subseteq A\mbox{{\rm-mod}}_{\mathscr{P}}$. Then $G$ induces an
action $\overline{G}$ on $\Gamma_{n-1}$, namely, for $M\in\Gamma_{n-1}$,
$\overline{G}(M)$ is the unique module in $\Gamma_{n-1}$ such that
$\overline{G}(M)\simeq G(M)$ in $A\mbox{{\rm-mod}}$. Clearly,
$\overline{\Omega_{A}}(M(i))=M(n-i)$, where $\Omega_{A}$ is the syzygy
operator of $A$.

######  Lemma 2.9.

Let $n\geq 2$. If $G$ is a stable equivalence from $A$ to itself, then the
induced action $\overline{G}$ on $\Gamma_{n-1}$ is either
$\overline{\Omega_{A}}$ or the identity action.

Proof. If $n=2$, the conclusion is clear. Let $n\geq 3$. Since $A$ is a local,
symmetric and Nakayama algebra, there are almost split sequences in
$A\mbox{{\rm-mod}}$:

$0\longrightarrow M(1)\longrightarrow M(2)\longrightarrow M(1)\longrightarrow
0$

and

$0\longrightarrow M(j)\longrightarrow M(j-1)\oplus M(j+1)\longrightarrow
M(j)\longrightarrow 0$

for $2\leq j\leq n-1.$ Let Irr$(X,Y)$ denote the $R$-space ${\rm
rad}_{A}(X,Y)/{\rm rad}_{A}^{2}(X,Y)$. By a general result on stable
equivalences (see [4, Lemma 1.2, p. 336]), we have
Irr$(X,Y)\simeq$Irr$(G(X),G(Y))$ as $R$-spaces for $X,Y\in A\mbox{{\rm-
mod}}_{\mathscr{P}}$ . It then follows that $G(M(1))\simeq M(1)$ or
$G(M(1))\simeq M(n-1)=\Omega_{A}(M(1))$. If $G(M(1))\simeq M(1)$, we can show
that $G(M(i))\simeq M(i)$ for $i\in[n-1]$. If $G(M(1))\simeq
M(n-1)=\Omega_{A}(M(1))$, then we consider the stable equivalence
$\Omega_{A}^{-1}G$. Since $\Omega_{A}:A\mbox{{\rm-{\text@underline{mod}}}}\to
A\mbox{{\rm-{\text@underline{mod}}}}$ is a stable equivalence, it follows from
$(\Omega_{A}^{-1}G)(M(1))\simeq M(1)$ that $(\Omega^{-1}_{A}G)(M(i))\simeq
M(i)$ for $i\in[n-1]$. Hence $\overline{G}$ is the identity map or equals
$\overline{\Omega_{A}}$. $\square$

A non-projective, non-injective simple module over an Artin algebra is called
a node if the middle term of the almost split sequence starting at the simple
module is projective (see [21]).

By [21, Lemma 1], a non-injective simple module $S$ of an Artin algebra is a
node if and only if $S$ is not a composition factor of ${\rm rad}(Q)/{\rm
soc}(Q)$ for any indecomposable projective module $Q$. Thus an Artin algebra
has no nodes if and only if every non-projective, non-injective simple module
is a composition factor of ${\rm rad}(P)/{\rm soc}(P)$ for some indecomposable
projective module $P$.

Given an Artin algebra $\Lambda$, let $I$ be the trace of the direct sum of
all non-isomorphic nodes in $\Lambda$, and $J$ be the left annihilator of $I$
in $\Lambda.$ Martínez-Villa showed in [21, Theorem 2.10] that an Artin
algebra $\Lambda$ with nodes is stably equivalent to the triangular matrix
algebra

$\Lambda^{\prime}:=\left(\begin{array}[]{cc}\Lambda/I&0\\\ I&\Lambda/J\\\
\end{array}\right)$

without nodes. It is shown that $\Lambda$ and $\Lambda^{\prime}$ have the same
numbers of non-isomorphic, non-projective simples (see [36, Lemma 2.10 (3)]).
We often say that $\Lambda^{\prime}$ is obtained from $\Lambda$ by eliminating
nodes.

######  Lemma 2.10.

Let $M$ be a generator for $A\mbox{{\rm-mod}}$.

$(1)$ If $n\geq 2$, then every simple ${\rm End}_{A}(M)$-module is neither
projective nor injective.

$(2)$ ${\rm End}_{A}(M)$ has nodes if and only if $n=2$.

Proof. Recall that $A:=R[x]/(f(x)^{n})$ with $f(x)$ an irreducible polynomial
in $R[x]$. Set $E:={\rm End}_{A}(M)$. Since $A$ is a local, symmetric and
Nakayama algebra, any indecomposable direct summand of $M$ is isomorphic to a
submodule of ${}_{A}A$. Let $e_{M}$ denote the Hom-functor ${\rm
Hom}_{A}(M,-):{\rm add}(_{A}M)\to E\mbox{{\rm-proj}}$ and $P:=e_{M}(_{A}A).$
Clearly, $e_{M}$ is an equivalence and $P$ is an indecomposable projective-
injective $E$-module with ${\rm soc}(P)\simeq{\rm top}(P)$ since $A$ is a
local symmetric Nakayama algebra. The left exactness of $e_{M}$ implies that
any indecomposable projective $E$-module is isomorphic to a submodule of $P$.

(1) Suppose $n\geq 2$. Then the indecomposable, projective-injective
$E$-module $P$ is not simple. To show that every indecomposable projective
$E$-module is not simple, we show that ${\rm Hom}_{E}(X,Y)\neq 0$ for any
indecomposable projective $E$-modules $X$ and $Y$. In fact, let $X^{\prime}$
and $Y^{\prime}$ be indecomposable direct summands of ${}_{A}M$ such that
$e_{M}(X^{\prime})=X$ and $e_{M}(Y^{\prime})=Y$. Since $A$ is a local algebra,
we always have ${\rm Hom}_{A}(X^{\prime},Y^{\prime})\neq 0$. Thus

${\rm Hom}_{E}(X,Y)={\rm
Hom}_{E}(e_{M}(X^{\prime}),e_{M}(Y^{\prime}))\simeq{\rm
Hom}_{A}(X^{\prime},Y^{\prime})\neq 0.$

This implies that $E$ has no simple projective modules. Note that the Nakayama
functor $\nu:E\mbox{{\rm-proj}}\to E$-inj is an equivalence. Hence, for any
indecomposable injective $E$-modules $U,V$, we also have

${\rm Hom}_{E}(U,V)\simeq{\rm Hom}_{E}(\nu^{-1}(U),\nu^{-1}(V))\neq 0.$

This shows that $E$ has no simple injective modules.

(2) If $n=1$, then the algebra $A$ is simple and therefore the algebra $E$ is
semisimple. Thus $E$ has no nodes by definition.

If $n=2$, then ${\rm End}_{A}(M)$ is Morita equivalent to either $A$ or the
Auslander algebra $C$ of $A$. Note that $C$ is a Nakayama algebra with $2$
indecomposable projective $C$-modules $P_{1}$ and $P_{2}=P$ of lengths $2$ and
$3$, respectively. There is an almost split sequence $0\to{\rm top}(P_{2})\to
P_{1}\to{\rm top}(P_{1})\to 0$, which shows that ${\rm End}_{A}(M)$ has a
node.

Finally, we consider $n\geq 3$. Let $S$ be a non-projective, non-injective
simple $E$-module, we show that $S$ is a composition factor of ${\rm
rad}(P)/{\rm soc}(P).$ Actually, let $Q$ be the projective cover of $S$ with
$Q={\rm Hom}_{A}(M,Y)$ for $Y\in{\rm add}(M)$. Then $Q$ is a submodule of $P$
with ${\rm soc}(Q)={\rm soc}(P)$. By (1), $Q$ is not simple. If $Q\neq P$,
then $Q\subset{\rm rad}(P)$ and $0\neq Q/{\rm soc}(Q)$ is a submodule of ${\rm
rad}(P)/{\rm soc}(P)$. Thus $S$ is a composition factor of ${\rm rad}(P)/{\rm
soc}(P)$. If $Q=P$, then the multiplicity of ${\rm top}(P)$ in $P$ is at least
$LL(A)\geq 3$, thus $S={\rm top}(P)$ is a composition factor of ${\rm
rad}(P)/{\rm soc}(P)$. Hence $E$ has no nodes. $\square$

######  Remark 2.11.

If $n=2$ and $C$ is the Auslander algebra of $A:=R[x]/(f(x)^{2})$, then $C$
has $2$ indecomposable projective modules $P_{1}$ and $P_{2}$, the non-
projective indecomposable $C$-modules are $S_{1}:={\rm top}(P_{1})$, $S_{2}:$
= top$(P_{2})$ and the injective envelope $I(S_{1})$ of $S_{1}$. Moreover,
there are only two almost split sequences of $C$-modules

$0\longrightarrow S_{1}\longrightarrow I(S_{1})\longrightarrow
S_{2}\longrightarrow 0,\quad\quad 0\longrightarrow S_{2}\longrightarrow
P_{2}\longrightarrow I(S_{1})\longrightarrow 0.$

By eliminating nodes, $A$ and $C$ are stably equivalent to the corresponding
triangular matrix algebras $A^{\prime}$ and $C^{\prime}$, respectively. The
only non-injective, indecomposable projective $A^{\prime}$-module is simple,
while $C^{\prime}$ has two non-injective, indecomposable projective modules
one of which is simple. Note that the Frobenius parts of both $A^{\prime}$ and
$C^{\prime}$ are zero.

######  Lemma 2.12.

Given $M,N\in A\mbox{{\rm-mod}}$, the endomorphism algebras ${{\rm
End}}_{A}(M)$ and ${\rm End}_{A}(N)$ are Morita equivalent if and only if the
basic modules $\mathcal{B}(M)$ and $\mathcal{B}(N)$ of $M$ and $N$ are
isomorphic.

Proof. Suppose that ${{\rm End}}_{A}(M)$ and ${\rm End}_{A}(N)$ are Morita
equivalent. Then there is an $R$-linear equivalence $G:{{\rm
add}}(M)\rightarrow{{\rm add}}(N)$. For $j\in[n]$, ${\rm
End}_{A}(R[x]/(f(x)^{j})\simeq R[x]/(f(x)^{j})$ as algebras. Thus for
indecomposable $A$-modules $X$ and $Y$, ${\rm End}_{A}(X)\simeq{\rm
End}_{A}(Y)$ if and only if $X\simeq Y.$ It then follows from ${\rm
End}_{A}(C)\simeq{\rm End}_{A}(G(C))$ for $C\in{{\rm add}}(M)$ that $X\simeq
G(X)$ for any indecomposable module $X\in{\rm add}(M)$. Therefore
$\mathcal{B}(M)\simeq\mathcal{B}(N)$ as $A$-modules. $\square$

An algebra is said to be _representation-finite_ if it has only finitely many
non-isomorphic indecomposable modules. Consequently, if ${\rm
add}(N)\subseteq{\rm add}(M)$ and ${\rm End}_{A}(M)$ is representation-finite,
then ${\rm End}_{A}(N)$ is representation-finite.

If $R$ is perfect or $f(x)\leq x^{r}-1$ for some $r\in\mathbb{Z}_{>0}$ (for
instance, $f(x)\leq m_{d}(x)$ for an invertible matrix $d\in M_{m}(R)$ of
finite order) , then $f(x)$ is a separable polynomial in $R[x]$.

######  Lemma 2.13.

If the polynomial $f(x)$ is separable and $K:=R[x]/(f(x))$, then $A$ can be
viewed as a $K$-algebra and $A\simeq K[x]/(x^{n})$ as algebras over $K$.

Proof. Since $f(x)$ is separable and ${\rm rad}(A)=(f(x))/(f(x)^{n})$, we know
that $A/{\rm rad}(A)\simeq K$ is a separable $R$-algebra. By Wedderburn-Malcev
Theorem [32, Theorems 24 and 28], there exists a subalgebra $S$ of $A$ such
that $A=S\oplus{\rm rad}(A)$ as $R$-vector spaces. Consequently, $S\simeq
A/{\rm rad}(A)\simeq K.$ So $A$ can be viewed as a $K$-algebra. Since $A$ is a
finite-dimensional, elementary, local $K$-algebra of representation-finite
type, there is a natural number $m$ such that $A\simeq K[x]/(x^{m})$. By
comparing the $K$-dimensions of the algebras in this isomorphism, we get
$m=n$. $\square$

######  Corollary 2.14.

If the polynomial $f(x)$ is separable and $g(x)\in R[x]$ is irreducible such
that $A$ is stably equivalent to $R[x]/(g(x)^{m})$ for an integer $m\geq 2$,
then $A\simeq R[x]/(g(x)^{m})$.

Proof. Since stably equivalent algebras of representation-finite type have the
same number of non-projective, indecomposable modules, we get $n-1=m-1$, and
therefore $n=m$. Set $B:=R[x]/(g(x)^{m})$. Let
$F:A\mbox{{\rm-{\text@underline{mod}}}}\rightarrow B$-mod be a stable
equivalence and $S$ the only simple $A$-module. Thanks to $n=m\geq 2$, $S$ is
not projective and ${\rm End}_{A}(S)\simeq\underline{{\rm End}}_{A}(S)$. Thus
$F(S)$ is indecomposable and

${\rm End}_{A}(S)\simeq\underline{{\rm End}}_{A}(S)\simeq\underline{{\rm
End}}_{B}(F(S))={\rm End}_{B}(F(S))/\mathcal{P}(F(S),F(S))$

is a division ring. Since $\mathcal{P}(F(S),F(S))\subset{\rm rad}({\rm
End}_{B}(F(S)))$, it follows that $\mathcal{P}(F(S),F(S))={\rm rad}({\rm
End}_{B}(F(S)))$. This yields the following isomorphism of algebras

$R[x]/(f(x))\simeq A/{\rm rad}(A)\simeq\underline{{\rm
End}}_{A}(S)\simeq\underline{{\rm End}}_{B}(F(S))\simeq B/{\rm rad}(B)\simeq
R[x]/(g(x)).$

In particular, $g(x)$ is also a separable polynomial. Let $K:=R[x]/(f(x))$.
Then Lemma 2.13 implies that $A\simeq K[x]/(x^{n})\simeq B$ as algebras.
$\square$

For $c\in M_{n}(R)$, let $R[c]$ be the unitary subalgebra of $M_{n}(R)$
generated by $c$. Then there is a surjective homomorphism
$\varphi:R[x]\rightarrow R[c]$, sending $x$ to $c$. Let $R^{n}$ be the
$n$-dimensional vector space over $R$ consisting of column vectors. Then
$R^{n}$ is naturally an $R[c]$-module and can be viewed as an $R[x]$-module
via $\varphi$. By the classification of modules over principal ideal domains,
there exist irreducible polynomials $f_{1}(x),\cdots,f_{s}(x)\in R[x]$ and
positive integers $t_{ij}$ such that

$(\star)\quad\quad
R^{n}\simeq\bigoplus^{s}_{i=1}\bigoplus^{l_{i}}_{j=1}\;R[x]/(f_{i}(x)^{t_{ij}})$

as $R[x]$-modules.

######  Definition 2.15.

The polynomials $f_{i}(x)^{t_{ij}}$ in $(\star)$ are called the _elementary
divisors_ of $c$ (over $R$).

Note that $\mathcal{E}_{c}$ is the set of pairwise distinct elementary
divisors of $c$ and $\mathcal{M}_{c}$ is the set of maximal elements of
$\mathcal{E}_{c}$ with respect to “$\leq$”. In particular, $m_{c}(x)$ is the
product of elementary divisors in $\mathcal{M}_{c}$ and ${\rm Ker}(\varphi)$
is exactly the ideal of $R[x]$ generated by $m_{c}(x)$. Let $A_{c}:=R[x]/({\rm
Ker}(\varphi))\simeq R[c].$

######  Lemma 2.16.

There is a bijection $\pi$ from $\mathcal{E}_{c}$ to the set of non-isomorphic
indecomposable direct summands of the $A_{c}$-module $R^{n}$, sending $h(x)$
to the $A_{c}$-module $R[x]/(h(x))$ for $h(x)\in\mathcal{E}_{c}$.

Suppose that $R$ is a field of characteristic $p\geq 0$. For a positive
integer $m$, there exist unique determined $s,m^{\prime}\in\mathbb{N}$ such
that $m=p^{s}m^{\prime}$ and $p\nmid m^{\prime}$, we define $\nu_{p}(m):=s$.
Here, we understand $\nu_{p}(m):=0$ if $p=0$.

In the following, we denote by $e_{ij}$ the matrix units in $M_{n}(R)$ and by
$c_{\sigma}:=\sum^{s}_{i=1}e_{i,(i)\sigma}$ the permutation matrix of
$\sigma\in\Sigma_{n}$.

######  Lemma 2.17.

Let $R$ be a field of characteristic $p\geq 0$ and $\sigma\in\Sigma_{n}$ a
permutation of cycle type $(\lambda_{1},\cdots,\lambda_{k})$. Then
$\mathcal{E}_{c_{\sigma}}=\\{f(x)^{p^{\nu_{p}(\lambda_{i})}}\mid
i\in[k],f(x)~{}\mbox{is irreducible},f(x)\leq x^{\lambda_{i}}-1\\}$.

Proof. For conjugate permutations in $\Sigma_{n}$, their corresponding
permutation matrices are similar, and therefore have the same elementary
divisors. Thus, without loss of generality, we may assume that
$\sigma=(1,\cdots,\lambda_{1})(\lambda_{1}+1,\cdots,\lambda_{1}+\lambda_{2})\cdots(\sum^{k-1}_{j=1}\lambda_{j}+1,\cdots,n)$.
Then $c_{\sigma}$ is a diagonal block-matrix of the form
$c_{\sigma}=c_{\sigma_{1}}\oplus c_{\sigma_{2}}\oplus\cdots\oplus
c_{\sigma_{k}}$, where $\sigma_{i}$ is a $\lambda_{i}$-cycle in
$\Sigma_{\lambda_{i}}$ for $i\in[k]$. In particular,
$\mathcal{E}_{c_{\sigma}}=\bigcup_{i\in[k]}\mathcal{E}_{c_{\sigma_{i}}}$. For
a matrix $d\in M_{m}(R)$, let $p_{d}(x)$ denote the characteristic polynomial
of $d$ in $R[x]$. For $i\in[k]$, we write
$\lambda_{i}=p^{\nu_{p}(\lambda_{i})}\lambda_{i}^{\prime}$ with
$p\nmid\lambda^{\prime}_{i}$. Then
$x^{\lambda^{\prime}_{i}}-1=\prod^{h_{i}}_{j=1}f_{ij}(x)$ is a product of
distinct irreducible polynomials $f_{ij}(x)$ in $R[x]$. It follows from
$p_{c_{\sigma_{i}}}(x)=x^{\lambda_{i}}-1=x^{p^{\nu_{p}(\lambda_{i})}\lambda_{i}^{\prime}}-1=(x^{\lambda^{\prime}_{i}}-1)^{p^{\nu_{p}(\lambda_{i})}}=\prod^{h_{i}}_{j=1}f_{ij}(x)^{p^{\nu_{p}(\lambda_{i})}}$
that $p_{c_{\sigma_{i}}}(x)=m_{c_{\sigma_{i}}}(x)=x^{\lambda_{i}}-1$. Hence
$\mathcal{E}_{c_{\sigma_{i}}}=\mathcal{M}_{c_{\sigma_{i}}}$. This implies
$\mathcal{E}_{c_{\sigma}}=\\{f(x)^{p^{\nu_{p}(\lambda_{i})}}\mid
i\in[k],f(x)~{}\mbox{is irreducible},f(x)\leq x^{\lambda_{i}}-1\\}$. $\square$

Now, we prove a result on congruences of matrices that appear as the Cartan
matrices of the endomorphism rings of modules over polynomial algebras. Two
multisets $\\{\\{x_{1},\cdots,x_{s}\\}\\}$ and
$\\{\\{y_{1},\cdots,y_{s}\\}\\}$ are equal if and only if there exists a
$\sigma\in\Sigma_{s}$ such that
$(y_{1},\cdots,y_{s})^{\sigma}:=(y_{(1)\sigma},\cdots,y_{(s)\sigma})=(x_{1},\cdots,x_{s}).$

######  Lemma 2.18.

For an integer $s\geq 2$, let $m_{1}>m_{2}>\cdots>m_{s}\geq 1$ and
$n_{1}>n_{2}>\cdots>n_{s}\geq 1$ be two series of integers with $m_{1}=n_{1}$.
Set $X:=\sum^{s}_{k=1}(\sum^{k}_{l=1}m_{k}(e_{kl}+e_{lk})-m_{k}e_{kk})\in
M_{s}(\mathbb{Z})$ and
$Y:=\sum^{s}_{k=1}(\sum^{k}_{l=1}n_{k}(e_{kl}+e_{lk})-n_{k}e_{kk})\in
M_{s}(\mathbb{Z})$. Then $X$ and $Y$ are congruent in $M_{s}(\mathbb{Z})$ if
and only if there is $\sigma\in\Sigma_{s}$ such that
$(n_{1}-n_{2},\cdots,n_{s-1}-n_{s},n_{s})=(m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s})^{\sigma}$.

Proof. We define $U:=I-\sum^{s-1}_{t=1}e_{t,t+1},D_{1}:={\rm
diag}(m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s})$ and $D_{2}:={\rm
diag}(n_{1}-n_{2},\cdots,n_{s-1}-n_{s},n_{s}).$ Then $U^{tr}XU=D_{1}$ and
$U^{tr}YU=D_{2}$ Thus $X$ and $Y$ are congruent in $M_{s}(\mathbb{Z})$ if and
only if $D_{1}$ and $D_{2}$ are congruent in $M_{s}(\mathbb{Z}).$ Now, we show
that $D_{1}$ and $D_{2}$ are congruent in $M_{s}(\mathbb{Z})$ if and only if
there is an element $\sigma\in\Sigma_{s}$ such that
$(n_{1}-n_{2},\cdots,n_{s-1}-n_{s},n_{s})=(m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s})^{\sigma}.$
Indeed, if
$(n_{1}-n_{2},\cdots,n_{s-1}-n_{s},n_{s})=(m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s})^{\sigma},$
then $c^{tr}_{\sigma}D_{1}c_{\sigma}=D_{2}$. This means that $D_{1}$ and
$D_{2}$ are congruent in $M_{s}(\mathbb{Z}).$ Conversely, suppose that $D_{1}$
and $D_{2}$ are congruent in $M_{s}(\mathbb{Z})$. Then there is an invertible
matrix $H=(a_{ij})_{1\leq i,j\leq s}\in M_{n}(\mathbb{Z})$ such that
$H^{tr}D_{1}H=D_{2}.$ This implies

$(*)\quad\sum^{s-1}_{r=1}(\sum^{s}_{k=1}(a^{2}_{kr}))(m_{r}-m_{r+1})+(\sum^{s}_{k=1}(a^{2}_{ks}))m_{s}=n_{1}=m_{1}.$

Since $H$ is invertible in $M_{s}(\mathbb{Z})$, each column of $H$ has a
nonzero element, and therefore $\sum^{s}_{k=1}(a^{2}_{kr})\geq 1$ for
$r\in[s].$ Now it follows from ($*$) that $\sum^{s}_{k=1}(a^{2}_{kr})=1$ for
all $r\in[s]$. Thus each row and column of $H$ has only one nonzero entry
which is either $1$ or $-1$. This implies that $H=\epsilon c_{\tau}$ for
$\tau\in\Sigma_{s}$ and $\epsilon$ a diagonal matrix with the entries in
$\\{1,-1\\}$. Hence $H^{tr}=H^{-1}.$ This shows that $D_{1}$ and $D_{2}$ are
similar. So
$\\{\\{m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s}\\}\\}=\\{\\{n_{1}-n_{2},\cdots,n_{s-1}-n_{s},n_{s}\\}\\}$
as multisets, and therefore
$(n_{1}-n_{2},\cdots,n_{s-1}-n_{s},n_{s})=(m_{1}-m_{2},\cdots,m_{s-1}-m_{s},m_{s})^{\sigma}$
for some $\sigma\in\Sigma_{s}.$ $\square$

###  2.3 Centralizer matrix algebras of representation-finite type

We first mention a few properties of centralizer matrix algebras.

######  Lemma 2.19.

For $c\in M_{n}(R)$, the followings hold.

$(1)$ There are isomorphisms of $R$-algebras:

$S_{n}(c,R)\simeq S_{n}(c^{tr},R)\simeq S_{n}(c,R)^{{}^{\rm op}}\simeq{\rm
End}_{A_{c}}(R^{n}),$

where $c^{tr}$ denotes the transpose of the matrix $c$.

$(2)$ Let $p_{c}(x)$ be the characteristic polynomial of $c$. Then
$S_{n}(c,R)=R[c]$ if and only if $m_{c}(x)=p_{c}(x)$.

Proof. (1) The first isomorphism follows from the fact that any matrix over a
field is similar to its transpose [17, Theorem 66, p.76], the second
isomorphism is given by sending a matrix in $S_{n}(c^{tr},R)$ to its transpose
in $S_{n}(c,R)^{{}^{\rm op}}$, and the last isomorphism follows by
interpreting $c$ as a linear transformation on the $n$-dimensional $R$-space
$R^{n}$ with respect to a basis.

(2) Note that $m_{c}(x)$ is the product of polynomials in $\mathcal{M}_{c}$
and that $p_{c}(x)$ is the product of all elementary divisors (counting
multiplicity) of $c$. Thus $m_{c}(x)=p_{c}(x)$ if and only if
$\mathcal{M}_{c}$ consists of all elementary divisors of $c$. Now, suppose
$m_{c}(x)=p_{c}(x)$. Then $\mathcal{M}_{c}=\mathcal{E}_{c}$ and it follows
from the decomposition of $R^{n}$ in $(\star)$ that $R^{n}\simeq R[c]$ as
$R[c]$-modules. Thus $S_{n}(c,R)\simeq{\rm End}_{R[c]}(R^{n})\simeq{\rm
End}_{R[c]}(R[c])\simeq R[c]$, and therefore the subalgebra $R[c]$ of
$S_{n}(c,R)$ has the same $R$-dimension as $S_{n}(c,R)$ does. Hence
$R[c]=S_{n}(c,R)$.

Suppose $S_{n}(c,R)=R[c]$. Then $S_{n}(c,R)=R[c]\simeq R[x]/(m_{c}(x))$. This
shows that $S_{n}(c,R)$ is a symmetric algebra. Hence $R^{n}$ is a projective
generator for $R[c]\mbox{{\rm-mod}}$ (see, for example, the argument of the
proof of Lemma 3.5(2)). Since $R[c]\simeq
R[x]/(m_{c}(x))\simeq\prod_{f(x)\in\mathcal{M}_{c}}R[x]/(f(x))$ is a basic
algebra, we have $R^{n}\simeq R[c]\oplus M$ for a projective $R[c]$-module
$M$. Thus $R[c]=S_{n}(c,R)\simeq{\rm End}_{R[c]}(R^{n})\simeq{\rm
End}_{R[c]}(R[c]\oplus M)$. This implies $M=0$, that is, $R^{n}\simeq R[c]$ as
$R[c]$-modules. In particular, ${\rm deg}(m_{c}(x))={\rm dim}_{R}(R[c])={\rm
dim}_{R}(R^{n})=n$. Then ${\rm deg}(p_{c}(x))=n={\rm deg}(m_{c}(x))$. But
$m_{c}(x)$ is a divisor of $p_{c}(x)$, and therefore $m_{c}(x)=p_{c}(x)$.
$\square$

In general, $S_{n}(c,R)$ has neither to equal $R[c]$, nor to be
representation-finite. Next, we point out a condition for $S_{n}(c,R)$ to be
representation-finite in terms of elementary divisors.

######  Lemma 2.20.

Suppose $R$ is a perfect field. For $c\in M_{n}(R)$ and
$f(x)\in\mathcal{M}_{c}$, let $b_{f(x)}:=\mbox{max}\\{3,P_{c}(f(x))\\}$. Then
$S_{n}(c,R)$ is representation-finite if and only if
$P_{c}(f(x))\subseteq\\{1,b_{f(x)}-1,b_{f(x)}\\}$ for all
$f(x)\in\mathcal{M}_{c}$.

Proof. Since $R$ is perfect, all irreducible factors of $m_{c}(x)$ are
separable over $R$. Clearly, $S_{n}(c,R)$ is representation-finite if and only
if every block of $S_{n}(c,R)$ is representation-finite. The blocks of
$S_{n}(c,R)$ are parameterized by $\mathcal{M}_{c}$. Let $g(x)\in R[x]$ be an
irreducible polynomial such that $g(x)^{s}\in\mathcal{M}_{c}$ with
$s\in\mathbb{N}$. Then $b_{g(x)^{s}}=\max\\{3,s\\}$. By Lemmas 2.2, 2.16 and
2.19, the block of $S_{n}(c,R)$ related to $g(x)^{s}$ is Morita equivalent to
the algebra

$E:={\rm End}_{R[x]/(g(x)^{s})}\big{(}\bigoplus_{t\in
P_{c}(g(x)^{s})}R[x]/(g(x)^{t})\big{)}.$

Since $R$ is a perfect field, the algebraic closure $\overline{R}$ of $R$ is a
separable extension of $R$. By [16, Theorem 3.3] which says that, for a
separable extension $L/R$, a finite-dimensional $R$-algebra $A$ is
representation-finite if and only if so is the $L$-algebra $L\otimes_{R}A$.
Hence it suffices to consider when $\overline{R}\otimes_{R}E$ is
representation-finite. Since $R$ is perfect, $g(x)$ has only simple roots in
$\overline{R}$. Let $\alpha_{1},\cdots,\alpha_{m}$ be the roots of $g(x)$ in
$\overline{R}$. Then

$\overline{R}\otimes_{R}E\simeq{\rm
End}_{\overline{R}\otimes_{R}R[x]/(g(x)^{s})}\big{(}\overline{R}\otimes_{R}\bigoplus_{t\in
P_{c}(g(x)^{s})}R[x]/(g(x)^{t})\big{)}\simeq{\rm
End}_{\overline{R}[x]/(\prod^{m}_{i=1}(x-\alpha_{i})^{s})}\big{(}\bigoplus_{t\in
P_{c}(g(x)^{s})}\overline{R}[x]/(\prod^{m}_{i=1}(x-\alpha_{i})^{t})\big{)}.$

Thus each block of $\overline{R}\otimes_{R}E$ is isomorphic to ${\rm
End}_{\overline{R}[x]/(x^{s})}\big{(}\bigoplus_{t\in
P_{c}(g(x)^{s})}\overline{R}[x]/(x^{t})\big{)}.$ By [11, Theorem 2.1 (i)] (see
also [10]), the endomorphism algebra ${\rm
End}_{\overline{R}[x]/(x^{s})}\big{(}\bigoplus_{t\in
P_{c}(g(x)^{s})}\overline{R}[x]/(x^{t})\big{)}$ is representation-finite if
and only if either $s\leq 3$ and $P_{c}(g(x)^{s})\subseteq\\{1,2,3\\}$ or
$s\geq 4$ and $P_{c}(g(x)^{s})\subseteq\\{1,s-1,s\\}$. This is equivalent to
saying that $P_{c}(g(x)^{s})\subseteq\\{1,b_{g(x)^{s}}-1,b_{g(x)^{s}}\\}$.
$\square$

As a corollary of Lemma 2.20, we have the following.

######  Corollary 2.21.

Let $R$ be a perfect field of characteristic $p$ and $c\in M_{n}(R)$ be a
permutation matrix such that the associated permutation of $c$ is of the cycle
type $\lambda=(\lambda_{1},\cdots,\lambda_{s})$. Then $S_{n}(c,R)$ is
representation-finite if and only if there exists an positive integer $t$ such
that $\nu_{p}(\lambda_{i})\in\\{0,t\\}$ for all $i\in[s]$.

Proof. By Lemma 2.17, for $g(x)\in\mathcal{M}_{c}$, all the integers in
$P_{c}(g(x))$ are $p$-powers and $(x-1)^{p^{\nu_{p}(\lambda_{i})}}$ is an
elementary divisor of $c$ for $i\in[s]$. In particular,
$(x-1)^{p^{m}}\in\mathcal{M}_{c}$ with $m=\max\\{\nu_{p}(\lambda_{i})\mid
i\in[s]\\}$.

Now, suppose that $S_{n}(c,R)$ is representation-finite. By Lemma 2.20, we
deduce that $P_{c}((x-1)^{p^{m}})$ does not contain two different $p$-powers
$p^{a}>1$ and $p^{b}>1$. Since $p^{\nu_{p}(\lambda_{i})}\in
P_{c}((x-1)^{p^{m}})$ for $i\in[s]$, there do not exist $\lambda_{i}$ and
$\lambda_{j}$ with $i,j\in[s]$ such that
$\nu_{p}(\lambda_{i})>\nu_{p}(\lambda_{j})\geq 1$, that is, there exists an
integer $t>0$ such that $\nu_{p}(\lambda_{i})\in\\{0,t\\}$ for all $i\in[s]$.

Conversely, suppose that there exists an integer $t>0$ such that
$\nu_{p}(\lambda_{i})\in\\{0,t\\}$ for all $i\in[s]$. Then, for each
$g(x)\in\mathcal{M}_{c}$, we deduce from Lemma 2.17 that
$P_{c}(g(x))\subseteq\\{1,p^{t}\\}$. Thus it follows from Lemma 2.20 that
$S_{n}(c,R)$ is representation-finite. $\square$

##  3 Derived and stable equivalences of centralizer matrix algebras

This section is devoted to proving all results mentioned in the introduction.

Assume that $R$ is a field of characteristic $p\geq 0$ unless stated
otherwise. For $c\in M_{n}(R)$, let $m_{R,c}(x)$ or $m_{c}(x)$ be the minimal
polynomial of $c$ over $R$ and $A_{c}:=R[x]/(m_{c}(x))$. Now, let $d\in
M_{m}(R)$. We write

$m_{c}(x)=\prod^{l_{c}}_{i=1}f_{i}(x)^{n_{i}}\mbox{ for }n_{i}\geq 1\;\mbox{
and }\;m_{d}(x)=\prod^{l_{d}}_{j=1}g_{j}(x)^{m_{j}}\;\mbox{ for }m_{j}\geq 1,$
$U_{i}:=R[x]/(f_{i}(x)^{n_{i}})\mbox{ for }i\in[l_{c}]\;\mbox{ and
}\;\;V_{j}:=R[x]/(g_{j}(x)^{m_{j}})\mbox{ for }j\in[l_{d}],$

where $f_{i}(x)$ and $g_{j}(x)$ are irreducible polynomials in $R[x]$. Thus
$U_{i}$ and $V_{j}$ are local, symmetric Nakayama $R$-algebras, and

$A_{c}\simeq U_{1}\times U_{2}\times\cdots\times U_{l_{c}}\mbox{ and
}\;A_{d}\simeq V_{1}\times V_{2}\times\cdots\times V_{l_{d}}.$

Recall that $A_{c}\simeq R[c]$ and
$R^{n}=\\{(a_{1},a_{2},\cdots,a_{n})^{tr}\mid a_{i}\in R,1\leq i\leq n\\}$ is
viewed as an $A_{c}$-module. According to these blocks of $A_{c}$ and $A_{d}$,
we decompose the $A_{c}$-module $R^{n}$ and the $A_{d}$-module $R^{m}$ as

$R^{n}=\bigoplus^{l_{c}}_{i=1}M_{i}\;\mbox{ and
}\;R^{m}=\bigoplus^{l_{d}}_{j=1}N_{j},$

where $M_{i}$ (respectively, $N_{j}$) is the sum of indecomposable direct
summands of $R^{n}$ (respectively, $R^{m}$) belonging to the block $U_{i}$
(respectively, $V_{j}$). Then
$\mathcal{B}(M_{i})\simeq\bigoplus_{r\in{P_{c}(f_{i}(x)^{n_{i}})}}R[x]/(f_{i}(x)^{r})$
as $U_{i}$-modules and
$\mathcal{B}(N_{j})\simeq\bigoplus_{s\in{P_{d}(g_{j}(x)^{m_{j}})}}R[x]/(g_{j}(x)^{s})$
as $V_{j}$-modules. Since $R^{n}$ is a faithful $M_{n}(R)$-module, $R^{n}$ is
also a faithful $R[c]$-module, and therefore $M_{i}$ is a faithful
$U_{i}$-module for $i\in[l_{c}]$. Similarly, $N_{j}$ is a faithful
$V_{j}$-module for $j\in[l_{d}]$. Further, we set

$A_{i}:={\rm End}_{U_{i}}(M_{i})\;\mbox{ and }\;B_{j}:={\rm
End}_{V_{j}}(N_{j})$

for $i\in[l_{c}]$ and $j\in[l_{d}]$. Then $A_{i}$ and $B_{j}$ are
indecomposable as algebras for $i\in[l_{c}]$ and $j\in[l_{d}]$. Clearly,
$A_{i}$ (respectively, $B_{j}$) is semisimple if and only if $n_{i}=1$
(respectively, $m_{j}=1$). In this case, $A_{i}\simeq M_{k}(R[x]/(f_{i}(x)))$
for some $k\in\mathbb{N}$ (respectively, $B_{j}\simeq M_{t}(R[x]/(g_{j}(x)))$
for some $t\in\mathbb{N}$). By Lemma 2.19,

$S_{n}(c,R)\simeq\prod^{l_{c}}_{i=1}{{\rm
End}}_{U_{i}}(M_{i})=\prod^{l_{c}}_{i=1}A_{i}\,\mbox{ and
}\;S_{m}(d,R)\simeq\prod^{l_{d}}_{j=1}{{\rm
End}}_{V_{j}}(N_{j})=\prod^{l_{d}}_{i=1}B_{j}.$

As the $R[c]$-module $R^{n}$ is a generator, we see that the bimodule
${}_{R[c]}R^{n}_{S_{n}(c,R)}$ has the double centralizer property. In
particular, ${\rm End}_{S_{n}(c,R)}(R^{n}_{{}_{S_{n}(c,R)}})=R[c]$.

###  3.1 Characterizations of Morita and derived equivalences: Proof of
Theorem 1.2

In this section we prove the main result, Theorem 1.2.

######  Lemma 3.1.

$(1)$ $\mathcal{M}_{c}=\\{f_{i}(x)^{n_{i}}\mid 1\leq i\leq l_{c}\\}$ and
$|\mathcal{M}_{c}|=l_{c}$.

$(2)$ If $A_{i}$ and $B_{j}$ are derived equivalent, then $U_{i}\simeq V_{j}$
and $n_{i}=m_{j}$.

Proof. (1) follows by definition. (2) is a consequence of Lemma 2.5. $\square$

######  Lemma 3.2.

Let $c\in M_{n}(R)$ and $d\in M_{m}(R)$. Then $c\stackrel{{\scriptstyle
M}}{{\sim}}d$ if and only if there is an isomorphism $\varphi:R[c]\simeq R[d]$
of algebras such that $\mathcal{B}(R^{n})\simeq\mathcal{B}(R^{m})$, where
$R^{m}$ is viewed as an $R[c]$-module via $\varphi$.

Proof. Suppose $c\stackrel{{\scriptstyle M}}{{\sim}}d$. Then, by definition,
there is a bijection $\pi$ between $\mathcal{M}_{c}$ and $\mathcal{M}_{d}$
such that $R[x]/(f(x))\simeq R[x]/((f(x))\pi)$ as algebras and
${P_{c}(f(x))}={P_{d}((f(x))\pi)}$ for all $f(x)\in\mathcal{M}_{c}.$ It
follows from

$R[c]\simeq\prod_{f(x)\in\mathcal{M}_{c}}R[x]/(f(x))\mbox{ and
}\;R[d]\simeq\prod_{g(x)\in\mathcal{M}_{d}}R[x]/(g(x))$

that there is an isomorphism $\varphi:R[c]\simeq R[d].$ After reordering the
factors, we may assume that $(f_{i}(x)^{n_{i}})\pi=g_{i}(x)^{m_{i}}$ for
$i\in[l_{c}].$ Then the condition ${P_{c}(f(x))}={P_{d}((f(x))\pi)}$ implies
that $\mathcal{B}(M_{i})\simeq\mathcal{B}(N_{i})$ for $i\in[l_{c}]$. Here
$N_{i}$ is viewed as an $R[c]$-module via $\varphi$. Hence
$\mathcal{B}(R^{n})\simeq\mathcal{B}(R^{m})$, where $R^{m}$ is viewed as an
$R[c]$-module via $\varphi$.

Conversely, suppose that there is an isomorphism $\varphi:R[c]\simeq R[d]$
such that $\mathcal{B}(R^{n})\simeq\mathcal{B}(R^{m})$ when $R^{m}$ is
regarded as an $R[c]$-module via $\varphi$. We may assume that $\varphi$
restricts to an isomorphism $\varphi_{i}:U_{i}\simeq V_{i}$ for $i\in[l_{c}].$
Then the condition $\mathcal{B}(R^{n})\simeq\mathcal{B}(R^{m})$ implies that
$\mathcal{B}(M_{i})\simeq\mathcal{B}(N_{i})$ for $i\in[l_{c}].$ Since
$\mathcal{B}(M_{i})\simeq\bigoplus_{r\in{P_{c}(f_{i}(x)^{n_{i}})}}R[x]/(f_{i}(x)^{r})$
as $U_{i}$-modules and
$\mathcal{B}(N_{i})\simeq\bigoplus_{s\in{P_{d}(g_{i}(x)^{m_{i}})}}R[x]/(g_{i}(x)^{s})$
as $V_{i}$-modules, we have
${P_{c}(f_{i}(x)^{n_{i}})}={P_{d}(g_{i}(x)^{m_{i}})}.$ Now we define a map
$\pi:\mathcal{M}_{c}\to\mathcal{M}_{d}$ by $f_{i}(x)^{n_{i}}\mapsto
g_{i}(x)^{m_{i}}$ for $i\in[l_{c}].$ Then $\pi$ defines an $M$-equivalence
$c\stackrel{{\scriptstyle M}}{{\sim}}d$. $\square$

Recall that for an algebra $A$ and an idempotent $e\in A$, the Schur functor
$Ae\otimes_{eAe}-:eAe\mbox{{\rm-mod}}\to A\mbox{{\rm-mod}}$ is fully faithful.
Following [14, Section 2], we say that a stable equivalence
$\Phi:A\mbox{{\rm-{\text@underline{mod}}}}\rightarrow B$-mod of Morita type
lifts to a Morita equivalence if there is a Morita equivalence $F:A\mbox{{\rm-
mod}}{}\rightarrow B\mbox{{\rm-mod}}$ such the following diagram of functors
is commutative (up to natural isomorphism)

$\textstyle{A\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{B\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{A\mbox{{\rm-
mod}}}$$\textstyle{B\mbox{{\rm-
mod}}}$$\scriptstyle{can.}$$\scriptstyle{can.}$$\scriptstyle{\Phi}$$\scriptstyle{F}$

The following is proved in [14, Lemma 3.1, Propositions 3.3 and 3.5, and
Remark 4.6].

######  Lemma 3.3.

Let $A$ and $B$ be two algebras without nonzero semisimple direct summands
such that $A/{\rm rad}(A)$ and $B/{\rm rad}(B)$ are separable, and let $e$ and
$f$ be $\nu$-stable idempotent in $A$ and $B$ such that $eAe$ and $fAf$ are
their Frobenius parts. Suppose there is a stable equivalence
$\Phi:A\mbox{{\rm-{\text@underline{mod}}}}\rightarrow B$-mod of Morita type.
Then the followings hold.

$(1)$ If $\Phi(S)$ is isomorphic in $B$-mod to a simple $B$-module for each
simple $A$-module $S$, then $\Phi$ lifts to a Morita equivalence.

$(2)$ The functor $\Phi$ restricts to a stable equivalence
$\Phi_{1}:eAe\mbox{{\rm-{\text@underline{mod}}}}\rightarrow fBf$-mod of Morita
type such that the following diagram is commutative (up to natural
isomorphism)

$\textstyle{A\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{B\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{eAe\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{fBf\mbox{{\rm-{\text@underline{mod}}}}}$$\scriptstyle{\lambda}$$\scriptstyle{\lambda}$$\scriptstyle{\Phi}$$\scriptstyle{\Phi_{1}}$

where $\lambda$ stands for the corresponding Schur functor. Moreover, if
$\Phi_{1}$ lifts to a Morita equivalence, then so is $\Phi$.

Proof of Theorem 1.2. If $S_{n}(c,R)$ and $S_{m}(d,R)$ are Morita (or derived,
or almost $\nu$-stable derived) equivalent, then they have the same number of
blocks, that is, $l_{c}=l_{d}$. Further, we may assume that $A_{i}$ and
$B_{i}$ are Morita (or derived, or almost $\nu$-stable derived)) equivalent
and that $F_{i}$ is such an equivalence for $i\in[l_{c}]$. By Lemma 3.1(2),
there is an isomorphism $\varphi_{i}:U_{i}\simeq V_{i}$ of algebras and
$n_{i}=m_{i}$ for $i\in[l_{c}].$

(1) Suppose $c\stackrel{{\scriptstyle M}}{{\sim}}d$. Then it follows from
Lemmas 2.2 and 3.2 that $S_{n}(c,R)$ and $S_{m}(d,R)$ are Morita equivalent.
Conversely, suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are Morita equivalent.
Then it follows from Lemma 2.12 that
$\mathcal{B}(M_{i})\simeq\mathcal{B}(N_{i})$ if $N_{i}$ is regarded as a
$U_{i}$-module via $\varphi_{i}$. Let $\psi_{c}:A_{c}\rightarrow R[c]$ and
$\psi_{d}:A_{d}\rightarrow R[d]$ be the canonical isomorphisms of algebras.
Then $\varphi:=\psi^{-1}_{c}(\prod^{l_{c}}_{i=1}\varphi_{i})\psi_{d}$ is an
isomorphism between the algebras $R[c]$ and $R[d]$. Hence
$\mathcal{B}(R^{n})\simeq\mathcal{B}(R^{m})$ where $R^{m}$ is viewed as an
$R[c]$-module via $\varphi$. By Lemma 3.2, we have $c\stackrel{{\scriptstyle
M}}{{\sim}}d.$

(2) Suppose $c\stackrel{{\scriptstyle D}}{{\sim}}d$. By the definition of
$D$-equivalences, $A_{c}\simeq A_{d}$ as algebras and there is a map
$\pi:\mathcal{M}_{c}\to\mathcal{M}_{d}$ such that
$\mathcal{H}_{P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{H}_{P_{d}((f_{i}(x)^{n_{i}})\pi)}$
for $f_{i}(x)^{n_{i}}\in\mathcal{M}_{c}.$ Without loss of generality, we
assume $(f_{i}(x)^{n_{i}})\pi=g_{i}(x)^{m_{i}}$ for $i\in[l_{c}].$ Then, for
$i\in[l_{c}]$, $R[x]/(f_{i}(x)^{n_{i}})\simeq R[x]/(g_{i}(x)^{m_{i}})$ as
algebras and
$\mathcal{H}_{P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{H}_{P_{d}(g_{i}(x)^{m_{i}})}$.
Recall that
$\mathcal{B}(M_{i})\simeq\bigoplus_{r\in{P_{c}(f_{i}(x)^{n_{i}})}}R[x]/(f_{i}(x)^{r})$
as $U_{i}$-modules and
$\mathcal{B}(N_{i})\simeq\bigoplus_{s\in{P_{d}(g_{i}(x)^{m_{i}})}}R[x]/(g_{i}(x)^{s})$
as $V_{i}$-modules. It follows from Remark 2.8 that ${\rm
End}_{U_{i}}(\mathcal{B}(M_{i}))$ and ${\rm End}_{V_{i}}(\mathcal{B}(N_{i}))$
are derived equivalent. Thanks to Lemma 2.2, $A_{i}$ and $B_{i}$ are also
derived equivalent. Thus $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent.

Conversely, suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent.
Let $i\in[l_{c}]$. Then $A_{i}$ and $B_{i}$ are derived equivalent, and there
is an isomorphism $U_{i}\simeq V_{i}$ of algebras such that $U_{i}/{\rm
rad}(U_{i})\simeq V_{i}/{\rm rad}(V_{i})$, that is, $R[x]/(f_{i}(x))\simeq
R[x]/(g_{i}(x)).$ Let $K_{i}$ be a splitting field for $f_{i}(x)g_{i}(x).$
Since $K_{i}\otimes_{R}A_{i}\simeq{\rm
End}_{K_{i}\otimes_{R}U_{i}}(K_{i}\otimes_{R}M_{i})$ and
$K_{i}\otimes_{R}B_{i}\simeq{\rm
End}_{K_{i}\otimes_{R}V_{i}}(K_{i}\otimes_{R}N_{i})$, the two algebras ${\rm
End}_{K_{i}\otimes_{R}U_{i}}(K_{i}\otimes_{R}M_{i})$ and ${\rm
End}_{K_{i}\otimes_{R}V_{i}}(K_{i}\otimes_{R}N_{i})$ are derived equivalent.

For the irreducible polynomial $f_{i}(x)$, there is a separable irreducible
polynomial $u_{i}(x)\in R[x]$ and an integer $s_{i}\in\mathbb{N}$ such that
$f_{i}(x)=u_{i}(x^{p^{s_{i}}})$. Here, for $p=0$, we understand $p^{s_{i}}=1$.
Similarly, there is a separable irreducible polynomial $v_{i}(x)$ and an
integer $t_{i}\in\mathbb{N}$ such that $g_{i}(x)=v_{i}(x^{p^{t_{i}}})$. It
follows from $K_{i}\otimes_{R}\big{(}R[x]/(f_{i}(x))\big{)}\simeq
K_{i}\otimes_{R}\big{(}R[x]/(g_{i}(x))\big{)}$ that $s_{i}=t_{i}$ and that
$u_{i}(x)$ and $v_{i}(x)$ have the same number of roots. Therefore
$f_{i}(x),g_{i}(x),u_{i}(x)$ and $v_{i}(x)$ have the same number of distinct
roots in $K_{i}$. Let $w_{i}$ be the number of roots of $u_{i}(x)$ in $K_{i}$.
Suppose that $\alpha_{i1},\alpha_{i2},\cdots,\alpha_{iw_{i}}$ are the roots of
$f_{i}(x)$ in $K_{i}$ and that $\beta_{i1},\beta_{i2},\cdots,\beta_{iw_{i}}$
are the roots of $g_{i}(x)$ in $K_{i}$. Then
$K_{i}\otimes_{R}U_{i}=K_{i}\otimes_{R}\big{(}R[x]/(f_{i}(x)^{n_{i}})\big{)}\simeq\prod^{w_{i}}_{q=1}K_{i}[x]/((x-\alpha_{iq})^{n_{i}\cdot
p^{s_{i}}})$. Similarly,
$K_{i}\otimes_{R}V_{i}=K_{i}\otimes_{R}\big{(}R[x]/(g_{i}(x)^{m_{i}})\big{)}\simeq\prod^{w_{i}}_{q=1}K_{i}[x]/((x-\beta_{iq})^{m_{i}\cdot
p^{s_{i}}}).$ Now, we shall show
$\mathcal{H}_{P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{H}_{P_{d}(g_{i}(x)^{m_{i}})}$.
Indeed, given a $U_{i}$-module $R[x]/(f_{i}(x)^{r})$, we have the isomorphism

$K_{i}\otimes_{R}\big{(}R[x]/(f_{i}(x)^{r})\big{)}\simeq\bigoplus^{w_{i}}_{q=1}K_{i}[x]/\big{(}(x-\alpha_{iq})^{rp^{s_{i}}}\big{)}$

as
$\prod^{w_{i}}_{q=1}K_{i}[x]/\big{(}(x-\alpha_{iq})^{n_{i}p^{s_{i}}}\big{)}$-modules.
Since ${\rm Hom}_{U_{i}}(M_{i},-):{\rm add}(M_{i})\to A_{i}\mbox{-proj}$ is an
equivalence, we see that $|P_{c}(f_{i}(x)^{n_{i}})|$ equals the number of
indecomposable projective $A_{i}$-modules, hence equals the number of simple
$A_{i}$-modules. Since derived equivalent algebras have the same number of
simple modules, we get $|P_{c}(f_{i}(x)^{n_{i}})|=|P_{d}(g_{i}(x)^{n_{i}})|.$
Put $h_{i}=|P_{c}(f_{i}(x)^{n_{i}})|.$ For $h_{i}=1$, we have
$\mathcal{H}_{P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{H}_{P_{d}(g_{i}(x)^{m_{i}})}$.
So we assume that $h_{i}\geq 2$ and
$P_{c}(f_{i}(x)^{n_{i}})=\\{u_{i1},\cdots,u_{ih_{i}}\\}$ with
$u_{i1}>\cdots>u_{ih_{i}}.$ Since $A_{i}={\rm End}_{U_{i}}(M_{i})$ is Morita
equivalent to ${\rm End}_{U_{i}}(\mathcal{B}(M_{i}))$, the algebra
$K_{i}\otimes_{R}A_{i}$ is Morita equivalent to $K_{i}\otimes_{R}{\rm
End}_{U_{i}}(\mathcal{B}(M_{i}))\simeq{\rm
End}_{K_{i}\otimes_{R}U_{i}}(K_{i}\otimes_{R}\mathcal{B}(M_{i}))$. As
$\mathcal{B}(M_{i})\simeq\bigoplus_{k\in[h_{i}]}R[x]/(f_{i}(x)^{u_{ik}})$ as
$U_{i}$-modules, we get

$K_{i}\otimes_{R}\mathcal{B}(M_{i})\simeq\bigoplus^{w_{i}}_{q=1}\bigoplus_{k\in[h_{i}]}K_{i}[x]/((x-\alpha_{iq})^{u_{ik}p^{s_{i}}})$

as
$\prod^{w_{i}}_{q=1}K_{i}[x]/\big{(}(x-\alpha_{iq})^{n_{i}p^{s_{i}}}\big{)}$-modules.
Then a block of $K_{i}\otimes_{R}A_{i}$ is Morita equivalent to the algebra

$E_{c,i}:={\rm
End}_{K_{i}[x]/((x-\alpha_{iq})^{n_{i}p^{s_{i}}})}\big{(}\bigoplus_{k\in[h_{i}]}K_{i}[x]/((x-\alpha_{iq})^{u_{ik}p^{s_{i}}})\big{)}$

for some $q\in[w_{i}].$ Similarly,

$K_{i}\otimes_{R}\big{(}R[x]/(g_{i}(x)^{r})\big{)}\simeq\bigoplus^{w_{i}}_{q=1}K_{i}[x]/\big{(}(x-\beta_{iq})^{rp^{s_{i}}}\big{)}$

as
$\prod^{w_{i}}_{q=1}K_{i}[x]/\big{(}(x-\beta_{iq})^{n_{i}p^{s_{i}}}\big{)}$-modules.
We write $P_{d}(g_{i}(x)^{n_{i}})=\\{v_{i1},\cdots,v_{ih_{i}}\\}$ with
$v_{i1}>\cdots>v_{ih_{i}}$. Then a block of $K_{i}\otimes_{R}B_{i}$ is Morita
equivalent to an algebra of the form

$E_{d,i}:={\rm
End}_{K_{i}[x]/((x-\beta_{iq^{\prime}})^{n_{i}p^{s_{i}}})}\big{(}\bigoplus_{k\in[h_{i}]}K_{i}[x]/((x-\beta_{iq^{\prime}})^{v_{ik}p^{s_{i}}})\big{)}$

for some $q^{\prime}\in[w_{i}].$ Remind that $u_{i1}=n_{i}=m_{i}=v_{i1}$,
$E_{c,i}\simeq{\rm
End}_{K_{i}[x]/(x^{n_{i}p^{s_{i}}})}(\oplus_{k\in[h_{i}]}K_{i}[x]/(x^{u_{ik}p^{s_{i}}}))$
and $E_{d,i}\simeq{\rm
End}_{K_{i}[x]/(x^{n_{i}p^{s_{i}}})}(\oplus_{k\in[h_{i}]}K_{i}[x]/(x^{v_{ik}p^{s_{i}}})).$
Thus the Cartan matrices of $E_{c,i}$ and $E_{d,i}$ (as $K_{i}$-algebras) are
the $h_{i}\times h_{i}$ matrices

$H_{i}:=p^{s_{i}}\sum^{h_{i}}_{k=1}(\sum^{k}_{l=1}u_{ik}(e_{kl}+e_{lk})-u_{ik}e_{kk})\mbox{
and
}J_{i}:=p^{s_{i}}\sum^{h_{i}}_{k=1}(\sum^{k}_{l=1}v_{ik}(e_{kl}+e_{lk})-v_{ik}e_{kk}),$

respectively. Then there exists an invertible matrix $\Phi_{i}\in
M_{h_{i}}(\mathbb{Z})$ such that $\Phi^{tr}_{i}H_{i}\Phi_{i}=J_{i}.$ This
follows from [37, Chapter 6, Proposition 6.8.9] which says that the Cartan
matrices of derived equivalent, split algebras are congruent by an invertible
matrix with integral entries. Thanks to Lemma 2.18, we have
$\mathcal{H}_{P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{H}_{P_{d}(g_{i}(x)^{m_{i}})}$
as multisets.

Now let $\pi:\mathcal{M}_{c}\to\mathcal{M}_{d}$ be the map given by
$f_{i}(x)^{n_{i}}\mapsto g_{i}(x)^{m_{i}}$ for $i\in[l_{c}].$ Then $\pi$ gives
rise to a $D$-equivalence $c\stackrel{{\scriptstyle D}}{{\sim}}d$.

(3) Suppose $c\stackrel{{\scriptstyle AD}}{{\sim}}d.$ Then $U_{i}\simeq V_{i}$
for $i\in[l_{c}]$ by definition. The condition
${P_{c}(f_{i}(x)^{n_{i}})}={P_{d}(g_{i}(x)^{m_{i}}))}$ or
${P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{J}_{P_{d}(g_{i}(x)^{m_{i}}))}$ implies
that either
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\mathcal{B}(N_{i})_{\mathscr{P}}$ or
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\Omega_{V_{i}}(\mathcal{B}(N_{i})_{\mathscr{P}})$
as $U_{i}$-modules. It then follows from Lemma 2.3 that $A_{i}$ and $B_{i}$
are almost $\nu$-stable derived equivalent. Hence $S_{n}(c,R)$ and
$S_{m}(d,R)$ are almost $\nu$-stable derived equivalent.

Conversely, suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are almost $\nu$-stable
derived equivalent. Thanks to Lemma 2.4, the almost $\nu$-stable derived
equivalence $F_{i}$ induces a stable equivalence, say $\overline{F_{i}}$,
between $U_{i}$ and $V_{i}$ such that
$\overline{F_{i}}(\mathcal{B}(M_{i})_{\mathscr{P}})\simeq\mathcal{B}(N_{i})_{\mathscr{P}}$
and $m_{i}=n_{i}$ for $i\in[l_{c}]$. The isomorphism $\varphi_{i}$ induces a
natural stable equivalence $\Phi_{i}$ between $V_{i}$ and $U_{i}$. The
composition $\Phi_{i}\circ\overline{F_{i}}$ is then a stable equivalence from
$U_{i}$ to itself. Now, by Lemma 2.9, we deduce either
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\mathcal{B}(N_{i})_{\mathscr{P}}$ or
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\Omega_{V_{i}}(\mathcal{B}(N_{i})_{\mathscr{P}})$
as $U_{i}$-modules, where $N_{i}$ is viewed as a $U_{i}$-module via
$\varphi_{i}$. Note that
$\mathcal{B}(M_{i})\simeq\bigoplus_{r\in{P_{c}(f_{i}(x)^{n_{i}})}}R[x]/(f_{i}(x)^{r})$
as $U_{i}$-modules and
$\mathcal{B}(N_{j})\simeq\bigoplus_{s\in{P_{d}(g_{j}(x)^{m_{j}})}}R[x]/(g_{j}(x)^{s})$
as $V_{j}$-modules. Thus
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\mathcal{B}(N_{i})_{\mathscr{P}}$ is
equivalent to ${P_{c}(f_{i}(x)^{n_{i}})}={P_{d}(g_{i}(x)^{m_{i}})}$, and
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\Omega_{V_{i}}(\mathcal{B}(N_{i})_{\mathscr{P}}$
is equivalent to
${P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{J}_{P_{d}(g_{i}(x)^{m_{i}})}$ for
$i\in[l_{c}]$. Now we define a map $\pi:\mathcal{M}_{c}\to\mathcal{M}_{d}$ by
$f_{i}(x)^{n_{i}}\mapsto g_{i}(x)^{m_{i}}$ for $i\in[l_{c}].$ Then $\pi$
defines an $AD$-equivalence: $c\stackrel{{\scriptstyle AD}}{{\sim}}d$.

(4) Assume that either $R$ is perfect or both $c$ and $d$ are invertible
matrices of finite order. Then all irreducible factors of $m_{c}(x)$ and
$m_{d}(x)$ are separable polynomials over $R$. Let $A_{i}={\rm
End}_{U_{i}}(M_{i})$ be a block in $S_{n}(c,R)$ and $e_{M_{i}}:={\rm
Hom}_{U_{i}}(M_{i},-):U_{i}\mbox{{\rm-mod}}{}\rightarrow A_{i}\mbox{{\rm-
mod}}$ be an evaluation functor. For any indecomposable projective
$A_{i}$-module $e_{M_{i}}(X)$ with $X$ an indecomposable direct summand of
$M_{i}$, we have

${\rm End}_{A_{i}}\big{(}{\rm top}(e_{M_{i}}(X))\big{)}\simeq{\rm
End}_{A_{i}}(e_{M_{i}}(X))/{\rm rad}({\rm
End}_{A_{i}}(e_{M_{i}}(X)))\simeq{\rm End}_{U_{i}}(X)/{\rm rad}({\rm
End}_{U_{i}}(X))\simeq R[x]/(f_{i}(x)).$

Thus ${\rm End}_{A_{i}}\big{(}{\rm top}(e_{M_{i}}(X))\big{)}$ is separable.
This implies that the semisimple quotient $A_{i}/{\rm rad}(A_{i})$ of $A_{i}$
is separable. Hence all the semisimple quotients of blocks of $S_{n}(c,R)$ and
$S_{m}(d,R)$ are separable. In particular, all the semisimple blocks of
$S_{n}(c,R)$ and $S_{m}(d,R)$ are separable.

Now, suppose that $c$ and $d$ are $SM$-equivalent. By Lemma 2.3 and the proof
of (3), there is an almost $\nu$-stable derived equivalence between the non-
semisimple blocks of $S_{n}(c,R)$ and $S_{m}(d,R)$ . Further, by [13, Theorem
1.1], there is a stable equivalence $F$ of Morita type between the non-
semisimple blocks $S_{n}(c,R)$ and $S_{m}(d,R)$. Since the semisimple blocks
of $S_{n}(c,R)$ and $S_{m}(d,R)$ are separable algebras, $F$ can be extended
to a stable equivalence of Morita type between $S_{n}(c,R)$ and $S_{m}(d,R)$.

Conversely, suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are stably equivalent
of Morita type, and that the stable equivalence is given by a functor $F$. Let
$A_{1},\cdots,A_{s}$ be the non-semisimple blocks of $S_{n}(c,R)$, and let
$B_{1},\cdots,B_{s}$ be the non-semisimple blocks of $S_{m}(d,R)$. By [18,
Theorem 2.2] and [19, Lemma 4.8], we may assume that $F$ induces a stable
equivalence $F_{i}$ of Morita type, between $A_{i}$ and $B_{i}$ for $1\leq
i\leq s$.

To show $c\stackrel{{\scriptstyle SM}}{{\sim}}d$, we consider the generator
$M_{i}$ for $U_{i}\mbox{{\rm-mod}}$. It follows from $\nu_{A_{i}}{\rm
Hom}_{U_{i}}(M_{i},U_{i})\simeq{\rm Hom}_{U_{i}}(M_{i},\nu_{U_{i}}U_{i})$ (see
[14, Remark 2.9 (2)]) that the Frobenius parts of $A_{i}$ and $B_{i}$ are
Morita equivalent to $U_{i}$ and $V_{i}$, respectively. Since $A_{i}/{\rm
rad}(A_{i})$ and $B_{i}/{\rm rad}(B_{i})$ are separable, it follows from Lemma
3.3(2) that $F_{i}$ restricts to a stabe equivalence $G_{i}$ of Morita type
between $U_{i}$ and $V_{i}$. As $f_{i}(x)$ is separable and both $A_{i}$ and
$B_{i}$ are non-semisimple, Corollary 2.14 implies that $U_{i}\simeq V_{i}$,
that is, $R[x]/(f_{i}(x)^{n_{i}})\simeq R[x]/(g_{i}(x)^{m_{i}})$, and
$n_{i}=m_{i}$.

Now we regard $V_{i}$-modules as $U_{i}$-modules via this isomorphism. Let
$\overline{A_{i}}:={\rm
End}_{U_{i}}(U_{i}\oplus\mathcal{B}(M_{i})_{\mathscr{P}})$,
$\overline{B_{i}}:={\rm
End}_{V_{i}}(V_{i}\oplus\mathcal{B}(N_{i})_{\mathscr{P}})$ and
$\overline{C_{i}}:={\rm
End}_{V_{i}}(V_{i}\oplus\Omega_{V_{i}}(\mathcal{B}(N_{i})_{\mathscr{P}}))$,
and let $e,f$ and $g$ be the $\nu$-stable idempotents of
$\overline{A_{i}},\overline{B_{i}}$ and $\overline{C_{i}}$, defining their
Frobenius parts, respectively. Then
$A_{i},\overline{A_{i}},B_{i},\overline{B_{i}}$ and $\overline{C_{i}}$ are
stably equivalent of Morita type, and there is the following commutative (up
to natural isomorphism) diagram by Lemma 3.3(2):

$\textstyle{\overline{A_{i}}\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{\overline{B_{i}}\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{\overline{C_{i}}\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{e\overline{A_{i}}e\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{f\overline{B_{i}}f\mbox{{\rm-{\text@underline{mod}}}}}$$\textstyle{g\overline{C_{i}}g\mbox{{\rm-{\text@underline{mod}}}}}$$\scriptstyle{\lambda}$$\scriptstyle{\lambda}$$\scriptstyle{\lambda}$$\scriptstyle{\Phi}$$\scriptstyle{\Phi_{1}}$$\scriptstyle{\Psi}$$\scriptstyle{\Psi_{1}}$

where $\lambda$ is the full embedding of stable module categories induced by
the corresponding Schur functor and where $\Phi$ and $\Psi$ define stable
equivalences of Morita type between $\overline{A_{i}}$ and $\overline{B_{i}}$,
and between $\overline{B_{i}}$ and $\overline{C_{i}}$, respectively, while
$\Phi_{1}$ and $\Psi_{1}$ are the restrictions of $\Phi$ and $\Psi$,
respectively. Note that $e\overline{A_{i}}e\simeq U_{i}\simeq V_{i}\simeq
f\overline{B_{i}}f\simeq g\overline{C_{i}}g$. Identifying $f\overline{B_{i}}f$
with $g\overline{C_{i}}g$, we can choose $\Psi$ so that $\Psi_{1}$ is the
syzygy functor on $f\overline{B_{i}}f$-mod (see the arguments in [20,
Proposition 3.3 and Corollary 3.4]). Let $S$ be the simple
$e\overline{A_{i}}e$-module. Then it follows from Lemma 2.9 that either
$\Phi_{1}(S)$ or $\Psi_{1}\circ\Phi_{1}(S)$ is simple. By Lemma 3.3(1), either
$\Phi_{1}$ or $\Psi_{1}\circ\Phi_{1}$ can be lifted to a Morita equivalence,
and therefore either $\Phi$ or $\Psi\circ\Phi$ can be lifted to a Morita
equivalence. It then follows from Lemma 2.12 that either
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\mathcal{B}(N_{i})_{\mathscr{P}}$ or
$\mathcal{B}(M_{i})_{\mathscr{P}}\simeq\Omega_{V_{i}}(\mathcal{B}(N_{i})_{\mathscr{P}})$.
Therefore ${P_{c}(f_{i}(x)^{n_{i}})}={P_{d}(g_{i}(x)^{m_{i}}))}$ or
${P_{c}(f_{i}(x)^{n_{i}})}=\mathcal{J}_{P_{d}(g_{i}(x)^{m_{i}}))}.$ Now we
define a map $\pi:\mathcal{R}_{c}\to\mathcal{R}_{d}$ by
$f_{i}(x)^{n_{i}}\mapsto g_{i}(x)^{m_{i}}$ for
$f_{i}(x)^{n_{i}}\in\mathcal{R}_{c}$. Then $\pi$ defines an $SM$-equivalence
$c\stackrel{{\scriptstyle SM}}{{\sim}}d$ of matrices. $\square$

Instead of $R$ being a field, we can prove the following for noetherian
domains.

######  Remark 3.4.

Suppose that $R$ is a noetherian domain, $c\in M_{n}(R)$ and $d\in M_{m}(R)$.
If $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent, then
$c\stackrel{{\scriptstyle D}}{{\sim}}d$ as matrices over the fraction field of
$R$.

Proof. Assume that $R$ is a noetherian domain with $K$ its fractional field.
Then it follows from $S_{n}(c,R)\subseteq M_{n}(R)$ that $S_{n}(c,R)$ is a
finitely generated $R$-algebra. Thus $S_{n}(c,R)$ is a noetherian algebra,
$S_{n}(c,R)\mbox{{\rm-mod}}$ is an abelian category and
${\mathscr{D}}^{b}(S_{n}(c,R))$ is well defined.

Regarding $K$ as an $R$-algebra, we have the isomorphism of $K$-algebras

$\displaystyle\varphi:K\otimes_{R}M_{n}(R)$ $\displaystyle\longrightarrow
M_{n}(K),\;\;\sum^{s}_{i=1}a_{i}\otimes b_{i}$
$\displaystyle\mapsto\,\sum^{s}_{i=1}(a_{i}I_{n})b_{i}$

where $I_{n}$ is the identity matrix in $M_{n}(K).$ Further, $K$ is a flat
$R$-module and there is the commutative diagram of $K$-algebras

$\textstyle{K\otimes_{R}S_{n}(c,R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu}$$\textstyle{S_{n}(c,K)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K\otimes_{R}M_{n}(R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sim}$$\scriptstyle{\varphi}$$\textstyle{M_{n}(K)}$

where $\mu$ is the restriction of $\varphi$. Remark that ${\rm Im}(\mu)$
belongs to $S_{n}(c,K)$. Since $K$ is the fractional field of $R$, we can find
an element $0\neq r\in R$ for each matrix $a\in M_{n}(K)$ such that $ra\in
M_{n}(R)$. This implies that $\mu$ is surjective, and therefore an
isomorphism. Thus $K\otimes_{R}S_{n}(c,R)\simeq S_{n}(c,K)$ as $K$-algebras.

Suppose that the $R$-algebras $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived
equivalent. Then there is a tilting complex $T$ for $S_{n}(c,R)$ such that
${\rm End}_{{\mathscr{D}}^{b}(S_{n}(c,R))}(T)\simeq S_{m}(d,R)$ as
$R$-algebras. Since $K$ is a flat $R$-module, ${\rm
Tor}^{R}_{i}(S_{n}(c,R),K)=0$ and ${\rm Tor}^{R}_{i}(S_{m}(d,R),K)=0$ for all
$i\geq 1$. It then follows from [28, Theorem 2.1] that $K\otimes_{R}T$ is a
tilting complex for $K\otimes_{R}S_{n}(c,R)$ with ${\rm
End}_{{\mathscr{D}}^{b}(K\otimes_{R}S_{n}(c,R))}(K\otimes_{R}T)\simeq
K\otimes_{R}S_{m}(d,R)$ as $K$-algebras. Thus the $K$-algebras $S_{n}(c,K)$
and $S_{m}(d,K)$ are derived equivalent. By Theorem 1.2, there holds
$c\stackrel{{\scriptstyle D}}{{\sim}}d$. $\square$

###  3.2 Relations among derived, Morita and stable equivalences: Proof of
Corollary 1.3

Let $\Lambda$ be an Artin algebra and $M$ a generator-cogenerator for
$\Lambda$-mod. Then the _rigidity dimension_ ${\rm rd}(M)$ of $M$ is defined
by

${\rm rd}(M):=\operatorname{sup}\\{n\in\mathbb{N}\mid{{\rm
Ext}}^{i}_{\Lambda}(M,M)=0,\forall\ 1\leq i\leq n\\}.$

If no such $n$ exists, we define ${\rm rd}(M)=0$. The dominant dimension of
the algebra $\Lambda$, denoted by ${\rm dom.dim}(\Lambda)$, is the maximal
$t\in\mathbb{N}$ (or $\infty$) such that all the terms
$I_{0},I_{1},\cdots,I_{t-1}$in a minimal injective resolution

$0\longrightarrow\Lambda\longrightarrow I_{0}\longrightarrow
I_{1}\longrightarrow\cdots\longrightarrow I_{t}\longrightarrow\cdots$

of ${}_{\Lambda}\Lambda$ are projective. By [23, Lemma 3], ${\rm dom.dim}({\rm
End}_{\Lambda}(M))={\rm rd}(M)+2$.

The following lemma describes the dominant dimensions of principal centralizer
matrix algebras.

######  Lemma 3.5.

$(1)$ ${\rm dom.dim}(A_{i})\in\\{2,\infty\\}$. Particularly, ${\rm
dom.dim}(S_{n}(c,R))\in\\{2,\infty\\}$.

$(2)$ ${\rm dom.dim}(A_{i})=\infty$ if and only if $A_{i}$ is a symmetric
Nakayama algebra if and only if $P_{c}(f_{i}(x)^{n_{i}})$ is a singleton set.
Thus ${\rm dom.dim}(S_{n}(c,R))=\infty$ if and only if $S_{n}(c,R)$ is a
symmetric Nakayama algebra if and only if $P_{c}(f_{i}(x)^{n_{i}})$ is a
singleton set for all $i\in[l_{c}].$

Proof. If $\Lambda$ is an Artin algebra and $L\in\Lambda\mbox{{\rm-mod}}$,
then it follows from the Auslander-Reiten formula $D{\rm
Ext}^{1}_{\Lambda}(L,L)\simeq\overline{{\rm Hom}}_{\Lambda}(L,\tau L)$ that
${\rm Ext}^{1}_{\Lambda}(L,L)\neq 0$ if $\tau L\simeq L$, where $D$ is the
usual duality of an Artin algebra and $\tau:=D{\rm Tr}$ denotes the Auslander-
Reiten translation.

Let $i\in[l_{c}]$. For the $U_{i}$-module $M_{i}$,
$\tau(M_{i})_{\mathscr{P}}\simeq(M_{i})_{\mathscr{P}}$, and therefore ${\rm
rd}(M_{i})=\infty$ if $M_{i}$ is projective, and $0$, otherwise. Since ${\rm
dom.dim}(A_{i})={\rm dom.dim}({\rm End}_{U_{i}}(M_{i}))={\rm rd}(M_{i})+2,$ we
deduce that ${\rm dom.dim}(A_{i})\in\\{2,\infty\\}$ and that ${\rm
dom.dim}(A_{i})=\infty$ if and only if $M_{i}$ is projective if and only if
$A_{i}$ is a symmetric Nakayama algebra if and only if
$P_{c}(f_{i}(x)^{n_{i}})$ is a singleton set. Hence ${\rm
dom.dim}(S_{n}(c,R))\in\\{2,\infty\\}$. Moreover, ${\rm
dom.dim}(S_{n}(c,R))=\infty$ if and only if $S_{n}(c,R)$ is a symmetric
Nakayama algebra if and only if $P_{c}(f_{i}(x)^{n_{i}})$ is a singleton set
for all $i\in[l_{c}].$ $\square$

Let $\Lambda$ be an Artin algebra. We denote by
$\mathscr{P}(\Lambda)_{\mathscr{I}}$ the set of all isomorphism classes of
projective $\Lambda$-modules without any nonzero injective summands. In [22,
Proposition 1.5 and Theorems 1.7 and 2.6], Martínez-Villa proved the
following.

######  Lemma 3.6.

[22] Let $F:\Lambda$-mod $\rightarrow\bar{\Lambda}$-mod be a stable
equivalence of Artin algebras $\Lambda$ and $\bar{\Lambda}$ both with neither
nodes nor semisimple summands.

$(1)$ The functor $F$ provides a bijection
$F^{\prime}:\mathscr{P}(\Lambda)_{\mathscr{I}}\rightarrow\mathscr{P}(\bar{\Lambda})_{\mathscr{I}}$,
which preserves simple projective modules in
$\mathscr{P}(\Lambda)_{\mathscr{I}}$.

$(2)$ The functor $F$ induces a stable equivalence between the Frobenius parts
of $\Lambda$ and $\bar{\Lambda}$.

$(3)$ Let $0\rightarrow X\oplus Q_{1}\stackrel{{\scriptstyle
f}}{{\rightarrow}}Y\oplus Q_{2}\oplus P\stackrel{{\scriptstyle
g}}{{\rightarrow}}Z\rightarrow 0$ be an exact sequence of $\Lambda$-modules
without any split exact sequences as its direct summands, where
$X,Y,Z\in\Lambda\mbox{{\rm-mod}}_{\mathscr{P}}$,
$Q_{1},Q_{2}\in\mathscr{P}(\Lambda)_{\mathscr{I}}$ and $P$ is a projective-
injective $\Lambda$-module. Then there is a short exact sequence

$0\longrightarrow F(X)\oplus F^{\prime}(Q_{1})\stackrel{{\scriptstyle
f^{\prime}}}{{\longrightarrow}}F(Y)\oplus F^{\prime}(Q_{2})\oplus
P^{\prime}\stackrel{{\scriptstyle
g^{\prime}}}{{\longrightarrow}}F(Z)\longrightarrow 0$

in $\bar{\Lambda}\mbox{{\rm-mod}}$ such that $P^{\prime}$ is projective-
injective and that no split exact sequences are its direct summands.

We say that a stable equivalence
$F:\Lambda\mbox{{\rm-{\text@underline{mod}}}}\rightarrow\bar{\Lambda}$-mod of
Artin algebras $\Lambda$ and $\bar{\Lambda}$ _preserves non-semisimple blocks_
if for non-projective indecomposable modules $M,N\in\Lambda\mbox{{\rm-
mod}}_{\mathscr{P}}$, $F(M)$ and $F(N)$ lie in the same block of
$\bar{\Lambda}$ if and only if $M$ and $N$ lie in the same block of $\Lambda$.
In general, a stable equivalence may not preserve the numbers of non-
semisimple blocks of algebras. This can be seen by [36, Example 3.12], for
instance.

In the following, we will show that stable equivalences between principal
centralizer matrix algebras over a field do have this property.

As $S_{n}(c,R)\simeq{\rm End}_{R[c]}(R^{n})\simeq\prod^{l_{c}}_{i=1}{{\rm
End}}_{U_{i}}(M_{i})=\prod^{l_{c}}_{i=1}A_{i}$, we consider the non-semisimple
blocks of $S_{n}(c,R)$ and denote by $A$ the direct sum of its non-semisimple
blocks. Now, we partition these blocks of $A$ in the following way such that

(1) $\mathcal{S}_{A,\geq 2}:=\\{A_{1},A_{2},\cdots,A_{a_{1}}\\}$ consists of
the blocks $A_{i}$ with $n_{i}\geq 3$ and having at least 2 non-injective,
indecomposable projective modules for $1\leq i\leq a_{1}$;

(2) $\mathcal{S}_{A,1}:=\\{A_{a_{1}+1},A_{a_{1}+2},\cdots,A_{a_{2}}\\}$
consists of the blocks $A_{i}$ with $n_{i}\geq 3$ and having only 1 non-
injective, indecomposable projective module for $a_{1}<i\leq a_{2}$;

(3) $\mathcal{S}_{A,0}:=\\{A_{a_{2}+1},A_{a_{2}+2},\cdots,A_{a_{3}}\\}$
consists of the blocks $A_{i}$ with $n_{i}\geq 3$ and having no non-injective,
indecomposable projective modules for $a_{2}<i\leq a_{3}$;

(4) $\mathcal{S}_{A}:=\\{A_{a_{3}+1},A_{a_{3}+2},\cdots,A_{a_{4}}\\}$ consists
of the blocks $A_{i}$ with $n_{i}=2$ for $a_{3}<i\leq a_{4}$, where $0\leq
a_{1}\leq a_{2}\leq a_{3}\leq a_{4}\leq l_{c}.$

Note that $n_{i}$ is the Loewy length of both $U_{i}$ and the center of
$A_{i}$ by Lemma 2.5. For an Artin algebra $\Lambda$, we have denoted by
$\Lambda^{\prime}$ the triangular matrix algebra obtained from $\Lambda$ by
eliminating all nodes of $\Lambda$. Thus $\Lambda$ and $\Lambda^{\prime}$ are
stably equivalent, but the latter has no nodes.

Let $\widetilde{A}:=\prod^{a_{3}}_{i=1}A_{i}\;\times\prod_{a_{3}<i\leq
a_{4}}(A_{i})^{\prime}.$ Then we have the following result.

######  Lemma 3.7.

The algebra $\widetilde{A}$ has neither nodes nor semisimple direct summands.
Moreover, there exists a stable equivalence
$F_{A}:A\mbox{{\rm-{\text@underline{mod}}}}\rightarrow\widetilde{A}$-mod such
that $F_{A}$ preserves non-semisimple blocks of algebras and that the
restriction of $F_{A}$ to the block $A_{i}$ is induced by the identity functor
for all $i\in[a_{3}]$.

Proof. Recall that $U_{i}=R[x]/(f_{i}(x)^{n_{i}})$ and $A_{i}={\rm
End}_{U_{i}}(M_{i})$. For $i\in[a_{3}],$ it follows by Lemma 2.10 that the
block $A_{i}$ has neither nodes nor projective simple modules. For
$a_{3}<j\leq a_{4}$, the triangular matrix algebra $(A_{j})^{\prime}$ is
stably equivalent to $A_{j}$ and has no nodes. Clearly, $\widetilde{A}$ does
not have any semisimple direct summands. Now it is easy to get a desired
stable equivalence $F_{A}$ between $A$ and $\widetilde{A}$. $\square$

Let $S_{m}(d,R)$ be another centralizer matrix algebra and $B$ be the sum of
its non-semisimple blocks. Similarly, we have a partition of blocks for $B$.
This is given by the natural numbers $0\leq b_{1}\leq b_{2}\leq b_{3}\leq
b_{4}\leq l_{d}$, namely the partition
$\\{B_{1},\cdots,B_{b_{1}}\\}\cup\\{B_{b_{1}+1},\cdots,B_{b_{2}}\\}\cup\\{B_{b_{2}+1},\cdots,B_{b_{3}}\\}\cup\\{B_{b_{3}+1},\cdots,B_{b_{4}}\\}$
has the corresponding properties as the blocks of $A$.

By Lemma 3.7,
$\widetilde{B}:=\prod^{b_{3}}_{j=1}B_{j}\;\times\prod_{b_{3}<j\leq
b_{4}}(B_{j})^{\prime}$ has neither nodes nor semisimple direct summands, and
there is a stable equivalence
$F_{B}:B\mbox{{\rm-{\text@underline{mod}}}}\rightarrow\widetilde{B}$-mod
preserving non-semsimple blocks such that the restriction of $F_{B}$ to
$B_{j}$ is induced by the identity functor for $j\in[b_{3}]$.

Now, we assume that there is a stable equivalence $F$ between $S_{n}(c,R)$ and
$S_{m}(d,R)$. Then $F$ restricts to a stable equivalence between $A$ and $B$.
Thus $H:=F_{B}\circ F\circ F_{A}^{-1}:\widetilde{A}$-mod $\to$
$\widetilde{B}$-mod is a stable equivalence. Let $J$ be a quasi-inverse of
$H$. As defined in Lemma 3.6(1),
$H^{\prime}:\mathscr{P}(\widetilde{A})_{\mathscr{I}}\rightarrow\mathscr{P}(\widetilde{B})_{\mathscr{I}}$
and
$J^{\prime}:\mathscr{P}(\widetilde{B})_{\mathscr{I}}\rightarrow\mathscr{P}(\widetilde{A})_{\mathscr{I}}$
are the bijections induced by $H$ and $J$, respectively.

######  Lemma 3.8.

The correspondence $H^{\prime}$ induces a bijection between
$\mathcal{S}_{A,\geq 2}\cup\mathcal{S}_{A,1}$ and $\mathcal{S}_{B,\geq
2}\cup\mathcal{S}_{B,1}$ such that the corresponding blocks have the same
number of non-injective, indecomposable projective modules. In particular,
$a_{l}=b_{l}$ for $1\leq l\leq 2$.

Proof. By the proof in Lemma 2.10, for $i\in[a_{2}]$, the block $A_{i}$ of
$\widetilde{A}$ has a unique projective-injective indecomposable module, say
$P_{i}$. We have to discuss the following $2$ cases.

(i) $H$ induces a bijection from $\mathcal{S}_{A,\geq 2}$ to
$\mathcal{S}_{B,\geq 2}$, such that the corresponding blocks have the same
number of non-injective, indecomposable projective modules.

In fact, let $A_{i}\in\mathcal{S}_{A,\geq 2}$, that is, $A_{i}$ has at least
$2$ non-injective, indecomposable projective modules $P_{i1}$ and $P_{i2}$
which are non-simple by Lemma 2.10. Then there exist $2$ indecomposable direct
summands $M_{i1}$ and $M_{i2}$ of the $U_{i}$-module $M_{i}$ such that
$P_{ir}\simeq{\rm Hom}_{U_{i}}(M_{i},M_{ir})$ as $A_{i}$-modules for $1\leq
r\leq 2$. As $U_{i}$ is a local Nakayama algebra, we may assume that $M_{i1}$
is isomorphic to a proper submodule of $M_{i2}$. It then follows from the left
exactness of the Hom-functor ${\rm Hom}_{U_{i}}(M_{i},-)$ that $P_{i1}$ is
isomorphic to a submodule of $P_{i2}$. Hence there is an exact sequence of
$A_{i}$-modules

$0\longrightarrow P_{i1}\longrightarrow P_{i2}\longrightarrow
P_{i2}/P_{i1}\longrightarrow 0$

with $P_{i2}/P_{i1}$ indecomposable. By Lemma 3.6(3), there is an exact
sequence of $\widetilde{B}$-modules

$(*)\quad 0\longrightarrow H^{\prime}(P_{i1})\longrightarrow
H^{\prime}(P_{i2})\oplus P^{\prime}_{1}\longrightarrow
H(P_{i2}/P_{i1})\longrightarrow 0,$

with $P^{\prime}_{1}$ being projective-injective, such that this sequence does
not contain split exact sequences as its direct summands. Then ${\rm
Hom}_{\widetilde{B}}(H^{\prime}(P_{i1}),H^{\prime}(P_{i2}))\neq 0$, and
therefore $H^{\prime}(P_{i1})$ and $H^{\prime}(P_{i2})$ lie in the same block
of $\widetilde{B}$. Otherwise, the sequence $(*)$ would contain a split
sequence $0\to 0\to H^{\prime}(P_{i2})\stackrel{{\scriptstyle
1}}{{\to}}H^{\prime}(P_{i2})\to 0$ as its summand. Let $O$ be the block of
$\widetilde{B}$ to which $H^{\prime}(P_{i1})$ and $H^{\prime}(P_{i2})$ belong.
Note that a block $(B_{j})^{\prime}$ of $\widetilde{B}$, with $b_{3}<j\leq
b_{4}$, has at most $2$ non-injective, indecomposable projective modules, one
of which is a simple module by Remark 2.11. It then follows from Lemma 3.6(1)
that $O\not\simeq(B_{j})^{\prime}$ as algebras for $b_{3}<j\leq b_{4}$. Hence
$O$ is a block of the form $B_{j}$ for $j\in[b_{1}]$ and $H^{\prime}$ sends
all the non-injective, indecomposable projective $A_{i}$-modules to the ones
belonging to the block $B_{j}.$ Similarly, the non-injective, indecomposable
projective $B_{j}$-modules are mapped by $J^{\prime}$ into modules belonging
to the block $A_{i}.$ Thus $H^{\prime}$ restricts to a bijection between the
set of non-injective, indecomposable projective $A_{i}$-modules and the one of
non-injective, indecomposable projective $B_{j}$-modules. This implies that
$H$ induces a bijection from the set of the blocks in $\mathcal{S}_{A,\geq 2}$
to the set of blocks in $\mathcal{S}_{B,\geq 2}$, such that the corresponding
blocks have the same number of non-injective, indecomposable projective
modules. Clearly, $a_{1}=b_{1}$.

(ii) $H$ induces a bijection between $\mathcal{S}_{A,1}$ and
$\mathcal{S}_{B,1}$.

Actually, let $A_{i}\in\mathcal{S}_{A,1}$, that is, $A_{i}$ has only $1$ non-
injective, indecomposable projective module, say $Q_{i}$. If
$H^{\prime}(Q_{i})$ lies in some block $(B_{j})^{\prime}$ for $b_{3}<j\leq
b_{4}$, then, by Remark 2.11, $H^{\prime}(Q_{i})$ has a simple projective
submodule, say $\widetilde{P_{j}}$. With a similar argument as in (i), we
deduce that $Q_{i}$ and the simple projective module
$J^{\prime}(\widetilde{P_{j}})$ lie in the same block $A_{i}$. Note that
$Q_{i}$ is not simple and $Q_{i}\not\simeq J^{\prime}(\widetilde{P_{j}})$.
This implies that the block $A_{i}$ contains $2$ non-injective, indecomposable
projective modules, a contradiction. Thus it follows from (i) that
$H^{\prime}(Q_{i})$ belongs to a block $B_{j}\in\mathcal{S}_{B,1}$. So $H$
induces a bijection from the set of blocks in $\mathcal{S}_{A,1}$ to the set
of blocks in $\mathcal{S}_{B,1}$, and therefore $a_{2}-a_{1}=b_{2}-b_{1}$ and
$a_{2}=b_{2}$. $\square$

######  Lemma 3.9.

Let $\widetilde{A_{1}}$ and $\widetilde{B_{1}}$ be the sum of blocks in
$\mathcal{S}_{A,\geq 2}\cup\mathcal{S}_{A,1}$ and $\mathcal{S}_{B,\geq
2}\cup\mathcal{S}_{B,1}$, respectively. Then the functor $H$ restricts to a
stable equivalence between $\widetilde{A_{1}}$ and $\widetilde{B_{1}}$
preserving non-semisimple blocks.

Proof. By Lemma 3.8, we assume $H^{\prime}(A_{i})=B_{i}$ for $i\in[a_{2}]$.
Let $i\in[a_{2}]$ and $A_{i}$ be a block in $\widetilde{A_{1}}$. Suppose that
$M$ is a non-projective, indecomposable $A_{i}$-module. Then $H(M)$ is
indecomposable. Further, we show that $H(M)$ lies in the block $B_{i}$ of
$\widetilde{B_{1}}$.

(a) By the proof of Lemma 2.10, any non-injective, indecomposable projective
$A_{i}$-module is isomorphic to a submodule of the unique projective-injective
$A_{i}$-module $P_{i}$. This implies that ${\rm rad}(P_{i})$ is indecomposable
and each non-injective, indecomposable projective $A_{i}$-module is isomorphic
to a submodule of ${\rm rad}(P_{i})$. If ${\rm rad}(P_{i})$ is projective,
then $H^{\prime}({\rm rad}(P_{i}))$ lies in the block $B_{i}$. If ${\rm
rad}(P_{i})$ is not projective, then there exists a non-injective,
indecomposable projective $A_{i}$-module $P$ and an exact sequence

$\quad 0\longrightarrow P\stackrel{{\scriptstyle\iota}}{{\longrightarrow}}{\rm
rad}(P_{i})\stackrel{{\scriptstyle\eta}}{{\longrightarrow}}{\rm
rad}(P_{i})/P\longrightarrow 0$

without split direct summands. Applying Lemma 3.6(3) to this sequence, we get
an exact sequence of $\widetilde{B}$-modules

$0\longrightarrow
H^{\prime}(P)\stackrel{{\scriptstyle\iota^{\prime}}}{{\longrightarrow}}H({\rm
rad}(P_{i}))\oplus
P^{\prime}_{2}\stackrel{{\scriptstyle\eta^{\prime}}}{{\longrightarrow}}H({\rm
rad}(P_{i})/P)\longrightarrow 0,$

where $P^{\prime}_{2}$ is projective-injective. We show that $H({\rm
rad}(P_{i}))$ lies in the block $B_{i}.$ Suppose contrarily that $H({\rm
rad}(P_{i}))$ does not belong to the block $B_{i}.$ Then ${\rm
Hom}_{\widetilde{B}}(H^{\prime}(P),H({\rm rad}(P_{i})))=0$, and therefore
${\rm Im}(\iota^{\prime})\subset P^{\prime}_{2}.$ Thus $H({\rm
rad}(P_{i})/P)\simeq(H({\rm rad}(P_{i}))\oplus P^{\prime}_{2})/{\rm
Im}(\iota^{\prime})\simeq P^{\prime}_{2}/{\rm Im}(\iota^{\prime})\oplus H({\rm
rad}(P_{i}))$. This yields that ${\rm rad}(P_{i})$ is isomorphic to a direct
summand of ${\rm rad}(P_{i})/P$, a contradiction.

(b) Let $P(M)$ be the projective cover of $M$. Then there is an exact sequence
of $A_{i}$-modules

$0\longrightarrow\Omega(M)\stackrel{{\scriptstyle
f}}{{\longrightarrow}}P(M)\stackrel{{\scriptstyle
g}}{{\longrightarrow}}M\longrightarrow 0,$

which has no split direct summands. Write $\Omega(M)=L_{1}\oplus L_{2}$ for
$L_{1}\in{A_{i}\mbox{{\rm-mod}}}_{\mathscr{P}}$ and
$L_{2}\in\mathscr{P}(A_{i})_{\mathscr{I}}$, and $P(M)=Q\oplus(P^{\oplus
t}_{i})$ with $Q\in\mathscr{P}(A_{i})_{\mathscr{I}}$ and $t\in\mathbb{N}.$ By
Lemma 3.6(3), we get an exact sequence of $\widetilde{B}$-modules

$({\ddagger})\quad 0\longrightarrow H(L_{1})\oplus
H^{\prime}(L_{2})\stackrel{{\scriptstyle
f^{\prime}}}{{\longrightarrow}}H^{\prime}(Q)\oplus
P^{\prime}_{3}\stackrel{{\scriptstyle
g^{\prime}}}{{\longrightarrow}}H(M)\longrightarrow 0,$

such that $P^{\prime}_{3}$ is projective-injective and (${\ddagger}$) has no
split direct summands.

(1)Assume $Q\neq 0$. Then $H(M)$ has to lie in $B_{i}$. Otherwise, we would
have ${\rm Hom}_{\widetilde{B}}(H^{\prime}(Q),H(M))=0$, and therefore ${\rm
Im}(f^{\prime})=H^{\prime}(Q)\oplus({\rm Im}(f^{\prime})\cap P^{\prime}_{3}).$
Since $H^{\prime}(Q)$ is projective, there exists a homomorphism
$h^{\prime}:H^{\prime}(Q)\rightarrow H(L_{1})\oplus H^{\prime}(L_{2})$ such
that $h^{\prime}f^{\prime}$ is the identity on $H^{\prime}(Q)$, and therefore

$0\longrightarrow h^{\prime}(H^{\prime}(Q))\stackrel{{\scriptstyle
f^{\prime}}}{{\longrightarrow}}H^{\prime}(Q)\longrightarrow 0\longrightarrow
0$

is a split direct summand of (${\ddagger}$), a contradiction. Hence $H(M)$
belongs to the block $B_{i}$.

(2) Assume $Q=0$. Then $H^{\prime}(Q)=0.$ Suppose that
$P^{\prime}_{3}=\bigoplus^{l}_{k=1}C_{k}$, where $l\geq 1$ and $C_{k}$ is a
sum of indecomposable direct summands of $P^{\prime}_{3}$ belonging to the
same block. We shall show that all indecomposable direct summands of
$P^{\prime}_{3}$ lie in the same block of $\widetilde{B}$, that is, $l=1$.
Now, suppose contrarily that $l\geq 2$. By $({\ddagger})$, ${\rm
Im}(f^{\prime})=\bigoplus^{l}_{k=1}D_{k}$, where $D_{k}$ is a submodules of
$C_{k}$ for $k\in[l]$. Since $H(L_{1})\oplus H^{\prime}(L_{2})$ contains no
projective-injective direct summands, each $D_{k}$ is a proper submodules of
$C_{k}$. Therefore $H(M)\simeq P^{\prime}_{3}/{\rm
Im}(f^{\prime})\simeq\bigoplus^{l}_{k=1}C_{k}/D_{k}$ is decomposable, a
contradiction. Thus $l=1$ and all indecomposable direct summands of
$P^{\prime}_{3}$ lie in the same block of $\widetilde{B}$. This also imply
that $P^{\prime}_{3}$ belongs to the same block.

Suppose that $\Omega(M)$ has a direct summand isomorphic to ${\rm
rad}(P_{i}).$ Then $f^{\prime}$ in $({\ddagger})$ restricts to an injective
homomorphism from $H({\rm rad}(P_{i}))$ if ${\rm rad}(P_{i})$ is not
projective (or from $H^{\prime}({\rm rad}(P_{i}))$ if ${\rm rad}(P_{i})$ is
projective) to $P^{\prime}_{3}$. In particular, $H({\rm rad}(P_{i}))$ (or
$H^{\prime}({\rm rad}(P_{i}))$) and $P^{\prime}_{3}$ lie in the same block
$B_{i}$ of $\widetilde{B}$. Also, $H^{\prime}(Q)=0$ implies ${\rm
Hom}_{\widetilde{B}}(P^{\prime}_{3},H(M))\neq 0$ in $({\ddagger})$. This
implies that $H(M)$ and $P^{\prime}_{3}$ lie in the same block $B_{i}$.

Suppose that $\Omega(M)$ has no direct summands isomorphic to ${\rm
rad}(P_{i}).$ Recall that $P(M)=P_{i}^{\oplus t}$ under the assumption $Q=0$.
We consider the exact sequence of $A_{i}$-modules

$(\sharp)\quad 0\longrightarrow\Omega(M)\stackrel{{\scriptstyle
f}}{{\longrightarrow}}({\rm rad}(P_{i}))^{\oplus t}\stackrel{{\scriptstyle
g}}{{\longrightarrow}}{\rm rad}(M)\longrightarrow 0.$

Deleting the split direct summands of ($\sharp$), we obtain an exact sequence
of $A_{i}$-modules

$0\longrightarrow\Omega(M)\stackrel{{\scriptstyle
f_{0}}}{{\longrightarrow}}({\rm rad}(P_{i}))^{\oplus r}\stackrel{{\scriptstyle
g_{0}}}{{\longrightarrow}}X\longrightarrow 0$

for some $r\leq t$ and a submodule $X$ of ${\rm rad}(M)$. Thanks to Lemma
3.6(3), there is an indecomposable direct summand $L$ of $\Omega(M)$ such that
$H(L)$ (or $H^{\prime}(L))$ lies in the block $B_{i}$. Otherwise $H(X)$ would
contain a direct summand isomorphic to $H(({\rm rad}(P_{i}))^{\oplus r})$ (or
$H^{\prime}(({\rm rad}(P_{i}))^{\oplus r})$) (see the argument in (a)), a
contradiction to that $X$ contains no direct summands isomorphic to $({\rm
rad}(P_{i}))^{\oplus r}).$ Now, we see from (${\ddagger}$) that the modules
$H(M)$, $P^{\prime}_{3}$ and $H(L)$ lie in the same block $B_{i}$ of
$\widetilde{B}$ or the modules $H(M)$, $P^{\prime}_{3}$ and $H^{\prime}(L)$
lie in the same block $B_{i}$ of $\widetilde{B}$.

Thus we have proved that $H(M)$ lies in the block $B_{i}$ of
$\widetilde{B_{1}}$. Similarly, for $i\in[a_{2}]$ and a non-projective,
indecomposable $B_{i}$-module $N$, we see that $J(N)$ belongs to the block
$A_{i}$ of $\widetilde{A_{1}}$. Hence $H$ induces a stable equivalence between
$\widetilde{A_{1}}$ and $\widetilde{B_{1}}$, which preserves non-semisimple
blocks. $\square$

For an Artin algebra $\Lambda$, let $\Gamma_{\Lambda}$ denote the Auslander-
Reiten quiver of $\Lambda$ and $\Gamma^{s}_{\Lambda}$ the stable Auslander-
Reiten quiver of $\Gamma_{\Lambda}$ obtained by removing all projective
vertices from $\Gamma_{\Lambda}$. For a local, symmetric Nakayama algebra
$\Lambda_{0}:=R[x]/(f(x)^{n})$, $\Gamma^{s}_{\Lambda_{0}}$ is a connected
quiver such that there are two arrows between any two vertices if they are
connected by an irreducible map.

######  Lemma 3.10.

Let $\widetilde{A_{2}}$ and $\widetilde{B_{2}}$ be the sum of blocks in
$\mathcal{S}_{A,0}\cup\mathcal{S}_{A}$ and
$\mathcal{S}_{B,0}\cup\mathcal{S}_{B}$, respectively. The functor $H$
restricts to a stable equivalence between $\widetilde{A_{2}}$ and
$\widetilde{B_{2}}$, which preserves non-semisimple blocks.

Proof. Given a non-projective, indecomposable $\widetilde{A_{2}}$-module $Y$,
the module $H(Y)$ belongs to a block of $\widetilde{B_{2}}$ by Lemma 3.9. Let
$A_{i}\in\mathcal{S}_{A,0}$ be a block, that is, all indecomposable projective
$A_{i}$-modules are injective. Then the $U_{i}$-module $M_{i}$ is projective
and $A_{i}={\rm End}_{U_{i}}(M_{i})$ is a symmetric Nakayama algebra which is
Morita equivalent to $U_{i}=R[x]/(f_{i}(x)^{n_{i}})$. We show that $H$
restricts to a stable equivalence between $A_{i}$ and a block in
$\mathcal{S}_{B,0}.$ Let $\\{K_{l}\mid l\in[n_{i}-1]\\}$ be the set of all (up
to isomorphism) non-projective, indecomposable $A_{i}$-modules. Since
$\Gamma^{s}_{A_{i}}$ is a connected quiver with two arrows between any two
connected vertices, it follows from [4, Lemma 1.2(d), p. 336] that all modules
$H(K_{l}),l\in[n_{i}-1]$, lie in the same Auslander-Reiten component of a
block $W$ in $\widetilde{B_{2}}$. For a block $B_{k}$ in $\mathcal{S}_{B}$, we
consider the node-eliminated block $(B_{k})^{\prime}$. Since $B_{k}$ and
$(B_{k})^{\prime}$ are stably equivalent, the quiver $\Gamma^{s}_{B_{k}}$ and
$\Gamma^{s}_{(B_{k})^{\prime}}$ are isomorphic as translation quivers by [4,
Corollary 1.10, p.342]. According to Remark 2.11, for $b_{3}<k\leq b_{4}$,
$\Gamma^{s}_{B_{k}}$ either contains only $1$ vertex or is of the form
$\bullet\rightarrow\bullet\rightarrow\bullet$. This implies that $W$ can not
be a block $(B_{k})^{\prime}$ in $\widetilde{B_{2}}$ with $b_{3}<k\leq b_{4}.$
Thus $W$ is a block in $\mathcal{S}_{B,0}$, say $B_{j}$. Clearly, $B_{j}$ is a
symmetric Nakayama algebra and all $H(K_{l}),l\in[n_{i}-1]$, are precisely the
non-projective, indecomposable $B_{j}$-modules (up to isomorphism). Similarly,
for each block $B_{j}\in\mathcal{S}_{B,0}$, there is a unique block $A_{i}$ in
$\mathcal{S}_{A,0}$ such that $J$ restricts to a stable equivalence between
$B_{j}$ and $A_{i}$. In this way, $H$ induces not only a one-to-one
correspondence but also a stable equivalence between the blocks in
$\mathcal{S}_{A,0}$ and $\mathcal{S}_{B,0}.$

Let $A_{i}\in\mathcal{S}_{A}$ be a block, that is, $a_{3}<i\leq a_{4}.$ The
quiver $\Gamma^{s}_{(A_{i})^{\prime}}$ for $a_{3}<i\leq a_{4}$ (respectively,
$\Gamma^{s}_{(B_{j})^{\prime}}$ for $b_{3}<j\leq b_{4}$) is connected with
either $1$ or $3$ vertices. Clearly, $H$ restricts to a stable equivalence
between $(A_{i})^{\prime}$ and some block $(B_{j})^{\prime}$ with $b_{3}<j\leq
b_{4}$. Hence $H$ induces not only a one-to-one correspondence but also a
stable equivalence between the the blocks in $\mathcal{S}_{A}$ and
$\mathcal{S}_{B}.$ $\square$

######  Lemma 3.11.

Let $c\in M_{n}(R)$ and $d\in M_{m}(R)$. Suppose that there is a stable
equivalence $F$ between $S_{n}(c,R)$ and $S_{m}(d,R)$. Then $F$ preserves non-
semisimple blocks. Moreover, if $A_{i}$ and $B_{j}$ are stably equivalent,
then $n_{i}=m_{j}$.

Proof. We keep all notions introduced previously. Suppose that there is a
stable equivalence $F$ between $S_{n}(c,R)$ and $S_{m}(d,R)$. Recall that $A$
and $B$ are the sum of non-semisimple blocks of $S_{n}(c,R)$ and $S_{m}(d,R)$,
respectively, and
$F_{A}:A\mbox{{\rm-{\text@underline{mod}}}}\rightarrow\widetilde{A}$-mod and
$F_{B}:B\mbox{{\rm-{\text@underline{mod}}}}\rightarrow\widetilde{B}$-mod are
stable equivalences.

By Lemmas 3.9 and 3.10, the stable equivalence $H=F_{B}\circ F\circ
F_{A}^{-1}:\widetilde{A}\mbox{{\rm-{\text@underline{mod}}}}\rightarrow\widetilde{B}$-mod
preserves non-semisimple blocks. Since $F_{A}$ and $F_{B}$ preserve non-
semisimple blocks, we infer that $F$ preserves non-semisimple blocks.

Suppose that the blocks $A_{i}$ and $B_{j}$ are stably equivalent. Then
$A_{i}$ is semisimple if and only if $B_{j}$ is semisimple. In this case,
$n_{i}=m_{j}=1$. So we may assume that $n_{i}\geq 2$ and $m_{j}\geq 2$. We
first show that $n_{i}=2$ if and only if $m_{j}=2$. Suppose contrarily that
either $n_{i}=2$ and $m_{j}\geq 3$ or $n_{i}\geq 3$ and $m_{j}=2$. We only
deal with the situation $n_{i}=2$ and $m_{j}\geq 3$. The other case can be
done similarly. It follows from Lemma 2.10 that $A_{i}$ has nodes but $B_{j}$
does not have nodes. We replace $A_{i}$ by a stably equivalent algebra
$(A_{i})^{\prime}$ without nodes. By Lemma 3.6(2), the Frobenius parts of
$(A_{i})^{\prime}$ and $B_{j}$ are stably equivalent, while the Frobenius
parts of $(A_{i})^{\prime}$ is zero by Remark 2.11 and the Frobenius parts of
$B_{j}$ is $V_{j}$ by the proof of Theorem 1.2(4). Thus $V_{j}$-mod is zero.
This is a contradiction. Thus $n_{i}=2$ if and only if $m_{j}=2$. Now assume
$n_{i}\geq 3$ and $m_{j}\geq 3.$ With a similar argument, we deduce that the
Frobenius parts of $A_{i}$ and $B_{j}$ are stably equivalent, that is, $U_{i}$
and $V_{j}$ are stably equivalent. In particular, $U_{i}$ and $V_{j}$ have the
same number of non-projective, indecomposable modules, that is,
$n_{i}-1=m_{j}-1.$ Hence $n_{i}=m_{j}$.$\square$

Proof of Corollary 1.3. We keep all notion introduced previously. Let $c\in
M_{n}(R)$ and $d\in M_{m}(R)$.

(1) Assume that $c$ and $d$ are permutation matrices and that $S_{n}(c,R)$ and
$S_{m}(d,R)$ are derived equivalent. Then $S_{n}(c,R)$ and $S_{m}(d,R)$ have
the same number of blocks, that is, $l_{c}=l_{d}$. So we may assume that
$A_{i}$ and $B_{i}$ are derived equivalent for $i\in[l_{c}].$ By Lemma 3.1,
$U_{i}\simeq V_{i}$ and $n_{i}=m_{i}$ for $i\in[l_{c}].$ By Theorem 1.2(1), it
suffices to show that $P_{c}(f_{i}(x)^{n_{i}})=P_{d}(g_{i}(x)^{m_{i}})$ for
$i\in[l_{c}].$ By Lemma 2.17, the integers in $P_{c}(f_{i}(x)^{n_{i}})$ are
$p$-powers for $i\in[l_{c}]$. Similarly, the integers in
$P_{c}(g_{i}(x)^{m_{i}})$ are $p$-powers for $i\in[l_{c}]$. Let
$t_{i}:=|P_{c}(f_{i}(x)^{n_{i}})|=|P_{d}(g_{i}(x)^{n_{i}})|$ for
$i\in[l_{c}]$. If $t_{i}=1$, then
$P_{c}(f_{i}(x)^{n_{i}})=\\{n_{i}\\}=\\{m_{i}\\}=P_{d}(g_{i}(x)^{m_{i}})$. For
instance, if $p=0$, then $t_{i}=1$. Now, we may assume that $t_{i}\geq 2$ and
$p>0$. Let $P_{c}(f_{i}(x)^{n_{i}}):=\\{p^{u_{1}},\cdots,p^{u_{t_{i}}}\\}$
with $u_{1}>\cdots>u_{t_{i}}$ and
$P_{d}(g_{i}(x)^{m_{i}}):=\\{p^{v_{1}},\cdots,p^{v_{t_{i}}}\\}$ with
$v_{1}>\cdots>v_{t_{i}}$. By Theorem 1.2(2), we get
$\\{p^{u_{1}}-p^{u_{2}},\cdots,p^{u_{t_{i}-1}}-p^{u_{t_{i}}},p^{u_{t_{i}}}\\}=\\{p^{v_{1}}-p^{v_{2}},\cdots,p^{v_{t_{i}-1}}-p^{v_{t_{i}}},p^{v_{t_{i}}}\\}.$
For positive integers $a>b$ and $s>t$, the number $p^{a}-p^{b}$ is a $p$-power
if and only if $p=2$ and $a=b+1$; and the equality $p^{a}-p^{b}=p^{s}-p^{t}$
holds if and only if $a=s$ and $b=t$. By considering the cases $p=2$ and
$p\geq 3$ separately, we get $u_{k}=v_{k}$ for all $k\in[t_{i}].$ Thus
$P_{c}(f_{i}(x)^{n_{i}})=P_{d}(g_{i}(x)^{m_{i}})$ for $i\in[l_{c}].$ This
implies that $A$ and $B$ are Morita equivalent by Theorem 1.2(1).

(2) Let $R$ be a perfect field. Suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are
representation-finite and stably equivalent. Let $F$ be a stable equivalence
between $S_{n}(c,R)$ and $S_{m}(d,R).$ Then it follows from Lemma 2.20 that

$P_{c}(f_{i}(x)^{n_{i}})\subseteq\\{1,\max\\{n_{i},3\\}-1,\max\\{n_{i},3\\}\\}\mbox{
and
}P_{d}(g_{j}(x)^{m_{j}})\subseteq\\{1,\max\\{m_{j},3\\}-1,\max\\{m_{j},3\\}\\}$

for all $f_{i}(x)^{n_{i}}\in\mathcal{M}_{c}$ and
$g_{j}(x)^{m_{j}}\in\mathcal{M}_{d}.$ Reordering the blocks of $S_{n}(c,R)$
and $S_{m}(d,R)$, we may assume that there are natural numbers $a\leq
a^{\prime}$ and $b\leq b^{\prime}$ such that $n_{i}\geq 3$ (respectively,
$m_{j}\geq 3$) if and only if $i\in[a]$ (respectively, $j\in[b]$), and that
$n_{i}=2$ (respectively, $m_{j}=2$) if and only if $a<i\leq a^{\prime}$
(respectively, $b<j\leq b^{\prime}$). Let $A$ and $B$ be the sums of non-
semisimple blocks in $S_{n}(c,R)$ and $S_{m}(d,R)$, respectively. By Lemma
3.11, $F$ preserves non-semisimple blocks. Moreover, if $A_{i}$ and $B_{j}$
are stably equivalent, then $n_{i}=m_{j}$. Thus $a=b$ and
$a^{\prime}=b^{\prime}.$ Therefore we may assume that $F$ restricts to a
stable equivalence $F_{i}$ between $A_{i}$ and $B_{i}$ for $i\in[a^{\prime}]$.
In particular, $n_{i}=m_{i}$ for $i\in[a^{\prime}].$

Let $i\in[a].$ Then $n_{i}=m_{i}\geq 3$. By Lemma 2.10(2), $A_{i}$ and $B_{i}$
have no nodes. By Lemma 3.6(2), $F_{i}$ induces a stable equivalence between
the Frobenius parts of $A_{i}$ and $B_{i}$, that is, $U_{i}$ and $V_{i}$ are
stably equivalent. Let $K_{i}:=U_{i}/{\rm rad}(U_{i})$. Since $f_{i}(x)$ is
separable by our assumption on the ground field, it follows from Corollary
2.14 that $U_{i}\simeq K_{i}[x]/(x^{n_{i}})\simeq V_{i}$ as algebras.

According to Lemma 3.6(1), $A_{i}$ and $B_{i}$ have the same number of non-
injective, indecomposable projective modules, and hence the same number of
indecomposable projective modules. It then follows from
$\mathcal{B}(M_{i})\simeq\bigoplus_{r\in{P_{c}(f_{i}(x)^{n_{i}})}}R[x]/(f_{i}(x)^{r})$
as $U_{i}$-modules and
$\mathcal{B}(N_{j})\simeq\bigoplus_{s\in{P_{d}(g_{j}(x)^{m_{j}})}}R[x]/(g_{j}(x)^{s})$
as $V_{j}$-modules that

$|P_{c}(f_{i}(x)^{n_{i}})|=|P_{d}(g_{i}(x)^{n_{i}})|.$

Thus, due to the inclusions $\\{n_{i}\\}\subseteq
P_{c}(f_{i}(x)^{n_{i}})\subseteq\\{1,n_{i}-1,n_{i}\\}$ and
$\\{n_{i}\\}\subseteq P_{d}(g_{i}(x)^{n_{i}})\subseteq\\{1,n_{i}-1,n_{i}\\}$,
we obtain $P_{c}(f_{i}(x)^{n_{i}})=P_{d}(g_{i}(x)^{n_{i}})$ or
$P_{c}(f_{i}(x)^{n_{i}})=\mathcal{J}_{P_{d}(g_{i}(x)^{n_{i}})}.$ Hence $A_{i}$
and $B_{i}$ are almost $\nu$-stable derived equivalent by Lemma 2.3 if we
identify $U_{i}$ with $V_{i}$, and therefore $A_{i}$ and $B_{i}$ are stably
equivalent of Morita type by [13, Theorem 1.1].

Suppose $a<i\leq a^{\prime}.$ Then $n_{i}=m_{i}=2$. As the argument in the
foregoing case $i\in[a]$, we can show that $U_{i}\simeq V_{i}$ as algebras and
that $A_{i}$ and $B_{i}$ are Morita equivalent. Thus $A$ and $B$ are stably
equivalent of Morita type. Since $R$ is perfect, the semisimple blocks of
$S_{n}(c,R)$ and $S_{m}(d,R)$ are separable $R$-algebras. Hence $S_{n}(c,R)$
and $S_{m}(d,R)$ are stably equivalent of Morita type.

(3) Suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent. By
Lemma 3.5(1), ${\rm dom.dim}(S_{n}(c,R))\in\\{2,\infty\\}$. Thus, to prove
that $S_{n}(c,R)$ and $S_{m}(d,R)$ have the same dominant dimension, we show
that ${\rm dom.dim}(S_{n}(c,R))=\infty$ if and only if ${\rm
dom.dim}(S_{m}(d,R))=\infty.$ However, this follows from Theorem 1.2(2) and
Lemma 3.5(2) immediately. Thus $S_{n}(c,R)$ and $S_{m}(d,R)$ have the same
dominant dimension.

Now, suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are stably equivalent. Assume
${\rm dom.dim}(S_{n}(c,R))=\infty.$ Then $S_{n}(c,R)$ is a symmetric Nakayama
algebra by Lemma 3.5(2). By [26, Corollary 1.2], every non-simple projective
$S_{m}(d,R)$-module is injective, while the indecomposable projective modules
are of the form ${\rm Hom}_{V_{j}}(N_{j},R[x]/(g_{j}(x)^{s}))$ for $s\in
P_{d}(g_{j}(x)^{m_{j}})$ and $j\in[l_{d}]$. Thus $P_{d}(g_{j}(x)^{m_{j}})$ is
a singleton set for all $g_{j}(x)^{m_{j}}\in\mathcal{M}_{d}$, and therefore
${\rm dom.dim}(S_{m}(d,R))=\infty$ by Lemma 3.5(2). $\square$

For representation-finite, self-injective algebras over an algebraically
closed field, Asashiba proved in [3] that stable equivalences lift to stable
equivalences of Morita type. His proof uses classification of representation-
finite, self-injective algebras under derived equivalences. In general,
principal centralizer matrix algebras do not have to be self-injective. As
shown in the above, our proof uses a completely different strategy.

######  Corollary 3.12.

Let $R$ be a noetherian domain of characteristic $p>0$ and
$\sigma\in\Sigma_{n}$ be of cycle type
$\lambda:=(\lambda_{1},\cdots,\lambda_{s})$, and let $\sigma^{+}$ be a
permutation in $\Sigma_{n+1}$ of cycle type
$\lambda^{+}:=(\lambda_{1},\cdots,\lambda_{s},1)$. Then the following are
equivalent

$(a)$ $S_{n}(c_{\sigma},R)$ and $S_{n+1}(c_{\sigma^{+}},R)$ are derived
equivalent.

$(b)$ $S_{n}(c_{\sigma},R)$ and $S_{n+1}(c_{\sigma^{+}},R)$ are Morita
equivalent.

$(c)$ There is an $i\in[s]$ such that $p\nmid\lambda_{i}$.

Proof. Let $K$ be the fraction field of $R$ and $F_{p}$ be the prime field of
$K$. Since $c_{\sigma^{+}}$ is just the diagonal block-matrix
diag$(c_{\sigma},1)$, we have
$\mathcal{E}_{c_{\sigma^{+}}}=\mathcal{E}_{c_{\sigma}}\cup\\{x-1\\}$ when
$c_{\sigma}$ and $c_{\sigma^{+}}$ are viewed as matrices over either $K$ or
$F_{p}$. Note that all $\lambda_{i}$ are exactly the orbit lengths of
$\langle\sigma\rangle$ on $[n]$.

$(a)\Rightarrow(c)$ Suppose $S_{n}(c_{\sigma},R)$ and
$S_{n+1}(c_{\sigma^{+}},R)$ are derived equivalent. Then it follows from
Remark 3.4 that $S_{n}(c_{\sigma},K)$ and $S_{n+1}(c_{\sigma^{+}},K)$ are
derived equivalent, and hence Morita equivalent by Corollary 1.3. Further, by
Lemma 2.17, $p\nmid\lambda_{i}$ for some number $i$.

$(c)\Rightarrow(b)$ Assume (c). Then it follows from Lemma 2.17 that
$x-1\in\mathcal{E}_{c_{\sigma}}.$ By Theorem 1.2, $S_{n}(c_{\sigma},F_{p})$
and $S_{n+1}(c_{\sigma^{+}},F_{p})$ are Morita equivalent. With an argument
similar to the one in Remark 3.4, we obtain the isomorphisms of $R$-algebras

$R\otimes_{F_{p}}S_{n}(c_{\sigma},F_{p})\simeq S_{n}(c_{\sigma},R)\;\mbox{ and
}\;R\otimes_{F_{p}}S_{m}(c_{\sigma^{+}},F_{p})\simeq S_{m}(c_{\sigma^{+}},R).$

Hence $S_{n}(c_{\sigma},R)$ and $S_{n+1}(c_{\sigma^{+}},R)$ are Morita
equivalent. $\square$

Finally, we consider the case of nilpotent matrices. Let $c\in M_{n}(R)$ be a
nilpotent matrix. Then the Jordan canonical form $c_{0}$ of $c$ is unique up
to the ordering of its Jordan blocks. Further, $c_{0}$ has a Jordan block of
size $t$ if and only if ${\rm rank}(c^{t+1})+{\rm rank}(c^{t-1})-2{\rm
rank}(c^{t})>0$. We set $I_{c}:=\\{t\geq 1\mid c_{0}~{}\mbox{ has a Jordan
block of size t}\\}.$ Note that $\mathcal{M}_{c}$ consists of only one
polynomial of the form $x^{r}$ with $r$ being the maximal number in $I_{c}$.
Thus $I_{c}=P_{c}(x^{r}).$

######  Corollary 3.13.

Let $c\in M_{n}(R)$ be a nilpotent matrix and $d\in M_{m}(R)$. Then
$S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent if and only if $d=\lambda
I_{m}+b$ with $\lambda\in R$ and $b$ being a nilpotent matrix such that
$\mathcal{H}_{I_{b}}=\mathcal{H}_{I_{c}}.$

Proof. Suppose $d=\lambda I_{m}+b$ with $\lambda\in R$ and $b\in M_{m}(R)$ a
nilpotent matrix, such that $\mathcal{H}_{I_{b}}=\mathcal{H}_{I_{c}}.$ Then
$S_{m}(d,R)=S_{m}(b,R).$ Let $x^{s}$ be the unique polynomial in
$\mathcal{M}_{b}$. Clearly, the assumption
$\mathcal{H}_{I_{c}}=\mathcal{H}_{I_{b}}$ implies
$\mathcal{H}_{P_{c}(x^{r})}=\mathcal{H}_{P_{b}(x^{s})}.$ It then follows from
Theorem 1.2(2) that $S_{n}(c,R)$ and $S_{m}(b,R)$ are derived equivalent.

Conversely, suppose that $S_{n}(c,R)$ and $S_{m}(d,R)$ are derived equivalent.
Then it follows from Theorem 1.2(2) that $\mathcal{M}_{d}$ consists of only
one polynomial, say $f(x)^{s}$ with an irreducible polynomial $f(x)\in R[x]$
and $s\in\mathbb{N}$, and that $R[x]/(x^{r})\simeq R[x]/(f(x)^{s})$ as
algebras. Thus $r=s$ and $f(x)=x-\lambda$ for some $\lambda\in R$. Set
$b:=\lambda I_{m}-d$. Then $m_{b}(x)=x^{s}$, that is, $b$ is a nilpotent
matrix. Clearly, $P_{d}(f(x)^{s})=P_{b}(x^{s}).$ Therefore
$\mathcal{H}_{P_{c}(x^{r})}=\mathcal{H}_{P_{d}(f(x)^{s})}=\mathcal{H}_{P_{b}(x^{s})}.$
Hence $\mathcal{H}_{I_{c}}=\mathcal{H}_{I_{b}}.$ $\square$

###  3.3 Restrictions of derived equivalences

In this section we investigate the relation between a derived equivalence of
centralizer algebras of permutation matrices and the one of their $p$-parts.

Let $\sigma=\sigma_{1}\cdots\sigma_{s}\in\Sigma_{n}$ be the product of
disjoint cycle-permutations $\sigma_{i}$ of cycle type
$\lambda=(\lambda_{1},\cdots,\lambda_{s})$ with $\lambda_{i}\geq 1$ for $1\leq
i\leq s$. For a prime number $p>0$, a cycle $\sigma_{i}$ is said to be _$p$
-regular_ if $p\nmid\lambda_{i}$, and _$p$ -singular_ if $p\mid\lambda_{i}$.
If $p=0$, all cycles are $p$-regular. Let $r(\sigma)$ (respectively,
$s(\sigma)$) be the product of the $p$-regular (respectively, $p$-singular)
cycles of $\sigma$. Now, we consider $r(\sigma)$ and $s(\sigma)$ as elements
in $\Sigma_{n}$.

Recall that $\nu_{p}(n)$ denotes the largest non-negative integer such that
$p^{\nu_{p}(n)}$ divides $n$.

######  Proposition 3.14.

Let $R$ be a field of characteristic $p\geq 0$, $\sigma\in\Sigma_{n}$ and
$\tau\in\Sigma_{m}$. If $S_{n}(c_{\sigma},R)$ and $S_{m}(c_{\tau},R)$ are
derived equivalent, then

$(1)$ $S_{n}(c_{r(\sigma)},R)$ and $S_{m}(c_{r(\tau)},R)$ are derived
equivalent, and

$(2)$ $S_{n}(c_{s(\sigma)},R)$ and $S_{m}(c_{s(\tau)},R)$ are derived
equivalent.

Proof. By Theorem 1.2(1), we show the following: If
$c_{\sigma}\stackrel{{\scriptstyle D}}{{\sim}}c_{\tau}$, then
$c_{r(\sigma)}\stackrel{{\scriptstyle D}}{{\sim}}c_{r(\tau)}$ and
$c_{s(\sigma)}\stackrel{{\scriptstyle D}}{{\sim}}c_{s(\tau)}$.

Indeed, let $\lambda=(\lambda_{1},\cdots,\lambda_{k})$ be the cycle type of
$\sigma$. For $i\in[k]$ and an irreducible factor $f(x)$ of
$x^{\lambda_{i}}-1$, we define $q_{f(x)}:=max\\{\nu_{p}(\lambda_{j})\mid
j\in[k],f(x)\leq x^{\lambda_{j}}-1\\}$. Then it follows from Lemma 2.17 that
$\mathcal{E}_{c_{\sigma}}=\\{f(x)^{p^{\nu_{p}(\lambda_{i})}}\mid
i\in[k],f(x)\mbox{ is irreducible and }f(x)\leq x^{\lambda_{i}}-1\\}$ and

$\mathcal{M}_{c_{\sigma}}=\\{f(x)^{p^{q_{f(x)}}}\mid i\in[k],f(x)\mbox{ is
irreducible and }f(x)\leq x^{\lambda_{i}}-1\\}.$

This shows that $\mathcal{M}_{c_{r(\sigma)}}=\\{f(x)\mid
f(x)\in\mathcal{E}_{c_{\sigma}}~{}\mbox{ is irreducible}\\}$ and
$\mathcal{M}_{c_{s(\sigma)}}=\\{g(x)\mid
g(x)\in\mathcal{M}_{c_{\sigma}}~{}\mbox{ is reducible }\\}$.

Let $a$ denote the order of $r(\sigma)$. Then it follows from the definition
of $r(\sigma)$ that $p\nmid a$ and therefore $m_{r(\sigma)}(x)\leq x^{a}-1$ is
a product of distinct irreducible polynomials. Thus
$\mathcal{E}_{c_{r(\sigma)}}=\mathcal{M}_{c_{r(\sigma)}}$, consisting only of
some irreducible polynomials. By the definition of $s(\sigma)$, we see that
$\mathcal{E}_{c_{\sigma}}=(\mathcal{E}_{c_{r(\sigma)}}\setminus\\{x-1\\})\cup\mathcal{E}_{c_{s(\sigma)}}$
and that polynomials in $\mathcal{E}_{c_{s(\sigma)}}\setminus\\{x-1\\}$ are
reducible. Hence

$\mathcal{E}_{c_{s(\sigma)}}=\begin{cases}\\{u(x)\in\mathcal{E}_{c_{\sigma}}\mid
u(x)~{}\mbox{is reducible in}~{}R[x]\\}&\mbox{ if }r(\sigma)=1,\\\
\\{u(x)\in\mathcal{E}_{c_{\sigma}}\mid u(x)~{}\mbox{is reducible
in}~{}R[x]\\}\cup\\{x-1\\}&\mbox{ if }r(\sigma)\neq 1.\end{cases}$

Suppose $c_{\sigma}\stackrel{{\scriptstyle D}}{{\sim}}c_{\tau}$ . Then there
is a bijection $\pi:\mathcal{M}_{c_{\sigma}}\to\mathcal{M}_{c_{\tau}}$ such
that $R[x]/(h(x))\simeq R[x]/((h(x)\pi)$ as algebras and
$P_{c_{\sigma}}(h(x))=P_{c_{\tau}}((h(x))\pi)$ for
$h(x)\in\mathcal{M}_{c_{\sigma}}$. For irreducible polynomials $w(x),z(x)\in
R[x]$, if $R[x]/(w(x)^{a})\simeq R[x]/(z(x)^{b})$ as algebras, then $a=b$ and
$R[x]/(w(x)^{e})\simeq R[x]/(z(x)^{e})$ as algebras for all $e\leq a$. Thus we
may extend $\pi$ to a bijection between $\mathcal{E}_{c_{\sigma}}$ and
$\mathcal{E}_{c_{\tau}}$ such that $R[x]/(h(x))\simeq R[x]/((h(x)\pi)$ as
algebras for $h(x)\in\mathcal{E}_{c_{\sigma}}$. Since
$\mathcal{E}_{c_{\sigma}}=(\mathcal{E}_{c_{s(\sigma)}}\setminus\\{x-1\\})\dot{\cup}\mathcal{E}_{c_{r(\sigma)}}$,
the restriction of $\pi$ on $\mathcal{M}_{c_{r(\sigma)}}$ (respectively,
$\mathcal{M}_{c_{s(\sigma)}}$) maps onto $\mathcal{M}_{c_{r(\tau)}}$
(respectively, $\mathcal{M}_{c_{s(\tau)}}$). For
$v(x)\in\mathcal{M}_{c_{r(\sigma)}}$, there holds
$P_{c_{r(\sigma)}}(v(x))=P_{c_{r(\tau)}}((v(x))\pi)=\\{1\\}$. Thus
$c_{r(\sigma)}$ and $c_{r(\tau)}$ are $D$-equivalent.

Particularly, $r(\sigma)=1$ if and only if $r(\tau)=1$. This yields that
$x-1\in\mathcal{E}_{c_{s(\sigma)}}$ if and only if
$x-1\in\mathcal{E}_{c_{s(\tau)}}$. Let $a$ and $b$ be nonnegative integers
such that $(x-1)^{p^{a}}\in\mathcal{M}_{c_{\sigma}}$ and
$(x-1)^{p^{b}}\in\mathcal{M}_{c_{\tau}}$ are the only polynomials divisible by
$x-1$, and that $((x-1)^{p^{a}})\pi=j(x)$ and $((x-1)^{p^{b}})\pi^{-1}=k(x)$.
Then it follows from $R[x]/(h(x))\simeq R[x]/((h(x))\pi)$ for
$h(x)\in\mathcal{M}_{c_{\sigma}}$ that $j(x)=(x+u)^{p^{a}}$ and
$k(x)=(x+v)^{p^{b}}$ for $u,v\in R$. Since $P_{c_{\sigma}}(k(x))\subseteq
P_{c_{\sigma}}((x-1)^{p^{a}})$ and $P_{c_{\tau}}(j(x))\subseteq
P_{c_{\tau}}((x-1)^{p^{b}})$, we obtain

$P_{c_{\sigma}}(k(x))=P_{c_{\sigma}}((x-1)^{p^{a}})=P_{c_{\tau}}(j(x))=P_{c_{\tau}}((x-1)^{p^{b}}).$

Hence $a=b$. We may assume that $\pi$ maps $(x-1)^{p^{a}}$ in
$\mathcal{M}_{c_{\sigma}}$ to $(x-1)^{p^{b}}$ in $\mathcal{M}_{c_{\tau}}$. By
the above calculations of $\mathcal{M}_{c_{s(\sigma)}}$ and
$\mathcal{E}_{c_{s(\sigma)}}$, we get
$P_{c_{s(\sigma)}}(h(x))=P_{c_{\sigma}}(h(x))\setminus\\{1\\}$ for
$h(x)\in\mathcal{M}_{c_{s(\sigma)}}$ with $h(x)\neq(x-1)^{p^{a}}$. For
$(x-1)^{p^{a}}$ in $\mathcal{M}_{c_{\sigma}}$, we have
$P_{c_{\sigma}}((x-1)^{p^{a}})=P_{c_{s(\sigma)}}((x-1)^{p^{a}})$. Thus the
matrices $c_{s(\sigma)}$ and $c_{s(\tau)}$ are $D$-equivalent. $\square$

For a counterexample to the converse of Proposition 3.14, we refer to Example
4.5 in the next section.

##  4 Examples and questions

In this section we display a few examples to illustrate our results in the
previous sections.

The following example shows that the centralizer matrix algebras of non-
conjugate matrices may be Morita equivalent.

######  Example 4.1.

Let $R$ be a field and $J_{n}(\lambda)$ the $n\times n$ Jordan matrix with the
eigenvalue $\lambda\in R$. We take $c=J_{3}(1)\oplus J_{4}(1)\oplus
J_{3}(0)\oplus J_{2}(0)$ and $d=J_{3}(0)\oplus J_{4}(0)\oplus J_{3}(1)\oplus
J_{2}(1)$. In general, we have $m_{c\oplus d}(x)=[m_{c}(x),m_{d}(x)]$, where
$[f(x),g(x)]$ stands for the least common multiple of $f(x)$ and $g(x)$ in
$R[x]$. Then $m_{c}(x)=x^{3}(x-1)^{4}$,
$\mathcal{E}_{c}=\\{x^{2},x^{3},(x-1)^{3},(x-1)^{4}\\}$,
$\mathcal{M}_{c}=\\{x^{3},(x-1)^{4}\\}$,
$P_{c}(x^{3})=\\{2,3\\},P_{c}((x-1)^{4})=\\{3,4\\}$, and
$m_{d}(x)=x^{4}(x-1)^{3}$,
$\mathcal{E}_{d}=\\{x^{3},x^{4},(x-1)^{2},(x-1)^{3}\\}$,
$\mathcal{M}_{d}=\\{x^{4},(x-1)^{3}\\}$,$P_{d}(x^{4})=\\{3,4\\}$,
$P_{d}((x-1)^{3})=\\{2,3\\}$. Let $\pi:\mathcal{M}_{c}\to\mathcal{M}_{d}$ be
the map: $x^{3}\mapsto(x-1)^{3},(x-1)^{4}\mapsto x^{4}$. Then it follows from
Theorem 1.2(1) that $S_{12}(c,R)$ and $S_{12}(d,R)$ are Morita equivalent,
while $c$ and $d$ are not conjugate since they have different minimal
polynomials.

The next example shows that the existence of a Morita equivalence between
principal centralizer matrix algebras depends on the ground field.

######  Example 4.2.

Let
$\sigma:=(1~{}2~{}3~{}4~{}5)(6~{}7~{}8\cdots~{}17~{}18),\tau:=(1~{}2~{}3~{}4~{}5~{}6~{}7)(8~{}9\cdots~{}17~{}18)\in\Sigma_{18}$.
The minimal polynomials of $c_{\sigma}$ and $c_{\tau}$ over $\mathbb{Q}$ are
$(x-1)(x^{4}+x^{3}+x^{2}+x+1)(x^{12}+x^{11}+\cdots+x+1)$ and
$(x-1)(x^{10}+x^{9}+\cdots+x+1)(x^{6}+x^{5}+\cdots+x+1)$, respectively. In
this case,
$\mathcal{M}_{c_{\sigma}}=\\{x-1,x^{4}+x^{3}+x^{2}+x+1,x^{12}+x^{11}+\cdots+x+1\\}$
and
$\mathcal{M}_{c_{\tau}}=\\{x-1,x^{10}+x^{9}+\cdots+x+1,x^{6}+x^{5}+\cdots+x+1\\}$.
By Theorem $\ref{main1}(1)$, $S_{18}(c_{\sigma},\mathbb{Q})$ and
$S_{18}(c_{\tau},\mathbb{Q})$ are not Morita equivalent, while
$S_{18}(c_{\sigma},\mathbb{C})$ and $S_{18}(c_{\tau},\mathbb{C})$ are Morita
equivalent (see also [35, Theorem 1.2(2)]).

Now, we show that even for centralizer matrix algebras, almost $\nu$-stable
derived equivalences may not always arise from Morita equivalences.

######  Example 4.3.

Let $a=J_{5}(0)\oplus J_{4}(0)\oplus J_{2}(0)$ and $b=J_{5}(0)\oplus
J_{3}(0)\oplus J_{1}(0)$. Then the centralizer algebras $S_{11}(a,R)$ and
$S_{9}(b,R)$ are not Morita equivalent, but they are almost $\nu$-stable
derived equivalent by Theorem 1.2(3).

We point out that even in the class of principal centralizer matrix algebras,
derived equivalences do not have to preserve representation-finiteness, while
almost $\nu$-stable derived equivalences always preserve representation-
finiteness for arbitrary algebras.

######  Example 4.4.

Let $R$ be an algebraically closed field, $c:=J_{5}(0)\oplus J_{4}(0)\oplus
J_{1}(0)\in M_{10}(R)$ and $d:=J_{5}(0)\oplus J_{2}(0)\oplus J_{1}(0)\in
M_{8}(R)$. Then $S_{10}(c,R)$ and $S_{8}(d,R)$ are derived equivalent by
Theorem 1.2(2), while $S_{10}(c,R)$ is representation-finite, but $S_{8}(d,R)$
is not by Lemma 2.20.

Generally, the converse of Proposition 3.14 may be false, as we can see by the
following example.

######  Example 4.5.

Let $R$ be an algebraically closed field of characteristic $5$. We take
$\sigma\in\Sigma_{19}$ with the cycle type $(15,4)$, and $\tau\in\Sigma_{20}$
with the cycle type $(15,3,2)$. In this case, $r(\sigma)$ is a permutation of
the cycle type $(4,1^{15})$ and $s(\sigma)$ is a permutation of cycle type
$(15,1^{4})$, while $r(\tau)$ has the cycle type $(3,2,1^{15})$ and $s(\tau)$
has the cycle type $(15,1^{5})$. Clearly, $S_{19}(c_{s(\sigma)},R)$ and
$S_{20}(c_{s(\tau)},R)$ are derived equivalent by Corollary 3.12. Since
$\mathcal{M}_{c_{r(\sigma)}}=\\{x-1,x+1,x-\eta,x+\eta\\}$ and
$\mathcal{M}_{c_{r(\tau)}}=\\{x-1,x+1,x+\epsilon,x-\epsilon^{2}\\}$, where
$\eta$ and $\epsilon$ are $4$-th and $3$-th primitive roots of unity,
respectively, it follows from Theorem 1.2(2) that $S_{19}(c_{r(\sigma)},R)$
and $S_{20}(c_{r(\tau)},R)$ are derived equivalent.

By Lemma 2.17,
$\mathcal{M}_{c_{\sigma}}=\\{(x-1)^{5},(x-\epsilon)^{5},(x-\epsilon^{2})^{5},x+1,x-\eta,x+\eta\\}$
and
$\mathcal{M}_{c_{\tau}}=\\{(x-1)^{5},(x-\epsilon)^{5},(x-\epsilon^{2})^{5},x+1\\}$.
Clearly, $|\mathcal{M}_{c_{\sigma}}|=6\neq 4=|\mathcal{M}_{c_{\tau}}|$. Hence
no bijections between $\mathcal{M}_{c_{\sigma}}$ and
$\mathcal{M}_{c_{\sigma}}$ exist, and therefore $S_{19}(c_{\sigma},R)$ and
$S_{20}(c_{\tau},R)$ cannot be derived equivalent by Theorem 1.2.

This shows that derived equivalences of the centralizer matrix algebras of
$p$-regular parts and $p$-singular parts of permutations do not have to
guarantee a derived equivalence of the ones of the permutations themselves.

Having described derived equivalences of principal centralizer matrix
algebras, we propose the following questions for further study. In the
following, $R$ stands for a field.

Question 1. Under which necessary and sufficient conditions on matrices $c\in
M_{n}(R)$ and $d\in M_{m}(R)$ are $S_{n}(c,d)$ and $S_{m}(d,R)$ stably
equivalent?

Question 2. For which permutations $\sigma\in\Sigma_{n}$ and
$\tau\in\Sigma_{m}$ do derived equivalences between $S_{n}(c_{s(\sigma)},R)$
and $S_{m}(c_{s(\tau)},R)$, and between $S_{n}(c_{r(\sigma)},R)$ and
$S_{m}(c_{r(\tau)},R)$ ensure a derived equivalence between
$S_{n}(c_{\sigma},R)$ and $S_{m}(c_{\tau},R)$ ?

Related to general consideration of the centralizers of matrices, we mention
the following.

Question 3. Describe structural and homological properties of $S_{n}(C,R)$ for
$|C|\geq 2$.

Acknowledgements. The research work was supported partially by the National
Natural Science Foundation of China (Grants 12031014 and 12226314). The