Biserial minor degenerations of matrix algebras over a field
Let n≥2 be a positive integer, K an arbitrary field, and q=[q⁽¹⁾|…|q⁽ⁿ⁾] an n-block matrix of n×n square matrices q⁽¹⁾,…,q⁽ⁿ⁾ with coefficients in K satisfying the conditions (C1) and (C2) listed in the introduction. We study minor degenerations Mqn(K) of the full matrix algebra Mn(K) in the sense o...
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irk-123456789-1545332019-06-16T01:30:55Z Biserial minor degenerations of matrix algebras over a field Wlodarska, A. Let n≥2 be a positive integer, K an arbitrary field, and q=[q⁽¹⁾|…|q⁽ⁿ⁾] an n-block matrix of n×n square matrices q⁽¹⁾,…,q⁽ⁿ⁾ with coefficients in K satisfying the conditions (C1) and (C2) listed in the introduction. We study minor degenerations Mqn(K) of the full matrix algebra Mn(K) in the sense of Fujita-Sakai-Simson [7]. A characterisation of all block matrices q=[q⁽¹⁾|…|q⁽ⁿ⁾] such that the algebra Mqn(K) is basic and right biserial is given in the paper. We also prove that a basic algebra Mqn(K) is right biserial if and only if Mqn(K) is right special biserial. It is also shown that the K-dimensions of the left socle of Mqn(K) and of the right socle of Mqn(K) coincide, in case Mqn(K) is basic and biserial. 2010 Article Biserial minor degenerations of matrix algebras over a field / A. Wlodarska // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 2. — С. 125–137. — Бібліогр.: 18 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:16G10, 16G60, 14R20, 16S80. http://dspace.nbuv.gov.ua/handle/123456789/154533 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Let n≥2 be a positive integer, K an arbitrary field, and q=[q⁽¹⁾|…|q⁽ⁿ⁾] an n-block matrix of n×n square matrices q⁽¹⁾,…,q⁽ⁿ⁾ with coefficients in K satisfying the conditions (C1) and (C2) listed in the introduction. We study minor degenerations Mqn(K) of the full matrix algebra Mn(K) in the sense of Fujita-Sakai-Simson [7]. A characterisation of all block matrices q=[q⁽¹⁾|…|q⁽ⁿ⁾] such that the algebra Mqn(K) is basic and right biserial is given in the paper. We also prove that a basic algebra Mqn(K) is right biserial if and only if Mqn(K) is right special biserial. It is also shown that the K-dimensions of the left socle of Mqn(K) and of the right socle of Mqn(K) coincide, in case Mqn(K) is basic and biserial. |
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Wlodarska, A. Biserial minor degenerations of matrix algebras over a field Algebra and Discrete Mathematics |
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Wlodarska, A. |
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Wlodarska, A. |
| title |
Biserial minor degenerations of matrix algebras over a field |
| title_short |
Biserial minor degenerations of matrix algebras over a field |
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Biserial minor degenerations of matrix algebras over a field |
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Biserial minor degenerations of matrix algebras over a field |
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Biserial minor degenerations of matrix algebras over a field |
| title_sort |
biserial minor degenerations of matrix algebras over a field |
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Інститут прикладної математики і механіки НАН України |
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2010 |
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Biserial minor degenerations of matrix algebras over a field / A. Wlodarska // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 2. — С. 125–137. — Бібліогр.: 18 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
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AT wlodarskaa biserialminordegenerationsofmatrixalgebrasoverafield |
| first_indexed |
2025-07-14T06:36:28Z |
| last_indexed |
2025-07-14T06:36:28Z |
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1837603213686079488 |
| fulltext |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 9 (2010). Number 2. pp. 125 – 137
c© Journal “Algebra and Discrete Mathematics”
Biserial minor degenerations of matrix algebras
over a field
Anna W lodarska
Abstract. Let n ≥ 2 be a positive integer, K an arbitrary
field, and q = [q(1)| . . . |q(n)] an n-block matrix of n × n square
matrices q(1), . . . , q(n) with coefficients in K satisfying the condi-
tions (C1) and (C2) listed in the introduction. We study minor
degenerations Mq
n(K) of the full matrix algebra Mn(K) in the sense
of Fujita-Sakai-Simson [7]. A characterisation of all block matrices
q = [q(1)| . . . |q(n)] such that the algebra M
q
n(K) is basic and right
biserial is given in the paper. We also prove that a basic algebra
M
q
n(K) is right biserial if and only if Mq
n(K) is right special biserial.
It is also shown that the K-dimensions of the left socle of Mq
n(K)
and of the right socle of Mq
n(K) coincide, in case M
q
n(K) is basic
and biserial.
Introduction
Throughout this paper, n ≥ 2 is an integer and K an arbitrary field. We
denote by Mn(K) the K-algebra of all square n× n matrices with coefficients
in K. Following [7], by a minor constant structure matrix of size n × n2
with coefficients in K we mean any n-block matrix q = [q(1)|q(2)| . . . |q(n)], where
q(1) = [q
(1)
ij ], . . . , q(n) = [q
(n)
ij ] ∈ Mn(K) are n×n matrices satisfying the following
two conditions:
(C1) q
(r)
rj = 1 and q
(r)
jr = 1, for all j, r ∈ {1, . . . , n}.
(C2) q
(r)
ij q
(j)
is = q
(r)
is q
(j)
rs , for all i, j, r, s ∈ {1, . . . , n}.
We call q basic if, in addition, the following condition is satisfied:
(C3) q
(r)
jj = 0, for r = 1, . . . , n and all j ∈ {1, . . . , n} such that j 6= r.
The set of all minor constant structure matrices q of size n × n2, with
coefficients in K is denoted by STn(K) ⊆ Mn×n2(K). A matrix q in STn(K) is
An author is supported by Polish Research Grant N N201/2692/35/2008-2011.
2000 Mathematics Subject Classification: 16G10, 16G60, 14R20, 16S80.
Key words and phrases: right special biserial algebra, biserial algebra, Gabriel
quiver.
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126 biserial minor degenerations matrix algebras
called (0, 1)-matrix, if each entry q
(r)
ij is either 0 or 1. Throughout this paper,
any matrix q in STn(K) will be simply called a structure matrix.
Given q ∈ STn(K), a minor q-degeneration M
q
n(K) of the full matrix K-
algebra Mn(K) is defined in [7] to be the K-vector space Mn(K) equipped with
the multiplication
(1) ·q : Mn(K)⊗K Mn(K) //Mn(K)
given by the formula λ′ ·q λ′′ = [λij ], where λij =
n∑
s=1
λ′
isq
(s)
ij λ′′
sj , for i, j ∈
{1, . . . , n} and λ′ = [λ′
ij ], λ
′′ = [λ′′
ij ] ∈ Mn(K). It is easy to see that ·q defines a
K-algebra structure on Mn(K) and the unity matrix E is the identity element of
the algebra M
q
n(K). If n ≥ 2 and q is basic then the global homological dimension
of the algebra M
q
n(K) is infinite.
We recall that a class of algebras of type M
q
n(K) were studied by Fujita in
[5] (called full matrix algebras with structure systems) as a framework for a
study of factor algebras of tiled R-orders Λ, in relation with the results of the
papers [4], [11], [14] (see also [6] and [8]), where R is a discrete valuation domain.
The results in [7] show that one can treat the algebras M
q
n(K) by an elementary
algebraic geometry technique and study them in a deformation theory context.
Note also that the authors in [7] follow an old idea of the skew matrix ring
construction by Kupisch in [12], see also Oshiro and Rim [13].
The minor degenerations M
q
n(K) of the algebra Mn(K) and their modules
are investigated in [7] by means of the properties of the coefficients of the matrix
q and by applying quivers with relations. In particular, the Gabriel quiver of
M
q
n(K) is described and conditions for q to be M
q
n(K) a Frobenius algebra are
given.
In the present paper we give necessary and sufficient conditions for coefficients
of q ∈ STn(K) to be M
q
n(K) a right biserial algebra or a right special biserial
algebra, see [9], [18] and Sections 2 and 3 for definitions. One of the main results
of the paper is the following theorem.
Theorem 1. Assume that K is a field, n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is
a basic structure matrix and, given j, l ∈ {1, . . . , n}, we set
(2) M(j,l) = {p ∈ {1, . . . , n}; q
(l)
jp 6= 0} and m(j,l) = |M(j,l)|.
The following four conditions are equivalent.
(a) The algebra M
q
n(K) is right biserial (see Section 2).
(b) The algebra M
q
n(K) is right special biserial (see Section 3).
(c) For each i ∈ {1, . . . , n}, at least one of the following two conditions is
satisfied.
(c1) There exists a permutation τi : {1, . . . , n} → {1, . . . , n} such that
the equality q
(τi(l))
ip = 0 implies the equality q
(τi(j))
ip = 0, for l < j and each
p ∈ {1, . . . , n}.
(c2) There are two indices si < ri such that the sets M(i,si) and M(i,ri)
have the following properties:
(c21) |M(i,si) ∪M(i,ri)| = n− 1,
(c22) the set M(i,si) ∩M(i,ri) is empty or has precisely one element,
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A. W lodarska 127
(c23) there exist two bijections
τ(i,si) : {1, . . . ,m(i,si)} → M(i,si) and τ(i,ri) : {1, . . . ,m(i,ri)} → M(i,ri) such
that, given τ ∈ {τ(i,si), τ(i,ri)}, the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and all p ∈ {1, . . . , n}.
(d) For any i ∈ {1, . . . , n}, each of the following two conditions is satisfied.
(d1) There is one or two indices ri ∈ {1, . . . , n} such that ri 6= i and
q
(t)
iri
= 0, for all t 6∈ {i, ri}.
(d2) For any s 6= i such that q
(t′)
is = 0, for all t′ 6∈ {i, s}, there is at
most one index l(i,s) ∈ {1, . . . , n} such that l(i,s) 6= s, q
(s)
il(i,s)
6= 0 and q
(p′)
sl(i,s)
= 0,
for all p′ 6∈ {s, l(i,s)}.
The equivalence of (a) and (c) is proved in Section 2, and the equivalence
of the statements (a), (b), and (d) is proved in Section 3, where we also collect
basic facts on the algebras M
q
n(K) that are special biserial. In Corollary 3 we
show that dimKsoc(AqAq
) = dimKsoc(Aq
Aq), for any biserial and basic algebra
Aq = M
q
n(K). Moreover, we give an example of a non-biserial algebra Aq such
that dimKsoc(AqAq
) 6= dimKsoc(Aq
Aq).
Throughout this paper we use the standard terminology and notation in-
troduced in [1], [2], [3], [15], [17]. Given a ring R with an identity element, we
denote by J(R) the Jacobson radical of R, and by mod(R) the category of finitely
generated right R-modules, and by pr(R) the full subcategory of mod(R) of
right projective R-modules. For any homomorphism h : M −→ N in mod(R),
we denote by Imh the image of h. Given n ≥ 1, we denote by eij the matrix
unit in Mn(K) with 1 on the (i, j) entry, end zeros elsewhere. We fix n ≥ 2 and
we set
(3) Aq = M
q
n(K) = e1Aq ⊕ . . .⊕ enAq,
for q ∈ STn(K). Obviously, e1 = e11, . . . , en = enn is a complete set of pairwise
orthogonal primitive idempotents of Aq. We recall that Aq is said to be basic, if
eiAq 6∼= ejAq, for i 6= j. The paper contains part of author’s doctoral disserta-
tion written in Department of Algebra and Geometry of Nicolaus Copernicus
University.
1. Preliminaries
Throughout, we use the notation M(j,l) and m(j,l) as defined in (2) and, given
λ, λ′ ∈ M
q
n(K), we often write simply λλ′ instead of λ ·q λ
′.
Note that, in view of the definition (1) of ·q, we have
(4) ers ·q ejl =
{
q
(s)
rl erl, for s = j,
0, otherwise,
Recall that a right module M over a ring R is called serial (or uniserial), if
M has a unique composition series, see [1].
The following lemma collects elementary properties of the algebra Aq =
M
q
n(K) which we frequently use in the paper.
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128 biserial minor degenerations matrix algebras
Lemma 1. Assume that n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is a basic structure
matrix, Aq = M
q
n(K) and i, r, s ∈ {1, . . . , n}.
(a) ersAq =
∑
l∈M(r,s)
erlK.
(b) eirAq ⊆ eisAq if and only if q
(s)
ir 6= 0. Moreover, eirAq 6= eisAq, for
r 6= s.
(c) If L is a right submodule of eiAq, then L = eii1Aq + . . . + eiisAq, for
some i1, . . . , is ∈ {1, . . . , n}. If, in addition, L is serial, then L = eitAq, for
some t ∈ {1, . . . , n}.
(d) A right ideal S of Aq is simple if and only if S has the form S = ersK,
where ers is a matrix unit such that r 6= s and q
(s)
rl = 0, for all l 6= s.
(e) The Jacobson radical J(Aq) of Aq consists of all matrices λ = [λij ] ∈
Mn(K) such that λ11 = . . . = λnn = 0.
(f) Assume that q ∈ STn(K) is an arbitrary structure matrix. The algebra
Aq is basic if and only if the matrix q satisfies the condition (S3).
Proof. (a) If λ =
∑
j,l
λjlejl ∈ Aq, where λjl ∈ K, then (4) implies
ers ·q λ =
∑
j,l
λjlers ·q ejl =
n∑
l=1
λslq
(s)
rl erl =
∑
l∈M(r,s)
λslq
(s)
rl erl
and we get ersAq ⊆
∑
l∈M(r,s)
erlK. The inverse inclusion holds, because the
equality (4) yields erl =
1
q
(s)
rl
ers ·q esl, for all l ∈ M(r,s).
(b) By (a) and (4), we get q
(s)
ir 6= 0 if and only if eirAq ⊆ eisAq. This proves
the first part of (b). To prove the second part assume, to the contrary, that
r 6= s and eirAq = eisAq. Then, in view of (a), we have M(i,r) = M(i,s), and
consequently, we get the contradiction 0 6= q
(s)
ir q
(r)
is = q
(s)
is q
(r)
ss = 0, because of
(C2) and (C3). This finishes the proof of (b).
(c) Assume that L is a right submodule of eiAq. If λ ∈ L, then λ =
m∑
p=1
(
n∑
j=1
µ
(p)
ij eij) ·q λp, for some m ≥ 1, µ
(p)
ij ∈ K and λp = [λ
(p)
r′l′ ] ∈ Aq. Then,
according to (4), we get
λ ·q el =
m∑
p=1
(
n∑
j=1
µ
(p)
ij eij) ·q λp ·q el =
m∑
p=1
n∑
j=1
µ
(p)
ij λ
(p)
jl q
(j)
il eil,
for any l ∈ {1, . . . , n}. Hence, given l such that λ ·q el 6= 0, the element
eil = (
m∑
p=1
n∑
j=1
µ
(p)
ij λ
(p)
jl q
(j)
il )−1λ ·q el
belongs to L, and consequently L = eii1Aq + . . .+ eiisAq, for some i1, . . . , is ∈
{1, . . . , n}. Hence, if L is serial, then the right modules
eii1Aq, . . . , eiisAq form a chain and there is an index t ∈ {i1, . . . , is} such that
L = eitAq.
For the proof of (d), (e), and (f) we refer to [7].
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A. W lodarska 129
Recall from [1] that to any basic and connected finite dimensional K-algebra
A, with a complete set of primitive orthogonal idempotents {e1, e2, . . . , en}, we
associate the Gabriel quiver QA = (QA
0 , Q
A
1 ) as follows, see [10]. The set
QA
0 = {1, . . . , n} is the set of points of QA, which elements are in bijective
correspondence with the idempotents e1, e2, . . . , en. Given two points i, j ∈ QA
0 ,
the arrows β : i → j in QA
1 are in bijective correspondence with the vectors
in a fixed basis of the K-vector space ei[J(A)/J(A)
2]ej . The following simple
observation was made in [7, Corolary 2.20].
Lemma 2. Assume that n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is a basic structure
matrix and let Aq = M
q
n(K).
(a) Q
Aq
0 = {1, . . . , n}.
(b) Given i, j ∈ Q
Aq
0 , there exists an arrow i → j in Q
Aq
1 if and only if i 6= j
and q
(r)
ij = 0, for all r 6∈ {i, j}. In this case, there is a unique arrow βij : i → j
that corresponds to the coset qeij ∈ ei[J(Aq)/J(Aq)
2]ej of the matrix unit eij.
(c) The quiver QAq
is connected and has no loops.
2. When Aq = M
q
n
(K) is a biserial algebra?
One of the aims of this section is to give a characterisation of the right biserial
algebras M
q
n(K) in terms of the coefficients of the structure matrix q.
Now, we describe serial submodules of the projective Aq-modules eiAq in
terms of the coefficients of q.
Lemma 3. Assume that K is a field, n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is a
basic structure matrix, given i, r ∈ {1, . . . , n}. Let M(i,r) be the set (2).
(a) A right Aq-module eirAq is serial if and only if there exists a bijection
τ : {1, . . . ,m(i,r)} → M(i,r) such that the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and each p ∈ {1, . . . , n}.
(b) A right Aq-module eiAq is serial if and only if there exists a permutation
τ : {1, . . . , n} → {1, . . . , n} such that the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and each p ∈ {1, . . . , n}.
Proof. (a) Fix i, r ∈ {1, . . . , n}. Note that, by Lemma 1(a),(b), the module eirAq
is serial if and only if the submodules eitAq of eirAq, with t ∈ M(i,r), form a
chain, or equivalently (by Lemma 1(a)) if and only if there exists a bijection
τ : {1, . . . ,m(i,r)} → M(i,r) such that the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and each p ∈ {1, . . . , n}. Consequently, (a) follows.
(b) By applying (a) to ei = eii, we get eiAq = eiiAq, m(i,i) = n and
M(i,i) = {1, . . . , n}. Thus, by the arguments given above, eiAq is serial if
and only if there exists a permutation τ : {1, . . . , n} → {1, . . . , n} such that
the equality q
(τ(l))
ip = 0 implies the equality q
(τ(j))
ip = 0, for l < j and each
p ∈ {1, . . . , n}.
In the following two lemmata we study the structure of the Jacobson radical
J(eiAq) of eiAq in terms of the coefficients of q.
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130 biserial minor degenerations matrix algebras
Lemma 4. Assume that K is a field, n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is a
basic structure matrix and i ∈ {1, . . . , n}. Then the Jacobson radical J(eiAq) of
eiAq is a direct sum of two serial proper submodules if and only if there are two
indices s < r such that the sets M(i,s), M(i,r) (2) have the following properties:
• |M(i,s) ∪M(i,r)| = n− 1,
• the set M(i,s) ∩M(i,r) is empty,
• there exist two bijections
τ(i,s) : {1, . . . ,m(i,s)} → M(i,s) and τ(i,r) : {1, . . . ,m(i,r)} → M(i,r)
such that, given τ ∈ {τ(i,s), τ(i,r)}, the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and each p ∈ {1, . . . , n}.
Proof. Fix i ∈ {1, . . . , n}. By Lemma 1(c), J(eiAq) is a direct sum of two
serial proper submodules if and only if there are two indices s < r such that
J(eiAq) = eisAq ⊕ eirAq and eisAq, eirAq are serial. According to Lemma 1(a),
eisAq ∩ eirAq = 0 if and only if the set M(i,s) ∩M(i,r) is empty. Moreover, by [1,
Proposition 4.5(c)] and Lemma 1(a),(e), we have J(eiAq) = eisAq + eirAq if and
only if |M(i,s) ∪M(i,r)| = n− 1. By Lemma 3(a), the right modules eisAq, eirAq
are serial if and only if there exist two bijections τ(i,s) : {1, . . . ,m(i,s)} → M(i,s)
and τ(i,r) : {1, . . . ,m(i,r)} → M(i,r) such that, given τ ∈ {τ(i,s), τ(i,r)} the
equality q
(τ(l))
ip = 0 implies the equality q
(τ(j))
ip = 0, for l < j and each p ∈
{1, . . . , n}. Hence, the required equivalence follows.
Lemma 5. Assume that K is a field, n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is a
basic structure matrix, given i ∈ {1, . . . , n}. The Jacobson radical J(eiAq) of
eiAq is a sum of two serial submodules L′ and L′′ such that L′ ∩ L′′ is a simple
module if and only if there are two indices s < r such that the sets M(i,s),M(i,r)
(2) have the following properties:
• |M(i,s) ∪M(i,r)| = n− 1,
• the set M(i,s) ∩M(i,r) has precisely one element,
• there exist two bijections
τ(i,s) : {1, . . . ,m(i,s)} → M(i,s) and τ(i,r) : {1, . . . ,m(i,r)} → M(i,r)
such that, given τ ∈ {τ(i,s), τ(i,r)} the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and each p ∈ {1, . . . , n}.
Proof. Fix i ∈ {1, . . . , n}. By Lemma 1(c), there exist serial submodules L′ and
L′′ of J(eiAq) such that J(eiAq) = L′ + L′′ and the module L′ ∩ L′′ is simple
if and only if there exit two indices s < r such that J(eiAq) = eisAq + eirAq,
the module eisAq ∩ eirAq is simple and eisAq, eirAq are serial. According to
Lemma 1(a) and (d), the module eisAq ∩ eirAq is simple if and only the set
M(i,s) ∩M(i,r) has precisely one element. Hence the equivalence follows as in
the proof of Lemma 4.
We recall from [9] that a finite dimensional K-algebra A is right (resp. left)
biserial if every indecomposable projective right (resp. left) A-module P is
serial, or the Jacobson radical J(P ) of P is a sum of two serial submodules P1
and P2 such that the module P1 ∩ P2 is zero or simple. An algebra A is said to
be biserial, if it is both left and right biserial.
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The following corollary proves the equivalence of (a) and (c) in Theorem 1.
Corollary 1. Assume that K is a field, n ≥ 2 and q = [q(1)| . . . |q(n)] ∈ STn(K)
is a basic structure matrix. Then the following conditions are equivalent.
(a) The algebra M
q
n(K) is right biserial.
(b) For each i ∈ {1, . . . , n}, at least one of the following two conditions is
satisfied:
(b1) there exists a permutation τi : {1, . . . , n} → {1, . . . , n} such that
the equality q
(τi(l))
ip = 0 implies the equality q
(τi(j))
ip = 0, for l < j and each
p ∈ {1, . . . , n},
(b2) there are two indices si < ri such that the sets M(i,si), M(i,ri)
have the following properties:
(b21) |M(i,si) ∪M(i,ri)| = n− 1,
(b22) the set M(i,si) ∩M(i,ri) is empty or has precisely one element,
(b23) there exist two bijections
τ(i,si) : {1, . . . ,m(i,si)} → M(i,si) and τ(i,ri) : {1, . . . ,m(i,ri)} → M(i,ri) such
that, given τ ∈ {τ(i,si), τ(i,ri)}, the equality q
(τ(l))
ip = 0 implies the equality
q
(τ(j))
ip = 0, for l < j and each p ∈ {1, . . . , n}.
Proof. Apply [1, Corollary 5.17] and Lemmata 3, 4, and 5.
As an immediate consequence of Corollary 1 we get the following corollary.
Corollary 2. Assume that q ∈ STn(K) and q ∈ STn(K) is its (0, 1)-limit in
the sense of [7]. The algebra Aq is right biserial if and only if the algebra Aq is
right biserial.
3. Special biserial algebras M
q
n
(K)
In this section we study basic special biserial minor degenerations M
q
n(K) of
Mn(K) and we prove that the algebra M
q
n(K) is right special biserial if and only
if the algebra M
q
n(K) is right biserial.
We recall from [18] (see also [16]) that a K-algebra of the form KQ/Ω, where
Q is an quiver and Ω is an admissible ideal of the path K-algebra KQ of Q is
called a right special biserial, if the following two conditions are satisfied:
(a) any vertex of Q is a starting point of at most two arrows, and
(b) given an arrow β : i → j in Q, there is at most one arrow γ : j → r in Q
such that βγ 6∈ Ω.
Lemma 6. Assume that K is a field, n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is a
basic structure matrix and Aq = M
q
n(K). Let QAq
= (Q
Aq
0 , Q
Aq
1 ) be the Gabriel
quiver of Aq and i ∈ {1, . . . , n} is viewed as a vertex of QAq
.
(a) If eiAq is serial, then i is a starting point of precisely one arrow in QAq
.
(b) If J(eiAq) = L′ + L′′, where L′ 6= L′′ are serial proper submodules of
J(eiAq) and the module L′ ∩ L′′ is simple or zero, then i is a starting point of
precisely two arrows in QAq
.
(c) If Aq is right biserial, then each vertex of QAq
is a starting point of at
most two arrows.
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132 biserial minor degenerations matrix algebras
Proof. Let Aq = M
q
n(K) and let QAq
= (Q
Aq
0 , Q
Aq
1 ). Fix i, l in Q
Aq
0 such that
i 6= l. Note that, by Lemma 1(e),(f), [1, Lemma I.4.2(a) ] and [1, Appendix
3.5(b)], we have
(5) eiAqel/eiJ(Aq)
2el ∼= HomAq
(elAq, eiAq)/rad2
pr(Aq)(elAq, eiAq).
where rad2
pr(Aq) is the square of the Jacobson radical radpr(Aq) of the category
pr(Aq). A homomorphism f : elAq → eiAq is irreducible in the category pr(Aq) if
and only if f is a non-isomorphism and f 6∈ rad2
pr(Aq)(elAq, eiAq), or equivalently,
there is an arrow βil : i → l in QAq
.
(a) Assume that the module eiAq is serial. Then J(eiAq) contains a unique
maximal submodule J(eiAq)
′, the module J(eiAq)/J(eiAq)
′ is simple and, hence,
the projective cover of J(eiAq) has the form h′ : ejAq → J(eiAq), for some
j 6= i. Moreover, the composite homomorphism h = (ejAq
h′
−→ J(eiAq) ⊂ eiAq) is
irreducible in pr(Aq). To show it, assume to the contrary that h is not irreducible.
It follows that h ∈ rad2
pr(Aq)(ejAq, eiAq). Hence, there are two non-zero non-
isomorphisms
ejAq
f ′
//esAq
f ′′
//eiAq,
for some s 6∈ {i, j}, such that f ′′ ◦ f ′ 6= 0. It follows that Im f ′′ ⊆ J(eiAq) and
there is g : esAq → ejAq such that f ′′ = h ◦ g, that is, the diagram
ejAq
h
// eiAq
esAq,
g
cc
f ′′
OO
is commutative. Hence, we get g ◦ f ′ 6= 0, because h ◦ g ◦ f ′ = f ′′ ◦ f ′ 6= 0. Since
0 6= g ◦ f ′ ∈ End(ejAq) ∼= K, then g ◦ f ′ = µ · id, for some non-zero µ ∈ K. It
follows that f ′ is an isomorphism and we get a contradiction. Consequently, h is
an irreducible homomorphism and there is an arrow βij : i → j in QAq
.
Assume that there is an arrow βip : i → p in QAq
starting from i. Then
there is an irreducible homomorphism g′ : epAq → eiAq and, by the arguments
used earlier, there is a commutative diagram
ejAq
h
// eiAq
epAq,
u
cc
g′
OO
It follows that u is an isomorphism and, hence, p = j and βip = βij . This finishes
the proof of (a).
(b) Assume that J(eiAq) = L′ + L′′, where L′ 6= L′′ are serial proper
submodules of J(eiAq) and the module L′ ∩ L′′ is simple or zero. One can show,
as in the proof of (a), that there are homomorphisms h′ : ejAq → J(eiAq) and
h′′ : erAq → J(eiAq), for some j 6= r (because dimK HomK(ejAq, eiAq) = 1),
such that the homomorphism (h′, h′′) : ejAq ⊕ erAq −→L′ + L′′ = J(eiAq) is a
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projective cover of J(eiAq) (because L′ 6= L′′ are serial proper submodules of
J(eiAq)), and that the composite homomorphism
h = (ejAq ⊕ erAq
(h′,h′′)
−→ L′ + L′′ = J(eiAq) ⊂ eiAq)
is irreducible in pr(Aq). It follows that the composite homomorphisms h̃′ =
(ejAq → ejAq ⊕ erAq
h
−→ eiAq), h̃
′′ = (erAq → ejAq ⊕ erAq
h
−→ eiAq) are irre-
ducible homomorphisms in pr(Aq). Since j 6= r, then in view of the isomorphism
(5), the irreducible homomorphisms h̃′ and h̃′′ correspond to two different arrows
βij : i → j and βir : i → r in QAq
starting from i.
To finish the proof of (b), assume that there is an arrow βit : i → t in QAq
starting from i. Then there is an irreducible homomorphism g : etAq → eiAq
and, by the arguments used earlier, there is a commutative diagram
ejAq ⊕ erAq
h
// eiAq
etAq,
u
ff
g
OO
where u = (uj , ur) and uj : etAq → ejAq, ur : etAq → erAq. Since g is irreducible
and h̃′, h̃′′ belong to the Jacobson radical of the category pr(Aq), then one of the
maps uj , ur is an isomorphism, see [1, Appendix 3.5(b)]. If uj is an isomorphism,
then t = j and βit = βij . If ur is an isomorphism, then t = r and βit = βir. This
finishes the proof of (b).
Since (c) is a consequence of (a) and (b), the proof is complete.
Now we describe the matrices q ∈ STn(K) such that the algebra M
q
n(K) is
right special biserial.
Theorem 2. Assume that K is a field, n ≥ 2 and q = [q(1)| . . . |q(n)] ∈ STn(K)
is a basic structure matrix. The following conditions are equivalent.
(a) The algebra M
q
n(K) is right biserial.
(b) The algebra M
q
n(K) is right special biserial.
(c) For any i ∈ {1, . . . , n}, each of the following two conditions is satisfied:
(c1) there is one or two indices ri ∈ {1, . . . , n} such that ri 6= i and
q
(t)
iri
= 0, for all t 6∈ {i, ri},
(c2) for any s 6= i such that q
(t′)
is = 0, for all t′ 6∈ {i, s}, there is at
most one index l(i,s) ∈ {1, . . . , n} such that l(i,s) 6= s, q
(s)
il(i,s)
6= 0 and q
(p′)
sl(i,s)
= 0,
for all p′ 6∈ {s, l(i,s)}.
Proof. Let A = M
q
n(K) and let QA = (QA
0 , Q
A
1 ). By the proof of Gabriel’s
Theorem given in [1, Chapter III], the map h : KQA −→ A defined on arrows
βij : i → j by the formula h(βij) = eij uniquely extends to a K-algebra surjective
homomorphism h : KQA −→ A, such that h(ω) = q
(i2)
i1i3
q
(i3)
i1i4
· · . . . · q
(il−1)
i1il
ei1il ,
for any path ω = βi1i2βi2i3 . . . βil−1il . Moreover, the ideal Ω = Kerh is an
admissible and h induces a K-algebra isomorphism KQA/Ω ∼= A. Hence, in
view of the assumption that q is the basic structure matrix, the ideal Ω contains
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134 biserial minor degenerations matrix algebras
the elements βii1βi1i2 . . . βimi, for any cycle and the path βijβjl′ , if q
(j)
il′ = 0,
for i, j, l′ ∈ {1, . . . , n}. Throughout the proof, we view A as the bound quiver
algebra A ∼= KQA/Ω.
(a)⇒(b) Assume that A = M
q
n(K) ∼= KQA/Ω is right biserial. By Lemma
6(c), every vertex of QA is a starting point of at most two arrows. It remains
to show that, for any arrow βij : i → j in QA, there exists at most one arrow
βjr : j → r in QA such that βijβjr 6∈ Ω, or equivalently, q
(j)
ir 6= 0.
If n = 2, then according to [7, Example 2.8] and Lemma 1(f), up to isomor-
phism, there exists precisely one basic algebra, namely the algebra A ∼= KQA/Ω,
given by the quiver
1
β12
//2
β21
oo
and the relations β12β21 and β21β12. Thus, in case n = 2, our claim follows.
If n = 3, then there are precisely five such algebras listed in [7, Theorem 4.1
], up to isomorphism, and described by means of quivers with relations. A case
by case inspection shows that the implication (a)⇒(b) holds, for each of the five
algebras listed in [7, Theorem 4.1].
Assume that n ≥ 4 and there exists an arrow βij : i → j in QA. Suppose, to
the contrary, that there exist two different arrows βjr : j → r and βjp : j → p
in QA such that βijβjr 6∈ Ω and βijβjp 6∈ Ω. The arguments given above yield
q
(j)
ir 6= 0 and q
(j)
ip 6= 0. By our assumption and Lemma 2, q
(p)
jr = q
(r)
jp = 0. Hence
we conclude that
q
(j)
ir q
(r)
ip = q
(j)
ip q
(r)
jp = 0 and q
(j)
ip q
(p)
ir = q
(j)
ir q
(p)
jr = 0,
because of (C2). Hence, in view of (C1), we get
(6) q
(r)
ip = q
(p)
ir = 0 and q
(p)
ip = q
(r)
ir = 1.
It follows that there is no permutation τi satisfying the condition (b1) of Corollary
1. Because A is a right biserial, for the algebra A the condition (b2) of Corollary
1 is satisfied and we have two sets M(i,si), M(i,ri) such that |M(i,si)∪M(i,ri)| =
n − 1, for si < ri. Hence we get j ∈ M(i,si) or j ∈ M(i,ri), because q is basic.
Without loss of generality, we can assume that j ∈ M(i,si). Thus by (2) and
Lemma 1(b), we have eisiA ⊇ eijA ⊃ eirA and eijA ⊃ eipA. According to
Lemma 1(b), this implies q
(si)
ir 6= 0 and q
(si)
ip 6= 0. Hence, in view of (6), the there
is no bijection τ(i,si) : {1, . . . ,m(i,si)} → M(i,si) such that the condition (b23) of
Corollary 1 is satisfied and we get a contradiction. Consequently, the algebra A,
with q ∈ STn(K) and n ≥ 4 is right special biserial. This finishes the proof of
the implication (a)⇒(b).
The implication (b)⇒(a) holds, for any basic algebra A, see [18, Lemma 1].
(b)⇔(c) Recall that A ∼= KQA/Ω and fix i ∈ {1, . . . , n}. By Lemma 2(b),
the condition (c1) is satisfied if and only if the vertex i in QA is a starting
point of at most two arrows in QA. Moreover, according to Lemma 2(b) and the
property βii1βi1i2 ∈ Ω, if q
(i1)
ii2
= 0, the condition (c2) is satisfied if and only if
for any arrow βis : i → s in QA there is at most one arrow βsl(i,s) : s → l(i,s)
in QA such that βisβsl(i,s) 6∈ Ω. Consequently, the equivalence of (b) and (c) is
proved and the proof is complete.
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Note that, together with Corollary 1, Theorem 2 completes the proof of
Theorem 1. We recall from [18] that the implication (a)⇒(b) does not hold, for
arbitrary basic algebra.
Now we prove an interesting property of the socle of the algebras Aq =
M
q
n(K). For this purpose, we recall from [7] that the transpose of q ∈ STn(K)
is defined to be the n-block matrix qtr = q̃ = [q̃(1)| . . . |q̃(n)], where q̃(j) = [q(j)]tr
is the transpose of q(j), for j = 1, . . . , n.
Corollary 3. Assume that K is a field, n ≥ 2 and the structure matrix q =
[q(1)| . . . |q(n)] ∈ STn(K) is basic. If the algebra Aq = M
q
n(K) is biserial, then
dimKsoc(AqAq
) = dimKsoc(Aq
Aq).
Proof. Assume that n ≥ 2, q = [q(1)| . . . |q(n)] ∈ STn(K) is basic and the algebra
Aq = Mn(K) is biserial. It follows from Lemma 1(d) that
(7) dimKsoc(ejAq) ∈ {1, 2},
for any j ∈ {1, . . . , n} Note that, according to [7, Lemma 2.15(a)], there is an
algebra isomorphism (Aq)
op ∼= Aqtr . It follows that the algebra Aqtr is biserial
and, by (7), we have
(8) dimKsoc(Aqej) ∈ {1, 2},
for any j ∈ {1, . . . , n}.
Fix j ∈ {1, . . . , n}. First, we prove that, dimK soc(Aqei) = 1, if soc(ejAq) =
ejiK, for some i ∈ {1, . . . , n}. Assume that soc(ejAq) = ejiK, for some i ∈
{1, . . . , n}. Then, by Lemma 1(b), we get q
(s)
ji 6= 0, for all s ∈ {1, . . . , n}. Moreover,
the definition of qtr yields (qtr)
(s)
ij 6= 0, for each s ∈ {1, . . . , n}.
The condition (C1) yields the equality (qtr)
(s)
is = 1, for each s ∈ {1, . . . , n}.
Hence and from Lemma 1(d), we conclude that the equality (qtr)
(j)
ip = 0, for p 6= j
holds only for j ∈ {1, . . . , n}. Equivalently, by Lemma 1(d) and the isomorphism
(Aq)
op ∼= Aqtr , we have dimKsoc(Aqei) = dimKsoc(eiAqtr ) = 1.
In the sequel, we denote by lqr (resp. lql ) the number of indecomposable
projective right (resp. left) Aq-modules of the form etAq (resp. Aqet) with
simple socle. Note that, if etAq 6∼= ep′Aq and the modules soc(etAq), soc(ep′Aq)
are simple, then soc(etAq) 6∼= soc(ep′Aq), because the modules etAq, ep′Aq are
injective.
By the argument applied above and Lemma 1(f), we obtain lqr ≤ lql and lq
tr
r ≤
lq
tr
l . Since, in view of the isomorphism (Aq)
op ∼= Aqtr , we have lql = lq
tr
r and
lqr = lq
tr
l , then lql ≤ lqr , that is, lqr = lql . Because Aqe1 ⊕ . . . ⊕ Aqen = Aq =
e1Aq ⊕ . . .⊕ enAq and lqr = lql , then the formulae (7) and (8) yield the required
equality dimKsoc(Aq
Aq) = dimKsoc(AqAq
).
We end the paper by an example of a non-biserial basic minor degeneration
Aq of Mn(K) such that dimKsoc(Aq
Aq) 6= dimKsoc(AqAq
).
Example 1. Assume that n = 4 and let Aq = M
q
4(K) is given by the basic
structure matrix
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136 biserial minor degenerations matrix algebras
q =
1111 0100 0010 0001
1011 1111 0010 0001
1000 0100 1111 0001
1000 0100 0010 1111
It follows from Lemma 1(d) that we have
soc(e1Aq) = e12K ⊕ e13K ⊕ e14K,
soc(e2Aq) = e23K ⊕ e24K,
soc(e3Aq) = e31K ⊕ e32K ⊕ e34K, and
soc(e4Aq) = e41K ⊕ e42K ⊕ e43K
and hence dimKsoc(AqAq
) = 11. Moreover, in view of (7), the algebra Aq is
not biserial, because dimK soc(e1Aq) = 3. On the other hand, since the algebra
Aqtr
∼= Aop is given by the structure matrix
qtr =
1111 0100 0010 0001
1000 1111 0010 0001
1100 0100 1111 0001
1100 0100 0010 1111
we get
soc(e1Aqtr ) = e12K ⊕ e13K ⊕ e14K,
soc(e2Aqtr ) = e21K ⊕ e23K ⊕ e24K,
soc(e3Aqtr ) = e32K ⊕ e34K,
soc(e4Aqtr ) = e42K ⊕ e43K.
This shows that the dimension of the left socle of the algebra Aq equals
dimKsoc(Aq
Aq) = dimKsoc(AqtrAqtr
) = 10. Consequently, we have shown that
dimKsoc(AqAq
) 6= dimKsoc(Aq
Aq).
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Contact information
A. W lodarska Faculty of Mathematics and Computer Sci-
ence, Nicolaus Copernicus University,
87-100 Toruń, Poland
E-Mail: anna@mat.uni.torun.pl
Received by the editors: ????
and in final form ????.
Anna Włodarska
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