In this section, we show that nAmitsur 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 $(\gamma (A[{X}_{n}])\cap A)[{X}_{n}]\subseteq \gamma (A[{X}_{n}])$ for any ring A.
For any radical γ, consider the radical class
$${\gamma}_{n}=\left\{A:A\text{isaringwith}A\left[{X}_{n}\right]\in \gamma \right\}\text{,}$$
defined by Tumurbat and Wisbauer in [13]. It is clear that, for any ring A and any radical $\gamma ,{\gamma}_{n}\left(A\right)=\gamma \left(A\right)$ if and only if $\gamma \left(A\right)\left[{X}_{n}\right]\in \gamma $ 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 ${\gamma}_{n}\left(A\right)=\gamma \left(A\left[{X}_{n}\right]\right)\cap A$ for any ring A.
Proposition 2.3.
If A is an nAmitsur (respectively, hereditary nAmitsur) ring, then
$${\gamma}_{n}\left(A\right)=\gamma \left(A\left[{X}_{n}\right]\right)\cap A\subseteq \gamma \left(A\right)\text{,}$$
for any radical (respectively, hereditary radical) γ.
Proof. We prove the result for the case when A is an nAmitsur ring. Let γ be an arbitrary radical. We have $\gamma \left(A\left[{X}_{n}\right]\right)=\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)\left[{X}_{n}\right]$. Notice that since ${\gamma}_{n}\left(A\right)\in {\gamma}_{n}$, it is clear that ${\gamma}_{n}\left(A\right)\left[{X}_{n}\right]\in \gamma $. Therefore ${\gamma}_{n}\left(A\right)\left[{X}_{n}\right]\subseteq \gamma \left(A\left[{X}_{n}\right]\right)$.
Now
$$\begin{array}{l}\gamma \left(\left(A/{\gamma}_{n}\left(A\right)\right)\left[{X}_{n}\right]\right)\cong \gamma \left(A\left[{X}_{n}\right]/{\gamma}_{n}\left(A\right)\left[{X}_{n}\right]\right)\\ \text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}=\gamma \left(A\left[{X}_{n}\right]\right)/{\gamma}_{n}\left(A\right)\left[{X}_{n}\right]\\ \text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}=\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)\left[{X}_{n}\right]/{\gamma}_{n}\left(A\right)\left[{X}_{n}\right]\\ \text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\cong \left(\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)/{\gamma}_{n}\left(A\right)\right)\left[{X}_{n}\right]\end{array}$$
Hence $\left(\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)/{\gamma}_{n}\left(A\right)\right)\left[{X}_{n}\right]\in \gamma $ and so $\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)/{\gamma}_{n}\left(A\right)\in {\gamma}_{n}$. Thus $\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)/{\gamma}_{n}\left(A\right)\subseteq {\gamma}_{n}\left(A/{\gamma}_{n}\left(A\right)\right)=\overline{0}$, whence $\gamma \left(A\left[{X}_{n}\right]\right)\cap A={\gamma}_{n}\left(A\right)$.
The assertion $\gamma \left(A\left[{X}_{n}\right]\right)\cap A\subseteq \gamma \left(A\right)$ now follows, taking into account that ${\gamma}_{n}\subseteq \gamma $.
Let I be an ideal of a ring A. If there exists a radical (respectively, hereditary radical) γ such that $I=\gamma (A)$, then I is called a radical ideal (respectively, hradical ideal) of A. From Proposition 2.3, it is clear that if A is an nAmitsur (respectively, hereditary nAmitsur) ring, then $\gamma \left(A\left[{X}_{n}\right]\right)\cap A$ is a radical ideal (respectively, hradical ideal) of A, for any radical (respectively, hereditary radical) γ.
Proposition 2.4.
Let A be an nAmitsur (respectively, hereditary nAmitsur) ring. Then $A\in S{\gamma}_{n}$ implies that $A\left[{X}_{n}\right]\in S\gamma $, for any radical (respectively, hereditary radical) γ
Proof. Assume that A is an nAmitsur ring and let γ be an arbitrary radical Suppose that $A\in S{\gamma}_{n}$. Since
$$\left(\gamma \left(A\left[{X}_{n}\right]\right)\cap A\right)\left[{X}_{n}\right]=\gamma \left(A\left[{X}_{n}\right]\right)\in \gamma \text{,}$$
it is clear that $\gamma \left(A\left[{X}_{n}\right]\right)\cap A\in {\gamma}_{n}$. Thus $\gamma \left(A\left[{X}_{n}\right]\right)\cap A\subseteq {\gamma}_{n}\left(A\right)=0$. Hence $\gamma \left(A\left[{X}_{n}\right]\right)=0$; that is $A\left[{X}_{n}\right]\in S\gamma $. The proof is similar when A is a hereditary nAmitsur ring.
In what follows, for a radical γ and a ring A, we put
$$\begin{array}{l}{A}_{\gamma}=\gamma (A[{X}_{n}])\cap A\text{,}\\ {\overline{A}}_{\gamma}=A/{A}_{\gamma}\text{,}\\ \overline{\gamma (A[{X}_{n}])}=\gamma (A[{X}_{n}])/{A}_{\gamma}\left[{X}_{n}\right]\text{.}\end{array}$$
Lemma 2.5.
$\gamma ({\overline{A}}_{\gamma}[{X}_{n}])\cap {\overline{A}}_{\gamma}=\overline{0}$, for any ring A and any radical γ.
Proof. Suppose that there exists a radical γ and a ring A such that $\gamma (({\overline{A}}_{\gamma}[{X}_{n}])\cap {\overline{A}}_{\gamma}\ne \overline{0}$. Then there is a nonzero element $\overline{a}\in \gamma ({\overline{A}}_{\gamma}[{X}_{n}])\cap {\overline{A}}_{\gamma}$. Hence there exists $0\ne a\notin A\cap \gamma (A[{X}_{n}]$, which is a preimage of $\overline{a}$. By Proposition 2.1, $((\gamma (A[{X}_{n}])\cap A)[{X}_{n}]\subseteq \gamma (A[{X}_{n}])$. Taking into account that radical classes are closed under extensions, we have $\gamma ({\overline{A}}_{\gamma}[{X}_{n}])\cong \overline{\gamma (A[{X}_{n}])}$. Consequently, $a\in \gamma (A[{X}_{n}]$ and $a\in A$, which is a contradiction.
Theorem 2.6.
For a ring A, the following conditions are equivalent:

(i) A is a nAmitsur ring;

(ii) $\gamma ({\overline{A}}_{\gamma}[{X}_{n}])=\overline{0}$ for every radical γ;

(iii) ${\gamma}_{0}({\overline{A}}_{{\gamma}_{0}}[{X}_{n}])=\overline{0}$, where ${\gamma}_{0}=L(H)$ and H is any ideal of $A[{X}_{n}]$;

(iv) every radical ideal I of $A[{X}_{n}]$ is a polynomial ring;

(v) $\gamma \left(A[{X}_{n}]\right)={\gamma}_{n}\left(A\right)[{X}_{n}]$ for every radical γ.
Proof. (ii) implies (i). If γ is a radical, then we know that $(\gamma (A[{X}_{n}])\cap A)[{X}_{n}]\subseteq \gamma (A[{X}_{n}]$. Since radical classes are closed under extensions, $\gamma ({\overline{A}}_{\gamma}[{X}_{n}])\cong \overline{\gamma (A[{X}_{n}])}$. Hence, if $\gamma ({\overline{A}}_{\gamma}[{X}_{n}])=\overline{0}$, then $\overline{\gamma (A[{X}_{n}])}=\overline{0}$ and so $\gamma (A[{X}_{n}])\subseteq (\gamma (A[{X}_{n}])\cap A)[{X}_{n}]$. Therefore A is a nAmitsur ring.
(iii) implies (ii). Let γ be an arbitrary radical. Since $\gamma (A[{X}_{n}])\u22b4A[{X}_{n}]$, we take $H=\gamma (A[{X}_{n}])$ and ${\gamma}_{0}=L(H)$. Then ${\gamma}_{0}(A[{X}_{n}])=H=\gamma (A[{X}_{n}])$. Thus, if (iii) holds, then $\gamma ({\overline{A}}_{\gamma}[{X}_{n}])=\overline{0}$.
(i) implies (iii). This is clear.
(i) implies (iv). Let I be radical ideal of A[X_{n}]. Then $I=\gamma (A[{X}_{n}])$ for some radical γ. If A is a nAmitsur ring, then $\gamma (A[{X}_{n}])=(\gamma (A[{X}_{n}])\cap A)[{X}_{n}]$ and so I is a polynomial ring.
(iv) implies (i). Let γ be a radical such that $I=\gamma (A[{X}_{n}])$ , that is, I is a radical ideal of the ring A[X_{n}]. Then, by condition
(iv), I is a polynomial ring. Hence there exists ${I}^{\prime}\u22b4A$ such that ${I}^{\prime}[{X}_{n}]=\gamma (A[{X}_{n}])$. Now it is easy to see that ${I}^{\prime}=A\cap \gamma (A[{X}_{n}]).$
(i) implies (v). If A is an nAmitsur ring, then (v) follows from Proposition 2.3.
(v) implies (i). Suppose that $\gamma \left(A[{X}_{n}]\right)={\gamma}_{n}\left(A\right)[{X}_{n}]$ for any radical γ. Then
$$\begin{array}{l}{\gamma}_{n}\left(A\right)[{X}_{n}]=\left({\gamma}_{n}\left(A\right)\cap A\right)[{X}_{n}]\\ \subseteq \left({\gamma}_{n}\left(A\right)\left[{X}_{n}\right]\cap A\right)[{X}_{n}]=\left(\gamma \left(A[{X}_{n}]\right)\cap A\right)[{X}_{n}].\end{array}$$
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 ${I}_{0},{I}_{1},\mathrm{...},{I}_{n}$ of subrings such that $I={I}_{0}\u22b4{I}_{1}\u22b4\dots \text{}\u22b4{I}_{n}=A$.
Theorem 2.7.
For a ring A, the following statements are equivalent:

(i) A is a hereditary nAmitsur ring;

(ii) $\gamma ({\overline{A}}_{\gamma}[{X}_{n}])=\overline{0}$ for every hereditary radical γ;

(iii) ${\gamma}_{1}({\overline{A}}_{{\gamma}_{1}}[{X}_{n}])=\overline{0}$, where ${\gamma}_{1}={L}_{h}(H)$ and H is any accessible subring of $A[{X}_{n}]$;

(iv) every hradical ideal I of $A[{X}_{n}]$ is a polynomial ring.

(v) $\gamma \left(A[{X}_{n}]\right)={\gamma}_{n}\left(A\right)[{X}_{n}]$ for every hereditary radical γ.
Proposition 2.8.
If F is a finite field, then F is not a hereditary nAmitsur ring, for every natural number n.
Proof. If F is a finite field, then $\gamma =U\left(\left\{F\right\}\right)$ is a special radical. Now ${x}_{n}^{{p}^{m}}{x}_{n}\in \gamma (F[{X}_{n}])$, where p^{m} is the number of elements in F. Therefore $\gamma (F[{X}_{n}])\u2288(\gamma (F[{X}_{n}])\cap F)[{X}_{n}]$ and hence F is not a hereditary nAmitsur 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 X_{n} of cardinality 1, where n>1. If A is a ring such that $A[{X}_{n}\backslash Y]$is an Amitsur ring (respectively, hereditary Amitsur ring), then $\gamma (A[{X}_{n}])\ne 0$ implies that $A\cap \gamma (A[{X}_{n}])\ne 0$, for any radical (respectively, hereditary radical) γ.
Proof. Let γ be an arbitray radical and A be a ring such that $A[{X}_{n}\backslash Y]$ is a an Amitsur ring. Assume that $\gamma (A[{X}_{n}])\ne 0$. Since $A[{X}_{n}\backslash Y]$ is an Amitsur ring,
$$\gamma (A[{X}_{n}])=\gamma ((A[{X}_{n}\backslash Y])[Y])=(A[{X}_{n}\backslash Y]\cap \gamma (A[{X}_{n}]))[Y]\text{.}$$
For any $0\ne a\in \gamma (A[{X}_{n}])$, there exist elements ${x}_{{i}_{1}},\mathrm{...},{x}_{{i}_{n(a)}}\in {X}_{n}$ and ${a}_{{\alpha}_{1},\mathrm{...}{,}_{{\alpha}_{n(a)}}}\in A$ such that
$$a=\sum {a}_{{\alpha}_{1},\mathrm{...}{,}_{{\alpha}_{n(a)}}}{x}_{{i}_{1}}^{{\alpha}_{1}}\mathrm{...}{x}_{{i}_{n(a)}}^{{\alpha}_{n(a)}}$$
where, for each ${x}_{{i}_{j}}$ there exists an exponent ${\alpha}_{j}\ne 0$ such that
$${a}_{{\alpha}_{1},\mathrm{...}{,}_{{\alpha}_{n(a)}}}{x}_{{i}_{1}}^{{\alpha}_{1}}\mathrm{...}{x}_{{i}_{n(a)}}^{{\alpha}_{n(a)}}\ne 0$$
or a ∈ A. The number of nonzero summands of a is called the length of a and is denoted by ℓ (a).
Suppose that $A\cap \gamma (A[{X}_{n}])=0$. For each $0\ne a\in \gamma (A[{X}_{n}])$, $\mathcal{l}(a)\u2a7e1$. Choose $0\ne a\in \gamma (A[{X}_{n}])$ with $\mathcal{l}(a)$ minimal. If a depends on only one indeterminate, then the coefficients of a belong to $A\cap \gamma (A[{X}_{n}])=0$ and so a=0, which is a contradiction. Therefore a depends on at least two indeterminates, that is, $n(a)\u2a7e2$. We can write
$$0\ne a={\displaystyle \sum {f}_{{\alpha}_{n(a)}}}({x}_{{i}_{1}}\mathrm{...}{x}_{{i}_{n(a)1}}){x}_{{i}_{n(a)}}^{{\alpha}_{n(a)}}\in (A[{X}_{n}\backslash Y]\cap \gamma (A[{X}_{n}]))[Y]\text{,}$$
where $Y=\left\{{x}_{{i}_{n(a)}}\right\}$ and ${f}_{{\alpha}_{n(a)}}={f}_{{\alpha}_{n(a)}}({x}_{{i}_{1}}\mathrm{...}{x}_{{i}_{n(a)1}})\in A[{X}_{n}\backslash Y]\cap \gamma (A[{X}_{n}])$. If $\mathcal{l}({f}_{{\alpha}_{n(a)}})=\mathcal{l}(a)$, take ${f}_{{\alpha}_{n(a)}}$ instead of a. The number of indeterminates in ${f}_{{\alpha}_{n(a)}}$ is less than the number of indeterminates in a. Continuing with this procedure, we can find $0\ne {f}_{k}\in \gamma (A[{X}_{n}])$ such that either ${f}_{k}\in A[Y])$ for some subset Y of X_{n} of cardinality 1, or $\mathcal{l}({f}_{k})<\mathcal{l}(a)$. In the first case, ${f}_{k}\in A\cap \gamma (A[{X}_{n}])=0$, which is a contradiction. The second case contradicts the minimality of $\mathcal{l}(a)$. Thus we conclude that $A\cap \gamma (A[{X}_{n}])\ne 0$, as desired.
Let n be a natural number. We recall that a class α of rings said to be npolynomially extensible if $A[{X}_{n}]\in \alpha $ for all $A\in \alpha $. 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 npolynomially extensible, for any natural number n.
For any ring A, let A^{0} be the zeroring on the additive group of A, that is, A^{0} has the additive group of A and multiplication defined by ab=0 for all $a,b\in A$. Clearly, the class of all zerorings is npolynomially extensible, for any natural number n. Moreover, in [9], it was proved that every zeroring is an Amitsur ring. We are now in a position to prove the following proposition.
Proposition 2.10.
For any ring A, the ring A^{0} is an nAmitsur ring.
Proof. Let γ be an arbitrary radical. From Proposition 2.1,
$$({A}^{0}\cap \gamma ({A}^{0}[{X}_{n}]))[{X}_{n}]\subseteq \gamma ({A}^{0}[{X}_{n}])$$
Suppose that
$$\gamma ({A}^{0}[{X}_{n}])\u2288({A}^{0}\cap \gamma ({A}^{0}[{X}_{n}]))[{X}_{n}]\text{.}$$
Then $\gamma ({\overline{{A}^{0}}}_{\gamma}[{X}_{n}])\ne \overline{0}$. Hence, by the previous lemma, ${A}^{0}\cap \gamma ({A}^{0}[{X}_{n}]\ne \overline{0}$, 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 npolynomially 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 nAmitsur ring.
Proof. ${\overline{A}}_{\gamma}$ is a Baer radical ring and therefore ${\overline{A}}_{\gamma}\left[{X}_{n}\right]$ is also a Baer radical ring. If $\gamma ({\overline{A}}_{\gamma}\left[{X}_{n}\right])\ne \overline{0}$, then, by the previous lemma, ${\overline{A}}_{\gamma}\cap \gamma ({\overline{A}}_{\gamma}[{X}_{n}])\ne \overline{0}$. This is a contradiction with Lemma 2.5. Thus $\gamma ({\overline{A}}_{\gamma}\left[{X}_{n}\right])=\overline{0}$. Hence, by Theorem 2.7, A is a hereditary nAmitsur 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 $S=A[{x}_{i}:i\in \Lambda ]$ be the polynomial ring over A with Λ commuting indeterminates. For every radical class γ, $A\cap \gamma (S)=0$ if and only if $\gamma (S)=0$.
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 nAmitsur ring.
Proof. Let A be an infinite integral domain without proper prime homomorphic images. Let γ be a hereditary radical and suppose that $\gamma (A[{X}_{n}])\ne 0$. By the lemma above, $A\cap \gamma (A[{X}_{n}])\ne 0$. We also know that $(A\cap \gamma (A[{X}_{n}]))[{X}_{n}]\subseteq \gamma (A[{X}_{n}])$. If $\gamma (A[{X}_{n}])\u2288(A\cap \gamma (A[{X}_{n}]))[{X}_{n}]$, then we
have$\overline{0}\ne \overline{\gamma (A[{X}_{n}])}$ $=\gamma (A[{X}_{n}])/(A\cap \gamma (A[{X}_{n}]))[{X}_{n}]\u22b4A[{X}_{n}]/(A\cap \gamma (A[{X}_{n}]))[{X}_{n}]$. By assumption, $\overline{{A}_{\gamma}}$ is a Baer
radical ring and hence ${\overline{A}}_{\gamma}\left[{X}_{n}\right]$ and $\overline{\gamma (A[{X}_{n}])}\cong \gamma \left({\overline{A}}_{\gamma}\left[{X}_{n}\right]\right)$ are Baer radical rings. Then $\overline{\gamma (A[{X}_{n}])}$ is not a semiprime ring and so it has a nonzero ideal $\overline{I}$ such that ${\overline{I}}^{2}=\overline{0}$. Let ${\overline{I}}_{\gamma}$ be the ideal of ${\overline{A}}_{\gamma}$ generated by the
coefficients of $\overline{g\left({x}_{1},\mathrm{...},{x}_{n}\right)}\in \overline{I}$. Therefore
$$\overline{0}\ne \overline{I}\subseteq {\overline{I}}_{\gamma}\left[{X}_{n}\right]\cap \gamma \left({\overline{A}}_{\gamma}\left[{X}_{n}\right]\right)=\gamma \left({\overline{I}}_{\gamma}\left[{X}_{n}\right]\right),$$
since γ is hereditary Then, from the lemma above, ${\overline{I}}_{\gamma}\cap \gamma ({\overline{I}}_{\gamma}[{X}_{n}])\ne \overline{0}$, where ${\overline{I}}_{\gamma}\cap \gamma ({\overline{I}}_{\gamma}[{X}_{n}])\subseteq {\overline{A}}_{\gamma}\cap \gamma ({\overline{A}}_{\gamma}[{X}_{n}])$. However, from Lemma 2.5, we have ${\overline{A}}_{\gamma}\cap \gamma ({\overline{A}}_{\gamma}[{X}_{n}])=\overline{0}$, which is a contradiction.
Example 2.14.([4])

(i) The ring
$W=\left\{\frac{2x}{2y+1}:\text{}x\text{and}y\text{areintegersand}\left(2x,2y+1\right)=1\right\}$
is an infinite integral domain and every proper homomorphic image of W is a Baer radical ring. Therefore W is a hereditary nAmitsur ring.

(ii) Taking into account Lemma 2.12, it can be easily seen that any infinite field is an nAmitsur ring.
Next, we introduce the following classes of rings, which rely on the concept of nAmitsur ring:
${\mathcal{T}}_{n}=\{A:A$ is a ring whose nonzero prime homomorphic images are not hereditary nAmitsur rings
${\mathcal{T}}_{ns}=\{A:A$ is a ring whose prime homomorphic images have no nonzero ideals that are hereditary nAmitsur rings,
It is easy to see that each of the classes ${\mathcal{T}}_{n}$ and ${\mathcal{T}}_{ns}$ is homomorphically closed and that ${\mathcal{T}}_{ns}\subseteq {\mathcal{T}}_{n}$. Moreover, we will show that they are radical classes. For this purpose, we require the following class of rings:
${\tau}_{n}=\left\{A:A\text{isaprimeandhereditary}n\text{Amitsurring}\right\}$,
Lemma 2.15.
τ _{n} is a hereditary class of rings.
Proof. Suppose that $0\ne I\u22b4A\in {\tau}_{n}$. Let γ be a hereditary radical. Since A is a prime ring, $A[{X}_{n}]$ is also a prime ring. So, if $\gamma (A[{X}_{n}])\ne 0$, then $\gamma (I[{X}_{n}])=I[{X}_{n}]\cap \gamma (A[{X}_{n}])\ne 0$. Moreover, all the coefficients of any $g({x}_{1},\dots ,{x}_{n})\in \gamma (I[{X}_{n}])$ are in $I$ and also in $\gamma (I[{X}_{n}])$ and therefore $\gamma (I[{X}_{n}])\subseteq \left(I\cap \gamma (I[{X}_{n}])\right)[{X}_{n}]$. Taking into account Proposition 2.1, $\gamma (I[{X}_{n}])=\left(I\cap \gamma (I[{X}_{n}])\right)[{X}_{n}]$, 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 $\u03f5{\tau}_{n}$ is a special class of rings.
Proof. Since ${\tau}_{n}$ is a hereditary class of prime rings, $\u03f5{\tau}_{n}$ is also a hereditary class of prime rings. It remains to show that the class $\u03f5{\tau}_{n}$ is essentially closed. Let $A{\u22b4}^{\u2022}B$ and $A\in \u03f5{\tau}_{n}$. Then there exists a nonzero ideal I of A such that $I{\u22b2}^{\u2022}A$ and $I\in {\tau}_{n}$. 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 I_{B} be the ideal of B generated by I. By Andrunakievich's Lemma, ${I}_{B}^{3}\subseteq I$, where $0\ne {I}_{B}^{3}{\u22b4}^{\u2022}B$. By Lemma 2.15, ${I}_{B}^{3}$ is a hereditary nAmitsur ring. Thus $B\in \u03f5{\tau}_{n}$.
Corollary 2.17.
$\u03f5{\tau}_{n}=\{A:A$ is a prime ring having a nonzero ideal I that is a hereditary nAmitsur ring.
Theorem 2.18.
${\mathcal{T}}_{n}$ is a radical class, and ${\mathcal{T}}_{ns}$ is a special radical class.
Proof. We claim that $U({\tau}_{n})={\mathcal{T}}_{n}$. If $A\in {\mathcal{T}}_{n}$, then every nonzero prime homomorphic image of $A$ is not a hereditary nAmitsur ring, that is, $A\in U({\tau}_{n})$. Therefore ${\mathcal{T}}_{n}\subseteq U({\tau}_{n})$. Let $A\in U({\tau}_{n})\backslash {\mathcal{T}}_{n}$. Since $A\notin {\mathcal{T}}_{n}$, there is a nonzero prime homomorphic image $\overline{A}$ of A which is a hereditary nAmitsur ring. Hence $\overline{A}\in {\tau}_{n}\cap U({\tau}_{n})$ and so $\overline{A}=\overline{0}$, which is a contradiction. Therefore $U({\tau}_{n})\subseteq {\mathcal{T}}_{n}$. Lemma 2.15, now yields that $U({\tau}_{n})={\mathcal{T}}_{n}$ is a radical class.
Next, we show that $U(\u03f5{\tau}_{n})={\mathcal{T}}_{ns}$. Suppose that $U(\u03f5{\tau}_{n})\u2288{\mathcal{T}}_{ns}$. Then there exists $0\ne A\in U(\u03f5{\tau}_{n})$ such that $A\notin {\mathcal{T}}_{ns}$. Since $A\notin {\mathcal{T}}_{ns}$, there is a homomorphic image $\overline{A}$ of A such that $\overline{0}\ne \overline{A}\in \u03f5{\tau}_{n}$. Therefore $\overline{0}\notin \overline{A}\in \u03f5{\tau}_{n}\cap U(\u03f5{\tau}_{n})=0$, which is a contradiction. Hence $U(\u03f5{\tau}_{n})\subseteq {\mathcal{T}}_{ns}$. Now suppose that ${\mathcal{T}}_{ns}\u2288U(\u03f5{\tau}_{n})$. Then there exists $0\ne A\in {\mathcal{T}}_{ns}$ such that $A\notin U(\u03f5{\tau}_{n})$. Since $A\notin U(\u03f5{\tau}_{n})$, there is a nonzero prime homomorphic image $\stackrel{}{A}$ of A such that $\overline{A}\in \u03f5{\tau}_{n}$. Since ${\mathcal{T}}_{ns}$ is homomorphically closed, $\overline{A}\in {\mathcal{T}}_{ns}$. On the other hand, every homomorphic image of A is not in $\u03f5{\tau}_{n}$, which is a contradiction. Thus ${\mathcal{T}}_{ns}\subseteq U(\u03f5{\tau}_{n})$. By Lemma 2.16, $\u03f5{\tau}_{n}$ is a special class of rings and so ${\mathcal{T}}_{ns}=U(\u03f5{\tau}_{n})$ is a special radical.
Theorem 2.19.
The semisimple class $S{\mathcal{T}}_{ns}$ is npolynomially extensible.
Proof. Let $A\in S{\mathcal{T}}_{ns}$. By Theorem 2.18, ${\mathcal{T}}_{ns}$ is a special radical. Therefore there exist ideals H_{i} $(i\in \Lambda )$ of $A$ such that, for each $i\in \Lambda $, ${B}_{i}=A/{H}_{i}\in \u03f5{\tau}_{n}$ and ${\cap}_{i\in \Lambda}{H}_{i}=0$. We claim that ${\mathcal{T}}_{ns}(A[{X}_{n}])\subseteq {H}_{i}[{X}_{n}]$, for any $i\in \Lambda $. If ${\mathcal{T}}_{ns}(A[{X}_{n}])\overline{)\subseteq}{H}_{i}[{X}_{n}]$ for some $i\in \Lambda $, then
$${\mathcal{T}}_{ns}((A/{H}_{i})[{X}_{n}])\cong {\mathcal{T}}_{ns}\left(\frac{A[{X}_{n}]}{{H}_{i}[{X}_{n}]}\right)\ne \overline{0}.$$
Thus $0\ne {\mathcal{T}}_{ns}({B}_{i}[{X}_{n}])$. Since $\u03f5{\tau}_{n}$ is a special class of rings, $A/{H}_{i}$ is a prime ring. On the other hand, there exists an ideal J_{i} of B_{i} such that $0\ne {J}_{i}{\u22b4}^{\u2022}{B}_{i}$ and J_{i} is a prime hereditary nAmitsur ring. We have, for any $i\in \Lambda $
$$0\ne {\mathcal{T}}_{ns}({J}_{i}[{X}_{n}])=({J}_{i}\cap {\mathcal{T}}_{ns}({J}_{i}[{X}_{n}])[{X}_{n}]\in {\mathcal{T}}_{ns}\text{.}$$
Thus $0\ne {J}_{i}\cap {\mathcal{T}}_{ns}({J}_{i}[{X}_{n}])\in {\mathcal{T}}_{ns}$, where ${J}_{i}\cap {\mathcal{T}}_{ns}({J}_{i}[{X}_{n}])\u22b4{B}_{i}$. But ${J}_{i}\cap {\mathcal{T}}_{ns}({J}_{i}[{X}_{n}])\subseteq {\mathcal{T}}_{ns}({B}_{i})=0$, which is contradiction. Thus ${B}_{i}[{X}_{n}]\in S{\mathcal{T}}_{ns}$ and consequently ${\mathcal{T}}_{ns}(A[{X}_{n}]\subseteq {H}_{i}[{X}_{n}]$, for any $i\in \Lambda $. If $0\ne {\mathcal{T}}_{ns}(A[{X}_{n}])$, then there exists $0\ne g({X}_{n})\in {\mathcal{T}}_{ns}(A[{X}_{n}])$. But all the coefficients a_{i} of of $g({X}_{n})$ are in ${\cap}_{i\in \Lambda}{H}_{i}=0$. Hence ${a}_{i}=0$ and so ${\mathcal{T}}_{n}(A[{X}_{n}])=0$, that is, $A[{X}_{n}]\in S{\mathcal{T}}_{ns}$.
We recall from [5] that a radical γ is called small if $\gamma \vee {\gamma}^{\prime}\ne {A}_{ss}$ for each proper radical ${\gamma}^{\prime}$, where ${A}_{ss}$ denotes the class of all associative rings. Dually, a nonzero radical γ is large if $\gamma \cap {\gamma}^{\prime}\ne 0$ for each proper radical ${\gamma}^{\prime}$.
Let ${\mathbb{L}}_{s}$ denote the collection of all strongly hereditary (that is, closed under subrings) and large radicals, and let $\mathbb{L}$ denote the collection of all radicals γ such that $\gamma \cap {\gamma}_{\alpha}\ne 0$ for every ${\gamma}_{\alpha}\in {\mathbb{L}}_{s}$. In [10], S. Tumurbat et al. proved that $\mathbb{L}$ is a complete sublattice in the lattice of all radicals. Now we have the following:
Proposition 2.20.
Each of the radicals ${\mathcal{T}}_{n}$ and ${\mathcal{T}}_{ns}$ belongs to $\mathbb{L}$
Proof. The radical ${\mathcal{T}}_{ns}$ contains all fields ${\mathbb{Z}}_{p}$ and all rings ${\mathbb{Z}}_{p}^{0}$ of prime order p (where ${\mathbb{Z}}_{p}^{0}$ denotes the ring with the additive group of ${\mathbb{Z}}_{p}$ and with multiplication defined by ab=0 for all $a,b\in {\mathbb{Z}}_{p}$). In [10], it was shown that every nonzero strongly hereditary radical contains a prime field or a simple zero ring of prime order. Hence ${\mathcal{T}}_{ns}\cap {\gamma}_{\alpha}\ne 0$ for every ${\gamma}_{\alpha}\in {\mathbb{L}}_{s}$. Thus ${\mathcal{T}}_{ns}\in \mathbb{L}$ and also ${\mathcal{T}}_{n}\in \mathbb{L}$.
Definition 2.21.
We call a radical γ an nbad radical if, for every prime ring $A\in \gamma $, there exists a hereditary radical ${\gamma}_{A}$ such that ${\gamma}_{A}\left(A\left[{X}_{n}\right]\right)\ne \left({\gamma}_{A}\left(A\left[{X}_{n}\right]\right)\cap A\right)\left[{X}_{n}\right]$.
In what follows, ${\mathbb{L}}_{b}$ and ${\mathbb{L}}_{sb}$ denote, respectively, the class of all nbad radicals and the class of all special radicals that are nbad radicals.
Proposition 2.22.
The classes ${\mathbb{L}}_{b}$ and ${\mathbb{L}}_{sb}$ satisfy the following conditions:

(i) If ${\gamma}_{i}\in {\mathbb{L}}_{b}$ for each $i\in \Lambda $, then $\underset{i\in \Lambda}{\cap}{\gamma}_{i}\in {\mathbb{L}}_{b}$;

(ii) If ${\gamma}_{i}\in {\mathbb{L}}_{sb}$ for each $i\in \Lambda $, then $\underset{i\in \Lambda}{\cap}{\gamma}_{i}\in {\mathbb{L}}_{sb}$.
Proof. (i) Let A be a prime ring in $\underset{i\in \Lambda}{\cap}{\gamma}_{i}$. For each $i\in \Lambda $, there exists a hereditary radical ${\left({\gamma}_{i}\right)}_{A}$ such that
$${\left({\gamma}_{i}\right)}_{A}\left(A\left[{X}_{n}\right]\right)\ne \left({\left({\gamma}_{i}\right)}_{A}\left(A\left[{X}_{n}\right]\right)\cap A\right)\left[{X}_{n}\right]\text{.}$$
Then we may choose any one of ${\left({\gamma}_{i}\right)}_{A}$, $i\in \Lambda $.
Statement (ii) may be proved in a similar way.
Theorem 2.23.
Both ${\mathbb{L}}_{b}$ and ${\mathbb{L}}_{sb}$ are complete sublattices in the lattice of all radicals.
Proof. We have ${\gamma}_{i}\subseteq {\mathcal{T}}_{n}$ for each nbad radical ${\gamma}_{i}$. Therefore $\mathcal{L}\left(\cup {\gamma}_{i}\right)\subseteq {\mathcal{T}}_{n}$ and $\mathcal{L}\left(\cup {\gamma}_{i}\right)\in {\mathbb{L}}_{b}$. By Proposition 2.22, $\cap {\gamma}_{i}\in {\mathbb{L}}_{b}$. Thus ${\mathbb{L}}_{b}$ is a complete sublattice in the lattice of all radicals. The proof that ${\mathbb{L}}_{sb}$ 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 $S\gamma $ such that γ does not have the Amitsur property. If ${\mathcal{T}}_{1s}$ does not have the Amitsur property, then we have a positive answer to this question.