# Verification of Computation of Galois group

Let $f = x^8-10x^4 + 1$. We want the Galois group over $\mathbb{Q}$
The roots are $\pm \sqrt{ \pm \sqrt{5 \pm 2\sqrt6}}$. I'm not sure about the splitting field,but I think that it would just be $\mathbb{Q}(i, \sqrt{5 + 2\sqrt6})$, but I can't figure out if this would cover the root $\sqrt{5-2\sqrt6}$.
Then $|Gal_f{Q}| = [\mathbb{Q}(i, \sqrt{5 + 2\sqrt6}): \mathbb{Q}] = [\mathbb{Q}(i, \sqrt{5 + 2\sqrt6}) : \mathbb{Q}(\sqrt{5 + 2\sqrt6})][\mathbb{Q}( \sqrt{5 + 2\sqrt6}):\mathbb{Q}]$.
I claim that $[\mathbb{Q}(\sqrt{5 + 2\sqrt6}):\mathbb{Q}] = 8$ because of the minimal polynomial $f = x^8-10x^4 + 1$ and $[\mathbb{Q}(i, \sqrt{5 + 2\sqrt6}) : \mathbb{Q}(\sqrt{5 + 2\sqrt6})] = 2$ since the minimal polynomial x^2 + 1 is of degree 2. This means that $Gal_f(\mathbb{Q})| = 2\times8=16$ so the Galois group is isomorphic to some subgroup of $S_8$ of order $16$

Is this correct? Could someone please help me figure out what group of order $16$ it would be isomorphic to? Thanks

• @Arthur Yes I fixed it! Thank you – user559412 May 21 '18 at 3:02
• $i=\sqrt{1}=\sqrt{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}$ – sharding4 May 21 '18 at 3:14
• @sharding4 Are you saying that I don't need to include $i$ in the extension? Should the splitting field be $\mathbb{Q}(\sqrt{5 + 2 \sqrt6}, \sqrt{5-2\sqrt6})$ or am I still wrong? – user559412 May 21 '18 at 3:19
• Actually I was thinking that you could get $\sqrt{5-2\sqrt6}$ from something like $i/\sqrt{5+2\sqrt6}$, so you were correct about the splitting field being $\mathbb{Q}(i, \sqrt{5 + 2\sqrt6})$ – sharding4 May 21 '18 at 3:25
• @sharding4 Ok that's wonderful. Thank you. Do you have any idea if the rest of what I say is correct? – user559412 May 21 '18 at 3:26

Note that the Galois group is non-abelian as complex conjugation will not commute with any automorphism that sends $\sqrt{5 + 2\sqrt6} \to i\sqrt{5 + 2\sqrt6}$. Now $\sqrt{5 + 2\sqrt6} = \sqrt{\sqrt{2}+\sqrt{3}}=\alpha$, say, so we see that the splitting field contains at least 7 quadratic subfields, $\Bbb{Q}[\sqrt{6}],\Bbb{Q}[\sqrt{2}],\Bbb{Q}[\sqrt{3}],\Bbb{Q}[\sqrt{i}],\Bbb{Q}[\sqrt{-2}],\Bbb{Q}[\sqrt{-3}],\Bbb{Q}[\sqrt{-6}]$. I believe that limits us to only $D_4\times C_2$ as having the required 7 seven normal subgroups of order 8 (using the subgroup structure for groups of order 16 that can be found here).
If the Galois group is the direct product, we should also be able to show that using the result here. That is, we need to find 2 Galois subfields with intersection $\Bbb{Q}$ that generate the splitting field. For $D_4\times C_2$ we'll need to find a dihedral and a quadratic extension of $\Bbb{Q}$ contained in $\mathbb{Q}[i, \sqrt{5 + 2\sqrt6}]=K$.
The dihedral extension we want will turn out to be the splitting field of $x^4-2x^2-2$, one root of which is $\sqrt{1+\sqrt{3}}$. It's a tedious but not too difficult check that \begin{align}\frac{1}{4}\left(\alpha^7+\alpha^5-9\alpha^3-9\alpha\right)&= \frac{\alpha}{4}\left(\alpha^6+\alpha^4-9\alpha^2-9\right) \\ &= \frac{\alpha}{4}\left(11\sqrt{2} + 9\sqrt{3} +5 + 2\sqrt{6}-9\sqrt{2}-9\sqrt{3} -9\right) \\ &= \frac{\alpha}{4}\left(-4 + 2\sqrt{2}+ 2\sqrt{6} \right) = \frac{\alpha}{2}\left(-2 + \sqrt{2}+ \sqrt{6} \right) \\ &= \sqrt{\left(\sqrt{2}+\sqrt{3}\right)\left(3 - \sqrt{2} + \sqrt{3} - \sqrt{6}\right) } \\ &= \sqrt{1+\sqrt{3}} \end{align} $1+\sqrt{3}$ satisfies $g(x)=x^4-2x^2-2$ which is Eisenstein at 2, so it's clear that $1+\sqrt{3}$ isn't a square in $\Bbb{Q}[\sqrt{3}]$. Since $1-\sqrt{3}<0$ two of the roots of $g$ are complex and the degree of the splitting field, call it $F$, must then be 8. Thus the Galois group must be $D_4$. Now $F$ contains the quadratic subfields $\Bbb{Q}[\sqrt{3}],\Bbb{Q}[\sqrt{-2}]$ and $\Bbb{Q}[\sqrt{-6}]$, but any of the other 4 quadratic subfields of $K$ along with $F$ will generate $K$, so $\Bbb{Q}[i]$, say. Thus we get $\operatorname{Gal}(K/\Bbb{Q})\cong D_4\times C_2$ as desired.