Exchange Principle

Theorem

Let $D$ be a strongly minimal set in $\mathcal{M}$.

Let $A$ be a subset of $D$.

Let $b, c \in D$.

If $b$ is algebraic over $A\cup\{c\}$ but not over $A$, then $c$ is algebraic over $A\cup\{b\}$.

Proof

Let $D(x)$ be a formula defining $D$, which exists since by definition, strongly minimal sets are definable.

To simplify exposition, we will assume below that all further mentioned formulas are $\mathcal{L}$-formulas with parameters from $A$, in addition to whatever other parameters are supplied.

Suppose $b$ is algebraic over $A\cup\{c\}$ but not over $A$.

By definition of being algebraic, there is some $\phi(x,c)$ with $x$ free such that $\mathcal{M}\models \phi(b,c)$ and $\{m\in \mathcal{M} : \mathcal{M}\models \phi(m, c)\}$ is finite.

Consequently, the subset $\{m\in \mathcal{M} : \mathcal{M}\models D(m) \wedge \phi(m, c)\} = \{d\in D : \mathcal{M}\models \phi(d, c)\}$ must have some finite $n\in \mathbb{N}$ many elements.

Let $\psi(y)$ be the formula

$\displaystyle D(y) \wedge \exists x_1 \cdots \exists x_n \left( \left( \bigwedge_{i=1,\dots,n} D(x_i) \wedge \phi(x_i, y) \right) \wedge \forall z \left( D(z) \wedge \phi(z, y) \rightarrow \left( \bigvee_{i=1,\dots,n} z = x_i \right) \right) \right)$

which asserts that

$\{d\in D : \mathcal{M}\models \phi(d, y)\}$ has $n$ many elements.

Note the dependence on the free variable $y$.

We will argue that $\phi(b, x) \wedge \psi(x)$ demonstrates the algebraicity of $c$ over $A\cup \{b\}$.

Suppose (with the intent of deriving a contradiction) that $\{d\in D : \mathcal{M}\models \phi(b, d) \wedge \psi(d)\}$ is infinite.

Since $D$ is strongly minimal, $\{d\in D : \mathcal{M}\models \phi(b, d) \wedge \psi(d)\}$ is cofinite in $D$.

Thus $D - \{d\in D : \mathcal{M}\models \phi(b, d) \wedge \psi(d)\}$ has some finite $k\in \mathbb N$ many elements.

Define $\chi(x)$ to be a formula which asserts that $D - \{d\in D : \mathcal{M}\models \phi(x, d) \wedge \psi(d)\}$ has $k$ many elements. This can be done similarly to how $\psi(y)$ was defined above.

$\chi$ cannot define a finite subset of $D$, since $\chi$ involves only parameters from $A$ and $\mathcal{M}\models \chi(b)$, and so this would imply that $b$ is algebraic over $A$.

Thus, $\chi$ defines an infinite subset of $D$.

So, we may let $b_1, \dots, b_{n+1}$ be distinct elements of $D$ such that $\mathcal{M}\models \chi(b_i)$ for each $i = 1,\dots,n+1$.

For each $i=1,\dots,n+1$, define

$C_i = \{d\in D : \phi(b_i, d) \wedge \psi(d)\}$

Then each $C_i$ is cofinite in $D$, since $\mathcal{M}\models \chi(b_i)$ and hence $D - \{d\in D : \mathcal{M}\models \phi(b_i, d) \wedge \psi(d)\}$ has $k$ many elements.

It follows that $\displaystyle \bigcap_{i=1,\dots,n+1} C_i$ is nonempty, since $D$ is infinite and the intersection excludes at most $k \cdot (n+1)$ elements of $D$.

Let $\displaystyle \hat c \in \bigcap_{i=1,\dots,n+1} C_i$.

By definition of each $C_i$, this means that $\mathcal{M}\models \psi(\hat c)$ and $\mathcal{M}\models \phi(b_i, \hat c)$ for $i=1,\dots,n+1$.

But this is a contradiction, since the definition of $\psi$ gives us that $\mathcal{M}\models \phi(d, \hat c)$ for only $n$ many $d\in D$.

Thus $\{d\in D : \mathcal{M}\models \phi(b, d) \wedge \psi(d)\} = \{m\in \mathcal{M} : \mathcal{M}\models \phi(b, m)\wedge \psi(m)\}$ is finite.

Since $\mathcal{M}\models \phi(b, c) \wedge \psi(c)$, this means that $c$ is definable over $A\cup \{b\}$.

$\blacksquare$