Constants are Definable

Let $K/k$ be a geometric function field, i.e. a finitely generated extension of an algebraically closed field $k$. We claim that there exists a one-parameter formula $\Theta$ such that
$$K \models \Theta(a) \iff a \in k.$$

Let $\mathcal{P}^c(k)$ denote the set of finite subsets of $k$ of odd cardinality greater than $c$. Let $\mathcal{P} = \mathcal{P}^0(k)$ denote the set of finite subsets of $k$ of odd cardinality.

Pairing $K \times \mathcal{P}'(k) \to k[t].$

For every $S \in \mathcal{P}(k)$ we can associate a polynomial $P_S(t) = \prod_{a \in S} (t-a) \in k[t]$.

For $(S, x) \in  K \times \mathcal{P}(k)$ we can associate the polynomial $p_{S,x}(T)$ defined as

• $T^2 - P_S(x)$ if $\text{char} k \neq 2$;
• $T^2 - T - P_S(x)$ if $\text{char} k = 2$. 

If $p_{S,x}$ has a root in $K$, we say that $(S, x)$ is a good pair.

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There exists $\mathcal{P}^c$ that capture $k$.

Lemma. Let $K/k$ be geometric function field. Then there exists $c= c_{K/k}$ such that for all $x \in K$,
if $(S, x)$ is a good pair for some $S \in \mathcal{P}^c(k)$ then $x \in k$.

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Application. $k$ is definable.

Let $c= c_{K/k}$ be as in the above lemma. Let $S$ in $\mathcal{P}^c$ be any set of absolute algebraic elements.
L et $\Theta(a)$ be the following formula in the language of fields:
$$\exists T, p_{S,a}(T) = 0.$$

Proposition. $k = \{ x \in K \mid \Theta(x)\}.$

Proof.  Clearly by the above lemma, if $x$ is a root of $\Theta$ then $x \in k$.
Conversely, let $x \in k$. Then?
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Proof of Lemma.

Observation. Let $x \in K$ and $S$ be a finite subset of $k$. Then $P_{S,x}(T)$ has a root iff $K_S$ has a $k(x)$-embedding in $K$ where $K_S$ is an extension if $k(t)$ by the roots of $P_{S,t}(T)$ in $K$. This is equivalent to a dominant rational $k$-map $X \to \dashrightarrow C_S$ where $X\to k$ is a projective normal model of $K/k$ and $C_S \to \mathbb{P}^1_t$ is the normalization of $\mathbb{P}^1_t$ in the Galois extension $K_S/k(t)$.