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.
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. Let R be a ring. The group of units in R is denoted by . The polynomial ring with an indeterminate x over R is denoted by , and let denote the set of all coefficients of given a polynomial . Use , , , and 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 and .
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 implies for , 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 for implies . 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 called Abelian if every idempotent is central. Following the literature, a ring R is said to be directly finite (or Dedekind finite) if for implies . 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 .
Theorem 1. Let K be a field and S be a noncommutative set of cardinality . Let , , and . Set I be the ideal of generated by with , and J be the ideal of generated by . Let , and . Then, we have the following:(1)R is a unit-IFP ring with where is the subalgebra of of all polynomials of zero constant term.(2) and .(3)R is a prime ring.
Proof. First, note that is a domain (hence Abelian) and is Abelian by Lemma 1 to follow.(1)We apply the method in the proof of [6, Theorem 1.2] to here. Note that . Identify and with their images in R for simplicity, respectively. Then . Note that R is isomorphic to that is the ring coproduct of and over K.By [7, Corollary 2.16(i)] (refer to [1, Lemma 4.4(a)] also), is generated by the units of and , together with elements of the form , where and for some i, such that . Note that is a domain withbecause every polynomial of zero constant term in is nilpotent. So, if both γ and ϵ are nonzero, then they are contained in such that and with . Therefore,where is the subalgebra of of all polynomials of zero constant term, and , . Here, note .
Next suppose that for . Then, by [7, Corollary 2.16(ii)] (refer to [1, Lemma 4.4(b)] also), there exist and sets in some with such that and . When are nonzero, implies . Furthermore, implies thatThen α and β are sums of nonzero elements of the formswhere and .
We will show for all . Here, , say . By the argument above, we have with , , , , for some with . Then we obtainnoting that because and . But and , so or . Note that every term of contains an element in and every term of contains an element in . Thus, , proving that . Therefore, R is a unit-IFP ring. Notice that R is not IFP, since but for some .(2)We apply the proof of [6, Theorem 1.3]. Let and . Since R is unit-IFP by (1), and by the argument in the proof of [6, Theorem 1.3]. So, by (1), there exist with such that for some , , , , . Then, c is of the form , and hence, .For the converse inclusion, consider the expansion of and let e be a term of it. Note that because .
If does not occur in e, then .
If two or more ’s occur in e, then e is of the form with and ; hence, e is zero because every term of contains a monomial in .
Assume that exactly one occurs in e, i.e., with and . Then every term of e is of the form with , , , and . But and so or ; hence, e is zero.
This implies , and we get . Therefore,We next show . Consider an element in of the formwith for all i. Then , , , for some with . Suppose that for all i. Since , for any choice . Let E be a product of n times of ’s and ’s in the expansion of F.
Suppose that any does not occur in E, then .
Suppose that exactly one occurs in E, i.e., . Then every term of E contains a product of the formBut , and so or ; hence, E is zero.
Suppose that two or more ’s occur in E. Then E is of the formwith . So every term of E contains a product of the formwhere and . Assume here that for all . But . This yields ; hence, we have that or . So or . This yields .
These results imply , and follows.(3)Suppose that for . Assume on the contrary that both α and β are nonzero. Note that . Then, by the argument above, α is a sum of nonzero elements of the form and β is a sum of nonzero elements of the form , where with , , and . Let . Then, is the sum of nonzero elements of the form .Next we use the method in [8, Example 14]. There exists a finite subset of S such that is contained in the subring of R generated by , say . Let , where , , and .
Take such that . Observe that nonzero monomials in can be embedded into the set of natural numbers through the corresponding σ say. Then they are totally ordered via σ (for example, and because and , where when ). We identify f with for a nonzero monomial f in .
We can writewith , , where ’s and ’s, are nonzero monomials of coefficient 1, and . Then every is of the form , hence nonzero. Furthermore, is the largest in the set ; hence, this must be zero because , contrary to . So or . Therefore, R is prime.
In the following, we further study the structure of zero divisors in the unit-IFP ring R in Theorem 1.
Proposition 1. Let R be the ring in Theorem 1. Let for . Then and for some with .
Proof. Let R be the unit-IFP ring in Theorem 1. Suppose that for . Then, α is a sum of nonzero elements of the form and β is a sum of nonzero elements of the form , where with , , and . Write , where , , , , with . Then, with , , , , . Note that or .
Suppose . Then,where . Here, we have because .
Suppose . Then,where . Here, we have because .
Therefore, and for some with .
3. Armendariz Property
In this section we study Armendariz property of the ring R in Theorem 1. Following Rege and Chhawchharia [9], a ring R (possibly without identity) is said to be Armendariz provided that if for then for all A ring R is usually called reduced if . 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 ring in Theorem 1 is Armendariz. Furthermore, is a local ring with .
Proof. Suppose that for . We again use the method in the proof of Theorem 1(3). There exists a finite subset of T such that and are contained in the subring of generated by , say. Let . Observe that nonzero monomials in 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 . We identify f with for a nonzero monomial f in .
We can write and for some and some monomials in , where every monomial in can be assumed to be of coefficient 1. In this situation, we can assume thatThen is smaller than other ’s. But , and so . It then follows that . Since is smallest in , other ’s are also contained in . This implies for all i and j. From this result, we can say that for all and . Therefore, is Armendariz.
and ; hence, . Especially, is a local ring.
Theorem 2. The ring R 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 . Suppose that and for . Then, because every element in is invertible in . So we have that and for some with and . It then follows that or ; hence,Suppose that for . Note that is Armendariz (hence Abelian) by Lemma 1; hence, is directly finite. is clearly a domain, so directly finite. Then, by [1, Proposition 4.5], there exist , subsets , and such that , , and . Furthermore, by [1, Proposition 4.6], we may assume that and .
Since is a domain, U and V are contained in . But is Armendariz. So, since , we have for any and , where and . Write and . Then, we obtain for any and .
Observe that any coefficient of is a sum of elements of the form where , , and and that the coefficients in are sums of elements of the form where , , and . Hence the product of the coefficients of and will be the sums of elements of the form .
We next observe the structure of . Note that are products of units in , units in , and elements of the form where , such that and . Since is Armendariz, implies that for all . Hence, will be in the subring generated by the coefficients of polynomials in , and elements of the form where and such that .
Consider the element such that with and . Consider next the product . From and , we obtain that (then ) or , by the property (). Hence, the elements of the preceding form are closed under multiplication.
Next, since is a domain, we have and . By Lemma 1, ; hence, by [12, Theorem 3]. This yields . But , and so . Thus, .
Hence, will be a sum of elements in K, , and elements of where such that .
Now, since and , we have that or by the property (). In either case, because or also by applying the property (). Therefore, R is Armendariz.
Data Availability
No data were used to support this work.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2019R1F1A1040405).