Is $\mathbb{Q}(5^{1/3})$ a Galois extension over $\mathbb{Q}$? I am trying to prove or disprove that the simple extension $\mathbb{Q}(5^{1/3})$ is Galois over $\mathbb{Q}$.  
I suspect that this extension is not Galois, because an extension if Galois over $\mathbb{Q}$ if and only if it is the splitting field of an irreducible polynomial over $\mathbb{Q}$.  We know that $5^{1/3}$ is a root of $x^3-5$ over $\mathbb{Q}$, but $\mathbb{Q}(5^{1/3})$ is not the full splitting field of $x^3-5$.  However, I am not sure how to prove that this extension is in fact not Galois.
 A: If an extension is Galois it must be normal and $K/E$ being normal is equivalent to saying that if an irreducible polynomial in $E[x]$ has one root in $K$, then all its roots must lie in $K$.
So, to show that $\mathbb{Q}(5^{1/3})/\mathbb{Q}$ isn't Galois, all you need to do is find an irreducible polynomial with one root in your extension field that doesn't split, which it sounds like you have done.
A: Galois $\equiv$ Vegas. What happens $\in$ Vegas, stays $\in$ Vegas.
Proposition. Suppose $L/K$ is Galois and intermediate in $M/L/K$. Any $K$-automorphism of $M$ fixes $L$ setwise ((though rarely pointwise, which is a stronger condition)).
Proof guide. Let $L$ be the splitting field of $f(x)$ over $K$. That is, it is generated over $K$ by the full set of roots of $K$. Check that any $\sigma\in{\rm Aut}_KM$ fixes $K$ and fixes the set of roots of $f(x)$.
If $\alpha$ is a root of $f(x)$ over $\Bbb Q$ then $\Bbb Q(\alpha)$ sits inside the splitting field $S$ of $f(x)$. To check that $\Bbb Q(\alpha)$ is not Galois it suffices to find a conjugate of $\alpha$ not contained in $\Bbb Q(\alpha)$. You can do that with $\Bbb Q(2^{1/5})$.
A: A somewhat direct way: Note that if $\sigma : \mathbb{Q}(\sqrt[3]{5}) \to \mathbb{Q}(\sqrt[3]{5})$  were a morphism (fixing $\mathbb{Q}$), then $\sigma(\sqrt[3]{5})$ is also a root of $x^3-5$, however $\mathbb{Q}(\sqrt[3]{5})\subseteq {\mathbb R}$ and hence it can only be $\sigma(\sqrt[3]{5}) = \sqrt[3]{5}$, because the other roots are not real and so do not belong to $\mathbb{Q}(\sqrt[3]{5})$. This implies that $\sigma = id$ (why?), and so this is the only such morphism. Hence the fixed field of ${\text Gal}({\mathbb Q}(\sqrt[3]{5})/{\mathbb Q})$ is ...
A: All you need to realise is that the spliting field of $x^3-5$ is not contained in $\mathbb{R}$ and on the other hand $\mathbb{Q}(5^{\frac13}) \subset \mathbb{R}$.
