Why doesn't the fundamental theorem of Galois theory apply to the extension $\mathbb{Q}(\sqrt[5]{3})$? I understand that this field is not a splitting field for any polynomial, as it does not contain the roots of unity.  If we had something like $\mathbb{Q}(\sqrt[5]{3}, \gamma)$, where $\gamma$ is a primitive fifth root of unity, the theorem would apply, as then the field would be a splitting field of $x^5 - 3$.  So basically, we need a root of unity to be there, but I'm not sure how to put this in formal terms.  
 A: As mentioned in the comments, $\mathbb{Q}(\sqrt[5]{3}) \supset \mathbb{Q}$ is not a Galois extension because it is not a splitting field of $x^5-3$, the minimal polynomial of $\sqrt[5]{3}$. You need the extension field to be a Galois extension, or equivalently a normal and separable extension, for the fundamental theorem of Galois theory to apply. Your extension $\mathbb{Q}(\sqrt[5]{3}) \supset \mathbb{Q}$ is separable but not normal. To show this, you need only to find a single polynomial that is irreducible over $\mathbb{Q}$ and has a root in $\mathbb{Q}(\sqrt[5]{3})$, but doesn't split completely over $\mathbb{Q}(\sqrt[5]{3})$. The polynomial $x^5-3$ is your example of this.
You can also realize that the Galois correspondence can't really apply here be because since $\mathbb{Q}(\sqrt[5]{3}) \supset \mathbb{Q}$ is not Galois, if we were to consider $\operatorname{Aut}_{\mathbb{Q}} \mathbb{Q}(\sqrt[5]{3})$, the fixed field is of these automorphisms is more than just $\mathbb{Q}$ by definition so the corresponding towers of intermediate fields and subgroups of $\operatorname{Aut}_{\mathbb{Q}} \mathbb{Q}(\sqrt[5]{3})$ would be off. 
