# Extension of Antoine’s Ring Construction

Authors: Tai Keun Kwak
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

This article concerns a kind of extension of Antoine’s ring construction which is unit-IFP and Armendariz. Such extensions are also shown to be unit-IFP and Armendariz, by which we may extend the classes of unit-IFP and Armendariz rings.


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## Body

## 1. Introduction

Antoine constructed a kind of ring coproduct and proved that such a ring is Armendariz in [1]. Kim et al. showed that such an Armendariz ring is also unit-IFP [2]. We consider an extended situation of Antoine’s construction and show that such construction also provides examples of unit-IFP rings and Armendariz rings.Throughout this note, every ring is an associative ring with identity unless otherwise stated. LetR be a ring. The group of units in R is denoted by UR. The polynomial ring with an indeterminate x over R is denoted by Rx, and let Cfx denote the set of all coefficients of given a polynomial fx. Use JR, N∗R, N∗R, and NR to denote the Jacobson radical, the prime radical, the upper nilradical (i.e., sum of all nil ideals), and the set of all nilpotent elements in R, respectively. It is well known that N∗R⊆N∗R⊆NR and N∗R⊆JR.


## 2. Unit-IFP Property

In this section, we observe the unit-IFP property of a kind of ring which extends the situation of Antoine’s construction in [1, Theorem 4.7].Due to Bell [3], a ring R is said to have the Insertion-of-Factors-Property (simply, be an IFP ring) if ab=0 implies aRb=0 for a,b∈R, while Narbonne [4] and Shin [5] used the terms semicommutative and SI for the IFP, respectively. Following Kim et al. [2], a ring R is said to satisfy Insertion-of-Factors-Property on units (simply, be a unit-IFP ring) if ab=0 for a,b∈R implies aURb=0. They proved that Köthe’s conjecture (i.e., the upper nilradical contains every nil left ideal) holds for unit-IFP rings in [2, Theorem 1.3(1)]. IFP rings are clearly unit-IFP but not conversely by [2, Example 1.1].A ring is usually calledAbelian if every idempotent is central. Following the literature, a ring R is said to be directly finite (or Dedekind finite) if ab=1 for a,b∈R implies ba=1. Abelian rings are clearly directly finite. Unit-IFP rings are Abelian by [2, Lemma 1.2].In the following which is generalizations of [6, Theorem 1.2 and Theorem 1.3], the free algebra generated by a set A over a field K is denoted by KA.Theorem1.
LetK be a field and S be a noncommutative set of cardinality ≥2. Let n≥2, T⊊S, and S0=S\T. Set I be the ideal of KS generated by t1t2,…,tn with t1,t2,…,tn∈T, and J be the ideal of KT generated by Tn. Let R=KS/I, R1=KS0 and R2=KT/J. Then, we have the following:(1)
R is a unit-IFP ring with UR=k+g+γδϵk∈K\0,g∈R2′,γ∈R2TpR2,δ∈R,ϵ∈R2TqR2for somep,q≥1withp+q≥n, where R2′ is the subalgebra of R2 of all polynomials of zero constant term.(2)
NR=g+γδϵg∈R2′,γ∈R2TpR2,δ∈R,ϵ∈R2TqR2for somep,q≥1withp+q≥n and NRn=0.(3)
R is a prime ring.Proof.
First, note thatR1 is a domain (hence Abelian) and R2 is Abelian by Lemma 1 to follow.(1)
We apply the method in the proof of [6, Theorem 1.2] to here. Note that Tn⊂I. Identify k∈K and s∈S with their images in R for simplicity, respectively. Then Tn=0. Note that R is isomorphic to R1∗KR2 that is the ring coproduct of R1 and R2 over K.
By [7, Corollary 2.16(i)] (refer to [1, Lemma 4.4(a)] also), UR is generated by the units of R1 and R2, together with elements of the form 1−γδϵ, where δ∈R and γ,ϵ∈Ri for some i, such that ϵγ=0. Note that R1 is a domain with(1)UR1=UKS0=K0,UR2=f∈R2the constant term offis nonzero,because every polynomial of zero constant term in R2 is nilpotent. So, if both γ and ϵ are nonzero, then they are contained in R2 such that γ∈R2TpR2 and ϵ∈R2TqR2 with p+q≥n. Therefore,(2)UR=k+g+γδεk∈K0,g∈R2′,δ∈R,γ∈R2TpR2,ε∈R2TqR2,where R2′ is the subalgebra of R2 of all polynomials of zero constant term, and p,q≥1, p+q≥n. Here, note g+γδϵ2n=0.
Next suppose thatαβ=0 for α,β∈R\0. Then, by [7, Corollary 2.16(ii)] (refer to [1, Lemma 4.4(b)] also), there exist u∈UR and sets C,D in some Ri with CD=0 such that α∈RCu and β∈u−1DR. When C,D are nonzero, CD=0 implies C,D⊆R2. Furthermore, CD=0 implies that(3)C⊆R2TlR2andD⊆R2TmR2for somel,m≥1withl+m≥n.
Thenα and β are sums of nonzero elements of the forms(4)r1t1t2⋯tluandu−1t1′t2′⋯tm′r2,respectively,where ti,tj′∈T and r1,r2∈R.
We will showαvβ=0 for all v∈UR. Here, uvu−1∈UR, say w=uvu−1. By the argument above, we have w=k+g+γδε with k∈K\0, g∈R2′, δ∈R, γ∈R2TpR2, ϵ∈R2TqR2 for some p,q≥1 with p+q≥n. Then we obtain(5)r1t1t2⋯tluvu−1t1′t2′⋯tm′r2=r1t1t2⋯tlwt1′t2′⋯tm′r2=r1t1t2⋯tlk+g+γδϵt1′t2′⋯tm′r2=r1t1t2⋯tlk+gt1′t2′⋯tm′r2+r1t1t2⋯tlγδϵt1′t2′⋯tm′r2=r1t1t2⋯tlγδϵt1′t2′⋯tm′r2,noting that r1t1t2⋯tlk+gt1′t2′⋯tm′r2=0 because l+m≥n and Tn=0. But l+m≥n and p+q≥n, so l+p≥n or q+m≥n. Note that every term of r1t1t2⋯tlγ contains an element in Tl+p and every term of ϵt1′t2′⋯tm′r2 contains an element in Tq+m. Thus, r1t1t2⋯tlγδϵt1′t2′⋯tm′r2=0, proving that αURβ=0. Therefore, R is a unit-IFP ring. Notice that R is not IFP, since t1t2⋯tn=0 but t1s0t2⋯tn≠0 for some s0∈S0.(2)
We apply the proof of [6, Theorem 1.3]. Let c∈NR and u∈UR. Since R is unit-IFP by (1), uc∈NR and u+c∈UR by the argument in the proof of [6, Theorem 1.3]. So, by (1), there exist p,q≥1 with p+q≥n such that u+c=k+g+γδϵ for some k∈K\0, g∈R2′, δ∈R, γ∈R2TpR2, ϵ∈R2TqR2. Then, c is of the form g+γδϵ, and hence, NR⊆g+γδϵkK\0,g∈R2′,δ∈R,γ∈R2TpR2,ϵ∈R2TqR2.
For the converse inclusion, consider the expansion ofg+γδϵn and let e be a term of it. Note that gn=0 because Tn=0.
Ifγδϵ does not occur in e, then e=gn=0.
If two or moreγδϵ’s occur in e, then e is of the form r1γδϵghγδϵr2 with h≥0 and ri∈R; hence, e is zero because every term of γδϵ contains a monomial in Tp+q.
Assume that exactly oneγδϵ occurs in e, i.e., e=ghγδϵgk with h,k≥0 and h+k=n−1. Then every term of e is of the form r1′s1bh+pf′s2r2′ with s1∈Th+p, s2∈Tq+k, h≥0, and f′,ri′∈R. But h+p+q+k≥2n−1 and so h+p≥n or q+k≥n; hence, e is zero.
This impliesg+γδϵn=0, and we get NR⊇g+γδϵk∈K\0,g∈R2′,δ∈R,γ∈R2TpR2,ϵ∈R2TqR2. Therefore,(6)NR=g+γδϵg∈R2′,δ∈R,γ∈R2TpR2,ϵ∈R2TqR2for somep,q≥1withp+q≥n.
We next showNRn=0. Consider an element in NRn of the form(7)F=g1+γ1δ1ϵ1g2+γ2δ2ϵ2⋯gn+γnδnϵn,with gi+γiδiϵi∈NR for all i. Then gi∈R2′, δi∈R, γi∈R2TpiR2, ϵi∈R2TqiR2 for some pi,qi≥1 with pi+qi≥n. Suppose that gi+γiδiϵi≠0 for all i. Since Tn=0, gi1⋯gin=0 for any choice i1,…,in. Let E be a product of n times of gi’s and γiδiϵi’s in the expansion of F.
Suppose that anyγiδiϵi does not occur in E, then E=g1⋯gn=0.
Suppose that exactly oneγiδiϵi occurs in E, i.e., E=gi1⋯gisγiδiϵigis+1⋯gin−1. Then every term of E contains a product of the form(8)r1b1⋯bs+pif′b1′⋯bqi+n−1−s′r2withf′,r1,r2∈R,bj,bl′∈T.
Buts+pi+qi+n−1−s=pi+qi+n−1≥2n−1, and so s+pi≥n or qi+n−1−s≥n; hence, E is zero.
Suppose that two or moreγiδiϵi’s occur in E. Then E is of the form(9)gi1⋯gis1γj1δj1ϵj1gis1+1⋯gis2γj2δj2ϵj2gis2+1⋯gis3⋯gist−1+1⋯gistγjtδjtϵjtgist+1⋯gin−t,with 2≤t≤n. So every term of E contains a product of the form(10)b11⋯b1s1+pj1f1′b21⋯b2qj1+s2−s1+pj2f2′b31⋯b3qj2+s3−s2+pj3f3′bt1⋯btqjt−1+st−st−1+pjtft′bt+11⋯bt+1qjt+n−t−st,where fj′∈R and bhk∈T. Assume here that ml=qjl+sl+1−sl+pjl+1≤n−1 for all l=1,2,…,t−1. But s1+pj1+m1+⋯+mt−1+qjt+n−t−st≥tn+n−tand∑l=1t−1ml≤t−1n−t−1. This yields s1+pj1+qjt+n−t−st≥tn+n−t−t−1n−t−1=2n−1; hence, we have that s1+pj1≥n or qjt+n+1−t−st≥n. So bs1+pj1=0 or bqjt+n−t−st=0. This yields E=0.
These results implyF=0, and NRn=0 follows.(3)
Suppose thatαRβ=0 for α,β∈R. Assume on the contrary that both α and β are nonzero. Note that αβ=0. Then, by the argument above, α is a sum of nonzero elements of the form r1t1t2⋯tlu and β is a sum of nonzero elements of the form u−1t1′t2′⋯tm′r2, where l,m≥1 with l+m≥n, ti,tj′∈T, and r1,r2∈R. Let a∈S0. Then, αaβ is the sum of nonzero elements of the form r1t1t2⋯tluau−1t1′t2′⋯tm′r2.
Next we use the method in [8, Example 14]. There exists a finite subset S1 of S such that αaβ is contained in the subring of R generated by S1, say R3. Let S1=a1,a2,…,am1,bm1+1,bm1+2,…,bm1+m2, where a=a1, ai∈S0, and bj∈T.
Takeu≥1 such that 10u>m1+m2. Observe that nonzero monomials in R3 can be embedded into the set of natural numbers through the corresponding a1⟶10u+1,a2⟶10u+2,…,am1⟶10u+m1,bm1+1⟶10u+m1+1,bm1+2⟶10u+m1+2,…,bm1+m2⟶10u+m1+m2, σ say. Then they are totally ordered via σ (for example, a1<bm1+1 and a1a2<bm1+1a2 because 10u+1<10u+m1+1 and 10u+1,10u+2<10u+m1+1,10u+2, where α,β=α⋅10f+1+β when 10f+1>β≥10f). We identify f with σf for a nonzero monomial f in R3.
We can write(11)α=c1d1+⋯+cv1dv1,aβ=c1′d1′+⋯+cw1′dw1′,with d1<⋯<dv1, d1′<⋯<dw1′, where di’s and dj′’s, are nonzero monomials of coefficient 1, and ci,cj′∈K\0. Then every cidicj′dj′=cicj′didj′ is of the form r1t1t2⋯tluau−1t1′t2′⋯tm′r2, hence nonzero. Furthermore, dv1dw1′ is the largest in the set didj′; hence, this must be zero because αaβ=0, contrary to dv1dw1′≠0. So α=0 or β=0. Therefore, R is prime.
In the following, we further study the structure of zero divisors in the unit-IFP ringR in Theorem 1.Proposition 1.
LetR be the ring in Theorem 1. Let αβ=0 for 0≠α,β∈R. Then α∈RTl′ and β∈Tm′R for some l′,m′≥1 with l′+m′≥n.Proof.
LetR be the unit-IFP ring in Theorem 1. Suppose that αβ=0 for 0≠α,β∈R. Then, α is a sum of nonzero elements of the form r1t1t2⋯tlu and β is a sum of nonzero elements of the form u−1t1′t2′⋯tm′r2, where l,m≥1 with l+m≥n, ti,tj′∈T, and r1,r2∈R. Write u=k+g+γδϵ, where k∈K\0, g∈R2′, δ∈R, γ∈R2TpR2, ϵ∈R2TqR2 with p+q≥n. Then, u−1=k′+g′+γ′δ′ϵ′ with k′∈K\0, g′∈R2′, δ′∈R, γ′∈R2TpR2, ϵ′∈R2TqR2. Note that l+p≥n or q+m≥n.
Supposel+p≥n. Then,(12)r1t1t2⋯tlu=r1t1t2⋯tlk+g+γδϵ=kr1t1t2⋯tl+r1t1t2⋯tlg+r1t1t2⋯tlγδϵ=kr1t1t2⋯tl+r1t1t2⋯tlg∈RTl,u−1t1′t2′⋯tm′r2=k′+g′+γ′δ′ϵ′t1′t2′⋯tm′r2=kt1′t2′⋯tm′r2+g′t1′t2′⋯tm′r2+γ′δ′ϵ′t1′t2′⋯tm′r2∈Tm1R,where m1=minm,p. Here, we have l+m1≥n because l+m,l+p≥n.
Supposeq+m≥n. Then,(13)r1t1t2⋯tlu=r1t1t2⋯tlk+g+γδϵ=kr1t1t2⋯tl+r1t1t2⋯tlg+r1t1t2⋯tlγδϵ∈RTl1,u−1t1′t2′⋯tm′r2=k′+g′+γ′δ′ϵ′t1′t2′⋯tm′r2=kt1′t2′⋯tm′r2+g′t1′t2′⋯tm′r2+γ′δ′ϵ′t1′t2′⋯tm′r2=kt1′t2′⋯tm′r2+g′t1′t2′⋯tm′r2∈TmR,where l1=minl,q. Here, we have l1+m≥n because l+m,q+m≥n.
Therefore,α∈RTl′ and β∈Tm′R for some l′,m′≥1 with l′+m′≥n.


## 3. Armendariz Property

In this section we study Armendariz property of the ringR in Theorem 1. Following Rege and Chhawchharia [9], a ring R (possibly without identity) is said to be Armendariz provided that if fxgx=0 for fx,gx∈Rx then ab=0 for all a∈Cfx,b∈Cgx. A ring R is usually called reduced if NR=0. Reduced rings are Armendariz by [10, Lemma 1]. Armendariz rings are Abelian by the proof of [11, Theorem 6]. Some conditions under which Armendariz rings are unit-IFP are provided in [2].Lemma 1.
The ringR2 in Theorem 1 is Armendariz. Furthermore, R2 is a local ring with JR2=N∗R2=N∗R2=NR2.Proof.
Suppose thatfxgx=0 for fx,gx∈R2x\0. We again use the method in the proof of Theorem 1(3). There exists a finite subset T1 of T such that Cfx and Cgx are contained in the subring of R2 generated by T1, R4 say. Let T1=b1,b2,…,bm. Observe that nonzero monomials in R4 can be embedded into the set of natural numbers through a similar corresponding to one in the proof of Theorem 1(3). Express this corresponding by σ1. We identify f with σ1f for a nonzero monomial f in R4.
We can writefx=∑i=1Mhiki and gx=∑j=1Ntjlj for some ki,lj∈Kx and some monomials hi,tj in R2, where every monomial in hi,tj can be assumed to be of coefficient 1. In this situation, we can assume that(14)h1<h2<⋯<hM,t1<t2<⋯<tN.
Thenh1t1 is smaller than other hitj’s. But fxgx=0, and so h1t1k1l1=0. It then follows that h1t1k1l1∈TnKx. Since h1t1 is smallest in hitj, other hitj’s are also contained in Tn. This implies hitj=0 for all i and j. From this result, we can say that αβ=0 for all α∈Cfx and β∈Cgx. Therefore, R2 is Armendariz.
R
2
\
R
2
′
≅
K and R2′n=0; hence, R2′=NR2=N∗R2=N∗R2=JR2. Especially, R2 is a local ring.Theorem2.
The ringR in Theorem 1 is Armendariz.Proof.
We apply the proof of [1, Theorem 4.7] to this extended situation. First, we check a property of the ring R2. Suppose that αβ=0 and γδ=0 for α,β,γ,δ∈R2. Then, αβ,γδ∈R2′ because every element in R2\R2′ is invertible in R2. So we have that α∈R2TpR2,β∈R2TqR2 and γ∈R2Tp′R2,δ∈R2Tq′R2 for some p,q,p′,q′≥1 with p+q≥n and p′+q′≥n. It then follows that p+q′≥n or q+p′≥n; hence,(15)“βα=0andδγ=0” and“αδ=0orγβ=0”.
Suppose thatfxgx=0 for fx,gx∈Rx\0. Note that R2 is Armendariz (hence Abelian) by Lemma 1; hence, R2 is directly finite. R1 is clearly a domain, so directly finite. Then, by [1, Proposition 4.5], there exist px∈Kx, subsets U,V⊆Rix, and αx,βx∈Rx such that UV=0, pxfx∈RxUαx, and pxgx∈βxVRx. Furthermore, by [1, Proposition 4.6], we may assume that Cfx⊆Cpxfx and Cgx⊆Cpxgx.
SinceR1 is a domain, U and V are contained in R2x. But R2 is Armendariz. So, since UV=0, we have uv=0 for any u∈Ct1x and v∈Ct2x, where t1x∈U and t2x∈V. Write CU=Ct1xt1x∈U and CV=Ct2xt2x∈V. Then, we obtain uv=0 for any u∈CU and v∈CV.
Observe that any coefficient ofpxfx is a sum of elements of the form ruα∗ where r∈R, u∈CU, and α∗∈Cαx and that the coefficients in pxgx are sums of elements of the form β∗vs where s∈R, v∈CV, and β∗∈Cβx. Hence the product of the coefficients of fx and gx will be the sums of elements of the form ruα∗β∗vs.
We next observe the structure ofα∗β∗. Note that αx,βx are products of units in R1x, units in R2x, and elements of the form hx−γxδxϵx where hx∈Kx\0, γx,ϵx∈R2x such that ϵxγx=0 and δx∈Rx. Since R2 is Armendariz, ϵxγx=0 implies that ϵγ=0 for all ϵ∈Cγx,γ∈Cϵx. Hence, α∗β∗ will be in the subring generated by the coefficients of polynomials in UR1x,UR2x, and elements of the form γδϵ where δ∈R and γ,ϵ∈R2 such that ϵγ=0.
Consider the elementγ′δ′ε′ such that γ′,ε′∈R2 with ϵ′γ′=0 and δ′∈R. Consider next the product γδϵγ′δ′ϵ′. From ϵγ=0 and ϵ′γ′=0, we obtain that ϵγ′=0 (then γδϵγ′δ′ϵ′=0) or ϵ′γ=0, by the property (∗). Hence, the elements of the preceding form γδϵ are closed under multiplication.
Next, sinceR1 is a domain, we have UR1=K\0 and UR1x=K\0. By Lemma 1, R2′=N∗R2; hence, N∗R2x=N∗R2x=R2′x by [12, Theorem 3]. This yields K\0+R2′x⊆UR2x. But R2x\R2′x≅Kx, and so UR2x⊆K\0+R2′x. Thus, UR2x=K\0+R2′x.
Hence,α∗β∗ will be a sum of elements in K, R2, and elements of R2γδϵR2 where γ,ϵ∈R2 such that ϵγ=0.
Now, sinceuv=0 and ϵγ=0, we have that uγ=0 or ϵv=0 by the property (∗). In either case, ruR2γδϵR2vs=0 because uR2γ=0 or ϵR2v=0 also by applying the property (∗). Therefore, R is Armendariz.



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