Quasi-duo Partial skew polynomial rings
In this paper we consider rings R with a partial action α of Z on R. We give necessary and sufficient conditions for partial skew polynomial rings and partial skew Laurent polynomial rings to be quasi-duo rings and in this case we describe the Jacobson radical. Moreover, we give some examples to sho...
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Інститут прикладної математики і механіки НАН України
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| Цитувати: | Quasi-duo Partial skew polynomial rings / W. Cortes, M.Ferrero, L.Gobbi // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 53–63. — Бібліогр.: 8 назв. — англ. |
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irk-123456789-1548682019-06-17T01:31:15Z Quasi-duo Partial skew polynomial rings Cortes, W. Ferrero, M. Gobbi, L. In this paper we consider rings R with a partial action α of Z on R. We give necessary and sufficient conditions for partial skew polynomial rings and partial skew Laurent polynomial rings to be quasi-duo rings and in this case we describe the Jacobson radical. Moreover, we give some examples to show that our results are not an easy generalization of the global case. 2011 Article Quasi-duo Partial skew polynomial rings / W. Cortes, M.Ferrero, L.Gobbi // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 53–63. — Бібліогр.: 8 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:16S36; 16S35. http://dspace.nbuv.gov.ua/handle/123456789/154868 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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In this paper we consider rings R with a partial action α of Z on R. We give necessary and sufficient conditions for partial skew polynomial rings and partial skew Laurent polynomial rings to be quasi-duo rings and in this case we describe the Jacobson radical. Moreover, we give some examples to show that our results are not an easy generalization of the global case. |
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Cortes, W. Ferrero, M. Gobbi, L. |
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Cortes, W. Ferrero, M. Gobbi, L. Quasi-duo Partial skew polynomial rings Algebra and Discrete Mathematics |
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Cortes, W. Ferrero, M. Gobbi, L. |
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Cortes, W. |
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Quasi-duo Partial skew polynomial rings |
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Quasi-duo Partial skew polynomial rings |
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Quasi-duo Partial skew polynomial rings |
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Quasi-duo Partial skew polynomial rings |
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Quasi-duo Partial skew polynomial rings |
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quasi-duo partial skew polynomial rings |
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Інститут прикладної математики і механіки НАН України |
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2011 |
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http://dspace.nbuv.gov.ua/handle/123456789/154868 |
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Quasi-duo Partial skew polynomial rings / W. Cortes, M.Ferrero, L.Gobbi // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 53–63. — Бібліогр.: 8 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT cortesw quasiduopartialskewpolynomialrings AT ferrerom quasiduopartialskewpolynomialrings AT gobbil quasiduopartialskewpolynomialrings |
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2025-07-14T06:56:11Z |
| last_indexed |
2025-07-14T06:56:11Z |
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1837604454438797312 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 12 (2011). Number 2. pp. 53 – 63
c© Journal “Algebra and Discrete Mathematics”
Quasi-duo Partial skew polynomial rings
Wagner Cortes, Miguel Ferrero and Luciane Gobbi
Communicated by V. V. Kirichenko
Abstract. In this paper we consider rings R with a partial
action α of Z on R. We give necessary and sufficient conditions for
partial skew polynomial rings and partial skew Laurent polynomial
rings to be quasi-duo rings and in this case we describe the Jacobson
radical. Moreover, we give some examples to show that our results
are not an easy generalization of the global case.
Introduction
Partial actions of groups have been introduced in the theory of operator
algebras giving powerful tools of their study (see [3] and the literature
quoted therein). In [3], the authors introduced partial actions on rings in
a pure algebraic context and studied partial skew group rings. In [2], the
authors defined a partial action as follows: let R be a ring with an identity
1R and let Z be the additive group of integers. A partial action α of Z on
R is a collection of ideals Si, i ∈ Z, isomorphisms of rings αi : S−i → Si
and the following conditions hold:
(i) S0 = R and α0 is the identity map of R;
(ii) S−(i+j) ⊇ α−1
i (Si ∩ S−j),
(iii) αj ◦ αi(a) = αj+i(a), for any a ∈ α−1
i (Si ∩ S−j).
The second named author was partially supported by Conselho Nacional de De-
senvolvimento Cientfico e Tecnolgico (CNPq, Brazil) and the third named author was
partially supported by Capes (Brazil)
2000 Mathematics Subject Classification: 16S36; 16S35.
Key words and phrases: partial action; quasi-duo; Jacobson radical; partial
skew polynomial rings.
54 Quasi-duo Partial skew polynomial rings
The above properties easily imply that αj(S−j ∩ Si) = Sj ∩ Si+j , for
all i, j ∈ Z, and that α−i = α−1
i , for every i ∈ Z.
Following [2], the partial skew Laurent polynomial ring R〈x;α〉 in an
indeterminate x is the set of all finite formal sums
∑m
i=−n aix
i, ai ∈ Si,
where the addition is defined in the usual way and the multiplication is
defined by (aix
i)(ajx
j) = αi(α−i(ai)aj)x
i+j , for any i, j ∈ Z. The partial
skew polynomial ring R[x;α] is the subring of R〈x;α〉 whose elements are
the polynomials
∑n
i=0 aix
i, ai ∈ Si.
Given a partial action α of Z on R, an enveloping action is a ring T
containing R together with a global action β = {σi : i ∈ Z} on T , where
σ is an automorphism of T such that the partial action αi is given by
the restriction of σi ([3], Definition 4.2). Note that T does not necessarily
have an identity, since the group acting on R is infinite. It is shown in
([3], Theorem 4.5) that a partial action α has an enveloping action if and
only if all the ideals Si are generated by central idempotents of R.
When α has an enveloping action (T, σ), where σ is an automorphism
of T , we may consider that R is an ideal of T and the following properties
hold:
(i) T =
∑
i∈Z σ
i(R);
(ii) Si = R ∩ σi(R), for every i ∈ Z;
(iii) αi(a) = σi(a), for all i ∈ Z and a ∈ S−i.
In order to have associative rings and apply the results which are
known for skew polynomial rings and skew Laurent polynomial rings, we
assume throughout the paper that all ideals Si are generated by central
idempotents of R. The idempotent corresponding to Si will be denoted by
1i and the enveloping action of α by (T, σ), where σ is an automorphism
of T . By condition (ii) above we have that 1i = 1Rσ
i(1R). This fact and
conditions (i) and (iii) above will be used freely in the paper. Also the
following remark will be used without further mention: if I is an ideal
of R, then I is also an ideal of T . In fact, if a ∈ I and t ∈ T we have
ta = t1Ra ∈ Ra ⊆ I, and similarly at ∈ I.
The skew Laurent polynomial ring T 〈x;σ〉 is the set of formal finite
sums
∑q
i=p aix
i, ai ∈ T , with usual sum and the multiplication is given
by xa = σ(a)x, for all a ∈ T . The partial skew Laurent polynomial ring
R〈x;α〉 is a subring of T 〈x;σ〉. Moreover, R[x;α] is a subring of the skew
polynomial ring T [x;σ].
We recall some terminology from [2]. We say that an ideal I of R
is an α-ideal ( α-invariant ideal) if αi(I ∩ S−i) ⊆ I ∩ Si, for all i ≥ 0
(αi(I ∩ S−i) = I ∩ Si, for all i ∈ Z). Note that I is an α-ideal of R if and
only if the set of all polynomials
∑
i≥0 aix
i, where ai ∈ I ∩ Si, is an ideal
W. Cortes, M. Ferrero, L. Gobbi 55
of R[x;α]. A similar result holds in R〈x;α〉 if I is an α-invariant ideal
of R.
A ring R is called right (left) quasi-duo if every maximal right (left)
ideal of R is two-sided or, equivalently, every right (left) primitive ho-
momorphic image of R is a division ring [7]. We refer [7] for further
information on quasi-duo rings.
Let J(R) be the Jacobson radical of R. Then from the definition we
have that R is right (left) quasi-duo if and only if R/J(R) is right (left)
quasi-duo and in this case R/J(R) is a reduced ring. We will use this
property in the paper without further mention.
Next we recall some terminology and definitions on Z-graded rings
(see [8] for further details). A ring R is a Z-graded ring if R =
⊕
n∈ZRn,
where each Rn is an additive subgroup of R such that RnRm ⊆ Rn+m, for
all n,m ∈ Z. It is known that 1R ∈ R0. An ideal I of a Z-graded ring R
is called homogeneous if I =
⊕
n∈Z(I ∩Rn). Note that the rings R[x;α]
and R〈x;α〉 are naturally Z-graded rings.
The main purpose of this paper is to study partial skew polynomial
rings and partial skew Laurent polynomial rings which are quasi-duo. In
Section 1 we give necessary and sufficient conditions for a partial skew
polynomial rings to be quasi-duo and in this case we give an explicit
description of the Jacobson radical.
In Section 2 we consider a partial action of finite type α of Z on R
(we recall this definition in the beginning of the section). Then we give
necessary and sufficient conditions for a partial skew Laurent polynomial
ring to be quasi-duo. We also give an explicit description of the Jacobson
radical in this case.
In Section 3 we give some examples to show that our results are not
easy generalizations of the global case.
1. Quasi-duo partial skew polynomial rings
We begin with the following.
Proposition 1. Let (R,α) be a partial action of Z on R. Then R is right
quasi-duo if and only if T is right quasi-duo.
Proof. If T is right quasi-duo, then R is right quasi-duo by ([5], Corollary 2),
since the natural mapping ϕ : T → R defined by ϕ(t) = t.1R, for all t ∈ T ,
is a surjective homomorphism.
Conversely, suppose that R is right quasi-duo and let M be a maximal
right ideal of T . Then there exists s ∈ Z such that M ∩ σs(R) is a
proper right ideal of σs(R). Let L be a maximal right ideal of σs(R) with
56 Quasi-duo Partial skew polynomial rings
M ∩ σs(R) ⊆ L. Put X = {t ∈ T : tσs(1R) ∈ L} and note that X is a
right ideal of T . Also, since σs(R) ≃ R is right quasi duo it follows that L
is two-sided and hence X is also a two-sided ideal of T . Finally, if x ∈M ,
then xσs(1R) ∈ M ∩ σs(R) ⊆ L and so M ⊆ X. Therefore M = X is a
two-sided ideal of T .
A skew polynomial ring S[x;σ] of automorphism type is a commutative
ring if and only if S is a commutative ring and σ = idS .
For the partial case, we have the following result.
Proposition 2. Let (R,α) be a partial action of Z on R. Then the
following conditions are equivalent:
(i) R is commutative and αi = idSi
, for all i ∈ Z.
(ii) R[x;α] is commutative.
(iii) R〈x;α〉 is commutative.
Proof. (ii) ⇒ (i). We clearly have that R is commutative. Take any a ∈
S−i. We have a1ix
i = 1ix
ia = αi(1−ia)x
i = αi(a)x
i and so αi(a) = 1ia.
Thus Si ⊆ S−i, for any i > 0. Also, 1ix
i = 1ix
i1i = αi(1−i1i)x
i and we
have 1i = αi(1−i1i). Applying α−i to this relation we obtain 1−i = 1−i1i,
hence S−i ⊆ Si and now (i) follows easily.
(i) ⇒ (ii). By assumption we easily have that aix
i1jx
j = 1jx
jaix
i
and raix
i = aix
ir, for every j ∈ Z and r ∈ R. So, R[x;α] is commutative.
The proof (i) ⇔ (iii) is similar with the proof of (i) ⇔ (ii).
Recall that if S is a ring and σ : S → S is an automorphism, an
element a ∈ S is said to be σ-nilpotent if for every m ≥ 1 there exists
n ≥ 1 such that aσm(a)σ2m(a)...σmn(a) = 0 (see [7] for more details). A
subset B of S is σ-nil if every element of B is σ-nilpotent. Now we extend
this notion to partial actions.
Definition 1. Let (R,α) be a partial action of Z on R. An element a ∈ R
is said to be α-nilpotent if for every m ≥ 1 there exists n ≥ 1 such that
aαm(a1−m)α2m(a1−2m)...αmn(a1−mn) = 0. A subset I of R is called α-nil
if every element of I is α-nilpotent.
We write N i
α(R) = {a ∈ R : ∃n ≥ 1, aαi(a1−i)...αni(a1−ni) = 0} and
Nα(R) = ∩i≥1N
i
α(R). Also N i(T ) = {a ∈ T : ∃n ≥ 1, aσi(a)...σni(a) =
0}, for any i ≥ 1, and N(T ) = ∩i≥1N
i(T ).
Lemma 1. (i) Nα(R) contains all α-nil subsets I of R.
(ii) For any n > 0 we have Nn
α (R) = Nn(T )∩R. In particular, Nα(R) =
N(T ) ∩R.
W. Cortes, M. Ferrero, L. Gobbi 57
(iii) Nα(R) is an α-invariant subset of R.
Proof. (i) is clear. (iii) follows from (ii) since N(T ) is a σ-invariant subset
of T . Thus we only proof (ii). Assume that a ∈ R. Since 1−i = 1Rσ
−i(1R),
for any i, there exists m > 0 with aαn(a1−n)...αnm(a1−nm) = 0 if and
only if for such m we have aσn(a)σn(1R)1R...σ
nm(a)σnm(1R)1R = 0. This
is equivalent to aσn(a)...σnm(a) = 0 and so a ∈ Nn(T ). Thus a ∈ Nn
α (R)
if and only if a ∈ Nn(T ) ∩R.
The Jacobson radical of a skew polynomial ring and a skew Laurent
polynomial ring are described in ([1], Theorem 3.1). Now we obtain
similar results for partial skew polynomial rings and partial skew Laurent
polynomial rings.
Proposition 3. Let (R,α) be a partial action of Z on R. Then there exist
α-nil α-invariant ideals K ⊆ J(R) and I of R such that J(R〈x;α〉) =
K〈x;α〉 and J(R[x;α]) = J(R) ∩ I +
∑
i≥1(Si ∩ I)x
i.
Proof. By ([4], Proposition 6.1) we have J(R〈x;α〉) = J(T 〈x;σ〉) ∩
R〈x;α〉.
By ([1], Theorem 3.1) J(T 〈x;σ〉) = L〈x;σ〉, where L ⊆ J(T ) is σ-nil
σ-invariant ideal of T . So J(R〈x;α〉) = L〈x;σ〉 ∩R〈x;α〉 = (L∩R)〈x;α〉,
where L ∩ R ⊆ J(T ) ∩ R = J(R) is an α-nil α-invariant ideal of R. For
R[x;α] the proof is similar.
As in [8] we denote by A be the set of all maximal right ideals M
of R[x;α] such that Six
i * M , for some i ≥ 1, and by B the set of all
remaining maximal right ideals of R[x;α]. Since R[x;α] is naturally a
Z-graded ring, using ([8], Proposition 3) we have that
A(R[x;α]) =
⋂
M∈A
M = {f ∈ R[x;α]; fSix
i ⊆ J(R[x;α]), for all i ≥ 1}.
Also, we easily see that
B(R[x;α]) = (
⋂
M∈B
M ∩R)⊕
∑
i≥1
Six
i.
Note that, J(R[x;α]) = A(R[x;α]) ∩ B(R[x;α]).
The next result gives a characterization of Nα(R) when R[x;α] is right
quasi-duo.
Lemma 2. If R[x;α] is a right quasi-duo ring, then
Nα(R) = A(R[x, α]) ∩R = {a ∈ R | a1ix
i ∈ J(R[x;α]), for all i ≥ 1}.
Moreover, Nα(R) is an α-invariant ideal of R.
58 Quasi-duo Partial skew polynomial rings
Proof. Let a ∈ Nα(R). Then for all i ≥ 1 there exists n ≥ 1
such that aαi(a1−i)α2i(a1−2i)...αni(a1−ni) = 0. Consider u = a1ix
i +
J(R[x;α]) ∈ R[x, α]/J(R[x, α]). By the above we have un = 0 and since
R[x, α]/J(R[x, α]) is reduced we obtain a1ix
i ∈ J(R[x;α]). It follows that
a ∈ A(R[x, α]) ∩R.
On the other hand, let a ∈ R be such that a1ix
i ∈ J(R[x;α]), for all
i ≥ 1. We fix such an i. Then by Proposition 1.5 there exists an α-nil
ideal I of R such that a1i ∈ I. Hence there exists m with
a1iαi(a1i1−i)α2i(a1i1−2i)...αmi(a1i1−mi) = 0.
This easily gives
a1Rσ
i(1R)αi(a1−i)σ
2i(1R)α2i(a1−2i)...σ
mi(1R)αmi(a1−mi)σ
(m+1)i(1R) = 0
and it follows that
aαi(a1−i)α2i(a1−2i)...αmi(a1−mi)α(m+1)i(a1−(m+1)i) = 0.
Hence a ∈ N i
α(R), for all i > 0, i.e., a ∈ Nα(R). The rest is clear.
As a consequence of Lemma 1.6 we have the following:
Corollary 1. Suppose that R[x;α] is right quasi-duo. Then Nα(R) is an
α-invariant ideal of R and J(R[x;α]) ⊆ Nα(R)[x;α] = A(R[x;α]).
Proof. The inclusion Nα(R)[x;α] ⊆ A(R[x;α]) is immediate from Lemma
1.6. Assume that axi ∈ A(R[x;α]), where i > 0. Then by Proposition 3
of [8] we have that axi ∈ J(R[x;α]) and again by Lemma 1.6 we obtain
a ∈ Nα(R).
Next we will describe the Jacobson radical of R[x;α], when R[x;α] is
quasi-duo.
Recall that a ring S is a subdirect product of the rings {Si : i ∈ Ω} if
for any i ∈ Ω there exists a surjective homomorphism ϕi : S → Si such
that
⋂
i∈Ω kerϕi = 0.
Lemma 3. Let U and V be ideals of R such that U ⊆ V and V is α-
invariant. Then U +
∑
i≥1(V ∩ Si)x
i is a two-sided ideal of R[x;α] and
R[x;α]/(U +
∑
i≥1(V ∩Si)x
i) is a right quasi-duo ring if and only if R/U
and R[x;α]/V [x;α] are right quasi-duo rings.
Proof. We clearly have that U +
∑
i≥1(V ∩ Si)x
i is an ideal of R[x;α].
Since V is α-invariant α induces a partial action α of Z on R/V . Then
there exists an isomorphism (R/V )[x;α] ≃ R[x;α]/V [x;α] and note that
(U +
∑
i≥1
Six
i) ∩ V [x;α] = U +
∑
i≥1
(V ∩ Si)x
i.
W. Cortes, M. Ferrero, L. Gobbi 59
We have an isomorphism
ϕ : R[x;α]/(U +
∑
i≥1
(V ∩ Si)x
i) ≃ R/U +
∑
i≥1
Six
i
defined by ϕ(r + axi) = (r + U) + (a + V )xi, where Si = (Si + V )/V ,
i > 1.
Consider the natural homomorphisms
ψ1 : R[x;α]/(U +
∑
i≥1
(V ∩ Si)x
i) → R/U and
ψ2 : R[x;α]/(U +
∑
i≥1
(V ∩ Si)x
i) → (R/V )[x;α].
It is easy to see that ker(ψ1)∩ker(ψ2) = 0 and so R[x;α]/(U +
∑
i≥1(V ∩
Si)x
i) is a subdirect product of R/U and R/V [x;α]. So the result follows
from ([6], Corollary 3.6(2)).
Theorem 1. R[x;α] is right quasi-duo if and only if R is right quasi-duo,
J(R[x;α]) = J(R) ∩Nα(R) +
∑
i≥1(Nα(R) ∩ Si)x
i and (R/Nα(R))[x;α]
is commutative, where α is the partial action induced by α on R/Nα(R).
Proof. Suppose that R[x;α] is right quasi-duo. Then, by Corollary 1.7,
we have that Nα(R)[x;α] = A(R[x;α]) and so
R[x;α]/A(R[x;α]) ≃ (R/Nα(R))[x;α].
Since the partial skew polynomial ring R[x;α] is Z-graded, then by Theo-
rem 5 of [8] we have that (R/Nα(R))[x;α] is commutative and R is right
quasi-duo. LetM ∈ B. Then we easily obtain thatM =M∩R+
∑
i≥1 Six
i,
where M∩R is a maximal ideal of R. Thus B(R[x;α]) = J(R)+
∑
i≥1 Six
i.
Hence,
J(R[x;α]) = A(R[x;α])∩B(R[x;α]) = Nα(R)∩J(R)+
∑
i≥1
(Nα(R)∩Si)x
i.
Conversely, assume that R is right quasi-duo, J(R[x;α]) = J(R) ∩
Nα(R)+
∑
i≥1(Nα(R)∩Si)x
i and (R/Nα(R))[x;α] is commutative. Then
R[x;α]/J(R[x;α]) = R[x;α]/(J(R) ∩Nα(R) +
∑
i≥1
(Nα(R) ∩ Si)x
i)).
Thus applying Lemma 1.8 with U = J(R) ∩Nα(R) and V = Nα(R) we
easily conclude that R[x;α]/J(R[x;α]) is right quasi-duo and so R[x;α]
is right quasi-duo.
60 Quasi-duo Partial skew polynomial rings
2. Quasi-duo partial skew Laurent polynomial rings
In this section we study quasi-duo partial skew Laurent polynomial
rings. Let A be the set of all maximal right ideals M of R〈x;α〉 such that
1nx
n /∈M , for some 0 6= n ∈ Z, and B the set of maximal right ideals M
of R〈x;α〉 such that 1ix
i ∈M , for all 0 6= i ∈ Z. Then for any M ∈ B we
easily have that M = (M ∩ R) ⊕
∑
i 6=0 Six
i, with Si ⊆ (M ∩ R) for all
i 6= 0. Also we write A(R〈x;α〉) =
⋂
M∈AM and B(R〈x;α〉) = ∩M∈BM .
We begin with the following easy remark.
Remark 1. Suppose thatM is an ideal ofR < x;α > such that 1jx
j /∈M ,
for some 0 6= j ∈ Z. Then 1−jx
−j /∈M .
Lemma 4. Suppose that R < x;α > is a right quasi-duo ring. Then
A(R < x;α >) = Nα(R) < x;α >.
Proof. First we show that Nα(R) = A(R〈x;α〉) ∩ R. Suppose r ∈
Nα(R) and take any i ≥ 1. Then there exists n ≥ 1 such that
rαi(r1−i)...αni(r1−ni) = 0 and hence r1ix
i ∈ R〈x;α〉 is a nilpotent
element. Since R〈x;α〉/J(R〈x;α〉) is reduced it follows that r1ix
i ∈
J(R〈x;α〉). Hence r1ix
i ∈ M , for all M ∈ A. Note that for each
M ∈ A there exists nM ≥ 1 such that 1nM
xnM /∈ M and since
r1ix
i ∈ M , for all i ≥ 0, then we have that r ∈ M , for all M ∈ A.
So r ∈ ∩M∈AM = A(R〈x;α〉).
On the other hand, let a ∈ A(R〈x;α〉)∩R. Then we have that a1ix
i ∈
J(R〈x;α〉) = K〈x;α〉, for all i ≥ 1, where K is an α-nil ideal of R. Thus
we easily obtain that a ∈ Nα(R).
From the first part we conclude that Nα(R)〈x;α〉 ⊆ A(R〈x;α〉).
Conversely, let f =
∑n
j=p ajx
j ∈ A(R〈x;α〉). Then we have that
f1ix
i ∈ J(R〈x;α〉), for all 0 6= i ∈ Z. Fix any i 6= 0 with p ≤ i ≤ n.
Then the coefficient of degree 0 in f1−ix
−i is ai. Since J(R〈x;α〉) is a
homogeneous ideal it follows that ai ∈ J(R〈x;α〉) ⊆ A(〈x;α〉), for any
p ≤ i ≤ n. So aix
i ∈ A(R〈x;α〉), for p ≤ i ≤ n. Now arguing as before we
have that ai ∈ Nα(R) and we are done.
The following definition was given in [4].
Definition 2. Let R be a ring and α a partial action of Z on R. We say
that α is of finite type if there exists j1, ..., jn ∈ Z such that for any k ∈ Z
we have that R = S−k+j1 + ...+ S−k+jn .
In the following lemma we show that when α is a partial action of finite
type do not exist maximal right ideals of R〈x;α〉 in B. So to compute the
Jacobson radical of R〈x;α〉 it is enough to compute A(R〈x;α〉).
W. Cortes, M. Ferrero, L. Gobbi 61
Lemma 5. If α is a partial action of finite type of Z on R, then B = ∅.
In particular J(R〈x;α〉) = A(R〈x;α〉).
Proof. Let I be a maximal right ideal in B. Then I = (I ∩R)⊕
∑
i 6=0 Six
i,
where I ∩R is a right ideal of R which contains Si, for all i 6= 0. By the
fact that α is of finite type we have that R =
⊕n
i=1 S1+i ⊂ I ∩R, for some
n ≥ 0. It follows that I = R < x;α >, which is a contradiction.
From now on α is a partial action of finite type of Z on R and (T, σ)
is the enveloping action of (R,α), where σ is an automorphism of T .
Now we give a precise description of the Jacobson radical of R〈x;α〉,
when R〈x;α〉 is a quasi-duo ring.
Proposition 4. If R〈x;α〉 is right quasi-duo, then
J(R〈x;α〉) = Nα(R)〈x;α〉.
In particular, Nα(R) is an α-invariant ideal of R.
Proof. The result follows from Lemmas 2.2 and 2.4.
Finally we give the main result of this section which extends ([8],
Corollary 10). The proof is an easy consequence of the previews results.
Theorem 2. R〈x;α〉 is right quasi-duo if and only if Nα(R) is an α-
invariant ideal of R,
J(R〈x;α〉) = Nα(R)〈x;α〉
and (R/Nα(R))〈x;α〉 is a commutative ring, where α is the partial action
induced by α on R/Nα(R).
3. Examples
In this section we give examples to answer some natural questions
which can be risen after the results we obtained in the paper.
Example 1. LetK be a field, T = Ke1⊕Ke2⊕Ke3, where {e1, e2, e3} are
orthogonal central idempotents. We define an automorphism σ : T → T
as follows: σ(e1) = e2, σ(e2) = e3, σ(e3) = e1 and σ|K = idk.
Now take R = Ke1 ⊕Ke2 and consider the partial action α of Z on
R defined as the restriction of σ. This means that we take Si = Ke1,
for all i ≡ 2(mod3), and Sj = Ke2, for all j ≡ 1(mod3), Sl = R, for all
l ≡ 0(mod3). Thus α is given by α1(e1) = e2, α2(e2) = e1, α3 = idR,
and so on. We clearly have that (T, σ) is the enveloping action of (R,α).
62 Quasi-duo Partial skew polynomial rings
Note that N1
α(R) = R because, for all r = a1e1 + a2e2 ∈ R, we have
that rα1(re1)α2(re2) = 0. Since α3i = idR, for all i ∈ Z, we have that
Nα(R) =
⋂
i≥1N
i(R) = 0. Moreover, we easily have that 0 = Nσ(T )
N(T ). Then, in this case,
N1
α(R) ) Nα(R) = Nσ(T ) N(T ).
The next example shows that R[x;α] may be right quasi-duo even
when T [x;σ] is not right quasi-duo.
Example 2. Let K, T and σ be as in the last example. We consider
R = Ke1 and we have a natural partial action α of Z on R as follows:
Si = R and αi = idR, for all i ≡ 0(mod3); Sj = 0 and αj = 0 otherwise.
Thus, R[x;α] =
⊕
i≥0Ke1x
3i is right quasi-duo because is commutative.
We easily have that eiσ(ei) = 0, for i = 1, 2, 3. Thus ei ∈ N(T ), for
i = 1, 2, 3. Note that
(e1 + e2)σ(e1 + e2)σ
2(e1 + e2) = (e1 + e2)(e2 + e3)(e3 + e1) = 0
and we obtain that e1 + e2 ∈ N(T ) but 1 = e1 + e2 + e3 /∈ N(T ). Hence
N(T ) is not an ideal of T and by ([7], Proposition 2.3) T [x;σ] is not right
quasi-duo.
The next example shows that the Theorem 2.5 is does not hold when
α is not of finite type.
Example 3. Let K be a field and R = Ke1 ⊕Ke2. We define a partial
action of Z on R as follows: S0 = R, Si = Ke1 for i 6= 0, α0 = idR and
αi = idSi
for i 6= 0. Note that in this case R < x;α > is a right quasi-duo
ring. We claim that Nα(R) = Ke2. In fact, e2αi(e1e2) = 0, for any i 6= 0
and we obtain that e2 ∈ Nα(R). Since αi(e1) = e1, for any i 6= 0, we
have that e1 /∈ Nα(R). Thus Nα(R) = Ke2. It is not difficult to see that
M = Ke1 +
∑
i 6=0 Six
i = Ke1〈x;α〉 is a maximal ideal of R〈x;α〉 and
B(R〈x;α〉) = {M}. Since A(R〈x;α〉) = Nα(R)〈x;α〉, then J(R〈x;α〉 =
A(R〈x;α〉) ∩ B(R〈x;α〉) = Ke2〈x;α〉 ∩Ke1〈x;α〉 = (0) 6= Nα〈x;α〉.
The next example shows that the converse of Lemma 2.4 is not true,
in general.
Example 4. Let R = ⊕n
i=−n,i 6=0Kei be a ring, where K is a field and
{ei : 1 ≤ i ≤ n, i 6= 0} is a set of orthogonal idempotents. We define a
partial action of Z of R as follows: the ideals are Si = 0 for all |i| > n,
Si = Kei for i 6= 0 and −n ≤ i ≤ n and S0 = R. The isomorphisms αi are
the zero application for all |i| > n, αi(e−i) = ei, for i 6= 0 and −n ≤ i ≤ n
and α0 = idR. We easily have that α is not of finite type. We see that
W. Cortes, M. Ferrero, L. Gobbi 63
even in this case the set of maximal ideals in B is empty. In fact, if M ∈ B,
then M ∩R contains Si for all i 6= 0 and it follows that R ⊆M ∩R. The
result follows.
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Contact information
Wagner Cortes,
Miguel Ferrero
Instituto de Matematica Universidade Federal
do Rio Grande do Sul 91509-900, Porto Alegre,
RS, Brazil
E-Mail: cortes@mat.ufrgs.br,
mferrero@mat.ufrgs.br
Luciane Gobbi Centro de Ciências Exatas e Naturais Universi-
dade Federal de Santa Maria 97105-900, Santa
Maria, RS, Brazil
E-Mail: lucianegobbi@yahoo.com
Received by the editors: 13.10.2011
and in final form 13.10.2011.
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