# A Note on Primitivity of Ideals in Skew Polynomial Rings of Automorphism Type

Authors: Edilson Soares Miranda
Journal: International Journal of Mathematics and Mathematical Sciences
 (2016)

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## Abstract

We extend results about primitive ideals in polynomial rings over nil rings originally proved by Smoktunowicz (2005) forσ-primitive ideals in skew polynomial rings of automorphism type.


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## Body

## 1. Introduction

Throughout this paperR denotes an associative ring but does not necessarily have an identity element and σ:R→R an automorphism of R, unless otherwise stated. We denote by R[x;σ] the skew polynomial rings of automorphism type whose elements are polynomials ∑i=0naixi, ai∈R, for every i≥0, with usual addition and the following multiplication: xa=σ(a)x for all a∈R.A ringR is said to be a Jacobson ring if every prime ideal of R is an intersection of (either left or right) primitive ideals of R. In [1], Smoktunowicz proved that if R is a nil ring and I an ideal of R[x], then R[x]/I is Jacobson radical if and only if R[x]/I′[x] is Jacobson radical, where I′ is the ideal of R generated by coefficients of polynomial from I. Also if R is a nil ring and I is a primitive ideal of R[x], then I=M[x] for some ideal M of R and affirmative answer to this question is equivalent to the Köthe conjecture. Our main results state that if R is a nil ring and I an ideal of R[x;σ], then R[x;σ]/I is σ-Jacobson radical if and only if R[x;σ]/I′[x;σ] is σ-Jacobson radical, where I′ is the ideal of R generated by coefficients of polynomial from I. Also if R is a nil ring and I is a σ-primitive ideal of R[x;σ], then I=M[x;σ] for some ideal M of R. This result includes, as particular cases, all the above results.Now we recall some terminology and results; see [2–4]. A right ideal Q of a ring R is called modular in R if and only if there exists an element b∈R such that a-ba∈Q for every a∈R. An ideal I of a ring R is said to be a σ-invariant if and only if σ(I)=I. An ideal P of R is said to be a right σ-primitive in R if and only if there exists a modular maximal right ideal σ-invariant Q of R such that P is the maximal ideal contained in Q. For f∈R[x;σ], deg⁡(f) denotes the degree of f and lc(f) the leading coefficient of f.


## 2. Results

We begin with the following results that extend ([1, Lemma  1]) and the proof is also similar to the one in the paper.Lemma 1.
LetR be a ring, J a right ideal of R, f∈J[x;σ], Q a right ideal of R[x;σ], and b∈R[x;σ] such that a-ba∈Q for every a∈R[x;σ]. If b-fx∈Q, then, for every i≥1, there are fi∈J[x;σ] such that b-fixi∈Q and deg⁡(fi)≤deg⁡(f).Proof.
We proceed by induction onn. If n=1, we put f1=f. Suppose the lemma holds for some n≥1. Let (1)fn=a0+a1x+⋯+akxk∈Jx;σ,with b-fnxn∈Q and k≤deg⁡(f). Consider (2)fn+1=fσa0+a1+a2x+⋯+akxk-1∈Jx;σ.Since b-fx∈Q, then fx=b+q, q∈Q. Thus (3)b-fn+1xn+1=b-fnxn+a0-ba0xn-qa0xn∈Q.We denote byR1 the usual extension of R to a ring with identity and by σ again the natural extension of σ to R1.The next lemma extends ([1, Lemma  2]).Lemma 2.
LetI be an ideal of R[x;σ] with σ(I)=I and J a right ideal of R with σ(J)=J. Consider p=a0+a1x+⋯+akxk∈I, k>0, and (4)U=∑i∈ZJx;σσiakR1x;σ.(i)
Ifh∈Ul, l≥1, and deg⁡(h)≥k, then there exists g∈Ul-1 such that h-g∈I and deg⁡(g)<deg⁡(h).(ii)
LetQ be a right ideal of R[x;σ], b∈R[x;σ] such that a-ba∈Q for every a∈R[x;σ], and I⊂Q. If b-fx∈Q with f∈J[x;σ], deg⁡(f)≥1, and b-g∈Q, where g∈Udeg⁡(f), then, for every i>deg⁡(g), there exists gi∈J[x;σ] such that b-gixi∈Q and deg⁡(gi)<k.Proof.
(i) Leth=c0+c1x+⋯+ctxt∈Ul, ct≠0, and k≤t. We can write (5)ct=∑j=0m′αjβj,αj∈U,βj∈Ul-1.Then (6)αj=∑i=0njtjiσqjiakuji,tji∈J,uji∈R1,qij∈Z.Hence (7)ct=∑j=0m′∑i=0njtjiσqjiakujiβj=∑i=0mpiσliakeiqiwith pi∈J, ei,qi∈R1, qi∈Ul-1, and li∈Z. Put (8)g=h-ctxt+∑i=0mpiσlip-σliakxkσ-keiqixt-k.Therefore (9)g-h=∑i=0mpiσlipσ-keiσ-kqixt-k∈I.Since J[x;σ]Ul-1⊂Ul-1 and h∈Ul-1, then g∈Ul-1 and deg⁡(g)<deg⁡(h).
(ii) By Lemma1, for every i≥1, there exists fi∈J[x;σ] such that (10)b-fixi∈Q,deg⁡fi≤deg⁡f.Consider (11)g=∑j=0mcjxj,cj∈Udeg⁡f.For every n>m denote (12)hn=∑j=0mfn-jσn-jcj∈Jx;σ∩Udeg⁡f.Note that deg⁡(hn)≤deg⁡fn-j≤deg⁡(f); thus for every i≥1(13)fixi=b+qi,qi∈Q.Hence (14)b-hnxn=b-b∑j=0mcjxj+∑j=0mqn-jcjxj∈Q.Because b-g∈Q and g-bg∈Q, then b-bg∈Q. We have that, for every n>t, there exists hn∈Udeg⁡(f)⊆J[x;σ] such that b-hnxn∈Q. If deg⁡(hn)<k, then hn is the gn required. If deg⁡(hn)≥k, by first part of this lemma, there exists (15)λn1∈Udeg⁡f-1⊆Jx;σsuch that hn-λn1∈I and deg⁡(λn1)<deg⁡(hn). Thus b-λn1xn∈Q for all n>m. If deg⁡(λn1)<k, then λn1 is the gn required. If deg⁡(λn1)≥k, using similar arguments as above, we can find s∈N such that (16)λns∈Udeg⁡f-s⊆Jx;σwith b-λnsxn∈Q and deg⁡(λns)≤deg⁡(f)-s<k for every n>m. Hence λns is the gn required.Letr∈R, Q a right ideal of R[x;σ], σ(Q)=Q, and b∈R[x;σ] such that a-ba∈Q for all a∈R[x;σ]. Following [1] we have the following. We say that v is a “good number for r,” if, for all sufficiently large n, there are fn∈R[x;σ] such that b-rfnxn∈Q with deg⁡(fn)≤v. Let A⊆R; we denote (17)A~=a∈A-Q∣σa-a∈Q.Lemma 3.
LetQ be a right ideal of R[x;σ] maximal in the set of all right ideals σ-invariants with b∈R[x;σ] such that a-ba∈Q for all a∈R[x;σ]. Suppose f∈R[x;σ] with b-fxj∈Q for some j≥1. If there is no right ideal J of R with σ(J)=J, J⊈Q, and J≠R, then there exists a positive integer v and r∈R~ such that if w∈rR[x;σ] with b-wxm∈Q, m≥0, and deg⁡(w)≤v, then lc(w)∈rR~, lc(w)(Q∩R)⊆Q, and v is a good number for all a∈rR~.Proof.
Letv be minimal positive integer such that there exists w′=i0+i1x+⋯+rxv∈R[x;σ] and m≥1 with b-w′xm∈Q and deg⁡(w′)=v. It is clear that r∉Q. If c=σ(r)-r∉Q, put g=σ(w′)-w′ and (18)A=∑i∈ZσicR1.Thus A is a right ideal of R with σ(A)=A and A⊈Q. By assumption A=R, then r=∑i=0sσqi(c)li, where li∈R1 and qi∈Z. Put (19)t=w′-∑i=0sσqigσ-vli.Comparing the leading coefficients of w′ and ∑i=0sσqi(g)σ-v(li), we have that (20)b-txm∈Q,deg⁡t≤v-1,which contradicts the minimality of v. Therefore σ(r)-r∈Q; consequently r∈R~.
Suppose thatrq∉Q for some q∈R∩Q. Put g′=w′σ-v(q)∈Q; using similar arguments as above we can have a contradiction. Hence r(Q∩R)⊆Q.
If there existsw∈rR[x;σ] with b-wxj∈Q, j≥0, and deg⁡(w)≤v, then using similar arguments as above we can show that lc(w)∈rR~ and lc(w)(Q∩R)⫅Q. Moreover, if a∈rR~, put B=aR+Q∩R; we have that B is a right ideal of R with σ(B)=B and B⊈Q.
By assumptionB=aR+Q∩R=R. Thus w′=aw′′+q′, where w′′∈R[x;σ], deg⁡(w′′)≤deg⁡(w′), and q′∈Q. Therefore b-aw′′xm∈Q. Consequently v is a good number for all a∈rR~.Lemma 4.
LetJ be a right ideal R with σ(J)=J, J⊈Q, and J≠R such that for all sufficiently large n there are fn∈J[x;σ] such that b-fnxn∈Q and deg⁡(fn)≤k, where Q is a right ideal of R[x;σ] and b∈R[x;σ] such that a-ba∈Q for every a∈R[x;σ]. Then there exists a positive integer v and r∈R~ such that if w∈rR[x;σ] with b-wxm∈Q, m≥0, and deg⁡(w)≤v one has that lc(w)∈rR~, lc(w)(Q∩R)⊆Q, and v is a good number for all a∈rR~.Proof.
Letv be minimal positive integer such that for all sufficiently large n there are fn∈J[x;σ] such that b-fnxn∈Q and deg⁡(fn)≤v. Put (21)w′=i0+i1x+⋯+iv-1xv-1+rxv∈Jx;σwith b-w′xm∈Q, m≥0, and deg⁡(w′)≤v. By Lemma 1 and minimality of v we have that r∉Q. Using the same ideas of Lemma 3, we have that r∈R~ and r(Q∩R)⊆Q. Since rR⊆J, we have that the first part of lemma is satisfied.
Leta∈rR~⊆J~; we denote by B the right ideal of R: (22)B=∑i∈ZσiaR1,σB=B,B⊆J,B⊈Q.For sufficiently large n there are gn∈B[x;σ]⊆J[x;σ] such that b-gnxn∈Q and deg⁡(gn)≤v. Put (23)gn=cn0+cn1x+⋯+cnvxv∈Bx;σ.For every 0≤j≤v we have that cnj=∑i=0mjσqni(a)lni, where lni∈R1 and qni∈Z. Consequently (24)cnj=∑i=0mjσqnia-alni+a∑i=0mjlni.Since a∈rR~, we can write (25)cnj=snj+rnj,snj∈Q∩R,rnj∈R.Put hn=rn0+rn1x+⋯+rnvxv; thus b-ahnxn∈Q. Therefore v is a good number for all a∈rR~.Lemma 5.
LetQ be a right ideal of R[x;σ], b∈R[x;σ], such that a-ba∈Q for all a∈R[x;σ] and v is good number for all a∈rR~, where r∈R~. Assume that for every w∈rR[x;σ] with b-wxm∈Q, m≥0, and deg⁡(w)≤v one has that lc(w)∈rR~ and lc(w)(Q∩R)⊆Q. If there are p and p′∈rR~ with (26)pR~+Q∩rR∩p′R~+Q∩rR⊆Q,then v-1 is a good number for r.Proof.
Sincev is a good number for p and p′, then for every sufficiently large n there are gn∈pR[x;σ] and gn′∈p′R[x;σ] such that (27)b-gnxn∈Q,(28)b-gn′xn∈Qwith deg⁡(gn), deg⁡(gn′)≤v. Consider (29)gn=pn0+pn1x+⋯+pnvxv,pnv∈pR~,gn′=pn0′+pn1′x+⋯+pnv′xv,pnv′∈p′R~.Since pnv-pnv′∈Q, then(30)pnv∈pR~+Q∩rR∩p′R~+Q∩rR⊆Q,a contradiction.
Thus there exists sufficiently largei∈N such that c=piv-piv′∈rR~; hence v is a good number for c. Then for all sufficiently large n there are hn∈R[x;σ] such that b-chnxn∈Q and deg⁡(hn)≤v. We denote (31)hn=rn0+rn1x+⋯+rnvxv.Consider (32)kn=chn+gi′-giσ-vrnv∈rRx;σ.Since gi′-gi∈Q, then b-knxn∈Q. Moreover (33)kn=chn-crnvxv+∑j=0v-1pij′-pijxjσ-vrnv.Consequently v-1 is a good number for r.The following theorem extends ([1, Theorem  1]).Theorem 6.
LetR be a nil ring and let I be a σ-primitive ideal in R[x;σ]. Then I=I′[x;σ], where I′ is an ideal σ-invariant of R.Proof.
Assume by contradiction that there area0,a1,…,ak∈R with (34)a0+a1x+⋯+akxk∈I,ak∉I.Since I is a σ-primitive ideal in R[x;σ], there is a right ideal Q of R[x;σ] with σ(Q)=Q and b∈R[x;σ] such that a-ba∈Q for all a∈R[x;σ]. Moreover Q is a maximal in the set of right ideals σ-invariants and I is the maximal ideal contained in Q. We have that R[x;σ]x⫅Q; otherwise b∈R, which is impossible because R is a nil ring. By definition of Q it follows that R[x;σ]x+Q=R[x;σ].
Ifb-hxi∈Q for some i≥0 with h∈R[x;σ], then deg⁡(h)≥1. In fact, if h∈R, let t≥1 be the minimal positive integer with respect to ht∈Q. Thus (b-hxi)σ-i(ht-1)∈Q. Then bσ-i(ht-1)∈Q; hence σ-i(ht-1)∈Q. Consequently ht-1∈Q, a contradiction.
LetJ be a right ideal of R with σ(J)=J and J⊈Q. We have that J[x;σ]x+Q=R[x;σ]. There exists f∈J[x;σ] such that b-fx∈Q. Consider (35)U=∑i∈ZJx;σσiakR1x;σ.Since I is an ideal σ-prime and ak∉I, then U⊈I. Consequently U⊈Q, because I is the maximal ideal contained in Q. Then Udeg⁡(f)+Q=R[x;σ]. There exists g′∈Udeg⁡(f) such that b-g′∈Q. By Lemma 2, for every i≥deg⁡(g′), there are gi′∈J[x;σ] such that b-gi′xi∈Q and deg⁡(gi′)<k. Lemmas 3 and 4 imply that there are r′∈R~ and v′≥1 such that if w∈r′R[x;σ] with b-wxm∈Q, m≥1, and deg⁡(w)≤v′, then lc(w)∈r′R~ and lcw(Q∩R)⊆Q. Moreover v′ is a good number for all a∈r′R~. Let v be minimal such that v is a good number for all a∈r′R~. We have that v≤v′. Let r∈r′R~. Since v is a good number for r, then for sufficiently large n there are hn∈R[x;σ], such that (36)b-rhnxn∈Q,deg⁡hn≤v.Consider fn=rhn, then b-fnxn∈Q and deg⁡(fn)≤v. For some i∈N, there are fi, fi+1,…,fi+k∈rR[x;σ], such that b-fjxj∈Q, deg⁡(fj)≤v, and i≤j≤i+k. Put (37)fj=raj0+raj1x+⋯+rajvxv=gj+cjxv,where gj=raj0+raj1x+⋯+rajv-1xv-1∈rR[x;σ] and cj=rajv. Since deg⁡(fj)≤v≤v′, then cj∉Q. Moreover, (38)σcj-cj∈Q,cjQ∩R⊆Q.Since R is a nil ring, consider ej=cjnj, where nj is a minimal with respect to the condition cjnj∉Q. Thus σ(ej)-ej∈Q for all i≥0. We have that (39)fjσjej=gjσjej+cjσj+vejxv=gjσjej+cjσj+vej-ejxv+cjejxv.Put tj=gjσj(ej)∈rR[x;σ]. Thus, (40)fjσjej-tj∈Qdeg⁡tj≤v-1for every i≤j≤i+k. Since ej∈rR~⊆r′R~, if v-1 is not a good number for r, then Lemma 5 implies that(41)⋂j=1i+kejR~+Q∩rR⫅Q.In this case, there exists s∈⋂j=1i+k(ejR~+Q∩rR) such that s∉Q. Consequently s-ejdj∈Q∩rR, dj∈R, and ejdj∈ejR~. Then s∈rR~⊆r′R~. Therefore v is a good number for s. Then for sufficiently large n there are f¯n∈sR[x;σ], such that (42)b-f¯nxn∈Q,deg⁡f¯n≤v.Let(43)f¯n=∑j=0vsbjnxj.Since b-f¯jxj∈Q, s-ejdj∈Q, and ejdj-bejdj∈Q, then (b-fjxj)ejdj∈Q. Thus bejdj-fjxjejdj∈Q; hence s-fjxjejdj∈Q for every i≤j≤i+k.
Let(44)g¯n=∑j=0vfi+v-jσi+v-jei+v-jdi+v-jbjnxi+v∈rRx;σ. We have that f¯n-g¯n∈Q. Thus b-g¯nxn=(b-f¯nxn)+(f¯n-g¯n)xn∈Q. Put (45)h¯n=∑j=0vti+v-jσi+v-jdi+v-jbjn∈rRx;σ. We can write b-h¯nxi+v+n as (46)b-∑j=0vti+v-j-fi+v-jσi+v-jei+v-jσi+v-jdi+v-jbjnxi+v+n-g¯nxn. Thus for all sufficient large n(47)b-h¯nxi+v+n∈Q,deg⁡h¯n≤v-1. Then v-1 is a good number for all r∈r′R~. This contradicts the minimality of v.Recall that theσ-Jacobson radical Jσ(R) of a ring R is defined as the intersection of all σ-primitive ideals of R. A ring R is a σ-Jacobson radical if Jσ(R)=R.Theorem 7.
LetR be a nil ring and let I be an ideal of R[x;σ]. Consider I¯ the ideal of R generated by coefficients of polynomial from I. Then R[x;σ]/I¯[x;σ] is σ-Jacobson radical if and only if R[x;σ]/I is σ-Jacobson radical.Proof.
Assume by contradiction thatR[x;σ]/I is not σ-Jacobson radical. Then there is a σ-primitive ideal P of R[x;σ]/I such that P≠R[x;σ]/I. We have that there is an ideal K of R[x;σ] such that P=K/I. Therefore K is a σ-primitive ideal of R[x;σ]. By Theorem 6, there is an ideal P¯ of R such that K=P¯[x;σ]. It is clear that I¯⊆P¯. Since (48)Rx;σ/I¯x;σP¯x;σ/I¯x;σ≃Rx;σK,then P¯[x;σ]/I¯[x;σ] is a σ-primitive ideal, a contradiction. Using the fact that I⊆I¯[x;σ], the converse follows.Corollary 8.
IfR is a nil ring, then the polynomial ring of type automorphism R[x;σ] can not be homomorphically mapped onto a σ-simple σ-primitive ring.



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