Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 38 (1997), No. 1, 149-159.

Radicals Induced by the Total of Rings

Dedicated to Prof. Dr. F. Kasch on his 75^{th} birthday

K. I. Beidar R. Wiegandt^{∗}

Department of Mathematics, National Cheng Kung University Tainan 70101, Taiwan, ROC t14270@sparc1.cc.ncku.edu.tw

Mathematical Institute, Hungarian Academy of Sciences P.O.Box 127, H-1364 Budapest wiegandt@math-inst.hu

Abstract. The upper radicals K and K_{p} of the classes of rings with zero total and of prime rings with zero total, respectively, are introduced and discussed. K is a supernilpotent normal but nonspecial radical and K_{p} is a special normal radical. Crucial is the existence of the ring constructed in Theorem 3.1 which shows that K ≠ K_{p}. Also the position of K and K_{p} is localized in the lattice of all radicals. MSC 1991: 16N80 Keywords: supernilpotent, normal and special radicals, total

1. Introduction

The total Tot(A) of a ring A has been introduced by Kasch [8] and it reminds to a radical property, though the shortcoming is that Tot(A) need not be an ideal because it is not closed under addition. To overcome this difficulty, the following idea is plausible: the upper radical K of all rings A with Tot(A) = 0, may be a decent radical. We call K the Kasch radical, and show that K is a supernilpotent normal radical. Simultaneously we consider also the upper radical K_{p} of all prime rings A with Tot(A) = 0. K_{p} is a special normal radical containing the Kasch radical K. One of the benefits of radical theoretical investigations is to explore rings with interesting but unusual properties. This is featured in this note by constructing a biregular ring G such that Tot(G) = 0 and Tot(G/P) ≠ 0 for each prime ideal P of G. The existence of such a ring is crucial, it shows that K ≠ K_{p}. Thus, in a natural way we have got a nonspecial supernilpotent normal radical, namely K. We also position the radicals K and K_{p} in the lattice of all radicals. The Kasch radical is bigger than the Behrens and the Jacobson radicals, and is incomparable with the Brown–McCoy radical, the generalized nil-radical, and the strongly prime radical. Rings with zero total have been studied in the forthcoming paper [2].

∗Research partially supported by the Hungarian National Foundation for Scientific Research Grant #T16432 0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1997 Heldermann Verlag

2. The radicals K and K_{p}

All rings considered will be associative, not necessarily with unity element. We shall make use of the following notation: [x, y] = xy − yx. As usual, I ◁ A will denote the fact that I is an ideal in a ring A. An ideal I of a ring A is said to be essential (or large) in A, if I ∩ K ≠ 0 for every nonzero ideal K of A. This fact will be denoted by I ◁ ·A. A nonzero element a ∈ A is said to be partially invertible, if there exists an element b ∈ A such that ab is a nonzero idempotent. Following Kasch [8] the total Tot(A) of a ring A may be defined as the set

Tot(A) = {a ∈ A | aA does not contain a nonzero idempotent}
= {a ∈ A | a is not partially invertible}.

Although Tot(A) is regretfully not closed under addition, it enjoys interesting properties, for instance

(a) Tot(A) is left and right symmetric,

(b) Tot(A) is a semi-ideal: aTot(A)b ⊆ Tot(A) for every a, b ∈ A,

(c) J(A) + Tot(A) ⊆ Tot(A) where J(A) is the Jacobson radical of A. A class C of rings is said to be a weakly special class, if

(i) C consists of semiprime rings, (ii) C is hereditary: I ◁ A ∈ C implies I ∈ C, (iii) C is closed under essential extensions: I ◁ ·A and I ∈ C imply A ∈ C. As is well-known (see [11] and [5]) the upper radical

UC = {A | A has no homomorphic image in C}

of a weakly special class C is a supernilpotent radical, and every supernilpotent radical is the upper radical of a weakly special class.

A special class C of rings is a weakly special class consisting of prime rings. The upper radical UC of a special class C is called a special radical. As known from [1], for a special radical γ = UC a ring A has zero γ-radical, γ(A) = 0, exactly when A is a subdirect sum of rings from C.

For details on radical theory the reader is referred to the books [4] and [14].

Given a ring A with a subset S, we set

l_{A}(S) = {x ∈ A | xS = 0},
r_{A}(S) = {x ∈ A | Sx = 0}.

Proposition 2.1 The class M = {A | Tot(A) = 0} is a weakly special class, and hence the Kasch radical K = UM is a supernilpotent radical.

Proof. Since J(A) ⊆ Tot(A) for every ring, Tot(A) = 0 implies J(A) = 0. Thus a ring A with Tot(A) = 0 is semiprimitive, and so also is semiprime.

Next, we show that the class M is hereditary. Assume that I ◁ A and Tot(A) = 0. For any element a ∈ I, a ≠ 0, there exists an element b ∈ A such that ab = (ab)^{2} ≠ 0. Choosing b′ = bab, we get

(ab′)^{2} = (ab)^{4} = ab = (ab)^{2} = ab′

and so the element a is partially invertible also in I.

To prove that the class M is closed under essential extensions, let us consider I ◁ ·A such that Tot(I) = 0, and let us assume that there exists a nonzero element a ∈ Tot(A). Then for any nonzero element b ∈ I we have ab ∈ I, and hence by Tot(I) = 0 there exists an element b′ ∈ I such that abb′ = (abb′)^{2} ≠ 0, provided that ab ≠ 0. This is, however, impossible because 0 ≠ a ∈ Tot(A). Thus ab = 0 for all b ∈ I, that is, 0 ≠ a ∈ l_{A}(I). Since 0 ≠ l_{A}(I) ◁ A and I ◁ ·A, we conclude that I ∩ l_{A}(I) ≠ 0 which contradicts the semiprimeness of I ∈ M.

Lemma 2.2 Let R be a right ideal of a ring A. Suppose that P is a semiprime proper ideal of the ring R. Then:

(i) P^{4} ⊆ AP^{2}A ∩ R ⊆ P. (ii) If Q is an ideal of A containing AP^{2}A and maximal with respect to the property

Q ∩ R ⊆ P, then:

(a) Q is a semiprime ideal of the ring A,

(b) if \overline{A} = A/Q, \overline{R} = (R + Q)/Q, and \overline{P} = (P + Q)/Q, then r\overline{A}(\overline{R}) = 0, l\overline{R}(\overline{R}) = \overline{P} and \overline{R}/\overline{P} ≅ R/P.

Proof. Set I = AP^{2}A. Then

(I ∩ R)^{2} ⊆ IR = (AP)P (AR) ⊆ P.

Since P is a semiprime ideal of R, we conclude that I ∩ R ⊆ P. Obviously P^{4} ⊆ AP^{2}A and so P^{4} ⊆ I ∩ R which proves (i). Further, by Zorn’s Lemma there exists an ideal Q of A containing I and maximal with respect to the property Q ∩ R ⊆ P. If Q is not a semiprime ideal of A, then there exists an ideal J of A such that J ⊃ Q and J^{2} ⊆ Q. By the choice of Q we have that J ∩ R ⊈ P. Further,

(J ∩ R)^{2} ⊆ J^{2} ∩ R ⊆ Q ∩ R ⊆ P,

which contradicts the semiprimeness of the ideal P of the ring R. Therefore Q is a semiprime ideal of the ring A and so \overline{A} is a semiprime ring. Now one can easily show that l\overline{R}(\overline{R}) is a semiprime ideal of the ring \overline{R}. Since \overline{P}^{2} = 0, we conclude that \overline{P} ⊆ l\overline{R}(\overline{R}). Next, the relation Q ∩ R ⊆ P implies that \overline{R}/\overline{P} ≅ R/P and so \overline{P} is a semiprime ideal of the ring \overline{R}. As (l\overline{R}(\overline{R}))^{2} = 0, it follows that l\overline{R}(\overline{R}) ⊆ \overline{P} and hence l\overline{R}(\overline{R}) = \overline{P}. Finally, let K = r\overline{A}(\overline{R}). If K ≠ 0, we conclude from the choice of Q that K ∩ \overline{R} ≠ 0. Hence K ∩ \overline{R} is a nonzero nilpotent right ideal of the semiprime ring \overline{A}, a contradiction. Thus r\overline{A}(\overline{R}) = 0.

Lemma 2.3 Let A be a semiprime ring with right ideal R. Then:

(i A ∈ M implies that R/l_{R}(R) ∈ M. (ii) R/l_{R}(R) ∈ M and r_{A}(R) = 0 imply that A ∈ M.

Proof. (i) Let a ∈ R \ l_{R}(R). We claim that aR contains a nonzero idempotent. Indeed, since a ∉ l_{R}(R), ar ≠ 0 for some r ∈ R. As Tot(A) = 0, there exists b ∈ A such that arb = (arb)^{2} ≠ 0. Clearly rb ∈ R and so our claim is proved. It follows that R/l_{R}(R) ∈ M.

(ii) Suppose that A ∉ M. Then Tot(A) ≠ 0. Choose any nonzero element a ∈ Tot(A). If aAR = 0, then (RaA)^{2} = 0 which implies that RaA = 0 and so aA ⊆ r_{A}(R) = 0. Therefore aA = 0 which contradicts the semiprimeness of A. Hence axR ≠ 0 for some x ∈ A and so (axR)^{3} ≠ 0 because axR is a nonzero right ideal of the semiprime ring A. It follows that rax ∉ l_{R}(R) for some r ∈ R. As R/l_{R}(R) ∈ M, we conclude that there exists an element b ∈ R such that raxb + l_{R}(R) is a nonzero idempotent of the ring R/l_{R}(R). In particular

(raxb)^{3} − raxb ∈ l_{R}(R) and (raxb)^{2} ≠ 0.

It follows that (raxb)^{4} − (raxb)^{2} = [(raxb)^{3} − raxb](raxb) = 0. Setting s = xbraxb, we conclude that (ras)^{2} = ras and ras ≠ 0. We have that (asr)^{4} = (asr)^{2} and r(asr)^{2}as = (ras)^{3} = ras ≠ 0, a contradiction to a ∈ Tot(A). Hence Tot(A) = 0 and A ∈ M.

A radical α is said to be right strong if for every right ideal R of any ring A the relation α(R) = R implies R ⊆ α(A). Next, a radical α is called right hereditary if for every right ideal R of any ring A the relation α(A) = A implies that α(R) = R. The concepts of a left strong radical and a left hereditary radical are defined analogously (see [13]). Normal radicals were introduced in [6] in relation to Morita contexts. For us it is enough to know that the right strong and right hereditary supernilpotent radicals are precisely the supernilpotent normal radicals (see the remark after Theorem 4 in [12]).

Theorem 2.4 The Kasch radical K is a right and left strong, right and left hereditary radical, hence a normal radical.

Proof. Let R be a right ideal of a ring A such that K(R) = R. We show that R ⊆ K(A). Suppose that R ⊈ K(A). Then there exists an ideal I of A such that A/I ∈ M and R ⊈ I. Replacing A and R by A/I and (R+I)/I respectively, we may assume that A ∈ M and R is a nonzero K-radical right ideal of A. Taking into account that R/l_{R}(R) ∈ M by Lemma 2.3(i), we conclude that R = l_{R}(R) and so R^{2} = 0. Since A is semiprime, we infer that R = 0, a contradiction. Hence K is a right strong radical.

Next, let R be a right ideal of a K-radical ring A. We show that R is a K-radical ring as well. Suppose that K(R) ≠ R. Then there exists an ideal P of the ring R such that 0 ≠ R/P ∈ M. Note that P is a semiprime ideal of R. Let Q, \overline{A}, \overline{R}, and \overline{P} be as in Lemma 2.2. Then \overline{R}/l\overline{R}(\overline{R}) ≅ R/P ∈ M and r\overline{A}(\overline{R}) = 0. It follows from Lemma 2.3(ii) that \overline{A} ∈ M contradicting \overline{A} ∈ K. Thus K is a right hereditary radical.

By symmetry K is a left strong left hereditary radical, and by the remark done before Theorem 2.4, K is a normal radical.

Proposition 2.5 The class

M_{p} = {A | Tot(A) = 0 and A is a prime ring}

is a special class. Hence K_{p} = UM_{p} is a special radical and K ⊆ K_{p}.

Proof. The class P of all prime rings is a special class, as is well-known. Thus so is the class M_{p} = M ∩ P. Since M_{p} ⊆ M, it follows K ⊆ K_{p}.

An analogous proof to that of Theorem 2.4 yields

Proposition 2.6 K_{p} is a right and left strong, right and left hereditary radical, and also a normal radical.

3. An example showing K ≠ K_{p}

Let R be a ring with a family of subrings {R_{i} | i ∈ I} such that (i) R = ∪_{i∈I}R_{i}; (ii) Each R_{i} is a simple Ore domain; (iii) For any i, j ∈ I there exists k ∈ I such that R_{i} ∪ R_{j} ⊆ R_{k}. Then one can easily show that

R is a simple Ore domain.  (1)

Recall that a ring A is called biregular, if each principal ideal of A is generated by a central idempotent. Further, a ring A is called reduced if A has no nonzero nilpotent elements.

Theorem 3.1 There exists a biregular reduced ring G such that Tot(G) = 0 but Tot(G/P) ≠ 0 for all prime ideals P of G. Moreover, G/P is a simple Ore domain which is not a division ring.

Proof. Let Z_{2} be a two element field. We set U = ∏_{i=1}^{∞} U_{i} and V = ⊕_{i=1}^{∞}U_{i} where U_{i} = Z_{2} for all i = 1, 2, . . .. Next we denote by B the factor ring U/V. Given u, v ∈ B, we write u < v if u = uv ≠ v. Clearly < is a partial order on B, and B has no nonzero minimal elements (i.e., for each 0 ≠ v ∈ B there exists 0 ≠ u ∈ B with u < v). Given a maximal ideal M of B, the subset τ = B \ M is called an ultrafilter. We denote by T the set of all ultrafilters of B. For e ∈ B, we set

T (e) = {τ ∈ T | e ∈ τ}.

Note that T (e) = ∅ if and only if e = 0. It is known that T is a compact Hausdorff space with base of closed-open subsets {T (e) | e ∈ B}. Furthermore, every closed-open subset of T is of the form T (e), e ∈ B. Clearly T (u) ∩ T (v) = T (uv) for all u, v ∈ B. Since B has no nonzero minimal idempotents, T has no isolated points (i.e., points which are open subsets).

A finite subset e_{1}, e_{2}, …, e_{n} ∈ B is said to be a partition of 1 if e_{1} + e_{2} + … + e_{n} = 1 and e_{i}e_{j} = 0 for all i ≠ j.

Let Q be the rational number field. Given any division Q-algebra D, the Q-algebra D{x_{i}, y_{i} | i = 1, 2, …, n} generated by D and x_{i}, y_{i}, i = 1, 2, …, n, with relations

[x_{i}, x_{j}] = [y_{i}, y_{j}] = 0, [x_{i}, y_{j}] = δ_{ij}

is a simple Noetherian domain by [9, Theorem 3.8].

We set R = Q{x_{τ}, y_{τ} | τ ∈ T }. Clearly R is a simple Ore domain according to the above result and (1). Note that R is a Weyl algebra in {x_{τ}, y_{τ} | τ ∈ T }. Given a subset S ⊆ T, we denote by R_{S} the subalgebra of R generated by {x_{τ}, y_{τ} | τ ∈ T \ S}. Obviously R_{S} = Q{x_{τ}, y_{τ} | τ ∈ T \ S}. Let Δ be the classical division ring of quotients of R. Clearly Δ ⊇ Δ_{S}, the classical division ring of quotients of R_{S}. We denote by A_{S} the subalgebra of Δ generated by Δ_{S} and {x_{τ}, y_{τ} | τ ∈ S}. Clearly A_{S} = Δ_{S}{x_{τ}, y_{τ} | τ ∈ S} and so A_{S} is a simple Ore domain. Note that A_{S} is a division ring if and only if S = ∅. Further, if U, V ⊆ T, then the subalgebra of Δ generated by A_{U} and A_{V} is contained in A_{U∩V}. In particular A_{U} ⊆ A_{V} if and only if V ⊆ U.

Consider Δ as a discrete topological space. Now let H be the Q-algebra of all continuous functions f : T → Δ. Given f ∈ H, f(T) is a compact subspace of the discrete topological space Δ because T is a compact space and f is continuous. Therefore |f(T)| < ∞ and so there exists a partition e_{1}, e_{2}, …, e_{n} of 1 such that f|_{T (e_{i})} = const. The function f is said to be special if f(T (e_{i})) ⊆ A_{T (e_{i})} for all i = 1, 2, …, n. We denote by G the subset of all special functions and claim that G is a subalgebra. Indeed, let g, h ∈ G. Then there exist partitions u_{1}, u_{2}, …, u_{m} and v_{1}, v_{2}, …, v_{n} such that g(T (u_{i})) ⊆ A_{T (u_{i})} and g(T (v_{j})) ⊆ A_{T (v_{j})} for all i, j. Clearly (g +h)|_{T (u_{i})∩T (v_{j})} = const and (g ·h)|_{T (u_{i})∩T (v_{j})} = const for all i, j such that u_{i}v_{j} ≠ 0. Setting K = {(i, j) | u_{i}v_{j} ≠ 0}, we note that {u_{i}v_{j} | (i, j) ∈ K} is a partition. Since T (u_{i}) ∩ T (v_{j}) = T (u_{i}v_{j}), the subalgebra of Δ generated by A_{T (u_{i})} and A_{T (v_{j})} is contained in A_{T (u_{i}v_{j})} for all (i, j) ∈ K. Therefore (g+h)(T (u_{i}v_{j})) ⊆ A_{T (u_{i}v_{j})} and (g·h)(T (u_{i}v_{j})) ⊆ A_{T (u_{i}v_{j})} which proves our claim. Given v ∈ B, we denote by \overline{v} the function given by the rule

\overline{v}(τ) = \{ \begin{matrix}1 & if & v ∈ τ \\ 0 & if & v ∉ τ. \end{matrix}

Clearly \overline{v}, v ∈ B, are the only central idempotents of G. Furthermore, if e_{1}, e_{2}, …, e_{n} is a partition of 1 in B, then \overline{e_{1}}, \overline{e_{2}}, …, \overline{e_{n}} is a partition of 1 in G.

We claim that Tot(G) = 0. Indeed, let 0 ≠ f ∈ G. Then a = f(τ) ≠ 0 for some τ ∈ T. Clearly f^{−1}(a) is a closed-open subset of T and so f^{−1}(a) = T (e) for some e ∈ B^{∗}. Since a ∈ Δ involves only a finite number of the x_{σ} and y_{σ}, there exists a finite subset S ⊆ T such that a ∈ Δ_{T \S} ⊆ A_{T \S}. Recalling that T has no isolated points, we conclude that T (e) ⊈ S. Let σ ∈ T (e) \ S. Choose v ∈ B^{∗} such that σ ∈ T (v) ⊆ T (e) (i.e., v ∈ σ and v ≤ e) and T (v) ∩ S = ∅. Therefore a ∈ Δ_{T \S} ⊆ A_{T (e)}, and it follows that a^{−1} ∈ A_{T (e)}. Define the function g : T → Δ by the rule

g(τ) = \{ \begin{matrix}a^{−1} & if & τ ∈ T (e) \\ 0 & if & τ ∉ T (e). \end{matrix}

Then g ∈ G and fg = \overline{e} ∈ fG. Hence f ∉ Tot(G) and thus Tot(G) = 0.

Let P be any prime ideal of G. We set

M_{P} = \{v ∈ B | \overline{v} ∈ P \}

and claim that M_{P} is a maximal ideal of B and G/P ≅ A_{τ_{P}} where τ_{P} = B \ M_{P}. Indeed, the first statement follows from the fact that \{\overline{v} | v ∈ B\} is the Boolean ring of central idempotents of G which is isomorphic to B. Consider now the map α : G → Δ given by the rule α(f) = f(τ_{P}), f ∈ G. First we show that α(G) ⊆ A_{τ_{P}}. To this end, let f ∈ G and e_{1}, e_{2}, …, e_{n} be a partition of 1 such that f|_{T (e_{i})} ⊆ A_{T (e_{i})}. Clearly e_{j} ∈ τ_{P} for some j. Therefore,

α(f) = f(τ_{P}) ∈ A_{T (e_{j})} ⊆ A_{τ_{P}}

because τ_{P} ∈ T (e_{i}) and so α(G) ⊆ A_{τ_{P}}. Next we show that α(G) = A_{τ_{P}}. Let g ∈ A_{τ_{P}}. Since g involves only a finite number of x_{τ} and y_{τ}, there exists a finite subset S ⊆ T such that τ_{P} ∉ S and g ∈ A_{T \S} ⊆ A_{τ_{P}}. Since T is a compact Hausdorff space, there exists e ∈ B such that τ_{P} ∈ T (e) and T (e) ∩ S = ∅. Define the function \overline{g} : T → Δ by the rule

\overline{g}(τ) = \{ \begin{matrix}g & if & τ ∈ T (e) \\ 0 & if & τ ∉ T (e). \end{matrix}

Since S ⊆ T \ T (e), T (e) ⊆ T \ S and so A_{T \S} ⊆ A_{T (e)}. Hence \overline{g} ∈ G. Clearly \overline{g}(τ_{P}) = g which proves our statement. Finally we show that ker(α) = P. Indeed, let f ∈ ker(α). Then f(τ_{P}) = 0 and so τ_{P} ∈ f^{−1}(0). Being a closed-open subset, f^{−1}(0) = T (v) for some v ∈ B. Since τ_{P} ∈ f^{−1}(0), we conclude that v ∈ τ_{P} and so v ∉ M_{P}. It follows that \overline{v} ∉ P. We already know that \overline{v} is a central idempotent of G. As \overline{v}f(τ) = 0 for all τ ∈ T, we see that \overline{v}f = 0 and so \overline{v}(GfG) = 0. Recalling that P is prime, we conclude that f ∈ P. Hence ker(α) ⊆ P. On the other hand let f ∈ P. Suppose that f ∉ ker(α). Then a = f(τ_{P}) ≠ 0. We already know that f^{−1}(a) = T (e) for some e ∈ B and a ∈ A_{T (e)}. Note that

e ∈ τ_{P} .  (2)

Since A_{T (e)} is a simple ring with 1, there exist b_{1}, c_{1}, …, b_{m}, c_{m} ∈ A_{T (e)} such that ∑_{i=1}^{m} b_{i}ac_{i} = 1. Define functions g_{i}, h_{i} : T → Δ by the rules

g_{i}(τ) = \{ \begin{matrix}b_{i} & if & e ∈ τ \\ 0 & if & e ∉ τ. \end{matrix}
h_{i}(τ) = \{ \begin{matrix}c_{i} & if & e ∈ τ \\ 0 & if & e ∉ τ. \end{matrix}

Clearly g_{i}, h_{i} ∈ G, i = 1, 2, …, m and ∑_{i=1}^{m} g_{i}fh_{i} = \overline{e} ∈ P. Hence e ∈ M_{P} and so e ∉ τ_{P} contradicting (2). Hence it follows that ker(α) = P. Since G/P ≅ A_{τ_{P}} is a simple Ore domain which is not a division ring, we conclude that Tot(G/P) ≠ 0 for all prime ideals P of G. Thus G is a ring without nonzero nilpotent ideals all of whose prime ideals are maximal. Hence it follows from [3, Corollary 4.4] that G is a biregular ring. The proof is complete.

Corollary 3.2 K ≠ K_{p} and K is a supernilpotent nonspecial radical.

Proof. The ring G in Theorem 3.1 has no nonzero homomorphic image in M_{p}, whence G ∈ K_{p}. Since Tot(G) = 0 and the class M is hereditary by Proposition 2.1, it follows that K(G) = 0. This proves K ≠ K_{p}. According to Theorem 3.1, G is a semiprime ring. Thus G is a subdirect sum of the prime rings G/P where P ranges through all prime ideals of G. By Theorem 3.1, G/P is a simple prime ring, so either K(G/P) = 0 or K(G/P) = G/P. If K(G/P) = 0, then by K = UM the ring G/P has a nonzero homomorphic image in M. Hence the simplicity of G/P implies G/P ∈ M, that is, Tot(G/P) = 0, contradicting Theorem 3.1. Thus K(G/P) = G/P holds for every prime ideal P of G. This means that G cannot be represented as a subdirect sum of prime rings with zero K-radical, and so K is a nonspecial radical.

The first example for a supernilpotent nonspecial radical was given by Rjabuhin [11]. An example for a supernilpotent normal nonspecial radical was constructed by Jaegermann and Sands (see p.345 in [7]).

4. The position of the radicals K and K_{p}

In this section we shall make clear the relation of the radicals K and K_{p} to other well-known radicals.

Theorem 4.1 The radical K_{p} is properly contained in the upper radical T of all matrix rings over division rings and in the upper radical Uν of the class ν of all von Neumann regular rings. The radicals K and K_{p} are incomparable with the Brown–McCoy radical G, with the generalized nil-radical N_{g}, with the Groenewald-Heyman strongly prime radical s. The Kasch radical K contains properly the Behrens radical B and the Jacobson radical J.

Proof. As is well-known, the class ν of all von Neumann regular rings is a hereditary radical class, so we may talk about the upper radical class Uν. If A is a von Neumann regular ring, then the elements of A are obviously partially invertible, whence Tot(A) = 0. Hence ν ⊆ M follows implying

K_{p} = UM_{p} ⊆ U(ν ∩ P)

where P stands for the class of all prime rings. Clearly Uν ⊆ U(ν ∩ P). Let A ∈ ν. Being a von Neumann regular ring, A is semiprime and so is a subdirect sum of prime rings. As ν is homomorphically closed, each of these prime rings is von Neumann regular and hence A is U(ν ∩ P)-semisimple. Therefore Uν ⊇ U(ν ∩ P) and so Uν = U(ν ∩ P). Thus it follows that K_{p} ⊆ Uν.

Let V be a countably infinite dimensional vector space with a basis {x_{i} | i=1, 2 …, n, …}, T the ring of all linear transformations of V. Let us consider the subring S of T generated by the subring H of all linear transformations having finite rank and by the linear transformation t defined by tx_{i} = x_{i+1}, i = 1, 2, …, n, …. It is well-known that H is a simple von Neumann regular ring (without unity element) and H is the only proper ideal of T. Clearly H is a minimal proper ideal of S and thus S is a subdirectly irreducible prime ring. Moreover, the factor ring S/H and also S are not von Neumann regular. Thus ν(S) = H and S ∈ Uν are valid. For any nonzero element a ∈ S there exists obviously an element b ∈ H such that 0 ≠ ab ∈ H. Since H is von Neumann regular, there exists an element x ∈ H such that ab = (ab)x(ab), and so 0 ≠ abx = (abx)^{2}. Thus every nonzero element a ∈ S is partially invertible, that is, Tot(S) = 0. Since S is a prime ring, also S ∈ M_{p} holds. Hence we have established that S ∈ Uν \ K_{p}, proving K_{p} ≠ Uν.

The containment U ⊆ T is obvious, and the ring H is in T but not in Uν.

In [10] it has been shown that the ring H is in the strongly prime radical class which is properly contained in the class G of all Brown–McCoy radical rings. Hence H ∈ s \ Uν is valid implying s ⊈ Uν and G ⊈ Uν.

In [4, Example 8] a ring A is given such that A is a simple ring without zero divisors, has a unity element e, and e is its only nonzero idempotent, but A is not a division ring. Thus A ∉ G. Assume that A ∉ K. Since A is simple, necessarily Tot(A) = 0, that is, for every nonzero element a ∈ A there exists an element x ∈ A such that ax = e. Hence A is a division ring, contradicting the properties of A. Thus A ∈ K holds proving that A ∈ K \ G, that is, K ⊈ G.

Since s ⊈ Uν, s ⊆ G, K ⊆ K_{p} ⊆ Uν and K ⊈ G, we conclude that the radicals s and G are not comparable with K, K_{p}, and Uν.

Since the ring A is simple and has no zero divisors, it follows that A is not in the upper radical determined by the class of all rings without zero divisors, that is, not in the generalized nil-radical N_{g}. Hence A ∈ K\N_{g} which yields K ⊈ N_{g}. It is well-known (cf. [10]) that s ⊂ N_{g}, so by s ⊈ Uν we get that N_{g} ⊈ Uν. Thus also N_{g} is not comparable with K, K_{p}, and Uν.

Consider now the Behrens radical B, that is, the upper radical determined by the class of all subdirectly irreducible rings whose hearts contain nonzero idempotents. If B ⊈ K, then there exists a ring D which is in B but not in K. Then D has a nonzero homomorphic image E in M. Clearly E ∈ B and Tot(E) = 0. In particular E contains a nonzero idempotent e. Let I be a ideal of E maximal with respect to the property e ∉ I. Then E/I is a subdirectly irreducible ring whose heart contains a nonzero idempotent (namely, e + I), contradicting E ∈ B. Thus B ⊆ K. Since B ⊆ G necessarily B ≠ K.

The place of the Kasch radical in the lattice of radicals is summarized in the following diagram where L and F stand for the Levitzki locally nilpotent radical and for the upper radical determined by the class of all division rings, respectively. The containments not explained in the proof of Theorem 4.1 are well-known.

[figure: F; T; G; B; J; L; N_{g}; s; Uν; K; K_{p}]

References

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[8] Kasch, F.: Partiell invertierbare Homomorphismen und das Total. Algebra Berichte 60, Verlag Reinhard Fischer, München 1988.

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[10] Parmenter, M. M.; Stewart, P. N.; Wiegandt, R.: On the Groenewald-Heyman strongly prime radical. Quaest. Math. 7 (1984), 225–240.

K. I. Beidar, R. Wiegandt: Radicals Induced by the Total of Rings 159

[11] Rjabuhin, Ju. M.: On overnilpotent and special radicals. Issled. Alg. Mat. Anal., Kishinev 1965, 65–72, in Russian.

[12] Sands, A. D.: On normal radicals. J. London Math. Soc. (2), 11 (1975), 361–365.

[13] Sands, A. D.: On relation among radical properties. Glasgow Math. J. 18 (1977), 17–23.

[14] Szász, F. A.: Radicals of Rings. John Wiley & Sons, 1981.

Received February 27, 1996