Does there exist an infinite Boolean ring $R$ (not assume unital, only associative) with the property that for any nonzero $x,y\in R$, there is a ring automorphism $\varphi\colon R\to R$ such that $\varphi(x)=y$?
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2$\begingroup$ Let $\mathcal{R}$ be the subring of $\mathcal{P}(\omega)/\mathit{Fin}$ consisting of (classes containing) sets of asymptotic density zero. Does this work? $\endgroup$– Noah SchweberCommented Oct 20 at 23:35
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2$\begingroup$ 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). $\endgroup$– YCorCommented Oct 21 at 13:07
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2$\begingroup$ 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. $\endgroup$– Emil JeřábekCommented Oct 23 at 9:51
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1$\begingroup$ 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. $\endgroup$– YCorCommented Oct 23 at 11:02
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3 Answers
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Let $X$ be Cantor minus 1 point, and $R$ be the set of compact open subsets of $X$. This works because any two compact open subsets of $X$ are sent one to another by a self-homeomorphism of $X$.
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1$\begingroup$ (Note that this example of $R$ is countable, and it can be shown to be the only one up to isomorphism.) $\endgroup$– YCorCommented Oct 21 at 11:17
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1$\begingroup$ A variant using $A=P(\omega)/$fin is: for any $f\in \mathrm{Hom}(A,\mathbf{Z}/2\mathbf{Z})$, $R_f=\mathrm{Ker}(f)$. Topologically, this corresponds to: given the point $x$ determined by $f$ in $\beta^*\omega$ (Stone-Cech remainder), $R_f$ is the set of clopen subsets of $\beta^*\omega$ not containing $x$. $\endgroup$– YCorCommented Oct 21 at 11:23
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The Boolean ring of measurable subsets of $\mathbb{R}$ with finite Lebesgue measures, modulo a.e. equivalence, also works, since any two non-null subsets of $\mathbb{R}$ with finite measures are related to each other by a non-singular automorphism of $\mathbb{R}$.
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Very similar to David Gao's and YCor's answers: let $\lambda<\kappa$ be infinite cardinals, and let $R$ be the collection of equivalence classes of subsets of $\kappa$ having cardinality at most $\lambda$ determined by the equivalence relation $A\sim B$ iff $|A\triangle B|<\lambda$. It's easy to prove that for any two sets $A, B\subset \kappa$ having cardinality $\lambda$, there is a permutation $\pi$ of $\kappa$ with $\pi(A)=B$, and this $\pi$ induces an automorphism of $R$ taking the equivalence class of $A$ to the equivalence class of $B$.
