# Semi-rigid boolean algebras

A boolean algebra is rigid if it has no nontrivial automorphisms. Call it semi-rigid if none of its nontrivial automorphisms has any fixed points other than 0 and 1.* The four-element algebra $$\{0, b, \neg b, 1\}$$ is a simple example of a semi-rigidity. Preliminary question: Are there semi-rigid complete atomless boolean algebras (CABA's)? I suspect so: Let $$B$$ be a CABA with an element $$b$$ such that the principal ideals $$B \upharpoonright b$$ and $$B \upharpoonright \neg b$$ (having greatest elements $$b$$ and $$\neg b$$) are isomorphic copies of the same rigid CABA $$C$$. I suspect this $$B$$ has exactly one nontrivial automorphism, which interchanges $$B \upharpoonright b$$ and $$B \upharpoonright \neg b$$.

Even if this $$B$$ can be proved semi-rigid, though, it would be an uninteresting example because its semi-rigidity would reduce to rigidity (of principal ideals). So my question is: Are there semi-rigid CABA's none of whose principal ideals are rigid?

*The rationale for the term "semi-rigid" is this: Knowing where an automorphism of a semi-rigid boolean algebra maps one element -- any element, other than 0 and 1 -- determines where it maps all other elements; there is no further flexibility. For if $$\phi$$ and $$\phi'$$ were distinct automorphisms that mapped some $$b$$ to the same element, then $$\phi^{-1} \circ \phi'$$ would be a nontrivial automorphism with fixed point $$b$$.

There is no such algebra. In fact, suppose that $$B$$ satisfies the indicated condition. Choose $$a$$ in $$B$$ with $$a$$ not equal to $$0$$ or $$1$$. Let $$f$$ be a nontrivial automorphism of the principal ideal determined by $$a$$. Define $$g(x)=f(x \wedge a)\vee(x \wedge-a)$$. Then $$g$$ is a nontrivial automorphism of $$B$$ with fixed point $$-a$$.