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An analogue of Mathias forcing is studied in connection of free Boolean algebras and nowhere dense ultraÿlters. Some applications to rigid Boolean algebras are given.

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Abstract. It is proved that the following conditions are equivalent:(a) there exists a complete, atomless, σ-centered Boolean algebra, which does not contain any regular, atomless, countable subalgebra, (b) there exists a nowhere dense ultrafilter on ω. Therefore the existence of such algebras is undecidable in ZFC. In "forcing language" condition (a) says that there exists a nontrivial σ-centered forcing not adding Cohen reals A subalgebra B of a Boolean algebra A is called regular whenever for every X ⊆ B, sup B X = 1 implies sup A X = 1; see e.g. Heindorf and Shapiro [6]. Clearly, every dense subalgebra is regular. Although every complete Boolean algebra contains a free Boolean algebra of the same size (see the Balcar-Franek Theorem; [2]), not always such an embedding is regular. For instance, if B is a measure algebra, then it contains a free subalgebra of the same cardinality as B, but B cannot contain any infinite free Boolean algebra as a regular subalgebra. Indeed, measure algebras are weakly σ-distributive but free Boolean algebras are not, and a regular subalgebra of a weakly σ-distributive one is again σ-distributive. Thus B does not contain any free Boolean algebra. On the other hand, measure algebras are not σ-centered. So, a natural question arises whether there exists a σ-centered, complete, atomless Boolean algebra B without regular free subalgebras. Since countable atomless Boolean algebras are free and every free Boolean algebra contains a countable regular free subalgebra, it is enough to ask whether B contains a countable regular subalgebra. In the paper we prove that such an algebra exists iff there exists a nowhere dense ultrafilter.

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