# If $[\overline F : F] = \infty$, does $F$ have extension of degree $n$ for any $n \geq 1$?

Let $F$ be a field and assume that $[\overline F : F] = \infty$. Does it imply that for any $n \geq 1$, there is a field extension $K/F$ of degree $n$?

Notice that if $[\overline F : F] < \infty$ then $F$ is a real closed field so the answer is no (for $n > 2$). Moreover, it is not interesting to ask for extensions $\overline F/K$ such that $[\overline F : K] = [\overline K : K] = n$ because, by Artin-Schreier, this implies $n \leq 2$.

I know that there exist extensions $K/F$ with arbitrarily large degree (take $x_0 \in \overline F \setminus F$ then $K_0 = F(x_0)$ has finite degree over $F$, so we can find $x_1 \in \overline F \setminus K_0$, then $K_1=K_0(x_1)$ has finite degree over $F$ and so on).

Clearly, there are $F$-vector subspaces of $\overline F$ of dimension $n$ over $F$, but they might not be subfields.

I know that $L = \Bbb Q(\sqrt p \mid p \text{ prime}) / \Bbb Q$ has no sub-extension $K/\Bbb Q$ of degree $3$. But here $L$ is not the algebraic closure of $\Bbb Q$.

• Moreover, $L(i)/\Bbb Q$ has no subextension $L(i)/K$ of degree $3$, since $\prod_{n \geq 1} \Bbb Z/2\Bbb Z$ has exponent $2$, so has no subgroup of order $3$. – Watson Feb 3 '17 at 10:47

No. Let $p$ be any prime and let $F$ be any field whose absolute Galois group is the profinite completion $\bar{\mathbf Z}$, and consider the compositum $L$ of all finite extensions of $F$ of degree prime to $p$ in some fixed algebraic closure $\bar{F}$. Then, $\textrm{Gal}(\bar{L}/L) \cong \mathbf Z_p$, and any finite extension of $L$ has degree a power of $p$. Since the Galois group is infinite, it follows that the extension is also infinite.
Concrete examples include $F = \mathbb F_q$ for any prime $q$, and $F = \mathbb C((T))$.