(⇒) clear.

(⇐) Suppose that the subrings S, S₁ of A fulfills S ⊈ S₁ and S₁ ⊈ S. Then there exist elements a, b of A such that a ∉ S, a ∈ S₁ and b ∉ S₁, b ∈ S. By the assumption, we have either 〈a〉 ⊆ 〈b〉 or 〈b〉 ⊆ 〈a〉. If 〈a〉 ⊆ 〈b〉 then 〈a〉 ⊆ 〈b〉 ⊆ S. Hence a ∈ S. This is a contradiction. Therefore, 〈b〉 ⊆ 〈a〉 ⊆ S₁. Thus b ∈ S₁, again a contradiction. Hence we have either S ⊆ S₁ or S₁ ⊆ S. It shows that A is a ring with the chain property.

We consider elements a, b ∈ A. Then either 〈a〉 ⊆ 〈b〉 or 〈b〉 ⊆ 〈a〉 by Proposition 2.2. Suppose that 〈a〉 ⊆ 〈b〉. Then 〈b〉 is a commutative ring we have [a, b] = 0.

Let CH be the class of rings defined by

A class ℳ of rings said to be matrix-extensible if A ∈ ℳ if and only if the matrix ring M_n(A) ∈ ℳ for any natural number n.

It is easy to see that ℤ_p ∈ CH, where p is a prime number. If M_n(ℤ_p) ∈ CH, where n ≥ 2, then by Corollary 2.3, M_n(ℤ_p) is a commutative ring. But M_n(ℤ_p) is not commutative. Thus M_n(ℤ_p) ∉ CH.

We recall that a class ℳ of rings said to be strongly hereditary if it satisfies: If A is a ring in ℳ, then every subring S of A is in ℳ.

We shall show that CH is strongly hereditary. Let A ∈ CH and S is a subring of A. Since A has the chain property, for any a, b ∈ S ⊆ A, we have either 〈a〉 ⊆ 〈b〉 or 〈b〉 ⊆ 〈a〉. Thus, by Proposition 2.2, S has the chain property. This shows that CH is a strongly hereditary. In particular, CH is hereditary class of rings. Now we claim that CH is closed class under homomorphic images. Let Ā = A/H be a homomorphic image of A ∈ CH. We consider any subrings S̄, S1¯ of Ā. Then there exist subrings S, S₁ of A such that S¯=SH,S1¯=S1H, where H ⊆ S ∩ S₁. Since A is in CH, we have either S ⊆ S₁ or S₁ ⊆ S. If S ⊆ S₁, then S¯=SH⊆S1H=S1¯. Therefore S¯⊆S1¯. The other case gives S1¯⊆S¯.

Let A = ∪I_α be a ring, where

with each I_α ⊴ A and I_α ∈ CH. We consider any elements a, b ∈ A. Then there exists I_α, such that a, b ∈ I_α. Since I_α ∈ CH, we have either 〈a〉 ⊆ 〈b〉 or 〈b〉 ⊆ 〈a〉. Therefore, by Proposition 2.2, A ∈ CH.

We shall show that ℒ(CH) is strongly hereditary and by Proposition 1.1, ℒ(CH) is strongly hereditary in the special case σ =“subring of”. Now we claim that ℒ(CH) is a large radical in the lattice of all strongly hereditary radicals.

First of all, we will be see that every non zero strongly hereditary radical γ contains a prime field ℤ_p or a simple zero ring ℤp0 with prime order. Let us consider a ring A ∈ γ and a nonzero element a ∈ A. Since γ is strongly hereditary, the subring 〈a〉 ∈ γ. Using Zorn’s lemma, there exists an ideal I of 〈a〉 which is maximal respect to a ∉ I. Then the factor ring 〈a〉¯=〈a〉/I is a simple ring and 〈a〉¯∈γ.

If 〈a〉¯2=0¯, then by the simplicity of 〈a〉¯, 〈a〉¯ is a zero ring of prime order. If 〈a〉¯2≠0¯ then by the commutativity of 〈a〉¯, 〈a〉¯ is a field. Thus the subring of 〈a〉¯ generated by the unit element of 〈a〉¯ is isomorphic to the ring ℤ of integers or to the prime field ℤ_p of p elements. By the strong hereditariness of γ the relation 〈a〉¯∈γ implies ℤ ∈ γ or ℤ_p ∈ γ holds and in both cases ℤ_p ∈ γ. Thus every strongly hereditary radical γ contains either a finite prime field or a simple zero-ring with prime order. But it is clear that CH contains all finite prime fields and all simple zero-rings with prime order. Thus ℒ(CH) ∩ γ ≠ 0, for every strongly hereditary radical 0 ≠ γ. Hence ℒ(CH) is a large radical in the lattice of all strongly hereditary radicals.

From the above, every atom γ₀ in the lattice of all strongly hereditary radicals is generated by either a finite prime field or a simple zero-ring of prime order. Thus γ₀ ⊆ ℒ(CH).

We denote by the collection of all strongly hereditary and large radicals.

We consider radicals γ₁, …, γ_α … such that . Since and each γ_s is large in the lattice of all strongly hereditary radicals, each γ_α contains all simple zero-rings with prime order and all prime fields. By Proposition 1.1, ℒ(∪γ_α) is strongly hereditary. It is clear that ∩γ_α is strongly hereditary. Hence ℒ(∪γ_α) and ∩γ_α contain all simple zero-rings with prime order and all prime fields. Therefore ℒ(∪γ_α) and ∩γ_α are large radicals in the class of all strongly hereditary radicals. We denote by γ₀ the lower radical generated by all simple zero-rings with prime order and all prime fields. Then it is clear that γ₀ is an atom in .

Let X = {x₁, …, x_λ, …} be an infinite set of symbols. Then by Proposition 2.8 in [2], the lower radical ℒ(F[X]) determined by the free ring F[X] is strongly hereditary. It is also σ-hereditary and small in the lattice of all radicals. Moreover, ℒ(F[X]) is large in the lattice of all strongly hereditary radicals. Suppose that γ⁰ is a coatom in . Then there exists a free ring F[X] such that F[X] ∉ γ⁰. Since ℒ(F[X]) is small in the lattice of all radicals, we have

Thus, is not coatomic.

We denote by , the collection of all radicals γ such that γ ∩ γ_α ≠ 0 for every .

Let A be a simple zero-ring with prime order or a prime field. Then ℒ(A) is strongly hereditary and an atom in the lattice of all hereditary radicals. Let us consider . Then ℒ(A) ∩ γ_α ≠ 0 and also ℒ(A) ⊆ γ_α. Hence ∩γ_α contains all simple zero-rings with prime order and all prime fields. Thus (∩γ_α) ∩ γ_β ≠ 0 and also ℒ(∪γ_α) ∩ γ_β ≠ 0 for every .

This can be proved in a similar way as the proof of Proposition 2.8.

We recall from [4] the definition of an (hereditary) Amitsur ring and the definition of the radicals and . A ring A is said to be an (hereditary) Amitsur ring if γ(A[x]) = (γ(A[x]) ∩ A)[x], for all (hereditary) radicals γ, respectively. Let us recall and as follows:

and

We shall show that a ∈ aA for every element a ∈ A. By Corollary 2.3, we have aA ⊴ A. Let a ∉ aA, for an element a ∈ A. Then Ā = A/aA ≠ 0̄. It is clear that (a + aA)² ⊆ aA. Therefore Ā has a nonzero nilpotent element. By condition (h), A has a nonzero nilpotent element, which is a contradiction. Thus a ∈ Aa for every a ∈ A. There exists an element e such that ae = ea = a. It is clear that a ∈ a²A. Thus there exists x ∈ A such that ax = e. Hence A is a field. Suppose that charA = 0 and let e be the unit element of A. Then there exists a subring S of A which is isomorphic to ℤ. Therefore S does not have the chain property which is a contradiction.

By Proposition 2.2, A has a nonzero nilpotent element. Put

It is clear that , where is the nil radical. Moreover,

and S has a nonzero nilpotent element. Therefore, since A is commutative ring , which is a contradiction. Thus .

First of all, we claim that 0 ≠ β(A) for any ring A ∈ ch which has a nonzero zero-divisor. Note that by Lemma 2.13, A is a nil ring. Let us consider the case β(A) ≠ A. Then there exists an element a ∈ A such that aⁿ = 0 and aA ≠ 0. If aA = A, then 0 ≠ A = aA = a²A = … = aⁿA = 0. This is impossible. Hence aA ⊊ A. Therefore there exists a non-zero element b ∈ A and b ∉ aA with b^m = 0 for some natural m. It is clear that aA ⊊ 〈b〉. Thus aA is a nilpotent ideal of A. Therefore 0 ≠ β(A), for any ring A ∈ ch which has a nonzero zero-divisor. Since β(A) ≠ A there exists c ∈ A such that β(A) ⊊ ℤc+cA⊴A. It is clear that ℤc+cA is a nilpotent ideal of A. Thus β(A/β(A)) ≠ 0. It is a contradiction.

It follows from Lemma 2.12, Proposition 2.14.

A radical γ has the Amitsur property if

Clear.

For a radical γ, let γ_x = {A | A[x] ∈ γ}.

By Corollary 2.15, ch = C ∪ D and C ∩ D = ∅︀, where C is the class of Baer radical rings with condition (h) and D is the class of fields with the chain property. By Proposition 2.17 ℒ(C) is prime-like and it is not hard to check that ℒ(C)(A[x]) = 0, for all prime rings A. Hence ℒ(C ∪ D) = ℒ(ℒ(C) ∪ ℒ(D)). Thus ℒ(ch) is prime-like and also γ is prime-like. Hence by Proposition 2.18 we have β = γ_x and γ has the Amitsur property.

We put ℱ = {all fields}. Let denote the upper radical class generated by a class ℳ of rings.

Let be a nonzero semiprime ring. Then A has a nonzero accessible subring B ∈ D, where D is the class of fields with chain property. Since B is a field, we have B² = B⊴A and also B is direct sumand of A. Then there exists a ring B′ such that A = B⊕B′. Therefore we have . Since B is a field we have

which is a contradiction. Hence . Therefore, Theorem 2.16 implies that A is a hereditary Amitsur ring.