Difficulty in Proving Elementary Equiv of Geometric Function Fields implies Isomorphism

We proved that elementary equivalence between arithmetic function fields implies isomorphism. We will explain why we cannot apply this proof directly to the case of geometric function fields.

Let $k$ and $l$ be algebraically closed fields. Let $K/k $ and $L/l$ be finitely generated extensions. Suppose $K$ and $L$ are elementarily equivalent. Then

• $k$ and $l$ have the same prime field $F$;
• $\text{td}(K/k) = \text{td}(L/l)$

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Let $(t_1, \ldots, t_d)$ be a separable transcendence basis of $K/k$.

Notice: Unlike in the arithmetic case, we cannot in general write $K = F(t_1, \ldots, t_d)[x]$. (Why?)

Suppose $K = k(t_1, \ldots, t_d)[x]$ and $P(\underline{T}) \in k[\underline{T}]$ is an irreducible polynomial such that $P(t_1, \ldots, t_d, x) = 0.$ (In particular, $K$ is the function field of the irreducible variety $\text{Spec} k[\underline{T}]/(P)$.)

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Cutting out $x$ by a polynomial with coefficient in $F$.

As $P$ does not have coefficients in $F$, we cannot use $P$ to cut out $x$. Our idea is to try to replace $P$ with a polynomial $Q$ with coefficients in $F$. 

Let $\alpha_i$'s be the coefficients of $P$, then $P$ is defined over $F[\underline{\alpha}] \subset k$, i.e. we can view it as an element of $F[\underline{\alpha}][\underline{T}].$ Now $F[\underline{\alpha}= F[\underline{Z}]/(\underline{f})$ for some finite system $\underline{f}$ of relations $f_j$ of the variables $Z_i$.
So we can view $P$ as the equivalence class of an irreducible (why?) polynomial $Q \in F[\underline{Z}, \underline{T}].$

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Cutting out $\underline{\alpha}, \underline{t}, x$ by a formula.

Let $\Psi(\underline{\zeta},\underline{\xi}, z)$ be the conjunction of

• $\Xi (\underline{\xi})$, i.e. $\underline{\xi}$ is a transcendental basis;
• $\underline{f}(\underline{\zeta})$;
• $P(\underline{\zeta}, z).$

Define $\iota: K \to L$.

As $\Psi$ has roots in $K$ and $K \equiv L$, $\Psi$ must have roots $(\underline{\beta}, \underline{u},y)$ in $L$. Thus we define $\iota: F[\underline{\alpha}, \underline{t}, x] \to L$ by

$\alpha_i \mapsto \beta_i$, $t_i \mapsto u_i$ and $x \mapsto y$.

Then as $\underline{u}$ is a root of $\Xi$, it must be a separable transcendence basis of $L/l$. 

What is the problem?

We want to extend $\iota$ to a function field embedding $K/k \to L/l$ where $K = k[t_0, \ldots, t_d][x]$. To this end, it suffices to extend 

$\iota': R=F[\underline{\alpha}] \to F[\underline{\beta}]$ to a field extension $k \to l$. 

The problem is:

1. Image of $\iota'$ might not lie in $l$
2. $\iota'$ might not be injective

Even when these two problems are resolved, we could only guarantee an embedding of $k_0 = \text{Frac} R$ into $l$. This would extend to a function field embedding $K_0/k_0 \to L/l$ where $K_0 = k_0(\underline{t},x).$

We do not know if (2) can be resolved, but (1) can be resolved by replacing $\Psi$ with a more complicated formula. This is because the constant  are definable. In other words, there exists a formula $\Theta(a)$ such that $\Theta(a)$ holds in $K$ iff $a \in k$.

Take $\Psi'$ to be $\Psi(\underline{\zeta},\underline{\xi}, x) \vee \Theta(\zeta_i).$

Then the roots $\beta_i$ have to all satisfy $\Theta(\beta_i)$ in $L$ and thus must lie in $l$.