Find two non isomorphic galois extensions of degree three of $Q(z_6)$ As it is in the title, I am looking for two Galois extensions of degree 3 of $Q(z_6)$ where $z_6$ is the primitive 6th root of unity.
My idea: Take $Q(z_6,\sqrt[3]{2})$ and $Q(z_6,\sqrt[3]{3})$ These are extensions of degree 3 from $Q(z_6)$ and are splitting fields of $(x^2-x+1)(x^3-2)$ and $(x^2-x+1)(x^3-2)$ over $Q$ so they are Galois. Thus by FT of Galois, $Q(z_6,\sqrt[3]{2})/Q(z_6)$ is Galois. Hence, the only problem is to show these two are not isomorphic which i do not know how to do. My idea would be to show there is not thing in $Q(z_6,\sqrt[3]{3})$ that if you $^3$ you get $2$. But showing that form a basis is an impossible task. Is there a shorter way? Perhaps these are isomorphic? Are two extensions isomorphic iff they are roots of the same polynomial?
Thank you
 A: It's a little strange to take a sixth root of unity $z_6 = \zeta_6$, since $\mathbf{Q}(\zeta_6) = \mathbf{Q}(\zeta_3) = \mathbf{Q}(\sqrt{-3})$. 
As to the answer, of course @lord-shark-the-unknown is correct, but here is an elementary solution that uses neither Kummer theory nor anything about ramification. 
Let's take the following two fields:
$$L = \mathbf{Q}(\zeta_{9}), \qquad K = \mathbf{Q}(\zeta_3,\sqrt[3]{2}).$$
Using material in a standard course on Galois theory, one has 
$$\mathrm{Gal}(L/\mathbf{Q}) = (\mathbf{Z}/9 \mathbf{Z})^{\times} \simeq \mathbf{Z}/6 \mathbf{Z},$$
$$\mathrm{Gal}(K/\mathbf{Q}) \simeq S_3,$$
Certainly both $L$ and $K$ contain $E$, and so $E$ is the field corresponding to the unique surjections  onto $\mathbf{Z}/2\mathbf{Z}$ in either case. Galois theory again tells you that $\mathrm{Gal}(L/E) = \mathbf{Z}/3 \mathbf{Z} \subset \mathbf{Z}/6 \mathbf{Z}$ and $\mathrm{Gal}(K/E) =  \mathbf{Z}/3 \mathbf{Z} \subset S_3$, so these are both Galois extensions of $E$ of degree $3$. On the other hand, clearly $K$ and $L$ are not isomorphic fields because they have different Galois groups over $\mathbf{Q}$.
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