Galois extension of $\mathbb{Q}(i)$ with Galois group $D_8$

Let $\mathbb{Q}(i)$ be given, where $i^2=-1$. I want to know:

Whether there is a Galois extension $K/\mathbb{Q}(i)$ such that $\mathrm{Gal}(K/\mathbb{Q}(i))\cong D_8$, where $D_8$ is the dihedral group of size 8?

I know the Fundamental Theorem of Galois theory, and how to calculate Galois groups.

I tried several guesses, such as $\mathbb{Q}(2^{1/4},i)$, $\mathbb{Q}(2^{1/8},i)$, but none worked.

• Take a dihedral extension of the rationals not containing i and lift it. – franz lemmermeyer May 17 '17 at 14:47
• But @franzlemmermeyer, all the examples of extensions with group $D_8$ that we see in basic Galois Theory classes are $\Bbb Q(m^{1/4},i)$ over $\Bbb Q$. – Lubin May 17 '17 at 15:14
• I'm trying to prove $\mathbb{Q}(\sqrt{3+\sqrt{2}}, \sqrt{3-\sqrt{2}})/\mathbb{Q}$ has Galois group $D_8$ – MaudPieTheRocktorate May 17 '17 at 15:17
• Nevermind, I found one. $\mathbb{Q}(\sqrt{3+5\sqrt{2}}, \sqrt{3-5\sqrt{2}})/\mathbb{Q}$, the splitting field of $x^4-6x^2-41$. – MaudPieTheRocktorate May 17 '17 at 16:34
• um... what's wrong with $\Bbb Q(2^{1/4}, i)$ ?? – mercio May 18 '17 at 9:30

Instead, you could think only about the Galois groups. You can use the Fundamental Theorem of Galois Theory to show that such an extension exists if and only if there exists a group $G$ such that $G\cong D_8$ and $\operatorname{Gal}(K/\mathbb{Q}(i))$ is a normal subgroup of $G$. If you show that, and find such a group (up to isomorphism), then you're done.