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 $x \in S$.

  1. 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 $\Bbb Q$...

  2. 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 $n \geq 1$.

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    $\begingroup$ What about the totally real subfield of $\overline {\mathbb Q}$? That has a unique extension of degree $2$. $\endgroup$ – lulu Feb 2 '17 at 13:41
  • $\begingroup$ @lulu : thank you. But $\Bbb R \cap \overline{\Bbb Q}$ has no extension of degree $n \geq 3$ (it is a real closed field), I think. $\endgroup$ – Watson Feb 2 '17 at 13:57
  • $\begingroup$ You fixed $n$ and then asked for an $F$ with a unique extension of degree $n$. Did you mean $F$ should have a unique extension of degree $n$ for all $n$? $\endgroup$ – lulu Feb 2 '17 at 14:01
  • $\begingroup$ @lulu : no, I don't necessarily want that (but of course, this would solve the problem – i.e. finding a quasi-finite field $F \subset \overline{\Bbb Q}$). But I want for every $n \geq 2$ to find some algebraic extension $F=F_n$... Does this make sense? $\endgroup$ – Watson Feb 2 '17 at 14:06
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    $\begingroup$ Oh, ok. Then my simple construction won't work. Artin (?) showed that the only elements of finite order in $Gal(\overline {\mathbb Q}/\mathbb Q)$ are the identity and complex conjugation (up to conjugation). I think that means that the example I gave is the only one of its kind (but I haven't thought that through carefully). $\endgroup$ – lulu Feb 2 '17 at 14:16

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

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  • $\begingroup$ I'm very late for accepting and commenting, sorry. Anyway, thank you for your answer! $\endgroup$ – Watson Aug 17 '18 at 14:07
  • $\begingroup$ (Just for clarity, I denoted by $F = F_n$ what you wrote $K$ in your answer). $\endgroup$ – Watson Aug 17 '18 at 14:08

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