Algebraic extension of $\Bbb Q$ with exactly one extension of given degree $n$

Question

Let $n \geq 2$ be any integer. Is there an algebraic extension $F_n$ of $\Bbb Q$ such that $F_n$ has exactly one field extension $K/F_n$ of degree $n$?

Here I mean "exactly one" in a strict sense, i.e. I don't allow "up to (field / $F$-algebra) isomorphisms". But a solution with "exactly one up to field (or $F$-algebra) isomorphisms" would also be welcome.

I'm very interested in the case where $n$ has two distinct prime factors.

---

My thoughts:

1. This answer provides a construction for $n=2$. I was able to generalize it for $n=p^r$ where $p$ is an odd prime. Let $S = \left\{\zeta_{p^r}^j\sqrt[p^r]{2} \mid 0 \leq j < p^r \right\}$. Then $$\mathscr F_S = \left\{L/\Bbb Q \text{ algebraic extension} \mid \forall x \in S,\; x \not \in L \text{ and } \zeta_{p^r} \in L \right\} =\left\{L/\Bbb Q \text{ algebraic extension} \mid \sqrt[p^r]{2} \not \in L \text{ and } \zeta_{p^r} \in L \right\} $$ has a maximal element $F$, by Zorn's lemma.

In particular, we have $$ F \subsetneq K \text{ and } K/\Bbb Q \text{ algebraic extension} \implies \exists x \in S,\; x \in K \implies \exists x \in S,\; F \subsetneq F(x) \subseteq K $$ But $X^{p^r}-2$ is the minimal polynomial of any $x \in S$ over $F$ : it is irreducible over $F$ because $2$ is not a $p$-th power in $F$. Therefore $F(x)$ has degree $p^r$ over $F$ and using the implications above, we conclude that $F(x) = F(\sqrt[p^r]{2})$ is the only extension of degree $p^r$ of $F$, when $x \in S$.

2. Assume now that we want to build a field $F$ with the desired property for some $n=\prod_{i=1}^r p_i^{n_i}$. I tried to do some kind of compositum, without any success. I have some trouble with the irreducibility over $F$ of the minimal polynomial of some $x \in S$ ($S$ suitably chosen) over $\Bbb Q$...

3. I know that $\mathbf C((t))$ is quasi-finite and embeds abstractly in $\bf C$, so there is an uncountable subfield of $\bf C$ having exactly one field extension of degree $n$ for any $n \geq 1$.

Answer 1

If you choose a random $\sigma \in \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ and consider the fixed field $K\subset \bar{\mathbb{Q}}$ of $\sigma$, then $\bar{\mathbb{Q}}/K$ will be Galois. The Galois group $G$ will almost always be $\hat{\mathbb{Z}}$, the profinite completion of the integers, and this is a group that has exactly one finite index subgroup of each index. Then $K$ will solve your problem for each $n$. There will be infinitely many such fields.

The problem is that it is hard to write down any concrete element of the absolute Galois group (except complex conjugation, as lulu referred to), and then also hard to write down its fixed field. So I'm afraid this answer is very nonconstructive.

See here for details: https://mathoverflow.net/questions/273224/what-is-the-probability-of-generating-a-given-procyclic-subgroup-in-mathrmgal