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


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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.