If $a = \sqrt{1+\sqrt{2}}$ and $E = \mathbb{Q}(a)$, find the minimal polynomial of $a$ over $\mathbb{Q}$, $[E:\mathbb{Q}]$, and $Gal(E / \mathbb{Q})$. I have found the minimal polynomial for this problem to be $m(x) = x^4 -2x^2 -1$. So then $[E:\mathbb{Q}]$ is $4$ because that's the degree of the minimal polynomial. The next step (finding the Galois group) is where I'm struggling. My first instinct was to look at the roots and what they would map to, those being:
$\sqrt{1+\sqrt{2}} \rightarrow \sqrt{1+\sqrt{2}}$ and $\sqrt{1+\sqrt{2}} \rightarrow -\sqrt{1+\sqrt{2}}$
So I thought that the Galois group would have order $2$ leading me to believe it was $\mathbb{Z}_2$, however I remembered that $|Gal(E/\mathbb{Q})| = [E:\mathbb{Q}]$. So I knew something was wrong. Any tips on how to go about this problem from here? 
Thanks in advance for your help!
 A: In this case the splitting field is generated by $\sqrt{1+\sqrt2}$
and $\sqrt{1-\sqrt2}$. These generate a degree $4$ extension
of $\Bbb Q(\sqrt2)$ (why?). There will be Galois automorphisms taking
$\sqrt{1+\sqrt2}$ to $\pm\sqrt{1-\sqrt2}$.
A: Sorry about my earlier mistake (in a deleted comment), but on revisiting this page I noticed that there's something worth adding. Your question asks for the Galois group for $E : \mathbb Q$, not the Galois group for the splitting field of $x^4 - 2x^2 - 1$ over $\mathbb Q$. Since the extension $E : \mathbb Q$ is not normal, the statement $|{\rm Gal}(E : \mathbb Q)| = [E: \mathbb Q]$ is false, and your claim that $|{\rm Gal} (E : \mathbb Q) | = 2$ is correct!.
I'll explain in more detail, but first, we should agree on some notation:
$$ a = \sqrt{1 + \sqrt{2}},  \ \ \ \ \ \ b = \sqrt{1 - \sqrt{2}}.$$
As Lord Shark explained, the splitting field of $x^4 - 2x^2 - 1$ is not $E$. It  is actually $E(b)$, which is strictly larger than $E$ since $b$ is imaginary whereas everything in $E$ is real. Therefore, $E : \mathbb Q$ is not a normal extension.
[One characterisation of what it means for an extension $L : K$ to be normal is that every irreducible polynomial $p(x) \in K[x]$ with one root in $L$ splits completely over $L$. Here, the polynomial $x^4 - 2x^2 - 1$ is irreducible in $\mathbb Q[x]$, but it has two roots $\pm a$ in $E$ and two roots $\pm b$ outside of $E$.]
Now for the key point: given a general finite field extension $L : K$, it is not true that $|{\rm Gal}(L : K)| = [L:K]$! This equality holds if and only if the extension $L : K$ is normal and separable, but in our example, $E : \mathbb Q$ is not normal.
[The correct general statement is that $|Gal(L:K)| = [L : \Phi({\rm Gal}(L:K))]$, where $\Phi({\rm Gal}(L:K))]$ is the subfield of $L$ consisting of those elements that are fixed by all automorphisms in ${\rm Gal}(L:K))]$. Only in the case where $L : K$ is normal and separable do we have $\Phi({\rm Gal}(L:K)) = K$.]
Anyway, you're right that $|{\rm Gal}(E : \mathbb Q)| = 2$. Since $E = \mathbb Q (a)$, every automorphism in ${\rm Gal}(E : \mathbb Q)$ is uniquely determined by where it sends $a$. As you pointed out, the image of $a$ under any automorphism in ${\rm Gal}(E : \mathbb Q)$ must be either $\pm a$. This is because automorphisms in ${\rm Gal}(E : \mathbb Q)$ must map roots of $x^4 - 2x^2 - 1$ to roots of $x^4 - 2x^2 - 1$. Furthermore, there really does exist an automorphism sending $a \mapsto -a$. (This is more obvious if you think of $\mathbb Q(a)$ as $\mathbb Q[x]/(x^4 - 2x^2 - 1)$, with $x$ representing $a$.)
Finally, let's check that the formula $|Gal(E:\mathbb Q)| = [E : \Phi({\rm Gal}(E:\mathbb Q))]$ holds here. The subfield $\Phi({\rm Gal}(E:\mathbb Q))$, which by definition consists of all elements of $E$ fixed under $a \mapsto \pm a$, is the subfield $\mathbb Q(a^2) = \mathbb Q(\sqrt 2)$. And yes, $[E : \mathbb Q(\sqrt 2)] = 2 = |{\rm Gal}(E : \mathbb Q)|$, as expected.
