On partial skew Armendariz rings
In this paper we consider rings R with a partial action α of an infinite cyclic group G on R. We introduce the concept of partial skew Armendariz rings and partial α-rigid rings. We show that partial α-rigid rings are partial skew Armendariz rings and we give necessary and sufficient conditions f...
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irk-123456789-1547602019-06-16T01:31:25Z On partial skew Armendariz rings Cortes, W. In this paper we consider rings R with a partial action α of an infinite cyclic group G on R. We introduce the concept of partial skew Armendariz rings and partial α-rigid rings. We show that partial α-rigid rings are partial skew Armendariz rings and we give necessary and sufficient conditions for R to be a partial skew Armendariz ring. We study the transfer of Baer property, a.c.c. on right annhilators property, right p.p. property and right zip property between R and R[x;α]. We also show that R[x;α] and R⟨x;α⟩ are not necessarily associative rings when R satisfies the concepts mentioned above. 2011 Article On partial skew Armendariz rings / W. Cortes // Algebra and Discrete Mathematics. — 2011. — Vol. 11, № 1. — С. 23–45. — Бібліогр.: 39 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:16S36; 16S35 http://dspace.nbuv.gov.ua/handle/123456789/154760 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 an infinite cyclic group G on R. We introduce the concept of partial skew Armendariz rings and partial α-rigid rings. We show that partial α-rigid rings are partial skew Armendariz rings and we give necessary and sufficient conditions for R to be a partial skew Armendariz ring. We study the transfer of Baer property, a.c.c. on right annhilators property, right p.p. property and right zip property between R and R[x;α].
We also show that R[x;α] and R⟨x;α⟩ are not necessarily associative rings when R satisfies the concepts mentioned above. |
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Cortes, W. |
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Cortes, W. On partial skew Armendariz rings Algebra and Discrete Mathematics |
| author_facet |
Cortes, W. |
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Cortes, W. |
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On partial skew Armendariz rings |
| title_short |
On partial skew Armendariz rings |
| title_full |
On partial skew Armendariz rings |
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On partial skew Armendariz rings |
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On partial skew Armendariz rings |
| title_sort |
on partial skew armendariz 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/123456789/154760 |
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On partial skew Armendariz rings / W. Cortes // Algebra and Discrete Mathematics. — 2011. — Vol. 11, № 1. — С. 23–45. — Бібліогр.: 39 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT cortesw onpartialskewarmendarizrings |
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2025-07-14T06:51:56Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 11 (2011). Number 1. pp. 23 – 45
c© Journal “Algebra and Discrete Mathematics”
On partial skew Armendariz rings
Wagner Cortes
Communicated by V. V. Kirichenko
Abstract. In this paper we consider rings R with a partial
action α of an infinite cyclic group G on R. We introduce the concept
of partial skew Armendariz rings and partial α-rigid rings. We show
that partial α-rigid rings are partial skew Armendariz rings and we
give necessary and sufficient conditions for R to be a partial skew
Armendariz ring. We study the transfer of Baer property, a.c.c. on
right annhilators property, right p.p. property and right zip property
between R and R[x;α].
We also show that R[x;α] and R〈x;α〉 are not necessarily asso-
ciative rings when R satisfies the concepts mentioned above.
1. Introduction
Partial actions of groups have been introduced in the theory of operator
algebras giving powerful tools of their study (see [15] and [17] and the
literature quoted therein). Also in [15] the authors introduced partial
actions on algebras in a pure algebraic context. Let G be a group and
R a unital k-algebra, where k is a commutative ring. A partial action α
of G on R is a collection of ideals Sg of R, g ∈ G, and isomorphisms of
(non-necessarily unital) k-algebras αg : Sg−1 → Sg such that:
(i) S1 = R and α1 is the identity map of R;
(ii) S(gh)−1 ⊇ α−1
h (Sh ∩ Sg−1);
(iii) αg ◦ αh(x) = αgh(x), for every x ∈ α−1
h (Sh ∩ Sg−1).
2000 Mathematics Subject Classification: 16S36; 16S35.
Key words and phrases: partial actions, Armendariz rings, Baer rings and P.P
rings.
24 On partial skew Armendariz rings
The property (ii) easily implies that αg(Sg−1 ∩ Sh) = Sg ∩ Sgh, for all
g, h ∈ G. Also αg−1 = α−1
g , for every g ∈ G.
Given a partial action α of a group G on R, an enveloping action is
an algebra T together with a global action β = {σg | g ∈ G} of G on T ,
where σg is an automorphism of T , such that the partial action is given by
restriction of the global action ([15], Definition 4.2). From ([15], Theorem
4.5) we know that a partial action α has an enveloping action if and only
if all the ideals Sg are unital algebras, i.e., Sg is generated by a central
idempotent of R, for every g ∈ G. In this case the partial skew group
ring R ⋆α G is an associative algebra (this is not true in general, see ([15],
Example 3.5)).
When α has an enveloping action (T, β) we may consider that R is an
ideal of T and the following properties hold:
(i) the subalgebra of T generated by
⋃
g∈G σg(R) coincides with T and
we have T =
∑
g∈G σg(R);
(ii) Sg = R ∩ σg(R), for every g ∈ G;
(iii) αg(x) = σg(x), for every g ∈ G and x ∈ S−1
g .
The authors in [13] introduced the concepts of partial skew Laurent
polynomial rings and partial skew polynomial rings. In these cases, the
authors studied prime and maximal ideals with the assumption that the
partial action has an enveloping action. Moreover, the authors in [14]
studied Goldie property in partial skew polynomial rings and partial skew
Laurent polynomial rings.
Let R be an associative ring with identity element 1R, G an infinite
cyclic group generated by σ and {ασi : Sσ−i → Sσi , i ∈ Z} a partial action
of G on R. We simplify the notation by putting αi = ασi and Si = Sσi , for
every i ∈ Z. Then the partial skew group ring can be identified with the set
of all finite formal sums
∑m
i=−n aix
i, ai ∈ Si for every −n ≤ i ≤ m, where
the sum is usual and the multiplication rule is axibxj = αi(α−i(a)b)x
i+j .
We denote this ring by R〈x;α〉 and call it partial skew Laurent polynomial
ring. The partial skew polynomial ring R[x;α] is defined as the subring of
R〈x;α〉 whose elements are the polynomials
∑n
i=0 bix
i, bi ∈ Si for every
0 ≤ i ≤ n with usual sum of polynomials and multiplication rule as before.
In this paper, we assume that R is an associative ring and α is a partial
action of an infinite cyclic group G on R such that R[x;α] and R〈x;α〉
are not necessarily associative rings.
In the Section 2, we study McCoy’s result in partial skew Laurent
polynomial rings when a partial action α of an infinite cyclic group G on
R has an enveloping action (T, σ), where σ is an automorphism of T .
In the Section 3, we introduce the concept of partial skew Armendariz
W. Cortes 25
rings and we show that R is a partial skew Armendariz ring if and only
if the canonical map from the set of right annihilators of R to the set
of right annihilators of R[x;α] is bijective and with this result we study
the transfer of Baer property , p.p-property, ascending chain condition on
right annihilators property and right zip property between R and R[x;α].
We study necessary and sufficient conditions for R[x;α] to be reduced and
as a consequence of this result we introduce the concept of partial α-rigid
rings. We show that partial α-rigid rings are partial skew Armendariz
rings.
In the Section 4, we give examples to show that Baer rings, p.q. Baer
rings, p.p-rings and quasi-Baer rings do not imply the associativity of
the partial skew polynomial rings and partial skew Laurent polynomial
rings. When a partial action α of a group G on R has an enveloping
action (T, β), we study conditions to show the following equivalences: R
is Baer ⇔ T is Baer, R is p.q. Baer ⇔ T is p.q. Baer, R is quasi-Baer ⇔
T is quasi-Baer and R is p.p ⇔ T is p.p. When (R,α) has an enveloping
action (T, σ), where σ is an automorphism of T , we show that T is σ-rigid
⇒ R is partial α-rigid and T is skew Armendariz ⇒ R is partial skew
Armendariz. Moreover, we give an example to show that the converse of
each arrow is not true.
Throughout this article, for a non-empty subset Y of a ring S, we
denote rS(Y ) = {a ∈ S : Y a = 0} (lR(Y ) = {a ∈ S : aY = 0}) the right
(left) annihilator of Y in S.
2. Generalization of McCoy’s result
in partial skew Laurent polynomial rings
In [33], McCoy proved that if R is a commutative ring, then whenever
g(x) is a zero-divisor in R[x] there exists a non-zero element c ∈ R such
that cg(x) = 0, and in [26] it was proved that if rR[x](f(x)R[x]) 6= (0),
then rR[x](f(x)R[x]) ∩R 6= (0), where f(x) ∈ R[x] and rR[x](f(x)R[x]) =
{h(x) ∈ R[x] : f(x)R[x]h(x) = 0}. Moreover, the author in ([12], The-
orems 2.3 and 2.4), generalized these results for skew polynomial rings
of automorphism and derivation type. We study this situation in partial
skew Laurent polynomial rings but we are unable to prove an analogue
result for partial skew polynomial rings.
Throughout this section we assume that R is not necessarily a com-
mutative ring and α is a partial action of an infinite cyclic group G on R
with enveloping action (T, σ), where σ is an automorphism of T .
26 On partial skew Armendariz rings
Lemma 1. Let f(x) =
∑n
i=p bix
i and g(x) =
∑m
i=q aix
i be two elements
of R〈x;α〉. Then f(x)R
∑m
i=q αj(ai1−j)x
i+j = 0, for every j ∈ Z, if and
only if f(x)R〈x;α〉g(x) = 0.
Proof. Suppose that f(x)R
∑m
i=q αj(ai1−j)x
i+j = 0, for every j ∈ Z.
Then, we have that
f(x)rxjg(x) = f(x)r1jx
jg(x) = f(x)r
m
∑
i=q
αj(ai1−j)x
i+j = 0.
So, f(x)R〈x;α〉g(x) = 0.
The converse is trivial.
Let h(x) =
∑n
i=s aix
i ∈ R〈x;α〉. The length of h(x) is the number
len(h(x)) = n− s+ 1 and we use this number below.
Theorem 1. Let f(x) ∈ R〈x;α〉. If rR〈x;α〉(f(x)R〈x;α〉) 6= 0 then
rR〈x;α〉(f(x)R〈x;α〉) ∩ R 6= 0, where rR〈x;α〉(f(x)R〈x;α〉) = {h(x) ∈
R〈x;α〉 : f(x)R〈x;α〉h(x) = 0}.
Proof. We freely use Lemma 1 without mention. Let f(x) = apx
p + ....+
aqx
q. If either f(x) is constant or f(x) = 0 or f(x) = aqx
q, then the
assertion is clear. So, assume that q 6= 0, p < q and
rR〈x;α〉(f(x)R〈x;α〉) ∩R = (0).
Let g(x) = btx
t + ... + bmxm ∈ rR〈x;α〉(f(x)R〈x;α〉) of minimal length
with bm 6= 0. Then
f(x)R〈x;α〉g(x) = 0,
and we have that f(x)1−qx
−qR
∑m
i=t αj(bi1−j)x
i+j = 0, for every j ∈ Z.
Thus, aqRαj(1−jbm) = 0, for every j ∈ Z. Hence, aqR〈x;α〉bm = 0 and
we have that
aqR〈x;α〉g(x) = aqR〈x;α〉(btx
t + ...+ bm−1x
m−1).
So,
f(x)R〈x;α〉aqR〈x;α〉(btx
t + ...+ bm−1x
m−1)) =
= f(x)(R〈x;α〉aqR〈x;α〉)g(x) = 0,
for every j ∈ Z, and we obtain that
f(x)R〈x;α〉aqR(αj(bt1−j)x
t+j + ...+ αj(bm−11−j)x
m−1+j) = 0,
W. Cortes 27
for every j ∈ Z. Note that by choice of g(x), we have that aqR(αj(bt1−j)x
t+j+
...+ αj(bm−11−j)x
m−1+j)) = 0. Thus,
aqR〈x;α〉(αj(bt1−j)x
t+j + ...+ αj(bm−11−j)x
m−1+j) = 0
and it follows that
aq ∈ lR(R〈x;α〉(αj(bt1−j)x
t+j + ...+ αj(bm−11−j)x
m−1+j)+
+R〈x;α〉(bmxm)),
for every j ∈ Z. Hence,
(apx
p + ....+ aq−1x
q−1)R〈x;α〉g(x) = (0),
and we obtain that aq−1Rαj(1−jbm) = 0, for every j ∈ Z. Consequently,
(apx
p + ....+ aq−1x
q−1)R〈x;α〉aq−1R〈x;α〉
∑m−1
i=t bix
i =
f(x)(R〈x;α〉aq−1R〈x;α〉)g(x) = (0).
By choice of g(x), we have that
aq−1R〈x;α〉g(x) = (0)
and we get
aq−1 ∈ lR(R〈x;α〉(αj(bt1−j)x
t+j + ...+ αj(bm−11−j)x
m−1+j)+
+R〈x;α〉(bmxm))
for every j ∈ Z. Now, repeating this process we obtain that
as ∈ lR(R〈x;α〉(αj(bt1−j)x
t+j + ...+ αj(bm−11−j)x
m−1+j)+
+R〈x;α〉(bmxm))
for every p ≤ s ≤ q and j ∈ Z. Since asR〈x;α〉bmxm = (0), for every p ≤
s ≤ q, then (apx
p+....+aqx
q)R〈x;α〉bm = (0). This is a contradiction.
28 On partial skew Armendariz rings
3. On partial skew Armendariz ring
Rege and Chhawchharia introduced the notion of an Armendariz ring,
see [9]. A ring R is called Armendariz if whenever polynomials
n
∑
i=0
aix
i,
m
∑
i=0
bix
i ∈ R[x] satisfy f(x)g(x) = 0, then aibj = 0, for every 0 ≤ i ≤ n and
0 ≤ j ≤ m. The name Armendariz ring was chosen because Armendariz,
showed that a reduced ring (i.e., a ring without nonzero nilpotent elements)
satisfies this condition, see [2]. Some properties of Armendariz rings have
been studied by many authors, see [2], [1], [30] and the literature quoted
therein. The authors in [27], introduced the notion of skew Armendariz
rings, they gave examples and investigated the properties of these rings.
In this section, we assume that R is an associative ring and α is a
partial action of an infinite cyclic group G on R such that (R,α) does not
necessarily have an enveloping action, unless otherwise stated.
Let S be a ring with an automorphism τ . Following [27], S is said to
be τ -rigid if aτ(a) = 0 implies that a = 0. Suppose that a partial action α
of an infinite cyclic group G on R has an enveloping action (T, σ), where
σ is an automorphism of T . In this case, a natural generalization of the
concept mentioned above it would be aα1(a1−1) = 0 ⇒ a = 0 and this is
equivalent to aσ(a) = 0 ⇒ a = 0. The next result shows that this natural
generalization implies that the partial action is a global action.
Proposition 1. Let α be a partial action of an infinite cyclic group G on
R and (T, σ) the enveloping action of (R,α), where σ is an automorphism
of T . If for every a ∈ R, aσ(a) = 0 ⇒ a = 0, then R = T .
Proof. We claim that for each t ∈ T such that tσ(t) = 0 we have that
t = 0. In fact, we easily have that t1Rα1(t1R1−1) = 0, which implies that
t1R = 0. Note that for each i > 0 we have that
(tσi(1R))σ(tσ
i(1R)σ
i−1(1R)σ
i(1R)) = 0.
Thus, we obtain that σ−i(tσi(1R))σ(σ
−i(tσi(1R)))1Rσ(1R) = 0, which
implies that σ−i(tσi(1R))σ(σ
−i(tσi(1R))) = 0 and by assumption we have
that σ−1(tσi(1R)) = 0. Hence, tσi(1R) = 0, for every i ≥ 0.
Now, let b = tσ−i(1R), for i > 0. Then bσ(bσ−i(1R)σ
−i−1(1R)) = 0.
Proceeding with similar method as above we obtain that tσ−i(1R) = 0,
for every i > 0. So, tT = 0 and by Remark 2.5 of [16] we have that t = 0.
Let T 1 = T ⊕ Z with usual sum and multiplication defined by the
rule (a, z)(b, z1) = (ab+ az1 + bz, zz1). Note that (0, 1) is the identity of
T 1. We can extend the automorphism σ of T to T 1 by the rule: σ(a, z) =
W. Cortes 29
(σ(a), z). We claim that T 1 is a σ-rigid ring, i.e, if (a, z) ∈ T 1 such that
(a, z)σ(a, z) = 0, then we have that (a, z) = 0. In fact, for each (a, z) ∈ T 1
such that (a, z)σ(a, z) = (0, 0) we have that aσ(a) = 0. Thus, a = 0 and
we obtain that (a, z) = (0, 0). Hence, by ([27], Corollary 4) and ([24],
Lemma 4) we have that all the idempotents of T 1 are invariant by σ. Since
for any idempotent e ∈ T , the element (e, 0) is an idempotent in T 1, then
σ(e) = e. So, σ(1R) = 1R and we have that R = T .
Remark 1. Let α be a partial action of an infinite cyclic group G on R
with an enveloping action (T, σ), where σ is an automorphism of T . It is
natural to ask if R[x;α] were reduced then we would have that R = T .
This is not true in general as show the following example: Let K be a field,
{ei : i ∈ Z} a set of central orthogonal idempotents, T = ⊕i∈ZKei with
an automorphism σ defined by σ(ei) = ei+1, for every i ∈ Z and R = Ke0.
We clearly have that the automorphism σ induces a partial action of the
group G =< σ > as follows: S0 = R, Si = 0, for every i ∈ Z\{0} and the
maps α0 = idR and αi = 0, for every i ∈ Z\{0}. Note that R[x;α] = R
and R & T .
In the next proposition we give necessary and sufficient conditions for
partial skew polynomial rings to be reduced rings and this generalizes
([27], Proposition 3).
Proposition 2. R[x;α] is reduced if and only if for every an ∈ Sn with
α−n(an)an = 0, n ≥ 0 implies that an = 0.
Proof. Suppose that R[x, α] is reduced. Let aj ∈ Sj such that α−j(aj)aj =
0, for j ≥ 0. Then ajx
jajx
j = αj(α−j(aj)aj)x
2j = 0. Hence, by assump-
tion we have that axj = 0. So, aj = 0.
Conversely, let f(x) =
∑n
i=0 aix
i be a polynomial in R[x;α] such that
(f(x))2 = 0. Then, αn(α−n(an)an) = 0 and by assumption we have that
an = 0. Proceeding in a similar way we obtain that a0 = a1 = ... = an = 0.
So, f = 0.
Using the Proposition 2 we can generalize the notion of σ-rigid rings
as follows.
Definition 1. Let G be an infinite cyclic group and α a partial action of
G on R. We say that R is a partial α-rigid ring if for every a ∈ Sn with
α−n(a)a = 0, n ≥ 0, we have that a = 0.
Remark 2. Note that if R is partial α-rigid, then R is reduced.
The following definition generalizes ([27], Definition).
30 On partial skew Armendariz rings
Definition 2. We say that R is a partial skew Armendariz ring if given
f(x) =
∑n
i=0 aix
i and g(x) =
∑m
i=0 bix
i in R[x;α] such that f(x)g(x) = 0,
then α−i(ai)bj = 0, for every 0 ≤ i ≤ n and 0 ≤ j ≤ m.
Proposition 3. Suppose that (R,α) has an enveloping action (T, σ),
where σ is an automorphism of T . If R is a partial α-rigid ring then R is
a partial skew Armendariz ring.
Proof. Let f(x) =
∑n
i=0 aixi and g(x) =
∑n
j=0 bjxj be polynomi-
als in R[x;α] such that f(x)g(x) = 0. Then a0b0 = 0. Note
that for degree 1 we have that a0b1 + a1α1(b01−1) = 0 and
we obtain that α−1(a1)b0a1α1(b01−1) = 0. Thus, we have that
α−1(a1α1(b01−1))a1α1(b01−1)) = α−1(a1)b0a1α1(b01−1) = 0. Since R
is partial α-rigid then 0 = a1α1(b01−1) = α−1(a1)b0 and consequently
a0b1 = 0. Now, for degree 2 we have that
a0b2 + a1α1(b11−1) + a2α2(b01−2) = 0 (*)
Multiplying (*) on the right side by b1 and using the equality a0b1 = a0b0 =
0 we have that b0a2α2(b01−2) = 0. Thus, α−2(a2)b0a2α2(b01−2) = 0 and
we obtain that α−2(a2α2(b01−2))a2α2(b01−2) = α−2(a2)b0a2α2(b01−2) =
0. Hence, a2α2(b01−2) = α−2(a2)b0 = 0. So, b1a1α1(b11−1) = 0. Pro-
ceeding with similar method as before we have that 0 = a1α1(b11−1) =
α−1(a1)b1 and it follows that a0b2 = 0. Now, using a standard induction
we have the desired result.
Now, we recall some information on rings of quotients of a semiprime
ring R′ that we need in the sequel. For background on this subject we
refer the reader to [34], Section 24; [38], Chap. 9.
An ideal H of a semiprime ring R′ is essential as a two sided ideal if
rR′(H) = 0. The set of all essential ideals of R′ will be denoted by E =
E(R′). Note that E is a filter of ideals which is closed under multiplication
and intersection. If I is an ideal of R′, then I ⊕AnnR′(I) ∈ E .
Denote by Q = QE the ring of right quotients of R′ with respect to
the filter E , i.e., the Martindale ring of right quotients of R′. Recall that
the elements of Q arise from right R′-homomorphisms from H ∈ E to R′:
for any H ∈ E and a right R′-homomorphism f : H → R′ there exists
q ∈ Q such that qh = f(h), for every h ∈ H, and conversely, if q ∈ Q
there exists F ∈ E such that qF ⊆ R′. Also, elements q, p ∈ Q are equal
if and only if they coincide on some essential ideal of R′ and if qH = 0,
for some q ∈ Q and H ∈ E , then q = 0.
W. Cortes 31
Given an ideal I of R′, the closure of I in R′ as defined in [22] is
[I] = {x ∈ R′ | there existsH ∈ E(R′) withxH ⊆ I} =
{x ∈ R′ | there existsH ∈ E(R′) withHx ⊆ I}.
Note that [I] = rR′(rR′(I)) and if [I] = I, then I is a closed ideal.
Let R be a partial α-rigid ring. Then R is reduced and we have that R
has a Martindale ring of right quotients Q constructed as before. Following
[21] the partial action α can be extended to a partial action α∗ of G on Q,
where the ideals S∗
i are the extension of [Si] to Q. The ideal S∗
i for every
i ∈ Z is generated by a central idempotent 1∗i . Let Qs be the Martindale’s
right symmetric ring of quotients of R, i.e, according ([31], Proposition
5.14.7) we have that Qs = {q ∈ Q : qJ ⊆ R and Jq ⊆ R for some J ∈
E}. Now, let (I∗)s = {q ∈ Q : qJ ∪ Jq ⊆ I for some J ∈ E} and using
similar methods of ([21], Proposition 2.2) we have that Q(I)s ≃ (I∗)s,
where Q(I)s is the Martindale’s right symmetric ring of quotients of I.
Moreover, if I∗ is generated by a central idempotent, then we have that
(I∗)s = I∗ ∩Qs(R) and we denote (I∗)s = I∗∗.
By ([31], Proposition 5.14.17) we have that 1∗i ∈ Z(Q) = Z(Qs) and
1∗i ∈ (I∗)s, for every i ∈ Z. Thus S∗∗
i = Qs1∗i , for every i ∈ Z. Now, using
similar methods of ([21], Proposition 2.3 and Theorem 3.1), we can extend
the partial action α to a partial action α∗∗ = {α∗∗
i : S∗∗
−i → S∗∗
i : i ∈ Z} of
G on Qs and α∗∗
g = α∗
g|S∗∗
−i
, for every i ∈ Z. Moreover, by ([15], Theorem
4.5) we have that (Qs, α∗∗) has an enveloping action (T ′′, σ′′), where σ′′
is an automorphism of T ′′. We use these facts in the next result.
Proposition 4. R is partial α-rigid if an only if Qs is partial α∗∗-rigid.
Proof. Suppose that R is partial α-rigid. Let q ∈ S∗∗
n such that α∗∗
−n(q)q =
0. Since q ∈ S∗∗
−n ⊆ S∗
n, then there exists an essential ideal L of R such
that qL ⊆ Sn. Thus, for every l ∈ L we have that α∗∗
−n(q)ql = 0. By ([31],
Exercise 14.4) Qs is reduced, since R is reduced. Hence, qlα∗∗
−n(q) = 0,
which implies that 0 = qlα∗∗
−n(q)α
∗∗
−n(l1
∗
n) = qlα∗∗
−n(ql). Since α∗∗ = α∗|S∗∗
n
,
then by ([21], Theorem 3.1) we have that α∗
−n restricted to S−n is α−n and
we obtain that α∗∗ restricted to Sn is αn. So, 0 = qlα∗∗
−n(ql) = qlα−n(ql)
and by assumption we have that ql = 0, for every l ∈ L. Consequently,
qL = 0. Therefore, q=0.
The converse is trivial.
In the next result we show that partial α-rigid rings are partial skew
Armendariz rings and it generalizes ([27], Corollary 4).
32 On partial skew Armendariz rings
Theorem 2. If R is partial α-rigid, then R is partial skew Armendariz.
Proof. By Proposition 4, Qs is partial α∗∗-rigid and by Proposition 3, Qs
is partial skew Armendariz. Since R is a subring of Qs and α∗∗
i |Si
= αi,
we have that R is partial skew Armendariz.
Remark 3. By Example 4.5 the converse of the Theorem 3.9 is not true.
In the next two results we give examples of partial skew Armendariz
rings. The next result generalizes ([27], Proposition 10) with similar proof.
Proposition 5. Let D be a domain and α a partial action of an infinite
cyclic group G on D. Then D is partial skew Armendariz.
One may ask if there exits a partial action of a group G on a domain
S. The next example shows that such partial action exists.
Example. Assume that D is a domain, that is not a division ring,
with identity element, σ is an automorphism of D such that σi 6= idD,
for any i 6= 0 and I a two-sided ideal of D. For any integer i we define
Si = I ∩ σi(I) and αi : S−i → Si as the restriction of σ−i to S−i. Then it
is easy to see that α = {αi | i ∈ Z} is a partial action of the infinite cyclic
group G =< σ > on I.
Let M be a (R,R)-bimodule. Then the trivial extension of R by M is
the ring T (R,M) = R⊕M with usual sum and the following multiplication:
(r,m)(s, n) = (rs, rn + ms). This ring is isomorphic to the ring of all
matrices
[
r m
0 r
]
, where r ∈ R and m ∈ M with usual matrix sum and
multiplication.
Let α = {αi : S−i → Si} be a partial action of an infinite cyclic group
G on R and T (R,R) the trivial extension of R. Then we can extend the
partial action α to T (R,R) as follows: S̄i = T (Si, Si) and ᾱi : S̄−i → S̄i
with ᾱi(a, b) = (αi(a), αi(b)), for every i ∈ Z. Since T (R, 0) is isomorphic
to R, we can identify the restriction of ᾱi to T (R, 0) with αi, for every
i ∈ Z. The next proposition provides examples of partial skew Armendariz
rings that are not semi-prime rings and it generalizes ([27], Proposition
15).
Proposition 6. If R is a partial α-rigid ring, then T (R,R) is a partial
skew Armendariz ring.
Proof. Let f(x) =
∑n
i=0(ai, bi)x
i and g(x) =
∑n
j=0(cj , dj)x
j be elements
in T (R,R)[x;α] such that f(x)g(x) = 0. We easily have that f(x) =
(p0, p1), g(x) = (q0, q1) and 0 = f(x)g(x) = (p0q0, p0q1 + p1q0), where
p0 =
∑n
i=0 aix
i, p1 =
∑n
i=0 bix
i, q0 =
∑n
j=0 cjx
j and q1 =
∑n
j=0 djx
j .
W. Cortes 33
Note that q0p0 = 0 and using similar ideas of ([27] Proposition 15)
we obtain that p0q1 = p1q0 = 0. Hence, by Theorem 2, α−i(ai)cj =
α−i(ai)dj = α−i(bi)cj = 0, for every 0 ≤ i ≤ n and 0 ≤ j ≤ n. So,
(α−i(ai), α−i(bi))(cj , dj) = (0, 0), for every 0 ≤ i ≤ n and 0 ≤ j ≤ n.
A right annihilator of a non-empty subset X of a not necessarily
associative ring S′ is defined by rS′(X) = {a ∈ S : Xa = 0}. Since S′ is
not necessarily associative, then rS′(X) is not necessarily a right ideal. If
we have an associative ring S1 contained in S, then for every non-empty
subset Y of S1 we have rS1
(Y ) = rS(Y ) ∩ S1. We put rAnnR(2
R) =
{rR(U) : U ⊆ R} and for a not necessarily associative ring S′ we put
analogously rAnnS′(2S
′
) = {rS′(U) : U ⊆ S′}.
Lemma 2. Let U be a non-empty subset of R. Then
rR[x;α](U) = rR(U)R[x;α].
Proof. Let f(x) =
∑n
i=0 aix
i ∈ R[x;α] such that Uf(x) = 0. Then
Uai = 0 for every 0 ≤ i ≤ n and it follows that ai ∈ rR(U) for every
0 ≤ i ≤ n. Thus, f(x) =
∑n
i=0 aix
i ∈ rR(U)R[x;α]. Hence, rR[x;α](U) ⊆
rR(U)R[x;α] and we easily have that rR(U)R[x;α] ⊆ rR[x;α](U). So,
rR[x;α](U) = rR(U)R[x;α].
From Lemma 2 we have the maps
φ : rAnnR(2
R) → rAnnR[x;α](2
R[x;α])
defined by φ(I) = IR[x;α], for every I ∈ rAnnR(2
R) and
Ψ : rAnnR[x;α](2
R[x;α]) → rAnnR(2
R)
defined by Ψ(J) = J ∩R, for every J ∈ rAnnR[x;α](2
R[x;α]). Obviously, φ
is injective and Ψ is surjective. Clearly φ is surjective if and only if Ψ is
injective, and in this case φ and Ψ are the inverses of each other.
The following result generalizes ([11], Proposition 3.2).
Lemma 3. The following conditions are equivalent:
(i) R is a partial skew Armendariz ring.
(ii) φ : rAnnR(2
R) → rAnnR[x;α](2
R[x;α]) is bijective.
Proof. Suppose that R is a partial skew Armendariz ring. It is only
necessary to show that φ is surjective. For every f(x) =
∑n
i=0 aix
i ∈
R[x;α] we define Cf(x) = {α−i(ai), 0 ≤ i ≤ n} and for a subset S of
R[x;α] we denote the set ∪
f(x)∈S
Cf(x) by CS . We claim that rR[x;α](f(x)) =
34 On partial skew Armendariz rings
rR[x;α](Cf(x)). In fact, let g(x) =
∑m
i=0 bix
i ∈ rR[x;α](f(x)). Then, we have
that f(x)g(x) = 0. By assumption, we have that α−i(ai)bj = 0, for every
0 ≤ i ≤ n and 0 ≤ j ≤ m. Thus, g(x) ∈ rR[x;α](Cf(x)).
On the other hand, let h(x) =
∑p
i=0 cix
i be an element in R[x;α] such
that Cf(x)h(x) = 0. Then we have that α−i(ai)ck = 0, for every 0 ≤ i ≤ n
and 0 ≤ k ≤ p, which implies that f(x)h(x) = 0.
Since R is partial skew Armendariz, then rR[x;α](S) =
rR[x;α]( ∪
f(x)∈S
Cf(x)) and by Lemma 2 we have that rR[x;α](Cf(x)) =
rR(Cf(x))R[x;α]. Hence,
rR[x;α](S) = ∩
f(x)∈S
rR[x;α](f(x)) = ∩
f(x)∈S
rR[x;α](Cf(x)) =
= ( ∩
f(x)∈S
rR(Cf(x)))R[x;α] = rR(CS)R[x;α].
So, φ is surjective.
Conversely, let f(x) =
∑n
i=0 aix
i and g(x) =
∑m
i=0 bix
i be ele-
ments in R[x;α] such that f(x)g(x) = 0. Then, by assumption, g(x) ∈
rR[x,α](f(x)) = BR[x;α], for some right ideal B of R. Thus, we have that,
bi ∈ B ⊂ rR[x;α](f(x)), for every 0 ≤ i ≤ m. Hence, α−i(ai)bj = 0, for
every 0 ≤ i ≤ n and 0 ≤ j ≤ m. So, R is a partial skew Armendariz
ring.
The next lemma generalizes ([27], Corollary 19) and the proof is similar.
Lemma 4. Let R be a partial skew Armendariz ring and e =
∑n
i=0 eix
i ∈
R[x;α]. If e2 = e, then e = e0.
The following definition appears in [26].
Definition 3. A ring R is called a Baer ring, if the left annihilator of
each subset of R is generated by an idempotent.
Remark 4. Note that the definition of a Baer ring is left-right symmetric.
The notion of right ideals in non-associative rings is the same as in
associative rings. Moreover, if S′ is not necessarily associative, we have
for each a ∈ S′ the set aS′ : {as : s ∈ S′}. We use these facts in the next
results.
The proofs of the next four theorems are similar with ([11], Theorem
3.6), ([11], Theorem 3.8), ([11], Proposition 3.9) and ([12], Theorem 2.8)
respectively, and we put their proofs here for the reader’s convenience.
The next theorem generalizes ([11], Theorem 3.6).
W. Cortes 35
Theorem 3. Suppose that R is a partial skew Armendariz ring. Then the
following conditions are equivalent:
(i) R is a Baer ring.
(iii) R[x;α] satisfies the following property: for every non-empty subset
U of R[x;α] we have that rR[x;α](U) = eR[x;α], for some idempotent
e ∈ R.
Proof. Suppose that R is a Baer ring. Let ∅ 6= U be a subset of R[x;α] and
f(x) =
∑n
i=0 aix
i ∈ U . We define Cf(x) = {α−i(ai), for every 0 ≤ i ≤
n}, and CU = ∪
f(x)∈U
Cf(x). By assumption, we have that rR(CU ) = eR,
with e2 = e. It is easy to see that eR[x;α] ⊂ rR[x;α](U). By Lemma 3, we
have
rR[x;α](U) = rR(CU )R[x;α] = eR[x;α].
Conversely, let ∅ 6= A be a subset of R. Then, by assumption we
have that rR[x;α](A) = eR[x;α], where e2 = e. By Lemma 4, e = e0 is a
constant polynomial. Thus,
rR[x;α](A) = eR[x;α] = e0R[x;α].
Hence, rR[x;α](A)∩R = e0R[x;α]∩R = e0R. So, R is a Baer ring.
Definition 4. A ring R is called left (right) p.p if the left (right) anni-
hilator of every element is generated by an idempotent as a left (right)
ideal.
The next theorem generalizes ([11], Theorem 3.8).
Theorem 4. Suppose that R is a partial skew Armendariz ring. Then the
following conditions are equivalent:
(i) R is a right p.p-ring.
(ii) R[x;α] satisfies the following property: for every polynomial f(x)
of R[x;α] we have rR[x;α](f(x)) = eR[x;α], for some idempotent e ∈ R.
We recall that a ring S satisfies the ascending chain condition on
right annihilator ideals if for any chain rS(Y1) ⊆ rS(Y2) ⊆ ..., there exists
i ≥ 1 such that rS(Yi) = rS(Yi+p), for all p ≥ 0, where Yj are non-empty
subsets of S for every j ≥ 0. Moreover, for a not necessarily associative
rings S′ satisfies the ascending chain conditions on right annihilators if
rS′(Y ′
1) ⊆ rS′(Y ′
2) ⊆ ..., there exists i ≥ 1 such that rS′(Y ′
i ) = rS′(Y ′
i+p),
for all p ≥ 0, where Y ′
j are non-empty subsets of S′ for every j ≥ 0. In the
next proposition, we suppose that R is a partial skew Armendariz ring
36 On partial skew Armendariz rings
to study the transfer of ascending chain condition on right annihilators
property between R and R[x;α] and this generalizes ([11], Proposition
3.9).
Theorem 5. Suppose that R is a partial skew Armendariz ring. Then the
following conditions are equivalent:
(i) R satisfies ascending chain condition on right annihilator ideals.
(ii) R[x;α] satisfies the ascending chain condition on right annihilators.
Proof. Suppose that R[x;α] satisfies the ascending chain condition on
right annihilators. We consider the chain, rR(U1) ⊂ rR(U2) ⊂ .., where
Ui ⊂ R, for every i ≥ 1. We claim that rR[x;α](Ui) ⊂ rR[x;α](Ui+1). In fact,
let g(x) =
∑n
i=0 cix
i be an element of R[x;α], such that Uig(x) = (0).
Then, Uicj = (0), and we obtain that Ui+1cj = (0). Hence, by assumption
there exists l ≥ 1 such that rR[x;α](Ul) = rR[x;α](Ul+p), for every p ∈ N .
So,
rR(Ul) = rR[x;α](Ul) ∩R = rR[x;α](Ul+p) ∩R = rR(Ul+p),
for every p ∈ N.
Conversely, suppose that rR[x;α](V1) ⊂ rR[x;α](V2) ⊂ .... By Lemma 3,
rR[x;α](Vi) = rR(CVi
)R[x;α], where CVi
= ∪
f(x)∈Vi
Cf(x),
Cf(x) = {α−i(ai), for every 0 ≤ i ≤ m}
and f(x) =
∑m
i=0 aix
i. We claim that rR(CVi
) ⊂ rR(CVi+1
). In fact,
let y ∈ rR(CVi
). Then we have that CVi
y = 0. Thus, f(x)y = 0, for
every f(x) ∈ Vi. Hence, CVi+1
y = 0 and by assumption, there exists
n ≥ 1 such that rR(CVn) = rR(CVn+k
), for every k ∈ N. So, rR[x;α](Vn) =
rR[x;α](Vn+k), for every k ∈ N.
In [18], Faith called a ring R right zip if the right annihilator rR(U)
of a non-empty subset U of R is zero implies that rR(Y ) = 0 for every
non-empty finite subset Y ⊆ U . Equivalently, for a left ideal L of R with
rR(L) = 0, there exists a finitely generated left ideal L1 ⊆ L such that
rR(L1) = 0. R is zip if it is right and left zip. The concept of zip rings
was initiated by Zelmanowitz [39] and appeared in various papers [3], [6],
[7], [8], [18] [19], and references therein. Zelmanowitz stated that any ring
satisfying the descending chain condition on right annihilators ideals is
a right zip ring (although not so-called at that time), but the converse
does not hold. Extensions of zip rings were studied by several authors.
Beachy and Blair [3] showed that if R is a commutative zip ring, then
the polynomial ring R[x] over R is zip. The authors in [25] proved the
W. Cortes 37
following result: suppose that R is Armendariz ring. Then R is a right
(left) zip ring if and only if R[x] is a right (left) zip ring. The next result
generalizes ([12], Theorem 2.8).
Theorem 6. Suppose that R is a partial skew Armendariz ring. Then the
following conditions are equivalent:
(i) R is a right zip ring.
(ii) R[x;α] satisfies the following property: for every non-empty subset
U of R[x;α] such that rR[x;α](U) = 0, we have that rR[x;α](Y ) = 0 for any
non-empty finite subset Y of X.
Proof. Suppose that R is a right zip ring. Let U be a non-empty subset
of R[x;α] such that rR[x;α](U) = 0. Then, by Lemma 3, rR[x;α](U) =
rR(CU )R[x;α], where CU = ∪f(x)∈UCf(x) and Cf(x) = {α−i(ai) : 0 ≤ i ≤
n} with f(x) =
∑n
i=0 aix
i ∈ U . Thus, rR(CU ) = 0 and by assumption for
any non-empty finite subset W = {α−i1(ai1), ..., α−in(ain)} of CU we have
that rR(W ) = 0. For every α−ij (aij ) ∈ W there exists gaij (x) ∈ U such
that some of the coefficients of gaij (x) are aij , for every 1 ≤ j ≤ n. Let
U0 be a minimal subset of U such that gaij (x) ∈ U0, for every 1 ≤ j ≤ n.
Then U0 is non-empty finite subset of U . We denote W0 = ∪f(x)∈U0
(Cf(x))
and note that W ⊆ W0. Hence, rR(W0) ⊆ rR(W ) = 0 and, by Lemma 3,
rR[x;α](U0) = rR(W0)R[x;α]. So, rR[x;α](U0) = 0.
Conversely, let Y be a non-empty subset of R such that rR(Y ) = 0
and f(x) =
∑n
i=0 aix
i ∈ rR[x;α](Y ). Then we have that ai ∈ rR(Y ) = 0
and it follows that f(x) = 0. Thus, by assumption for any non-empty
finite subset Y1 = {y0, ..., yn} of Y we have that rR[x;α](Y1) = 0. Hence,
rR(Y1) = rR[x;α](Y1) ∩R = (0).
4. Examples
In this section, we give an example to show that the partial skew Laurent
polynomial rings and the partial skew polynomial rings are not necessarily
associative rings if the base ring satisfy either Baer or quasi-Baer or p.q.
Baer or p.p. Moreover, for a ring R with a partial action α of a group
G with enveloping action (T, β) we study the transfer of Baer property,
quasi-Baer property, p.q. Baer property and p.p property between R
and T .
We begin with the following proposition.
Proposition 7. Let α be a partial action of an infinite cyclic group G on
R and (T, σ) its enveloping action, where σ is an automorphism of T . If
T is σ-rigid, then R is partial α-rigid.
38 On partial skew Armendariz rings
Proof. Let a ∈ Sn such that α−n(a)a = 0. Then σ−n(a)a = 0, which
implies that aσn(a) = 0. Thus, by ([28], Lemma 4) we have that a2 = 0.
So, a = 0, since T is reduced.
The next example shows that the converse of the proposition above is
not true.
Example 1. Let K be a field and T = ⊕i∈ZKei, where {ei, i ∈ Z} are
orthogonal central idempotents. We define an action of an infinite cyclic
group G generated by σ as follows: σ|K = idK and σ(ei) = ei+1. Assume
that R = Ke0 and we have a partial action α of G on R. Note that R
is partial α-rigid, but T is not σ-rigid, because e2σ(e2) = e2e3 = 0 and
e2 6= 0.
Let T ′ be a ring and σ′ an endomorphism of T ′. The skew polyno-
mial ring of endomorphism type T ′[x;σ′] is the set of all finite formal
sums
∑n
i=0 aix
i with usual sum and the multiplication rule is axibxj =
aσi(b)xi+j . Next, we recall the definition that appears in ([27], Definition).
Definition 5. Let T ′ be a ring and σ′ an endomorphism of T ′. The ring
T ′ is said to be skew Armendariz ring if given f(x) =
∑n
i=0 aix
i and
g(x) =
∑m
i=0 bix
i in T [x;σ′] such that f(x)g(x) = 0, then aiσ
i(bj) = 0,
for every 0 ≤ i ≤ n and 0 ≤ j ≤ m.
When a partial action α of G on R has an enveloping action (T, σ)
one may ask if R were a partial skew Armendariz ring, then T would be
a skew Armendariz ring, but the next example shows that this is not true
in general.
Example 2. Let K, T , σ, R and α as in the Example 1. We clearly have
that R is a partial skew Armendariz ring, but T is not skew Armendariz
since for f = e−1 + e−1x and g = −e−2 + e−1x in T [x;σ] we obtain that
fg = 0 and e−1e−1 6= 0.
The next example shows that the partial skew Armendariz property
does not imply the associativity of the partial skew polynomial rings and
partial skew Laurent polynomial rings. Moreover, the next example shows
that the converse of Theorem 3.9 is not true.
Example 3. Let R = K[X,Y ]/(X2, Y 2), where K is a field. Then R =
K1 +Kx+Ky +Kxy, where x and y represent the classes of X and Y
in R. Note that R is a four dimensional K-vector space. Let G =< σ >
be an infinite cyclic group generated by σ and we define the partial action
α of G on R as follows: let I = Kx + Kxy, Si = I and αi : S−i → Si
W. Cortes 39
defined by αi(x) = xy, αi(xy) = x and αi|K = idK , for every i ∈ Z \ {0}
(by definition S0 = R and α0 = idR). We claim that R is a partial skew
Armendariz ring. In fact, let f(z) = (γ0+γ1x+γ2y+γ3xy)+(a0x+b0xy)z
and g(z) = (θ0+θ1x+θ2y+θ3xy)+(a1x+b1xy)z be polynomials of degree
1 in R[z;α] such that f(z)g(z) = 0. Then (γ0 + γ1x+ γ2y + γ3xy)(θ0 +
θ1x+ θ2y+ θ3xy) = 0, which implies γ0θ0 = 0. Thus, we have the followin
g cases:
Case 1: If γ0 = 0 and θ0 6= 0, then we have that f(z) = 0.
Case 2: If γ0 6= 0 and θ0 = 0, then we have that g(z) = 0.
Case 3: If γ0 = θ0 = 0, then we have that (γ1θ2 + γ2θ1)xy = 0 and
γ2a1x = b0θ2xy = 0. Thus, γ2a1 + b0θ2 = 0, γ2a1 = 0 and b0θ2 = 0, which
implies either γ2 = 0 or a1 = 0 and either b0 = 0 or θ2 = 0. Hence, we
have the following cases:
Case 3.1: If γ2 = 0, a1 6= 0, b0 = 0 and θ2 6= 0, then we have that
[(γ1x + γ3xy) + (a0x)z][(θ1x + θ2y + θ3xy) + (a1x + b1xy)z] = 0. Thus,
(γ1x+ γ3xy)(θ1x+ θ2y + θ3xy) = 0, α1(α−1(a0x)(θ1x+ θ2y + θ3xy)) =
α1((a0xy)(θ1x+ θ2y + θ3xy)) = 0 and α1(α−1(a0x)(a1x+ b1xy)) = 0.
Case 3.2: If γ2 = 0, a1 6= 0, b0 6= 0, θ2 = 0, then we have that
[(γ1x+ γ3xy) + (a0x+ b0xy)z][(θ1x+ θ3xy) + (a1x+ b1xy)z] = 0. Thus,
α1(α−1((a0x + b0xy))(θ1x + θ3xy)) = α1((a0xy + b0x)(θ1x + θ3xy)) =
0, (γ1x + γ3xy)(a1x + b1xy) = 0, (γ1x + γ3xy)(θ1x + θ3xy) = 0 and
α1(α−1((a0x+ b0xy))(a1x+ b1xy)) = 0.
Case 3.3: If γ2 6= 0, a1 = 0, θ2 = 0, b0 6= 0, then we have that
[(γ1x + γ2y + γ3xy) + (a0x)z][(θ1x + θ2y + θ3xy) + (b1xy)z] = 0. Thus,
α1(α−1(a0x)(θ1x + θ2y + θ3xy)) = α1(a0xy(θ1x + θ2y + θ3xy)) = 0,
α1(α−1(a0x)(b1xy)) = 0, (γ1x+ γ2y + γ3xy)(b1xy) = 0 and (γ1x+ γ2y +
γ3xy)(θ1x+ θ2y + θ3xy) = 0.
Case 3.4: If γ2 6= 0, a1 = 0, θ2 6= 0, b0 = 0, then we have
that [(γ1x + γ2y + γ3xy) + a0xz][(θ1x + θ2y + θ3xy) + (b1xy)z] =
0. Thus, α1(α−1(a0x)(θ1x + θ3xy)) = α1((a0xy)(θ1x + θ3xy)) = 0,
α1(α−1(a0x)(b1xy)) = 0, (γ1x+ γ2y + γ3xy)(b1xy) = 0 and (γ1x+ γ2y +
γ3xy)(θ1x+ θ3xy) = 0.
Case 3.5: If γ2 = 0, a1 = 0, θ2 6= 0, b0 = 0, then we have that [(γ1x+
γ3xy)+(a0x)z][(θ1x+θ2+θ3xy)+(b1xy)z] = 0. Thus, α1(α−1(a0x)(θ1x+
θ2y+θ3xy)) = α1((a0xy)(θ1x+θ2y+θ3xy)) = 0, α1(α−1(a0x)(b1xy)) = 0,
(γ1x+ γ3xy)(b1xy) = 0 and (γ1x+ γ3xy)(θ1x+ θ3xy) = 0.
Case 3.6: If γ2 = 0, a1 = 0, θ2 = 0, b0 6= 0, then we have that
[(γ1x + γ3xy) + (a0x + b0xy)z][(θ1x + θ3xy) + (b1xy)z] = 0. Thus,
α1(α−1(a0x + b0xy)(θ1x + θ3xy)) = α1((a0xy + b0x)(θ1x + θ3xy)) = 0,
α1(α−1(a0x + b0xy)(b1xy)) = 0, (γ1x + γ3xy)(b1xy) = 0 and (γ1x +
40 On partial skew Armendariz rings
γ3xy)(θ1x+ θ3xy) = 0.
Case 3.7: If γ2 = 0, a1 6= 0, θ2 = 0, b0 = 0, then we have that [(γ1x+
γ3xy)+(a0x)z][(θ1x+θ3xy)+(a1x+b1xy)z] = 0. Thus, α1(α−1(a0x)(θ1x+
θ3xy)) = α1((a0xy)(θ1x + θ3xy)) = 0, α1(α−1(a0x)(a1x + b1xy)) = 0,
(γ1x+ γ3xy)(a1x+ b1xy) = 0 and (γ1x+ γ3xy)(θ1x+ θ3xy) = 0.
Case 3.8: If γ2 6= 0, a1 = 0, θ2 = 0, b0 = 0, then we have that [(γ1x+
γ2y+γ3xy)+(a0x)z][(θ1x+θ3xy)+(b1xy)z] = 0. Thus, α1(α−1(a0x)(θ1x+
θ3xy)) = α1((a0xy)(θ1x + θ3xy)) = 0, α1(α−1(a0x)(b1xy)) = 0, (γ1x +
γ2y + γ3xy)(b1xy) = 0 and (γ1x+ γ2y + γ3xy)(θ1x+ θ3xy) = 0.
Case 3.9: If γ2 = 0, a1 = 0, θ2 = 0, b0 = 0, then we have that [(γ1x+
γ3xy) + (a0x)z][(θ1x + θ3xy) + (b1xy)z] = 0. Thus, α1(α−1(a0x)(θ1x +
θ3xy)) = α1((a0xy)(θ1x + θ3xy)) = 0, α1(α−1(a0x)(b1xy)) = 0, (γ1x +
γ3xy)(b1xy) = 0 and (γ1x+ γ3xy)(θ1x+ θ3xy) = 0.
So, the result follows for polynomials of degree 1.
Next, let f(z) =
∑n
i=0 aiz
i and g(z) =
∑n
i=0 biz
i be polynomials of
degree n in R[z;α] such that f(z)g(z) = 0. Then a0b0 = 0. Since I2 = 0,
we have that αi(α−i(ai)bj) = 0, for every 1 ≤ i ≤ n and 1 ≤ j ≤ n.
Thus, we have that 0 = f(z)g(z) = a0b0+(a0+α1(α−1(a1)b0))z+(a0b2+
α2(α−2(a2)b0)z
2+ ....+(a0bn+αn(α−n(an)b0))z
n. Now, we can apply the
methods used to prove the result for polynomials of degree 1 to the degree
0 and 1, degree 0 and 2, and so on. So, R is partial skew Armendariz.
Moreover, the ring R[z;α] is not associative, because ((y)(xyz))y = 0 and
(y)((xyz)y) = (y)(xz) = xyz 6= 0.
The following definition appears in [26] and [5].
Definition 6. (i) A ring R is called a left (right) quasi-Baer ring, if
the left (right) annihilator of any left (right) ideal is generated by an
idempotent as a left (right) ideal.
(ii) A ring R is called a left (right) p.q.-Baer ring if the left (right)
annihilator of any principal left (right) ideal is generated by an idempotent
as a left (right) ideal.
Remark 5. It is well-known that Baer rings ⇒ left (right) quasi-Baer
rings ⇒ left (right) p.q-Baer rings and that Baer rings ⇒ left (right)
p.p-rings. In the articles [5], [27], [28] and the literature quoted therein we
can find examples where the converse of each arrow is not true in general.
In [23] ([35]), the author defined u.p. semigroups (groups) i.e, a semi-
group (group) G′ is an u.p semigroup (group) if for any non-empty subsets
A and B of G′, there exists at least one y ∈ G′ that has an unique repre-
sentation of the form y = ab with a ∈ A and b ∈ B. It is not difficult to
W. Cortes 41
show that any cyclic group is an u.p group. The author in [23] used u.p.
groups to study groups rings that are Baer and p.p rings. It is convenient
to remark that this concept is used to the characterization of zero divisors
in group rings and semigroups rings.
One may ask if Baer, quasi-Baer, p.q-Baer and p.p-rings would imply
the associativity of the partial skew Laurent polynomial ring, but the next
example shows that this is false in general and part of the example below
is contained in ([32], Proposition 2.1).
We denote the Jacobson radical of a ring S1 by J(S1).
Example 4. Let H be a cyclic group of order 5 and F a field of characteris-
tic 5. Then, by ([23], Theorem 2) we have that that FH is a Baer ring. Since
H is cyclic it follows that J(FH)3 has dimension 2 over F and we have
that J(FH) k J(FH)3. Note that {1, (y− 1), (y− 2)2, (y− 1)3, (y− 1)4}
is an F -basis of FH. It is easy to see that J(FH)3 = FH(y − 1)3 is the
subspace generated by (y − 1)3 and (y − 1)4. Let G be an infinite cyclic
group generated by σ, I = J(FH)3 and α the partial action of G on FH
defined as follows: Si = S−i = I for every 0 6= i ∈ Z and αi : S−i → Si
given by αi((y− 1)3) = (y− 1)4 and αi((y− 1)4) = (y− 1)3 (by definition
S0 = FH and α0 is the identity automorphism of FH). In this case
FH ∗α G = FH < x;α >. We claim that FH < x;α > is not associative.
In fact, let a = (y−1)+(y−1)3x (note that (y−1) ∈ J(FH) ⊆ FH = S0
and (y−1) /∈ J(FH)3 = I). Then we obtain that a.a = (y−1)2+(y−1)4x.
Thus, we have that (aa)a = (y − 1)3 + (y − 1)3x while a(aa) = (y − 1)3.
So, (FH) < x;α > is not associative and consequently (FH)[x;α] is not
associative by the same reasons.
Let G be a group and α a partial action of G on R. Suppose that
(R,α) has enveloping action (T, β). One may ask if R is either Baer or
quasi-Baer or p.q-Baer or p.p. implies that T is either Baer or quasi-Baer
or p.q-Baer or p.p, respectively. The next example shows this is not true
in general.
Example 5. Consider T , R, σ and α as in the Example 1. We clearly
have that R is Baer but T is not Baer, quasi-Baer , right (left) p.q.Baer
and right (left) p.p. In fact, let y = e1. Then rT (e1) = ⊕i∈Z, i 6=1Kei that
is not generated by an idempotent.
The proof of the following lemma is standard an so we omit it here.
Lemma 5. Let S be a ring and I an ideal of S generated by a central
idempotent. Then the following statements hold:
(i) If S is Baer, then I is Baer.
42 On partial skew Armendariz rings
(ii) If S is quasi-Baer, then I is quasi-Baer.
(iii) If S is right (left) p.q.-Baer, then I is right (left) p.q.-Baer.
(iv) If S is right (left) p.p, then I is right (left) p.p.
By ([20], Proposition 1.2) a partial action α of a group G on R with
enveloping action (T, β) is of finite type if there exists g1, ..., gn ∈ G such
that T =
∑n
i=1 βgi(R). Moreover, T has a unit 1T .
In the next result we suppose that the partial action α of G on R with
enveloping action (T, β) is of finite type.
Theorem 7. (i) R is Baer if and only if T is Baer.
(ii) R is quasi-Baer if and only if T is quasi-Baer.
(iii) R is right (left) p.q.-Baer if and only if T is right (left) p.q-Baer.
(iv) R is right (left) p.p if and only if T is right (left) p.p.
Proof. We only need to prove (i) because the other ones use similar ideas.
Suppose that R is Baer and let ∅ 6= Y be any subset of T . Then by ([20],
Proposition 1.10 and Remark 1.11) we have that T = R⊕A1 ⊕ ...⊕An
and 1T = 1R+f1+f2+ ...+fn, where 1R, f1, ..., fn are orthogonal central
idempotents with fi ∈ Ai for every 1 ≤ i ≤ n. Thus, for any y ∈ Y we have
that y = y1R+yf1+...+yfn ∈ Y 1R⊕Y f1⊕...⊕Y fn. Hence, for every a ∈
rT (Y ) and y ∈ Y , we have that ya = y1Ra1R + yf1af1 + ...+ yfnafn = 0,
which implies y1Ra1R = yf1af1 = ... = yfnafn = 0. So, a1R ∈ rR(Y 1R),
afi ∈ rAi
(Y fi) for every 1 ≤ i ≤ n and since a = a1R + af1 + ...+ afn,
we obtain that a ∈ rR(Y 1R)⊕ rA1
(Y f1)⊕ ...⊕ rAn(Y fn).
On the other hand, for every z ∈ rR(Y 1R)⊕ rA1
(Y f1)⊕ ...⊕ rAn(Y fn)
we have that yz = 0 for every y ∈ Y . Thus, rT (Y ) = rR(Y 1R) ⊕
rA1
(Y f1)⊕ ...⊕ rAn(Y fn). By the fact that the Baer property is preserved
by isomorphisms and using Lemma 5, we obtain that rR(Y 1R) = eR,
rAi
(Y fi) = eiAi, for 1 ≤ i ≤ n. Hence, rT (Y ) = eR ⊕ e1A1 ⊕ ...⊕ enAn.
So, rT (Y ) = (e+ e1+ ...+ en)T and we easily have that e+ e1+ ...+ en is
an idempotent since eei = ejek = 0, for every 1 ≤ i, j, k ≤ n with j 6= k.
By Lemma 5 the converse follows.
One may ask if a partial action α of an infinite cyclic group G on R
with enveloping action (T, σ) such that α is of finite type we would have:
R is partial α-rigid ⇔ T is σ-rigid and R is partial skew Armendariz ⇔
T is skew Armendariz. The next example shows that these facts are not
true in general.
Example 6. Assume that K is a field and T = ke1 ⊕Ke2 ⊕Ke3, where
{e1, e2, e3} are orthogonal central idempotents. We define an action of
an infinite cyclic group G generated by σ as follows: σ|k = idK and
W. Cortes 43
σ(ei) = ei+1. Let R = Ke1. We easily have a partial action α of G on R.
Note that (T, σ) is the enveloping action of (R,α), α is of finite type, R is
partial α-rigid and partial skew Armendariz, but T is not σ-rigid because
e2σ(e2) = 0 and e2 6= 0. We claim that T is not skew Armendariz. In fact,
let f(x) = e1+ e1x and g(x) = −e3+ e1x be polynomials in T [x;σ]. Then,
we have that f(x)g(x) = 0 and e1e1 6= 0. So, T is not a skew Armendariz
ring.
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W. Cortes 45
Contact information
W. Cortes Instituto de Matemática, UFRGS, 91509-900,
Porto Alegre-RS, Brazil
E-Mail: cortes@mat.ufrgs.br
Received by the editors: 09.02.2011
and in final form 09.02.2011.
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