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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. Apartial action α of GonR is a collection of idealsSg of R,g∈G, and isomorphisms of (non-necessarily unital)k-algebras αg:Sg1 →Sg such that:

(i) S1 =R andα1 is the identity map of R;

(ii)S(gh)1 ⊇α−1h (Sh∩Sg1);

(iii) αg◦αh(x) =αgh(x), for everyx∈α−1h (Sh∩Sg1).

2000 Mathematics Subject Classification:16S36; 16S35.

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

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The property (ii) easily implies that αg(Sg1 ∩Sh) =Sg∩Sgh, for all g, h∈G. Also αg1g−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 idealsSg 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 ⋆αGis 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∈Sg−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{ασi :Sσ−i →Sσi, i∈Z}a partial action ofGonR. We simplify the notation by puttingαiσi andSi =Sσi, for everyi∈Z. Then the partial skew group ring can be identified with the set of all finite formal sumsPm

i=−naixi,ai∈Si for every−n≤i≤m, where the sum is usual and the multiplication rule isaxibxji−i(a)b)xi+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=0bixi,bi ∈Si 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

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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 rR[x](f(x)R[x]) 6= (0), thenrR[x](f(x)R[x])∩R6= (0),wheref(x)∈R[x]andrR[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.

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Lemma 1. Let f(x) =Pn

i=pbixi and g(x) =Pm

i=qaixi be two elements of Rhx;αi. Thenf(x)RPm

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

Proof. Suppose that f(x)RPm

i=qαj(ai1−j)xi+j = 0, for every j ∈ Z. Then, we have that

f(x)rxjg(x) =f(x)r1jxjg(x) =f(x)r

m

X

i=q

αj(ai1−j)xi+j = 0.

So,f(x)Rhx;αig(x) = 0.

The converse is trivial.

Let h(x) = Pn

i=saixi ∈ 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 rRhx;αi(f(x)Rhx;αi) 6= 0 then rRhx;αi(f(x)Rhx;αi) ∩R 6= 0, where rRhx;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) =apxp+....+ aqxq. If either f(x) is constant or f(x) = 0 or f(x) = aqxq, then the assertion is clear. So, assume thatq 6= 0,p < q and

rRhx;αi(f(x)Rhx;αi)∩R= (0).

Let g(x) = btxt+...+bmxm ∈ rRhx;αi(f(x)Rhx;αi) of minimal length with bm6= 0. Then

f(x)Rhx;αig(x) = 0, and we have thatf(x)1−qx−qRPm

i=tαj(bi1−j)xi+j = 0, for every j∈Z. Thus, aqj(1−jbm) = 0, for everyj ∈Z. Hence,aqRhx;αibm = 0 and we have that

aqRhx;αig(x) =aqRhx;αi(btxt+...+bm−1xm−1).

So,

f(x)Rhx;αiaqRhx;αi(btxt+...+bm−1xm−1)) =

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

f(x)Rhx;αiaqR(αj(bt1−j)xt+j+...+αj(bm−11−j)xm−1+j) = 0,

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for everyj∈Z. Note that by choice ofg(x), we have thataqR(αj(bt1−j)xt+j+ ...+αj(bm−11−j)xm−1+j)) = 0. Thus,

aqRhx;αi(αj(bt1−j)xt+j+...+αj(bm−11−j)xm−1+j) = 0

and it follows that

aq∈lR(Rhx;αi(αj(bt1−j)xt+j+...+αj(bm−11−j)xm−1+j)+

+Rhx;αi(bmxm)),

for every j∈Z. Hence,

(apxp+....+aq−1xq−1)Rhx;αig(x) = (0),

and we obtain thataq−1j(1−jbm) = 0, for every j∈Z. Consequently, (apxp+....+aq−1xq−1)Rhx;αiaq−1Rhx;αiPm−1

i=t bixi = f(x)(Rhx;αiaq−1Rhx;αi)g(x) = (0).

By choice ofg(x), we have that

aq−1Rhx;αig(x) = (0)

and we get

aq−1∈lR(Rhx;αi(αj(bt1−j)xt+j+...+αj(bm−11−j)xm−1+j)+

+Rhx;αi(bmxm))

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

as∈lR(Rhx;αi(αj(bt1−j)xt+j+...+αj(bm−11−j)xm−1+j)+

+Rhx;αi(bmxm))

for everyp≤s≤q andj∈Z. Since asRhx;αibmxm= (0),for everyp≤ s≤q, then(apxp+....+aqxq)Rhx;αibm= (0). This is a contradiction.

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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

aixi,

m

P

i=0

bixi ∈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. Letα be a partial action of an infinite cyclic group Gon 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 thatt1Rα1(t1R1−1) = 0, which implies that t1R= 0. Note that for eachi >0 we have that

(tσi(1R))σ(tσi(1Ri−1(1Ri(1R)) = 0.

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

Now, let b =tσ−i(1R), for i >0. Thenbσ(bσ−i(1R−i−1(1R)) = 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 T1 = 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 T1. We can extend the automorphismσ ofT to T1 by the rule:σ(a, z) =

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(σ(a), z). We claim thatT1 is aσ-rigid ring, i.e, if(a, z)∈T1 such that (a, z)σ(a, z) = 0, then we have that(a, z) = 0. In fact, for each(a, z)∈T1 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 ofT1 are invariant byσ. Since for any idempotente∈T, the element(e,0)is an idempotent in T1, 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∈ZKei 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 mapsα0 =idR andαi = 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. Letaj ∈Sj such thatα−j(aj)aj = 0, forj ≥0. Thenajxjajxjj−j(aj)aj)x2j = 0. Hence, by assump- tion we have thataxj = 0. So,aj = 0.

Conversely, let f(x) =Pn

i=0aixi be a polynomial inR[x;α] such that (f(x))2= 0. Then, αn−n(an)an) = 0and by assumption we have that an= 0. Proceeding in a similar way we obtain thata0 =a1=...=an= 0.

So,f = 0.

Using the Proposition 2 we can generalize the notion ofσ-rigid rings as follows.

Definition 1. Let Gbe an infinite cyclic group and α a partial action of Gon 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 ifR is partial α-rigid, thenR is reduced.

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

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

i=0aixi andg(x) =Pm

i=0bixi inR[x;α]such thatf(x)g(x) = 0, then α−i(ai)bj = 0, for every0≤i≤nand 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(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 byb1and using the equalitya0b1 =a0b0 = 0 we have thatb0a2α2(b01−2) = 0. Thus,α−2(a2)b0a2α2(b01−2) = 0and 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 that0 =a1α1(b11−1) = 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 rR(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⊕AnnR(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.

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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 idealsSi are the extension of[Si] toQ. The ideal Si for every i∈Zis generated by a central idempotent 1i. LetQsbe the Martindale’s right symmetric ring of quotients of R, i.e, according ([31], Proposition 5.14.7) we have thatQs ={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∩Qs(R) and we denote(I)s=I∗∗.

By ([31], Proposition 5.14.17) we have that 1i ∈Z(Q) =Z(Qs) and 1i ∈(I)s, for every i∈Z. Thus Si∗∗=Qs1i, 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∗∗ →Si∗∗:i∈Z}of Gon Qs andα∗∗gg|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 ofT′′. 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 thatR is partialα-rigid. Letq ∈Sn∗∗such thatα∗∗−n(q)q = 0. Since q ∈S−n∗∗ ⊆Sn, 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) Qs is reduced, since R is reduced. Hence,qlα∗∗−n(q) = 0, which implies that0 =qlα∗∗−n(q)α∗∗−n(l1n) =qlα∗∗−n(ql). Sinceα∗∗|S∗∗n , then by ([21], Theorem 3.1) we have thatα−nrestricted toS−nisα−nand 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).

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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 ofQs and α∗∗i |Sii, 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 andα 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 i6= 0 andI a two-sided ideal of D. For any integer iwe define Si =I∩σi(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 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 partialα-rigid ring, then T(R, R) is a partial skew Armendariz ring.

Proof. Letf(x) =Pn

i=0(ai, bi)xi andg(x) =Pn

j=0(cj, dj)xj 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 = Pn

i=0aixi, p1 = Pn

i=0bixi, q0 = Pn

j=0cjxj and q1 = Pn

j=0djxj.

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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≤nand 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}. SinceS is not necessarily associative, thenrS(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(2R) = {rR(U) : U ⊆ R} and for a not necessarily associative ring S we put analogouslyrAnnS(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) = Pn

i=0aixi ∈ 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=0aixi∈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(2R)→rAnnR[x;α](2R[x;α]) defined byφ(I) =IR[x;α], for everyI ∈rAnnR(2R) and

Ψ :rAnnR[x;α](2R[x;α])→rAnnR(2R)

defined byΨ(J) =J∩R, for every J ∈rAnnR[x;α](2R[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(2R)→rAnnR[x;α](2R[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=0aixi ∈ 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)∈SCf(x)byCS. We claim thatrR[x;α](f(x)) =

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rR[x;α](Cf(x)).In fact, letg(x) =Pm

i=0bixi ∈rR[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)∈rR[x;α](Cf(x)).

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

i=0cixi be an element inR[x;α]such thatCf(x)h(x) = 0.Then we have that α−i(ai)ck= 0, for every0≤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)∈SCf(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)∈SrR[x;α](f(x)) = ∩

f(x)∈SrR[x;α](Cf(x)) =

= ( ∩

f(x)∈SrR(Cf(x)))R[x;α] =rR(CS)R[x;α].

So,φis surjective.

Conversely, let f(x) = Pn

i=0aixi and g(x) = Pm

i=0bixi 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 idealB ofR. Thus, we have that, bi ∈ B ⊂ rR[x;α](f(x)), for every0 ≤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. LetR be a partial skew Armendariz ring ande=Pn

i=0eixi ∈ 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 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).

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Theorem 3. Suppose thatR 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 thatRis a Baer ring. Let∅6=U be a subset ofR[x;α]and f(x) = Pn

i=0aixi ∈U. We define Cf(x) = {α−i(ai), f or every 0≤ i≤ n}, andCU = ∪

f(x)∈UCf(x). By assumption, we have thatrR(CU) = eR, with e2 =e. It is easy to see thateR[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 thatrR[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 thatR 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 haverR[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 rS(Yi) =rS(Yi+p), for all p≥0, where Yj 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 rS(Y1) ⊆rS(Y2) ⊆..., there exists i≥1 such thatrS(Yi) = rS(Yi+p ), for allp≥0, whereYj are non-empty subsets ofS for everyj ≥0. In the next proposition, we suppose thatR is a partial skew Armendariz ring

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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 thatR 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 everyi≥1. We claim thatrR[x;α](Ui)⊂rR[x;α](Ui+1).In fact, let g(x) = Pn

i=0cixi 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 thatrR[x;α](Ul) = rR[x;α](Ul+p), for everyp∈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],whereCVi = ∪

f(x)∈Vi

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

and f(x) = Pm

i=0aixi. 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 thatrR(CVn) =rR(CVn+k),for everyk∈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 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

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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 thatR 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 thatrR[x;α](U) = 0, we have thatrR[x;α](Y) = 0for any non-empty finite subsetY of X.

Proof. Suppose thatR is a right zip ring. LetU 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], whereCU =∪f(x)∈UCf(x) andCf(x)={α−i(ai) : 0≤i≤ n}with f(x) =Pn

i=0aixi ∈U. Thus, rR(CU) = 0 and by assumption for any non-empty finite subsetW ={α−i1(ai1), ..., α−in(ain)}ofCU we have that rR(W) = 0. For every α−ij(aij) ∈W there existsgaij(x)∈U such that some of the coefficients ofgaij(x) are aij, for every1≤j ≤n. Let U0 be a minimal subset ofU such thatgaij(x)∈U0, for every 1≤j≤n.

ThenU0 is non-empty finite subset ofU. We denoteW0=∪f(x)∈U0(Cf(x)) and note thatW ⊆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 andf(x) =Pn

i=0aixi ∈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 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. Letα be a partial action of an infinite cyclic group Gon 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.

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Proof. Let a ∈ Sn such that α−n(a)a = 0. Then σ−n(a)a = 0, which implies thataσn(a) = 0. Thus, by ([28], Lemma 4) we have thata2= 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∈ZKei, 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, becausee2σ(e2) =e2e3 = 0 and e26= 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=0aixi 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 ofT. The ring T is said to be skew Armendariz ring if given f(x) = Pn

i=0aixi and g(x) = Pm

i=0bixi in T[x;σ] such that f(x)g(x) = 0, then aiσi(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]/(X2, Y2), 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

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

(by definitionS0 =R andα0=idR). We claim that R is a partial skew Armendariz ring. In fact, letf(z) = (γ01x+γ2y+γ3xy)+(a0x+b0xy)z andg(z) = (θ01x+θ2y+θ3xy)+(a1x+b1xy)zbe polynomials of degree 1 in R[z;α]such that f(z)g(z) = 0. Then (γ01x+γ2y+γ3xy)(θ0+ θ1x+θ2y+θ3xy) = 0, which impliesγ0θ0 = 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 γ0 = θ0 = 0, then we have that (γ1θ22θ1)xy = 0 and γ2a1x=b0θ2xy= 0. Thus,γ2a1+b0θ2= 0, γ2a1 = 0andb0θ2 = 0, which implies either γ2 = 0or a1 = 0and either b0 = 0or 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,α1−1(a0x)(θ1x+θ2y+θ3xy)) =

α1((a0xy)(θ1x+θ2y+θ3xy)) = 0and α1−1(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, α1−1((a0x+b0xy))(θ1x+θ3xy)) = α1((a0xy+b0x)(θ1x+θ3xy)) = 0, (γ1x +γ3xy)(a1x+b1xy) = 0, (γ1x +γ3xy)(θ1x+θ3xy) = 0 and α1−1((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, α1−1(a0x)(θ1x+ θ2y +θ3xy)) = α1(a0xy(θ1x+ θ2y +θ3xy)) = 0, α1−1(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, α1−1(a0x)(θ1x + θ3xy)) = α1((a0xy)(θ1x + θ3xy)) = 0, α1−1(a0x)(b1xy)) = 0, (γ1x+γ2y+γ3xy)(b1xy) = 0 and(γ1x+γ2y+ γ3xy)(θ1x+θ3xy) = 0.

Case 3.5: Ifγ2= 0,a1 = 0,θ2 6= 0,b0 = 0, then we have that [(γ1x+ γ3xy) + (a0x)z][(θ1x+θ23xy) + (b1xy)z] = 0. Thus,α1−1(a0x)(θ1x+ θ2y+θ3xy)) =α1((a0xy)(θ1x+θ2y+θ3xy)) = 0,α1−1(a0x)(b1xy)) = 0, (γ1x+γ3xy)(b1xy) = 0and (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, α1−1(a0x+b0xy)(θ1x+θ3xy)) =α1((a0xy+b0x)(θ1x+θ3xy)) = 0, α1−1(a0x+b0xy)(b1xy)) = 0, (γ1x+ γ3xy)(b1xy) = 0 and (γ1x+

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γ3xy)(θ1x+θ3xy) = 0.

Case 3.7: Ifγ2 = 0,a16= 0,θ2 = 0,b0= 0, then we have that[(γ1x+ γ3xy)+(a0x)z][(θ1x+θ3xy)+(a1x+b1xy)z] = 0. Thus,α1−1(a0x)(θ1x+

θ3xy)) = α1((a0xy)(θ1x+θ3xy)) = 0,α1−1(a0x)(a1x+b1xy)) = 0, (γ1x+γ3xy)(a1x+b1xy) = 0 and(γ1x+γ3xy)(θ1x+θ3xy) = 0.

Case 3.8: Ifγ2 6= 0,a1= 0,θ2 = 0,b0= 0, then we have that[(γ1x+ γ2y+γ3xy)+(a0x)z][(θ1x+θ3xy)+(b1xy)z] = 0. Thus,α1−1(a0x)(θ1x+

θ3xy)) = α1((a0xy)(θ1x+θ3xy)) = 0, α1−1(a0x)(b1xy)) = 0,(γ1x+ γ2y+γ3xy)(b1xy) = 0and (γ1x+γ2y+γ3xy)(θ1x+θ3xy) = 0.

Case 3.9: Ifγ2 = 0,a1= 0,θ2 = 0,b0= 0, then we have that[(γ1x+ γ3xy) + (a0x)z][(θ1x+θ3xy) + (b1xy)z] = 0. Thus, α1−1(a0x)(θ1x+ θ3xy)) = α1((a0xy)(θ1x+θ3xy)) = 0, α1−1(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) = Pn

i=0aizi and g(z) = Pn

i=0bizi be polynomials of degree ninR[z;α]such that f(z)g(z) = 0. Then a0b0= 0. Since I2 = 0, we have that αi−i(ai)bj) = 0, for every 1 ≤ i ≤ n and 1 ≤ j ≤ n.

Thus, we have that0 =f(z)g(z) =a0b0+ (a01−1(a1)b0))z+ (a0b2+ α2−2(a2)b0)z2+....+ (a0bnn−n(an)b0))zn. 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

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