Elementary Equivalence of Arithmetic Function Fields of General Type implies Isomorphism

Let $K$ and $L$ be arithmetic function fields, i.e. finitely generated field extensions over their prime subfields. Let $k$ and $l$ be their absolute subfields, resp. Suppose $K$ and $L$ are elementarily equivalent. Then:

• $k \cong l$ and $\text{td}(K/k) = \text{td}(L/l)$
•  they have the same prime subfield $F$.

Theorem. There exists a field embedding $\iota: K \to L$ such that $L$ is finite separable over $\iota(K)$.

Proof. 

Let $(t_1, \ldots, t_d)$ be a separating transcendence basis of $K/k$, i.e. $K$ is algebraic separable over the polynomial ring $k[t_1, \ldots, t_d]$. Since $k = \overline{F} \cap K$ is algebraic separable over $F$, we must have $K$ is algebraic separable over $F[t_1,\ldots, t_d].$ Moreover, as $K$ is finitely generated over  $F$, it must be finitely generated over $F[t_1,\ldots, t_d].$ Thus $K = F(t_1, \ldots, t_d)[x]$ for some $x \in K$.
Question: Is the above the correct description of $x$?


To define a field embedding $\iota: K \to L$, it suffices to define $\iota(t_i)$ and $\iota(x)$. The idea is as follows. We want a formula $\Psi(\underline{\psi}, y)$ that holds for $(\underline{t}, x)$. Thus the sentence $\exists (\underline{\xi}, z), \Psi(\underline{\xi}, z)$ holds in $K$, so as $K \equiv L$, it must also holds in $L$. The "roots" of this sentence will be taken to be images of $t_i$ and of $x$ under $\iota$.

From before, we know that the exists a formula $\Xi(\underline{xi})$ such that  $\Xi(\underline{xi})$ holds in $K$ iff $\xi_1, \ldots, \xi_d$ is a transcendental basis of $K/k$. So this formula cuts out the $t_i$.

On the other hand, let $P(\underline{T}) \in k[\underline{T}]$ be an irreducible polynomial such that $P(t_1, \ldots, t_d, x) = 0$ (so that $K$ is canonically isomorphic to the functional field of the affine irreducible $F$-variety $\text{Spec} F[\underline{T}]/(P).$

Question: Since $x$ is algebraic over $F$, we know that $k= F(x)$. So why not just let $P \in F[T]$ (one-variable) be the minimal polynomial of $x$ over $F$? 

Then we can define the formula $\Psi(\underline{\xi}, z)$ to be the conjunction of:

• $\Xi(\underline{\xi})$
• $P(\underline{\xi}, z) = 0$ (notice as $P$ has coefficient in $F$, it is a first order sentence as $F$ is definable).

As $\Psi$ has roots $(\underline{t}, x)$ in $K$, it must have roots $(\underline{u}, y)$ in $L$. Define $\iota: K \to L$ by $t_i \mapsto u_i$ and $x \mapsto y$.

Then $(u_1, \ldots, u_d)$ is a separable transcendence basis of $L/l$ (being roots of $\Xi$), so $L$ is separable over $l(u_1, \ldots, u_d)$ and in particular over $\iota(K)$. 

Question: Does $j$ map $k$ isomorphically to $l$?

Proposition. If additionally, $K$ is of general type, then $K \cong L$ as fields.

Proof. By symmetry, we also have field embedding $\iota': L \to K$.

Let $j = \iota'\circ \iota: K \to K$.

Recall that if $K/k$ is a function field of general type, then every field $k$-embedding $K \hookrightarrow K$ is an isomorphism. Be careful: even though $j$ is a field embedding, it might not be a function field embedding! (It might not be identity on $k$).  It suffices then to show that some power $j^n$ of $j$ is a function field embedding. This is because then, as $K$ is of general type, $j^n$ must be an isomorphism. But then $j$ and thus $\iota$ itself must be isomorphisms. 

Indeed, notice that $j$ maps $k$ isomorphically to itself. (Why?) However $k$ is either a number field or a finite field, so it has only finitely many automorphisms. Thus for some $n$, $j^n$ is indeed identity on $k$.