Show that $E=\mathbb{Q}(a)$

Question

Let $f(x) \in \mathbb{Q}[x]$ an irreducible polynomial with the splitting field $E$ and let the group $Gal(E/\mathbb{Q})$ be abelian. If $a$ is a root of $f(x)$ then $E=\mathbb{Q}(a)$.

Could you give me some hints how to show this??

Answer 1

Note that $Gal(E|\mathbb{Q}(a))$ is a subgroup of $Gal(E|\mathbb{Q})$. As $Gal(E|\mathbb{Q})$ is Abelian, $Gal(E|\mathbb{Q}(a))$ is a normal subgroup. So, by the fundamental theorem of Galois theory, $\mathbb{Q}(a)/\mathbb{Q}$ is a normal extension. Thus, all roots of the minimal polynomial of $a$, which happens to be $f(x)$ lie in $\mathbb{Q}(a)$. Thus, the splitting field of $f(x)$ is in fact $\mathbb{Q}(a)$ i.e $E = \mathbb{Q}(a)$. Hope this helps.