Skew Polynomial Extensions over Zip Rings

Abstract

In this article, we study the relationship between left (right) zip property of 𝑅 and skew polynomial extension over 𝑅, using the skew versions of Armendariz rings.

1. Introduction

Throughout this paper 𝑅 denotes an associative ring with identity and 𝜎∶𝑅→𝑅 an automorphism of 𝑅, otherwise unless stated. We denote 𝑅[[𝑥;𝜎]] (𝑅[[𝑥,𝑥⁻¹;𝜎]]) the skew series rings (skew Laurent series rings) whose elements are the series ∑_𝑖≥0𝑎_𝑖𝑥^𝑖 (∑^∞_𝑗=𝑝𝑏_𝑗𝑥^𝑗), where the addition is defined as usual and the multiplication is defined by the rule, 𝑥𝑎=𝜎(𝑎)𝑥 (𝑥𝑎=𝜎(𝑎)𝑥 and 𝑥⁻¹𝑎=𝜎⁻¹(𝑎)𝑥), for any 𝑎∈𝑅. Note that the skew polynomial rings of automorphism type 𝑅[𝑥;𝜎] (skew Laurent of polynomial 𝑅[𝑥,𝑥⁻¹;𝜎]) are subrings of 𝑅[[𝑥;𝜎]] (𝑅[[𝑥,𝑥⁻¹;𝜎]]) whose elements are ∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖 (∑^𝑚_𝑗=𝑞𝑏_𝑗𝑥^𝑗) where the sum and multiplication are defined as before.

Rege and Chhawchharia in [1] introduced the notion of an Armendariz ring. A ring 𝑅 is called Armendariz if whenever polynomials ∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖, ∑^𝑚_𝑗=0𝑏_𝑗𝑥^𝑗∈𝑅[𝑥] satisfy 𝑓(𝑥)𝑔(𝑥)=0, then 𝑎_𝑖𝑏_𝑗=0 for each 0≤𝑖≤𝑛 and 0≤𝑗≤𝑚. The name Armendariz ring was chosen because Armendariz [2] had shown that a reduced ring (i.e., ring without nonzero nilpotent elements) satisfies this condition. Some properties of Armendariz rings have been studied by Rege and Chhawchharia [1], Armendariz [2], Anderson and Camillo [3], and Kim and Lee [4].

Faith in [5] called a ring 𝑅 right zip if the right annihilator 𝑟_𝑅(𝑋) of a subset 𝑋 of 𝑅 is zero, then 𝑟_𝑅(𝑌)=0 for a finite subset 𝑌⊆𝑋; equivalently, for a left ideal 𝐿 of 𝑅 with 𝑟_𝑅(𝐿)=0, there exists a finitely generated left ideal 𝐿₁⊆𝐿 such that 𝑟_𝑅(𝐿₁)=0. 𝑅 is zip if it is right and left zip. The concept of zip rings was initiated by Zelmanowitz [6] and appeared in various papers [5, 7–12], and references therein. Zelmanowitz stated that any ring satisfying the descending chain condition on right annihilators 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 [7] showed that if 𝑅 is a commutative zip ring, then the polynomial ring 𝑅[𝑥] over 𝑅 is zip. The authors in [13] proved that 𝑅 is a right (left) zip ring if and only if 𝑅[𝑥] is a right (left) zip ring when 𝑅 is an Armendariz ring.

In this paper, we study skew polynomial extensions over zip rings by using skew versions of Armendariz rings and we generalized the results of [13]. Our skew versions of Armendariz rings follow the ideas of [14, Definition]. Moreover, we provide some examples to display some of the phenomenas of Section 2.

2. Skew Polynomial Extensions over Zip Rings

Throughout this paper 𝜎 is an automorphism of 𝑅 unless otherwise stated and 𝑆 will denote one of the following rings: 𝑅[𝑥;𝜎], 𝑅[[𝑥;𝜎]], 𝑅[𝑥,𝑥⁻¹𝜎], and 𝑅[[𝑥,𝑥⁻¹;𝜎]]. A left (right) annihilator of a subset 𝑈 of 𝑅 is defined by 𝑙_𝑅(𝑈)={𝑎∈𝑅∶𝑎𝑈=0} (𝑟_𝑅(𝑈)={𝑎∈𝑅∶𝑈𝑎=0}). For a ring 𝑅, put 𝑟Ann_𝑅(2^𝑅)={𝑟_𝑅(𝑈)∶𝑈⊆𝑅} and 𝑙Ann_𝑅(2^𝑅)={𝑙_𝑅(𝑈)∶𝑈⊆𝑅}.

We begin with the following lemma and use it without further mention.

Lemma 2.1. Let 𝑆 be one of the rings above and 𝑈 a subset of 𝑅. The following statements hold:
(i) 𝑙_𝑆(𝑈)=𝑆𝑙_𝑅(𝑈),
(ii) 𝑟_𝑆(𝑈)=𝑟_𝑅(𝑈)𝑆.

Proof. (i) We only prove for the case 𝑆=𝑅[𝑥;𝜎] because the other cases are similar. Let 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑅[𝑥;𝜎] such that 𝑓(𝑥)𝑈=0. Then 𝜎^−𝑖(𝑎_𝑖)𝑈=0 for all 0≤𝑖≤𝑛 and it follows that 𝜎^−𝑖(𝑎_𝑖)∈𝑙_𝑅(𝑈) for all 0≤𝑖≤𝑛. Hence 𝑓(𝑥)=∑^𝑛_𝑖=0𝑥^𝑖𝜎^−𝑖(𝑎_𝑖)∈𝑅[𝑥;𝜎]𝑙_𝑅(𝑈). So 𝑙_{𝑅[𝑥;𝜎]}(𝑈)⊆𝑅[𝑥;𝜎]𝑙_𝑅(𝑈). We clearly have that 𝑅[𝑥;𝜎]𝑙_𝑅(𝑈)⊆𝑙_{𝑅[𝑥;𝜎]}(𝑈). Therefore, we have 𝑙_{𝑅[𝑥;𝜎]}(𝑈)=𝑅[𝑥;𝜎]𝑙_𝑅(𝑈).
(ii) We only prove for the case 𝑆=𝑅[𝑥;𝜎] because the other cases are similar. Let 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑅[𝑥;𝜎] such that 𝑈𝑓(𝑥)=0. Then 𝑈𝑎_𝑖=0 for all 0≤𝑖≤𝑛 and it follows that 𝑎_𝑖∈𝑟_𝑅(𝑈) for all 0≤𝑖≤𝑛. Hence 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑟_𝑅(𝑈)𝑅[𝑥;𝜎]. So 𝑟_{𝑅[𝑥;𝜎]}(𝑈)⊆𝑟_𝑅(𝑈)𝑅[𝑥;𝜎]. We clearly have that 𝑟_𝑅(𝑈)𝑅[𝑥;𝜎]⊆𝑟_{𝑅[𝑥;𝜎]}(𝑈). Therefore, we have 𝑟_{𝑅[𝑥;𝜎]}(𝑈)=𝑟_𝑅(𝑈)𝑅[𝑥;𝜎].

With the above lemma, we have maps 𝜙∶𝑟Ann_𝑅(2^𝑅)→𝑟Ann_𝑆(2^𝑆) defined by 𝜙(𝐼)=𝐼𝑆 for every 𝐼∈𝑟Ann_𝑅(2^𝑅) and

Ψ∶𝑙Ann_𝑅(2^𝑅)→𝑙Ann_𝑆(2^𝑆) (2.1) defined by Ψ(𝐼)=𝑆𝐼 for every 𝐼∈𝑙Ann_𝑅(2^𝑅). Moreover, we have maps Φ∶𝑟Ann_𝑆(2^𝑆)→𝑟Ann_𝑅(2^𝑅) defined by Φ(𝐽)=𝐽∩𝑅 for every 𝐽∈𝑟Ann_𝑆(2^𝑆) and Γ∶𝑙Ann_𝑆(2^𝑆)→𝑙Ann_𝑅(2^𝑅) defined by Γ(𝐽)=𝐽∩𝑅 for every 𝐽∈𝑙Ann_𝑆(2^𝑆). Obviously, 𝜙 is injective and Φ is surjective. Clearly, 𝜙 is surjective if and only if Φ is injective, and in this case 𝜙 and Φ are the inverses of each other. Note that Ψ and Γ satisfy the same relations as above. The first item of the definition below appears in [14, Definition].

Definition 2.2. (i) Suppose that 𝜎 is an endomorphism of 𝑅. A ring 𝑅 satisfies SA1^′ if for 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^𝑚_𝑗=0𝑏_𝑗𝑥^𝑗 in 𝑅[𝑥;𝜎], 𝑓(𝑥)𝑔(𝑥)=0 implies that 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 0≤𝑖≤𝑛 and 0≤𝑗≤𝑚. (ii)Suppose that 𝜎 is an endomorphism of 𝑅. A ring 𝑅 satisfies SA2^′ if for 𝑓(𝑥)=∑^∞_𝑖=0𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^∞_𝑗=0𝑏_𝑗𝑥^𝑗 in 𝑅[[𝑥;𝜎]], 𝑓(𝑥)𝑔(𝑥)=0 implies that 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 𝑖≥0, 𝑗≥0.(iii)Suppose that 𝜎 is an automorphism of 𝑅. A ring 𝑅 satisfies SA3^′ if for 𝑓(𝑥)=∑^𝑞_𝑖=𝑠𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^𝑛_𝑗=𝑡𝑏_𝑗𝑥^𝑗∈𝑅[𝑥,𝑥⁻¹;𝜎], 𝑓(𝑥)𝑔(𝑥)=0 implies that 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 𝑠≤𝑖≤𝑞 and 𝑡≤𝑗≤𝑛.(iv)Suppose that 𝜎 is an automorphism of 𝑅. A ring 𝑅 satisfies SA4^′ if for 𝑓(𝑥)=∑^∞_𝑖=𝑠𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^∞_𝑗=𝑡𝑏_𝑗𝑥^𝑗∈𝑅[[𝑥,𝑥⁻¹;𝜎]], 𝑓(𝑥)𝑔(𝑥)=0 implies that 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 𝑖≥𝑠 and 𝑗≥𝑡.

Note that if 𝑅 satisfies one of the conditions above, then all subrings 𝑆 of 𝑅 such that 𝜎(𝑆)⊆𝑆 satisfies the same property. The following implications are easy to verify: SA4^′⇒SA3^′ and SA2^′⇒SA1^′. Following [15, Example 2.1] when 𝜎=𝑖𝑑_𝑅, the last implication is not reversible.

Lemma 2.3. Let 𝜎 be an automorphism of 𝑅. Then
(i) 𝑅 satisfies 𝑆𝐴1^′ if and only if 𝑅 satisfies 𝑆𝐴3^′;
(ii) 𝑅 satisfies 𝑆𝐴2^′ if and only if 𝑅 satisfies 𝑆𝐴4^′.

Proof. Let 𝑓(𝑥),𝑔(𝑥)∈𝑅[𝑥,𝑥⁻¹;𝜎] such that 𝑓(𝑥)𝑔(𝑥)=0, where 𝑓(𝑥)=∑^𝑞_𝑖=−𝑝𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^𝑠_𝑗=−𝑡𝑏_𝑗𝑥^𝑗. We clearly have 𝑥^𝑝𝑓(𝑥)∈𝑅[𝑥;𝜎] and 𝑔(𝑥)𝑥^𝑡∈𝑅[𝑥;𝜎], then 𝑥^𝑝𝑓(𝑥)𝑔(𝑥)𝑥^𝑡=0. By assumption, 𝜎^𝑝(𝑎_𝑖)𝜎^𝑖+𝑝(𝑏_𝑗)=0 for all −𝑝≤𝑖≤𝑞 and −𝑡≤𝑗≤𝑠. Hence 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all −𝑝≤𝑖≤𝑞 and −𝑡≤𝑗≤𝑠. Since 𝑅[𝑥;𝜎]⊆𝑅[𝑥,𝑥⁻¹;𝜎], the converse follows.
The proof of the other statement is similar.

The following definition appears in [16, Definition 2.1].

Definition 2.4. Let 𝑅 be a ring and 𝜎 an endomorphism of 𝑅. Then 𝑅 is said 𝜎-compatible like right 𝑅-module, if 𝑎𝑟=0 if and only if 𝑎𝜎(𝑟)=0 for any 𝑎∈𝑅 and 𝑟∈𝑅.

Let 𝑅 be a ring and 𝛼 an endomorphism of 𝑅. Following [17], the endomorphism 𝛼 is said 𝛼-rigid if 𝑟𝛼(𝑟)=0, then 𝑟=0. A ring 𝑅 is said a rigid ring if it exists a rigid endomorphism 𝛼 of 𝑅.

Proposition 2.5. Let 𝜎 be an endomorphism of 𝑅. If 𝑅 is a reduced ring and 𝜎-compatible like right 𝑅-module, then 𝑅 is a 𝜎-rigid ring and hence satisfies 𝑆𝐴1^′ and 𝑆𝐴2^′.

Proof. We only prove the case of SA2^′ because the other are similar. We claim that 𝑅[[𝑥;𝜎]] is a reduced ring. In fact, let 𝑓(𝑥)=∑^∞_𝑖=0𝑎_𝑖𝑥^𝑖 such that (𝑓(𝑥))²=0. We have that 𝑎²₀=0. Since 𝑅 is reduced, then 𝑎₀=0. Next, we have 𝑎₁𝜎(𝑎₁)=0, since 𝑅 is 𝜎-compatible and reduced, then 𝑎₁=0. By induction, we get 𝑓(𝑥)=0. Hence 𝑅[[𝑥;𝜎]] is reduced. Using the same ideas of [14, Proposition 3], we have that 𝑅 is 𝜎-rigid and using similar ideas of [14, Corollary 4], we obtain that 𝑅 satisfies SA2^′.

Without the assumption that 𝑅 is 𝜎-compatible, Proposition 2.5 is not true. In fact, let 𝑅=ℤ₂⊕ℤ₂ and 𝜎∶𝑅→𝑅, defined by 𝜎((𝑎,𝑏))=(𝑏,𝑎). By [14, Example 2], 𝑅 does not satisfy SA2^′ because 𝑅 does not satisfy SA1^′. Observe that (1,0)(0,1)=(0,0) but (1,0)𝜎(0,1)≠(0,0) and so 𝑅 is not 𝜎-compatible. We have the following natural questions.

Questions
(i)Let 𝜎 be an endomorphism of 𝑅. Suppose that 𝑅 satisfies SA2^′. Is 𝑅𝜎-compatible like right 𝑅-module?(ii)Let 𝜎 be an endomorphism of 𝑅. Suppose that 𝑅 is 𝜎-compatible like right 𝑅-module. Does 𝑅 satisfy SA2^′?

The question (i) is false. Let 𝑅₀ be any domain and 𝑅=𝑅₀[𝑥]. Let 𝜎∶𝑅→𝑅 be defined by 𝜎(𝑡)=0 and 𝜎|_𝑅0=𝑖𝑑_𝑅0. By [16, Example 4.1], 𝑅 is not 𝜎-compatible and using the similar ideas of the proof of [14, Proposition 10], we have that 𝑅 satisfies SA2^′ and consequently 𝑅 satisfies SA1^′.

The question (ii) is false. Let 𝑅=𝐾[𝑥,𝑦]/(𝑥²,𝑦²), where 𝐾 is a field of characteristic 2, and consider 𝑇=𝑀₂(𝑅). In this case, take 𝜎=𝑖𝑑_𝑇. By [18, Example 3.6], 𝑆 does not satisfy SA2^′ because 𝑇 does not satisfy SA1^′. Moreover, 𝑇 is 𝜎-compatible like right 𝑇-module.

In [19] the authors introduced the following version of skew Armendariz rings. (i)Suppose that 𝜎 is an endomorphism of 𝑅. Let 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖,𝑔(𝑥)=∑^𝑚_𝑗=0𝑏_𝑗𝑥^𝑗∈𝑅[𝑥;𝜎] such that 𝑓(𝑥)𝑔(𝑥)=0 implies 𝑎_𝑖𝑏_𝑗=0 for all 0≤𝑖≤𝑛 and 0≤𝑗≤𝑚.(ii)Suppose that 𝜎 is an endomorphism of 𝑅. Let 𝑓(𝑥)=∑_𝑖≥0𝑎_𝑖𝑥^𝑖, 𝑔(𝑥)=∑_𝑗≥0𝑏_𝑗𝑥^𝑗∈𝑅[[𝑥;𝜎]] such that 𝑓(𝑥)𝑔(𝑥)=0 implies 𝑎_𝑖𝑏_𝑗=0 for all 𝑖≥0 and 𝑗≥0.

Note that the item (i) above in [20, Definition 1.1] the authors called it by 𝜎-Armendariz, the item (ii) above is similar with [20, Definition 1.1] and we call it here by 𝜎-power Armendariz.

In the next proposition, we give a relationship between the definition above and the skew versions of Armendariz rings used in this paper. Using [21, Lemma 2.1] and [20, Theorem 1.8], the proof of next proposition is easy to verify.

Proposition 2.6. Let 𝜎 be an endomorphism of 𝑅 and suppose that 𝑅 is 𝜎-compatible like right R-module. Then
(i) 𝑅 satisfies 𝑆𝐴1^′ if and only if 𝑅 is 𝜎-Armendariz;
(ii) 𝑅 satisfies 𝑆𝐴2^′ if and only if 𝑅 is 𝜎-power Armendariz.

The proposition above without the compatibility assumption is not true according to [20, Example 1.9] and the authors in [22, Theorem 2.2] obtained an approach of the result above without the compatibility assumption.

The following proposition is a generalization of [18, Proposition 3.4] and partially generalizes [15, Proposition 2.6].

Lemma 2.7. Let 𝑆 be any of the rings 𝑅[𝑥;𝜎] and 𝑅[[𝑥;𝜎]]. The following conditions are equivalent:
(i) 𝑅 satisfies 𝑆𝐴2^′(𝑆𝐴1^′);
(ii) 𝜙∶𝑟𝐴𝑛𝑛_𝑅(2^𝑅)→𝑟𝐴𝑛𝑛_𝑆(2^𝑆) defined by 𝜙(𝐽)=𝐽𝑆 is bijective;
(iii) Ψ∶𝑙𝐴𝑛𝑛_𝑅(2^𝑅)→𝑙𝐴𝑛𝑛_𝑆(2^𝑆) defined by Ψ(𝐽)=𝑆𝐽 is bijective.

Proof. We only prove the proposition in the case of SA2^′ because the equivalence of (i) and (ii) when 𝑅 satisfies SA1^′ was proved in [23, Proposition 3.2]. The equivalence between (i) and (iii) when 𝑅 satisfies SA1^′ has similar proof.
(i)→(ii). It is only necessary to show that 𝜙 is surjective. For an element 𝑓(𝑥)=∑^∞_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑅[[𝑥;𝜎]], define 𝐶_𝑓(𝑥)={𝜎^−𝑖(𝑎_𝑖),𝑖≥0}, and for a subset 𝑇 of 𝑅[[𝑥;𝜎]], we denote the set ∪_{𝑓(𝑥)∈𝑇}𝐶_𝑓(𝑥) by 𝐶_𝑇. We show that 𝑟_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥))=𝑟_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥)). In fact, given 𝑔(𝑥)=∑^∞_𝑗=0𝑏_𝑗𝑥^𝑗 in 𝑟_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥)), we have 𝑓(𝑥)𝑔(𝑥)=0. Since 𝑅 satisfies SA2^′, then 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 𝑖≥0 and 𝑗≥0. In particular, 𝜎^−𝑖(𝑎_𝑖)𝑏_𝑗=0 for all 𝑖≥0 and 𝑗≥0. Hence 𝑔(𝑥)∈𝑟_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥)).
On the other hand, let ℎ(𝑥)=∑^∞_𝑘=0𝑐_𝑘𝑥^𝑘 be an element in 𝑅[[𝑥;𝜎]] such that 𝐶_𝑓(𝑥)ℎ(𝑥)=0. It is clear that 𝑎_𝑖𝜎^𝑖(𝑐_𝑘)=0 for all 𝑖≥0 and 𝑘≥0. So 𝑓(𝑥)ℎ(𝑥)=(0). Since 𝑅 satisfies SA2^′ then 𝑟_{𝑅[[𝑥;𝜎]]}(𝑇)=𝑟_{𝑅[[𝑥;𝜎]]}(∪_{𝑓(𝑥)∈𝑇}𝐶_𝑓(𝑥)). Thus
𝑟_{𝑅[[𝑥;𝜎]]}(𝑇)=_{𝑓(𝑥)∈𝑇}𝑟_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥))=_{𝑓(𝑥)∈𝑇}𝑟_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥))=(_{𝑓(𝑥)∈𝑇}𝑟_𝑅(𝐶_𝑓(𝑥)))𝑅[[𝑥;𝜎]]=𝑟_𝑅(𝐶_𝑇)𝑅[[𝑥;𝜎]]. (2.2) Therefore, 𝜙 is surjective.
(ii)→(i). Let 𝑓(𝑥)=∑^∞_𝑖=0𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^∞_𝑗=0𝑏_𝑗𝑥^𝑗 be elements in 𝑅[[𝑥;𝜎]] such that 𝑓(𝑥)𝑔(𝑥)=0. By assumption, 𝑟_{𝑅[[𝑥,𝜎]]}(𝑓(𝑥))=𝐵𝑅[[𝑥;𝜎]], for some right ideal 𝐵 of 𝑅. Hence 𝑔(𝑥)∈𝐵𝑅[[𝑥;𝜎]] and we have that 𝑏_𝑗∈𝐵⊂𝑟_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥)) for all 𝑗≥0. So 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 𝑖≥0 and 𝑗≥0.
(iii)→(i). Let 𝑓(𝑥)=∑_𝑖≥0𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑_𝑗≥0𝑏_𝑗𝑥^𝑗 be elements in 𝑅[[𝑥;𝜎]] such that 𝑓(𝑥)𝑔(𝑥)=0. By assumption, 𝑙_{𝑅[[𝑥;𝜎]]}(𝑔(𝑥))=𝑅[[𝑥;𝜎]]𝐵 for some left ideal 𝐵 of 𝑅. We can write 𝑓(𝑥)=∑_𝑖≥0𝑥^𝑖𝜎^−𝑖(𝑎_𝑖)∈𝑅[[𝑥;𝜎]]𝐵. By the equality of the polynomials with the coefficients on the right side, we have that 𝜎^−𝑖(𝑎_𝑖)∈𝐵⊆𝑙_{𝑅[[𝑥;𝜎]]}(𝑔(𝑥)) for all 𝑖≥0. So 𝑎_𝑖𝜎^𝑖(𝑏_𝑗)=0 for all 𝑖≥0 and 𝑗≥0.
(i)→(iii). It is only necessary to show that Ψ is surjective. Let 𝑓(𝑥)=∑_𝑖≥0𝑎_𝑖𝑥^𝑖∈𝑅[[𝑥;𝜎]]. Define 𝐶_𝑓(𝑥)={𝑎_𝑖,𝑖≥0}, and for a subset 𝑇 of 𝑅[[𝑥;𝜎]], we denote the set ∪_{𝑓(𝑥)∈𝑇}𝐶_𝑓(𝑥) by 𝐶_𝑇. We show that
𝑙_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥))=𝑙_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥)). (2.3) In fact, given 𝑔(𝑥)=∑_𝑗≥0𝑏_𝑗𝑥^𝑗∈𝑙_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥)), we have 𝑔(𝑥)𝑓(𝑥)=0. Since 𝑅 satisfies SA2^′, then 𝑏_𝑗𝜎^𝑗(𝑎_𝑖)=0 for all 𝑖≥0 and 𝑗≥0. Hence 𝑔(𝑥)=∑_𝑗≥0𝑥^𝑗𝜎^−𝑗(𝑏_𝑗)∈𝑙_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥)).
On the other hand, let 𝑔(𝑥)∈𝑅[[𝑥;𝜎]] such that 𝑔(𝑥)𝐶_𝑓(𝑥)=0. Thus 𝑔(𝑥)𝑎_𝑖=0 for all 𝑖≥0. So 𝑔(𝑥)∑_𝑖≥0𝑎_𝑖𝑥^𝑖=𝑔(𝑥)𝑓(𝑥)=0, and we have that 𝑔(𝑥)∈𝑙_{𝑅[[𝑥;𝜎]]}(𝑓(𝑥)).
We easily have that for each subset 𝑇 of 𝑅[[𝑥;𝜎]],
𝑙_{𝑅[[𝑥;𝜎]]}(𝑇)=𝑙_{𝑅[[𝑥;𝜎]]}(_{𝑓(𝑥)∈𝑇}𝐶_𝑓(𝑥)). (2.4) We claim that 𝑙_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥))=𝑅[[𝑥;𝜎]]𝑙_𝑅(𝐶_𝑓(𝑥)). In fact, let 𝑔(𝑥)=∑_𝑗≥0𝑏_𝑗𝑥^𝑗 such that 𝑔(𝑥)𝐶_𝑓(𝑥)=0. Then we have that 0=𝑔(𝑥)𝑎_𝑖=∑_𝑗≥0𝑏_𝑗𝑥^𝑗𝑎_𝑖=∑_𝑗≥0𝑥^𝑗𝜎^−𝑗(𝑏_𝑗)𝑎_𝑖. Thus 𝜎^−𝑗(𝑏_𝑗)∈𝑙_𝑅(𝐶_𝑓(𝑥)), and it follows that
_𝑗≥0𝑥^𝑗𝜎^−𝑗(𝑏_𝑗)∈𝑅[[𝑥;𝜎]]𝑙_𝑅(𝐶_𝑓(𝑥)). (2.5) The other inclusion is trivial. So
𝑙_{𝑅[[𝑥;𝜎]]}(𝑇)=_{𝑓(𝑥)∈𝑇}𝑙_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥))=_{𝑓(𝑥)∈𝑇}𝑙_{𝑅[[𝑥;𝜎]]}(𝐶_𝑓(𝑥))=𝑅[[𝑥;𝜎]](_{𝑓(𝑥)∈𝑇}𝑙_𝑅(𝐶_𝑓(𝑥)))=𝑅[[𝑥;𝜎]]𝑙_𝑅(𝐶_𝑇). (2.6)
Therefore, Ψ is surjective.

Now we are able to prove the main results of this paper.

Theorem 2.8. Let 𝜎 be an automorphism of 𝑅.
(i)Suppose that 𝑅 satisfies 𝑆𝐴1^′. The following conditions are equivalent:(a)𝑅 is a right (left) zip ring;(b)𝑅[𝑥;𝜎] is a right (left) zip ring;(c)𝑅[𝑥,𝑥⁻¹,𝜎] is a right (left) zip ring.(ii)Suppose that 𝑅 satisfies 𝑆𝐴2^′. The following conditions are equivalent:(a)𝑅 is right (left) zip ring;(b)𝑅[[𝑥;𝜎]] is right (left) zip ring;(c)𝑅[[𝑥,𝑥⁻¹;𝜎]] is right (left) zip ring.

Proof. (i) We will show the right case because the left case is similar.
Suppose that 𝑅[𝑥;𝜎] is right zip. Let 𝑋 be a subset of 𝑅 such that 𝑟_𝑅(𝑋)=0, and 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑅[𝑥;𝜎] such that 𝑋𝑓(𝑥)=0. Thus 𝑎_𝑖∈𝑟_𝑅(𝑋)=0 and it follows that 𝑓(𝑥)=0. By assumption, there exists 𝑋₁={𝑥₀,…,𝑥_𝑛} such that 𝑟_{𝑅[𝑥;𝜎]}(𝑋₁)=0. Hence 𝑟_𝑅(𝑋₁)=𝑟_{𝑅[𝑥;𝜎]}(𝑋₁)∩𝑅=(0).
Conversely, let 𝑌⊆𝑅[𝑥;𝜎] such that 𝑟_{𝑅[𝑥;𝜎]}(𝑌)=0. By Lemma 2.7, 𝑟_{𝑅[𝑥;𝜎]}(𝑌)=𝑟_𝑅(𝑇)𝑅[𝑥;𝜎], where 𝑇=𝐶_𝑌=∪_{𝑓(𝑥)∈𝑌}𝐶_𝑓(𝑥) such that 𝐶_𝑓(𝑥)={𝜎^−𝑖(𝑎_𝑖)∶0≤𝑖≤𝑛} with 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑌. We have that 𝑟_𝑅(𝑇)=0 and, by assumption, there exists 𝑇₁={𝜎^−𝑖1(𝑎_𝑖1),…,𝜎^−𝑖𝑛(𝑎_𝑖𝑛)} such that 𝑟_𝑅(𝑇₁)=0. For each 𝜎^−𝑖𝑗(𝑎_𝑖𝑗)∈𝑇₁, there exists 𝑔_𝑎𝑖𝑗(𝑥)∈𝑌 such that some of the coefficients of 𝑔_𝑎𝑖𝑗(𝑥) are 𝑎_𝑖𝑗 for each 1≤𝑗≤𝑛. Let 𝑌₀ be a minimal subset of 𝑌 such that 𝑔_𝑎𝑖𝑗(𝑥)∈𝑌₀ for each 1≤𝑗≤𝑛. Then 𝑌₀ is nonempty finite subset of 𝑌. Set 𝑇₀=∪_{𝑓(𝑥)∈𝑌0}(𝐶_𝑓(𝑥)) and we have that 𝑇₁⊆𝑇₀. Hence 𝑟_𝑅(𝑇₀)⊆𝑟_𝑅(𝑇₁)=0. By Lemma 2.7, 𝑟_{𝑅[𝑥;𝜎]}(𝑌₀)=𝑟_𝑅(𝑇₀)𝑅[𝑥;𝜎] and it follows that 𝑟_{𝑅[𝑥;𝜎]}(𝑌₀)=0.
The proofs of (a)⇔(c) and of item (ii) follow similarly.

Let 𝜎 be an endomorphism of 𝑅 and 𝛿∶𝑅→𝑅 an additive map of 𝑅. The application 𝛿 is said to be a 𝜎-derivation if 𝛿(𝑎𝑏)=𝛿(𝑎)𝑏+𝜎(𝑎)𝛿(𝑏). The Ore extension 𝑅[𝑥;𝜎,𝛿] is the set of polynomials ∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖 with the usual sum, and the multiplication rule is 𝑥𝑎=𝜎(𝑎)𝑥+𝛿(𝑎).

Following [16], 𝑅 is said to be (𝜎,𝛿)-compatible, where 𝜎 is an endomorphism of 𝑅 and 𝛿 is a 𝜎-derivation of 𝑅 if 𝑎𝑏=0⇔𝑎𝜎(𝑏)=0 and 𝑎𝑏=0 implies that 𝑎𝛿(𝑏)=0.

In the next result we obtain a necessary and sufficient condition for 𝑅[𝑥;𝜎,𝛿] to be left zip, when 𝜎 is an endomorphism of 𝑅 using the skew version of Armendariz rings of [19].

Theorem 2.9. Let 𝜎 be an endomorphism of 𝑅 and 𝛿 a 𝜎-derivation of 𝑅. Suppose that if 𝑓(𝑥)𝑔(𝑥)=0 for 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^𝑚_𝑗=0𝑏_𝑗𝑥^𝑗∈𝑅[𝑥;𝜎,𝛿], then 𝑎_𝑖𝑏_𝑗=0 for all 0≤𝑖≤𝑛 and 0≤𝑗≤𝑚. Then 𝑅 is left zip if and only if 𝑅[𝑥;𝜎,𝛿] is left zip.

Proof. Let 𝑋 be any subset of 𝑅[𝑥;𝜎,𝛿] and 𝐶_𝑋=∪_{𝑓(𝑥)∈𝑋}𝐶_𝑓(𝑥), where 𝐶_𝑓(𝑥)={𝑎_𝑖,0≤𝑖≤𝑛} with 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖. Suppose that 𝑙_{𝑅[𝑥;𝜎,𝛿]}(𝑋)=0. We clearly have 𝑙_𝑅(𝐶_𝑋)=0. By assumption, there exists {𝑏₀,…,𝑏_𝑡}⊆𝐶_𝑋 such that 𝑙_𝑅(𝑌)=0. Let 𝑓_𝑏𝑖(𝑥)∈𝑋 be an element of 𝑋 with some of its coefficients are equal to 𝑏_𝑖 for all 1≤𝑖≤𝑡. Take 𝑋₀ be a minimal subset of 𝑋 with this property. We clearly have that 𝑋₀ is a finite set. We claim that 𝑙_{𝑅[𝑥;𝜎,𝛿]}(𝑋₀)=0. In fact, we easily have 𝑙_𝑅(𝐶_𝑋0)=0, where 𝐶_𝑋0=∪_{𝑓(𝑥)∈𝑋0}𝐶_𝑓(𝑥) with 𝐶_𝑓(𝑥) being defined as before. Next, let 𝑔(𝑥)=∑^𝑚_𝑗=0𝑏_𝑗𝑥^𝑗 such that 𝑔(𝑥)𝑋₀=0. Hence for any 𝑓(𝑥)=∑^𝑛_𝑖=0𝑎_𝑖𝑥^𝑖∈𝑋₀, 𝑔(𝑥)𝑓(𝑥)=0, and we have, by assumption, 𝑏_𝑗𝑎_𝑖=0 for all 0≤𝑗≤𝑚 and 0≤𝑖≤𝑛. Thus 𝑏_𝑗𝐶_𝑋0=0 for all 0≤𝑗≤𝑚 and it follows that 𝑔(𝑥)=0. So 𝑙_{𝑅[𝑥;𝜎,𝛿]}(𝑋₀)=0.
Using the methods of Theorem 2.8, the converse follows.

Remark 2.10. Let 𝑅 be a ring and 𝜎 an endomorphism of 𝑅. Suppose that 𝑅 is 𝜎-power Armendariz and left zip. Using similar methods of [20, Theorem 1.8], 𝑅 satisfies SA2^′ and with similar ideas of Theorem 2.9, we have that 𝑅 is a left zip ring if and only if 𝑅[[𝑥;𝜎]] is a left zip ring.

3. Examples

In this section, we present some examples of rings that satisfy SA1^′ and SA2^′, and they are zip rings. Moreover, an example of a 𝜎-rigid ring that is a zip ring is given.

Example 3.1.1. Let 𝐹 be any field and 𝜎∶𝐹→𝐹 any automorphism of 𝐹. Following [14, page 113], we consider the ring 𝑇(𝐹,𝐹) with automorphism 𝜎(𝑎,𝑏)=(𝜎(𝑎),𝜎(𝑏)) and we denote it by 𝜎. Note that
𝑇(𝐹,𝐹)={𝑎𝑏0𝑎∶𝑎,𝑏∈𝐹}. (3.1) By [14, Proposition 15], 𝑇(𝐹,𝐹) satisfies SA1^′, and using similar methods, we can prove that 𝑇(𝐹,𝐹) satisfies SA2^′. We claim that 𝑇(𝐹,𝐹) is a zip ring. In fact, the unique one-sided ideals of 𝑇(𝐹,𝐹) are {(⁰⁰₀₀)},
𝐼={0𝑏00∶𝑏∈𝐹}, (3.2) and 𝑇(𝐹,𝐹). Note that 𝑟_{𝑇(𝐹,𝐹)}(𝐼)≠{0} and 𝑙_{𝑇(𝐹,𝐹)}(𝐼)≠0. So we easily have that 𝑇(𝐹,𝐹) is a zip ring.

Example 3.2.2. Let 𝐹 be any field and 𝜎 a monomorphism of 𝐹, and let
𝑅={𝑎𝑏𝑐0𝑎𝑑00𝑎∶𝑎,𝑏,𝑐∈𝐹} (3.3) with usual addition and multiplication of matrix. Note that the monomorphism 𝜎 is naturally extended to 𝑅, and 𝑅 has the following one-sided ideals:
𝐼₁={00000𝑎000∶𝑎∈𝐹},𝐼₂={00𝑐000000∶𝑐∈𝐹}, (3.4)𝑅 and the zero ideal. We easily have 𝑟_𝑅(𝐼₂)≠0, 𝑙_𝑅(𝐼₂)≠0, 𝑟_𝑅(𝐼₁)≠0, and 𝑙_𝑅(𝐼₁)≠0. Now we clearly have that 𝑅 is a zip ring and by [14, Proposition 17], 𝑅 satisfies SA1^′, and with similar methods of [14, Proposition 17], we can prove that 𝑅 satisfies SA2^′.

Example 3.3.3.3. Let 𝐷 be any domain with identity, 𝑅=𝐷[𝑥], 𝜎 an endomorphism of 𝑅 defined by 𝜎(𝑓(𝑥))=𝑓(0). Since 𝑅 is a domain, then 𝑅 is right and left zip. Moreover, using similar methods of [14, Example 5], we have that 𝑅 satisfies SA1^′ and SA2^′.

Example 3.4.4. Let 𝐷 and 𝐷₁ be any domains, 𝜎 an monomorphism of 𝐷, and 𝜏 an monomorphism of 𝐷₁. Set 𝑅=𝐷×𝐷₁ with usual addition and multiplication, and we define an endomorphism 𝛾 of 𝑅 by 𝛾(𝑎,𝑏)=(𝜎(𝑎),𝜏(𝑏)). We easily have that 𝛾 is a monomorphism of 𝑅. Since 𝐷 is 𝜎-rigid and 𝐷₁ is 𝜏-rigid, we easily obtain that 𝑅 is 𝛾-rigid. We claim that 𝑅 is left and right zip. In fact, let 𝐼 be any left ideal of 𝑅. It is well known that 𝐼=𝐴×𝐵, where 𝐴 is a left ideal of 𝐷 and 𝐵 is a left ideal of 𝐷₁. Suppose that 𝑟_𝑅(𝐼)=0. Then 𝐴≠0 and 𝐵≠0. It is not difficult to show that 𝑟_𝐷(𝐴)=0 and 𝑟_𝐷1(𝐵)=0. Since 𝐷 and 𝐷₁ are left zip, then there exists a left finitely generated ideal 𝐿 of 𝐷 contained in 𝐴 such that 𝑟_𝐷(𝐿)=0 and a left finitely generated ideal 𝐿₁ of 𝐷₁ contained in 𝐵 such that 𝑟_𝐷1(𝐿₁)=0. Thus 𝑟_𝑅(𝐿×𝐿₁)=0 and 𝐿×𝐿₁ is a left finitely generated ideal of 𝑅 contained in 𝐴×𝐵. Hence 𝑅 is left zip. Using similar methods, we have that 𝑅 is right zip.

Example 3.5.5. Let 𝐹 be a field, 𝜎 an automorphism of 𝐹,
𝑅={𝑎𝑏𝑐0𝑎𝑑00𝑎∶𝑎,𝑏,𝑐∈𝐹}, (3.5) and 𝐷 a domain with automorphism 𝜏. Set 𝑇=𝑅×𝐷 and we define an endomorphism 𝛾 of 𝑇 by 𝛾(𝑎,𝑏)=(𝜎(𝑎),𝜏(𝑡)). It is clear that 𝛾 is an automorphism of 𝑇 and it is not difficult to show that 𝑇 satisfies SA1^′ and SA2' because 𝑅 and 𝐷 satisfy SA1^′ by [14,Proposition 17] and [14, Proposition 10], respectively, and using similar methods of [14,Proposition 17] and [14, Proposition 10], 𝑅 and 𝐷 satisfy SA2^′, respectively.
Using similar methods of Example 3.4, we have that 𝑇 is right and left zip and note that 𝑇 is not 𝛾-rigid, since 𝑇 is not a reduced ring.

Acknowledgment

The author is deeply indebted to the referees for many helpful comments and suggestions for the improvement of this paper.