Exchange Principle

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Theorem

Let $D$ be a strongly minimal set in $\MM$.

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

Let $b, c \in D$.

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

Proof

Let $\map D x$ 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 $\LL$-formulas with parameters from $A$, in addition to whatever other parameters are supplied.

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

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

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

Let $\map \psi y$ be the formula

$\ds \map D y \wedge \exists x_1 \cdots \exists x_n \paren {\paren {\bigwedge_{i \mathop = 1}^n \map D {x_i} \wedge \map \phi {x_i, y} } \wedge \forall z \paren {\map D z \wedge \map \phi {z, y} \rightarrow \paren {\bigvee_{i \mathop = 1}^n z = x_i} } }$

which asserts that:

$\set {d \in D : \MM \models \map \phi {d, y} }$ has $n$ many elements.

Note the dependence on the free variable $y$.

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

Aiming for a contradiction, suppose that $\set {d \in D: \MM \models \map \phi {b, d} \wedge \map \psi d}$ is infinite.

Since $D$ is strongly minimal, $\set {d \in D : \MM \models \map \phi {b, d} \wedge \map \psi d}$ is cofinite in $D$.

Thus $D \setminus \set {d \in D : \MM \models \map \phi {b, d} \wedge \map \psi d}$ has some finite $k \in \N$ many elements.

Define $\map \chi x$ to be a formula which asserts that $D \setminus \set {d \in D : \MM \models \map \phi {x, d} \wedge \map \psi d}$ has $k$ many elements.

This can be done similarly to how $\map \psi y$ was defined above.

$\chi$ cannot define a finite subset of $D$, since $\chi$ involves only parameters from $A$ and $\MM \models \map \chi b$, 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 $\MM \models \map \chi {b_i}$ for each $i = 1, \ldots, n + 1$.

For each $i = 1, \ldots, n + 1$, define:

$C_i = \set {d \in D : \map \phi {b_i, d} \wedge \map \psi d}$

Then each $C_i$ is cofinite in $D$, since $\MM \models \map \chi {b_i}$ and hence $D \setminus \set {d \in D : \MM \models \map \phi {b_i, d} \wedge \map \psi d}$ has $k$ many elements.

It follows that $\ds \bigcap_{i \mathop = 1}^{n + 1} C_i$ is nonempty, since $D$ is infinite and the intersection excludes at most $k \cdot \paren {n + 1}$ elements of $D$.

Let $\ds \hat c \in \bigcap_{i \mathop = 1}^{n + 1} C_i$.

By definition of each $C_i$, this means that $\MM \models \map \psi {\hat c}$ and $\MM \models \map \phi {b_i, \hat c}$ for $i = 1, \ldots, n + 1$.

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

Thus $\set {d \in D : \MM \models \map \phi {b, d} \wedge \map \psi d} = \set {m \in \MM : \MM \models \map \phi {b, m} \wedge \map \psi m}$ is finite.

Since $\MM \models \map \phi {b, c} \wedge \map \psi c$, this means that $c$ is definable over $A \cup \set b$.

$\blacksquare$