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I'm trying to determine how to show that the Galois group of $\mathbb{Q}(\sqrt[8]{2},i)/\mathbb{Q}(i\sqrt{2})$ is isomorphic to the quaternions. Since $i\sqrt{2}$ is the fixed field, I believe the imaginary component of the eighth root of unity must be fixed and that $\sqrt[8]{2}$ can only get mapped to either itself or its opposite. I also know that there must be three generators of the Galois group since the quaternions are generated by $i,j,k$ all of which have order $4$, and that there must be an automorphism "$-1$" which has order $2$. I believe all the automorphisms in the Galois group are determined by their action on $\sqrt[8]{2}$, and $i$. I have one automorphism $\sigma:\sqrt[8]{2}\mapsto \zeta\sqrt[8]{2},i\mapsto -i$ but I can't think of the other two generators or how $\sigma$ could relate to a "$-1$". Can anybody give me a hint on where to begin without giving me the answer?

Jeff M.
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    Possible duplicate of Prove that $\operatorname{Gal}(\mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{-2})) \cong Q_8$ – Stahl Dec 07 '16 at 03:53
  • Yes I looked at that question. I understand how the sigma mentioned in that question works, but I can't find the other two generators. I've been thinking about this problem for a while and have hit a road block. – Jeff M. Dec 07 '16 at 03:55
  • The automorphic image $\sigma(i)$ of $i$ obviously has the two usual alternatives, so to get all the eight you need four alternatives for the image $\sigma(u)$ of $u=\root8\of2$ (to a given image of $i$). Because $iu^4=i\sqrt2$ is in the fixed field knowing $\sigma(i)$ leaves you exactly four choices for $\sigma(u)$. – Jyrki Lahtonen Dec 07 '16 at 08:12
  • In other words, the second sentence may be the source of your difficulty. At least I don't see how that follows. Also, in the group $Q_8$ we have $ij=k$, so you don't need to worry about finding more than two generating automorphisms. Furthermore, any two non-commuting elements of $Q_8$ can play in the roles of $i$ and $j$. This is because the group $Q_8$ has a lot of automorphisms (inner and outer). That is not really needed here. As long as you locate two non-commuting automorphisms of order four you should be able to verify the $Q_8$ relations easily. – Jyrki Lahtonen Dec 07 '16 at 08:19
  • See http://math.stackexchange.com/a/2046653/300700 for a general method – nguyen quang do Dec 07 '16 at 14:21
  • Why does $\sigma:i \mapsto -i$? – Somatic Custard Jan 06 '18 at 16:25

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Consider the tower $$\mathbb{Q}(i\sqrt{2})\subseteq \mathbb{Q}(\sqrt{2}, i)\subseteq \mathbb{Q}(\sqrt[8]{2}, i)$$

Note that $\zeta_8\in \mathbb{Q}(\sqrt{2}, i)$. Now $$ \mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{2}, i)$$ is an extension of degree $4$ adding the fourth root of $\sqrt{2}$ and its Galois group is cyclic with generator $\sigma$ where

$$\sigma(\sqrt[8]{2})=i\sqrt[8]{2}$$ and of course lets state explicitly that $$\sigma(i)=i $$ Clearly $\sigma^4=e$.

Now define $$\tau(i)=-i\qquad \tau(\sqrt[8]{2})=\zeta_8 \sqrt[8]{2}$$

It follows that $\tau(\sqrt{2})=-\sqrt{2}$ and so $\tau$ fixes $i\sqrt{2}$

Thus it must be verified that

$$\tau^{-1}\sigma\tau =\sigma^{-1}$$ and $$\sigma^2=\tau^2$$

These relations follow by a straightforward calculation and I leave them to you. This verifies the generating relations of $Q_8$.

Rene Schipperus
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