# Shelah's uncountable categoricity theorem

Mathematics Asked by user826451 on January 12, 2021

Morley’s theorem says if $$T$$ is a complete theory in a countable language and $$T$$ is $$kappa$$-categorical for some uncountable $$kappa$$, then $$T$$ is $$kappa$$-categorical for any uncountable $$kappa$$. Shelah generalized this theorem to uncountable languages, but I can’t find the exact statement.

In particular, if $$T$$ is $$kappa$$-categorical for some $$kappa$$ greater (or also equal?) than $$|T|$$, then $$T$$ is $$kappa$$-categorical for every $$kappa$$ greater (or also equal) than $$|T|$$?

Saharon Shelah extended Morley's theorem to uncountable languages: if the language has cardinality $$kappa$$ and a theory is categorical in some uncountable cardinal greater than or equal to $$kappa$$ then it is categorical in all cardinalities greater than $$kappa$$.

Now this is actually a bit stronger than you might expect! Morley's theorem says that if a theory in a countable language is categorical in a cardinal greater than $$aleph_0$$, then it is categorical in all cardinalities greater than $$aleph_0$$. On the other hand, if a theory in a countable language is categorical in a cardinal equal to $$aleph_0$$ (i.e. the theory is countably categorical), this does not guarantee categoricity in any other cardinals.

Shelah's possibly surprising result, as stated above, puts together two theorems from Chapter IX of Classification Theory (pages 490 and 491). Theorem 1.16 is the natural generalization of Morley's theorem, and Theorem 1.19 deals separately with the case of a theory $$T$$ which is $$|T|$$-categorical, showing that this case trivializes when $$|T|$$ is uncountable.

THEOREM 1.16: Suppose $$T$$ is categorical in some $$lambda > |T|$$ or every model of $$T$$ of cardinality $$lambda$$ (for some $$lambda>|T|$$) is $$|T|^+$$-universal. Then $$T$$ is categorical in every $$mu > |T|$$, and every model of $$T$$ of cardinality $$>|T|$$ is saturated.

THEOREM 1.19: If $$T$$ is categorical in $$|T|>aleph_0$$, then $$T$$ is a definitional extension of some $$T'subseteq T$$, $$|T'|<|T|$$.

The point is that if $$T$$ is a definitional expansion of $$T'$$, then there is a one-to-one cardinality-preserving correspondence between the models of $$T$$ and the models of $$T'$$. If $$T$$ is $$|T|$$-categorical, then $$T'$$ is also $$|T|$$-categorical. Since $$|T'|<|T|$$, by Theorem 1.16, $$T'$$ is $$kappa$$-categorical for all $$kappa > |T'|$$. So also $$T$$ is $$kappa$$-categorical for all $$kappa>|T'|$$, and in particular for all $$kappa>|T|$$.

What's happened here is that the uncountable $$|T|$$-categorical theory $$T$$ is "secretly" just a theory of smaller cardinality. And this is really quite a silly situation to be in! To say that $$T$$ is a definitional expansion of $$T'$$ is to say that every symbol in the language of $$T$$ which is not in the language of $$T'$$ is defined by a formula in the language of $$T'$$. But there are only $$|T'|$$-many formulas in the language of $$T'$$, so while there are $$|T|$$-many new symbols, up to equivalence there are only $$|T'|$$-many!

To give an explicit example, we could take $$T'$$ to be the theory of algebraically closed fields and take $$T$$ to be the theory obtained by introducing uncountably many constant symbols $${c_alphamid alphain kappa}$$ and setting them all to $$0$$ by adding axioms $$c_alpha = 0$$ for all $$alpha$$. Theorem 1.19 says that every example has to be almost as trivial as this one. Upshot: the behavior of countable $$aleph_0$$-categorical theories is very special to the countable.

Answered by Alex Kruckman on January 12, 2021

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