검색
Article Search

JMB Journal of Microbiolog and Biotechnology

OPEN ACCESS eISSN 0454-8124
pISSN 1225-6951
QR Code

Article

Kyungpook Mathematical Journal 2020; 60(4): 711-721

Published online December 31, 2020

Copyright © Kyungpook Mathematical Journal.

On n-Amitsur Rings

Baatar Ochirbat, Deolinda I. C. Mendes* ,Sodnomkhorloo Tumurbat

Department of Mathematics, Mongolian University of Science and Technology, Ulaanbaatar, Mongolia
e-mail : ochirbat@must.edu.mn
Department of Mathematics, University of Beira Interior, Portugal
e-mail : imendes@ubi.pt
Department of Mathematics, National University of Mongoliaand School of Applied Science and Technology, Ulaanbaatar, Mongolia
e-mail : stumurbat@hotmail.com

Received: August 25, 2019; Revised: July 18, 2020; Accepted: July 21, 2020

The concepts of an Amitsur ring and a hereditary Amitsur ring, which were introduced and studied by S. Tumurbat in a recent paper, are generalized. For a positive integer n, a ring A is said to be an n-Amitsur ring if γ ( A [ X n ] ) = ( γ ( A [ X n ] ) A ) [ X n ] for all radicals γ, where A [ X n ] is the polynomial ring over A in n A satisfies the above equation for all hereditary radicals γ, then A is said to be a hereditary n-Amitsur ring. Characterizations and examples of these rings are provided. Moreover, new radicals associated with n-Amitsur rings are introduced and studied. One of these is a special radical and its semisimple class is polynomially extensible.

Keywords: Amitsur rings, hereditary Amitsur rings, radicals, radicals with the Amitsur property

Throughout this paper, all rings considered are associative and do not necessarily have an identity element. All classes of rings contain the one element ring and are closed under isomorphisms. Let us recall that a class γ of rings is called a radical class (in the sense of Kurosh and Amitsur) if γ is closed under homomorphisms, is closed under extensions (if I is an ideal of a ring A and both I and A/I (if I1⊆ Ii⊆ … ⊆ Ii⊆ … is an ascending chain of ideals of a ring A and if each Ii is in γ, then ∪ Ii is also in γ). The unique largest ideal of a ring A belonging to γ, denoted by γ (A), is called the γ-radical of A. In what follows, a radical class will be shortly called a radical. For a given radical γ, the semisimple class γ, denoted by Sγ, is the class of all rings A with γ (A)=0. If σ is a class of rings, then the smallest radical containing σ is called the lower radical determined by σ and is denoted by L(σ). The lower hereditary radical determined by σ, denoted by Lh(σ), is the smallest hereditary radical containing σ. A class α of rings is said to be hereditary if α is closed under ideals. If α is a hereditary class of rings, then U(α) denotes the upper radical determined by α, that is, the class of all rings that have no nonzero homomorphic image in α. To denote that I is an ideal of a ring A, we write IA. An ideal I of a ring A is said to be essential in A, denoted by IA, if I∩ J≠ 0 for any nonzero ideal J of A. A class α of rings is said to be essentially closed if IA and I ∈ α implies that A ∈ α . A hereditary class of prime rings that is essentially closed is called a special class. A radical γ is called special if it is the upper radical determined by a special class of rings. For other undefined terms and radical-theoretic properties used throughout this paper, we refer the reader to [4] and [14].

Let us recall that a radical γ is said to have the Amitsur property if

γ(A[x])=(γ(A[x])A)[x]

for all rings A. Radicals with this property have been studied in several papers, for example, in [1, 6, 7, 11, 12, 13]. Throughout this paper, n denotes an arbitrary but fixed natural number n. Let Xn denote a set of n commuting indeterminates x1,,xn. We introduce the following definition.

Definition 1.1.

A ring A is said to be an n-Amitsur ring if

γ(A[Xn])=(γA[Xn]A)[Xn]

for all radicals γ.

Definition 1.2.

A ring A is called a hereditary n-Amitsur ring if

γ(A[Xn])=(γA[Xn]A)[Xn]

for all hereditary radicals γ.

Clearly, every n-Amitsur ring is a hereditary n-Amitsur ring. If n=1, then a 1-Amitsur ring and a hereditary 1-Amitsur ring are just the concepts of Amitsur ring and hereditary Amitsur ring, respectively, which were introduced and investigated by S. Tumurbat in [9]. The purpose of this paper is to obtain generalizations of those results.

Let γ be a radical. If ASγ and γA[Xn]0, then A is not an n-Amitsur ring. Indeed, suppose that A is an n-Amitsur ring. Then 0γ(A[Xn])=(γ(A[Xn])A)[Xn] and hence 0γA[Xn]Aγ, which contradicts the assumption that ASγ. Hence, for example, p (the ring of integers modulo p, where p is a prime number) is not an n-Amitsur

ring. In fact, for γ=U({p}), it is clear that pSγ and we claim that p[Xn]Sγ. Since γ=U({p}) is a special radical,

γ(p[Xn])={Ip[Xn]:p[Xn]/Ip}

and so we can see that, for xiXn, 0xipxiγp[Xn], since every ap satisfies the polynomial equation xipxi=0 (see [2, 6]). Therefore the natural question arises as to whether n-Amitsur rings exist.

In this section, we show that n-Amitsur rings do indeed exist and we also provide some characterizations. Moreover, we shall introduce new radicals, associated with these rings, and study some of their properties.

We shall require the following known result:

Proposition 2.1. ([6, 13])

If γ is a radical, then (γ(A[Xn])A)[Xn]γ(A[Xn]) for any ring A.

For any radical γ, consider the radical class

γn=A:A is a ring with AXnγ,

defined by Tumurbat and Wisbauer in [13]. It is clear that, for any ring A and any radical γ,γnA=γA if and only if γAXnγ for every radical γ.

Using the method of proof of ([4] , Proposition 4.9.18), we may show the following:

Proposition 2.2.

If γ is a hereditary radical, then γnA=γAXnA for any ring A.

Proposition 2.3.

If A is an n-Amitsur (respectively, hereditary n-Amitsur) ring, then

γnA=γAXnAγA,

for any radical (respectively, hereditary radical) γ.

Proof. We prove the result for the case when A is an n-Amitsur ring. Let γ be an arbitrary radical. We have γAXn=γAXnAXn. Notice that since γnAγn, it is clear that γnAXnγ. Therefore γnAXnγAXn.

Now

γA/γnAXnγAXn/γnAXn=γAXn/γnAXn=γAXnAXn/γnAXnγAXnA/γnAXn

Hence γAXnA/γnAXnγ and so γAXnA/γnAγn. Thus γAXnA/γnAγnA/γnA=0¯, whence γAXnA=γnA.

The assertion γAXnAγA now follows, taking into account that γnγ.

Let I be an ideal of a ring A. If there exists a radical (respectively, hereditary radical) γ such that I=γ(A), then I is called a radical ideal (respectively, h-radical ideal) of A. From Proposition 2.3, it is clear that if A is an n-Amitsur (respectively, hereditary n-Amitsur) ring, then γAXnA is a radical ideal (respectively, h-radical ideal) of A, for any radical (respectively, hereditary radical) γ.

Proposition 2.4.

Let A be an n-Amitsur (respectively, hereditary n-Amitsur) ring. Then ASγn implies that AXnSγ, for any radical (respectively, hereditary radical) γ

Proof. Assume that A is an n-Amitsur ring and let γ be an arbitrary radical Suppose that ASγn. Since

γAXnAXn=γAXnγ,

it is clear that γAXnAγn. Thus γAXnAγnA=0. Hence γAXn=0; that is AXnSγ. The proof is similar when A is a hereditary n-Amitsur ring.

In what follows, for a radical γ and a ring A, we put

Aγ=γ(A[Xn])A, A¯γ=A/Aγ,γ(A[Xn])¯=γ(A[Xn])/AγXn.

Lemma 2.5.

γ(A¯γ[Xn])A¯γ=0¯, for any ring A and any radical γ.

Proof. Suppose that there exists a radical γ and a ring A such that γ((A¯γ[Xn])A¯γ0¯. Then there is a nonzero element a¯γ( A ¯ γ[Xn]) A ¯ γ. Hence there exists 0aAγ(A[Xn], which is a pre-image of a¯. By Proposition 2.1, ((γ(A[Xn])A)[Xn]γ(A[Xn]). Taking into account that radical classes are closed under extensions, we have γ(A¯γ[Xn])γ(A[Xn])¯. Consequently, aγ(A[Xn] and aA, which is a contradiction.

Theorem 2.6.

For a ring A, the following conditions are equivalent:

  • (i) A is a n-Amitsur ring;

  • (ii) γ(A¯γ[Xn])=0¯ for every radical γ;

  • (iii) γ0(A¯γ0[Xn])=0¯, where γ0=L(H) and H is any ideal of A[Xn];

  • (iv) every radical ideal I of A[Xn] is a polynomial ring;

  • (v) γA[Xn]=γnA[Xn] for every radical γ.

Proof. (ii) implies (i). If γ is a radical, then we know that (γ(A[Xn])A)[Xn]γ(A[Xn]. Since radical classes are closed under extensions, γ(A¯γ[Xn])γ(A[Xn])¯. Hence, if γ(A¯γ[Xn])=0¯, then γ(A[Xn])¯=0¯ and so γ(A[Xn])(γ(A[Xn])A)[Xn]. Therefore A is a n-Amitsur ring.

(iii) implies (ii). Let γ be an arbitrary radical. Since γ(A[Xn])A[Xn], we take H=γ(A[Xn]) and γ0=L(H). Then γ0(A[Xn])=H=γ(A[Xn]). Thus, if (iii) holds, then γ(A¯γ[Xn])=0¯.

(i) implies (iii). This is clear.

(i) implies (iv). Let I be radical ideal of A[Xn]. Then I=γ(A[Xn]) for some radical γ. If A is a n-Amitsur ring, then γ(A[Xn])=(γ(A[Xn])A)[Xn] and so I is a polynomial ring.

(iv) implies (i). Let γ be a radical such that I=γ(A[Xn]) , that is, I is a radical ideal of the ring A[Xn]. Then, by condition

(iv), I is a polynomial ring. Hence there exists IA such that I[Xn]=γ(A[Xn]). Now it is easy to see that I=Aγ(A[Xn]).

(i) implies (v). If A is an n-Amitsur ring, then (v) follows from Proposition 2.3.

(v) implies (i). Suppose that γA[Xn]=γnA[Xn] for any radical γ. Then

γnA[Xn]=γnAA[Xn]γnAXnA[Xn]=γA[Xn]A[Xn].

Taking into account Proposition 2.1, we have the desired result.

The next theorem may be proved in a similar way to the theorem above, and so we omit its proof. Recall that a subring I of a ring A is called an accessible subring of A if there exists a finite sequence I0,I1,...,In of subrings such that I=I0I1 In=A.

Theorem 2.7.

For a ring A, the following statements are equivalent:

  • (i) A is a hereditary n-Amitsur ring;

  • (ii) γ(A¯γ[Xn])=0¯ for every hereditary radical γ;

  • (iii) γ1(A¯γ1[Xn])=0¯, where γ1=Lh(H) and H is any accessible subring of A[Xn];

  • (iv) every h-radical ideal I of A[Xn] is a polynomial ring.

  • (v) γA[Xn]=γnA[Xn] for every hereditary radical γ.

Proposition 2.8.

If F is a finite field, then F is not a hereditary n-Amitsur ring, for every natural number n.

Proof. If F is a finite field, then γ=UF is a special radical. Now xnpmxnγ(F[Xn]), where pm is the number of elements in F. Therefore γ(F[Xn])(γ(F[Xn])F)[Xn] and hence F is not a hereditary n-Amitsur ring.

Following the reasoning in the proof of Proposition 2.6 of [13], we may prove the next result.

Lemma 2.9.

Let Y be any subset of Xn of cardinality 1, where n>1. If A is a ring such that A[Xn\Y]is an Amitsur ring (respectively, hereditary Amitsur ring), then γ(A[Xn])0 implies that Aγ(A[Xn])0, for any radical (respectively, hereditary radical) γ.

Proof. Let γ be an arbitray radical and A be a ring such that A[Xn\Y] is a an Amitsur ring. Assume that γ(A[Xn])0. Since A[Xn\Y] is an Amitsur ring,

γ(A[Xn])=γ((A[Xn\Y])[Y])=(A[Xn\Y]γ(A[Xn]))[Y].

For any 0aγ(A[Xn]), there exist elements xi1,...,xin(a)Xn and aα1,...,αn(a)A such that

a=aα1,...,αn(a)xi1α1...xin(a)αn(a)

where, for each xij there exists an exponent αj0 such that

aα1,...,αn(a)xi1α1...xin(a)αn(a)0

or a ∈ A. The number of nonzero summands of a is called the length of a and is denoted by ℓ (a).

Suppose that Aγ(A[Xn])=0. For each 0aγ(A[Xn]), l(a)1. Choose 0aγ(A[Xn]) with l(a) minimal. If a depends on only one indeterminate, then the coefficients of a belong to Aγ(A[Xn])=0 and so a=0, which is a contradiction. Therefore a depends on at least two indeterminates, that is, n(a)2. We can write

0a=fαn(a)(xi1...xin(a)1)xin(a)αn(a)(A[Xn\Y]γ(A[Xn]))[Y],

where Y=xin(a) and fαn(a)=fαn(a)(xi1...xin(a)1)A[Xn\Y]γ(A[Xn]). If l(fαn(a))=l(a), take fαn(a) instead of a. The number of indeterminates in fαn(a) is less than the number of indeterminates in a. Continuing with this procedure, we can find 0fkγ(A[Xn]) such that either fkA[Y]) for some subset Y of Xn of cardinality 1, or l(fk)<l(a). In the first case, fkAγ(A[Xn])=0, which is a contradiction. The second case contradicts the minimality of l(a). Thus we conclude that Aγ(A[Xn])0, as desired.

Let n be a natural number. We recall that a class α of rings said to be n-polynomially extensible if A[Xn]α for all Aα. If n=1, the class α of rings is said to be polynomially extensible. It is clear that if a class α of rings is polynomially extensible, then it is n-polynomially extensible, for any natural number n.

For any ring A, let A0 be the zero-ring on the additive group of A, that is, A0 has the additive group of A and multiplication defined by ab=0 for all a,bA. Clearly, the class of all zero-rings is n-polynomially extensible, for any natural number n. Moreover, in [9], it was proved that every zero-ring is an Amitsur ring. We are now in a position to prove the following proposition.

Proposition 2.10.

For any ring A, the ring A0 is an n-Amitsur ring.

Proof. Let γ be an arbitrary radical. From Proposition 2.1,

(A0γ(A0[Xn]))[Xn]γ(A0[Xn])

Suppose that

γ(A0[Xn])(A0γ(A0[Xn]))[Xn].

Then γ(A0¯γ[Xn])0¯. Hence, by the previous lemma, A0γ(A0[Xn]0¯, which is a contradiction with Lemma 2.5.

The Baer radical class is the upper radical determined by the class of all prime rings and also coincides with the upper radical determined by the class of all semiprime rings. Rings belonging to the Baer radical class are called Baer radical rings. It is well known that the Baer radical class is n-polynomially extensible, for any natural number n. In addition, it was proved in [9] that every Baer radical ring is a hereditary Amitsur ring. Hence we have:

Proposition 2.11.

If A is a Baer radical ring, then A is a hereditary n-Amitsur ring.

Proof. A¯γ is a Baer radical ring and therefore A¯γXn is also a Baer radical ring. If γ(A¯γXn)0¯, then, by the previous lemma, A¯γγ(A¯γ[Xn])0¯. This is a contradiction with Lemma 2.5. Thus γ(A¯γXn)=0¯. Hence, by Theorem 2.7, A is a hereditary n-Amitsur ring.

The following lemma is required in the proof of the next proposition.

Lemma 2.12. ([8])

Let A be an infinite integral domain, and let S=A[xi:iΛ] be the polynomial ring over A with Λ commuting indeterminates. For every radical class γ, Aγ(S)=0 if and only if γ(S)=0.

Proposition 2.13.

If A is an infinite integral domain whose every proper homomorphic image is a Baer radical ring, then A is a hereditary n-Amitsur ring.

Proof. Let A be an infinite integral domain without proper prime homomorphic images. Let γ be a hereditary radical and suppose that γ(A[Xn])0. By the lemma above, Aγ(A[Xn])0. We also know that (Aγ(A[Xn]))[Xn]γ(A[Xn]). If γ(A[Xn])(Aγ(A[Xn]))[Xn], then we

have0¯γ(A[Xn])¯ =γ(A[Xn])/(Aγ(A[Xn]))[Xn]A[Xn]/(Aγ(A[Xn]))[Xn]. By assumption, Aγ¯ is a Baer

radical ring and hence A¯γXn and γ(A[Xn])¯γ A¯ γXn are Baer radical rings. Then γ(A[Xn])¯ is not a semiprime ring and so it has a nonzero ideal I¯ such that I¯2=0¯. Let I¯γ be the ideal of A¯γ generated by the

coefficients of gx1,...,xn¯I¯. Therefore

0¯I¯ I ¯ γXnγ A ¯γXn=γ I ¯γXn,

since γ is hereditary Then, from the lemma above, I¯γγ(I¯γ[Xn])0¯, where I¯γγ(I¯γ[Xn])A¯γγ(A¯γ[Xn]). However, from Lemma 2.5, we have A¯γγ(A¯γ[Xn])=0¯, which is a contradiction.

Example 2.14.([4])

  • (i) The ring

    W=2x2y+1: x and y are integers and 2x,2y+1=1

    is an infinite integral domain and every proper homomorphic image of W is a Baer radical ring. Therefore W is a hereditary n-Amitsur ring.

  • (ii) Taking into account Lemma 2.12, it can be easily seen that any infinite field is an n-Amitsur ring.

Next, we introduce the following classes of rings, which rely on the concept of n-Amitsur ring:

Tn={A:A is a ring whose nonzero prime homomorphic images are not hereditary n-Amitsur rings

Tns={A:A is a ring whose prime homomorphic images have no nonzero ideals that are hereditary n-Amitsur rings,

It is easy to see that each of the classes Tn and Tns is homomorphically closed and that TnsTn. Moreover, we will show that they are radical classes. For this purpose, we require the following class of rings:

τn=A:A is a prime and hereditary n-Amitsur ring,

Lemma 2.15.

τ n is a hereditary class of rings.

Proof. Suppose that 0IAτn. Let γ be a hereditary radical. Since A is a prime ring, A[Xn] is also a prime ring. So, if γ(A[Xn])0, then γ(I[Xn])=I[Xn]γ(A[Xn])0. Moreover, all the coefficients of any g(x1,,xn)γ(I[Xn]) are in $I$ and also in γ(I[Xn]) and therefore γ(I[Xn])Iγ(I[Xn])[Xn]. Taking into account Proposition 2.1, γ(I[Xn])=Iγ(I[Xn])[Xn], as desired.

Let us recall that the essential cover of a class α of rings, denoted by ϵα, is the class of all rings that contain an essential ideal in α.

Lemma 2.16.

The essential cover ϵτn is a special class of rings.

Proof. Since τn is a hereditary class of prime rings, ϵτn is also a hereditary class of prime rings. It remains to show that the class ϵτn is essentially closed. Let AB and Aϵτn. Then there exists a nonzero ideal I of A such that IA and Iτn. Since I is a prime ring, A is a prime ring and hence B is also a prime ring, because the class of prime rings is essentially closed. Let IB be the ideal of B generated by I. By Andrunakievich's Lemma, IB3I, where 0IB3B. By Lemma 2.15, IB3 is a hereditary n-Amitsur ring. Thus Bϵτn.

Corollary 2.17.

ϵτn={A:A is a prime ring having a nonzero ideal I that is a hereditary n-Amitsur ring.

Theorem 2.18.

Tn is a radical class, and Tns is a special radical class.

Proof. We claim that U(τn)=Tn. If ATn, then every nonzero prime homomorphic image of $A$ is not a hereditary n-Amitsur ring, that is, AU(τn). Therefore TnU(τn). Let AU(τn)\Tn. Since ATn, there is a nonzero prime homomorphic image A¯ of A which is a hereditary n-Amitsur ring. Hence A¯τnU(τn) and so A¯=0¯, which is a contradiction. Therefore U(τn)Tn. Lemma 2.15, now yields that U(τn)=Tn is a radical class.

Next, we show that U(ϵτn)=Tns. Suppose that U(ϵτn)Tns. Then there exists 0AU(ϵτn) such that ATns. Since ATns, there is a homomorphic image A¯ of A such that 0¯A¯ϵτn. Therefore 0¯A¯ϵτnU(ϵτn)=0, which is a contradiction. Hence U(ϵτn)Tns. Now suppose that TnsU(ϵτn). Then there exists 0ATns such that AU(ϵτn). Since AU(ϵτn), there is a nonzero prime homomorphic image A- of A such that A¯ϵτn. Since Tns is homomorphically closed, A¯Tns. On the other hand, every homomorphic image of A is not in ϵτn, which is a contradiction. Thus TnsU(ϵτn). By Lemma 2.16, ϵτn is a special class of rings and so Tns=U(ϵτn) is a special radical.

Theorem 2.19.

The semisimple class STns is n-polynomially extensible.

Proof. Let ASTns. By Theorem 2.18, Tns is a special radical. Therefore there exist ideals Hi (iΛ) of $A$ such that, for each iΛ, Bi=A/Hiϵτn and iΛHi=0. We claim that Tns(A[Xn])Hi[Xn], for any iΛ. If Tns(A[Xn])Hi[Xn] for some iΛ, then

Tns((A/Hi)[Xn])TnsA[Xn ]Hi [Xn ]0¯.

Thus 0Tns(Bi[Xn]). Since ϵτn is a special class of rings, A/Hi is a prime ring. On the other hand, there exists an ideal Ji of Bi such that 0JiBi and Ji is a prime hereditary n-Amitsur ring. We have, for any iΛ

0Tns(Ji[Xn])=(JiTns(Ji[Xn])[Xn]Tns.

Thus 0JiTns(Ji[Xn])Tns, where JiTns(Ji[Xn])Bi. But JiTns(Ji[Xn])Tns(Bi)=0, which is contradiction. Thus Bi[Xn]STns and consequently Tns(A[Xn]Hi[Xn], for any iΛ. If 0Tns(A[Xn]), then there exists 0g(Xn)Tns(A[Xn]). But all the coefficients ai of of g(Xn) are in iΛHi=0. Hence ai=0 and so Tn(A[Xn])=0, that is, A[Xn]STns.

We recall from [5] that a radical γ is called small if γγAss for each proper radical γ, where Ass denotes the class of all associative rings. Dually, a nonzero radical γ is large if γγ0 for each proper radical γ.

Let Ls denote the collection of all strongly hereditary (that is, closed under subrings) and large radicals, and let L denote the collection of all radicals γ such that γγα0 for every γαLs. In [10], S. Tumurbat et al. proved that L is a complete sublattice in the lattice of all radicals. Now we have the following:

Proposition 2.20.

Each of the radicals Tn and Tns belongs to L

Proof. The radical Tns contains all fields p and all rings p0 of prime order p (where p0 denotes the ring with the additive group of p and with multiplication defined by ab=0 for all a,bp). In [10], it was shown that every nonzero strongly hereditary radical contains a prime field or a simple zero ring of prime order. Hence Tnsγα0 for every γαLs. Thus TnsL and also TnL.

Definition 2.21.

We call a radical γ an n-bad radical if, for every prime ring Aγ, there exists a hereditary radical γA such that γAAXnγAA Xn AXn.

In what follows, Lb and Lsb denote, respectively, the class of all n-bad radicals and the class of all special radicals that are n-bad radicals.

Proposition 2.22.

The classes Lb and Lsb satisfy the following conditions:

  • (i) If γiLb for each iΛ, then iΛγiLb;

  • (ii) If γiLsb for each iΛ, then iΛγiLsb.

Proof. (i) Let A be a prime ring in iΛγi. For each iΛ, there exists a hereditary radical γiA such that

γiAAXn γiAA XnAXn.

Then we may choose any one of γiA, iΛ.

Statement (ii) may be proved in a similar way.

Theorem 2.23.

Both Lb and Lsb are complete sublattices in the lattice of all radicals.

Proof. We have γiTn for each n-bad radical γi. Therefore LγiTn and LγiLb. By Proposition 2.22, γiLb. Thus Lb is a complete sublattice in the lattice of all radicals. The proof that Lsb is a complete sublattice in the lattice of all radicals is similar.

In [3], B.J. Gardner, J. Krempa and R. Wiegandt posed the question as to whether there exists a (hereditary) radical γ with polynomially extensible semisimple class Sγ such that γ does not have the Amitsur property. If T1s does not have the Amitsur property, then we have a positive answer to this question.

  1. N. Divinsky and A. Suliński, Kurosh radicals of rings with operators, Canad. J. Math., 17(1965), 278-280.
    CrossRef
  2. H. France-Jackson, T. Khulan and S. Tumurbat, On α-like radicals of rings, Bull. Aust. Math. Soc., 88(2)(2013), 331-339.
    CrossRef
  3. B. J. Gardner, J. Krempa and R. Wiegandt, Open problems in radical theory, Algebra Discrete Math., (3)(2007), 15-17.
  4. B. J. Gardner and R. Wiegandt, Radical theory of rings, Marcel Dekker, New York, 2004.
    CrossRef
  5. B. J. Gardner and L. Zhian, Small and large radical classes, Comm. Algebra, 20(1992), 2533-2551.
  6. J. Krempa, On radical properties of polynomial rings, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys., 20(1972), 545-548.
  7. N. V. Loi and R. Wiegandt, On the Amitsur property of radicals, Algebra Discrete Math., (3)(2006), 92-100.
  8. N. R. McConnell and T. Stokes, Radical ideals of radically simple rings and their ex-tensions, Theory of radicals, 185-196, Colloquia Mathematica Societatis Janos Bolyai 61, North-Holland, Amsterdam, 1993.
    CrossRef
  9. S. Tumurbat, On Amitsur rings, Quaest. Math., 42(5)(2019), 665-672.
    CrossRef
  10. S. Tumurbat, D. Dayantsolmon and T. Khulan, Notes on chain rings and radicals, Kyungpook Math. J., 58(2018), 473-479.
  11. S. Tumurbat and R. Wiegandt, Radicals of polynomial rings, Soochow J. Math., 29(4)(2003), 425-434.
  12. S. Tumurbat and R. Wiegandt. On radicals with Amitsur property, Comm. Algebra, 32(3)(2004), 1219-1227.
    CrossRef
  13. S. Tumurbat and R. Wisbauer, Radicals with the α-Amitsur property, J. Algebra Appl., 7(3)(2008), 347-361.
    CrossRef
  14. R. Wiegandt, Radical and semisimple classes of rings, Queen’s Papers in Pure and Applied Mathematics 37, Queen’s University, Kingston Ontario, 1974.