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International Journal of Mathematics and Mathematical Sciences Volume 2008, Article ID 835605,8pages

doi:10.1155/2008/835605

Research Article

Prime Ideals and Strongly Prime Ideals of

Skew Laurent Polynomial Rings

E. Hashemi

School of Mathematical Sciences, Shahrood University of Technology, P.O. Box 316-3619995161, Shahrood, Iran

Correspondence should be addressed to E. Hashemi,eb hashemi@yahoo.com

Received 26 February 2008; Accepted 28 April 2008

Recommended by Howard Bell

We first study connections between α-compatible ideals of R and related ideals of the skew Laurent polynomials ringRx, x−1;α, whereαis an automorphism ofR. Also we investigate the relationship ofPRandNrRofRwith the prime radical and the upper nil radical of the skew Laurent polynomial rings. Then by using Jordan’s ring, we extend above results to the case where αis not surjective.

Copyrightq2008 E. Hashemi. 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 the paper,Ralways denotes an associative ring with unity. We usePR,NrR,

and NR to denote the prime radical, the upper nil radical, and the set of all nilpotent elements ofR, respectively.

Recall that for a ring R with an injective ring endomorphism α : RR, Rx;α

is the Ore extension of R. The set{xj}

j≥0 is easily seen to be a left Ore subset of Rx;α,

so that one can localize Rx;α and form the skew Laurent polynomials ringRx, x−1;α.

Elements ofRx, x−1;αare finite sum of elements of the formxjrxi, wherer Rand i, j

are nonnegative integers. Multiplication is subject toxr αrxandrx−1 x−1αrfor all rR.

Now we consider Jordan’s construction of the ringAR, α see1, for more details. Let AR, α or Abe the subset {xirxi | r R , i 0} of the skew Laurent polynomial

ring Rx, x−1;α. For each j 0, xirxi xijαjrxij. It follows that the set of

all such elements forms a subring of Rx, x−1;α with xirxi xjsxj xijαjr

αisxij and xirxixjsxj xijαjrαisxij for r, s R and i, j 0. Note

that α is actually an automorphism of AR, α, given by xirxi to xiαrxi, for each

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maps xirxj to αirxji. For an α-ideal I of R, put ΔI

i≥0xiIxi. Hence ΔI is α-ideal of A. The constructions I → ΔI, JJR are inverses, so there is an order-preserving bijection between the sets of α-invariant ideals of R and α-invariant ideals ofA.

According to Krempa2, an endomorphismαof a ringRis called rigid ifaαa 0 impliesa 0 foraR.Ris called anα-rigid ring3if there exists a rigid endomorphism

αofR. Note that any rigid endomorphism of a ring is a monomorphism andα-rigid rings are reducedi.e.,Rhas no nonzero nilpotent elementby Hong et al.3. Properties ofα-rigid rings have been studied in Krempa2, Hirano4, and Hong et al.3,5.

On the other hand, a ringRis called 2-primal ifPR NR see6. Every reduced ring is obviously a 2-primal ring. Moreover, 2 primal rings have been extended to the class of rings which satisfyNrR NR, but the converse does not hold7, Example 3.3. Observe

thatRis a 2-primal ring if and only ifPR NrR NR, if and only ifPRis a completely

semiprime ideali.e.,a2 PRimplies thata PRfora RofR. We refer to612for

more detail on 2 primal rings.

Recall that a ringRis called strongly prime ifRis prime with no nonzero nil ideals. An idealP ofRis strongly prime ifR/P is a strongly prime ring. Allstronglyprime ideals are taken to be proper. We say an idealP of a ringRis minimal (strongly) prime ifP is minimal among strongly prime ideals ofR. Note thatsee 13NrR ∩ {P | P is a minimal

strongly prime ideal ofR}.

Recall that an idealP ofR is completely prime ifabP impliesaP or bP for

a, bR. Every completely prime ideal ofRis strongly prime and every strongly prime ideal is prime.

According to Hong et al. 5, for an endomorphism α of a ring R, an α-ideal I is called to be α-rigid ideal if aαaI implies that aI for aR. Hong et al. 5 studied connections betweenα-rigid ideals ofRand related ideals of some ring extensions. Also they studied relationship of PR and NrR of R with the prime radical and the

upper nil radical of the Ore extension Rx;α, δ of R in the cases when either PR or

NrR is anα-rigid ideal of R and obtaining the following result. LetPR resp.,NrR

be an α-rigid δ-ideal of R. Then PRx;α, δPRx;α, δ resp., NrRx;α, δ

NrRx;α, δ.

In14, the authors definedα-compatible rings, which are a generalization ofα-rigid rings. A ring R is called α-compatible if for each a, bR,ab 0 ⇔ aαb 0. In this case, clearly the endomorphismαis injective. In14, Lemma 2.2, the authors showed that

R is α-rigid if and only if R isα-compatible and reduced. Thus, the α-compatible ring is a generalization of α-rigid ring to the more general case where R is not assumed to be reduced.

Motivated by the above facts, for an endomorphism α of a ring R, we define α

-compatible ideals inRwhich are a generalization ofα-rigid ideals. For an idealI, we say thatI

is anα-compatible ideal ofRif for eacha, bR,abIaαbI. The definition is quite natural, in the light of its similarity with the notion ofα-rigid ideals, where inProposition 2.3, we will show thatIis anα-rigid ideal if and only ifIis anα-compatible ideal and completely semiprime.

In this paper, we first study connections betweenα-compatible ideals ofRand related ideals of the skew Laurent polynomial ringRx, x−1;α, whereαis an automorphism ofR.

Also we investigate the relationship ofPRandNrRofRwith the prime radical and the

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2. Prime ideals and strongly prime ideals of skew Laurent polynomial rings

Recall that an idealIofRis called anα-ideal ifαII;Iis calledα-invariant ifα−1I I. If Iis anα-ideal, thenα:R/IR/Idefined byαaI αa Iis an endomorphism. Then we have the following proposition.

Proposition 2.1. LetIbe an ideal of a ringR. Then the following statements are equivalent:

1Iis anα-compatible ideal;

2R/Iisα-compatible.

Proof. It is clear.

Proposition 2.2. LetIbe anα-compatible ideal of a ringR. Then

1Iisα-invariant;

2ifabI, thenaαnbI, αnabIfor every positive integern; conversely, ifkbor

αkabIfor some positive integerk, thenabI.

Proof. This follows from15, Lemma 2.2 and Proposition 2.3.

Recall from16that a one-sided idealIof a ringRhas the insertion of factors property

or simply, IFPifabIimpliesaRbIfora, bRBell in 1970 introduced this notion for

I0.

Proposition 2.3see15, Proposition 2.4. LetRbe a ring,I an ideal ofR, andα :RRan endomorphism ofR. Then the following conditions are equivalent:

1Iisα-rigid ideal ofR;

2Iisα-compatible, semiprime and has the IFP;

3Iisα-compatible and completely semiprime.

For anα-idealIofR, putΔIi≥0xiIxi.

Proposition 2.4. 1IfIis anα-compatible ideal ofR, thenΔIis anα-compatible ideal ofA.

2IfJis anα-compatible ideal ofA, thenJ ΔJ0andJ0is anα-compatible ideal ofR.

3 If P is a completely (semi)prime α-compatible ideal of R, then ΔP is a completely (semi)primeα-compatible ideal ofA.

4IfQis a completely (semi)primeα-compatible ideal ofA, thenQ ΔQ0andQ0 is a completely (semi)primeα-compatible ideal ofR.

5IfPis a primeα-compatible ideal ofR, thenΔPis a primeα-compatible ideal ofA.

Proof. 1SinceI is anα-ideal ofRIis an ideal ofA. Now, letxirxixjsxj ΔI.

Hencexijαjrαisxij ΔI and thatαjrαis I. Thusαjrαi1s I, sinceI is α-compatible. ConsequentlyxirxiαxjsxjΔI. ThereforeΔIisα-compatible.

2LetJ0 JJandrJ0. ThenαnrJ0for eachn≥0. Henceαnxnrxn rJ

for eachn ≥ 0. Thus xnrxn J, since J isα-compatible. Therefore ΔJ

0 ⊆ J0. Now, let xmrxm J. ThenαmxmrxmJ and thatr J, sinceJ isα-compatible. ThusJ ΔJ

0.

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3LetxirxixjsxjΔP. ThenxijαjrαisxijΔPand thatαjrαis

P. Hence rsP, by Proposition 2.2. Thus rP or sP, since P is completely prime. Consequently,xirxiΔPorxjsxjΔP.

4 By 2,Q ΔQ0 and Q0 is aα-compatible ideal of R. Since Q is completely

semiprime andQ ΔQ0, henceQ0 is completelysemiprime. LetxirxiAxjsxj

ΔP. ThenrRsP, byProposition 2.2. HencerPorsP, sincePis prime. Consequently

xirxiΔPorxjsxjΔP. ThereforeΔPis a prime ideal ofA.

Theorem 2.5. LetPbe a strongly primeα-compatible ideal ofR. ThenΔPis a strongly prime ideal ofA.

Proof. Since P is a prime α-compatible ideal of R, hence ΔP is a prime ideal of A, by

Proposition 2.4. We show that ΔP is a strongly prime ideal of A. Assume J I/ΔP

is a nil ideal of A/ΔP. Let aIi. Then xiaxiI. Since I/ΔP is a nil ideal, hence xiaxin ΔPfor somen 0. Hencexianxi xjpxj for somep P and j 0. Thus

αjan αipP. HenceanP, sinceP isα-compatible. ThenIiP/P is a nil ideal of

R/P for eachi≥0. HenceIiP, for eachi≥0. ThereforeI⊆ΔP. Consequently,ΔPis a

strongly prime ideal ofA.

Note that if I is an α-ideal of R, then Ix, x−1;α is an ideal of the skew Laurent

polynomials ringRx, x−1;α.

Theorem 2.6. Letαbe an automorphism ofR. LetIbe a semiprimeα-compatible ideal ofR. Assume fx niraixiandgx

m

jsbjxjRx, x−1;α. Then the following statements are equivalent: 1fxRx, x−1;αgxIx, x−1;α;

2aiRbjIfor eachi, j.

Proof. 1⇒2. AssumefxRx, x−1;αgxIx, x−1;α. Then

arxr· · ·anxn

cbsxs· · ·bmxm

Ix, x−1;α for eachcR.

HenceanαncbmI. ThusancbmI, sinceIisα-compatible. Next, replacingcbycbm−1dane,

where c, d, eR. Then arxr · · ·anxncbm−1danebsxs· · ·bm−1xm−1 ∈ Ix, x−1;α.

Henceanαncbm−1danebm−1∈Iand thatancbm−1danebm−1 ∈I, sinceIisα-compatible. Thus

RanRbm−12 ⊆ I. HenceRanRbm−1 ⊆ I, sinceI is semiprime. Continuing this process, we

obtainanRbkI, for k s, . . . , m. Hence fromα-compatibility ofI, we getarxr · · ·

anxnRx, x−1;αbsxs· · ·bm−1xm−1 ⊆ Ix, x−1;α. Using induction onnm, we obtain aiRbjIfor eachi, j.

2⇒1. It follows fromProposition 2.2.

Corollary 2.7. Letαbe an automorphism onR. IfI is a (semi)primeα-compatible ideal ofR, then Ix, x−1;αis a (semi)prime ideal ofRx, x−1;α.

Proof. Assume that I is a prime α-compatible ideal of R. Let fx niraixi and gx m

jsbjxjRx, x−1;αsuch thatfxRx, x−1;αgxIx, x−1;α. ThenaiRbjIfor each

i, j, byTheorem 2.6. Assumegx/Ix, x−1;α. Henceb

j /Ifor somej. ThusaiIfor each

ir, . . . , n, sinceIis prime. ThereforefxIx, x−1;α. Consequently,Ix, x−1;αis a prime

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Theorem 2.8. If each minimal prime ideal ofRisα-compatible, thenPRx, x−1;αΔPRx, x−1;α.

Proof. LetP be a minimal prime ideal ofR. ByProposition 2.4,ΔPis aα-compatible ideal ofA. AssumeairxiAxjsxj ΔP. ThenrRs P, sinceΔPisα-compatible. Hence

rP or sP. Thusairxi ΔP or xjsxj ΔP. Therefore ΔP is a prime ideal

of A. ThusΔPx, x−1;αis a prime ideal of Ax, x−1;α, by Corollary 2.7. Consequently, PRx, x−1;αΔPRx, x−1;α.

In14, the authors give some examples ofα-compatible rings however they are notα -rigid. Note that there exists a ringRfor which every nonzero proper ideal isα-compatible but

Ris notα-compatible. For example, letRF F

0F

, whereFis a field, and the endomorphism

αofRis defined byαa b

0c

a0 0c

fora, b, cF.

The following examples show that there existsα-compatible ideals which are notα -rigid.

Example 2.9see15, Example 2.5. LetF be a field. LetR {f f1 0 f

| f, f1 ∈Fx}, where Fxis the ring of polynomials overF. ThenRis a subring of the 2×2 matrix ring over the ringFx. Letα:RRbe an automorphism defined byαf f1

0 f

f uf1

0 f

, whereuis a fixed nonzero element ofF. Letpxbe an irreducible polynomial inFx. LetI {0f1

0 0

|

f1 ∈ px}, where px is the principal ideal ofFx generated bypx. Then I is an α-compatible ideal ofRbut is notα-rigid. Indeed, since0gx

0 0

α0gx

0 0

0 0

0 0

I, but

0gx

0 0

/

Iforgx/px. ThusIis notα-rigid.

Example 2.10 see 17, Example 2. Let Z2 be the field of integers modulo 2 and A

Z2a0, a1, a2, b0, b1, b2, c be the free algebra of polynomials with zero constant term in

noncommuting indeterminates a0, a1, a2, b0, b1, b2, c overZ2. Note thatA is a ring without

unity. Consider an ideal of Z2 A, say I, generated by a0b0, a1b2 a2b1, a0b1 a1b0, a0b2a1b1 a2b0,a2b2,a0rb0,a2rb2,a0 a1a2rb0b1 b2with rAand r1r2r3r4

with r1, r2, r3, r4 ∈ A. Then I has the IFP. Let α : RR be an inner automorphism i.e.,

there exists an invertible elementuR such thatαr u−1rufor eachr R. ThenI is α-compatible, sinceIhas the IFP. ButIis notα-rigid, sinceIis not completely semiprime.

Theorem 2.11. Letαbe an automorphism ofR. If each minimal prime ideal ofRisα-compatible, then PRx, x−1;αPRx, x−1;α.

Proof. It follows fromCorollary 2.7.

The following example shows that there exists a ringRsuch that all minimal prime ideals areα-compatible, but are notα-rigid.

Example 2.12see15, Example 2.11. LetR Mat2Z4be the 2×2 matrix ring over the

ring Z4. Then PR {

a

11a12

a21a22

| aij ∈ 2Z} is the only prime ideal of R. Letα : RR

be the endomorphism defined byαa11a12

a21a22

a11 −a12

a21 a22

. Thenαis an automorphism of

RandPRisα-compatible. However,PRis notα-rigid, since0 1 0 0

α0 1

0 0

PR, but

0 1

0 0

/

PR.

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Proof. Let P be a completely prime ideal ofR.R/P is domain, hence it is a reduced ring.

R/P is anα-compatible ring, henceR/P isα-rigid, by 14, Lemma 2.2. Let fx,gxR/Px, x−1;αsuch thatfxgx 0. Thenfx 0 orgx 0, by a same way as used in

3, Proposition 6. ThusRx, x−1;α/Px, x−1;αR/Px, x−1;αis domain andPx, x−1;α

is a completely prime ideal ofRx, x−1;α.

Corollary 2.14. Letαbe an automorphism onR. IfPRis anα-rigid ideal ofR, thenPRx, x−1; αPRx, x−1;α.

Proof. PR isα-rigid, hence PR is a completely semiprime α-compatible ideal of R, by

Proposition 2.3. ThereforePRx, x−1;αPRx, x−1;α, byTheorem 2.13.

Theorem 2.15. Letαbe an automorphism ofR. IfP is a strongly (semi)primeα-compatible ideal of R, thenPx, x−1;αis a strongly (semi)prime ideal ofRx, x−1;α.

Proof. By Corollary 2.7, Px, x−1;α is a prime ideal of Rx, x−1;α. Hence Rx, x−1;α/ Px, x−1;α R/Px, x−1;α is a prime ring. We claim that zero is the only nil ideal of

R/Px, x−1;α. Let J be a nil ideal of R/Px, x−1;α. Assume I be the set of all leading

coefficients of elements ofJ. First we show thatI is an ideal ofR/P. Clearly,Iis a left ideal ofR/P. LetaIandrR/P. Then there existsfx a0· · ·an−1xn−1axnJ. Hence

fxrm 0, for some nonnegative integersm. Thus a αnra· · ·αm−1nraαmnr 0, since it is the leading coefficient offxrm. Thereforearm0, sinceR/P isα-compatible. Consequently,Iis an ideal ofR/P. ClearlyI is a nil ideal ofR/P. HenceI 0 and soJ 0. ThereforePx, x−1;αis a strongly prime ideal ofRx, x−1;α.

Theorem 2.16. Let αbe an automorphism of R. If each minimal strongly prime ideal of R is α -compatible, thenNrRx, x−1;αNrRx, x−1;α.

Corollary 2.17. Letαbe an automorphism ofR. IfNrR is anα-rigid ideal of R, then NrRx,

x−1;αN

rRx, x−1;α.

Proof. NrR is α-rigid, hence NrR is a completely semiprime α-compatible ideal of

R, by Proposition 2.3, and that NrR is a strongly semiprime ideal of R. Therefore

NrRx, x−1;αNrRx, x−1;α, byTheorem 2.15.

Example 2.12also shows that there exists a ringRsuch that all minimal strongly prime ideals areα-compatible, but are notα-rigid.

Theorem 2.18. Assume each minimal prime ideal ofRisα-compatible. Then the following are equiv-alent:

1PRx, x−1;αis completely semiprime;

2PRx, x−1;α ΔPRx, x−1;αandPRis completely semiprime.

Proof. 1⇒2. Suppose thatPRx, x−1;αis a completely semiprime ideal ofRx, x−1;α.

It is enough to show thatΔPRx, x−1;α PRx, x−1;α, byTheorem 2.8. Let Q be a

minimal prime ideal of Rx, x−1;α and P AQ. Since PRx, x−1;α is a completely

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R. Since P is completely prime, P0 is a completely prime ideal of R. Let I be a minimal

prime ideal of R such that IP0. By assumption, I is α-compatible. Hence ΔI is a

prime α-compatible ideal of A. Thus ΔIx, x−1;α is a prime ideal of Rx, x−1;α and

ΔIx, x−1;αΔP

0x, x−1;αQ. SinceQis a minimal prime ideal ofRx, x−1;α, hence

ΔIx, x−1;α ΔP0x, x−1;α Q. ThereforeΔI ΔP0and thatI P0. Consequently

ΔPRx, x−1;α PRx, x−1;α and that PRx, x−1;α ΔPRx, x−1;α. Since PRx, x−1;α ΔPRx, x−1;αandPRx, x−1;αis completely semiprime, hencePR

is completely semiprime.

2⇒1. Since PR is α-compatible and completely semiprime, ΔPR is an α -compatible completely semiprime ideal of A. Hence ΔPRx, x−1;α is a completely

semiprime ideal of Ax, x−1;α Rx, x−1;α. Thus PRx, x−1;α ΔPRx, x−1;αis a completely semiprime ideal ofRx, x−1;α.

Corollary 2.19. Letαbe an automorphism ofR. Let each minimal prime ideal ofRbeα-compatible. Then the following are equivalent:

1PRx, x−1;αis completely semiprime;

2PRx, x−1;α PRx, x−1;αandPRis completely semiprime.

Acknowledgments

The author thanks the referee for his valuable comments and suggestions. This research is supported by the Shahrood University of Technology of Iran.

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1 D. A. Jordan, “Bijective extensions of injective ring endomorphisms,” Journal of the London

Mathemat-ical Society, vol. 25, no. 3, pp. 435–448, 1982.

2 J. Krempa, “Some examples of reduced rings,” Algebra Colloquium, vol. 3, no. 4, pp. 289–300, 1996.

3 C. Y. Hong, N. K. Kim, and T. K. Kwak, “Ore extensions of Baer and p.p.-rings,” Journal of Pure and

Applied Algebra, vol. 151, no. 3, pp. 215–226, 2000.

4 Y. Hirano, “On the uniqueness of rings of coefficients in skew polynomial rings,” Publicationes

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