Tiled orders of width 3
We consider projective cover over tiled order and calculate the kernel of epimorphism from direct sum of submodules of distributive module to their sum.
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Інститут прикладної математики і механіки НАН України
2009
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| Цитувати: | Tiled orders of width 3 / V. Zhuravlev, D. Zhuravlyov // Algebra and Discrete Mathematics. — 2009. — Vol. 8, № 1. — С. 111–123. — Бібліогр.: 7 назв. — англ. |
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irk-123456789-1533852019-06-15T01:26:36Z Tiled orders of width 3 Zhuravlev, V. Zhuravlyov, D. We consider projective cover over tiled order and calculate the kernel of epimorphism from direct sum of submodules of distributive module to their sum. 2009 Article Tiled orders of width 3 / V. Zhuravlev, D. Zhuravlyov // Algebra and Discrete Mathematics. — 2009. — Vol. 8, № 1. — С. 111–123. — Бібліогр.: 7 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 16P40. http://dspace.nbuv.gov.ua/handle/123456789/153385 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We consider projective cover over tiled order and calculate the kernel of epimorphism from direct sum of submodules of distributive module to their sum. |
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Zhuravlev, V. Zhuravlyov, D. |
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Zhuravlev, V. Zhuravlyov, D. Tiled orders of width 3 Algebra and Discrete Mathematics |
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Zhuravlev, V. Zhuravlyov, D. |
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Zhuravlev, V. |
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Tiled orders of width 3 |
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Tiled orders of width 3 |
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Tiled orders of width 3 |
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Tiled orders of width 3 |
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Tiled orders of width 3 |
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tiled orders of width 3 |
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Інститут прикладної математики і механіки НАН України |
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2009 |
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http://dspace.nbuv.gov.ua/handle/123456789/153385 |
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Tiled orders of width 3 / V. Zhuravlev, D. Zhuravlyov // Algebra and Discrete Mathematics. — 2009. — Vol. 8, № 1. — С. 111–123. — Бібліогр.: 7 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT zhuravlevv tiledordersofwidth3 AT zhuravlyovd tiledordersofwidth3 |
| first_indexed |
2025-07-14T04:36:33Z |
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2025-07-14T04:36:33Z |
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1837595670056271872 |
| fulltext |
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.
Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 1. (2009). pp. 111 – 123
c© Journal “Algebra and Discrete Mathematics”
Tiled orders of width 3
Viktor Zhuravlev, Dmytro Zhuravlyov
Communicated by V. V. Kirichenko
Abstract. We consider projective cover over tiled order and
calculate the kernel of epimorphism from direct sum of submodules
of distributive module to their sum.
1. Tiled orders over discrete valuation rings
Recall [1] that a semimaximal ring is a semiperfect semiprime right
Noetherian ring A such that for each primitive idempotent e ∈ A the
ring eAe is a discrete valuation ring (not necessarily commutative).
Denote by Mn(B) the ring of all n× n matrices over a ring B.
Theorem 1 (see [1]). Each semimaximal ring is isomorphic to a finite
direct product of prime rings of the following form:
Λ =
O πα12O . . . πα1nO
πα21O O . . . πα2nO
. . . . . . . . . . . . . . . . . . . . . . . . . . .
παn1O παn2O . . . O
, (1)
where n ≥ 1, O is a discrete valuation ring with a prime element π, and
αij are integers such that
αij + αjk ≥ αik, αii = 0
for all i, j, k.
2000 Mathematics Subject Classification: 16P40.
Key words and phrases: tiled order, distributive module, projective cover.
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.112 Tiled orders of width 3
The ring O is embedded into its classical division ring of fractions D,
and (1) is the set of all matrices (aij) ∈Mn(D) such that
aij ∈ παijO = eiiΛejj ,
where e11, . . . , enn are the matrix units of Mn(D). It is clear that Q =
Mn(D) is the classical ring of fractions of Λ.
Obviously, the ring A is right and left Noetherian.
Definition 1. A module M is distributive if its lattice of submodules is
distributive, i.e.,
K ∩ (L+N) = K ∩ L+K ∩N
for all submodules K, L, and N .
Clearly, any submodule and any factormodule of a distributive module
are distributive modules.
A semidistributive module is a direct sum of distributive modules. A
ring A is right (left) semidistributive if it is semidistributive as the right
(left) module over itself. A ring A is semidistributive if it is both left and
right semidistributive (see [7]).
Theorem 2 (see [6]). The following conditions for a semiperfect semi-
prime right Noetherian ring A are equivalent:
• A is semidistributive;
• A is a direct product of a semisimple artinian ring and a semimax-
imal ring.
By a tiled order over a discrete valuation ring, we mean a Noetherian
prime semiperfect semidistributive ring Λ with nonzero Jacobson radi-
cal. In this case, O = eΛe is a discrete valuation ring with a primitive
idempotent e ∈ Λ.
Definition 2. An integer matrix E = (αij) ∈Mn(Z) is called
• an exponent matrix if αij + αjk ≥ αik and αii = 0 for all i, j, k;
• a reduced exponent matrix if αij + αji > 0 for all i, j, i 6= j.
We use the following notation: Λ = {O, E(Λ)}, where E(Λ) = (αij) is
the exponent matrix of the ring Λ, i.e.
Λ =
n∑
i,j=1
eijπ
αijO,
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.V. Zhuravlev, D. Zhuravlyov 113
in which eij are the matrix units. If a tiled order is reduced, i.e., Λ/R(Λ)
is the direct product of division rings, then αij + αji > 0 if i 6= j, i.e.,
E(Λ) is reduced.
We denote by M(Λ) the poset (ordered by inclusion) of all projective
right Λ-modules that are contained in a fixed simple Q-module U . All
simple Q-modules are isomorphic, so we can choice one of them. Note
that the partially ordered sets Ml(Λ) and Mr(Λ) corresponding to the
left and the right modules are anti-isomorphic.
The set M(Λ) is completely determined by the exponent matrix
E(Λ) = (αij). Namely, if Λ is reduced, then
M(Λ) = {pz
i | i = 1, . . . n, and z ∈ Z},
where
pz
i ≤ pz′
j ⇐⇒
{
z − z′ ≥ αij if M(Λ) = Ml(Λ),
z − z′ ≥ αji if M(Λ) = Mr(Λ).
Obviously, M(Λ) is an infinite periodic set.
Let P be an arbitrary poset. A subset of P is called a chain if any
two of its elements are related. A subset of P is called a antichain if no
two distinct elements of the subset are related.
Definition 3. The maximal number w(P ) of elements in an antichain
of P is called the width of P .
The width of Mr(Λ) is called the width of a tiled order Λ and denotes
by w(Λ).
Definition 4. A right (resp. left) Λ-module M (resp. N) is called a right
(resp. left) Λ-lattice if M (resp. N) is a finitely generated free O-module.
Given a tiled order Λ we denote Latr(Λ) (resp. Latl(Λ)) the category
of right (resp. left) Λ-lattices. We denote by Sr(Λ) (resp. Sl(Λ)) the
partially ordered by inclusion set, formed by all Λ-lattices contained in
a fixed simple Mn(D)-module W (resp. in a left simple Mn(D)-module
V ). Such Λ-lattices are called irreducible.
Let Λ = {O, E(Λ)} be a tiled order, W (resp. V ) is a simple right
(resp. left) Mn(D)-module with D-basis e1, . . . , en such that eiejk =
δijek (eijek = δjkei).
Then any right (resp. left) irreducible Λ-lattice M (resp. N), lying
in W (resp. in V ) is a Λ-module with O-basis (πα1e1, . . . , π
αnen), while
{
αi + αij ≥ αj , for the right case;
αij + αj ≥ αi, for the left case.
(2)
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.114 Tiled orders of width 3
Thus, irreducible Λ-lattices M can be identified with integer-valued
vector (α1, . . . , αn) satisfying (2). We shall write E(M) = (α1, . . . , αn)
or M = (α1, . . . , αn).
The order relation on the set of such vectors and the operations on
them corresponding to sum and intersection of irreducible lattices are
obvious.
Remark 1. Obviously, irreducible Λ-lattices M1 = (α1, . . . , αn) and
M2 = (β1, . . . , βn) are isomorphic if and only if αi = βi + z for i =
1, . . . , n and z ∈ Z.
2. Kernel of epimorphism from direct sum of modules to
their sum
Proposition 1. Let M be an irreducible and non-projective Λ-module,
X be a maximal submodule of M . Then there exists projective submodule
of M , which is not submodule of X.
Proof. Let
E(M) = (α1, . . . , αi−1, αi, αi+1, . . . , αn) and
E(X) = (α1, . . . , αi−1, αi + 1, αi+1, . . . , αn).
Since M is right Λ-module, αi + αik ≥ αk for all i, k. Consider the
projective module παiPi with
E(παiPi) = (αi + αi1, . . . , αi + αi,i−1, αi + αii, αi + αi,i+1, . . . , αi + αin).
Obviously, παiPi ⊂M , but παiPi * X.
Since X is maximal submodule of M , then X + παiPi = M . Besides,
E(X ∩ παiPi) = (αi + αi1, . . . , αi + αi,i−1, αi + 1,
αi + αi,i+1, . . . , αi + αin) = E(παiRi), (3)
i. e. X ∩ παiPi = Ri, where Ri = rad Pi.
Proposition 2. Let X1, . . . , Xs be the set of all maximal submodules of
irreducible and non-projective Λ-module M with E(M) = (α1, . . . , αn)
and E(Xi) = E(M) + eji
, where ek = (0, . . . , 0
︸ ︷︷ ︸
k−1
, 1, 0, . . . , 0). Then
P (M) =
s
⊕
i=1
παjiPji
and M =
s∑
i=1
παjiPji
.
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.V. Zhuravlev, D. Zhuravlyov 115
Proof. Since rad M =
s⋂
i=1
Xi, we have E(rad M) = E(M) +
s∑
i=1
eji
,
P (M) =
s
⊕
i=1
παjiPji
. Besides, παjiPji
⊂M for each i, whereas
s∑
i=1
παjiPji
⊂
M . Suppose that
s∑
i=1
παjiPji
6= M . Then there is the maximal submod-
ule Xk such that παjiPji
⊆ Xk. This contradicts to inclusion παjkPjk
*
Xk.
Lemma 1. Let M1, M2, M3 be submodules of distributive module M
and ϕ : M1 ⊕M2 ⊕M3 →M1 +M2 +M3 be epimorphism of their direct
sum on their sum defined by the rule (x1, x2, x3) 7→ x1 + x2 + x3. Then
kerϕ = {(m12 −m31,m23 −m12,m31 −m23) | m12 ∈M1 ∩M2,m23 ∈
M2 ∩M3,m31 ∈M3 ∩M1}.
Proof. Let us calculate the kernel of homomorphism ϕ. By definition
kerϕ = {(x1, x2, x3) ∈M1 ⊕M2 ⊕M3 | x1 + x2 + x3 = 0}.
Hence x1 = −(x2 + x3) and x1 ∈ (M2 + M3) ∩M1. Similarly, x2 ∈
(M1 + M3) ∩M2 and x3 ∈ (M2 + M1) ∩M3. Since modules M1, M2,
M3 are distributive, we have (Mi + Mj) ∩ Mk = (Mi ∩ Mk) + (Mj ∩
Mk). Therefore x1 = x12 + x13, x2 = x21 + x23, x3 = x31 + x32, where
xij ∈ Mi ∩Mj . Since x1 + x2 = (x12 + x21) + (x13 + x23) ∈ M3 and
x13 + x23 ∈ M3, then x12 + x21 ∈ M3. Given that x12, x21 ∈ M1 ∩M2,
we have x12 +x21 ∈M1 ∩M2 ∩M3. Similarly x23 +x32 ∈M1 ∩M2 ∩M3,
x31 + x13 ∈M1 ∩M2 ∩M3.
Therefore x12 +x21 = t3 ∈M1∩M2∩M3, x23 +x32 = t1 ∈M1∩M2∩
M3, x31 +x13 = t2 ∈M1∩M2∩M3. Hence x21 = t3−x12, x32 = t1−x23,
x13 = t2 − x31. Then (x1, x2, x3) = (x12 + t2 − x31, x23 + t3 − x12, x31 +
t1−x23). From the equality x1 +x2 +x3 = 0 implies that t1 + t2 + t3 = 0.
Therefore
(x1, x2, x3) = (x12 + t2 − x31, x23 − (t1 + t2) − x12, x31 + t1 − x23) =
= ((x12 + t2) − x31, (x23 − t1) − (t2 + x12), x31 − (x23 − t1)) .
Denoting by x12 + t2 = y12 ∈ M1 ∩ M2, x23 − t1 = y23 ∈ M2 ∩ M3,
x31 = y31, we obtain kerϕ = {(y12 − y31, y23 − y12, y31 − y23) | y12 ∈
M1 ∩M2, y23 ∈M2 ∩M3, y31 ∈M3 ∩M1}.
If M1 ∩M2 ⊂ M3, then x12 + x21 ∈ M3 for any x1 ∈ M1, x2 ∈ M2.
Therefore kerϕ = {(x1, x2,−(x1+x2)) | x1 ∈ (M2+M3)∩M1, x2 ∈ (M3+
M1) ∩ M2}. Hence kerϕ ≃ ((M2 +M3) ∩M1) ⊕ ((M3 +M1) ∩M2)}.
Since (M2+M3)∩M1 = M2∩M1+M3∩M1 = M3∩M1 and (M3+M1)∩
M2 = M3∩M2+M1∩M2 = M3∩M2, then kerϕ ≃ (M3∩M1)⊕(M3∩M2).
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.116 Tiled orders of width 3
Let us write formally the expression for the kernel of the homomor-
phism ϕ in the other way
(y12 − y31, y23 − y12, y31 − y23) =
= y12(1,−1, 0) + y23(0, 1,−1) + y31(−1, 0, 1).
Note that ((M1 ∩M2)(1,−1, 0)) ∩ ((M2 ∩M3)(0, 1,−1)) = 0, but
(((M1 ∩M2)(1,−1, 0)) + ((M2 ∩M3)(0, 1,−1)))∩((M3 ∩M1)(−1, 0, 1)) 6= 0.
Therefore, the sum of modules is not direct.
Consider the epimorphism ψ : (M1∩M2)⊕ (M2∩M3)⊕ (M3∩M1) →
kerϕ, defined by the equality
ψ(x12, x23, x31) = (x12 − x31, x23 − x12, x31 − x23) .
Then kerψ = {(x12, x23, x31) | x12 − x31 = x23 − x12 = x31 − x23 =
0} = {(x12, x23, x31) | x12 = x23 = x31}. By the fundamental theorem on
homomorphism of modules we have
kerϕ ≃ ((M1 ∩M2) ⊕ (M2 ∩M3) ⊕ (M3 ∩M1)/ kerψ,
i. e.
kerϕ ≃ ((M1 ∩M2) ⊕ (M2 ∩M3) ⊕ (M3 ∩M1)) /(M1 ∩M2 ∩M3).
Note that in the general case
kerϕ ≃ {(y12 − y31, y23 − y12) | y31 ∈M3 ∩M1,
y23 ∈M2 ∩M3, y12 ∈M1 ∩M2}
or
kerϕ ≃ (M3 ∩M1) ⊕ (M2 ∩M3) + (M1 ∩M2)(1,−1).
Let M1, . . . ,Mn be submodules of M such that Mi *
∑
j 6=i
Mj for all
i = 1, . . . , n and I1, I2 be nonempty subsets of the set I = {1, . . . , n}
such that I1 ∪ I2 = I, I1 ∩ I2 = ∅. We have the following exact sequences
0 → K → ⊕
i∈I
Mi →
∑
i∈I
Mi → 0,
0 → K1 → ⊕
i∈I1
Mi →
∑
i∈I1
Mi → 0,
0 → K2 → ⊕
i∈I2
Mi →
∑
i∈I2
Mi → 0,
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.V. Zhuravlev, D. Zhuravlyov 117
where K,K1,K2 are the kernels of epimorphisms from direct sum on the
sum of modules. Next commutative diagram
0
��
0
��
0
��
0 // KI1 ⊕KI2
//
��
KI1 ⊕KI2
//
��
0 //
��
0
0 // KI
//
��
n
⊕
i=1
Mi
//
��
n∑
i=1
Mi
//
��
0
0 //
(
∑
j∈I1
Mj
)
∩
(
∑
j∈I2
Mj
)
//
��
(
∑
j∈I1
Mj
)
⊕
(
∑
j∈I2
Mj
)
//
��
n∑
i=1
Mi
//
��
0
0 0 0
has exact rows and two columns exact. Therefore by lemma 3 × 3 first
column
0 → KI1 ⊕KI2 → KI →
(∑
j∈I1
Mj
)
∩
(∑
j∈I2
Mj
)
→ 0.
is also exact.
In particular, if I2 = {k}, I1 = I \ {k}, then KI2 = 0 and we have
from the commutative diagram
0
��
0
��
0
��
0 // KI1
//
��
KI1
//
��
0 //
��
0
0 // KI
//
��
n
⊕
i=1
Mi
//
��
n∑
i=1
Mi
//
��
0
0 //
(
∑
i6=j
Mi
)
∩Mj
//
��
(
∑
i6=j
Mi
)
⊕Mj
//
��
n∑
i=1
Mi
//
��
0
0 0 0
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.118 Tiled orders of width 3
the exact sequence
0 → KI1 → KI →
∑
j∈I1
Mj
∩Mk → 0.
Theorem 3. Let M1, . . . ,Mn be submodules of distributive module
M =
n∑
i=1
Mi and epimorphism ϕ :
n
⊕
i=1
Mi 7→ M operates by the rule
ϕ(m1, . . . ,mn) = m1 + . . . + mn. Then ker ϕ = {(y1, . . . , yn) | yi =
∑
j 6=i
sign(j − i) ·mij , mij ∈Mi ∩Mj}.
Proof. We use induction by n. It is well known that the kernel of epi-
morphism equals to {m12,−m12}, where m12 ∈ M1 ∩M2, that implies
the base of induction for n = 2.
Suppose that the kernel of epimorphism ϕ(m1, . . . ,mn−1) = m1 +
. . . + mn−1 is Kn = {(y1, . . . , yn−1) | yi =
∑
j 6=i
sign(j − i) ·mij , mij ∈
Mi∩Mj}. Denote by L = {(y1, . . . , yn) | yi =
∑
j 6=i
sign(j − i) ·mij , mij ∈
Mi ∩Mj}. Obviously, Kn ≃ {(y1, . . . , yn−1, 0)} ⊂ L. Then
n
⊕
i=1
Mi
/
L ≃
(
n
⊕
i=1
Mi
/
Kn
)/
(L/Kn).
By assumption we have
n
⊕
i=1
Mi
/
Kn ≃
(
∑
i6=n
Mi
)
⊕Mn.
Proposition 3. L/Kn ≃
(
∑
i6=n
Mi
)
∩Mn.
Proof. Indeed,
L/Kn = {(y1, . . . , yn) +Kn} =
= {(m1n,m2n, . . . ,mn−1n,−(m1n +m2n + . . .+mn−1n)) +Kn} .
Consider epimorphism ψ : L/Kn 7→
(
∑
i6=n
Mi
)
∩Mn, for which
ψ ((m1n,m2n, . . . ,mn−1n,−(m1n +m2n + . . .+mn−1n)) +Kn) =
= m1n +m2n + . . .+mn−1n.
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.V. Zhuravlev, D. Zhuravlyov 119
The kernel of this epimorphism
ker ψ = {(m1n,m2n, . . . ,mn−1n, 0) +Kn, where
m1n +m2n + . . .+mn−1n = 0} ≃ Kn.
Therefore ψ is isomorphism.
Hence,
n
⊕
i=1
Mi
/
L ≃
(
∑
i6=n
Mi
)
⊕Mn
/
(
∑
i6=n
Mi
)
∩Mn ≃
n∑
i=1
Mi. On
the other hand
n
⊕
i=1
Mi
/
K ≃
n∑
i=1
Mi. Therefore K ≃ L. Obviously, L ⊆
K = ker ϕ. Hence, L = K.
Corollary 1. Let M be irreducible Λ-module and P (M) =
s
⊕
i=1
παjiPji
,
M =
s∑
i=1
παjiPji
. Then the kernel of epimorphism ϕ : P (M) 7→M equals
to ker ϕ = {(y1, . . . , yn) | yi =
∑
k 6=i
sign(k − i) ·mik, mik ∈ Pji
∩ Pjk
}.
Proof. Tiled order Λ is semidistributive ring. Therefore every irreducible
Λ-module is distributive. According to preliminary theorems core epi-
morphism has specified above form.
The kernel K as submodule in
n
⊕
i=1
Mi can be formally written as
виглядi
K =
∑
i<j
Mi ∩Mj(ei − ej),where ek = (0, . . . , 0
︸ ︷︷ ︸
k−1
, 1, 0, . . . , 0).
3. Tiled order of width 3
Proposition 4. Modules P
(
Mi∩(
∑
k 6=i
Mk)
)
, i = 1, . . . , n, have a common
direct summand P ′ if and only if the modules P (Mi∩Mj), i, j = 1, . . . , n,
also have common direct summand P ′.
Proof. Let modules P
(
Mi ∩ (
∑
k 6=i
Mk)
)
, i = 1, . . . , n, have a common di-
rect summand P ′. This is equivalent to the fact that moduleMi∩(
∑
k 6=i
Mk)
has the maximal submodulesXi with E(Xi) = E(Mi∩(
∑
k 6=i
Mk))+e
′. Since
Mi∩Mj =
(
Mi∩(
∑
k 6=i
Mk)
)
∩
(
Mj∩(
∑
k 6=j
Mk)
)
for i 6= j, then the module
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.120 Tiled orders of width 3
Mi∩Mj also have maximal submodulesNij with E(Nij) = E(Mi∩Mj)+e
′.
Therefore, the modules P (Mi ∩Mj), i, j = 1, . . . , n, have also common
summand P ′.
Now let the modules P (Mi ∩ Mj), i, j = 1, . . . , n, have a common
summand P ′. This means that the module P (Mi ∩Mj) has the maximal
submodule Nij with E(Nij) = E(Mi ∩Mj) + e′. Therefore, module Mk ∩
(Mi + Mj) = Mi ∩Mk + Mj ∩Mk has maximal submodule Xijk with
E(Xijk) = E(Mk ∩ (Mi + Mj)) + e′. Similarly we get that the module
Mk∩(Mi+· · ·+Mj) = Mi∩Mk+· · ·+Mj∩Mk has maximal submoduleXk
with E(Xk) = E(Mk∩(Mi+· · ·+Mj))+e
′. In particular, the module Mi∩
(
∑
k 6=i
Mk) has the maximal submodule Yi with E(Yi) = E(Mi∩(
∑
k 6=i
Mk))+
e′. This is equivalent to the fact that modules P
(
Mi ∩ (
∑
k 6=i
Mk)
)
, i =
1, . . . , n, have a common direct summand P ′.
Let module M with E(M) = (α1, . . . , αn) has a projective cover
P (M) = παiPi ⊕ παjPj ⊕ παkPk and M = παiPi + παjPj + παkPk. Then
K = (παiPi ∩ π
αjPj) (ei − ej) + (παjPj ∩ π
αkPk) (ej − ek)+
+ (παkPk ∩ παiPi) (ek − ei).
Suppose that παiPi ∩ π
αjPj = παjPj ∩ π
αkPk. Then
(παiPi + παkPk) ∩ π
αjPj = παiPi ∩ π
αjPj
and
(παiPi + παjPj) ∩ π
αkPk = (παjPj + παkPk) ∩ π
αiPi.
From the equality (παiPi + παkPk) ∩ π
αjPj = παiPi ∩ π
αjPj we get
(παiPi + παjPj) ∩ π
αkPk =
= ((παiPi + παjPj) ∩ π
αkPk) ∩ ((παjPj + παkPk) ∩ π
αiPi) =
= παiPi ∩ π
αkPk.
So we have two exact sequences
0 → παiPi ∩ π
αjPj → K → (παiPi + παjPj) ∩ π
αkPk → 0,
0 → παiPi ∩ π
αkPk → K → (παiPi + παkPk) ∩ π
αjPj → 0.
Whereas (παiPi + παkPk)∩π
αjPj = παiPi∩π
αjPj and (παiPi + παjPj)∩
παkPk = παiPi ∩ π
αkPk, then the exact sequence splits:
K ≃ (παiPi ∩ π
αjPj) ⊕ (παiPi ∩ π
αkPk) .
Let the width of tiled order do not exceed 3.
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.V. Zhuravlev, D. Zhuravlyov 121
Proposition 5. Let irreducible Λ-module M have exactly two maximal
non-projective submodules X and Y with
E(M) = (α1, . . . , αi−1, αi, αi+1, . . . , αn),
E(X) = (α1, . . . , αj−1, αj + 1, αj+1, . . . , αn) and
E(Y ) = (α1, . . . , αi−1, αi + 1, αi+1, . . . , αn).
Then P (M) = παiPi ⊕ παjPj and we have the exact sequence
0 → παiPi ∩ π
αjPj → παiPi ⊕ παjPj →M → 0.
Proof. We have M = X + παiPi = Y + παjPj , π
αiPi + παjPj ⊆ M , but
παiPi +παjPj does not belong to any maximal submodule X or Y . Then
παiPi + παjPj = M.
Since M/X ≃ Uj та M/Y ≃ Ui, then M/(radM) = M/(X ∩ Y ) ≃
Ui ⊕ Uj . Therefore P (M) ≃ P (M/radM) ≃ P (Ui ⊕ Uj) ≃ P (Ui) ⊕
P (Uj) ≃ Pi ⊕ Pj . Obviously, the kernel of epimorphism ϕ : παiPi ⊕
παjPj →M coincides with παiPi ∩ π
αjPj = παiRi ∩ π
αjRj .
Consider the case when the module M = (α1, . . . , αn) has exactly
three maximal submodules X = (α1, . . . , αi−1, αi + 1, αi+1, . . . , αn),
Y = (α1, . . . , αj−1, αj + 1, αj+1, . . . , αn) and Z = (α1, . . . , αk−1, αk +
1, αk+1, . . . , αn).
Let module M with E(M) = (α1, . . . , αn) have a projective cover
P (M) = παiPi ⊕ παjPj ⊕ παkPk
and M = παiPi + παjPj + παkPk. Then
K = (παiPi ∩ π
αjPj) (ei − ej) + (παjPj ∩ π
αkPk) (ej − ek)+
+ (παkPk ∩ παiPi) (ek − ei).
Also we have three exact sequences
0 → παiPi ∩ π
αjPj → K → (παiPi + παjPj) ∩ π
αkPk → 0,
0 → παjPj ∩ π
αkPk → K → (παjPj + παkPk) ∩ π
αiPi → 0,
0 → παkPk ∩ παiPi → K → (παkPk + παiPi) ∩ π
αjPj → 0.
Let modules παiPi ∩ παjPj , π
αjPj ∩ παkPk and παkPk ∩ παiPi are
pairwise different.
Projective cover P (K) of module K is a direct summand of each
of the the direct sums P (παiPi ∩ π
αjPj) ⊕ P ((παiPi + παjPj) ∩ π
αkPk),
P (παjPj ∩ παkPk) ⊕ P ((παjPj + παkPk) ∩ π
αiPi), P (παkPk ∩ παiPi) ⊕
P ((παkPk + παiPi) ∩ π
αjPj).
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.122 Tiled orders of width 3
Suppose that the module P (K) contains 2 isomorphic direct sum-
mand P ′. Since modules παiPi ∩ π
αjPj and (παiPi + παjPj)∩ π
αkPk are
irreducible, their projective coverings do not contain isomorphic direct
summands. Therefore, the module P ′ is a direct summand of modules
P (παiPi ∩ π
αjPj), P ((παiPi + παjPj) ∩ π
αkPk). The module P ′ is a di-
rect summand of modules P ((παiPi + παjPj) ∩ π
αkPk),
P ((παjPj + παkPk) ∩ π
αiPi), P ((παkPk + παiPi) ∩ π
αjPj).
Module P (K) contains as direct summand each of the mod-
ules P ((παiPi + παjPj) ∩ π
αkPk), P ((παjPj + παkPk) ∩ π
αiPi),
P ((παkPk + παiPi) ∩ π
αjPj).
Hence, we obtain that P (K) with a pairwise different modules παiPi∩
παjPj , π
αjPj ∩ παkPk, π
αkPk ∩ παiPi has at least four non-isomorphic
direct summands.
Thus, P (K) contains only non-isomorphic direct summands.
So P (K) = παaPa ⊕ παbPb ⊕ παcPc.
Now we have 2 exact sequences
0 → L→ P (K) → K → 0
0 → K → P (M) →M → 0
Theorem 4. L ≃ παaPa ∩ π
αbPb ∩ π
αcPc.
Proof. Consider the homomorphism ϕ : P (K) 7→ P (M) with the image
K. For corresponding to ϕ matrix [ϕ] we have
[ϕ] ∈
HomΛ (παaPa, π
αiPi) HomΛ (παbPb, π
αiPi) HomΛ (παcPc, π
αiPi)
HomΛ (παaPa, π
αjPj) HomΛ (παbPb, π
αjPj) HomΛ (παcPc, π
αjPj)
HomΛ (παaPa, π
αkPk) HomΛ (παbPb, π
αkPk) HomΛ (παcPc, π
αkPk)
.
Since HomΛ (παaPa, π
αiPi) ≃ παi−αaeiΛea = παi−αa · παiaO, then
[ϕ] = (ϕml) ∈
παi−αa+αiaO παi−αb+αibO παi−αc+αicO
παj−αa+αjaO παj−αb+αjbO παj−αc+αjcO
παk−αa+αkaO παk−αb+αkbO παk−αc+αkcO
.
Let m1 ∈ παaPa, m2 ∈ παbPb, m3 ∈ παcPc. Then
ϕ(m1,m2,m3) = (m1ϕ11 +m2ϕ12 +m3ϕ13,m1ϕ21+
+m2ϕ22 +m3ϕ23,m1ϕ31 +m2ϕ32 +m3ϕ33).
Since K = {(y1, y2,−(y1 + y2)}, the rank of [ϕ] is 2. So the kernel of
kerϕ is obtained from the system of equations
m1ϕ11 +m2ϕ12 +m3ϕ13 = 0, m1ϕ21 +m2ϕ22 +m3ϕ23 = 0.
Hence, m1, m2 are expressed by m3, and then kerϕ is isomorphic to
παaPa ∩ π
αbPb ∩ π
αcPc.
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.V. Zhuravlev, D. Zhuravlyov 123
Conclusion
The results obtained in sections 2, 3, to build a projective resolution of
irreducible modules over tiled order of width 3 and calculate the global
dimension of the order.
References
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Contact information
V. Zhuravlev Department of Mechanics and Mathematics,
Kiyv National Taras Shevchenko University,
Volodymyrska, 64, Kyiv 01033, Ukraine
E-Mail: vshur@univ.kiev.ua
D. Zhuravlyov Department of Mechanics and Mathematics,
Kiyv National Taras Shevchenko University,
Volodymyrska, 64, Kyiv 01033, Ukraine
Received by the editors: 07.04.2009
and in final form 07.04.2009.
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