c

Journal “Algebra and Discrete Mathematics”

### On partial skew Armendariz rings

Wagner Cortes

Communicated by V. V. Kirichenko

A b s t r a c t . In this paper we consider ringsR with a partial actionαof an infinite cyclic groupGonR. We introduce the concept of partial skew Armendariz rings and partialα-rigid rings. We show that partialα-rigid rings are partial skew Armendariz rings and we give necessary and sufficient conditions forR 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 betweenR andR[x;α].

We also show thatR[x;α]andRhx;αiare not necessarily asso- ciative rings whenRsatisfies 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 Gbe a group and
R a unitalk-algebra, wherek is a commutative ring. A*partial action* α
of GonR is a collection of idealsSg of R,g∈G, and isomorphisms of
(non-necessarily unital)k-algebras αg:S_{g}−1 →Sg such that:

(i) S_{1} =R andα_{1} is the identity map of R;

(ii)S_{(gh)}−1 ⊇α^{−1}_{h} (S_{h}∩S_{g}−1);

(iii) αg◦αh(x) =αgh(x), for everyx∈α^{−1}_{h} (Sh∩S_{g}−1).

2000 Mathematics Subject Classification:16S36; 16S35.

Key words and phrases:partial actions, Armendariz rings, Baer rings and P.P rings.

The property (ii) easily implies that αg(S_{g}−1 ∩S_{h}) =Sg∩S_{gh}, for all
g, h∈G. Also α_{g}−1 =α_{g}^{−1}, for everyg∈G.

Given a partial action α of a groupG onR, an enveloping action is
an algebraT together with a global action β ={σg |g∈G} of GonT,
whereσg is an automorphism ofT, 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 idealsS_{g} are unital algebras, i.e., S_{g} is generated by a central
idempotent of R, for every g ∈ G. In this case the partial skew group
ring R ⋆_{α}Gis 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 thatR is an ideal ofT and the following properties hold:

(i) the subalgebra ofT generated byS

g∈Gσg(R)coincides withT and we have T =P

g∈Gσg(R);

(ii) Sg=R∩σg(R), for every g∈G;

(iii) αg(x) =σg(x), for every g∈G andx∈S_{g}^{−1}.

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 element1R,Gan infinite
cyclic group generated byσ and{ασ^{i} :S_{σ}^{−i} →S_{σ}^{i}, i∈Z}a partial action
ofGonR. We simplify the notation by puttingαi =α_{σ}^{i} andSi =S_{σ}^{i}, for
everyi∈Z. Then the partial skew group ring can be identified with the set
of all finite formal sumsPm

i=−naix^{i},ai∈Si for every−n≤i≤m, where
the sum is usual and the multiplication rule isax^{i}bx^{j} =αi(α−i(a)b)x^{i+j}.
We denote this ring byRhx;αiand call it partial skew Laurent polynomial
ring. The partial skew polynomial ringR[x;α] is defined as the subring of
Rhx;αi whose elements are the polynomialsPn

i=0b_{i}x^{i},b_{i} ∈S_{i} for every
0≤i≤nwith usual sum of polynomials and multiplication rule as before.

In this paper, we assume thatRis an associative ring andαis a partial action of an infinite cyclic groupG onR such that R[x;α] andRhx;αi are not necessarily associative rings.

In the Section 2, we study McCoy’s result in partial skew Laurent polynomial rings when a partial actionα of an infinite cyclic groupG on R has an enveloping action(T, σ), whereσ is an automorphism ofT.

In the Section 3, we introduce the concept of partial skew Armendariz

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 ofR[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 betweenR andR[x;α].

We study necessary and sufficient conditions forR[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), whereσis an automorphism of T, we show thatT isσ-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 denoterS(Y) ={a∈S :Y a= 0} (lR(Y) ={a∈S :aY = 0}) the right (left) annihilator of Y inS.

2. Generalization of McCoy’s result

in partial skew Laurent polynomial rings

In [33], McCoy proved that ifR is a commutative ring, then whenever
g(x) is a zero-divisor inR[x]there exists a non-zero elementc∈R such
that cg(x) = 0, and in [26] it was proved that if r_{R[x]}(f(x)R[x]) 6= (0),
thenr_{R[x]}(f(x)R[x])∩R6= (0),wheref(x)∈R[x]andr_{R[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 andα is a partial action of an infinite cyclic group Gon R with enveloping action (T, σ), whereσ is an automorphism ofT.

Lemma 1. *Let* f(x) =Pn

i=pbix^{i} *and* g(x) =Pm

i=qaix^{i} *be two elements*
*of* Rhx;αi. Thenf(x)RPm

i=qαj(ai1−j)x^{i+j} = 0, for everyj∈Z*, if and*
*only if* f(x)Rhx;αig(x) = 0.

*Proof.* Suppose that f(x)RPm

i=qα_{j}(a_{i}1_{−j})x^{i+j} = 0, for every j ∈ Z.
Then, we have that

f(x)rx^{j}g(x) =f(x)r1jx^{j}g(x) =f(x)r

m

X

i=q

αj(ai1_{−j})x^{i+j} = 0.

So,f(x)Rhx;αig(x) = 0.

The converse is trivial.

Let h(x) = Pn

i=saix^{i} ∈ Rhx;αi. The length of h(x) is the number
len(h(x)) =n−s+ 1and we use this number below.

Theorem 1. *Let* f(x) ∈ Rhx;αi. If r_{Rhx;αi}(f(x)Rhx;αi) 6= 0 *then*
r_{Rhx;αi}(f(x)Rhx;αi) ∩R 6= 0, where r_{Rhx;αi}(f(x)Rhx;αi) = {h(x) ∈
Rhx;αi:f(x)Rhx;αih(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 thatq 6= 0,p < q and

r_{Rhx;αi}(f(x)Rhx;αi)∩R= (0).

Let g(x) = btx^{t}+...+bmx^{m} ∈ r_{Rhx;αi}(f(x)Rhx;αi) of minimal length
with bm6= 0. Then

f(x)Rhx;αig(x) = 0,
and we have thatf(x)1_{−q}x^{−q}RPm

i=tαj(bi1_{−j})x^{i+j} = 0, for every j∈Z.
Thus, aqRαj(1_{−j}bm) = 0, for everyj ∈Z. Hence,aqRhx;αibm = 0 and
we have that

a_{q}Rhx;αig(x) =a_{q}Rhx;αi(btx^{t}+...+b_{m−1}x^{m−1}).

So,

f(x)Rhx;αiaqRhx;αi(btx^{t}+...+b_{m−1}x^{m−1})) =

=f(x)(Rhx;αiaqRhx;αi)g(x) = 0, for every j∈Z, and we obtain that

f(x)Rhx;αiaqR(αj(bt1_{−j})x^{t+j}+...+αj(b_{m−1}1_{−j})x^{m−1+j}) = 0,

for everyj∈Z. Note that by choice ofg(x), we have thataqR(αj(bt1_{−j})x^{t+j}+
...+αj(b_{m−1}1_{−j})x^{m−1+j})) = 0. Thus,

aqRhx;αi(αj(bt1_{−j})x^{t+j}+...+αj(b_{m−1}1_{−j})x^{m−1+j}) = 0

and it follows that

aq∈lR(Rhx;αi(αj(bt1−j)x^{t+j}+...+αj(bm−11−j)x^{m−1+j})+

+Rhx;αi(bmx^{m})),

for every j∈Z. Hence,

(apx^{p}+....+a_{q−1}x^{q−1})Rhx;αig(x) = (0),

and we obtain thata_{q−1}Rαj(1_{−j}bm) = 0, for every j∈Z. Consequently,
(apx^{p}+....+a_{q−1}x^{q−1})Rhx;αia_{q−1}Rhx;αiP_{m−1}

i=t bix^{i} =
f(x)(Rhx;αia_{q−1}Rhx;αi)g(x) = (0).

By choice ofg(x), we have that

aq−1Rhx;αig(x) = (0)

and we get

a_{q−1}∈l_{R}(Rhx;αi(αj(bt1_{−j})x^{t+j}+...+αj(b_{m−1}1_{−j})x^{m−1+j})+

+Rhx;αi(bmx^{m}))

for every j∈Z. Now, repeating this process we obtain that

as∈lR(Rhx;αi(αj(bt1−j)x^{t+j}+...+αj(bm−11−j)x^{m−1+j})+

+Rhx;αi(bmx^{m}))

for everyp≤s≤q andj∈Z. Since a_{s}Rhx;αibmx^{m}= (0),for everyp≤
s≤q, then(a_{p}x^{p}+....+a_{q}x^{q})Rhx;αibm= (0). This is a contradiction.

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

P

i=0

aix^{i},

m

P

i=0

bix^{i} ∈R[x]satisfyf(x)g(x) = 0, thenaibj = 0,for every0≤i≤nand
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 groupGonRsuch 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 beτ-rigid if aτ(a) = 0implies thata= 0. Suppose that a partial action α of an infinite cyclic groupG onR has an enveloping action (T, σ), where σ is an automorphism ofT. 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) = 0we have that
t= 0. In fact, we easily have thatt1Rα_{1}(t1R1_{−1}) = 0, which implies that
t1R= 0. Note that for eachi >0 we have that

(tσ^{i}(1R))σ(tσ^{i}(1R)σ^{i−1}(1R)σ^{i}(1R)) = 0.

Thus, we obtain that σ^{−i}(tσ^{i}(1R))σ(σ^{−i}(tσ^{i}(1R)))1Rσ(1R) = 0, which
implies thatσ^{−i}(tσ^{i}(1R))σ(σ^{−i}(tσ^{i}(1R))) = 0and by assumption we have
thatσ^{−1}(tσ^{i}(1R)) = 0. Hence, tσ^{i}(1R) = 0, for every i≥0.

Now, let b =tσ^{−i}(1_{R}), for i >0. Thenbσ(bσ^{−i}(1_{R})σ^{−i−1}(1_{R})) = 0.

Proceeding with similar method as above we obtain that tσ^{−i}(1R) = 0,
for everyi >0. So,tT = 0and by Remark 2.5 of [16] we have thatt= 0.

Let T^{1} = T ⊕Z with usual sum and multiplication defined by the
rule (a, z)(b, z_{1}) = (ab+az_{1}+bz, zz_{1}). Note that(0,1)is the identity of
T^{1}. We can extend the automorphismσ ofT to T^{1} by the rule:σ(a, z) =

(σ(a), z). We claim thatT^{1} is aσ-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 thataσ(a) = 0. Thus,a= 0and
we obtain that (a, z) = (0,0). Hence, by ([27], Corollary 4) and ([24],
Lemma 4) we have that all the idempotents ofT^{1} are invariant byσ. Since
for any idempotente∈T, the element(e,0)is an idempotent in T^{1}, then
σ(e) =e. So,σ(1R) = 1R and we have thatR=T.

Remark 1. Letα be a partial action of an infinite cyclic groupGon R
with an enveloping action (T, σ), whereσ is an automorphism ofT. It is
natural to ask ifR[x;α]were reduced then we would have that R =T.
This is not true in general as show the following example: LetK be a field,
{ei:i∈Z}a set of central orthogonal idempotents, T =⊕_{i∈Z}Kei with
an automorphismσdefined byσ(ei) =ei+1, for everyi∈ZandR=Ke0.
We clearly have that the automorphismσ induces a partial action of the
group G=< σ >as follows: S0=R, Si = 0, for everyi∈Z\{0} and the
mapsα_{0} =id_{R} andα_{i} = 0, for every i∈Z\{0}. Note that R[x;α] =R
andR &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 thatR[x, α]is reduced. Leta_{j} ∈S_{j} such thatα_{−j}(a_{j})a_{j} =
0, forj ≥0. Thena_{j}x^{j}a_{j}x^{j} =α_{j}(α_{−j}(a_{j})a_{j})x^{2j} = 0. Hence, by assump-
tion we have thatax^{j} = 0. So,aj = 0.

Conversely, let f(x) =Pn

i=0aix^{i} be a polynomial inR[x;α] such that
(f(x))^{2}= 0. Then, αn(α_{−n}(an)an) = 0and by assumption we have that
an= 0. Proceeding in a similar way we obtain thata_{0} =a_{1}=...=an= 0.

So,f = 0.

Using the Proposition 2 we can generalize the notion ofσ-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∈S_{n} *with*
α_{−n}(a)a= 0,n≥0, we have that a= 0.

Remark 2. Note that ifR is partial α-rigid, thenR is reduced.

The following definition generalizes ([27], Definition).

Definition 2. *We say that* R *is a partial skew Armendariz ring if given*
f(x) =Pn

i=0aix^{i} *and*g(x) =Pm

i=0bix^{i} *in*R[x;α]*such that*f(x)g(x) = 0,
*then* α_{−i}(ai)bj = 0, for every0≤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 thenR *is*
*a partial skew Armendariz ring.*

*Proof.* Let f(x) = Pn

i=0aixi and g(x) = Pn

j=0bjxj 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}(a_{1}α_{1}(b_{0}1_{−1}))a_{1}α_{1}(b_{0}1_{−1})) = α_{−1}(a_{1})b_{0}a_{1}α_{1}(b_{0}1_{−1}) = 0. Since R
is partial α-rigid then 0 = a_{1}α_{1}(b_{0}1_{−1}) = α_{−1}(a_{1})b_{0} and consequently
a_{0}b_{1}= 0. Now, for degree 2 we have that

a_{0}b_{2}+a_{1}α_{1}(b_{1}1_{−1}) +a_{2}α_{2}(b_{0}1_{−2}) = 0 (*)
Multiplying (*) on the right side byb_{1}and using the equalitya_{0}b_{1} =a_{0}b_{0} =
0 we have thatb_{0}a_{2}α_{2}(b_{0}1_{−2}) = 0. Thus,α_{−2}(a_{2})b_{0}a_{2}α_{2}(b_{0}1_{−2}) = 0and
we obtain that α_{−2}(a_{2}α_{2}(b_{0}1_{−2}))a_{2}α_{2}(b_{0}1_{−2}) =α_{−2}(a_{2})b_{0}a_{2}α_{2}(b_{0}1_{−2}) =
0. Hence, a_{2}α_{2}(b_{0}1_{−2}) = α_{−2}(a_{2})b_{0} = 0. So, b_{1}a_{1}α_{1}(b_{1}1_{−1}) = 0. Pro-
ceeding with similar method as before we have that0 =a_{1}α_{1}(b_{1}1_{−1}) =
α−1(a1)b1 and it follows thata0b2 = 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 ringR^{′} is essential as a two sided ideal if
r_{R}′(H) = 0. The set of all essential ideals ofR^{′} will be denoted by E =
E(R^{′}). Note thatE is a filter of ideals which is closed under multiplication
and intersection. If I is an ideal ofR^{′}, thenI⊕Ann_{R}′(I)∈ E.

Denote by Q=QE the ring of right quotients ofR^{′} with respect to
the filter E, i.e., the Martindale ring of right quotients ofR^{′}. Recall that
the elements ofQarise from right R^{′}-homomorphisms fromH ∈ 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, elementsq, p∈Qare equal
if and only if they coincide on some essential ideal of R^{′} and if qH = 0,
for some q∈Qand H∈ E, thenq= 0.

Given an ideal I ofR^{′}, the closure of I inR^{′} 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, thenI is a closed ideal.

LetR be a partialα-rigid ring. ThenR is reduced and we have thatR
has a Martindale ring of right quotientsQconstructed as before. Following
[21] the partial actionαcan be extended to a partial actionα^{∗} ofGonQ,
where the idealsS_{i}^{∗} are the extension of[Si] toQ. The ideal S_{i}^{∗} for every
i∈Zis generated by a central idempotent 1^{∗}_{i}. LetQ^{s}be the Martindale’s
right symmetric ring of quotients of R, i.e, according ([31], Proposition
5.14.7) we have thatQ^{s} ={q ∈Q:qJ ⊆R and Jq ⊆R f or some J ∈
E}. Now, let (I^{∗})^{s} ={q ∈Q:qJ ∪Jq⊆ I f or 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 ofI.
Moreover, ifI^{∗} is generated by a central idempotent, then we have that
(I^{∗})^{s}=I^{∗}∩Q^{s}(R) and we denote(I^{∗})^{s}=I^{∗∗}.

By ([31], Proposition 5.14.17) we have that 1^{∗}_{i} ∈Z(Q) =Z(Q^{s}) and
1^{∗}_{i} ∈(I^{∗})^{s}, for every i∈Z. Thus S_{i}^{∗∗}=Q^{s}1^{∗}_{i}, for everyi∈Z. Now, using
similar methods of ([21], Proposition 2.3 and Theorem 3.1), we can extend
the partial actionαto a partial actionα^{∗∗}={α^{∗∗}_{i} :S_{−i}^{∗∗} →S_{i}^{∗∗}:i∈Z}of
Gon Q^{s} andα^{∗∗}_{g} =α^{∗}_{g}|S_{−i}^{∗∗}, for every i∈Z. Moreover, by ([15], Theorem
4.5) we have that (Q^{s}, α^{∗∗}) has an enveloping action (T^{′′}, σ^{′′}), where σ^{′′}

is an automorphism ofT^{′′}. We use these facts in the next result.

Proposition 4. R *is partial* α-rigid if an only if Q^{s} *is partial* α^{∗∗}*-rigid.*

*Proof.* Suppose thatR is partialα-rigid. Letq ∈S_{n}^{∗∗}such thatα^{∗∗}_{−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 everyl∈L we have thatα^{∗∗}_{−n}(q)ql= 0. By ([31],
Exercise 14.4) Q^{s} is reduced, since R is reduced. Hence,qlα^{∗∗}_{−n}(q) = 0,
which implies that0 =qlα^{∗∗}_{−n}(q)α^{∗∗}_{−n}(l1^{∗}_{n}) =qlα^{∗∗}_{−n}(ql). Sinceα^{∗∗}=α^{∗}|S^{∗∗}_{n} ,
then by ([21], Theorem 3.1) we have thatα^{∗}_{−n}restricted toS_{−n}isα_{−n}and
we obtain that α^{∗∗} restricted to Sn isαn. 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).

Theorem 2. *If* R *is partial* α-rigid, then R *is partial skew Armendariz.*

*Proof.* By Proposition 4, Q^{s} is partial α^{∗∗}-rigid and by Proposition 3,Q^{s}
is partial skew Armendariz. Since R is a subring ofQ^{s} 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 σ^{i} 6= idD,
for any i6= 0 andI a two-sided ideal of D. For any integer iwe define
Si =I∩σ^{i}(I) andαi :S_{−i}→Si as the restriction of σ^{−i} to S_{−i}. Then it
is easy to see thatα={αi|i∈Z}is a partial action of the infinite cyclic
groupG=< σ > on I.

Let M be a(R, R)-bimodule. Then the trivial extension ofR byM is the ringT(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

, wherer∈R andm∈M with usual matrix sum and multiplication.

Let α={αi:S_{−i} →Si}be a partial action of an infinite cyclic group
Gon R and T(R, R) the trivial extension ofR. Then we can extend the
partial actionα to T(R, R) as follows: S¯i=T(Si, Si) andα¯i: ¯S−i→S¯i

with α¯i(a, b) = (αi(a), αi(b)), for every i∈Z. Since T(R,0) is isomorphic
to R, we can identify the restriction of α¯_{i} to T(R,0) with α_{i}, 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.* Letf(x) =Pn

i=0(a_{i}, b_{i})x^{i} andg(x) =Pn

j=0(c_{j}, d_{j})x^{j} be elements
in T(R, R)[x;α] such that f(x)g(x) = 0. We easily have that f(x) =
(p_{0}, p_{1}), g(x) = (q_{0}, q_{1}) and 0 = f(x)g(x) = (p_{0}q_{0}, p_{0}q_{1} +p_{1}q_{0}), where

p_{0} = Pn

i=0aix^{i}, p_{1} = Pn

i=0bix^{i}, q_{0} = Pn

j=0cjx^{j} and q_{1} = Pn

j=0djx^{j}.

Note that q_{0}p_{0} = 0 and using similar ideas of ([27] Proposition 15)
we obtain that p_{0}q_{1} = p_{1}q_{0} = 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≤nand 0≤j≤n.

A right annihilator of a non-empty subset X of a not necessarily
associative ring S^{′} is defined by r_{S}′(X) ={a∈S:Xa= 0}. SinceS^{′} is
not necessarily associative, thenr_{S}′(X)is not necessarily a right ideal. If
we have an associative ring S_{1} contained in S, then for every non-empty
subset Y of S_{1} we have rS1(Y) = rS(Y)∩S_{1}. We put rAnnR(2^{R}) =
{rR(U) : U ⊆ R} and for a not necessarily associative ring S^{′} we put
analogouslyrAnnS^{′}(2^{S}^{′}) ={rS^{′}(U) :U ⊆S^{′}}.

Lemma 2. *Let* U *be a non-empty subset of* R. Then
r_{R[x;α]}(U) =rR(U)R[x;α].

*Proof.* Let f(x) = Pn

i=0aix^{i} ∈ R[x;α] such that U f(x) = 0. Then
U ai = 0 for every 0 ≤ i ≤ n and it follows that ai ∈ rR(U) for every
0≤i≤n. Thus, f(x) =Pn

i=0aix^{i}∈rR(U)R[x;α]. Hence,r_{R[x;α]}(U) ⊆
rR(U)R[x;α] and we easily have that rR(U)R[x;α] ⊆ r_{R[x;α]}(U). So,
r_{R[x;α]}(U) =rR(U)R[x;α].

From Lemma 2 we have the maps

φ:rAnn_{R}(2^{R})→rAnn_{R[x;α]}(2^{R[x;α]})
defined byφ(I) =IR[x;α], for everyI ∈rAnnR(2^{R}) and

Ψ :rAnn_{R[x;α]}(2^{R[x;α]})→rAnnR(2^{R})

defined byΨ(J) =J∩R, for every J ∈rAnn_{R[x;α]}(2^{R[x;α]}). Obviously, φ
is injective andΨis surjective. Clearlyφ is surjective if and only ifΨis
injective, and in this case φand Ψare 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})→rAnn_{R[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) = Pn

i=0aix^{i} ∈
R[x;α] we define C_{f(x)} = {α−i(ai), 0 ≤ i ≤ n} and for a subset S of
R[x;α]we denote the set ∪

f(x)∈SC_{f}_{(x)}byCS. We claim thatr_{R[x;α]}(f(x)) =

r_{R[x;α]}(C_{f(x)}).In fact, letg(x) =Pm

i=0bix^{i} ∈r_{R[x;α]}(f(x)). Then, we have
thatf(x)g(x) = 0. By assumption, we have thatα_{−i}(ai)bj = 0, for every
0≤i≤n and0≤j ≤m.Thus, g(x)∈r_{R[x;α]}(C_{f}_{(x)}).

On the other hand, let h(x) =Pp

i=0c_{i}x^{i} be an element inR[x;α]such
thatC_{f(x)}h(x) = 0.Then we have that α_{−i}(ai)c_{k}= 0, for every0≤i≤n
and 0≤k≤p, which implies that f(x)h(x) = 0.

Since R is partial skew Armendariz, then r_{R[x;α]}(S) =
r_{R[x;α]}( ∪

f(x)∈SC_{f}_{(x)}) and by Lemma 2 we have that r_{R[x;α]}(C_{f(x)}) =
r_{R}(C_{f(x)})R[x;α]. Hence,

r_{R[x;α]}(S) = ∩

f(x)∈Sr_{R[x;α]}(f(x)) = ∩

f(x)∈Sr_{R[x;α]}(C_{f(x)}) =

= ( ∩

f(x)∈Sr_{R}(C_{f}_{(x)}))R[x;α] =r_{R}(C_{S})R[x;α].

So,φis surjective.

Conversely, let f(x) = Pn

i=0aix^{i} and g(x) = Pm

i=0bix^{i} be ele-
ments in R[x;α] such that f(x)g(x) = 0. Then, by assumption,g(x) ∈
r_{R[x,α]}(f(x)) =BR[x;α], for some right idealB ofR. Thus, we have that,
b_{i} ∈ B ⊂ r_{R[x;α]}(f(x)), for every0 ≤i ≤ m. Hence, α_{−i}(a_{i})b_{j} = 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=Pn

i=0eix^{i} ∈
R[x;α]. If e^{2} =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 eacha∈S^{′} the setaS^{′}:{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).

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* r_{R[x;α]}(U) = eR[x;α], for some idempotent
e∈R.

*Proof.* Suppose thatRis a Baer ring. Let∅6=U be a subset ofR[x;α]and
f(x) = Pn

i=0a_{i}x^{i} ∈U. We define C_{f(x)} = {α−i(a_{i}), f or every 0≤ i≤
n}, andC_{U} = ∪

f(x)∈UC_{f}_{(x)}. By assumption, we have thatr_{R}(C_{U}) = eR,
with e^{2} =e. It is easy to see thateR[x;α]⊂r_{R[x;α]}(U).By Lemma 3, we
have

r_{R[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 thatr_{R[x;α]}(A) =eR[x;α],where e^{2} =e. By Lemma 4, e=e0 is a
constant polynomial. Thus,

r_{R[x;α]}(A) =eR[x;α] =e0R[x;α].

Hence,r_{R[x;α]}(A)∩R=e_{0}R[x;α]∩R=e_{0}R.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*r_{R[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 chainrS(Y1)⊆rS(Y2)⊆..., there exists
i≥1 such that r_{S}(Y_{i}) =r_{S}(Y_{i+p}), for all p≥0, where Y_{j} are non-empty
subsets ofS for every j≥0. Moreover, for a not necessarily associative
rings S^{′} satisfies the ascending chain conditions on right annihilators if
r_{S}′(Y_{1}^{′}) ⊆r_{S}′(Y_{2}^{′}) ⊆..., there exists i≥1 such thatr_{S}′(Y_{i}^{′}) = r_{S}′(Y_{i+p}^{′} ),
for allp≥0, whereY_{j}^{′} are non-empty subsets ofS^{′} for everyj ≥0. In the
next proposition, we suppose thatR is a partial skew Armendariz ring

to study the transfer of ascending chain condition on right annihilators property between R andR[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(U_{1})⊂rR(U_{2}) ⊂.., where
Ui ⊂R,for everyi≥1. We claim thatr_{R[x;α]}(Ui)⊂r_{R[x;α]}(U_{i+1}).In fact,
let g(x) = Pn

i=0cix^{i} be an element of R[x;α], such that Uig(x) = (0).

Then,Uicj = (0),and we obtain thatUi+1cj = (0). Hence, by assumption
there existsl≥1 such thatr_{R[x;α]}(Ul) = r_{R[x;α]}(U_{l+p}), for everyp∈N.
So,

rR(U_{l}) =r_{R[x;α]}(U_{l})∩R=r_{R[x;α]}(U_{l+p})∩R=rR(U_{l+p}),
for every p∈N.

Conversely, suppose that r_{R[x;α]}(V_{1})⊂r_{R[x;α]}(V_{2})⊂.... By Lemma 3,
r_{R[x;α]}(Vi) =rR(CVi)R[x;α],whereCVi = ∪

f(x)∈Vi

C_{f(x)},
C_{f(x)} ={α_{−i}(ai), f or every 0≤i≤m}

and f(x) = Pm

i=0aix^{i}. We claim that rR(CVi) ⊂ rR(CVi+1). In fact,
let y ∈ rR(CVi). Then we have that CViy = 0. Thus, f(x)y = 0, for
every f(x) ∈ Vi. Hence, CVi+1y = 0 and by assumption, there exists
n≥1such thatr_{R}(C_{V}_{n}) =r_{R}(C_{V}_{n+k}),for everyk∈N.So, r_{R[x;α]}(V_{n}) =
r_{R[x;α]}(V_{n+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 subsetY ⊆U. Equivalently, for a left idealL ofR with rR(L) = 0, there exists a finitely generated left idealL1 ⊆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

following result: suppose that R is Armendariz ring. Then R is a right (left) zip ring if and only ifR[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*r_{R[x;α]}(U) = 0, we have thatr_{R[x;α]}(Y) = 0*for any*
*non-empty finite subset*Y *of* X.

*Proof.* Suppose thatR is a right zip ring. LetU be a non-empty subset
of R[x;α] such that r_{R[x;α]}(U) = 0. Then, by Lemma 3, r_{R[x;α]}(U) =
r_{R}(C_{U})R[x;α], whereC_{U} =∪f(x)∈UC_{f(x)} andC_{f}_{(x)}={α−i(ai) : 0≤i≤
n}with f(x) =Pn

i=0aix^{i} ∈U. Thus, rR(CU) = 0 and by assumption for
any non-empty finite subsetW ={α−i1(ai1), ..., α_{−i}_{n}(ain)}ofCU we have
that rR(W) = 0. For every α_{−i}_{j}(aij) ∈W there existsga_{ij}(x)∈U such
that some of the coefficients ofga_{ij}(x) are aij, for every1≤j ≤n. Let
U_{0} be a minimal subset ofU such thatg_{a}_{ij}(x)∈U_{0}, for every 1≤j≤n.

ThenU_{0} is non-empty finite subset ofU. We denoteW_{0}=∪f(x)∈U0(C_{f(x)})
and note thatW ⊆W_{0}. Hence,rR(W_{0})⊆rR(W) = 0 and, by Lemma 3,
r_{R[x;α]}(U_{0}) =rR(W_{0})R[x;α]. So,r_{R[x;α]}(U_{0}) = 0.

Conversely, let Y be a non-empty subset of R such that rR(Y) = 0 andf(x) =Pn

i=0aix^{i} ∈r_{R[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 r_{R[x;α]}(Y1) = 0. Hence,
r_{R}(Y_{1}) =r_{R[x;α]}(Y_{1})∩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 Gwith enveloping action (T, β) we study the transfer of Baer property, quasi-Baer property, p.q. Baer property and p.p property between R andT.

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.

*Proof.* Let a ∈ Sn such that α_{−n}(a)a = 0. Then σ^{−n}(a)a = 0, which
implies thataσ^{n}(a) = 0. Thus, by ([28], Lemma 4) we have thata^{2}= 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 andT =⊕_{i∈Z}Kei, where{ei, i∈Z} are
orthogonal central idempotents. We define an action of an infinite cyclic
group Ggenerated byσ as follows: σ|K =idK andσ(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, becausee_{2}σ(e_{2}) =e_{2}e_{3} = 0 and
e_{2}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 Pn

i=0aix^{i} with usual sum and the multiplication rule is ax^{i}bx^{j} =
aσ^{i}(b)x^{i+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) = Pn

i=0aix^{i} *and*
g(x) = Pm

i=0bix^{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, thenT would be a skew Armendariz ring, but the next example shows that this is not true in general.

Example 2. LetK,T, σ,R andαas in the Example 1. We clearly have thatR 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 inT[x;σ]we obtain that f g= 0 ande−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]/(X^{2}, Y^{2}), whereK is a field. ThenR =
K1 +Kx+Ky+Kxy, where x andy represent the classes of X andY
inR. Note that R is a four dimensionalK-vector space. Let G=< σ >

be an infinite cyclic group generated byσ and we define the partial action
α of G on R as follows: let I = Kx+Kxy,Si =I and αi :S_{−i} → Si

defined byαi(x) =xy, αi(xy) =x and αi|K =idK, for every i∈Z\ {0}

(by definitionS_{0} =R andα_{0}=idR). We claim that R is a partial skew
Armendariz ring. In fact, letf(z) = (γ_{0}+γ_{1}x+γ_{2}y+γ_{3}xy)+(a_{0}x+b_{0}xy)z
andg(z) = (θ_{0}+θ_{1}x+θ_{2}y+θ_{3}xy)+(a_{1}x+b_{1}xy)zbe polynomials of degree
1 in R[z;α]such that f(z)g(z) = 0. Then (γ_{0}+γ_{1}x+γ_{2}y+γ_{3}xy)(θ_{0}+
θ_{1}x+θ_{2}y+θ_{3}xy) = 0, which impliesγ_{0}θ_{0} = 0. Thus, we have the followin
g cases:

Case 1: If γ0 = 0 andθ06= 0, then we have that f(z) = 0.

Case 2: If γ0 6= 0 andθ0= 0, then we have that g(z) = 0.

Case 3: If γ_{0} = θ_{0} = 0, then we have that (γ_{1}θ_{2}+γ_{2}θ_{1})xy = 0 and
γ_{2}a_{1}x=b_{0}θ_{2}xy= 0. Thus,γ_{2}a_{1}+b_{0}θ_{2}= 0, γ_{2}a_{1} = 0andb_{0}θ_{2} = 0, which
implies either γ_{2} = 0or a_{1} = 0and either b_{0} = 0or θ_{2} = 0. Hence, we
have the following cases:

Case 3.1: If γ_{2} = 0,a_{1} 6= 0,b_{0} = 0 and θ_{2} 6= 0, then we have that
[(γ_{1}x+γ_{3}xy) + (a_{0}x)z][(θ_{1}x+θ_{2}y+θ_{3}xy) + (a_{1}x+b_{1}xy)z] = 0. Thus,
(γ_{1}x+γ_{3}xy)(θ_{1}x+θ_{2}y+θ_{3}xy) = 0,α_{1}(α_{−1}(a_{0}x)(θ_{1}x+θ_{2}y+θ_{3}xy)) =

α_{1}((a_{0}xy)(θ_{1}x+θ_{2}y+θ_{3}xy)) = 0and α_{1}(α_{−1}(a_{0}x)(a_{1}x+b_{1}xy)) = 0.

Case 3.2: If γ_{2} = 0, a_{1} 6= 0, b_{0} 6= 0, θ_{2} = 0, then we have that
[(γ_{1}x+γ_{3}xy) + (a_{0}x+b_{0}xy)z][(θ_{1}x+θ_{3}xy) + (a_{1}x+b_{1}xy)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 γ_{2} 6= 0, a_{1} = 0, θ_{2} = 0, b_{0} 6= 0, then we have that
[(γ_{1}x+γ_{2}y+γ_{3}xy) + (a_{0}x)z][(θ_{1}x+θ_{2}y+θ_{3}xy) + (b_{1}xy)z] = 0. Thus,
α_{1}(α_{−1}(a_{0}x)(θ_{1}x+ θ_{2}y +θ_{3}xy)) = α_{1}(a_{0}xy(θ_{1}x+ θ_{2}y +θ_{3}xy)) = 0,
α_{1}(α_{−1}(a_{0}x)(b_{1}xy)) = 0, (γ_{1}x+γ_{2}y+γ_{3}xy)(b_{1}xy) = 0 and(γ_{1}x+γ_{2}y+
γ_{3}xy)(θ_{1}x+θ_{2}y+θ_{3}xy) = 0.

Case 3.4: If γ_{2} 6= 0, a_{1} = 0, θ_{2} 6= 0, b_{0} = 0, then we have
that [(γ_{1}x + γ_{2}y + γ_{3}xy) + a_{0}xz][(θ_{1}x + θ_{2}y + θ_{3}xy) + (b_{1}xy)z] =
0. Thus, α_{1}(α_{−1}(a_{0}x)(θ_{1}x + θ_{3}xy)) = α_{1}((a_{0}xy)(θ_{1}x + θ_{3}xy)) = 0,
α_{1}(α_{−1}(a_{0}x)(b_{1}xy)) = 0, (γ_{1}x+γ_{2}y+γ_{3}xy)(b_{1}xy) = 0 and(γ_{1}x+γ_{2}y+
γ_{3}xy)(θ_{1}x+θ_{3}xy) = 0.

Case 3.5: Ifγ2= 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,
(γ_{1}x+γ_{3}xy)(b_{1}xy) = 0and (γ_{1}x+γ_{3}xy)(θ_{1}x+θ_{3}xy) = 0.

Case 3.6: If γ_{2} = 0, a_{1} = 0, θ_{2} = 0, b_{0} 6= 0, then we have that
[(γ_{1}x + γ_{3}xy) + (a_{0}x + b_{0}xy)z][(θ_{1}x + θ_{3}xy) + (b_{1}xy)z] = 0. Thus,
α_{1}(α_{−1}(a_{0}x+b_{0}xy)(θ_{1}x+θ_{3}xy)) =α_{1}((a_{0}xy+b_{0}x)(θ_{1}x+θ_{3}xy)) = 0,
α_{1}(α_{−1}(a_{0}x+b_{0}xy)(b_{1}xy)) = 0, (γ_{1}x+ γ_{3}xy)(b_{1}xy) = 0 and (γ_{1}x+

γ_{3}xy)(θ_{1}x+θ_{3}xy) = 0.

Case 3.7: Ifγ_{2} = 0,a_{1}6= 0,θ_{2} = 0,b_{0}= 0, then we have that[(γ_{1}x+
γ_{3}xy)+(a_{0}x)z][(θ_{1}x+θ_{3}xy)+(a_{1}x+b_{1}xy)z] = 0. Thus,α_{1}(α_{−1}(a_{0}x)(θ_{1}x+

θ_{3}xy)) = α_{1}((a_{0}xy)(θ_{1}x+θ_{3}xy)) = 0,α_{1}(α_{−1}(a_{0}x)(a_{1}x+b_{1}xy)) = 0,
(γ_{1}x+γ_{3}xy)(a_{1}x+b_{1}xy) = 0 and(γ_{1}x+γ_{3}xy)(θ_{1}x+θ_{3}xy) = 0.

Case 3.8: Ifγ_{2} 6= 0,a_{1}= 0,θ_{2} = 0,b_{0}= 0, then we have that[(γ_{1}x+
γ_{2}y+γ_{3}xy)+(a_{0}x)z][(θ_{1}x+θ_{3}xy)+(b_{1}xy)z] = 0. Thus,α_{1}(α_{−1}(a_{0}x)(θ_{1}x+

θ_{3}xy)) = α_{1}((a_{0}xy)(θ_{1}x+θ_{3}xy)) = 0, α_{1}(α_{−1}(a_{0}x)(b_{1}xy)) = 0,(γ_{1}x+
γ_{2}y+γ_{3}xy)(b_{1}xy) = 0and (γ_{1}x+γ_{2}y+γ_{3}xy)(θ_{1}x+θ_{3}xy) = 0.

Case 3.9: Ifγ_{2} = 0,a_{1}= 0,θ_{2} = 0,b_{0}= 0, then we have that[(γ_{1}x+
γ_{3}xy) + (a_{0}x)z][(θ_{1}x+θ_{3}xy) + (b_{1}xy)z] = 0. Thus, α_{1}(α_{−1}(a_{0}x)(θ_{1}x+
θ_{3}xy)) = α_{1}((a_{0}xy)(θ_{1}x+θ_{3}xy)) = 0, α_{1}(α_{−1}(a_{0}x)(b_{1}xy)) = 0,(γ_{1}x+
γ_{3}xy)(b_{1}xy) = 0 and (γ_{1}x+γ_{3}xy)(θ_{1}x+θ_{3}xy) = 0.

So, the result follows for polynomials of degree 1.

Next, let f(z) = Pn

i=0aiz^{i} and g(z) = Pn

i=0biz^{i} be polynomials of
degree ninR[z;α]such that f(z)g(z) = 0. Then a_{0}b_{0}= 0. Since I^{2} = 0,
we have that α_{i}(α_{−i}(a_{i})b_{j}) = 0, for every 1 ≤ i ≤ n and 1 ≤ j ≤ n.

Thus, we have that0 =f(z)g(z) =a_{0}b_{0}+ (a_{0}+α_{1}(α_{−1}(a_{1})b_{0}))z+ (a_{0}b_{2}+
α_{2}(α_{−2}(a_{2})b_{0})z^{2}+....+ (a_{0}bn+αn(α_{−n}(an)b_{0}))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 ringR[z;α] is not associative, because((y)(xyz))y = 0and (y)((xyz)y) = (y)(xz) =xyz6= 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 andB ofG^{′}, there exists at least one y∈G^{′} that has an unique repre-
sentation of the formy =abwith a∈A andb∈B. It is not difficult to