  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 $\exists^\infty$. It follows that very slim fields in
the sense of Junker and Koenigsmann are the same thing as geometric fields in
the sense of Hrushovski 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.
