Is the number of orbits of the automorphism group of infinite field with a finite characteristic acting of the field is finite?

Question

I am trying to solve some statement in Model theory. And if i can show that given an infinite field $\mathbb{F}$ with a finite character, then the number of orbits of $Aut(\mathbb{F})$ acting on $\mathbb{F}^n$ is finite, it will solve my problem. so any help will be appreciated!

Also if there is a canonic form for such a group it will help to check it by myself if you can tell me what it is...

I meant to write infinite number of orbits not finite.

Answer 1

Given an infinite structure $M$, we say that $\text{Aut}(M)$ is oligomorphic if the action of $\text{Aut}(M)$ on $M^n$ has finitely many orbits for all $n\in \omega$.

Now if $\text{Aut}(M)$ is oligomorphic, then there are only finitely many first-order $n$-types relative to $T = \text{Th}(M)$ for all $n\in \omega$, and as a consequence $T$ is countably categorical and there are only finitely formulas in $n$ free variables up to equivalence modulo $T$. This is what's called the Ryll-Nardzewski theorem.

I claim that no infinite field $F$ has a countably categorical theory $T$. Indeed, since $F$ is infinite, we can find an elementary extension $F'$ of $F$ containing an element $a$ which is transcendental over $F$ (use Löwenheim-Skolem, or write down the type $\{p(x) \neq 0\mid p\in F[x]\}$ and use compactness). Then $F'$ contains distinct elements $\{a^n\mid n\in \omega\}$, and we have infinitely many formulas in two free variables, $\{y = x^n\mid n\in\omega\}$, which are pairwise inequivalent modulo $T$ (since they have different realizations in $F'$, witnessed by the pairs $(a,a^n)$).

So no field satisfies the condition in your question, and the characteristic is irrelevant.

Answer 2

This is usually not true (in fact, off the top of my head, I don't know how to construct any example where it is true). For instance, if $\mathbb{F}$ is the algebraic closure of $\mathbb{F}_p$, then every automorphism maps $\mathbb{F}_{p^n}$ to itself for each $n$, so $\mathbb{F}_{p^n}\setminus\bigcup_{d\mid n, d<n}\mathbb{F}_{p^d}$ is a union of orbits for each $n$. This set is nonempty for every $n$, so there are infinitely many orbits. For another example, every automorphism of $\mathbb{F}_p(x)$ preserves the degree of rational functions, so there are infinitely many orbits (at least one for each degree).