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Hungerford defines a field, $E$ as being closed if $E=E''$ where $E'= \{ \sigma \in \mathrm{Aut}(F/K)|\sigma(u)=u \text{ for all } u\in E \} = \mathrm{Aut}(F/E)$ is a subgroup of $\mathrm{Aut}(F/K)$,
$F$ is an extension field of $K$.

I want to show that in the extension of $\mathbb{Q}$ by $\mathbb{Q}(x)$, the intermediate field $\mathbb{Q}(x^3)$ is not closed.

My Attempt. Let $u \in \mathbb{Q}(x) - \mathbb{Q}(x^3)$. Then I'd want to show that $u$ is fixed by $\sigma \in \mathrm{Aut}(\mathbb{Q}(x) / \mathbb{Q}(x^3))$. So I let $u=x^3+x$. Then $\sigma(u)=\sigma(x^3)+\sigma(x)=x^3+\sigma(x)$

Any help as usual would be appreciated.

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Thanks for the editing. – Nana Apr 1 '11 at 12:04
up vote 0 down vote accepted

I don't really understand the definition you wrote for closed field, because we always have that $Aut(F/E)=Stab(E)\leq Aut(F/K)$.

Any way, if $\sigma \in Aut(\mathbb{Q}(x)/ \mathbb{Q}(x^3))$ then we only need to ask where it sends x. Since x is the root of $t^3-x^3\in\mathbb{Q}(x^3)[t]$ then $\sigma$ must be send x to another root of this polynomial, but in this field the only root is x (because if you extend to the complex numbers then the roots are $x,wx,w^2x$ where w is a root of unit of order 3).

What we get is that the fixed field of $Aut(\mathbb{Q}(x)/ \mathbb{Q}(x^3))$ is actually $\mathbb{Q}(x)$ and not $\mathbb{Q}(x^3)$ as in the galois case.

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The definition is from Hungerford's book, page 245. The latex previewer is not working properly I think. But I totally understand what you did. Thanks. – Nana Apr 1 '11 at 1:39

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