Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication
Let R be a ring with an endomorphism σ. We introduce (σ, 0)-multiplication which is a generalization of the simple 0- multiplication. It is proved that for arbitrary positive integers m ≤ n and n ≥ 2, R[x; σ] is a reduced ring if and only if Sn,m(R) is a ring with (σ, 0)-multiplication.
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irk-123456789-1523572019-06-11T01:25:21Z Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication Abdioglu, C. Şahinkaya, S. KÖR, A. Let R be a ring with an endomorphism σ. We introduce (σ, 0)-multiplication which is a generalization of the simple 0- multiplication. It is proved that for arbitrary positive integers m ≤ n and n ≥ 2, R[x; σ] is a reduced ring if and only if Sn,m(R) is a ring with (σ, 0)-multiplication. 2014 Article Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication / C. Abdioglu, S. Şahinkay, A. KÖR // Algebra and Discrete Mathematics. — 2014. — Vol. 17, № 1. — С. 1–11. — Бібліогр.: 8 назв. — англ. 1726-3255 2000 MSC:16N60,16S36,16W60. http://dspace.nbuv.gov.ua/handle/123456789/152357 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Let R be a ring with an endomorphism σ. We introduce (σ, 0)-multiplication which is a generalization of the simple 0- multiplication. It is proved that for arbitrary positive integers m ≤ n and n ≥ 2, R[x; σ] is a reduced ring if and only if Sn,m(R) is a ring with (σ, 0)-multiplication. |
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Abdioglu, C. Şahinkaya, S. KÖR, A. |
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Abdioglu, C. Şahinkaya, S. KÖR, A. Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication Algebra and Discrete Mathematics |
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Abdioglu, C. Şahinkaya, S. KÖR, A. |
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Abdioglu, C. |
| title |
Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication |
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Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication |
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Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication |
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Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication |
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Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication |
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rigid, quasi-rigid and matrix rings with (σ,0)-multiplication |
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Інститут прикладної математики і механіки НАН України |
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2014 |
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http://dspace.nbuv.gov.ua/handle/123456789/152357 |
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Rigid, quasi-rigid and matrix rings with (σ,0)-multiplication / C. Abdioglu, S. Şahinkay, A. KÖR // Algebra and Discrete Mathematics. — 2014. — Vol. 17, № 1. — С. 1–11. — Бібліогр.: 8 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
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AT abdiogluc rigidquasirigidandmatrixringswiths0multiplication AT sahinkayas rigidquasirigidandmatrixringswiths0multiplication AT kora rigidquasirigidandmatrixringswiths0multiplication |
| first_indexed |
2025-07-13T02:54:13Z |
| last_indexed |
2025-07-13T02:54:13Z |
| _version_ |
1837498639033827328 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 17 (2014). Number 1. pp. 1 – 11
c© Journal “Algebra and Discrete Mathematics”
Rigid, quasi-rigid and matrix rings with
(σ, 0)-multiplication
Cihat Abdioĝlu, Serap Şahinkaya and Arda KÖR
Communicated by D. Simson
Abstract. Let R be a ring with an endomorphism σ. We
introduce (σ, 0)-multiplication which is a generalization of the simple
0- multiplication. It is proved that for arbitrary positive integers
m ≤ n and n ≥ 2, R[x; σ] is a reduced ring if and only if Sn,m(R)
is a ring with (σ, 0)-multiplication.
1. Introduction
Throughout this paper, we will assume that R is an associative ring
with non-zero identity, σ is an endomorphism of the ring R and the
polynomial ring over R is denoted by R[x] with x its indeterminate.
In [6], the authors introduced and studied the notion of simple 0-
multiplication. A subring S of the full matrix ring Mn(R) of n×n matrices
over R is with simple 0- multiplication if for arbitrary (aij), (bij) ∈ S
then (aij)(bij) = 0 implies that ailblj = 0, for arbitrary 1 ≤ i, j, l ≤ n.
This definition is not meaningless because of the [4, Lemma 1.2]. Let R
be a domain (commutative or not) and R[x] is its polynomial ring. Let
f(x) =
∑n
i=0 aix
i, g(x) =
∑n
j=0 bjxj be elements of R[x]. It is easy to see
that if f(x)g(x) = 0, then aibj = 0 for every i and j since f(x) = 0 or
g(x) = 0. Armendariz [1] noted that the above result can be extended
the class of reduced rings, i.e., if it has no non-zero nilpotent elements. A
ring R is said to have the Armendariz property or is an Armendariz ring
if whenever polynomials
f(x) = a0 + a1x + · · · + amxm, g(x) = b0 + b1x + · · · + bnxn ∈ R[x]
2000 MSC: 16N60,16S36,16W60.
Key words and phrases: simple 0-multiplication, quasi σ-rigid rings.
2 Rigid and quasi-rigid rings
satisfy f(x)g(x) = 0, then aibj = 0 for each i, j. In [6, Theorem 2.1], the
authors show that many matrix rings with simple 0-multiplication are
Armendariz rings.
Recall that an endomorphism σ of a ring R is said to be rigid if
aσ(a) = 0 implies a = 0 for a ∈ R. A ring R is σ-rigid if there exists a
rigid endomorphism σ of R. Note that σ-rigid rings are reduced rings,i.e.,
the rings contains no nonzero nilpotent elements.
An ideal I of a ring R is said to be a σ-ideal if I is invariant under
the endomorphism σ of the ring R, i.e., σ(I) ⊆ I. Now, let σ be an
automorphism of the ring R and I be a σ-ideal of R. I is called a quasi
σ-rigid ideal if aRσ(a) ⊆ I, then a ∈ I for any a ∈ R [3]. If the zero ideal
{0} of R is a quasi σ-rigid ideal, then R is said to be a quasi σ-rigid ring
[3]. In Section 2, we obtain some ring extensions of quasi σ-rigid rings.
We prove that; the class of quasi σ-rigid rings is closed under taking finite
direct products (see Corollary 2.4).
We denote RG the group ring of a group G over a ring R and, for
cyclic group order n, write Cn. We also prove that; if RG is quasi σ-rigid,
then R is a quasi σ-rigid ring (see Theorem 2.6), and R is quasi σ-rigid
if and only if RC2 is quasi σ-rigid where R is a ring with 2−1 ∈ R (see
Corollary 2.8).
Let R be a quasi σ- rigid ring with σ : R → R endomorphism. In
Example 2.1, it is shown that M2(R) is not a quasi σ-rigid ring. Again,
in Example 2.12, we proved that
S4 =
a a12 a13 a14
0 a a23 a24
0 0 a a34
0 0 0 a
|a, aij ∈ R
is not a quasi σ-rigid ring however R is a quasi σ-rigid ring. Naturally, these
examples are starting points of our study. In this article, we introduce
and study subrings with (σ, 0)-multiplication of matrix rings which is
a generalization of the simple 0-multiplication. They are related to σ-
skew Armendariz rings. Applying them, we obtain the following result in
Section 3. Let σ be an endomorphism of a ring R. For arbitrary positive
integers m ≤ n and n ≥ 2, the following conditions are equivalent.
(i) R[x; σ] is a reduced ring.
(ii) Sn,m(R) is a ring with (σ, 0)-multiplication.
(iii) Sn,m(R) is a σ-skew Armendariz ring (see Theorem 3.3).
See Example 4 for the definition of the ring Sn,m(R).
C. Abdioĝlu, S. Şahinkaya, A. KÖR 3
2. Extensions of quasi σ-rigid rings
The following example shows that the class of quasi σ-rigid rings is
not closed under taking subrings.
Example 1. Let R be a quasi σ-rigid ring with σ : R → R endomorphism
defined by σ(
(
a b
c d
)
) =
(
a −b
−c d
)
. We take the nonzero element
a =
(
1 0
−1 0
)
. Since
aM2(R)σ(a) =
(
1 0
−1 0
) (
1 −1
0 0
) (
1 0
1 0
)
= 0,
M2(R) is not a quasi σ-rigid ring.
This example is one of the starting point of our study. First, we prove
that the finite direct product of quasi σ-rigid rings is again a quasi σ-rigid
ring.
Proposition 1. Let σ1 and σ2 be automorphisms of rings R1 and R2,
respectively. Assume that I1 is a quasi σ1-rigid ideal of R1 and I2 is a
quasi σ2-rigid ideal of R2. Then I1 × I2 is a quasi σ-rigid ideal of R1 ×R2,
where σ is an automorphism of R1 × R2 such that σ(a, b) = (σ1(a), σ2(b))
for any a ∈ R1 and b ∈ R2.
Proof. We assume that (a, b)R1 × R2σ(a, b) ⊆ I1 × I2, equivalently,
(a, b)R1 × R2(σ1(a), σ2(b)) ⊆ I1 × I2.
Then we have (aR1σ1(a), 0) ⊆ I1 × I2 and (0, bR2σ2(b)) ⊆ I1 × I2. Thus
we obtain that aR1σ1(a) ⊆ I1 and bR2σ2(b) ⊆ I2. Since I1 is a quasi
σ1-rigid ideal of R1 and I2 is a quasi σ2-rigid ideal of R2, we have a ∈ I1
and b ∈ I2. Hence, (a, b) ∈ I1 × I2.
Theorem 1. Assume that each σi, 1 ≤ i ≤ n, is an automorphism of
rings Ri. Then the finite direct product of quasi σi-rigid ideals Ii of Ri,
1 ≤ i ≤ n, is a quasi σ-rigid ideal, where σ is an automorphism of
∏n
i=1 Ri
such that σ(a1, a2, · · · , an) = (σ1(a1), σ2(a2), · · · , σn(an)) for any ai ∈ Ri.
As a parallel result to Theorem 1, we have the following corollaries
for quasi σ-rigid rings.
Corollary 1. Assume that each σi, 1 ≤ i ≤ n, is an automorphism
of rings Ri. Then the finite direct product of quasi σi-rigid rings Ri,
4 Rigid and quasi-rigid rings
1 ≤ i ≤ n, is a quasi σ-rigid ring, where σ is an automorphism of
∏n
i=1 Ri
such that σ(a1, a2, · · · , an) = (σ1(a1), σ2(a2), · · · , σn(an)) for any ai ∈ Ri.
Lemma 1. Let R be a subring of a ring S such that both share the same
identity. Suppose that S is a free left R-module with a basis G such that
1 ∈ G and ag = ga for all a ∈ R ring. Let σ be an endomorphism of
R. Assume that the epimorphism σ : S → S defined by σ(
∑
g∈G rgg) =
∑
g∈G σ(rg)g extends σ. If S is a quasi σ-rigid ring, then R is a quasi
σ-rigid ring.
Proof. Suppose that rRσ(r) = 0 for r ∈ R. Then, by hypothesis, we can
obtain that r
∑
g∈G Rgσ(r) = 0. Hence r = 0, since S is a quasi σ-rigid
ring.
Theorem 2. Let R be a ring and G be a group. If the group ring RG is
quasi σ-rigid, then R is a quasi σ-rigid ring.
Proof. Since S = RG = ⊕g∈GRg is a free left R-module with a basis G
satisfying the assumptions of Lemma 1, the proof of theorem is clear.
Example 2. Let R be a ring. Note that if G is a semigroup or C2,
then clearly the epimorphism σ : S → S defined by σ(
∑
g∈G rgg) =
∑
g∈G σ(rg)g extends σ. If the semigroup ring RG or RC2 is quasi σ-rigid,
then R is a quasi σ-rigid ring by Theorem 2.
Corollary 2. Let R be a ring with 2−1 ∈ R. Then R is quasi σ-rigid if
and only if RC2 is quasi σ-rigid.
Proof. If 2−1 ∈ R, then the mapping RC2 → R × R which is given by
a + bg → (a + b, a − b), is a ring isomorphism. The rest is clear from
Example 2.7 and Corollary 1.
Let σ be an epimorphism of a ring R. Then σ : R[x] → R[x], defined
by σ(
∑n
i=0 aix
i) =
∑n
i=0 σ(ai)x
i, is an epimorphism of the polynomial
ring R[x], and σ extends to σ.
Corollary 3. R is a quasi σ-rigid ring if and only if R[x] is a quasi
σ-rigid ring.
Since, for an automorphism σ of R, every σ-rigid ring is a quasi σ-rigid
ring, Corollary 1 holds for quasi σ-rigid rings.
Now we investigate a sufficient condition for Corollary 1.
Proposition 2. Assume that σ is an automorphism of a ring R and e
is a central idempotent of R. If R is a quasi σ-rigid ring then eR is also
a quasi σ-rigid ring.
C. Abdioĝlu, S. Şahinkaya, A. KÖR 5
Proof. For ea ∈ eR, we assume that ea(eR)σ(ea) = 0, equivalently,
0 = ea(eR)σ(ea) = eaeRσ(ea) = (ea)Rσ(ea).
Since R is a quasi σ-rigid ring, we have ea = 0.
The following example show that the condition e is a central idempo-
tent of R" in Proposition 2 is necessary.
Example 3. Let F be a field with char(F ) 6= 2. It is easy to see that the
ring R = M2(F ) with an endomorphism σ(
(
a b
c d
)
) =
(
a −b
−c d
)
is
a quasi σ-rigid ring. Consider the idempotent element e =
(
0 1
0 1
)
of R.
Since
(
0 1
0 1
) (
0 0
1 1
)
6=
(
0 0
1 1
) (
0 1
0 1
)
,
the idempotent e is not central. Let a =
(
0 1
0 0
)
. Now it is easy to see
that ea 6= 0 and
(
1 1
1 1
) (
2 2
2 2
)
σ(
(
1 1
1 1
)
) = 0.
Example 3 shows that for a quasi σ-rigid ring R, Mn(R) or the full
upper triangular matrix ring Tn(R) is not necessarily quasi σ-rigid.
Example 4. Let R be a ring. We consider the following subrings of Tn(R)
for any n ≥ 2.
(1)
Rn = RIn +
∑n
i=1
∑n
k=i+1 REij
=
a a12 a13 · · · a1n
0 a a23 · · · a2n
0 0 a · · · a3n
...
...
...
. . .
...
0 0 0 · · · a
: a, aij ∈ R
,
where Eij is the matrix units for all i, j and In is the identity matrix.
(2)
T (R, n) =
a1 a2 a3 · · · an
0 a1 a2 · · · an−1
0 0 a1 · · · an−2
...
...
...
. . .
...
0 0 0 · · · a1
: ai ∈ R
.
6 Rigid and quasi-rigid rings
(3) Let m ≤ n be positive integers. Let Sn,m(R) be the set of all n × n
matrices (aij) with entiries in a ring R such that
(a) for i > j, aij = 0,
(b) for i ≤ j, aij = akl when i − k = j − l and either 1 ≤ i, j, k, l ≤ m
or m ≤ i, j, k, l ≤ n.
Clearly, Sn,1(R) = Sn,n(R) = T (R, n).
Let σ be an endomorphism of a ring R, then σ : Mn(R) → Mn(R),
defined by σ((aij)) = (σ(aij)), is an also endomorphism of Mn(R) and σ
extends to σ. Now assume that R is a quasi σ-rigid ring. It is easy to see
that Rn, T (R, n) and Sn,m(R) are not quasi σ-rigid rings for n ≥ 2. For
instance, we consider the ring:
S4 =
a a12 a13 a14
0 a a23 a24
0 0 a a34
0 0 0 a
|a, aij ∈ R
.
Although R is a quasi σ-rigid ring, S4 is not a quasi σ-rigid ring.
Let a ∈ S4 such that a =
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
6= 0. Since
aS4σ(a) =
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
S4σ
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
=
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
S4
0 σ(1) 0 0
0 0 0 0
0 0 0 0
0 0 0 0
= 0,
S4 is not a quasi σ-rigid ring.
3. On σ-skew Armendariz and (σ, 0)-multiplication rings
In Corollary 3, we proved that R is a quasi σ-rigid ring if and only if
R[x] is a quasi σ-rigid ring.
Theorem 3. Assume that σ is a monomorphism of a ring R and σ(1) = 1,
where 1 denotes the identity of R. Then the factor ring R[x]/(x2) is σ-
skew Armendariz if and only if R is a σ-rigid ring, where (x2) is an ideal
of R[x] generated by x2.
C. Abdioĝlu, S. Şahinkaya, A. KÖR 7
Proof. (:⇒) Assume that R[x]/(x2) is a σ-skew Armendariz ring. Let
r ∈ R with σ(r)r = 0. Then
(σ(r) − xy)(r + xy) = σ(r)r + (σ(r)x − xσ(r))y − σ(1)x2y2 = 0,
because σ(r)x = xσ(r) in (R[x]/(x2))[y; σ], where x = x + (x2) ∈
R[x]/(x2). Since R[x]/(x2) is σ-skew Armendariz, we can obtain that
σ(r)x = 0 so σ(r) = 0. The injectivity of σ implies r = 0, and so R is
σ-rigid.
(⇐:) Assume that R[x; σ] is reduced. Let h = h + (x2) ∈ R[x]/(x2).
Suppose that p.q = 0 in (R[x]/(x2))[y; σ], where p = f0 +f1y + ...+fmym
and q = g0 + g1y + ... + gnyn. Let f i = ai0
+ ai1
x, gj = bj0
+ bj1
x for
each 0 ≤ i ≤ m, and 0 ≤ j ≤ n, where x2 = 0. Note that xy = yx since
α(1) = 1, ax = xa for any a ∈ R. Thus p = h0 + h1x and q = k0 + k1x,
where h0 =
∑m
i=0 ai0
yi, h1 =
∑m
i=0 ai1
yi, k0 =
∑n
j=0 bj0
yj , k1 =
∑n
j=0 bj1
yj
in R[y]. Since p.q = 0 and x2 = 0, we have
0 = p.q = 0 = h0k0 + (h0k1 + h1k0)x + h1k1x2 = h0k0 + (h0k1 + h1k0)x.
We get h0k0 = 0 and h0k1 + h1k0 = 0 in R[y; σ]. Since R[y; σ] is reduced,
k0h0 = 0 and so 0 = k0(k0k1+h1k0)h1 = (k0h1)2. Thus k0h1 = 0, h1k0 = 0
and h0k1 = 0. Moreover, R is σ-skew Armendariz by [8, Corollary 4].
Thus a0i
σi(b0j
) = 0, a0i
σi(b1j
) = 0 and a1i
σi(b0j
) = 0 for all 0 ≤ i ≤ m,
0 ≤ j ≤ n. Hence fiσ
i(gj) = 0 for all 0 ≤ i ≤ m, 0 ≤ j ≤ n. Therefore
R[x]/(x2) is σ-skew Armendariz.
In [6], the author defined and studied Armendariz and simple 0-
multiplication rings. In other words, a subring S of the ring Mn(R)
of n × n matrices over R is with simple 0-multiplication if for arbitrary
(aij), (bij) ∈ S then (aij)(bij) = 0 implies that for arbitrary 1 ≤ i, j, l ≤ n,
ailblj = 0.
Let σ be an endomorphism of a ring R. As we mentioned before, σ :
Mn(R) → Mn(R), defined by σ((aij)) = (σ(aij)), is an also endomorphism
of Mn(R) and σ extends to σ. We shall say that a subring S of the ring
Mn(R) of n×n matrices over R is with (σ, 0)-multiplication if for arbitrary
(aij), (bij) ∈ S, (aij)σ((bij)) = 0 implies that for arbitrary 1 ≤ i, j, l ≤ n
ailσ(blj) = 0.
Let σ be an epimorphism of a ring R. We know that σ : R[x] → R[x],
defined by σ(
∑n
i=0 aix
i) =
∑n
i=0 σ(ai)x
i, is an also epimorphism of the
polynomial ring R[x], and σ extends to σ. So, the map
Mn(R)[x] −→ Mn(R)[x],
8 Rigid and quasi-rigid rings
defined by
m
∑
i=0
Aix
i 7→
m
∑
i=0
σ(Ai)x
i
is an endomorphism of the matrix ring Mn(R)[x] and clearly this map
extends σ. By the same notation of authors in [6], φ denotes the canonical
isomorphism of Mn(R)[x] onto Mn(R[x]). It is given by
φ(σ(A0) + σ(A1)x + ... + σ(Am)xm) = (fij(x)),
where
fij(x) = (σ(a
(0)
ij ) + σ(a
(1)
ij )x + ... + σ(a
(m)
ij )xm)
and σ(a
(k)
ij ) denotes the (i, j) entry of σ(Ak). In fact follows Eij will
denote the usual matrix unit.
Theorem 4. Let σ be an endomorphism of a ring R, and R be a σ-skew
Armendariz ring.
(1) If S is a subring of Mn(R) with (σ, 0)-multiplication and, for some
Ai, Bi ∈ S 0 ≤ i ≤ 1, (A0 + A1x)(B0 + B1x) = 0 then Atσt(Bu) = 0 for
0 ≤ t, u ≤ 1.
(2) If, for a subring S of Mn(R), φ(S[x]) is a subring of Mn(R[x])
with (σ, 0)-multiplication, then S is an σ-skew Armendariz ring.
Proof. (1) Assume that S is a subring of Mn(R) with (σ, 0)-multiplication
and, for some Ai, Bi ∈ S, 0 ≤ i ≤ 1. Then
0 = (A0 + A1x)(B0 + B1x)
= A0B0 + A0B1x + A1xB0 + A1xB1x
= A0B0 + [A0B1 + A1σ(B0)]x + A1σ(B1)x2
= a
(0)
il b
(0)
lj + (a
(0)
il b
(1)
lj + a
(1)
il σ(b
(0)
lj ))x + a
(1)
il σ(b
(1)
lj )x2.
Now, we set p =
∑1
t=0 a
(t)
il xt and q =
∑1
u=0 b
(u)
lj xu. Then pq = 0 and
a
(t)
il σt(b
(u)
lj ) = 0, since R is a σ-skew Armendariz ring.
(2) We prove only when n = 2. Other cases can be proved by the same
method. Suppose that
p(x) = σ(A0) + σ(A1)x + ... + σ(Am)xm ∈ S[x; σ]
q(x) = σ(B0) + σ(B1)x + ... + σ(Bm)xm ∈ S[x; σ]
C. Abdioĝlu, S. Şahinkaya, A. KÖR 9
such that p(x)q(x) = 0, where
σ(Ai) =
σ(a
(i)
11 ) σ(a
(i)
12 )
σ(a
(i)
21 ) σ(a
(i)
22 )
and σ(Bj) =
σ(b
(j)
11 ) σ(b
(j)
12 )
σ(b
(j)
21 ) σ(b
(j)
22 )
for 0 ≤ i ≤ m and 0 ≤ j ≤ m. We claim that σ(Ai)σ
i(σ(Bj)) = 0 for
0 ≤ i ≤ m and 0 ≤ j ≤ m. Then p(x)q(x) = 0 implies that
(σ(a
(0)
il ) + ... + σ(a
(m)
il )xm)σ((b
(0)
lj ) + ... + (b
(m)
lj )xm) = 0
since φ(S[x]) is (σ, 0)-multiplication. Now we can obtain that
σ(a
(t)
il )σt(σ(b
(u)
lj )) = 0 for all 0 ≤ i, j, u, t ≤ m since R is σ-skew Ar-
mendariz.
Now we return one of the important examples in the paper, the ring
Sn,m(R) that is not a (quasi) σ-rigid rings for n ≥ 2 by Example 2.12. We
consider our ring S4. Note that if R is an σ-rigid ring, then σ(e) = e for
e2 = e ∈ R. Let p = e12 +(e12 −e13)x and q = e34 +(e24 +e34)x ∈ S4[x; σ],
where eij ‘s are the matrix units in S4. Then pq = 0, but (e12−e13)σ(e34) 6=
0. Thus S4 is not σ-skew Armendariz. Similarly, for the case of n ≥ 5, we
have the same result.
Theorem 5. Let σ be an endomorphism of a ring R. For arbitrary positive
integers m ≤ n and n ≥ 2, the following conditions are equivalent.
(1) R[x; σ] is a reduced ring.
(2) Sn,m(R) is a ring with (σ, 0)-multiplication.
(3) Sn,m(R) is an σ-skew Armendariz ring.
Proof. To prove, we completely follow the proof of [6, Theorem 2.3].
(1) ⇒ (2) We will proceed by induction on n. Suppose that n ≥ 2 and
the result holds for smaller integers. Let A = (aij) , A = (aij) ∈ Sn,m(R)
and Aσ(A) = 0.
Note that the matrices obtained from A and A by deleting their first
rows and columns belong to Sn−1,m−1(R) when m > 1, and Sn−1,n−1(R)
when m = 1. The product of obtained matrices is equal to 0. So, applying
the induction assumption, we get that aijσi(ajl) = 0 for i ≥ 2 and all j, l.
Similarly, by deleting the last rows and columns, we get that aijσi(ajl) = 0
for l ≤ n − 1 and all i, j. Moreover,
a11σ(a1n) + a12σ(a2n) + · · · + a1nσ(ann) = 0. (1)
It is left to prove that a1jσ(ajn) = 0 for 1 ≤ j ≤ n. Let 1 ≤ j < k ≤ n.
10 Rigid and quasi-rigid rings
If k ≤ m, then aij = ak−j−1,k and k − j + 1 ≥ 2, so from the induction
conclusion, we get that a1jσ(akn) = ak−j−1,kσ(akn) = 0.
Similarly, we get that if m ≤ j (which is possible only when m < n),
then a1jσ(akn) = a1jσ(aj,j−k+n) = 0.
If j ≤ m < k, then a1jσ(akn) = am−j+1,mσ(am,n+m−k) = 0 (because
j < k implies that either m − j + 1 ≥ 2 or n + m − k ≤ n − 1).
Multiplying (1) on the left by a11 and the foregoing, we get that
a2
11σ(a1n) = 0. Hence, a11σ(a1n) = 0. Similarly, multiplying (1) (in
which now a11σ(a1n) = 0) on the left by a12, we get that a12σ(a2n) = 0.
Continuing in this way, we get a1jσ(ajn) = 0 for all j ≤ m − 1. These
results and (1) gives the result when m = n.
If m < n, then, same as above, multiplying (1) on the right by ann,
an−1,n, ..., am+1,n applying the foregoing relations , we get (successively)
that
a1nσ(ann) = a1,n−1σ(an−1,n)
= · · ·
= a1,m+1σ(am+1,n)
= 0.
Now (1) implies also that a1mσ(amn) = 0 and we are done.
(2) ⇒ (3) Note that φ((Sn,m(R))[x]) = Sn,m(R[x]). Now the rest
follows from Theorem 4.
(3) ⇒ (1) Clearly, Sn,m(R) contains a subring isomorphic to S2(R).
Hence (3) implies that S2(R) is a σ-skew Armendariz ring. Then R[x; σ]
is a reduced ring.
Theorem 6. Let σ be an endomorphism of a ring R with σ(1) = 1.
For arbitrary integers 1 < m < n, if T is a subring of Tn(R), which
properly contains Sn,m(R), then there are A0, A1, B0, B1 ∈ T such that
(A0 + A1x)(B0 + B1x) = 0 and A1σ(B0) 6= 0. In particular, T is not a
σ-skew Armendariz ring.
Proof. We proceed by induction on n. Let us observe first that if T is
an σ-skew Armendariz subring of the ring Tn(R), then by deleting in
every matrix from T the first(last) row and column, we get a σ-skew
Armendariz subring of the ring Tn−1(R).
To start the induction, assume that n = 3 and m = 3. Applying the
above observation, it suffices to show that no subring of T2(R), which
properly contains S2(R), is σ-skew Armendariz. It is clear that every
such subring S contains the matrices A = aE11, B = −aE22, for some
0 6= a ∈ R and C = E12.
C. Abdioĝlu, S. Şahinkaya, A. KÖR 11
Let σ(
(
a b
c d
)
) =
(
a −b
−c d
)
, p = A + Cx, q = B + Cx ∈ R[x; σ].
We have (A + Cx)(B + Cx) = 0 but Cσ((B)) 6= 0, so S is not a σ-skew
Armendariz ring.
Now, the rest of the proof is similar to the proof of [6, Theorem 2.4].
Corollary 4. For arbitrary integers 1 < m < n and every ring R with an
endomorphism σ, no subring of Tn(R), which properly contains Sn,m(R),
is with (σ, 0)-multiplication.
Acknowledgements
We would like to express our gratefulness to the referee for his/her valuable
suggestions and contributions. Special thanks to Tamer Koşan (from GIT).
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Contact information
C. Abdioĝlu Department of Mathematics, Karamanoĝlu
Mehmetbey University, Yunus Emre Campus,
Karaman, Turkey
E-Mail: cabdioglu@kmu.edu.tr
S. Şahinkaya,
A. KÖR
Department of Mathematics, Gebze Institute
of Technology, Çayirova Campus, 41400 Gebze-
Kocaeli, Turkey
E-Mail: ssahinkaya@gyte.edu.tr,
akor@gyte.edu.tr
Received by the editors: 26.04.2012
and in final form 19.12.2012.
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