Show that $\mathbb{Q}(2^{1/4}) / \mathbb{Q}$ is not Galois; prove that $\mathbb{Q}(2^{1/4}) / \mathbb{Q}(2^{1/2})$ is Galois I would like to show that $\mathbb{Q}(2^{1/4}) / \mathbb{Q}$ is not Galois.
Can I just say that it is not separable because $2^{1/4} \in \mathbb{Q}(2^{1/4})$ but its minimal polynomial in $\mathbb{Q}$ is $x^4 - 2$, which is not separable in $\mathbb{Q}$?
Meanwhile, $\mathbb{Q}(2^{1/4}) / \mathbb{Q}(2^{1/2})$ is an abelian extension and therefore its Galois group is normal, therefore $\mathbb{Q}(2^{1/4}) / \mathbb{Q}(2^{1/2})$ is Galois, right?
 A: Separable means that the polynomial has distinct roots, which $x^4-2$ does have. Note however, that for the extension to be Galois you need, separable and normal. This extension is not normal. One of the characterizations for normality is that $K/F$ is normal if for every $f(x)\in F[x]$ such that $K$ contains a root of $f$, then $f$ splits over $K$. 
This doesn't work here. Which roots of $x^4-2$ are you missing in $\mathbb{Q}(2^{1/4})$?. Also yes, the extension in the second paragraph is normal. 
By the way, in case you don't know this: every extension over a field of characteristic zero will be separable. Hence when you are working with this examples you have to check for normality. 
A: Since $\mathbb Q(2^{1/4})/\mathbb Q$ is a finite (hence algebraic) extension of a field of characteristic zero, it is separable.
Here is a way to see that $K = \mathbb Q(2^{1/4})$ is not Galois.
If $K$ were Galois, then $K$ would be invariant under any embedding $K \to \mathbb C$.
However, the field $K$ has 2 real embeddings $\sqrt[4]{2} \mapsto \pm \sqrt[4]2$ and 1 pair of complex embeddings $\sqrt[4]{2} \mapsto \pm i \sqrt[4]{2}$.
In general, a Galois extension $K/\mathbb Q$ must have all of its embeddings be either real or complex, not a mix of the two. You can tell how many real embeddings $K$ has by counting the number of real roots of the polynomial of a primitive element. In your example, the polynomial $x^4-2$ has two real roots $\pm \sqrt[4]2$ and two complex roots $\pm i \sqrt[4]2$, so adjoining a root of this polynomial to $\mathbb Q$ does not give a Galois extension.
