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Suppose $p$ is a prime number, $p\equiv1$ mod $3$ and $\mathbb{Q}(\zeta_p)$ is the $p$-th cyclotomic extension.

Prove that $\mathbb{Q}(\zeta_p)$ contains only one subfield $L$ such that $[L : \mathbb{Q}]=3$

$[L : \mathbb{Q}]=3 \implies L=\mathbb{Q}^H$ for some subgroup $H$ of $G=Gal(L/\mathbb{Q})$ of order $3$. The only possible Galois group with this order is $A_3$, so this would correspond to only one subfield $L$ as required. Is this correct?

Prove that for any rational number $A$, $L$ is not isomorphic to $\mathbb{Q}(\sqrt[3]{A})$

Not sure for this. I think I would need to find the corresponding Galois group for $\mathbb{Q}(\sqrt[3]{A})$ and show that this would not be isomorphic to $A_3$ (the alternating group).

$X^3-A=0$

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    $\begingroup$ The Galois group is cyclic of order $ p-1 $. Now can you finish it? $\endgroup$ – user175531 May 31 '16 at 11:47
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    $\begingroup$ Notice that $A_3=\mathbb{Z}/(3\mathbb{Z})$ and that if $A\in\mathbb{Q}$, $\mathbb{Q}(\sqrt[3]{A})$ is not a Galois extension, since it does not contain $\omega=e^{2\pi i/3}$ that belongs to the splitting field of $X^3-A$. $\endgroup$ – Jack D'Aurizio May 31 '16 at 11:50
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the Galois group $Gal(\Bbb{Q}(\zeta_p))$ is isomorphic to $(\Bbb{Z}/p\Bbb{Z})^*,\times)$ this also isomorphic to $(\Bbb{Z}/(p-1)\Bbb{Z}),+)$, and this last group is cyclic of degree $(p-1)$, so for every divisor $m$ of $(p-1)$, there are only an sub group of $(\Bbb{Z}/(p-1)\Bbb{Z}),+)$ of order $m$ . The hypotheses that $3$ divide $p-1$ give only an sub group of the cyclic $Gal(\Bbb{Q}(\zeta_p))$ of order $\frac {p-1}{3}$, so the fixed sub field is unique and of degree 3 over $\Bbb{Q}$.

If degre of $(\Bbb{Q}(^3\sqrt{A})/\Bbb{Q})$ is 3 thene this is not Galoisienne, so deferent to $L$, in this case the spliting field of $X^3-A$ is isomorphic to $S_3$.

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Note that $Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \cong \left(\mathbb{Z}_p\right)^* \cong \mathbb{Z}_{p-1}$ for any prime number $p$.

So our galois group is cyclic of order $p-1$. Since $p \equiv 1$ mod 3, $3 | (p-1)$. I.e. $\exists k \in \mathbb{N}$ such that $3k = p-1$. Or put another way, $\frac{p-1}{3} = k \in \mathbb{N}$. Since cyclic groups have exactly one subgroup for each number dividing its order, the galois group has exactly one subgroup of order $\frac{p-1}{3}$, call it $N$. Let $L$ be the fixed field of $N$. Using the basic facts listed in the Galois correspondence, $[L:\mathbb{Q}] = 3$.

Now since our Galois group is cyclic, all subgroups are normal. So $N$ is a normal subgroup of the Galois group. Again using facts from the Galois correspondence, this tells you that $L$ is a normal extension of $\mathbb{Q}$. That is, any irreducible polynomial $f(x) \in \mathbb{Q}[x]$ that has a root in $L$, splits over $L$.

Now as Jack mentioned in the comments, if $A \in \mathbb{Q}$, then $x^3-A$ does not split over $\mathbb{Q}(\sqrt[3]{A})$. However if $\mathbb{Q}(\sqrt[3]{A})$ were equal to $L$, then $x^3-A$ would have to split over it, since $\sqrt[3]{A} \in \mathbb{Q}(\sqrt[3]{A})$ and $\sqrt[3]{A}$ is a root of $x^3-A$. So $\mathbb{Q}(\sqrt[3]{A}) \neq L$.

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