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) $\mathbb{Q}(\sqrt{5}, \sqrt{7})$
(a) Clearly $\omega $ is the root of $x^2+x+1$, so if $\sigma$ is an automorphism of $\mathbb{Q}(\omega )$ in $\mathbb{Q}(\omega )$ then $\sigma$ is also the root of $x^2+x+1$, but the only root of this is $\omega $, then $[\mathbb{Q}(\omega ):\mathbb{Q}]=2$ and $|Gal(\mathbb{Q}(\omega )/\mathbb{Q})|=1$, where $\mathbb{Q}(\omega )$ is not galois on $\mathbb{Q}$
(b) $\sqrt[4]{2}$ is the root of $x^4-2$ and $x^4-2$ is irreducible in $\mathbb{Q}$ where $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=4$, in addition if $\sigma$ is an automorphism of $\mathbb{Q}(\sqrt[4]{2})$ then $\sigma$ also is the root of $x^4-2$, and as $x^4-2=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$ then $\sigma=\sqrt[4]{2}, -\sqrt[4]{2}, i\sqrt[4]{2}, -i\sqrt[4]{2}$ and thus $|Gal(\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q})|=4$, we conclude that $\mathbb{Q}(\sqrt[4]{2})$ is Galois copper $\mathbb{Q}$.
(c) I do not know how to do this, someone could explain to me how to find an irreducible polynomial that has as roots $\sqrt{5}$ and $\sqrt{7}$, I know that for $\sqrt{5}$ is $x^2-5$ and for $\sqrt{7}$ is $x^2-7$ but I do not know how to make this case. Thank you very much.
(a) every quadratic extension is Galois (except maybe in characteristic two).
(b) $\sqrt[4]2\in \Bbb R $ but its conjugate $i\sqrt[4]2\notin \Bbb R $.
(c) A compositum of Galois extensions is Galois.