THE ARMENDARIZ MODULE AND ITS APPLICATION TO
THE IKEDA-NAKAYAMA MODULE
M. TAMER KOS¸AN
Received 21 December 2005; Revised 5 July 2006; Accepted 5 September 2006
A ringRis called a right Ikeda-Nakayama (for short IN-ring) if the left annihilator of the intersection of any two right ideals is the sum of the left annihilators, that is, if(I∩
J)=(I) +(J) for all right idealsI andJofR.Ris called Armendariz ring if whenever polynomials f(x)=a0+a1x+···+amxm,g(x)=b0+b1x+···+bnxn∈R[x] satisfy f(x)g(x)=0, then aibj=0 for each i,j. In this paper, we show that ifR[x] is a right
IN-ring, thenRis a right IN-ring in caseRis an Armendariz ring.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
Throughout this work, all rings will be associative with identity. LetRbe a ring. A right (or left) annihilator of a subsetUofRis defined byrR(U)= {a∈R:Ua=0}(orR(U)= {a∈R:aU=0}).
Recall that, a ringRis called a right Ikeda-Nakayama ring if the left annihilator of the intersection of any two right ideals is the sum of the left annihilators, that is, if(I∩J)=
(I) +(J) for all right idealsIandJofR(cf. [6]). LetSMRbe an (S,R)-bimodule. Extend the notion of an IN-ring to module such asS(A∩B)=S(A) +S(B) for any submodules
A,BofMR(cf. [10]).
For a moduleMR, letM[x] be the set of all formal polynomials in indeterminatex with coefficients fromM. ThenM[x] becomes a rightR[x]-module under usual addition and multiplication of polynomials.
We prove that ifS[x]M[x]R[x]-bimodule andUandVareR[x]-submodules ofM[x]R[x],
then for anyt(x)∈S[x](U∩V), everyU+V
αt(x)
−→M[x] extends commutatively toM[x] byλ(s(x)) for somes(x)∈S[x], whereλ:S[x]→End(M[x]R[x]) if and only ifM[x] is an
IN-module.
Following [1],Ris called Armendariz ring if whenever polynomials f(x)=a0+a1x+
···+amxm andg(x)=b0+b1x+···+bnxn∈R[x] satisfy f(x)g(x)=0, then aibj=0 for eachi,j.
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 35238, Pages1–7
A moduleMis calledα-Armendariz if
(i) for anym∈Manda∈R,ma=0 if and only ifmα(a)=0; (ii) for anym(x)=ni=0mixi∈M[x] and f(x)=
s
j=0ajxj∈R[x], m(x)f(x)=0 impliesmiaj=0 for eachi,j(cf. [8,9]).
In [5, Proposition 3.1], Hirano showed that R is Armendariz ring if and only if rAnnR(2R)→rAnnR(2R[x]);A→AR[x] is bijective, whererAnnR(2R)= {rR(U) :U⊆R}. Using this proposition, in this paper, it is shown that ifR[x] is a right IN-ring, thenRis a right IN-ring, in caseRis an Armendariz ring.
2. Ikeda-Nakayama modules
LetS[x] andR[x] be the polynomial rings over ringsSandRand, for a moduleSMR, let
M[x] be the set of all formal polynomials in indeterminatexwith coefficients fromM. ThenM[x] becomes an (S[x],R[x])-bimodule under usual addition and multiplication of polynomials. Extend the notion of an IN-ring to module such as the following. Definition 2.1. Recall thatM[x] is called an Ikeda-Nakayama module (IN-module) if
S[x](U∩V)=S[x](U) +S[x](V) (2.1)
for anyR[x]-submodulesUandVofM[x]R[x]. Such modules and rings were studied by
many authors (cf. [4,6,10]). Professor Harmanci asked (private communication) for a description modulesM (ringsR) such thatM[x] (R[x]) are Ikeda-Nakayama modules (right Ikeda-Nakayama rings), respectively.
Note that there is a canonical ring homomorphismλ:S[x]→End(M[x]R[x]) given by λ(s(x))(m(x))=s(x)m(x) form(x)∈M[x] ands(x)∈S[x].
LetUandV beR[x]-submodules ofM[x]. Then, for anyt(x)∈S[x](U∩V),αt(x): U+V→M[x],u+v→t(x)uis well defined, whereu∈Uandv∈V.
Lemma 2.2. LetS[x]M[x]R[x]-bimodule andU and V beR[x]-submodules of M[x]R[x]. Then, for anyt(x)∈S[x](U∩V), everyU+V−→αt(x)M[x] extends commutatively toM[x]
byλ(s(x)) for somes(x)∈S[x] if and only ifM[x] is an IN-module. In particular, ifU∩V =0, then everyU+V α1
−→M[x] extends commutatively toM[x] byλ(s(x)) for somes(x)∈S[x] if and only ifS[x]=S[x](U) +S[x](V).
Proof. Lett(x)∈S[x](U∩V). Thenαt(x):U+V →M[x], u+v→t(x)u is a well de-finedR[x]-module homomorphism, whereu∈Uandv∈V. By assumption, there ex-istss(x)∈S[x] such thatλ(s(x)) extends toαt(x). Thus, for allu∈Uandv∈V,t(x)u=
αt(x)(u+v)=λ(s(x))(u+v)=s(x)(u+v) and so (t(x)−s(x))u+ (−s(x))v=0. It follows thatt(x)−s(x)∈S[x](U) and−s(x)∈S[x](V). Hencet(x)=(t(x)−s(x)) + (−s(x))∈ S[x](U) +S[x](V). The other inclusion is clear.
For converse, assume thatM[x] is an IN-module and, for any t(x)∈S[x](U∩V),
αt(x):U+V →M[x] defined as above. Fora(x)∈S[x](U) andb(x)∈S[x](V), write t(x)=a(x) +b(x). Then, for allu∈Uandv∈V,αt(x)(u+v)=t(x)u=(a(x) +b(x))u=
a(x)u+b(x)u=0 +b(x)u=b(x)u=b(x)u+b(x)v=b(x)(u+v)=λ(b(x)(u+v)).
Proposition 2.3. LetR[x] be the set of all polynomials in indeterminatexwith coefficients fromR. IfI andJ are right ideals of R[x] such that everyR[x]-linear map I+J→R[x] extends toR[x], then
R[x](I∩J)=R[x](I) +R[x](J). (2.2)
In particular, this holds ifI+J=R[x], in which caseR[x](I∩J)=R[x](I)⊕R[x](J).
Let N be an R[x]-submodule of M[x] and NC= {mi∈M:∃n∈Nwithn=m0+ m1x+···+mtxt}.
Theorem 2.4. LetMbe an Ikeda-Nakayama module and letNandKbeR[x]-submodules ofM[x] such thatS((N∩K)C)=S(NC∩KC). ThenM[x] is an IN-module.
Proof. LetUandV beR[x]-submodules ofM[x]. Let t(x)∈S[x](U∩V). Thenαt(x): U+V→M[x],u+v→t(x)uis a well definedR[x]-homomorphism, whereu∈U and v∈V . Similarly, for allt∈S(UC∩VC), theαt:UC+VC→M,u+v→tu is a well
definedR-homomorphism, whereu∈UC andv∈VC. SinceM is an IN-module, we haveS(U∩V)C)=S(NC∩KC)=S(UC) +S(VC) by assumption and definition. Hence
there exists a homomorphismht:M→M such thathti=αt, wherei:UC+VC→M is
the inclusion map by [10, Lemma 1]. We defineh:M[x]→M[x] such thatht(k0+k1x+
···+knxn)=ht(k0) +ht(k1)x+···+ht(kn)xn. It is clear thathtis well defined. Lett(x)=
t0+t1x+t2x2+···+tnxn∈S[x](U∩V). Thent0,t1,...,tn∈S((U∩V)C)=S(UC) +
S(VC). For eachtj,αtj:UC+VC→M,u+v→tuis a well definedR-homo-morphism, and then we define a maphtj:M→Msuch thathtji=αtj, wherei:UC+VC→Mis the inclusion map. We extend it by defininghtj:M[x]→M[x] such that, for j=0, 1, 2,...,n, h
j(k0+k1x+···+knxn)=(htj(k0) +htj(k1)x+···+htj(kn)xn)xj.
To complete the proof, we show that hi=αt(x), wherei:U+V→M[x] is the in-clusion map. Let h=nj=0hj andu=u0+u1x+···+urxr∈U andv(x)=v0+v1x+
···+vsxs∈V. Thenu0,u1,...,ur∈UCandv0,v1,...,vs∈VC. Sohj(u+v)=(htj(u0) + htj(u1)x+···+htj(ur)xr)xj=tjxj(u0+u1x+···+urxr) andh(u+v)=nj=0hj(u+v)=
t(x)(u+v). HenceM[x] is an IN-module byLemma 2.2.
Letαbe an endomorphism ofR, that is,αis a ring homomorphism fromRtoRwith α(1)=1. Following [9], a moduleMis calledα-Armendariz if
(1) for anym∈Manda∈R,ma=0 if and only ifmα(a)=0; (2) for any m(x)=ni=0mixi∈M[x] and f(x)=
s
j=0ajxj∈R[x], m(x)f(x)=0 impliesmiaj=0 for eachi,j.
Note that 1-Armendariz module is called Armendariz module.
We denoterAnnR(2M)= {rR(U)|U⊆M}andAnnR(2M)= {R(U)|U⊆M}. IfU
is a subset ofM, thenR[x](U)=R(U)[x] andrR[x](U)=rR(U)[x].Hence we have the maps
Φ:rAnnR2M−→rAnnR[x]
defined byΦ(rR(U))=rR[x](U)=rR(U)[x] for everyrR(U)∈rAnnR(2M) and
Φ:AnnR2M−→Ann R[x]
2M[x] (2.4)
defined byΦ(R(U))=R[x](U)=R(U)[x] for everyR(U)∈AnnR(2M).
For a polynomialm(x)∈M[x],Cm(x)denotes the set of coefficients ofm(x) and for a subsetVofM[x],CV denotes the setm(x)∈VCm(x). Then
rR[x](V)∩R=rR(V)=rRCV, R[x](V)∩R=R(V)=RCV. (2.5)
Hence we also have the maps
Ψ:rAnnR[x]
2M[x]−→rAnnR2M (2.6)
defined byΨ(rR[x](V))=rR[x](V)∩Rfor everyrR[x](V)∈rAnnR[x](2M[x]) and
Ψ:AnnR
[x]
2M[x]−→AnnR2M (2.7)
defined byΨ(R[x](V))=R[x](V)∩Rfor everyR[x](V)∈AnnR[x](2M[x]).
ObviouslyΦ(orΦ) is injective andΨ(orΨ) is surjective. Also,Φ(orΦ) is surjective if and only ifΨ(orΨ) is injective and in this caseΦandΨ(orΦandΨ) are the inverses of each other.
Proposition 2.5. LetMRbe a module. Then the following are equivalent.
(1)MRis an Armendariz module.
(2) The map Φ:rAnnR(2M)→rAnnR[x](2M[x]) defined by Φ(rR(U))=rR[x](U)= rR(U)[x], for everyrR(U)∈rAnnR(2M), is bijective.
(3) The map Φ:AnnR(2M)→AnnR[x](2M[x]) defined by Φ(R(U))=R[x](U)= R(U)[x], for everyR(U)∈AnnR(2M), is bijective.
Proof. (1)⇔(2). This is [3, Theorem 2.1].
(1)⇒(3). Assume M is an Armendariz module. Obviously Φ is injective. So it is enough to show Φ is surjective. Let R[x](V)∈AnnR[x](2M[x]) for some V ⊆M[x]. Then forR(CV)∈AnnR(2M), Φ(R(CV))=R[x](CV)=R[x](V). In fact, let f(x)∈ R[x](CV), where f(x)=a0+a1x+···+anxn. Then f(x)CV =0. Thus for allm∈CV,
f(x)m=a0m+a1mx+···+anmxn=0 and hence ajm=0 for all j. Let n(x)=n0+ n1x+···+ntxt∈Vbe arbitrary. Thenf(x)n(x)=0 sinceni∈CVfor alli. Hence f(x)∈ R[x](V). Conversely, let g(x)=b0+b1x+···+bkxk∈R[x](V). Then for all m(x)∈V,
g(x)m(x)=0, wherem(x)=m0+m1x+···+mlxl∈V. SinceMRis Armendariz,bjmi=
0 for alliand j. Henceg(x)mi=0 for alli. Sog(x)∈R[x](CV) sincem(x)∈V is arbi-trary. Consequently for eachR[x](V)∈AnnR[x](2M[x]) for someV⊆M[x] there exists
(3)⇒(1). Conversely, assume f(x)m(x)=0, wherem(x)=m0+m1x+···+mtxt∈
M[x] and f(x)=a0+a1x+···+akxk∈R[x]. By hypothesis,R[x](m(x))=R(U)[x] for
some U⊆M. Then f(x)∈R(U)[x] and henceaj∈R(U) for all j. Soaj∈R(U)⊆ R(U)[x]=R[x](m(x)) thenajm(x)=0. Consequently,ajmi=0 for alliandj. Therefore
MRis an Armendariz module.
ByProposition 2.5, we can obtain [5, Proposition 3.1].
Proposition 2.6. LetRbe a ring. The following statements are equivalent. (1)Ris Armendariz ring.
(2)rAnnR(2R)→rAnnR(2R[x]); A→AR[x] is bijective, where rAnnR(2R)= {rR(U) :
U⊆R}.
(3)AnnR(2R)→AnnR(2R[x]); B→R[x]B is bijective, where AnnR(2R)= {R(U) : U⊆R}.
Now, we give the main result of this work.
Theorem 2.7. LetRbe an Armendariz ring. IfR[x] is a right IN-ring, thenRis a right IN-ring.
Proof. LetIandJbe right ideals ofR. SinceRis an Armendariz ring, we haveR[x](I)=
R(I)[x] byProposition 2.6, for every right idealIofR. Note thatR[x](I)=R[x](I[x]). By assumption,R[x](I) +R[x](J)=R[x](I[x]) +R[x](J[x])=R[x](I[x]∩J[x])=R[x]((I∩ J)[x])=R[x](I∩J). Then R((I∩J)[x])=R(I[x]) +R(J[x])=(R(I) +R(J))[x]
im-plies thatR(I∩J)=R(I) +R(J). SoRis a right IN-ring.
Example 2.8. (i) SinceZis an Armendariz ring,Zis a right IN-ring if and only ifZ[x] is an IN-ring.
(ii) LetRbe a trivial extension ofZand theZ-moduleZ2∞, that is,R=Z⊕Z2∞ with the following addition and multiplication:
(n,a) + (m,b)=(n+m,a+b),
(n,a)(m,b)=(nm,nb+ma). (2.8)
AlsoRis the subringa n0a
:a∈Z, n∈Z2∞.Ris an IN-ring by [10]. As Lee and Zhou pointed out [8, Corollary 2.7],Ris an Armendariz ring. We consider the right idealsI andJofR[x]:
I= px
2 u(x) 0 px2
:u(x)∈Z2∞, pis prime
,
J= qx+qx
2 0 0 qx+qx2
:qis prime and (p,q)=1
.
(2.9)
Clearly,R[x](I∩J)=R[x] sincepandqare primes with (p,q)=1 and soI∩J=0. But R[x](I) andR[x](J) do not contain constant. Therefore,R[x](I) +R[x](J)=R[x](I∩J).
Recall that, a ringRis called reduced ring if it has no nonzero nilpotent elements, a ringRis called right p.p.-ring for alla∈R,rR(a)=eR, wheree2=e∈RandRis called
Baer ring, for allI≤RR,rR(I)=eR, wheree2=e∈R.
As a result ofTheorem 2.7, we can say the following corollary.
Corollary 2.9. LetR[x] be a right IN-ring. ThenRis a right IN-ring in each of the fol-lowing cases.
(1)R2=0.
(2)Ris a reduced ring.
(3)Ris an Abelian (if every idempotent ofRis central) and von Neumann regular ring. (4)Ris an Abelian right (left) p.p.-ring.
(5)Ris an Abelian Baer ring. Proof. AssumeR[x] is a right IN-ring.
(1) By [1], ifR2=0, thenRis an Armendariz ring. (2) By [2], reduced rings are Armendariz.
(3) Every Abelian von Neumann regular ring is a reduced ring.
(4) By [1, Theorem 6] or [7, Lemma 7], ifRis an Abelian right (left) p.p.-ring, then Ris an Armendariz (a Reduced and so Armendariz) ring.
(5) Every Abelian Baer ring is a reduced ring.
HenceRis a right IN-ring byTheorem 2.7.
Acknowledgments
The author wishes to express his sincere gratitude to his Ph.D. supervisor Professor Ab-dullah Harmanci (Hacettepe University, Turkey) for his encouragement and direction. The author would like to thank the referee and Professor Yiqiang Zhou (Memorial Uni-versity, Canada) for valuable comments and suggestions which improved the presentation of the paper.
References
[1] D. D. Anderson and V. Camillo, Armendariz rings and Gaussian rings, Communications in Alge-bra 26 (1998), no. 7, 2265–2272.
[2] E. P. Armendariz, A note on extensions of Baer and P.P.-rings, Journal of the Australian Mathe-matical Society, Series A 18 (1974), 470–473.
[3] M. Baser and A. Harmanci, Reduced and p.q.-bear modules, http://www.math.nthu.edu.tw/ ∼tjm/myweb/FrameToAppear.htm.
[4] V. Camillo, W. K. Nicholson, and M. F. Yousif, Ikeda-Nakayama rings, Journal of Algebra 226 (2000), no. 2, 1001–1010.
[5] Y. Hirano, On annihilator ideals of a polynomial ring over a noncommutative ring, Journal of Pure and Applied Algebra 168 (2002), no. 1, 45–52.
[6] M. Ikeda and T. Nakayama, On some characteristic properties of quasi-Frobenius and regular rings, Proceedings of the American Mathematical Society 5 (1954), no. 1, 15–19.
[7] N. K. Kim and Y. Lee, Armendariz rings and reduced rings, Journal of Algebra 223 (2000), no. 2, 477–488.
[9] , Reduced modules, Rings, Modules, Algebras, and Abelian Groups, Lecture Notes in Pure and Appl. Math., vol. 236, Marcel Dekker, New York, 2004, pp. 365–377.
[10] R. Wisbauer, M. F. Yousif, and Y. Zhou, Ikeda-Nakayama modules, Beitr¨age zur Algebra und Geometrie 43 (2002), no. 1, 111–119.
M. Tamer Kos¸an: Department of Mathematic, Faculty of Science-Literature, Kocatepe University, ANS Campus, Afyon 03200, Turkey
