# “Small” subfields of algebraically closed fields

Sufficient background:

Let $\mathcal{M}=(M,...)$ be an $\mathcal{L}$-structure and $X\subset M$.

Definition. $X$ is large if there exists a function $f:\mathcal{M}^n \overset {\leq k} \rightarrow \mathcal{M}$ definable in $\mathcal{M}$ such that $f(X^n)=M$ for some $n$, $k$. Otherwise, $X$ is small.

Here, $f$ is a kind of multi-valued function which targets some subsets of $M$ of less than or equal to $k$ elements. But this point is not rather of such importance, as in algebraically closed fields we do not need functions to be that "k-valued" (due to a lemma).

My question is...

In $ACF$, when exactly do we have that a subfield is small?

E.g. In $(\mathbb{C}, +, \cdot , 0, 1)$, $\mathbb{R}$ is large since there exists a definable function $f:\mathbb{R}^2 \rightarrow \mathbb{C}$ sending $(x, y)$ to $x+iy$ and thus $f(\mathbb{R}^2)=\mathbb{C}$. However, $\mathbb{Q}$ is small due to the general fact(which can easily be seen) that if $|X| < |M|$ and $|M|$ is infinite, then $|X|$ is small.

-

Let $X$ be a large subfield of an ACF $M$.

By quantifier elimination, every such definable function is piecewise algebraic, that is, there is a $d$ such that every element of $M$ is algebraic of degree at most $d$ over $X$. In particular, $M$ is the algebraic closure of $X$.

In characteristic $0$, every finite extension of $X$ is simple, hence the condition implies that every finite extension of $X$ has degree at most $d$, whence $[M:X]\le d$. It is well known that this is possible only if $X=M$, or if $X$ is real-closed and $M=X(\sqrt{-1})$.

In characteristic $p$, let $X\subseteq K\subseteq M$ be the separable closure of $X$. Then $M$ is a purely inseparable extension of $K$, and using the property above, every $a\in M$ satisfies $a^{p^k}\in K$ for $p^k\le d$. Since $M$ is perfect, this implies $K=M$. So, $M$ is separable over $X$, and we can use Artin primitive element theorem just like in the characteristic $0$ case to conclude that $X=M$.

Thus, $X$ is large iff either $X=M$ or $X$ is real-closed and $M=X(\sqrt{-1})$.

The argument above tacitly assumed that $f$ is definable without parameters. If you allow parameters $a_1,\dots,a_n\in M$, the same conclusion holds with $X(a_1,\dots,a_n)$ in place of $X$. However, a purely transcendental extension of any field is never real-closed or algebraically closed, hence in fact $a_1,\dots,a_n$ have to be algebraic over $X$, and we can forget about them. So again, $M=X$ or $M=X(\sqrt{-1})$.

-