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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.

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    $\begingroup$ For it to be Galois, it should also contain $\exp(2i\pi/3)$ $\endgroup$ Oct 19 '13 at 19:38
  • $\begingroup$ This was my intuition, but I was not sure how to prove that fact rigorously. $\endgroup$ Oct 19 '13 at 19:39
  • $\begingroup$ Both $5^{1/3}$ and $5^{1/3}\exp(2i\pi/3)$ are roots of $x^3-5$, so they are both in the Galois extension. $\endgroup$ Oct 19 '13 at 19:43
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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.

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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 $\Bbb Q(2^{1/5})$.

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  • $\begingroup$ Can't argue with that analogy. $\endgroup$
    – Dan Rust
    Oct 19 '13 at 20:42
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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 ...

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  • $\begingroup$ would you like to explain why $\sigma(5^{\frac{1}{3}})$ is a root of $x^3-5$? $\endgroup$
    – MANI
    Oct 4 '19 at 20:05

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