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\begin{document}

\title{Model Theory II Homework 7}
\author{Dakota Blair}
\maketitle

\setcounter{section}{15}

\begin{hw}[**]
Let $T$ be a countable, $\aleph_1$-categorical theory
such that $\st{N}$ and $\st{M}$ are models where
$\st{N} \models T$ and $\st{M} \models T$.
Further let $\varphi(x) \in L(\st{M}_0)$ be a strongly minimal formula
over the prime model $\st{M}_0$ of $T$.
If $\st{M} \subst \st{N}$
then $\dim \st{N} = \dim \st{M} + \dim \st{M}/\st{N}$
\end{hw}

\begin{lem*}[*]
Let $I$ be a basis of $\varphi(\st{M})$,
and $J$ an independent set in $\varphi(\st{N}) \setminus \st{M}$.
If $c \in \varphi(\st{N})$ then $c \in \acl IJ$.
\end{lem*}

\begin{dfn*}[$p_\varphi^A$]
Let $\varphi$ be a nonalgebraic formula over $A$.
Then
\[
p_\varphi^A =
\left\{
\varphi(x) \symand \symnot \psi(x)
\Big|
\psi \in L_1(A)
\text{ and }
|\psi(\st{M})| < \omega
\right\}
\]
is a type that says, ``$\varphi(x)$ and $x \not \in \acl A$''.
\end{dfn*}

\begin{remark*}
The formula $\varphi$ is strongly minimal
if and only if
$p_\varphi^A$ is the unique nonalgebraic type in $S_1(A)$ containing
$\varphi$.
\end{remark*}

\begin{proof}[Proof of (*)]
Proceed by induction on $|J|$.
If $|J| = 0$ then $\st{N} = \st{M}$
and therefore $c \in \varphi(\st{M}) \subset \acl I$.

It is also important to consider another base case,
in particular, the case when $J = \{b\}$ (so $|J| = 1$).
By Lemma 22(4) $M$ is atomic over $Ib$,
that is for each $\tuple{m} \in M$ the type $\tp(\tuple{m}/Ib)$ is isolated.
Since $c \in \varphi(\st{N})$ there is an $\tuple{m} \in M$
such that $c \in \acl I\tuple{m}b$.
Therefore $\tp(\tuple{m}/Ib)$ is isolated,
$\tp(c/I\tuple{m}b)$ is isolated,
and hence, by transitivity of isolation,
$\tp(c/Ib)$ is isolated.
But $\varphi(\st{M}) \subset \acl I \subset \acl Ib$
and $\varphi(\st{M})$ is infinite.
Therefore by Lemma 22(1)
the type $p_\varphi^{Ib}$ is not isolated hence
$c \not\models p_\varphi^{Ib}$, that is $c$ is algebraic over $Ib$.
Thus $c \in \varphi(\st{N})$ implies $c \in \acl Ib$.
Consequently if $|J| = 1$ and $c \in \varphi(\st{N})$
then $c \in \acl IJ$.

The induction hypothesis is therefore that
if $c \in \varphi(\st{N})$ then $c \in \acl IJ$
for any $|J| \leq \alpha$.

Now consider the successor case where
$J_{\alpha+1} = J_\alpha \union \{b\}$ and $b \not \in J_\alpha$.
Take $\st{N}'$ prime over $\st{M}J_\alpha$.
Note that $\dim \st{N}/\st{N'} = 1$,
for $\varphi(\st{N}) \subset \acl M J_\alpha b$, and $MJ_\alpha \subset N'$,
so $\varphi \subset \acl N'b = \acl_N b$.
Therefore
\[
\dim \st{N} = \dim \st{N}' + \dim \st{N}/\st{N}' = \dim \st{N}' + 1.
\]
By the induction hypothesis and Proposition 22(3)
\[
\dim \st{N}' = \dim \st{M} + \dim \st{N}'/\st{M}
= \dim \st{M} + |J_\alpha|.
\]
Thus
\[
\dim \st{N} = \dim\st{M} + |J_\alpha| + 1 = \dim \st{M} + |J_{\alpha+1}|.
\]
Thus if the induction hypothesis holds for
$|J| = \alpha$ it holds for $|J| = \alpha + 1$.

Now consider the limit case where the induction hypothesis
holds for all $\alpha < \beta$ and $\beta$ is a limit ordinal.
Then for any $J' \subset J$
let $\st{N}_{J'}$ be prime over $MJ'$.
If $|J'| < |J|$ we have that
$c \in \varphi(\st{N}_{J'})$ implies that $c \in \acl IJ'$.
If $c \in \varphi(\st{N})$ then in particular there is a $J'$ as above
such that $c \in \varphi(\st{N}_{J'})$.
Otherwise
\[
c \not\in \Union_{|J'|<|J|} \st{N}_{J'}
\hspace{1pc} \text{but} \hspace{1pc}
\st{N} = \Union_{\alpha < |J|} \st{N}_{J_\alpha}
\]
where $J = \{b_i\}_{i \leq |J|}$ and $J_\alpha = \{b_i\}_{i \leq \alpha}$.
Thus if $c \not\in \st{N}_{J_\alpha}$ for every $\alpha < |J|$
contradicts $c \in \varphi(N)$.
By the induction hypothesis this shows $c \in \acl IJ'$.
But $J' \subset J$ hence $\acl IJ' \subset \acl IJ$ implies $c \in \acl IJ$.
Thus $c \in \varphi(\st{N})$ implies $c \in \acl IJ$.
Further we may choose $J'$ large enough such that
$\dim \st{N}/\st{N}_{J'} < \dim \st{N}$.
Hence $\dim \st{N} = \dim \st{N}_{J'} + \dim \st{N}/\st{N}_{J'}$
by the induction hypothesis of $(**)$.
\end{proof}

\begin{hw}
$I(\Tfcp, \aleph_1) = \aleph_0$
\end{hw}

\begin{proof}[Proof sketch]
Each isomorphism class is uniquely identified by
the cardinality of $\aleph_0$ and $\aleph_1$ classes.
There are only $\aleph_0$ choices for these parameters
in $\aleph_1$ models of $\Tfcp$.
Therefore $I(\Tfcp, \aleph_1) = \aleph_0$.
\end{proof}

\begin{hw}
Let $T$ be a countable $\aleph_1$-categorical theory.
Show that $\st{M} \models T$ is atomic and $\omega$-saturated
if and only if
$\dim \varphi(\st{M})$ is infinite
for some (all) strongly minimal $\varphi$ over $\st{M}$.
%No vaughtian pairs
\end{hw}

\begin{hw}

\end{hw}

\begin{comment}
I have it down to one last detail. but
it hinges on the fact that all the non-algebraic types over a finite set A,
are of the form P_(phi), for some S.M. formula phi.
where P_(phi) is that type which says that
there is a guy in phi(M) which is not algebraic over A.
because phi is strongly minimal,
we know that P_(phi) is the unique non-algebraic type over A containing phi.
and if dim(phi) is infinite, then P_phi must be realized.
obviously the algebraic types are realized, because they are isolated.
so its done if you know that all the non-algebraic types over A
are of the form P_(phi)
\end{comment}

\end{document}
