Abelian Galois group of $f$ implies splitting is simple extensions by a root of $f$. Given an irreducible polynomial $f\in \mathbb{Q}[x]$ with Abelian Galois group, I would like to show that the splitting field $K$ over the rationals can be written as a simple extension $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $f$. 
I know that any finite extension can be written as a simple extension, but I fail to see how the commutativity of the Galois group allows this adjoined element to be a root of $f$. Can we use the fact that a normal subgroup of the Galois group has a fixed field that is Galois over $\mathbb{Q}$, and thus every intermediate field $\mathbb{Q}\subset E \subset K$ is Galois?
 A: As John Martin pointed out, you can take any root $\alpha \in K$ of $f$ and then you show that $\Bbb Q(\alpha)$ is Galois over $\Bbb Q$.
In particular, this means that every root of the minimal polynomial of $\alpha$ over $\Bbb Q$ is contained in $\Bbb Q(\alpha)$. Since $f$ is irreducible, the minimal polynomial of $\alpha$ over $\Bbb Q$ is actually $f$. Therefore, $\Bbb Q(\alpha)$ is the splitting field of $f$, which is $K$.
Why is $\Bbb Q(\alpha)$ is Galois over $\Bbb Q$ ? Well, you know that if $L/F$ is a Galois extension, then there is a correspondence between :


*

*the subextensions $M/F$ of $L$ such that $M/F$ is Galois

*the normal subgroups of $G = \text{Gal}(L/F)$


(If $H \trianglelefteq G\,$, then $\text{Gal}(L^H/F) = G/H\,$, where $L^H = \{ x \in L \;\mid\; \sigma(x)=x, \;\forall \sigma \in H \}$).
In our case, every subgroup of $\text{Gal}(K/\Bbb Q)$ is normal, so that every subextension $M/\Bbb Q$ of $K$ is Galois over $\Bbb Q$.

Here are some details about the correspondence.
Suppose that $F \subset M \subset L\;$ are field extensions, such that $L/F$ is Galois with abelian Galois group.
You want to prove that $M/F$ is Galois. It is not difficult to show that it is separable.
To show that it is a normal extension, let $\sigma \in \text{Gal}(L/F)$, $m \in M\;$ and let's show that $\sigma(m) \in M$.
Since $L/M$ is Galois, we know that $M=L^{\text{Gal}(L/M)}\;$, so it is sufficient to show that $\sigma(m)$ is fixed by every $\tau \in \text{Gal}(L/M) \subset \text{Gal}(L/F).\;$ 
You have $\tau(\sigma(m)) = \sigma(\tau(m)) = \sigma(m),\;$ because $\text{Gal}(L/F)$ is abelian and $\tau$ fixes $M$. QED.
