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So $f = x^3 +9x -2$, and $E$ is its splitting field. I need to show that there are exactly three intermediate field extensions $K$ such that $\mathbb{Q} \subset K$ is not normal.

By Descartes' rule of signs, $f$ has exactly one real root, which is positive. Thus it has two complex roots. Call them $\alpha$, $\omega_1$ and $\omega_2$. The minimal polynomial of $\alpha$ is $f$, while the minimal polynomial of both $\omega_1$ and $\omega_2$ is a second degree polynomial, say $g$. So $[E:\mathbb{Q}] = 6$. The only intermediate fields I can see are $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\omega_1)$ and $\mathbb{Q}(\omega_2)$. It's quite clear that $\mathbb{Q}(\alpha)$ is not a normal extension, but both $\mathbb{Q}(\omega_1)$ and $\mathbb{Q}(\omega_2)$ are the splitting fields of $g$, if I'm not mistaken, so they are normal.

I thought that $\mathbb{Q}(\alpha^2)$ might be one of the intermediate fields I'm looking for. This is where I'm scrambling for possible intermediate fields. I'm not even sure that this is a good approach to this problem. The actual roots of $f$ are quite hideous, so I stayed clear of them and just worked with them algebraically. Any help is much appreciated.

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Actually the intermediate extensions you are looking for are $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\omega_1)$ and $\mathbb{Q}(\omega_2)$. Your reasoning about $\mathbb{Q}(\omega_1)$ and $\mathbb{Q}(\omega_2)$ being normal is wrong, because you don't have the good minimal polynomial. Indeed, if $f$ is the minimal polynomial of $\alpha$ it is irreducible, and thus it is also the minimal polynomial of $\omega_1$ and $\omega_2$.

I think the simplest way to show that these extensions are the ones you are looking for is to use the fondamental theorem of Galois theory : (non)-normal extension are in bijection with (non)-normal subgroups of $Gal(E,\mathbb{Q})$. $Gal(E,\mathbb{Q})$ must contain an automorphism of order $3$ because its order is a multiple of $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$, and it also contains complex conjugation which is of order $2$; thus $Gal(E,\mathbb{Q})=S_3$. The non-normal subgroups of $S_3$ are each generated by a transposition of two roots, so the corresponding fixed subfields must each contain one root, and also have degree $3$; thus they must be equal to $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\omega_1)$ and $\mathbb{Q}(\omega_2)$.

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  • $\begingroup$ Oh my god. Of course $\mathbb{Q}(\omega_1)$ and $\mathbb{Q}(\omega_2)$ are not normal. I forgot about the modulus of the complex roots. Thank you, your answer really helped me out. $\endgroup$ – Auclair May 29 '16 at 15:11
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The key to this question is the Fundamental Theorem of Galois Theory. I suspect that trying to find the intermediate extensions will be a very tedious exercise; but you are only required to prove their existence!

As you quite rightly note, by Descartes's rule $f$ has precisely one real root and thus two non-real roots. Thus we write $f(x) = (x-\alpha)(x-\beta)(x-\overline{\beta})$, where $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{C}\backslash\mathbb{R}$. Also noting that $\alpha\in(0,1)$, and that $f$ is a cubic over $\mathbb{Z}$, we have that $f$ has no root in $\mathbb{Z}$ and is thus irreducible over $\mathbb{Z}$, and hence by Gauss's Lemma is irreducible over $\mathbb{Q}$, since it is monic.

Now a little group theory. The Galois group, $G:=\Gamma(E/\mathbb{Q})$, is naturally a subgroup of the symmetric group $S_3$, and a transitive subgroup as $f$ is irreducible. But $G$ also contains a element of order 2, coming from complex conjugation $\beta\mapsto\overline{\beta}$. The only transitive subgroup of $S_3$ containing an element of order 2 is $S_3$ itself. Thus $G\cong S_3$.

And now for the actual Galois Theory. As we are working over a field of characteristic zero, the splitting field $E / \mathbb{Q}$ must be a Galois extension. Thus by the Fundamental Theorem of Galois Theory, intermediate fields $M\subseteq E$ correspond bijectively to subgroups $H:=\Gamma(E/M)\leq G$, and the extension $M/\mathbb{Q}$ is Galois if and only if $H$ is a normal subgroup; i.e. $M/\mathbb{Q}$ is not Galois if and only if $H$ is a non-normal subgroup. Since $E/\mathbb{Q}$ is Galois, it is separable, and thus any intermediate extensions $E/M$ and $M/\mathbb{Q}$ are automatically separable. Thus $M/\mathbb{Q}$ is non-normal if and only if it is not Galois, if and only if $H$ is a non-normal subgroup of $G$. $G\cong S_3$ has precisely 3 non-normal subgroups, hence there are precisely 3 non-normal intermediate extensions $M/\mathbb{Q}$.

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