Galois extensions over $\mathbb{Q}$. 
Determine whether the following fields are Galois over $\mathbb{Q}$.
(a) $\mathbb{Q}(\omega )$, where $\omega  = exp(2\pi  i/3)$.
(b) $\mathbb{Q}(\sqrt[4]{2})$.
(c) $\mathbb{Q}(\sqrt{5}, \sqrt{7})$.

I have already shown that (a) is Galois over $\mathbb{Q}$, and (b) is not. For (c), I know that $\mathbb{Q}(\sqrt{5},\sqrt{7}) = \mathbb{Q}(\sqrt{5} + \sqrt{7})$, but I couldn't conclude. Any hint?
PS:  Use only results of automorphism and Galois extensions.

Definition. Let $K$ be an algebraic extension of $F$. Then $K$ is Galois over $F$ if $F = \mathcal{F}(Gal(K/F))$.

$\mathcal{F}$ is the fixed field.

Proposition. Let $K$ be a field extension of $F$. Then $K/F$ is Galois if and only if $|Gal(K/F)|=[K:F]$.


Hint: The minimal polynomial of $\sqrt{5} + \sqrt{7}$ is $x^4-24x^2+4$. Now show that $\mathbb{Q}(\sqrt{5} + \sqrt{7})$ is the splitting field of this polynomial.
Edit: Since you can't use splitting fields, here is another way to do it.
We can check pretty easily that $K=\mathbb{Q}(\sqrt{5}, \sqrt{7})$ has degree $4$ over $\mathbb{Q}$ with basis $(1, \sqrt{5}, \sqrt{7}, \sqrt{35} )$.
Let $\sigma \in Aut(K/\mathbb{Q})$. Then $\sigma$ is uniquely determined by its action on $\sqrt{5}$ and $\sqrt{7}$.
If $\sigma(\sqrt{5})=a+b\sqrt{5}+c\sqrt{7} +d\sqrt{35}$, then we have
$$
5=\sigma(\sqrt{5})^2=(a+b\sqrt{5}+c\sqrt{7} +d\sqrt{35})^2
$$
By equating coefficients we see that $b^2=1$, $a=c=d=0$ and thus $\sigma(\sqrt{5})=\pm \sqrt{5}$.
By the same reasoning, $\sigma(\sqrt{7})=\pm \sqrt{7}$.
We then get $4$ distinct automorphism and thus $K$ is Galois.
