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Oct 23 at 11:02 comment added YCor Still another one: inside the Boolean algebra of finite Boolean combinations of intervals of the real line, pick the bounded ones (this is an ideal) and mod out by the smaller ideal of finite subsets.
Oct 23 at 9:51 comment added Emil Jeřábek Boolean algebras where all elements $\ne0,1$ are in the same orbit are called homogeneous in forcing literature, which is replete with such examples. In fact, set theorists strongly prefer homogeneous forcings, because then the validity of statements in the generic extension $V[G]$ does not depend on the choice of the generic filter; if the forcing is not homogeneous, you’d better decompose it to pieces that are.
Oct 23 at 6:56 answer added John Griesmer timeline score: 1
Oct 21 at 13:07 comment added YCor Actually this is almost the same as asking about unital BAs $A$ in which the automorphism group has exactly 3 orbits: $\{0\}$, $\{1\}$ and the remainder. Which in Stone duality means Stone space in which all proper nonempty clopen subsets are equivalent modulo homeomorphism. Given such $A$, the kernel of every homomorphism to $\mathbf{Z}/2\mathbf{Z}$ provides an example of a non-unital BA with two orbits under automorphisms ($\{0\}$ and the remainder).
Oct 21 at 7:20 history became hot network question
Oct 21 at 5:12 answer added David Gao timeline score: 5
Oct 21 at 3:48 answer added YCor timeline score: 7
Oct 20 at 23:35 comment added Noah Schweber Let $\mathcal{R}$ be the subring of $\mathcal{P}(\omega)/\mathit{Fin}$ consisting of (classes containing) sets of asymptotic density zero. Does this work?
Oct 20 at 23:20 history asked Greg Oman CC BY-SA 4.0