# Problem with Morley's Theorem

Greets.

Morley's theorem states that a theory which is categorical for an uncountable cardinal is categorical in all uncountable cardinals. My problem with the theorem is that I haven't found a significant example in which this theorem can be applied, in which no other argument has been found.

The only examples I know are $RG$, $ACF_0$, and the theory of vector spaces over a finite field, but this examples are worked out with simple cardinal and ordinal arguments.

Maybe this question is useless, because maybe the importance of this theorem is just theoretic, if this is so, I would appreciate a reason why.

Thanks

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How do you define significant example? Does the fact that the algebraic closure of the metric completion of the algebraic closure of the $p$-adic numbers is isomorphic to the complex number mean nothing? – Asaf Karagila Oct 13 '12 at 17:19
I didn't know that!!!!, can you allow me the proof of that? – Camilo Arosemena Oct 13 '12 at 19:01
The $p$-adic numbers form a set of size continuum; so does the algebraic closure of such field, however by Baire category theorem this is not a complete metric space; it's completion, however, turns out to be complete and algebraically closed. If you run all the cardinality arguments you see that you never exceed the continuum. Well, now you have an algebraically closed field of size continuum and characteristic zero... what could that be? – Asaf Karagila Oct 13 '12 at 19:05
The fact that that field is isomorphic to $/mathbb{C}$ follows from the fact that the cardinal of $/mathbb{C}$ is uncountable, and this fact was known way before Morley's theorem – Camilo Arosemena Oct 13 '12 at 19:25
Yes, true. If you add the topological requirements then you have that already. Here is another quirk which is strange and follows from categoricity of ACF: Take $\mathbb C(t)$ (a transcendental extension of $\mathbb C$) which is clearly not algebraically closed. Cardinality arguments gives that $\mathbb C(t)$ is still of size continuum and properly extends $\mathbb C$. Close that field, and you have a proper extension of $\mathbb C$ which is algebraically closed. However by cardinality arguments this is again $\mathbb C$. – Asaf Karagila Oct 13 '12 at 19:27

One of the many contributions of Morley's work was to introduce a very general model-theoretic notion of dimension. A strongly minimal formula $\phi(x)$ has the property that given any model $M\models T$, we can assign a (possibly infinite) dimension $\kappa$ to $\phi(M)$, and $|\phi(M)| \leq \aleph_0 + \kappa$. Morley proved that models for uncountably categorical theories are completely controlled by the dimensions of their strongly minimal sets.

The proof of Morley's theorem goes like this:

Step 1 (the hard part): If a theory $T$ is categorical in some uncountable cardinal, then there is a strongly minimal formula $\phi(x)$ such that a model $M\models T$ is determined up to isomorphism by the dimension of $\phi(M)$, and $|M| = |\phi(M)|$.

Step 2 (an easy corollary): $T$ is $\kappa$-categorical for all uncountable $\kappa$. If $M,N\models T$ and $|M| = |N| = \kappa$, then $|\phi(M)| = |\phi(N)| = \kappa$, so $\phi(M)$ and $\phi(N)$ both have dimension $\kappa$, and $M\cong N$.

Now the reason that Morley's Theorem seems to add nothing new in each of the classic example cases you have in mind is that in each of these cases, Step 1 is already done, i.e. the strongly minimal set and the dimension notion are already familiar: linear dimension in the case of vector spaces, transcendence degree in the case of algebraically closed fields, cardinality in the case of the theory of infinite sets...

In fact, given any particular uncountably categorical theory, one can prove that it's uncountably categorical without appealing to Morley's theorem by doing Step 1 directly (exhibiting a dimension notion which determines models up to isomorphism) and then giving the argument for Step 2.

The value of Morley's theorem, of course, is that it guarantees that such a dimension notion exists. As such it's very important as a theorem of model theory. It increases our understanding of what the classes of models for first-order theories can look like.

EDIT: I also want point something out about your question. Morley's theorem has the structure "for all theories satisfying this property, the following is true". You complain that you can't find any examples of particular theories for which the conclusion can't be checked without appealing to Morley's theorem. This is a bit like complaining that the Pythagorean Theorem ("for all triangles satisfying the property of being right, the following is true") isn't useful, just because given any particular right triangle, you can do the arithmetic and check that $a^2 + b^2 = c^2$.

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