# 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

• These are Kummer extensions of $K=\Bbb Q(z_6)$. To show they are different, you need to show that neither $2\cdot3$ nor $2\cdot3^2$ are cubes in $K$. Commented Nov 28, 2019 at 17:23
• @LordSharktheUnknown I am not sure what those are, and it seems like an overkill, is there a simpler, perhaps more direct solution? I was never taught Kummer extensions Commented Nov 28, 2019 at 17:38
• Like it or not, these are examples of Kummer extensions, and you can read the basics here: en.wikipedia.org/wiki/Kummer_theory To solve this problem from scratch, you must inevitably reconstruct a small part of this theory. Commented Nov 28, 2019 at 17:42

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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