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Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)

Now, $\Bbb Q(\sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $\Bbb Q(\sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $\Bbb Q$?

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  • $\begingroup$ Is your question about a general $K$, or specifically $K=\mathbb Q(\sqrt[3]2)$? $\endgroup$ – BallBoy Nov 19 '18 at 14:39
  • $\begingroup$ Specifically $K=\Bbb Q(\sqrt[3]{2})$. let me edit $\endgroup$ – Eric Nov 19 '18 at 14:57
  • $\begingroup$ $\Bbb Q(\sqrt[3]{2})/\Bbb Q(\sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x \in \Bbb Q(\sqrt[3]{2})[x]$. For extensions of $\Bbb Q$ when not specified we mean $K/\Bbb Q$ is a normal extension, thus the splitting field of some $p(x) \in \Bbb Q[x]$. $\endgroup$ – reuns Nov 19 '18 at 15:11
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No.

First, such a polynomial $p(x)$ would have to be irreducible over $\mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $\mathbb Q$.

If $p(x)$ is such an irreducible polynomial, then we have that $\mathbb Q (^3 \! \sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).

So $\mathbb Q (^3 \! \sqrt2)$ becomes the splitting field for an irreducible polynomial over $\mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.

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They'd all have to be $\notin \mathbb Q$, as otherwise $p$ would be reducible.

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