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

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 $n \geq 1$.

• What about the totally real subfield of $\overline {\mathbb Q}$? That has a unique extension of degree $2$. – lulu Feb 2 '17 at 13:41
• @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. – Watson Feb 2 '17 at 13:57
• 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 $n$? – lulu Feb 2 '17 at 14:01
• @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? – Watson Feb 2 '17 at 14:06
• 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). – 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.
• (Just for clarity, I denoted by $F = F_n$ what you wrote $K$ in your answer). – Watson Aug 17 '18 at 14:08