When multiplication of topologizing filters is commutative. (English) Zbl 0949.16029
Let \(R\) be a ring with identity, and let \(\text{Fil }R\) denote the set of all right topologizing filters on \(R\). \(\text{Fil }R\) is isomorphic as a complete lattice and a semigroup to \(\text{torsp }R\), the set of torsion preradicals of \(R\). If \(R=K[x;\alpha]\) for a field \(K\) and an endomorphism \(\alpha\) of \(K\), then \(\text{Fil }R\) is commutative only if \(\alpha\) is an automorphism. If \(R\) is a right FBN ring such that \(IK=KI\) for every pair of ideals \(I,K\) of \(R\), then every finitely generated uniform right \(R\)-module is decisive in the sense of J. S. Golan [Torsion theories, Pitman Monogr. Surv. Pure Appl. Math. 29 (1986; Zbl 0657.16017)] and \(R\) satisfies the Jacobson Conjecture. If \(\text{Fil }R\) is commutative, then \(R\) satisfies the ACC on annihilator ideals, the prime radical of \(R\) is nilpotent, and \(I_R\) is finitely annihilated whenever \(I\) is an ideal of \(R\). A (semi) prime ring for which \(\text{Fil }R\) is commutative is right strongly (semi) prime. If \(R\) is a commutative domain and \(\text{Fil }R\) is commutative, then \(R\) satisfies the ACC on principal ideals. If \(R\) is commutative and \(\text{Fil }R\) is commutative, then \(R\) is semistable in the sense of Golan [op. cit.], \(\text{Ass }M\neq\emptyset\) for all nonzero modules \(M\), and every indecomposable injective module has the form \(E(R/P)\) for some prime ideal \(P\).
Reviewer: M.L.Teply (Milwaukee)
MSC:
16S90 | Torsion theories; radicals on module categories (associative algebraic aspects) |
16N60 | Prime and semiprime associative rings |
16N20 | Jacobson radical, quasimultiplication |
16P40 | Noetherian rings and modules (associative rings and algebras) |
16P60 | Chain conditions on annihilators and summands: Goldie-type conditions |
13C11 | Injective and flat modules and ideals in commutative rings |
13E99 | Chain conditions, finiteness conditions in commutative ring theory |