In an old exam of my Galois Theory class there is the following question which troubles me:

Let $p \neq 2$ be a prime number and $k \geq 1$ an integer. Give an example of a galois extension $L/K$ such that $Gal(L/K) = D_{2p^k}$ and $[K:\mathbb{Q}]<+\infty$.

My idea was to consider the $p^k$-th roots of unity on which $D_{2p^k}$ acts and then take $L=\mathbb{Q}(\mu _{p^k})$ and $K=\mathbb{Q}(\mu _{p^k})^{D_{2p^k}}$ which by (what we called in class) Artin's theorem would give us $Gal(L/K) = D_{2p^k}$.

What troubles me is that I want to stop here and say that I'm done but I haven't use the $p^k,p\neq2$ conditions (what i did would work all the same with $D_{2n}$ for all $n$) so I feel that I have very probably made a mistake.


$\mathbb{Q}(\mu _{p^k})$ is an abelian extension of $\mathbb{Q}$, i.e. Galois with abelian Galois group isomorphic to $(\mathbb{Z}/p^k\mathbb{Z})^\times$, so there are no dihedral groups to be found in there.

But the basic idea is correct: embed $D_{2p^k}$ into some other finite group that you know how to realise as a Galois group over $\mathbb{Q}$, and then realise this bigger Galois group and take $K$ to be the fixed field of $D_{2p^k}$. For example every finite group can be embedded into a suitable symmetric group.

Edit: If you want a more explicit description, embed $D_{2p^k}$ into $C_{p^k}\rtimes (\mathbb{Z}/p^k\mathbb{Z})^\times$, which is the Galois group of a polynomial of the form $x^{p^k}-a$, where $a\in \mathbb{Z}$ is $p$-power free. Then, you will be able to describe $K$ and $L$ absolutely explicitly.

Bonus: one can actually show that any $D_{2p^k}$ can be realised as a Galois group over $\mathbb{Q}$. There are several ways of doing this, all of them requiring much more machinery than you would expect in an undergraduate exam. But the basic idea is to take a quadratic field (let's actually say imaginary quadratic), and then produce an abelian extension on top of that with cyclic Galois group of order $p^k$, such that this extension turns out to be Galois over $\mathbb{Q}$ with dihedral Galois group. Such extensions are extensively studied in Iwasawa theory.

  • $\begingroup$ thanks a lot. A questions comes to me though, doesn't this work for any finite group ? $\endgroup$ – Zorba le Grec Dec 13 '12 at 14:06
  • $\begingroup$ @ortholle Yes, it does. I guess it depends on how explicitly you are supposed to describe the extension. See also my edit. $\endgroup$ – Alex B. Dec 13 '12 at 14:13
  • $\begingroup$ Yeah I just saw your edit, great answer :) thanks $\endgroup$ – Zorba le Grec Dec 13 '12 at 14:17

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.