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 $K_S/k(t)$.

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 $l$.  

 

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 $k$.