# The unique model of cardinal $\kappa$ of a $\kappa$-categorical countable theory is saturated.

Let $T$ be a $\kappa$-categorical ($\kappa \geq \aleph_1$) first-order theory in a countable language $\mathcal L$. I try to prove that its unique (up to isomorphism) model of cardinal $\kappa$ is saturated.

It is a theorem seen in a lecture, but the proof the teacher gave was kind of fuzzy, so I'm trying to retrieve it by myself. This is what I have done so far, but it seems a little bit convoluted.

Let $\mathfrak M$ be the model of $T$ of cardinal $\kappa$. First, recall a theorem that we have proved and that I will use :

Theorem 1. Let $\lambda > |\mathcal L|$ be a regular cardinal. If $T$ is $\lambda$-stable, then every model of $T$ of cardinal $\lambda$ admits an saturated elementary extension of cardinal $\lambda$.

Then consider the two following cases :

$\kappa$ is regular : $T$ being $\kappa$-categorical, it is $\omega$-stable so $\kappa$-stable, and so gets a saturated model of cardinal the regular cardinal $\kappa$ (cf. Theorem 1), which is $\mathfrak M$ by $\kappa$-categoricity.

$\kappa$ is singular : if we can show that $\mathfrak M$ is $\lambda$-saturated for all regular $\lambda < \kappa$, then we are done ; indeed, any $A \subseteq M$ has some cardinal $\mu < \kappa$, and as $\kappa$ is singular, the regular cardinal $\mu^+$ is stricly lesser than $\kappa$, hence the type of $S_1(A)$ are realized in the $\mu^+$-saturated model $\mathfrak M$.

So, all it remains is to show that $\mathfrak M$ is $\lambda$-saturated for all regular $\lambda < \kappa$. Take $A \subseteq M$ of cardinal $\lambda$. By Löwenheim-Skolem's theorem, there is $\mathfrak N \preceq \mathfrak M$ of cardinal $\lambda$ containing $A$. In particular, $\mathfrak N$ is a model of $T$ of regular cardinal $\lambda$, so Theorem 1 gives $\mathfrak N^\ast \succeq \mathfrak N$ saturated of cardinal $\lambda$. Applying Löwenheim-Skolem again, one gets $\mathfrak M^\ast \succeq \mathfrak N^\ast$ of cardinal $\kappa$. By $\kappa$-categoricity, it is $\mathfrak M$. But then $\mathfrak M$ is an elementary extension of $\mathfrak N^\ast$ which realized all types of $S_1(A)$. It concludes.

I have read on some thread that this theorem is "pretty easy". But my proof (if correct) uses some non trivial results (Theorem 1, and "$\omega$-stable $\implies$ $\kappa$-stable" for example). So maybe is there a more elementary proof ?

• The proof you give is more or less the proof I know. I believe it is the standard proof. I can also imagine contexts where Theorem 1 and $\omega$-stable implies $\kappa$-stable can be referred as "pretty easy". – Levon Haykazyan Mar 28 '14 at 0:41
• @LevonHaykazyan Ok thanks. For the "pretty easy" part (seen in this thread), I get that Theorem 1 can be easy (in fact, it is straightforward and it principally is the construction of a transfinite elementary chain of models). But "$\omega$-stable implies $\kappa$-stable" is kind of tricky (from the proof I know it is principally the fact that $S_1(A)$ does not have perfect closed set if $T$ is $\omega$-stable ; otherwise put, every type has a finite Cantor-Berdixon rank) : calling it easy is, I think, an overstatement. Thanks anyway. – Pece Mar 28 '14 at 9:22