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 $\Bbb Q$.

  • $\begingroup$ 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 $3$. $\endgroup$ – 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)) $.

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