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\begin{document}
\begin{center}
\textbf{Strongly minimal theories and Morley's categoricity theorem.
}\end{center}
\section{Motivation}
Let us begin by reminding ourselves of the final aim of this chapter
\footnote{
(unrelated and for your knowledge:) also for stability Shelah has proved the following:
\begin{thm} For $T$ complete in a countable $L$, one of the following holds:
\begin{enumerate}
\item
$T$ is not $k$-stable for any cardinal $\kappa$.
\item
$T$ is $\kappa$-stable for all $\kappa\geq 2^{\aleph_0}$
\item
$T$ is $\kappa$-stable if and only if $\kappa^{\aleph_0}=\kappa$.
\end{enumerate}
\end{thm}}
\begin{thm}[Morley]
If $T$ is a countable theory, then
$T$ is $\aleph_1$-categorical if and only if it is $\kappa$-categorical for all \textbf{uncountable} $\kappa$'s.
\end{thm}
\begin{proof}[Idea of the proof]
This may remind you of the vector spaces:
in some suitable sense every model of $T$ is generated by a base of
cardinality $\kappa$.
This base is in a suitable sense independent and every two
models whose bases have the same cardinality are isomorphic.
\end{proof}
In the rest, we are going to see what these suitable senses are! note that
the word uncountable is in bold for a reason.
For example ACF$_0$ is $\aleph_1$-categorical and not $\aleph_0$-categorical (there are
$\aleph_0$ many algebraically closed fields of cardinality $\aleph_0$
(find out why).
\par
The last ingredient we need for the proof of Morley's theorem is strong-minimality.
Note that there are two concepts (related to one another of course):
strongly minimal sets, and strongly minimal theories.
\\
Strongly minimal theories are similar
to vector spaces in the sense that they are concerned with the
concepts of dimension, independence, basis,
isomorphism of structures with similar bases,
etc.
You may now say, so they are also similar to algebraic closed fields, where
algebraic independence replaces linear independence and there is a concept of basis and
algebraic independence and transcendental elements in this sense.
Indeed there is famous trichotomy conjecture by Zilber that
(very crudely speaking) strongly minimal theories are of one of the following three
kinds:
either they are essentially trivial, or they are essentially like a vector space, or
they are essentially like an algebraic closed field (whatever essentially means!).
This conjuncture is disproved by Hrushovski.\\
Strongly minimal theories are uncountably categorical
and the reason is having an independent basis.
\section{Strongly minimal sets}
$T$ is complete, countable, possesses infinite models.
If $M\models T$, $\phi(x_1,\ldots,x_n,\bar{a})$ is a formula with $a\in M$,
by
$\phi(M,\bar{a})$ we mean $\{\bar{y}\in M^n:M\models \phi(\bar{y},\bar{a})\}$
\\
Sometimes it is easier to consider definable sets instead of
the formulae defining them:
\begin{defn}
Suppose that
$M\models T$, and
$X\subseteq M^n$ is an infinite definable set (defined with parameters in $M$).
\begin{enumerate}
\item
$X$ is called minimal in $M$ if for every definable (with parameters in
$M$) set $A$ either $X\cap A$ or $X\cap A^c$ is finite.
\\
\begin{tikzpicture}
\draw (0,0) rectangle (4,2)node[right]{$M$};
\draw (.5,.5) rectangle (3,1.5)node[right]{$X$};
\draw (2.2,.1)rectangle (3.5,1.1)node[right]{$A$};
\draw[fill] (2.5,.75)circle[radius=.1];
\draw[fill] (2.83,.85)circle[radius=.1];
\end{tikzpicture}
\begin{align*}
& X \text{ a definable set}\\
&\text{ for every definable } A \text{ either } A\cap X \text{ finite or } A^c\cap X
\end{align*}
\item
A formula
$\phi(\bar{x})$ (in $L(M)$) is strongly minimal if it defines a minimal set in all
elementary extensions of $M$.
\item
A non-algebraic type is strongly minimal if it contains a strongly minimal formula.
\item
$T$ is strongly minimal if every definable set in every model of
$T$ is either finite or cofinite (different wordings:
$T$ is strongly minimal if $x=x$ is a strongly minimal formula;
$T$ is strongly minimal if all sets are definable using only $\{=\}$)
\end{enumerate}
\end{defn}
\begin{ex}
Show that the following theories are strongly minimal and
in each case
determine the
closure of a given set
$X$ and make sense of the concepts of independence and bases:
\begin{enumerate}
\item
The theory of $ACF_0$
\item
The theory of $K$-vector spaces for a field $K$.
\end{enumerate}
\end{ex}
\begin{ex}
if $M$ is $\omega$-saturated then every minimal formula is strongly minimal.
\end{ex}
\begin{thm}
[Exchange Principle for strongly minimal sets, direct proof]
Suppose that
$M\models T$ and
$X$ is a strongly minimal subset of $M$ defined without parameters.
Let $A$ be a subset of $X$ and $a,b$ be two elements of $X$.
If $a\in \acl(A\cup \{b\})- \acl(A)$, then $b\in \acl(A\cup \{a\})$.
\end{thm}
\begin{proof}
I do the proof for $A=\emptyset$ and the general case is no different. I thank Michael LĂ¶sch
for having corrected and provided the following version of the proof:
we want to prove that for
all $a,b\in X$, if
$a\in \acl(b)$ then $b\in \acl(a)$ provided that
$a\not\in\acl(\emptyset)$.
\\
We need a formula $\eta(x,a)$ such that
$\eta(M,a)$ is finite and $b\in\eta(M,a)$. Let $\phi(x,b)$ be a
formula witnessing the assumption $a\in \acl(b)$, that is
$a\in\phi(M,b)$ and $|\phi(M,b)|=n$ for some finite $n$, and assume that
$X$ is defined by
the formula $\chi$, that is $X=\{x\in M\,:\, M\models\chi(x)\}$.
By our assumptions, the following hold for $b$:
\begin{itemize}
\item $b\in\phi(a,M)$,
\item $b\in\psi(M)$, where $\psi(x)$ is the equivalent formula to $|\phi(M,x)|=n$,
\item $b\in \chi(M)$.
\end{itemize}
Let us consider the formula
$\eta(x,a)=\phi(a,x)\land\psi(x)\land\chi(x)$.
By the items above, $b\in\eta(M,a)$ and it suffices to show that
$\eta(M,a)=X\cap\psi(M)\cap\phi(a,M)$ is finite.
To reach a contradiction, we assume that $\eta(M,a)$ is infinite.
By strongly minimality of $X$, we get \[|X-(\psi(M)\cap\phi(a,M))|=l\] for some finite $l$. Let
$\xi(x)$ be the formula expressing
\[
|X- (\psi(M)\cap\phi(x,M))|=l.
\]
We know that $a\in \xi(M)$.
If $\xi(M)$ is finite then $a\in \acl(\emptyset)$, and this is contradictory with
our assumptions. Hence $\xi(M)$ is infinite and we can find
$n+1$-many elements, say $a_1,\ldots a_{n+1}$, in $\xi(M)$. We define $B_i:=X\cap\psi(M)\cap\phi(a_i,M)$. Since $X$ is infinite
and $|X\backslash B_i|=l$, we have
\[\bigcap_{i=1}^{n+1}B_i=X\backslash(\bigcup_{i=1}^{n+1}X\backslash B_i)\] is infinite
and in particular nonempty. So, let $\hat b\in \bigcap B_i$. Then for each $i$,
$\phi(a_i,\hat{b})$ holds, that is $|\phi(M,\hat{b})|\geq n+1$. This contradicts the fact $\psi(\hat{b})$ holds.
\end{proof}
\begin{defn}
$A\subseteq X$ is independent if for every $a\in A$
\[
a\not \in \acl(A-\{a\}).
\]
Let $C\subseteq X$, then $A$ is independent over $C$ means that for all $a\in A$
\[
a\not \in \acl(C\cup (A-\{a\}))
\]
We say that $A$ is a basis for $Y\subseteq X$ if
$A\subseteq Y$, $A$ is independent, and $\acl(A)=Y$.
\end{defn}
As one expect from comparing the notion of basis with the notion of
basis in vector spaces and in Algebraic closed fields, any two bases have the cardinality.
\begin{lem}
If $A$ and $B$ are two bases for $Y\subseteq X$ then $|A|=|B|$.
\end{lem}
\begin{defn}
If $Y\subseteq X$, then the dimension of $Y$ is the cardinality of
a basis of $Y$. Let $\dim(Y)$ denote it.
\end{defn}
Every model of a strongly minimal theory is determined (up to isomorphism)
by its dimension. This is what have we have stated in the theorem below:
\begin{thm}\label{isomorph}
Suppose that $T$ is strongly minimal and $M,N\models T$.
Then $M\cong N$ if and only if $\dim(M)=\dim(N)$ (note that here $X=M$).
\end{thm}
\begin{proof}[Idea of the proof]
As we may expect, the bijection between the bases of $M$
and $N$ will extend to an isomorphism of $M$ and $N$.
\end{proof}
Let us prove a more general theorem instead:
\begin{thm}
Consider the following diagram:
\[
\xymatrix{
M&&N\\
&M_0\ar[ul]\ar[ur]\\
&A\ar[u]
}
\]
\begin{align*}
&M_0\prec M, M_0\prec N\\
& A \text{ subset of } M_0
\end{align*}
Let $\phi(x)$ be a strongly minimal formula with parameters from $A$.
Suppose that $\dim(\phi(M))=\dim(\phi(N))$. Then there is a partial
elementary map $f:\phi(M)\to \phi(N)$.
\end{thm}
The proof of this theorem will also remind you of
linear algebra. Let us
just remind ourselves of the following:
\begin{ob}
If $a$ is algebraic over $B$ then $\tp^M(a/B)$ is isolated.
\end{ob}
\begin{proof}[proof of the observation]
You have once proved this in your exercise sessions, but I still like to
mention it again. Whenever we say $\phi(x)$ isolates $P(x)$ ($P$ a type
in variable $x$) it means that every for every $x$, if $x$ satisfies
$\phi$ then it satisfies all formulae in $P$, that is
\[
\phi(M)=P(M)
\]
Now let $\phi$ be the formula that witnesses that $a$ is
algebraic over $B$. Then $\phi\in P$ and let's say
$\phi(M)=\{a_1,\ldots,a_n\}$. Then
$P(M)\subseteq \{a_1,\ldots,a_n\}$. We will throw away those $a_i$'s that
do not realise all formulae in $P$. Let's say
$a_{i1},\ldots,a_{im}$ do not respectively satisfy $\psi_1,\ldots,\psi_m$. Then
\[
(\phi\wedge \psi_1\wedge\ldots\psi_m)(M)=P(M)
\]
and this is what we need.
\end{proof}
\begin{proof}[proof of the theorem]
Let $B$ be a transcendence basis for $\phi(M)$ and
$C$ be a transcendence basis for $\phi(N)$. Because
$|B|=|C|$, there exists a bijection $f:B\to C$.
\begin{claim}
$f$ is an elementary map.
\end{claim}
\begin{proof}[hint on the proof of the claim]
You have seen before (in the previous lecture), and you will prove again in the exercise sessions
that in the diagram above, if $a_1,\ldots,a_n\in\phi(M)$ are independent over $A$
and $b_1,\ldots,b_n\in \phi(N)$ are independent over $A$, then
$\tp(\bar{a}/A)=\tp({\bar{b}}/A)$. Indeed if $B\subseteq \phi(M)$ is
infinite and
and $C\subseteq \phi(N)$ is infinite, then $B$ and $C$ are sets of
indiscernibles with the same type.
\end{proof}
Let
\[
I:=\{g:B'\to C':B\subseteq B'\subseteq \phi(M), C\subseteq C'\subseteq \phi(N),
f\subseteq g, g\text { elementary}
\}
\]
By Zorn's lemma $I$ has a maximal element $g:B'\to C'$. We now claim that
indeed $B'=\phi(M)$ and $C'=\phi(N)$.\par
So suppose otherwise; let $b\in \phi(M)-B'$. Since $\acl(B)=\phi(M)$, we have
$b$ is algebraic over $B$ and hence over $B'$. So $\tp(b/B')$ is isolated.\\
So there is $\psi(x,\bar{d})$ with $\bar{d}\in B'$ isolating $\tp(b/B')$.
Because $g$ is partial elementary, there is $c\in \phi(N)$ such that
$\psi(c,g(\bar{d}))$ holds. So \[
\tp^M(b/B')=\tp^N(c/C')
\]
and this means that we can extend $g$ by sending $b$ to $c$, and this is contradictory
with the fact $g$ is maximal. Thus $\phi(M)=B'$. Showing that $\phi(N)=C'$ is
similar.
\end{proof}
\begin{cor}
If a strongly minimal $T$ is countable, then it is $\kappa$-categorical for all
$\kappa> \aleph_0$.
\end{cor}
\begin{proof}
We use Theorem \ref{isomorph}.
Let $M\models T$.
Note that
$M=\acl(B)$ for $B$ its transcendence basis. Also
\[
acl(B)=\bigcup_{\phi\in L,b\in B, \phi(M,b)\text{ finite }}(\phi(M,b))
\]
so for every $B$
the size of $\acl(B)$ is $\leq |B|+|T|$.\\
If $M$ has size $\kappa\geq \aleph_1$ then $|B|=\kappa$. This means that
two models of size $\kappa$ have bases with the same dimension, so they are isomorphic.
\end{proof}
\begin{cor}
Let $T$ be a strongly minimal theory.
\begin{enumerate}
\item
$T$ is $\lambda$-stable for all
$\lambda\geq |T|$.
\item
$T$ is totally transcendental.
\item
$T$ has no Vaughtian pairs.
\end{enumerate}
\end{cor}
\begin{proof}
1.
First note that if $T$ is strongly minimal and $A$ is a parameter set, then there
can exactly one non-algebraic type be over $A$. Because if there are two types
$p_1$ and $p_2$ then there are $\psi_1\in p_1$ with $\neg \psi_1\in p_2$. So
both $\psi_1(M)$ and $\neg \psi(M)$ are infinite, which is not possible in strongly minima
theories.\par
So, for every set $A$ with $|A|=\lambda\geq |T|$
we have
\[
|S(A)|\leq |\acl(A)|+1 \text{(the number of algebraic and non-algberaic types)}
\]
furthermore $|\acl(A)|$ is no greater than $|A|$ because it
is a union of $\bigcup_{a\in A,\phi\in L}\phi(M,a)$ where each $\phi(M,a)$ is finite.
\par
2. Suppose that there is a binary tree. Suppose that
$\phi$ and $\neg \phi$ have appeared on top of the same node on the tree.
One of $\phi(M)$ and $\neg \phi(M)$ is finite, let's say $\phi(M)$ is finite. The tree cannot
contiune with infinite branches over $\phi$, contradiction.
\par
3. Suppose that $(M,N,\phi)$ form a Vaughtian pair with
$N\prec M$ and $\phi(M)=\phi(N)$ infinite. So there are only
finitely many elements in $N$ not satisfying $\phi$. Say there are $n$ elements.
Since the fact: `there are $n$ elements not satisfying $\phi$' is
expressible in the language, and $N\prec M$,
\[
\{x\in M|M\models \neg \phi(x)\}=\{x\in N:N\models \neg \phi(x)\}
\]
that is $M=N$, a contradiction.
\end{proof}
\begin{ex}
Consider the following diagram:
\[
\xymatrix{
M&&N\\
&M_0\ar[ul]\ar[ur]\\
&A\ar[u]
}
\]
\begin{align*}
&M_0\prec M, M_0\prec N\\
& A \text{ subset of } M_0
\end{align*}
\begin{enumerate}
\item
let $\bar{a}$ be a tuple in $A$
and $\phi(x,\bar{a})$ a formula.
Then show that the fact that
\begin{center}
$\phi(x,\bar{a})$ defines a strongly minimal set in $M$
\end{center}
is an elementary property of $\bar{a}$ contained in the $\tp^M(\bar{a})$.
It means that in the above diagram if $\phi(x,\bar{a})$ defines a strongly
minimal set in $M$ then it defines a strongly minimal set in $N$ too.
\item
Suppose that $a_1,\ldots,a_n$ in $\phi(M)$ are independent over ${A}$
and $b_1,\ldots b_n \in \phi(N)$ are independent over $A$. Then show that
$\tp^M(\bar{a}/A)=\tp^N(\bar{b}/A)$.
\item
Let $B\subseteq \phi(M)$ be infinite and independent over $A$. Show that
$B$ is a set of indiscernibles over $A$ (note that being a set of indiscernibles
is a stronger property than being a sequence of indiscernibles).
\item
Let $C\subseteq \phi(N)$ be infinite and independent over $A$. Show that
$C$ is a set of indiscernibles over $A$ with the same type as type of $B$.
\end{enumerate}
\end{ex}
\begin{ex}
\begin{enumerate}
\item
Let $T$ be $\omega$-stable. Show that if $M\models T$ then there is a minimal
formula in $M$.
\item
If $M\models T$ is $\aleph_0$-saturated and $\phi(\bar{x},\bar{a})$ is minimal in $M$,
then it is strongly minimal.
\end{enumerate}
\end{ex}
\section{Baldwin-Lachlan theorem and Morley's theorem}
\begin{thm}
Let $T$ be a complete theory with infinite models in a countable language.
Let $\kappa$ be an uncountable cardinal. Then
\begin{center}
$T$ is $\kappa$-categorical\\ if and only if\\ $T$ is $\omega$-stable and has
no Vaughtian pairs.
\end{center}
in particular (because the above characterisation does not depend on $\kappa$):
\begin{center}
$T$ is $\aleph_1$-categorical \\
if and only if\\
$T$ is $\kappa$-catergorical (for each uncountable $\kappa$).
\end{center}
\end{thm}
\textbf{Reminder}: the following items are all
proved in previous lectures and I have listed them only for our ease of reference.
\begin{enumerate}
\item \label{cat-to-stable}
if $T$ is $\kappa$-categorical (uncountable $\kappa$)
then $T$ is $\omega$-stable.
\item \label{cat-to-vaught}
if $T$ is $\kappa$-categorical (uncountable $\kappa$) then $T$ has no Vaughtian pairs.
\item
$T$ has a prime model if and only if the isolated types are dense,
\item
if $T$ has no binary tree of consistent formulae then the isolated types are dense,
\item
$\omega$-stable theories have no binary tree of consistent formula (they are totally
transcental)
\item \label{stable-to-prime}
(as a result of the facts above) if $T$ is $\omega$-stable, then it has a prime model (needed here).
\item
if $T$ is totally transcendental then there is a minimal formula in each $M\models T$,
\item
if $T$ eliminates $\exists^{\infty}$ then any minimal formula is strongly minimal,
\item
a theory without Vaughtian pairs eliminates $\exists^\infty$,
\item \label{this}
(as a result of the facts above) if $T$ is $\omega$-stable with no Vaughtian
pair, then there is a strongly-minimal formula (needed),
\item \label{prime}
if $T$ has no Vaughtian pairs and $M\models T$ and
$A\subseteq M$ and $\phi$ a formula with parameters in $A$, then
$M$ is a minimal extension of $\phi(M)\cup A$. (needed)
\end{enumerate}
\begin{proof}[proof of the theorem]
One direction is clear by items \ref{cat-to-stable} and \ref{cat-to-vaught}.
Let us do the more difficult direction: suppose that $T$ is
$\omega$-stable and it has no Vauhtian pairs.
then by item \ref{this} above, there exists a strongly minimal
formula $\phi(x)$. \\
Now let $M_1$ and $M_2$ be two models of cardinality $\kappa$.
By item \ref{stable-to-prime} above, there exists a prime model $M_0$:
\[
\xymatrix{
M_1&&M_2\\
&M_0\ar[ul]\ar[ur]\\
&&&M_0\prec M_1,M_0\prec M_2, \text{all three models of }T
}
\]
We now want to prove that $\dim(M_1/M)=\dim(M_2/M)=\kappa$ and then this implies
that $M_1$ and $M_2$ are isomorphic (by theorem \ref{isomorph}).\\
Since $T$ has no Vaughtian pairs, $M_1$ is a minimal extension of $\phi(M_1)\cup M_0$
and $M_2$ is a prime extension of $\phi(M_2)\cap M_0$ (item \ref{prime}). Therefore
$\phi(M_1)$ has cardinality $\kappa$ and hence
$\dim(M_1/M)=\dim(M_2/M)=\kappa$.
\end{proof}
\end{document}