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Let $D_n$ be the dihedral group of order $2n$. Then all the quotients of $D_n$ are dihedral as well, and of the form $D_k$ with $k \mid n$. So for a field $K/\mathbb{Q}$ with $\operatorname{Gal}(K/\mathbb{Q}) \cong D_n$, there exists, for any $k \mid n$, a subfield $F \subseteq K$ with $\operatorname{Gal}(F/\mathbb{Q}) \cong D_k$.

My question is about the reverse question. Given a number field $F/\mathbb{Q}$ with $\operatorname{Gal}(F/\mathbb{Q}) \cong D_k$, is there a field $K \supset F$ such that $\operatorname{Gal}(K/\mathbb{Q}) \cong D_n$ for any $n$ a multiple of $k$?

I've been told that this is called the "Galois embedding problem" and is not true for many types of groups. I was wondering if anyone could point me in the right direction for what is known about this in the dihedral case.

Thanks, MC

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The answer is "no", in general, since there may be local obstructions. Suppose, for example, that $k$ and $n$ are odd prime powers, and let $L/\mathbb{Q}$ be the unique intermediate quadratic in $F$. A necessary condition for the existence of $K$ is that the cyclic degree $k$ extension $F/L$ embeds into a cyclic degree $n$ extension $K/L$. Every prime $\mathfrak{p}$ of $L$ that is totally ramified in $F$ would have to be totally ramified in $K$, so if the residue characteristic of $\mathfrak{p}$ is coprime to $n$, then you need $n$ to divide the order of the multiplicative group of the residue field of $\mathfrak{p}$. This is a genuine restriction. For concreteness, take $k=3$, $n=9$. Then there are infinitely many $D_3$ extensions of $\mathbb{Q}$ in which $7$ is split in the quadratic and ramified in the cubic, but none of them embed inside a $D_{9}$ extension, because $7$ cannot be totally tamely ramified of degree $9$.

You can upgrade such local conditions to also ensure that the bottom cyclic group of order $2$ normalises the top group and acts on it by $-1$, so that the whole extension is dihedral. If all such local conditions are satisfied, then you can prove with class field theory that the sought-for embedding always exists. In particular if $k$ and $n/k$ are coprime, then the embedding will always exist. See [1, Section 3.1] for some examples of how such constructions work.

[1] A. Bartel, Large Selmer groups over number fields, Math. Proc. Cambridge Philos. Soc. 148 no. 1 (2010), pp. 73–86.

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  • $\begingroup$ Thanks, that makes a lot of sense. Thank you for the reference as well, that's precisely what I was looking for. $\endgroup$ – M C Jul 14 at 15:34
  • $\begingroup$ @MC: You are welcome! $\endgroup$ – Alex B. Jul 14 at 18:22

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