The semigroup of endomorphisms of a Boolean ring

Abstract

The family (R) of all endomorphisms of a ring R is a semigroup under composition. It follows easily that if R and T are isomorphic rings, then (R) and (T) are isomorphic semigroups. We devote ourselves here to the converse question: ‘If (R) and (T) are isomorphic, must R and T be isomorphic?’ As one might expect, the answer is, in general, negative. For example, the ring of integers has precisely two endomorphisms – the zero endomorphism and the identity automorphism. Since the same is true of the ring of rational numbers, the two endomorphism semigroups are isomorphic while the rings themselves are certainly not.

Equimorphy in varieties of distributive double p-algebras

“…The least nontrivial subvariety of R, the variety B of Boolean algebras, is 2-determined, see [12], [13] or [19].…”

Endomorphisms of monadic Boolean algebras

“…All of the above is in sharp contrast to the situation for Boolean algebras, where, as independently shown by Magill [10], Maxson [11], and Schein [17], for arbitrary non-trivial Boolean algebras B 0 and B 1 , B 0 ∼ = B 1 whenever End(B 0 ) ∼ = End(B 1 ). To put this in perspective, observe that, as shown by Monk [13], the lattice of subvarieties of M is an ω + 1 chain M 0 ⊂ M 1 ⊂ M 2 ⊂ · · · ⊂ M, where M 0 is the trivial variety and M 1 corresponds to the variety of Boolean algebras.…”

Representability and local representability of algebraic theories

“…Deleting {0, 1} from Bool (but keeping the singleton algebra {0= 1}, the terminal object), we obtain a guileless category Bool 0 closed under the formation of finite products. This category is also monoid-determining -this follows from the earlier mentioned fact that Boolean spaces are monoid-determining, but see also [13,15,19]. Theorem 2.6 thus applies to Bool 0 and we only need to deal with the two-point Boolean algebra {0, 1}.…”