Strongly Semicommutative Rings Relative to a Monoid
For a monoid M, we introduce strongly M-semicommutative rings obtained as a generalization of strongly semicommutative rings and investigate their properties. We show that if G is a finitely generated Abelian group, then G is torsion free if and only if there exists a ring R with |R| ≥ 2 such that R...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508201530687488 |
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| author | Nikmehr, M. J. Нікмер, М. Дж. |
| author_facet | Nikmehr, M. J. Нікмер, М. Дж. |
| author_sort | Nikmehr, M. J. |
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| datestamp_date | 2019-12-05T10:27:00Z |
| description | For a monoid M, we introduce strongly M-semicommutative rings obtained as a generalization of strongly semicommutative rings and investigate their properties. We show that if G is a finitely generated Abelian group, then G is torsion free if and only if there exists a ring R with |R| ≥ 2 such that R is strongly G-semicommutative. |
| first_indexed | 2026-03-24T02:21:27Z |
| format | Article |
| fulltext |
UDC 512.5
M. J. Nikmehr (K. N. Toosi Univ. Technology, Tehran, Iran)
STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID
СИЛЬНО НАПIВКОМУТАТИВНI КIЛЬЦЯ ВIДНОСНО МОНОЇДА
For a monoid M, we introduce strongly M -semicommutative rings, which are generalization of strongly semicommutative
rings and investigate their properties. We show that if G is a finitely generated Abelian group, then G is torsion free if and
only if there exists a ring R with |R| ≥ 2 such that R is strongly G-semicommutative.
Для моноїда M ми вводимо сильно M -напiвкомутативнi кiльця, що узагальнюють сильно напiвкомутативнi кiльця,
та вивчаємо їх властивостi. Показано, що якщо G — скiнченнопороджена абелева група, то G є вiльною вiд скруту
тодi i тiльки тодi, коли iснує кiльце R з |R| ≥ 2 таке, що R є сильно G-напiвкомутативним.
1. Introduction. Throughout this article, R and M denote an associative ring with identity and a
monoid, respectively. In [1] Cohn introduced the notion of reversible ring. A ring R is said to be
reversible, whenever a, b ∈ R satisfy ab = 0 then ba = 0. A ring R is called symmetric, whenever
abc = 0 implies acb = 0 for all a, b, c ∈ R. A ring R is called reduced, whenever a2 = 0 implies
a = 0 for all a ∈ R. A ring R is called semicommutative, whenever ab = 0 implies aRb = 0 for all
a, b ∈ R. The following implication holds:
reduced =⇒ symmetric =⇒ reversible =⇒ semicommutative.
In [13] Yang and Liu introduced the notion of strongly reversible. A ring R is called strongly
reversible, whenever polynomials f(x), g(x) ∈ R[x] satisfy f(x)g(x) = 0 implies g(x)f(x) = 0.
All reduced rings are strongly reversible but converse is not true. In [11] Singh and Juyal introduced
the notion of strongly reversible. A ring R is called strongly M -reversible, whenever αβ = 0
implies βα = 0 where α, β ∈ R[M ]. In [5] Huh and Lee showed that polynomial rings over
semicommutative rings need not be semicommutative. In [2] Gang and Ruijuan introduced the notion
of strongly semicommutative. A ring R is called strongly semicommutative, whenever polynomials
f(x), g(x) ∈ R[x] satisfy f(x)g(x) = 0 implies f(x)R[x]g(x) = 0. All reduced rings are strongly
semicommutative but converse is not true. Rege and Chhawchharia [10], introduced the notion of an
Armendariz ring. A ring R is called Armendariz, whenever polynomials f(x) = a0+a1x+a2x
2+. . .
. . .+anx
n, g(x) = b0+b1x+b2x
2+. . .+bmx
m ∈ R[x] satisfy f(x)g(x) = 0 then aibj = 0 for all i, j.
Some properties of Armendariz rings were given in [8, 9, 12]. In [7] Z. Liu studied a generalization
of Armendariz rings, which is called M -Armendariz rings, where M is monoid. A ring R is called
M -Armendariz, whenever α = a1g1+a2g2+. . .+angn, β = b1h1+b2h2+. . .+bmgm ∈ R[M ], with
gi, hj ∈M satisfy αβ = 0, then aibj = 0, for all i, j. A ring R is called strongly M-semicommutative,
whenever αβ = 0 implies αR[M ]β = 0, where α, β ∈ R[M ]. Let M = (N ∪{0},+). Then a ring R
is strongly M-semicommutative if and only if R is strongly semicommutative. Recall that a monoid
M is called a unique product monoid (u.p.-monoid) if for any two nonempty finite subsets A,B ⊆M
there exists an element g ∈ M uniquely in the form ab, where a ∈ A and b ∈ B. We investigate a
generalization of strongly semicommutative rings which we call strongly M -semicommutative rings.
It is proved that a ring R is strongly M -semicommutative if and only if its polynomial ring R[x]
c© M. J. NIKMEHR, 2014
1528 ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID 1529
is strongly M -semicommutative if and only if its Laurent polynomial ring R[x, x−1] is strongly
M -semicommutative. Also, we check the following questions:
(1) Does R being a strongly M -semicommutative imply R(+)R being strongly M -semicommu-
tative?
(2) R being a strongly M -semicommutative if and only if R is Abelian ring?
(3) R being strongly M -semicommutative if and only if R/I is strongly M -semicommutative?
2. Strongly M -semicommutative ring. We begin this section with the following definition
which have the main role in the whole work.
Definition 2.1. A ring R is called strongly M -semicommutative, whenever αβ = 0 implies
αR[M ]β = 0, where α, β ∈ R[M ].
Lemma 2.1 [6]. If R is a reduced ring, then
T3(R) =
a b c
0 a d
0 0 a
∣∣∣∣∣∣ a, b, c, d ∈ R
is a semicommutative ring.
Lemma 2.2 [7]. LetM be a monoid with |M | ≥ 2. Then the following conditions are equivalent:
(1) R is M -Armendariz and reduced.
(2) T3(R) is M -Armendariz.
Proposition 2.1. Let M be a monoid with |M | ≥ 2, and R is M -Armendariz and reduced. Then
T3(R) is strongly M -semicommutative.
Proof. Suppose that α = A0g1 + . . .+Angn, β = B0h1 + . . .+Bmhm ∈ T3(R)[M ], αβ = 0.
Since T3(R) is M -Armendariz by Lemma 2.2, so AiBj = 0. Also T3(R) is semicommutative by
Lemma 2.1, and hence AiT3(R)Bj = 0. Therefore αT3(R)[M ]β = 0. This means that T3(R) is
strongly M -semicommutative.
Before stating Proposition 2.2, we need the following lemmas.
Lemma 2.3 [11]. Let M be u.p.-monoid and R be a reduced ring. Then R is strongly M -
reversible.
Lemma 2.4 [11]. Let M be u.p.-monoid and R be a reduced ring. Then R[M ] is reduced.
Proposition 2.2. Let M be u.p.-monoid and R be a reduced ring. Then R is strongly M -
semicommutative.
Proof. Suppose α =
∑n
i=1
aigi, β =
∑m
j=1
bjhj are in R[M ] with ai, bj ∈ R and gi, hj ∈M
for all i, j. Take αβ = 0. So (αR[M ]β)2 = (αR[M ]β)(αR[M ]β) = αR[M ](βα)R[M ]β = 0, since
R is strongly M -reversible by Lemma 2.3. Also by Lemma 2.4, we have αR[M ]β = 0. Hence R is
strongly M -semicommutative ring.
Lemma 2.5. Subrings and direct products of strongly M -semicommutative ring are strongly
M -semicommutative.
Proof. Let Iλ(λ ∈ Λ) be ideals of R such that every
R
Iλ
is strongly M -semicommutative and
∩λ∈ΛIλ = 0. Suppose that α =
∑m
i=0
aigi, β =
∑n
j=0 bjhj ∈ R[M ], satisfy αβ = 0. For any
γ =
∑l
k=0
ckrk ∈ R[M ], we have that αγ β = 0 in
(
R
Iλ
)
[M ] for each λ ∈ Λ, since
R
Iλ
is strongly
M -semicommutative. So
∑
i+j+k=t
aickbj ∈ Iλ for t = 0, . . . ,m + n + l and any λ ∈ Λ, which
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1530 M. J. NIKMEHR
implies that
∑
i+j+k=t
aickbj = 0 for t = 0, . . . ,m + n + l, since ∩λ∈ΛIλ = 0. Thus we obtain
αR[M ]β = 0.
Proposition 2.3. Let M be a cancelative monoid and N an ideal of M. If R is strongly N -
semicommutative, then R is strongly M -semicommutative.
Proof. Suppose that α = a1g1 + a2g2 + . . . + angn, β = b1h1 + b2h2 + . . . + bmhm are
in R[M ] such that αβ = 0. Take g ∈ N. Then gg1, gg2, . . . , ggn, h1g, h2g, . . . , hmg ∈ N and
ggi 6= ggj and hig 6= hjg for all i 6= j. So α1β1 =
(∑n
i=1
aiggi
)(∑m
j=1
bjhjg
)
= 0. Since R
is strongly N -semicommutative, so α1R[N ]β1 = 0. Thus αR[M ]β = 0. Therefore R is strongly
M -semicommutative.
Lemma 2.6. Let M be a cyclic group of order n ≥ 2 and R a ring with unity. Then R is not
strongly M -semicommutative.
Proof. Suppose that M = e, g, g2, . . . , gn − 1. Let α =
(
1 0
0 0
)
e +
(
1 0
0 0
)
g + . . .
. . .+
(
1 0
0 0
)
gn−1 and β =
(
0 0
1 0
)
e+
(
0 0
1 0
)
g ∈ R[M ].
Then αβ = 0. But
(
0 0
1 0
)
R[M ]
(
1 0
0 0
)
6= 0, so αR[M ]β 6= 0. Thus R is not strongly
M -semicommutative.
Lemma 2.7. M be a monoid and N a submonoid of M. If R is strongly M -semicommutative
ring, then R is strongly N -semicommutative.
Lemma 2.8. Let M and N be u.p.-monoids. Then so is the monoid M ×N.
Proof. See [7] (Lemma 1.13).
Let T (G) be set of elements of finite order in an Abelian group G. Then T (G) is fully invariant
subgroup of G. G is said to be torsion-free if T (G) = {e}.
Theorem 2.1. Let G be a finitely generated Abelian group. Then the following conditions on G
are equivalent:
(1) G is torsion-free.
(2) There exists a ring R with |R| ≥ 2 such that R is strongly G-semicommutative.
Proof. (2) =⇒ (1). If g ∈ T (G) and g 6= e, then N = 〈g〉 is cyclic group of finite order. If a
ring R 6= 0 is strongly M -semicommutative. Then by Lemma 2.7 R is strongly N -semicommutative,
a contradiction with Lemma 2.6. Thus every ring R 6= 0 is not strongly M -semicommutative.
(1) =⇒ (2). Let G be a finitely generated Abelian group with T (G) = {e}. Then G =
= Z× Z× . . .× Z a finite direct product of group Z. By Lemma 2.8 G is u.p.-monoid. Let R be a
commutative reduced ring. Then by Proposition 2.2, R is strongly G-semicommutative.
It is natural to conjecture that R is a strongly semicommutative ring if for any nonzero proper
ideal I of R, R/I and I are strongly semicommutative, where I is considered as a strongly semi-
commutative ring without identity. Note that strongly semicommutative rings are Abelian, and so
every n by n upper (or lower) triangular matrix ring, for n ≥ 2, over any ring with identity can not
be strongly semicommutative.
Example 2.1 (see [13], Example 3.7). Let S be a division ring and
R =
a b c
0 a d
0 0 a
∣∣∣∣∣∣∣∣ a, b, c, d ∈ S
.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID 1531
Take an ideal I =
0 0 S
0 0 0
0 0 0
, which is strongly M -semicommutative nonzero proper ideal of
R. Take
α =
n∑
i=0
ai bj 0
0 ai ci
0 0 ai
gi, β =
m∑
j=0
uj vj 0
0 uj wj
0 0 uj
hj
are in R/I[M ] satisfying αβ = 0. Then we have that
n∑
i=0
aigi
n∑
i=0
bigi 0
0
n∑
i=0
aigi
n∑
i=0
cigi
0 0
n∑
i=0
aigi
m∑
j=0
ujhj
m∑
j=0
vjhj 0
0
m∑
j=0
ujhj
m∑
j=0
wjhj
0 0
m∑
j=0
ujhj
= 0
which implies
∑n
i=0
aigi
∑m
j=0
ujhj = 0, and hence
∑n
i=0
aigi = 0 or
∑m
j=0
ujhj = 0, since
S is division ring, and it is easy to prove that αR[M ]β = 0. There by we get that for any strongly
M -semicommutative nonzero proper ideal I of R, R/I is strongly M -semicommutative.
However we take a stronger condition I is reduced then we may have an affirmative answer as
in the following.
Proposition 2.4. For a ring R suppose that R/I is strongly M -semicommutative ring for some
ideal I of R. If I is reduced then R is strongly M -semicommutative.
Proof. Let αβ = 0 with α, β ∈ R[M ]. Then we have αR[M ]β ⊆ I[M ] and βI[M ]α = 0 since
βI[M ]α ⊆ I[M ], (βI[M ]α)2 = 0 and I[M ] is reduced. According
((αR[M ]β)I[M ])2 = αR[M ]βI[M ]αR[M ]βI[M ] = αR[M ](βI[M ]α)R[M ]βI[M ] = 0
and so αR[M ]βI[M ] = 0, and hence (αR[M ]β)2 ⊆ αR[M ]βI[M ] = 0 implies (αR[M ]β)2 = 0.
But αR[M ]β ⊆ I[M ] and so αR[M ]β = 0, therefore R is strongly M -semicommutative.
As a kind of converse of Proposition 2.4, we obtain the following situation.
Proposition 2.5. Let R be a strongly M -semicommutative ring and I be an ideal of R. If I is
an annihilator in R, then R/I is a strongly M -semicommutative ring.
Proof. Set I = rR(S) for some S ⊆ R and write t = t+I ∈ R
I
. Let αβ = 0, so S[M ]αR[M ]β =
= 0, since R is strongly M -semicommutative by hypothesis and we have rR(S)[M ] = rR[M ](S[M ]).
Thus αR[M ]β ∈ rR[M ](S[M ]) implies α
(
R
I
)
[M ]β = 0.
Lemma 2.9. For an Abelian ring R, R is strongly M -semicommutative if and only if eR and
(1−e)R are strongly M -semicommutative for every idempotent e of R if and only if eR and (1−e)R
are strongly M -semicommutative for some idempotent e of R.
Proof. Suppose that αβ = 0, since eR and (1 − e)R are strongly M -semicommutative, thus
eαeR[M ]eβe = 0 and (1− e)α(1− e)R[M ](1− e)β(1− e) = 0. So
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1532 M. J. NIKMEHR
αR[M ]β = eαR[M ]β + (1− e)αR[M ]β =
= eαeR[M ]eβe+ (1− e)α(1− e)R[M ](1− e)β(1− e) = 0,
and therefore R is strongly M -semicommutative.
For semicommutative rings relative to monoids, we have following results.
Proposition 2.6. Let M and N be a u.p.-monoid. If R is a reduced ring, then R[M ] is strongly
N -semicommutative.
Proof. By Lemma 2.4 R[M ] is reduced, sinceN is a u.p.-monoid and R[M ] is reduced, therefore
by Proposition 2.2, R[M ] is strongly N-semicommutative.
Proposition 2.7. Let M and N be a u.p.-monoid. If R is a reduced, then R is strongly M ×N -
semicommutative.
Proof. Suppose that
∑s
i=1
ai(mi, ni) is in R[M × N ]. Without loss of generality, we assume
that {n1, n2, . . . , ns} = {n1, n2, . . . , nt} with ni 6= nj when 1 ≤ i 6= j ≤ t. For any 1 ≤ p ≤ t,
denote Ap = {i | 1 ≤ i ≤ s, ni = np}. Then
∑t
p=1
(∑
i∈Ap
aimi
)
np ∈ R[M ][N ]. Note that
mi 6= mi′ for any i, i′ ∈ Ap with i 6= i′. Now it is easy to see that there exists an isomorphism of
rings R[M ×N ]→ R[M ][N ] defined by
s∑
i=1
ai(mi, ni) −→
t∑
p=1
∑
i∈Ap
aimi
np.
Suppose that
(∑s
i=1
ai(mi, ni)
)(∑s′
j=1
bj(m
′
j , n
′
j)
)
= 0 in R[M × N ]. Then from the above
isomorphism, it follows that t∑
p=1
∑
i∈Ap
aimi
np
t′∑
q=1
∑
j∈Bq
bjm
′
j
n′q
= 0
in R[M ][N ]. Therefore by Proposition 2.6 we have t∑
p=1
∑
i∈Ap
aimi
np
R[M ][N ]
t′∑
q=1
∑
j∈Bq
bjm
′
j
n′q
= 0,
so R is strongly M ×N -semicommutative.
LetMi, i ∈ I, be monoids. Denote
∐
i∈I
Mi =
{
(gi)i∈I | there exist only finite i’s such that gi 6=
6= ei, the identity of Mi
}
. Then
∐
i∈I
Mi is a monoid with the operation (gi)i∈I(g
′
i)i∈I = (gig
′
i)i∈I .
Corollary 2.1. Let Mi, i ∈ I be u.p.-monoids and R be a reduced ring. If R is strongly
Mi-semicommutative for some i0 ∈ I, then R is strongly
∐
i∈I
Mi-semicommutative.
Proof. Let α =
∑n
i=1
aigi, β =
∑m
j=1
bjhj ∈ R
[∐
i∈I
Mi
]
such that αβ = 0. Then α,
β ∈ R[M1 ×M2 × . . . ×Mn] for some finite subset {M1,M2, . . . ,Mn} ⊆ {Mi | i ∈ I}. Thus
α, β ∈ R[Mi0 ×M1 × . . . ×Mn}. The ring R, by Proposition 2.7 and by induction, is strongly
Mi0 ×M1 × . . .×Mn-semicommutative, so αR[Mi0 ×M1 × . . .×Mn]β = 0. Hence R is strongly∐
i∈I
Mi-semicommutative.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID 1533
Let R be an algebra over a commutative ring S. The Dorroh extension of R by S is the ring
R× S with operations (r1, s1) + (r2, s2) = (r1 + r2, s1 + s2) and (r1, s1)(r2, s2) = (r1r2 + s1r2 +
+ s2r1, s1s2), where ri ∈ R and si ∈ S. Let R be a commutative ring, M be an R-module, and σ
be an endomorphism of R. Rege and Chhawchharia [10] (Definition 1.3), give R ⊕M a (possibly
noncommutative) ring structure with multiplication (r1,m1)(r2,m2) = (r1r2, σ(r1)m2 + r2m1),
where ri ∈ R and mi ∈M. We shall call this extension the skewtrivial extension of R by M and σ.
Proposition 2.8. (1) Let R be an algebra over a commutative ring S, and D be the Dorroh
extension of R by S. If R is strongly M -semicommutative and S is a domain, then D is strongly
M -semicommutative.
(2) Let R be a commutative domain, and σ be an injective endomorphism of R. Then the
skewtrivial extension of R by R and σ is strongly M -semicommutative.
Proof. (1) Let α = (α1, α2) =
∑
(ri, si)gi, β = (β1, β2) =
∑
(sj , nj)hj ∈ D[M ] with
(α1, α2)(β1, β2) = 0. Then (α1β1 + α2β1 + β2α1, α2β2) = 0, so we have α1β1 + α2β1 + β2α1 = 0
and α2β2 = 0. Since S is a domain, α2 = 0 or β2 = 0. In the following computations we use freely the
condition that R is strongly M -semicommutative. Say α2 = 0, then 0 = α1β1 +β2α1 = α1(β1 +β2)
and since R is strongly M -semicommutative, we have α1(γ1 +γ2)(β1 +β2) = 0 such that γ1 +γ2 ∈
∈ R[M ] and so (α1γ1β1 +α2γ1β1 + γ2α1β1 +α2γ2β1 + β2α1γ1 + β2α2γ1 + β2γ2α1, α2γ2β2) = 0.
Also β2 = 0, then 0 = α1β1+α2β1 = (α1+α2)β1 and so we have (α1+α2)(γ1+γ2)β1 = 0 such that
γ1+γ2 ∈ R[M ] and so (α1γ1β1+α2γ1β1+γ2α1β1+α2h2β1+β2α1γ1+β2α2γ1+β2γ2α1, α2γ2β2) =
= 0. Therefore we obtain (α1, α2)(γ1, γ2)(β1, β2) = 0 for any γ = (γ1, γ2) ∈ D[M ], so in any case,
proving that D is strongly M -semicommutative.
(2) LetN be the skewtrivial extension ofR byR and σ. Set (α1, α2)(β1, β2) = 0 for (αi, βi) ∈ N
with i = 1, 2, 3. Then α1β1 = 0 and σ(α1)β2 + β1α2 = 0, so α1 = 0 and so β1 = 0, since R is a
domain. Say α1 = 0, then 0 = σ(α1)β2 + β1α2 = g1α2, therefore β1γ1α2 = 0 for any y1 ∈ N [M ],
since R is strongly semicommutative, and so 0 = (α1γ1β1, β1γ1α2) = (α1γ1β1, σ(α1)σ(γ1)β2 +
+ σ(α1)β1γ2 + β1γ1α2 = (α1, α2)(γ1, γ2)(β1, β2) for any γ = (γ1, γ2) ∈ N [M ]. Say β1 = 0, then
σ(α1)β2 = 0 and it follows that σ(α1) = 0, or β2 = 0, then α1 = 0 since σ is injective and R is a
domain. Hence we have (α1, α2)(γ1, γ2)(β1, β2) = 0 in any case.
Now we will study some conditions under which polynomial rings may be strongly M -semicom-
mutative. The Laurent polynomial ring with an indeterminate x over a ring R consists of all formal
sums
∑n
i=k
mix
i with obvious addition and multiplication, where mi ∈ R and k, n are (possibly
negative) integer; we denote it R[x;x−1].
Proposition 2.9. (1) Let R be a ring and ∆ be a multiplicatively closed subset of R consisting
of central regular elements. Then R is strongly M -semicommutative if and only if so is ∆−1R.
(2) For a ring R, R[x] is strongly M -semicommutative if and only if so is R[x;x−1].
Proof. (1) Let αβ = 0 with α =
∑n
i=0
(u−1ai)gi, β =
∑m
j=0
(v−1bj)hj , u, v ∈ ∆ and a, b ∈ R.
Since ∆ is contained in the center of R, we have 0 = αβ =
∑n
i=0
(u−1ai)gi
∑m
j=0
(v−1bj)hj =
=
∑
s=i+j
(aibj)(gihj)(uv)−1, so
n∑
i=0
aigi
m∑
j=1
bjhj = 0.
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1534 M. J. NIKMEHR
But R is strongly M -semicommutative by the condition, and hence for any
∑l
k=0
ckpk ∈ R[M ] we
have that
n∑
i=0
aigi
l∑
k=0
ckpk
m∑
j=0
bjhj =
∑
i+j+k=t
(aickbj)(gipkhj) = 0
for t = 0, 1, . . . ,m+ n+ l. Hence
αγβ =
n∑
i=0
(u−1ai)gi
l∑
k=0
(ω−1ck)pk
m∑
j=0
(v−1bj)hj =
∑
t=i+j+k
(aickbj)(gipkhj)(uωv)−1 = 0
for any γ =
∑l
k=0
(ω−1ck)pk ∈ ∆−1R[M ]. Hence ∆−1R is strongly M -semicommutative.
(2) Let ∆ = 1, x, x2, . . . . Then clearly ∆ is a multiplicatively closed subset of R[x]. Since
R[x;x−1] = ∆−1R[x], it follows that R[x;x−1] is strongly M -semicommutative by the result (1).
Given a ring R we denote the center of R by Z(R), i.e.,
Z(R) =
{
s ∈ R | sr = rs for all r ∈ R
}
.
Proposition 2.10. Let R be a ring and suppose that Z(R) contains an infinite subring every
nonzero element of which is regular in R. Then R is strongly M -semicommutative ring if and only
if R[x] is strongly M -semicommutative ring if and only if R[x;x−1] is strongly M -semicommutative
ring.
Proof. It suffices to prove that R[x] is strongly M -semicommutative ring when so is R, by
Lemma 2.5 and Proposition 2.9 (2). Since Z(R) contains an infinite subring every nonzero element
of which is regular in R by hypothesis, it follows that R[x] is a subdirect product of infinite number
of copies of R. Thus R[x] is strongly M -semicommutative by Lemma 2.5 because R is strongly
M -semicommutative ring by the assumption.
We study following proposition the connections between Armendariz rings and strongly M -
semicommutative rings. Recall that reduced rings, M is u.p.-monoid are both M -Armendariz and
strongly M -semicommutative rings Abelian. So it is natural to observe the relationships between
them.
Proposition 2.11. LetR[M ] be a Armendariz ring. Then the following statements are equivalent:
(1) R is a strongly M -semicommutative ring.
(2) R[x] is a strongly M -semicommutative ring.
(3) R[x, x−1] is a strongly M -semicommutative ring.
Proof. (1) ⇒ (2). It is easy to see that there exists an isomorphism of R[x][M ] −→ R[M ][x]
via
∑
i
(∑
p
aipx
p
)
gi −→
∑
p
(∑
i
aipgi
)
xp. Let
α =
∑
p
(∑
i
aipgi
)
xp, β =
∑
q
(∑
j
bjqhj
)
xq
be polynomial in R[M ][x], such that αβ = 0, where αi =
∑
p
aipgi and βj =
∑
q
bjqhj ∈ R[M ].
Since R[M ] is Armendariz, so R[M ][x] is a Armendariz ring, therefore αiβj = 0 for all i, j. Also
R is strongly M -semicommutative by the hypothesis, therefore αiγkβj = 0 for all i, j, k. Thus
αR[M ][x]β = 0.
(2)⇒ (3). By the Proposition 2.9 (2) is trivial.
(3)⇒ (1). It is clear.
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STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID 1535
Proposition 2.12. Let R be an M -Armendariz ring. If R is a semicommutative ring, then R is
strongly M -semicommutative.
Proof. Suppose that α =
∑m
i=0
aigi, β =
∑n
j=0
bjhj ∈ R[M ] satisfy αβ = 0. Since R is M -
Armendariz, and hence aibj = 0 for all i, j, also R is semicommutative, therefore aicbj = 0 for any
element c in R, for all i, j. Now it is easy to check that αγβ = 0 for any γ =
∑s
k=0
cklk ∈ R[M ].
Since reversible rings are semicommutative, the following corollary is clear.
Corollary 2.2. Let R be an M -Armendariz ring. If R is a reversible ring, then R is a strongly
M -Armendariz.
Let R be a commutative ring and M an R-module. The R-module R ⊕ M acquires a ring
structure where the product is defined by (a,m)(b, n) = (ab, an + bm). We shall use the notation
R(+)M for this ring. If M is not zero, this ring is not reduced, since M can be identified with the
ideal 0⊕M which has square zero. (It seems appropriate to call this ring as “R Nagata M”.)
Let R be a ring and A an ideal of R. The factor ring R = R/A has the natural structure of a left
R-, right R-bimodule. Denote a = a+A ∈ R for each a ∈ R. We use this structure to define a ring
structure on R⊕ (R/A) as follows:
(r, a)
(
r′, a′
)
=
(
rr′, ra′ + ar′
)
.
We denote this ring by R(+)R/A. Its properties are similar to those of R(+)M.
Proposition 2.13. Let R be a domain, A be an ideal of R. Suppose R/A is strongly M -
semicommutative. Then R(+)R/A is strongly M -semicommutative.
Proof. Let α, β be elements of {R(+)R/A}[M ], where
α =
m∑
i=0
(ai, ui)gi = (α0, α1)
and
β =
n∑
j=0
(bj , vj)hj = (β0, β1).
If αβ = 0, we have (α0, α1)(β0, β1) = 0. Thus we have the following equations:
α0β0 = 0, (2.1)
α0β1 + α1β0 = 0. (2.2)
Let α0 = 0. Then (2.2) becomes α1β0 = 0 over R/A. Since R/A is strongly M -semicommutative,
it follows that α1
(
R
A
)
[M ]β0 = 0. Also for any γ0 ∈ R[M ] implies that α1γ0β0 = 0. We conclude
that 0 =
(
α0γ0β0, α0γ0β1 + α0γ1β0 + α1γ0β0
)
= (α0, α1)(γ0, γ1)
(
β0, β1
)
. This case β0 = 0 is
similar.
Corollary 2.3. LetR be a domain,A be an ideal ofR. SupposeR/A is strongly semicommutative.
Then R(+)R/A is strongly semicommutative.
It follows from Proposition 2.13 that if R is a domain then R(+)R is strongly semicommutative.
This result can be extended to reduced rings. The following properties of these rings will be used:
(1) If a, b are elements of a reduced ring, then ab = 0 if and only if ba = 0.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1536 M. J. NIKMEHR
(2) Reduced rings are strongly semicommutative.
(3) If R is reduced, then so is the ring R[x]. We shall also identify {R(+)R}[x] with the ring
R[x](+)R[x] in a natural manner. Therefore if R is a reduced ring, then the ring R(+)R is strongly
semicommutative.
Proposition 2.14. Let M be u.p.-monoid and R be a reduced ring. Then the ring R(+)R is
strongly M -semicommutative.
Proof. Let α = (α0, α1), β = (β0, β1) be elements of {R(+)R}[M ], we claim that
α{R(+)R}[M ]β = 0. Write α =
∑m
i=0
(ai, ui)gi = (α0, α1) and β =
∑n
j=0
(bj , vj)hj = (β0, β1),
with corresponding representations for αk, βk (for k = 0, 1). Now we have
α0β0 = 0, (2.3)
α0β1 + α1β0 = 0. (2.4)
By Lemma 2.4 R[M ] is reduced, (2.3) implies
β0α0 = 0. (2.5)
Multiplying equation (2.4) by β0 on the left and using (2.5) we get β0α1β0 = 0. This implies that
(α1β0)2 = 0 and so (since R[M ] is reduced)
α1β0 = 0. (2.6)
This implies (on account of (2.4))
α0β1 = 0. (2.7)
Now (2.3), (2.6) and (2.7) yield (since R is strongly M -semicommutative)
α0R[M ]β0 = 0, α1R[M ]β0 = 0, and α0R[M ]β1 = 0.
Therefore (α0, α1)(γ0, γ1)(β0, β1) = (α0γ0β0, α0γ0β1 +α0γ1β0 +α1γ0β0) = 0 for each (γ0, γ1) of
{R(+)R}[M ].
The following theorem generalization of Proposition 2.14 has a similar proof.
Theorem 2.2. Let M be u.p.-monoid, R be a reduced ring and A an ideal of R such that R/A
is reduced. Then R(+)R/A is strongly M -semicommutative.
Remark 2.1. Recall that a ring R is strongly regular [3] if for each element a in R, there exists
an element b in R such that a = a2b. A ring is strongly regular, if and only if it is (von Neumann)
regular and reduced. If R is a strongly regular ring, then for each ideal A of R, R/A is strongly
regular and reduced. On applying Theorem 2.2 we get the following result: If R is a strongly regular
ring, then for each ideal A of R, then ring R(+)R/A is strongly M -semicommutative.
The ring R is called Abelian if every idempotent is central, that is, ae = ea for any e2 = e,
a ∈ R.
Recall that a ring R is a called right principally projective ring (or simples right p.p.-ring) if the
right annihilator of an element of R is generated by an idempotent.
Lemma 2.10. Let M be an monoid and R be strongly M -semicommutative. Then R is an
Abelian ring. The converse holds if R is a right p.p.-ring.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID 1537
Proof. If e is an idempotent in R, then e(1− e) = 0. Since R is strongly M -semicommutative,
we have eα(1− e) = 0 for any α ∈ R[M ] and so eα = eαe. On the other hand, (1− e)e = 0 implies
that (1− e)αe = 0, so we have αe = eαe. Therefore, αe = eα. For converse suppose now R is an
Abelian and right p.p.-ring. Let α, β ∈ R[M ] with αβ = 0. Then α ∈ Ann(β) = eR[M ] for some
e2 = e ∈ R and so βα = 0 and α = eα. Since R is Abelian, we have αγβ = eαγβ = αγβe = 0 for
any γ ∈ R[M ], so, αR[M ]β = 0. Therefore R is strongly M -semicommutative.
Before stating Example 2.2, we need the following lemmas.
Lemma 2.11 ([4], Lemma 1). Given a ring R we have the following assertion: R is an Abelian
ring if and only if R is a reduced ring if and only if R is a semicommutative ring, when R is a right
p.p.-ring.
Lemma 2.12 ([4], Lemma 2). Let S be an Abelian ring and define
a a12 a13 . . . a1n
0 a a23 . . . a2n
0 0 a . . . a3n
...
...
...
. . .
...
0 0 0 . . . a
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
a, aij ∈ S
= Rn
with n a positive integer ≥ 2. Then every idempotent in Rn is of the form
f 0 0 . . . 0
0 f 0 . . . 0
0 0 f . . . 0
...
...
...
. . .
...
0 0 0 . . . f
with f2 = f ∈ S and so Rn is Abelian.
Example 2.2. Let S be Abelian ring and
R =
a a12 . . . a1n
0 a . . . a2n
...
...
. . .
...
0 0 0 a
∣∣∣∣∣∣∣∣∣∣∣∣
a, aij ∈ S
.
Then R is Abelian by Lemma 2.12. Let M be a monoid with |M | ≥ 2. Take e, g ∈ M such that
e 6= g. Consider
α =
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
e+
0 1 −1 0
0 0 0 0
0 0 0 0
0 0 0 0
g ∈ R[M ],
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1538 M. J. NIKMEHR
β =
0 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0
e+
0 0 0 0
0 0 0 1
0 0 0 1
0 0 0 0
g ∈ R[M ].
Then αβ = 0, but
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
1 1 1 1
0 1 1 1
0 0 1 1
0 0 0 1
0 0 0 0
0 0 0 1
0 0 0 1
0 0 0 0
6= 0,
so R is not strongly M -semicommutative. Assuming that R is a right p.p.-ring, then R is reduced
by Lemma 2.11, a contradiction by the element
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
in R. Thus, R is not a right p.p.-ring. In fact there can not be an idempotent e ∈ R such that
AnnR
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
= eR.
Proposition 2.15. The direct limit of a direct system of strongly M -semicommutative rings is
also strongly M -semicommutative.
Proof. Let A = {Ri, αij} be a direct system of strongly M -semicommutative rings Ri for
i ∈ I and ring homomorphism αij : Ri → Rj for each i ≤ j satisfying αij(1) = 1, where I
is a directed partially ordered set. Let R = limRi be the direct limit of D with li : Ri → R and
ljαij = li, we will prove thatR is stronglyM -semicommutative ring. Take x, y ∈ R, then x = li(xi),
y = lj(yj) for some i, j ∈ I and there is k ∈ I such that i ≤ k, j ≤ k define x+ y = lk(αik(xi) +
+ αjk(yj)) and xy = lk(αik(xi)αjk(yj)), where αik(xi), αjk(yj) are in Rk. Then R forms a rings
with 0 = li(0) and 1 = li(1). Now suppose αβ = 0 for α =
∑m
s=1
asgs, β =
∑n
t=1
btht in
R[M ]−{0}. There exist is, jt, k ∈ I such that as = lis(ais), bt = ljt(bjt), is ≤ k, jt ≤ k. So asbt =
= lk(αisk(ais)αjtk(bjt)). Thus αβ =
(∑m
s=1
lk(αisk(ais))gs
)(∑n
t=0
lk(αjtk(bjt)ht
)
= 0. But Rk
is strongly M -semicommutative ring and so lk(αisk(ais)Rk[M ]αjtk(bjt)) = 0. Thus αR[M ]β = 0,
and hence R is strongly M -semicommutative ring.
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STRONGLY SEMICOMMUTATIVE RINGS RELATIVE TO A MONOID 1539
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Received 08.11.12
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
|
| id | umjimathkievua-article-2243 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:21:27Z |
| publishDate | 2014 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/95/50d84bb7dbaa5e51d21271ec004ea095.pdf |
| spelling | umjimathkievua-article-22432019-12-05T10:27:00Z Strongly Semicommutative Rings Relative to a Monoid Сильно напівкомутативні кільця вiдносно моноїда Nikmehr, M. J. Нікмер, М. Дж. For a monoid M, we introduce strongly M-semicommutative rings obtained as a generalization of strongly semicommutative rings and investigate their properties. We show that if G is a finitely generated Abelian group, then G is torsion free if and only if there exists a ring R with |R| ≥ 2 such that R is strongly G-semicommutative. Для моноїда M, ми вводимо сильно M,-напівкомутативні кільця, що узагальнюють сильно напівкомутативні кільця, та вивчаємо їх властивості. Показано, що якщо G — скінченнопороджена абелева група, то G є вільною від скруту тоді і тільки тоді, коли існує кільце R з |R| ≥ 2 таке, що R є сильно G-напівкомутативним. Institute of Mathematics, NAS of Ukraine 2014-11-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2243 Ukrains’kyi Matematychnyi Zhurnal; Vol. 66 No. 11 (2014); 1528–1539 Український математичний журнал; Том 66 № 11 (2014); 1528–1539 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2243/1487 https://umj.imath.kiev.ua/index.php/umj/article/view/2243/1488 Copyright (c) 2014 Nikmehr M. J. |
| spellingShingle | Nikmehr, M. J. Нікмер, М. Дж. Strongly Semicommutative Rings Relative to a Monoid |
| title | Strongly Semicommutative Rings Relative to a Monoid |
| title_alt | Сильно напівкомутативні кільця вiдносно моноїда |
| title_full | Strongly Semicommutative Rings Relative to a Monoid |
| title_fullStr | Strongly Semicommutative Rings Relative to a Monoid |
| title_full_unstemmed | Strongly Semicommutative Rings Relative to a Monoid |
| title_short | Strongly Semicommutative Rings Relative to a Monoid |
| title_sort | strongly semicommutative rings relative to a monoid |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2243 |
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