Galois-property extends across different roots?

Question

Suppose the extension $\mathbb{Q}(a)/\mathbb{Q}$ is Galois for some root $a$ of the polynomial $f(x)$. If $b$ is another root of $f(x)$, is $\mathbb{Q}(b)/\mathbb{Q}$ also Galois? I'm thinking it should be since $\mathbb{Q}(a) \cong \mathbb{Q}(b)$

Answer 1

Yes. One can use the mentioned isomorphism (isomorphism preserves normality and separability) or also observe that the assumption of normality gives $\mathbb Q(b)\subseteq \mathbb Q(a)$ while the fact that they are conjugate means they have the same degree over $\mathbb Q$, which forces equality. So not only is $\mathbb Q(b)$ also Galois, it equals $\mathbb Q(a)$.