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eISSN 0454-8124
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### Article

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

Published online December 31, 2020

### 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

### Abstract

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

### 1. Introduction

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 $I⊴A$. An ideal I of a ring A is said to be essential in A, denoted by $I⊴•A$, if I∩ J≠ 0 for any nonzero ideal J of A. A class α of rings is said to be essentially closed if $I⊴•A$ 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

$γ(A[Xn])=(γA[Xn]∩A)[Xn]$

### Definition 1.2.

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

$γ(A[Xn])=(γA[Xn]∩A)[Xn]$

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 $A∈Sγ$ and $γA[Xn]≠0$, then A is not an n-Amitsur ring. Indeed, suppose that A is an n-Amitsur ring. Then $0≠γ(A[Xn])=(γ(A[Xn])∩A)[Xn]$ and hence $0≠γA[Xn]∩A∈γ$, which contradicts the assumption that $A∈Sγ$. Hence, for example, $ℤp$ (the ring of integers modulo p, where p is a prime number) is not an n-Amitsur

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

$γ(ℤp[Xn])=∩{I⊴ℤp[Xn]:ℤp[Xn]/I≅ℤp}$

and so we can see that, for $xi∈Xn$, $0≠xip−xi∈γℤp[Xn]$, since every $a∈ℤp$ satisfies the polynomial equation $xip−xi=0$ (see [2, 6]). Therefore the natural question arises as to whether n-Amitsur rings exist.

### 2. n-Amitsur Rings and Radicals

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.

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 $γAXn∈γ$ 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 $γnA=γAXn∩A$ for any ring A.

### Proposition 2.3.

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

$γnA=γAXn∩A⊆γA,$

Proof. We prove the result for the case when A is an n-Amitsur ring. Let γ be an arbitrary radical. We have $γAXn=γAXn∩AXn$. Notice that since $γnA∈γn$, it is clear that $γnAXn∈γ$. Therefore $γnAXn⊆γAXn$.

Now

$γA/γnAXn≅γAXn/γnAXn =γAXn/γnAXn =γAXn∩AXn/γnAXn ≅γAXn∩A/γnAXn$

Hence $γAXn∩A/γnAXn∈γ$ and so $γAXn∩A/γnA∈γn$. Thus $γAXn∩A/γnA⊆γnA/γnA=0¯$, whence $γAXn∩A=γnA$.

The assertion $γAXn∩A⊆γA$ now follows, taking into account that $γn⊆γ$.

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 $γAXn∩A$ 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 $A∈Sγn$ implies that $AXn∈Sγ$, 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 $A∈Sγn$. Since

$γAXn∩AXn=γAXn∈γ,$

it is clear that $γAXn∩A∈γn$. Thus $γAXn∩A⊆γnA=0$. Hence $γAXn=0$; that is $AXn∈Sγ$. 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 $γ((A¯γ[Xn])∩A¯γ≠0¯$. Then there is a nonzero element $a¯∈γ( A ¯ γ[Xn])∩ A ¯ γ$. Hence there exists $0≠a∉A∩γ(A[Xn]$, 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, $a∈γ(A[Xn]$ and $a∈A$, 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 $I′⊴A$ such that $I′[Xn]=γ(A[Xn])$. Now it is easy to see that $I′=A∩γ(A[Xn]).$

(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

$γnA[Xn]=γnA∩A[Xn]⊆γnAXn∩A[Xn]=γA[Xn]∩A[Xn].$

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 .

### 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 $xnpm−xn∈γ(F[Xn])$, 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 $A∩γ(A[Xn])≠0$, 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 $0≠a∈γ(A[Xn])$, 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 $αj≠0$ 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 $A∩γ(A[Xn])=0$. For each $0≠a∈γ(A[Xn])$, $l(a)⩾1$. Choose $0≠a∈γ(A[Xn])$ with $l(a)$ minimal. If a depends on only one indeterminate, then the coefficients of a belong to $A∩γ(A[Xn])=0$ and so a=0, which is a contradiction. Therefore a depends on at least two indeterminates, that is, $n(a)⩾2$. We can write

$0≠a=∑fαn(a)(xi1...xin(a)−1)xin(a)αn(a)∈(A[Xn\Y]∩γ(A[Xn]))[Y],$

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 $0≠fk∈γ(A[Xn])$ such that either $fk∈A[Y])$ for some subset Y of Xn of cardinality 1, or $l(fk). In the first case, $fk∈A∩γ(A[Xn])=0$, which is a contradiction. The second case contradicts the minimality of $l(a)$. Thus we conclude that $A∩γ(A[Xn])≠0$, 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 $A∈α$. 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 $a,b∈A$. 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,

$(A0∩γ(A0[Xn]))[Xn]⊆γ(A0[Xn])$

Suppose that

$γ(A0[Xn])⊈(A0∩γ(A0[Xn]))[Xn].$

Then $γ(A0¯γ[Xn])≠0¯$. Hence, by the previous lemma, $A0∩γ(A0[Xn]≠0¯$, 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, $A¯γ∩γ(A¯γ[Xn])≠0¯$. 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 $S=A[xi:i∈Λ]$ be the polynomial ring over A with Λ commuting indeterminates. For every radical class γ, $A∩γ(S)=0$ 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, $A∩γ(A[Xn])≠0$. We also know that $(A∩γ(A[Xn]))[Xn]⊆γ(A[Xn])$. If $γ(A[Xn])⊈(A∩γ(A[Xn]))[Xn]$, then we

have$0¯≠γ(A[Xn])¯$ $=γ(A[Xn])/(A∩γ(A[Xn]))[Xn]⊴A[Xn]/(A∩γ(A[Xn]))[Xn]$. 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

$0¯≠I¯⊆ I ¯ γXn∩γ A ¯γXn=γ I ¯γXn,$

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

### Example 2.14.([4])

• (i) The ring

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 $Tns⊆Tn$. Moreover, we will show that they are radical classes. For this purpose, we require the following class of rings:

,

### Lemma 2.15.

τ n is a hereditary class of rings.

Proof. Suppose that $0≠I⊴A∈τn$. 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 $γ(I[Xn])⊆I∩γ(I[Xn])[Xn]$. Taking into account Proposition 2.1, $γ(I[Xn])=I∩γ(I[Xn])[Xn]$, 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 $A⊴•B$ and $A∈ϵτn$. Then there exists a nonzero ideal I of A such that $I⊲•A$ and $I∈τn$. 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, $IB3⊆I$, where $0≠IB3⊴•B$. By Lemma 2.15, $IB3$ is a hereditary n-Amitsur ring. Thus $B∈ϵτn$.

### 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 $A∈Tn$, then every nonzero prime homomorphic image of $A$ is not a hereditary n-Amitsur ring, that is, $A∈U(τn)$. Therefore $Tn⊆U(τn)$. Let $A∈U(τn)\Tn$. Since $A∉Tn$, there is a nonzero prime homomorphic image $A¯$ of A which is a hereditary n-Amitsur ring. Hence $A¯∈τn∩U(τn)$ 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 $0≠A∈U(ϵτn)$ such that $A∉Tns$. Since $A∉Tns$, there is a homomorphic image $A¯$ of A such that $0¯≠A¯∈ϵτn$. Therefore $0¯∉A¯∈ϵτn∩U(ϵτn)=0$, which is a contradiction. Hence $U(ϵτn)⊆Tns$. Now suppose that $Tns⊈U(ϵτn)$. Then there exists $0≠A∈Tns$ such that $A∉U(ϵτn)$. Since $A∉U(ϵτn)$, 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 $Tns⊆U(ϵτn)$. 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 $A∈STns$. By Theorem 2.18, $Tns$ is a special radical. Therefore there exist ideals Hi $(i∈Λ)$ of $A$ such that, for each $i∈Λ$, $Bi=A/Hi∈ϵτn$ and $∩i∈ΛHi=0$. We claim that $Tns(A[Xn])⊆Hi[Xn]$, for any $i∈Λ$. If $Tns(A[Xn])⊆Hi[Xn]$ for some $i∈Λ$, then

$Tns((A/Hi)[Xn])≅TnsA[Xn ]Hi [Xn ]≠0¯.$

Thus $0≠Tns(Bi[Xn])$. 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 $0≠Ji⊴•Bi$ and Ji is a prime hereditary n-Amitsur ring. We have, for any $i∈Λ$

$0≠Tns(Ji[Xn])=(Ji∩Tns(Ji[Xn])[Xn]∈Tns.$

Thus $0≠Ji∩Tns(Ji[Xn])∈Tns$, where $Ji∩Tns(Ji[Xn])⊴Bi$. But $Ji∩Tns(Ji[Xn])⊆Tns(Bi)=0$, which is contradiction. Thus $Bi[Xn]∈STns$ and consequently $Tns(A[Xn]⊆Hi[Xn]$, for any $i∈Λ$. If $0≠Tns(A[Xn])$, then there exists $0≠g(Xn)∈Tns(A[Xn])$. But all the coefficients ai of of $g(Xn)$ are in $∩i∈ΛHi=0$. 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 $γ∨γ′≠Ass$ for each proper radical $γ′$, where $Ass$ denotes the class of all associative rings. Dually, a nonzero radical γ is large if $γ∩γ′≠0$ 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 $γ∩γα≠0$ for every $γα∈Ls$. 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 $ℤp$ and all rings $ℤp0$ of prime order p (where $ℤp0$ denotes the ring with the additive group of $ℤp$ and with multiplication defined by ab=0 for all $a,b∈ℤp$). In [10], it was shown that every nonzero strongly hereditary radical contains a prime field or a simple zero ring of prime order. Hence $Tns∩γα≠0$ for every $γα∈Ls$. Thus $Tns∈L$ and also $Tn∈L$.

### Definition 2.21.

We call a radical γ an n-bad radical if, for every prime ring $A∈γ$, there exists a hereditary radical $γA$ such that $γAAXn≠γAA Xn ∩AXn$.

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 $γi∈Lb$ for each $i∈Λ$, then $∩i∈Λγi∈Lb$;

• (ii) If $γi∈Lsb$ for each $i∈Λ$, then $∩i∈Λγi∈Lsb$.

Proof. (i) Let A be a prime ring in $∩i∈Λγi$. For each $i∈Λ$, there exists a hereditary radical $γiA$ such that

$γiAAXn≠ γiAA Xn∩AXn.$

Then we may choose any one of $γiA$, $i∈Λ$.

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 $γi⊆Tn$ for each n-bad radical $γi$. Therefore $L∪γi⊆Tn$ and $L∪γi∈Lb$. By Proposition 2.22, $∩γi∈Lb$. 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.

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