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Question. Let $f$ be the minimal polynomial of $\sqrt{1+\sqrt{7}}$ over $\Bbb Q$. Then, let $L$ be its splitting field. Show that $\text{Gal}(L/\Bbb Q)$ is isomorphic to a dihedral group of order $8$.

So here's my attempt.

So firstly, we find the minimal polynomial. We have $$x=\sqrt{1+\sqrt{7}}\Rightarrow x^2=1+\sqrt{7}\Rightarrow x^2-1=\sqrt{7}\Rightarrow x^4-2x^2-6=0.$$ This is a monic polynomial that is irreducible by Eisenstein at $p=2$ and so, this is indeed the minimal polynomial. It follows that the splitting field over $\Bbb Q$ is $\Bbb Q(a,b)$ where $a=\sqrt{1+\sqrt{7}}$ and $b=\sqrt{1-\sqrt{7}}$. Now, this polynomial has four roots, namely $$x_1=\sqrt{1+\sqrt{7}},\quad x_2=-\sqrt{1+\sqrt{7}},\quad x_3=\sqrt{1-\sqrt{7}},\quad x_4=-\sqrt{1-\sqrt{7}}.$$ Now, two generators of this Galois group are $r$ and $s$, where $r$ fixes the sign of $\sqrt{7}$ and changes the sign of the outer part, with $s$ changing the sign of $\sqrt{7}$. In other words, we have $$r(x_1)=x_2,\quad r(x_2)=x_1,\quad r(x_3)=x_4,\quad r(x_4)=x_3,$$ $$s(x_1)=x_3,\quad s(x_2)=x_4,\quad s(x_3)=x_1,\quad s(x_4)=x_2.$$ As a permutation, we have $r$ acts on the roots like $(1 2)(3 4)$ and $s$ acts on the roots as $(13)(2 4)$. As the composition of two automorphisms is an automorphism, we must also have $rs=(1 4)(2 3)$.

Now of course the problem with this is that I've only got four permutations as opposed to the eight that the question seeks. I also have two permutations of cycle type $2,2$ as opposed to a $4$-cycle which a dihedral group of order $8$ must have. So it's clear to me that I've gone wrong, perhaps in the defining of my $r$ and $s$? But I can't quite see where I've gone wrong - I know my naming of $r$ and $s$ isn't too technical at this point but not quite sure as to what it can be. I did a similar example to this for $g(x):=x^4-2$ which I found to be a lot more straightforward as the roots involved $i$ so it was a lot clearer to me as to how I should define $r$ and $s$.

Anyway, thanks in advance for taking your time out to read this. Any clarification will be much appreciated.

thesmallprint
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    The extension is not Galois, we need the Galois closure $L$, see here. – Dietrich Burde May 28 '20 at 16:21
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    I think the minimal polynomial is rather $X^4-2X^2\color{red}{-6}$.. – rae306 May 28 '20 at 16:26
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    A priori, it is not clear that the $r$ and $s$ you write down actually exist as automorphisms in the Galois group. How do you justify this? – rae306 May 28 '20 at 16:45
  • @DietrichBurde Thank you, I will have a look at the post for sure. – thesmallprint May 28 '20 at 16:47
  • @rae306 As for the automorphisms, admittedly I was a bit skeptical myself - I was trying to closely follow this document where he tackles similar examples however fell a bit short admittedly with this one. Also thanks for the correction in the minimal polynomial. – thesmallprint May 28 '20 at 16:48
  • By the way, Dietrich Burde's comment is not correct. If $L$ is the splitting field of your polynomial, $L/\mathbf{Q}$ is Galois. – rae306 May 28 '20 at 16:54
  • I'm fairly sure that this is covered in this old thread as well as many others linked to it. – Jyrki Lahtonen May 28 '20 at 17:02
  • @thesmallprint I refer you to an answer of mine here: https://math.stackexchange.com/questions/3239143/galois-group-of-x4-2x2-6-generators – rae306 May 28 '20 at 17:06
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    The problem is that in order to show the existence of this "rotation" of order 4 and "reflection" of order 2, we must know that the Galois group is isomorphic to $D_4$ beforehand. This is always the same trick: (1) prove that $[L:\mathbf{Q}]=8$, (2) see that the only subgroup of $S_4$ of order 8 is $D_4$. – rae306 May 28 '20 at 17:08
  • The root $x_1$ is real, so we know that $x_3\notin \Bbb{Q}(x_1)$. Therefore the splitting field is a degree $8$ extension. You simply haven't found all the automorphifsms, yet. – Jyrki Lahtonen May 28 '20 at 17:13
  • My guess is that Dietrich Burde's comment is based on the fact that some biquadratic polynomials have $C_4$ as the Galois group. In that case your $r$ and $s$ won't come from automorphisms in spite of superficial similarity of the set of roots. See this oldie for a different look at the cyclic case. – Jyrki Lahtonen May 28 '20 at 17:16
  • Thank you for all your help. I have had a look at your answer rae306 and it is very helpful in determining the automorphisms. Also thank you Jyrki Lahtonen for pointing out that I haven't actually found the automorphisms. I guess this is where my confusion lies - in some cases we get that the Galois group is isomorphic to 4 permutations of $S_4$ whereas in this case we get $8$ permutations in $S_4$. Having found this it seems there are some subtleties which I seemingly haven't grasped yet. – thesmallprint May 28 '20 at 17:30

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