Semi-invariant subrings

Abstract

We say that a subring R0 of a ring R is semi-invariant if R0 is the ring of invariants in R under some set of ring endomorphisms of some ring containing R.We show that R0 is semi-invariant if and only if there is a ring S⊇R and a set X⊆S such that R0=CentR(X):={r∈R:xr=rx∀x∈X}; in particular, centralizers of subsets of R are semi-invariant subrings.We prove that a semi-invariant subring R0 of a semiprimary (resp. right perfect) ring R is again semiprimary (resp. right perfect) and satisfies Jac(R0)n⊆Jac(R) for some n∈N. This result holds for other families of semiperfect rings, but the semiperfect analogue fails in general. In order to overcome this, we specialize to Hausdorff linearly topologized rings and consider topologically semi-invariant subrings. This enables us to show that any topologically semi-invariant subring (e.g. a centralizer of a subset) of a semiperfect ring that can be endowed with a “good” topology (e.g. an inverse limit of semiprimary rings) is semiperfect.Among the applications: (1) The center of a semiprimary (resp. right perfect) ring is semiprimary (resp. right perfect). (2) If M is a finitely presented module over a “good” semiperfect ring (e.g. an inverse limit of semiprimary rings), then End(M) is semiperfect, hence M has a Krull–Schmidt decomposition. (This generalizes results of Bjork and Rowen; see Björk (1971) [5], Rowen (1986, 1987) [23,24].) (3) If ρ is a representation of a monoid or a ring over a module with a “good” semiperfect endomorphism ring (in the sense of (2)), then ρ has a Krull–Schmidt decomposition. (4) If S is a “good” commutative semiperfect ring and R is an S-algebra that is f.p. as an S-module, then R is semiperfect. (5) Let R⊆S be rings and let M be a right S-module. If End(MR) is semiprimary (resp. right perfect), then End(MS) is semiprimary (resp. right perfect).