Submission date: 1 August 2022

Abstract:

Let T be a complete theory of fields, possibly with extra structure. Suppose that model-theoretic algebraic closure agrees with field-theoretic algebraic closure, or more generally that model-theoretic algebraic closure has the exchange property. Then T has uniform finiteness, or equivalently, it eliminates the quantifier ∃^∞. It follows that very slim fields in the sense of Junker and Koenigsmann are the same thing as geometric fields in the sense of Hurshovski and Pillay. Modulo some fine print, these two concepts are also equivalent to algebraically bounded fields in the sense of van den Dries.

From the proof, one gets a one-cardinal theorem for geometric theories of fields: any infinite definable set has the same cardinality as the field. We investigate whether this extends to interpretable sets. We show that positive dimensional interpretable sets must have the same cardinality as the field, but zero-dimensional interpretable sets can have smaller cardinality. As an application, we show that any geometric theory of fields has an uncountable model with only countably many finite algebraic extensions.

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Full text arXiv 2208.00586: pdf, ps.