Simple rings whose lower radicals are atoms
Andrunakievich and Ryabukhin have asked for a description of all simple rings S such that the lower radical L(S) is an atom in the lattice of radicals. In his paper [Acta Math. Acad. Sci. Hung. 36, 57-62 (1980; Zbl 0464.16005)] \textit{E. R. Puczyłowski} has proved that L(S) is an atom whenever S is a simple ring with identity. The author shows that if the simple ring S satisfies \((*)\quad S\quad ideal\) in R and \(R/S\cong S\Rightarrow \exists I\quad ideal\) in R such that \(R=S\oplus I\) then L(S) is an atom. Of course, every simple ring with identity satisfies (*). Non-unital examples of simple rings satisfying (*) have been obtained by \textit{W. G. Leavitt} [Period. Math. Hung. 12, 129-140 (1981; Zbl 0439.16006)] and by \textit{W. G. Leavitt} and the reviewer in their paper [Publ. Math. 29, 95- 99 (1982; Zbl 0505.16007)]. Examples of rings which fail to satisfy (*), are the zero rings Z(p)\(\circ\) on cyclic groups of prime order p. Also L(Z(p)\(\circ)\) is not an atom, since \(L(Z(p^{\infty}))\varsubsetneq L(Z(p)\circ).\) However, it is unknown whether (*) is necessary in order that L(S) be an atom. The author shows that there are \(''Z(p^{\infty})\)-like'' rings R having a simple ring S as an ideal such that \(L(R)\varsubsetneq L(S),\) i.e. L(S) is not an atom. In fact, L(S) is an atom implies the non-existence of a ring R with a chain \(0=I_ 0\triangleleft I_ 1\triangleleft I_ 2\triangleleft...\triangleleft I_ n\triangleleft I_{n+1}\triangleleft...\) such that \(I_{n+1}/I_ n\cong S\) for each n, \(R=\cup_{n\geq 0}I_ n\) and R has no ideals but the \(I_ n.\) An example of a ring S (besides Z(p)\(\circ)\), is the ring F(K) of linear transformations of finite rank of an \(\aleph_ 0\)-dimensional K-vector space, where K is a field. Leavitt and the reviewer have shown that F(K) satisfies the above conditions for S [cf. \textit{W. G. Leavitt} and \textit{L. C. A. van Leeuwen}, Ring Theory, Proc. Antwerp Conf. 1978, Lect. Notes Pure Appl. Math. 51, 783-798 (1979; Zbl 0427.16019)]. Thus L(F(K)) is not an atom. For non- associative rings, the author constructs a \(''Z(p^{\infty})\)-like'' ring for a simple ring. Hence he obtains: let S be a simple ring. Then L(S) is not an atom in the lattice of all radicals.
scientific article; zbMATH DE number 3851305
scientific article