MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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 $M=X(\sqrt{-1})$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.