# m-categoricity of theory of algebraically closed fields of fixed characteristic

one more question on a point on page 89. At first glance, I accepted the result with respect ACF(n) shown in (iv). But now I wonder: Is Shoenfield referring to a result of model theory? I thought about all I know based on the texts of algebra I know, but I can not seem to recognize this result, so I think it concerns model theory. Is the demonstration particularly laborious? If, kindly, someone more experienced than me can give me a help ...

Addendum. I apologize for not having clarified the question, so I follow the suggestion that I was given. ACF(n) is the theory of algebraically closed fields of characteristic n ($0$ or prime). The axioms are given in the preceding pages of text, but these are not relevant to my question. Theories are countable (that's number of symbols $\le \aleph_0$). The point (iv) [on page 88] specifies one of four types of m-categoricity: the theory T is m-categorical for each m more than countable, but not for $\aleph_0$. It also adds that each ACF(n) has this property. The question is: where can I find proof of this fact?

• Can you give more detail for those without the book? – Elchanan Solomon Jan 21 '13 at 15:59

The required algebra is covered in many algebra textbooks, though they might not point out explicitly the model-theoretic implications. The basic idea is that an algebraically closed field is determined up to isomorphism by its characteristic and its transcendence degree. Now the cardinality of an infinite field of transcendence degree $\kappa$ is $\max\{\aleph_0,\kappa\}$. Thus (if we fix the characteristic) for any uncountable cardinal $\lambda$, there is just one ACF of cardinality $\lambda$, namely the one of transcendence degree $\lambda$. But there are countably many ACFs of cardinality $\aleph_0$, since the transcendence degree can be any natural number or $\aleph_0$.