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This is the setting I can use:

(P1) : If $q=\frac{r}{s}$ and $ q,r\in\mathbb{Q}$ then $s\in \mathbb{Q}$. This is because $\frac{r}{q} = \frac{s q}{q} =s \in \mathbb{Q}$ since $\mathbb{Q}$ is closed under division.

(P2) : If $F$ is a field and $p(x) \in F[x]$ is irreducible over $F$ and of degree $n$, and $\alpha$ a root of $p(x)$ then $$ F(\alpha)=\{a_0 + a_1\alpha + ...+ a_{n-1}\alpha^{n-1} : a_0, a_1,...,a_{n-1} \in F\}$$

(P3): $\zeta_3$ is a root of $x^2+x+1=0$ and a primitive 3rd root of unity.

Claim: $\mathbb{Q}(\zeta_3 \sqrt[3]{2}) \neq \mathbb{Q}(\zeta_3^2 \sqrt[3]{2})$.

Proof: Suppose $\mathbb{Q}(\zeta_3 \sqrt[3]{2}) = \mathbb{Q}(\zeta_3^2 \sqrt[3]{2})$.

$\zeta_3 \sqrt[3]{2}$ and $\zeta_3^2 \sqrt[3]{2}$ are roots of the polynomial $p^*(x)=x^3-2$, which is irreducible over $\mathbb{Q}$ by the fact that the third root is $\sqrt[3]{2}$ and none of the them lies in $\mathbb{Q}$. Thus $\text{degree}(\mathbb{Q}(\zeta_3^2 \sqrt[3]{2}))=3$ and $$ \mathbb{Q}(\zeta_3^2 \sqrt[3]{2}) = \{ a_0 + a_1 (\zeta_3^2 \sqrt[3]{2})+a_2(\zeta_3^2 \sqrt[3]{2})^2: a_0,a_1,a_2 \in \mathbb{Q}\}$$ by (P2). It follows that $$ \zeta \sqrt[3]{2} = a_0 + a_1 (\zeta_3^2 \sqrt[3]{2})+a_2(\zeta_3^2 \sqrt[3]{2})^2 \quad \text{with} \quad a_0,a_1,a_2 \in \mathbb{Q}$$

The imaginary part $\mathcal{Im}(*)$ of this equation gives $$ \mathcal{Im}(\zeta_3)\sqrt[3]{2}=a_1\mathcal{Im}(\zeta_3^2)\sqrt[3]{2} + a_2\mathcal{Im}(\zeta_3)\sqrt[3]{2}^2$$ since $\mathcal{Im}(\sqrt[3]{2})=\mathcal{Im}(a_0)=\mathcal{Im}(a_1)=\mathcal{Im}(a_2)=0$.

It follows from (P3) that $\mathcal{Im}(\zeta_3) = -\mathcal{Im}(\zeta_3)$ (since $\zeta_3^2+\zeta_3 +1 =0$), hence we have $$\mathcal{Im}(\zeta_3)\cdot (\sqrt[3]{2}+a_1\sqrt[3]{2}-a_2\sqrt[3]{2}^2)=0$$ $\mathcal{Im}(\zeta_3)\neq0$ by (P3) again, and finally $$a_2=\frac{1}{\sqrt[3]{2}}(1+a_1)\notin \mathbb{Q}$$ by (P1), which gives a contradiction since $a_2\in \mathbb{Q}$ by definition of $\mathbb{Q}(\zeta_3^2 \sqrt[3]{2})$.

Question: Is this proof correct? Do I use some non-trivial concept implicitly anywhere, as happens often to me?

Additionally: if anybody knows alternative proofs, please let me know.

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  • $\begingroup$ Just to be precise: in asking for alternative proofs I don't only think of proofs that are restricted to the prerequisites (P1) to (P3). $\endgroup$ – C. Moos Aug 4 '18 at 10:36
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Alternate proof, using Galois theory:

$\Bbb Q(\zeta_3,\sqrt[3]2)$ is Galois over $\Bbb Q$. Its Galois group is generated by the two automorphisms $$ \tau:\zeta_3\mapsto\zeta_3^2,\sqrt[3]2\mapsto \sqrt[3]2\\ \sigma:\zeta_3\mapsto\zeta_3, \sqrt[3]2\mapsto \zeta_3\sqrt[3]2 $$ The automorphism $\tau\sigma$ keeps one of your two fields fixed and not the other, so they can't be the same subfield of $\Bbb Q(\zeta_3,\sqrt[3]2)$.

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    $\begingroup$ Very nice proof, far more direct than my computational one and of course more inclined to the concepts that we intend to study in algebra. Thanks $\endgroup$ – C. Moos Aug 4 '18 at 11:19
  • $\begingroup$ I agree (+1).... $\endgroup$ – Dietrich Burde Aug 4 '18 at 16:10
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Another proof: if $K:=\mathbb{Q}(\zeta_3 \sqrt[3]{2}) = \mathbb{Q}(\zeta_3^2 \sqrt[3]{2})$, then $\zeta_3\in K$, so $$\mathbb Q \subset \mathbb{Q}(\zeta_3) \subseteq \mathbb{Q}(\zeta_3\sqrt[3]{2}).$$

But $[\mathbb{Q}(\zeta_3 \sqrt[3]{2}) : \mathbb{Q}]=3$, since $ \zeta_3 \sqrt[3]{2}$ is a root of $X^3-2$, and $[ \mathbb{Q}(\zeta_3): \mathbb{Q}]=2$ analogously, so the hypothesis would imply that 2 divides 3, which is absurd.

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  • $\begingroup$ Also a very nice proof. I tried quite some time thinking about these concepts, but could not tie the loose ends together. I'm baffled that there are so direct proofs at hand.Thank you. $\endgroup$ – C. Moos Aug 4 '18 at 11:21

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