# Normal closure $L$ of $\mathbb{Q}(\sqrt{7})$ and structure of $Gal(L/\mathbb{Q})$

I think I have done (a) but I need some guidance on (b), if possible

(a). Find a normal closure $L$ of $\mathbb{Q}(\sqrt{7})$

To construct the normal closure I could adjoin the roots of $\alpha=\sqrt{7}$ over $\mathbb{Q}$. $\alpha$ is a root of $X^4-7$ which is irreducible by Eisensrein's criterion using $p=7$. Hence it is the minimal polynomial. The roots are: {${\alpha, i\alpha, -\alpha, -i\alpha}$}. These roots are contained in field $K=\mathbb{Q}(\alpha, i)$ so the normal closure is contained in $K$.

Take $E=\mathbb{Q}(\alpha, i\alpha, -\alpha, -i\alpha)$ and $i, \alpha \in E$ so $\mathbb{Q}(\alpha, i) \subset K$. So $K=E$ which is the normal closure.

(b). Describe the structure of the group $Gal(L/\mathbb{Q})$

$L=\mathbb{Q(\alpha, i)}$, so:

$Gal(\mathbb{Q(\alpha, i)}/\mathbb{Q})$ Need to look at automorphisms.

First look at $H=Gal(L/\mathbb{Q}(i))$ cyclic group generated by automorphism: $\tau_1(\sqrt{7})=i\sqrt{7}$

Also, $H=Gal(L/\mathbb{Q}(\sqrt{7}))$ cyclic group generated by automorphism: $\tau_{2}(\sqrt{7})=\sqrt{7}$

So I think the structure of $Gal(L/\mathbb{Q})$ will consist of some combinations of these automorphisms, but I am not sure how to get at these.

Would really appreciate your help, many thanks

Once you get a lot of experience, you will expect that the splitting field of a polynomial $X^4-a$ will have for its Galois group the dihedral group of order eight. (There are exceptions to this general expectation.) This is the group of symmetries of the square, including the reflections.
I’m sure you’ve noticed that there are a couple of automorphisms of your field in plain sight: first, complex conjugation, which interchanges $i$ and $-i$ but leaves the real fourth root of $7$ fixed; and second the automorphism that leaves $i$ fixed but sends $\sqrt7$ to $i\sqrt7$. I think you can take things from there.