When multiplication of topologizing filters is commutative (Q1300655)

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 \textit{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 \(P\).