Skew Polynomial Extensions over Zip Rings

Volume 2008, Article ID 496720,9pages doi:10.1155/2008/496720

Research Article

Skew Polynomial Extensions over Zip Rings

Wagner Cortes

Instituto de Matem´atica, Universidade Federal do Rio Grande do Sul, 91509-900 Porto Alegre, RS, Brazil

Correspondence should be addressed to Wagner Cortes,[email protected] Received 17 September 2007; Revised 27 November 2007; Accepted 14 January 2008 Recommended by Francois Goichot

In this article, we study the relationship between leftrightzip property ofRand skew polynomial extension overR, using the skew versions of Armendariz rings.

Copyrightq2008 Wagner Cortes. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Throughout this paper R denotes an associative ring with identity andσ : R→R an auto- morphism ofR, otherwise unless stated. We denoteRx;σ Rx, x⁻¹;σthe skew series ringsskew Laurent series ringswhose elements are the series

i≥0a_ixⁱ _∞

jpb_jx^j, where the addition is defined as usual and the multiplication is defined by the rule,xa σax xa σax andx⁻¹a σ⁻¹ax, for any a ∈ R. Note that the skew polynomial rings of automorphism typeRx;σ skew Laurent of polynomialRx, x⁻¹;σare subrings ofRx;σ Rx, x⁻¹;σwhose elements are_n

i0a_ixⁱ_m

jqb_jx^jwhere the sum and multiplication are defined as before.

Rege and Chhawchharia in1introduced the notion of an Armendariz ring. A ringR is called Armendariz if whenever polynomials_n

i0a_ixⁱ,_m

j0b_jx^j ∈Rxsatisfyfxgx 0, thena_ib_j 0 for each 0≤i≤nand 0≤j ≤m. The name Armendariz ring was chosen because Armendariz2had shown that a reduced ringi.e., ring without nonzero nilpotent elements satisfies this condition. Some properties of Armendariz rings have been studied by Rege and Chhawchharia1, Armendariz2, Anderson and Camillo3, and Kim and Lee4.

Faith in 5called a ring R right zip if the right annihilatorrRXof a subset X ofR is zero, then rRY 0 for a finite subset Y ⊆ X; equivalently, for a left ideal L ofR with r_RL 0, there exists a finitely generated left idealL₁ ⊆Lsuch thatr_RL1 0.Ris zip if it is right and left zip. The concept of zip rings was initiated by Zelmanowitz6and appeared in various papers5,7–12, and references therein. Zelmanowitz stated that any ring satisfying

the descending chain condition on right annihilators is a right zip ringalthough not so-called at that time, but the converse does not hold. Extensions of zip rings were studied by several authors. Beachy and Blair7showed that ifRis a commutative zip ring, then the polynomial ringRxoverRis zip. The authors in13proved thatRis a rightleftzip ring if and only if Rxis a rightleftzip ring whenRis 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 of14, Definition. Moreover, we provide some examples to display some of the phenomenas ofSection 2.

2. Skew polynomial extensions over zip rings

Throughout this paperσ is an automorphism ofRunless otherwise stated andSwill denote one of the following rings: Rx;σ, Rx;σ, Rx, x⁻¹σ, and Rx, x⁻¹;σ. A left right annihilator of a subsetU ofRis defined byl_RU {a ∈ R : aU 0}rRU {a ∈ R : Ua0}. For a ringR, putrAnnR2^R {rRU:U⊆R}andlAnnR2^R {lRU:U⊆R}.

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

Lemma 2.1. LetSbe one of the rings above andUa subset ofR. The following statements hold:

il_SU Sl_RU, iir_SU r_RUS.

Proof. iWe only prove for the caseSRx;σbecause the other cases are similar. Letfx _n

i0aixⁱ ∈Rx;σsuch thatfxU 0. Thenσ⁻ⁱaiU 0 for all 0≤i ≤nand it follows that σ⁻ⁱai ∈ lRUfor all 0 ≤ i ≤ n. Hencefx _n

i0xⁱσ⁻ⁱai ∈ Rx;σlRU. SolRx;σU ⊆ Rx;σlRU. We clearly have thatRx;σlRU ⊆ l_Rx;σU. Therefore, we havel_Rx;σU Rx;σlRU.

iiWe only prove for the caseSRx;σbecause the other cases are similar. Letfx _n

i0aixⁱ ∈Rx;σsuch thatUfx 0. ThenUai 0 for all 0≤i ≤nand it follows thatai ∈ r_RUfor all 0 ≤i ≤ n. Hencefx _n

i0a_ixⁱ ∈ r_RURx;σ. Sor_Rx;σU ⊆r_RURx;σ.

We clearly have thatr_RURx;σ⊆rRx;σU. Therefore, we haverRx;σU r_RURx;σ.

With the above lemma, we have mapsφ:rAnnR2^R→rAnnS2^Sdefined byφI IS for everyI∈rAnnR2^Rand

Ψ:lAnn_R 2^R

−→lAnn_S 2^S

2.1 defined by ΨI SI for every I ∈ lAnnR2^R. Moreover, we have maps Φ : rAnn_S2^S→rAnnR2^R defined by ΦJ J ∩ R for every J ∈ rAnn_S2^S and Γ : lAnn_S2^S→lAnnR2^Rdefined byΓJ J∩Rfor everyJ∈lAnn_S2^S. Obviously,φis injective andΦis surjective. Clearly,φis surjective if and only ifΦis injective, and in this caseφandΦ are the inverses of each other. Note thatΨandΓsatisfy the same relations as above. The first item of the definition below appears in14, Definition.

Definition 2.2.iSuppose thatσis an endomorphism ofR. A ringRsatisfies SA1 if forfx _n

i0a_ixⁱandgx _m

j0b_jx^jinRx;σ, fxgx 0 implies thata_iσⁱbj 0 for all 0≤i≤n and 0≤j≤m.

iiSuppose thatσis an endomorphism ofR. A ringRsatisfies SA2 if forfx _∞

i0aixⁱ andgx _∞

j0bjx^jinRx;σ, fxgx 0 implies thataiσⁱbj 0 for alli≥0, j ≥0.

iiiSuppose thatσis an automorphism ofR. A ringRsatisfies SA3 if forfx q isa_ixⁱ and gx _n

jtbjx^j ∈ Rx, x⁻¹;σ, fxgx 0 implies that aiσⁱbj 0 for all s≤i≤qandt≤j≤n.

ivSuppose thatσis an automorphism ofR. A ringRsatisfies SA4 if forfx _∞

isa_ixⁱ andgx _∞

jtb_jx^j∈Rx, x⁻¹;σ,fxgx 0 implies thata_iσⁱbj 0 for alli≥s andj≥t.

Note that ifRsatisfies one of the conditions above, then all subringsSofRsuch that σS ⊆ Ssatisfies the same property. The following implications are easy to verify: SA4 ⇒ SA3 and SA2 ⇒SA1. Following15, Example 2.1whenσ id_R, the last implication is not reversible.

Lemma 2.3. Letσbe an automorphism ofR. Then iRsatisfies SA1 if and only ifRsatisfies SA3; iiRsatisfies SA2 if and only ifRsatisfies SA4.

Proof. Letfx, gx∈Rx, x⁻¹;σsuch thatfxgx 0, wherefx q

i−paixⁱandgx _s

j−tb_jx^j. We clearly havex^pfx∈Rx;σandgxx^t ∈Rx;σ, thenx^pfxgxx^t 0. By assumption,σ^paiσ^ipbj 0 for all−p ≤ i ≤ q and−t ≤ j ≤ s. Henceaiσⁱbj 0 for all

−p≤i≤qand−t≤j≤s. SinceRx;σ⊆Rx, x⁻¹;σ,the converse follows.

The proof of the other statement is similar.

The following definition appears in16, Definition 2.1.

Definition 2.4. LetRbe a ring andσan endomorphism ofR. ThenRis saidσ-compatible like rightR-module, ifar0 if and only ifaσr 0 for anya∈Randr∈R.

LetRbe a ring andαan endomorphism ofR. Following17, the endomorphismαis saidα-rigid ifrαr 0, thenr 0. A ringRis said a rigid ring if it exists a rigid endomorphism αofR.

Proposition 2.5. Letσ be an endomorphism ofR. IfRis a reduced ring andσ-compatible like right R-module, thenRis aσ-rigid ring and hence satisfies SA1 and SA2.

Proof. We only prove the case of SA2 because the other are similar. We claim thatRx;σis a reduced ring. In fact, letfx _∞

i0a_ixⁱsuch thatfx² 0. We have thata²₀0. SinceR is reduced, thena0 0. Next, we havea1σa1 0, sinceRisσ-compatible and reduced, then a10. By induction, we getfx 0. HenceRx;σis reduced. Using the same ideas of14, Proposition 3, we have thatRisσ-rigid and using similar ideas of14, Corollary 4, we obtain thatRsatisfies SA2.

Without the assumption thatRisσ-compatible,Proposition 2.5is not true. In fact, let RZ2⊕Z2andσ :R→R, defined byσa, b b, a. By14, Example 2,Rdoes not satisfy SA2 becauseRdoes not satisfy SA1. Observe that1,00,1 0,0but1,0σ0,1/ 0,0 and soRis notσ-compatible. We have the following natural questions.

Questions

iLetσbe an endomorphism ofR. Suppose thatRsatisfies SA2. IsR σ-compatible like rightR-module?

iiLetσbe an endomorphism ofR. Suppose thatRisσ-compatible like rightR-module.

DoesRsatisfy SA2?

The questioniis false. LetR₀be any domain andR R₀x. Letσ :R→Rbe defined byσt 0 andσ|R0 idR0. By16, Example 4.1,Ris notσ-compatible and using the similar ideas of the proof of14, Proposition 10, we have that Rsatisfies SA2 and consequentlyR satisfies SA1.

The questioniiis false. LetRKx, y/x², y², whereKis a field of characteristic 2, and considerT M2R. In this case, takeσ idT. By18, Example 3.6,Sdoes not satisfy SA2 becauseTdoes not satisfy SA1. Moreover,T isσ-compatible like rightT-module.

In19the authors introduced the following version of skew Armendariz rings.

iSuppose that σ is an endomorphism ofR. Letfx _n

i0a_ixⁱ, gx _m

j0b_jx^j ∈ Rx;σsuch thatfxgx 0 impliesaibj0 for all 0≤i≤nand 0≤j≤m.

iiSuppose that σ is an endomorphism ofR. Letfx

i≥0a_ixⁱ,gx

j≥0b_jx^j ∈ Rx;σsuch thatfxgx 0 impliesaibj0 for alli≥0 andj≥0.

Note that the itemiabove in20, Definition 1.1the authors called it byσ-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. Using21, Lemma 2.1and20, Theorem 1.8, the proof of next proposition is easy to verify.

Proposition 2.6. Letσ be an endomorphism ofR and suppose thatR isσ-compatible like right R- module. Then

iRsatisfies SA1 if and only ifRisσ-Armendariz;

iiRsatisfies SA2 if and only ifRisσ-power Armendariz.

The proposition above without the compatibility assumption is not true according to20, Example 1.9and the authors in22, Theorem 2.2obtained an approach of the result above without the compatibility assumption.

The following proposition is a generalization of 18, Proposition 3.4 and partially generalizes15, Proposition 2.6.

Lemma 2.7. LetSbe any of the ringsRx;σandRx;σ. The following conditions are equivalent:

iRsatisfies SA2 SA1;

iiφ:rAnn_R2^R→rAnnS2^Sdefined byφJ JSis bijective;

iii Ψ:lAnnR2^R→lAnnS2^Sdefined byΨJ SJis bijective.

Proof. We only prove the proposition in the case of SA2 because the equivalence ofiandii whenR satisfies SA1 was proved in23, Proposition 3.2. The equivalence betweeniand iiiwhenRsatisfies SA1 has similar proof.

i→ii. It is only necessary to show thatφis surjective. For an elementfx _∞

i0a_ixⁱ∈ Rx;σ, defineCfx {σ⁻ⁱai, i ≥ 0}, and for a subsetT ofRx;σ,we denote the set

∪fx∈TCfxbyCT. We show thatrRx;σfx rRx;σCfx.In fact, givengx _∞

j0bjx^j inr_Rx;σfx, we havefxgx 0. SinceRsatisfies SA2, thena_iσⁱbj 0 for alli≥0 and j≥0.In particular,σ⁻ⁱaibj0 for alli≥0 andj ≥0. Hencegx∈rRx;σCfx.

On the other hand, lethx _∞

k0ckx^kbe an element inRx;σsuch thatCfxhx 0.It is clear thataiσⁱck 0 for alli ≥0 andk ≥ 0. Sofxhx 0. SinceRsatisfies SA2 thenr_Rx;σT r_Rx;σ∪fx∈TC_fx. Thus

rRx;σT

fx∈T

rRx;σ fx

fx∈T

rRx;σ C_fx

fx∈T

rR

Cfx

Rx;σ rR

CT

Rx;σ. 2.2

Therefore,φis surjective.

ii→i. Let fx _∞

i0aixⁱ andgx _∞

j0bjx^j be elements inRx;σsuch that fxgx 0. By assumption,r_Rx,σfx BRx;σ, for some right idealB ofR. Hence gx∈BRx;σand we have thatb_j ∈B ⊂rRx;σfxfor allj ≥0. Soa_iσⁱbj 0 for all i≥0 andj≥0.

iii→i. Letfx

i≥0aixⁱ andgx

j≥0bjx^j be elements in Rx;σsuch that fxgx 0. By assumption, l_Rx;σgx Rx;σB for some left ideal B of R. We can write fx

i≥0xⁱσ⁻ⁱai ∈ Rx;σB. By the equality of the polynomials with the coefficients on the right side, we have that σ⁻ⁱai ∈ B ⊆ lRx;σgx for all i ≥ 0. So a_iσⁱbj 0 for alli≥0 andj≥0.

i→iii. It is only necessary to show thatΨis surjective. Letfx

i≥0a_ixⁱ∈Rx;σ.

DefineCfx{ai, i≥0}, and for a subsetTofRx;σ, we denote the set∪fx∈TCfxbyCT. We show that

lRx;σ fx

lRx;σ Cfx

. 2.3

In fact, givengx

j≥0bjx^j ∈lRx;σfx, we havegxfx 0. SinceRsatisfies SA2, thenb_jσ^jai 0 for alli≥0 andj≥0. Hencegx

j≥0x^jσ^−jbj∈l_Rx;σCfx.

On the other hand, letgx∈Rx;σsuch thatgxCfx 0. Thusgxai 0 for all i≥0. Sogx

i≥0aixⁱgxfx 0, and we have thatgx∈lRx;σfx.

We easily have that for each subsetTofRx;σ, l_Rx;σT l_Rx;σ

fx∈T

C_fx

. 2.4

We claim that l_Rx;σCfx Rx;σlRCfx. In fact, let gx

j≥0b_jx^j such that gxCfx 0. Then we have that 0 gxai

j≥0bjx^jai

j≥0x^jσ^−jbjai. Thusσ^−jbj∈ lRCfx, and it follows that

j≥0

x^jσ^−j b_j

∈Rx;σlR

Cfx

. 2.5

The other inclusion is trivial. So lRx;σT

fx∈T

lRx;σ Cfx

fx∈T

lRx;σ Cfx Rx;σ

fx∈T

lR

Cfx

Rx;σlR

CT

.

2.6 Therefore,Ψis surjective.

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

Theorem 2.8. Letσbe an automorphism ofR.

iSuppose thatRsatisfies SA1. The following conditions are equivalent:

aRis a right (left) zip ring;

bRx;σis a right (left) zip ring;

cRx, x⁻¹, σis a right (left) zip ring.

iiSuppose thatRsatisfies SA2. The following conditions are equivalent:

aRis right (left) zip ring;

bRx;σis right (left) zip ring;

cRx, x⁻¹;σis right (left) zip ring.

Proof. iWe will show the right case because the left case is similar.

Suppose thatRx;σis right zip. LetXbe a subset ofRsuch thatr_RX 0, andfx _n

i0a_ixⁱ ∈Rx;σsuch thatXfx 0. Thusa_i ∈r_RX 0 and it follows thatfx 0. By assumption, there existsX1{x0, . . . , xn}such thatrRx;σX1 0. HencerRX1 rRx;σX1∩ R 0.

Conversely, let Y ⊆ Rx;σ such that r_Rx;σY 0. By Lemma 2.7, r_Rx;σY r_RTRx;σ, where T C_Y ∪fx∈YCfx such that Cfx {σ⁻ⁱai : 0 ≤ i ≤ n}

with fx _n

i0aixⁱ ∈ Y. We have that rRT 0 and, by assumption, there exists T1 {σ⁻ⁱ¹ai1, . . . , σ⁻ⁱⁿain}such thatrRT1 0. For eachσ^−ijaij ∈ T1, there existsga_ijx ∈ Y such that some of the coefficients ofga_ijxareaij for each 1 ≤ j ≤ n. LetY0 be a minimal subset ofY such thatga_ijx∈Y0 for each 1 ≤j ≤n. ThenY0 is nonempty finite subset ofY. SetT₀ ∪fx∈Y0Cfxand we have thatT₁ ⊆ T₀. Hencer_RT0 ⊆ r_RT1 0. ByLemma 2.7, rRx;σY0 rRT0Rx;σand it follows thatrRx;σY0 0.

The proofs ofa⇔cand of itemiifollow similarly.

Letσbe an endomorphism ofRandδ:R→Ran additive map ofR. The applicationδis said to be aσ-derivation ifδab δabσaδb. The Ore extensionRx;σ, δis the set of polynomials_n

i0a_ixⁱwith the usual sum, and the multiplication rule isxaσaxδa.

Following16,Ris said to beσ, δ-compatible, whereσis an endomorphism ofRand δis aσ-derivation ofRifab0⇔aσb 0 andab0 implies thataδb 0.

In the next result we obtain a necessary and sufficient condition forRx;σ, δto be left zip, whenσis an endomorphism ofRusing the skew version of Armendariz rings of19.

Theorem 2.9. Letσbe an endomorphism ofRandδaσ-derivation ofR. Suppose that iffxgx 0 forfx _n

i0a_ixⁱandgx _m

j0b_jx^j∈Rx;σ, δ, thena_ib_j0 for all 0≤i≤nand 0≤j≤m.

ThenRis left zip if and only ifRx;σ, δis left zip.

Proof. LetX be any subset ofRx;σ, δandCX ∪fx∈XCfx, whereCfx {ai,0 ≤ i ≤ n}

withfx _n

i0aixⁱ. Suppose thatlRx;σ,δX 0. We clearly havelRCX 0. By assumption, there exists{b0, . . . , b_t} ⊆ C_X such thatl_RY 0. Let f_bix ∈ X be an element of X with some of its coefficients are equal tob_ifor all 1≤i≤t. TakeX₀be a minimal subset ofXwith this property. We clearly have thatX0is a finite set. We claim thatlRx;σ,δX0 0. In fact, we

easily havelRCX0 0, whereCX0∪fx∈X0CfxwithCfxbeing defined as before. Next, let gx _m

j0bjx^jsuch thatgxX00. Hence for anyfx _n

i0aixⁱ∈X0,gxfx 0, and we have, by assumption,b_ja_i0 for all 0≤j≤mand 0≤i≤n. Thusb_jC_X00 for all 0≤j≤m and it follows thatgx 0. SolRx;σ,δX0 0.

Using the methods ofTheorem 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 ofTheorem 2.9, we have thatRis a left zip ring if and only ifRx;σis 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 aσ-rigid ring that is a zip ring is given.

Example 3.1. LetFbe any field andσ :F→Fany automorphism ofF. Following14, page 113, we consider the ringTF, Fwith automorphismσa, b σa, σband we denote it byσ.

Note that

TF, F

a b 0 a

:a, b∈F

. 3.1

By14, Proposition 15,TF, Fsatisfies SA1, and using similar methods, we can prove that TF, Fsatisfies SA2. We claim thatTF, Fis a zip ring. In fact, the unique one-sided ideals of TF, Fare_{0 0}

0 0

,

I

0 b 0 0

:b∈F

, 3.2

andTF, F. Note thatrTF,FI/{0}andlTF,FI/0. So we easily have thatTF, Fis a zip ring.

Example 3.2. LetFbe any field andσa monomorphism ofF, and let

R

⎧⎨

⎩

⎛

⎝a b c 0 a d 0 0 a

⎞

⎠:a, b, c∈F

⎫⎬

⎭ 3.3 with usual addition and multiplication of matrix. Note that the monomorphismσis naturally extended toR, andRhas the following one-sided ideals:

I1

⎧⎨

⎩

⎛

⎝0 0 0 0 0 a 0 0 0

⎞

⎠:a∈F

⎫⎬

⎭, I2

⎧⎨

⎩

⎛

⎝0 0 c 0 0 0 0 0 0

⎞

⎠:c∈F

⎫⎬

⎭, 3.4 R and the zero ideal. We easily haver_RI2/0,l_RI2/0, r_RI1/0, andl_RI1/0. Now we clearly have thatRis a zip ring and by14, Proposition 17,Rsatisfies SA1, and with similar methods of14, Proposition 17, we can prove thatRsatisfies SA2.

Example 3.3. LetDbe any domain with identity,R Dx,σ an endomorphism ofRdefined byσfx f0. SinceRis a domain, thenRis right and left zip. Moreover, using similar methods of14, Example 5, we have thatRsatisfies SA1 and SA2.

Example 3.4. Let D and D1 be any domains, σ an monomorphism of D, and τ an monomorphism ofD₁. SetRD×D₁with usual addition and multiplication, and we define an endomorphismγ ofRbyγa, b σa, τb. We easily have thatγ is a monomorphism ofR. SinceDisσ-rigid andD1isτ-rigid, we easily obtain thatRisγ-rigid. We claim thatRis left and right zip. In fact, letIbe any left ideal ofR. It is well known thatI A×B, whereA is a left ideal ofDandBis a left ideal ofD₁. Suppose thatr_RI 0. ThenA /0 andB /0. It is not difficult to show thatrDA 0 andrD1B 0. SinceDandD1are left zip, then there exists a left finitely generated idealLofDcontained inAsuch thatrDL 0 and a left finitely generated idealL₁ofD₁contained inBsuch thatr_D1L1 0. Thusr_RL×L1 0 andL×L1is a left finitely generated ideal ofRcontained inA×B. HenceRis left zip. Using similar methods, we have thatRis right zip.

Example 3.5. LetFbe a field,σan automorphism ofF,

R

⎧⎨

⎩

⎛

⎝a b c 0 a d 0 0 a

⎞

⎠:a, b, c∈F

⎫⎬

⎭, 3.5

andDa domain with automorphismτ. SetT R×Dand we define an endomorphismγ of T byγa, b σa, τt. It is clear thatγ is an automorphism ofT and it is not difficult to show thatT satisfies SA1 and SA2’ becauseRandDsatisfy SA1 by14, Proposition 17and 14, Proposition 10, respectively, and using similar methods of14, Proposition 17and14, Proposition 10,RandDsatisfy SA2, respectively.

Using similar methods ofExample 3.4, we have thatTis right and left zip and note that Tis notγ-rigid, sinceT 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.

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23 W. Cortes, “Skew Armendariz rings and annihilator ideals of skew polynomial rings,” in Algebraic Structures and Their Representations, vol. 376 of Contemporary Mathematics, pp. 249–259, American Mathematical Society, Providence, RI, USA, 2005.