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I'm working on a big exercise from Dummit & Foote (p.584) with the end goal of constructing a polynomial with Galois group $Q_8$ (Quaternion group of order $8$). Take $$\alpha = \sqrt{(2+\sqrt{2})(3+\sqrt{3})}$$ I've shown $[\mathbb{Q}(\alpha):\mathbb{Q}] = 8$ and $\mathbb{Q}(\alpha)/\mathbb{Q}(\sqrt{2},\sqrt{3})$ is Galois. I can also use that $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ is Galois. Now, I'd like to show the Galois conjugates of $\alpha$ are given by $$\pm \sqrt{(2\pm\sqrt{2})(3\pm\sqrt{3})}$$ Obviously I could construct the minimal polynomial of $\alpha$ and from there the task is easy, but I was wondering if there's a more clever approach, i.e. can we use the information concerning the automorphisms in $\textrm{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}(\sqrt{2},\sqrt{3}))$ and $\textrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$ to obtain the conjugates. My intuition says yes, but of course the fact that being a Galois extension is not a transitive property may be a block in the road. Thanks.

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  • $\begingroup$ You'd want to look at $H_1 = \rm {Gal} (\mathbb Q(\alpha)/\mathbb Q(\sqrt 2))$ and $H_2 = \rm {Gal} (\mathbb Q(\alpha)/\mathbb Q(\sqrt 3))$, as they are both subgroups of $\rm{Gal}(\mathbb Q(\alpha)/\mathbb Q)$ - but only if this is something you have proven already. $\endgroup$ – TokenToucan Jan 16 '16 at 19:58
  • $\begingroup$ well, I haven't shown $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is Galois yet. Once I've shown these are indeed the Galois conjugates its easy to prove they lie in $\mathbb{Q}(\alpha)$ so we'll get that the extension will be Galois. $\endgroup$ – GiantTortoise1729 Jan 16 '16 at 20:36
  • $\begingroup$ If you can find $8$ distinct elements in $\Gal(\mathbb Q(\alpha)/\mathbb Q)$, then it will be Galois. $\endgroup$ – TokenToucan Jan 16 '16 at 20:44
  • $\begingroup$ Yes, I could do so, thanks, but this exercise seems to want me to prove its Galois by showing the extension is normal and separable. I was just wondering if there's a clever way to do it with the automorphisms I currently have. $\endgroup$ – GiantTortoise1729 Jan 16 '16 at 20:53
  • $\begingroup$ I got $144 - 288X^2 + 144X^4 - 24X^6 + X^8$ for the minimal polynomial, for what that’s worth. $\endgroup$ – Lubin Jan 17 '16 at 5:43
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$Gal((\sqrt2,\sqrt3):Q)$ is generated by $s(\sqrt2)=-\sqrt2, s(\sqrt3)=\sqrt3$, $t(\sqrt2)=\sqrt2, t(\sqrt3)=-\sqrt3$. Remark that $Gal(Q(\alpha):Q(\sqrt2,\sqrt3)=2$ so since the extension is separable, you two extension $s_1, s_2$ of $s$ and $t_1,t_2$ of $t$ to $Gal(Q(\alpha):Q)$.

You have $\alpha$ is the root of $X^2-(2+\sqrt2)(3+\sqrt3)$ so $s_i(\alpha), i=1,2$ is a root of $X^2 -(2-\sqrt2)(3+\sqrt3)$ this implies that $s_i(\alpha)\in\{\sqrt{(2-\sqrt2)(3+\sqrt3)},-\sqrt{(2-\sqrt2)(3+\sqrt3)}$

Use a similar argument to extend $Id$, $t$ and $st$ to $Gal(Q(\alpha):Q)$.

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