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New to Galois Theory, and I'm working on the "beginner" problem of finding the intermediate fields of $\mathbb{Q}(\zeta_3, \sqrt[3]{2})$. Questions like this have already been asked on this site, and despite reading the solutions, I am still a bit confused, so perhaps I can outline what I have done, and then if someone can point out where there may be a mistake, that would be great.

I understand that $\mathbb{Q}(\zeta_3, \sqrt[3]{2})$ is the splitting field of the polynomial $x^3 - 2$, which has roots $\sqrt[3]{2}, ~ \zeta_3\sqrt[3]{2}, ~ \zeta_3^2\sqrt[3]{2}$, and I understand $[\mathbb{Q}(\zeta_3, \sqrt[3]{2}) : \mathbb{Q}] = 6$, and has a basis $\{1, ~ \zeta_3, ~ \sqrt[3]{2}, ~ \sqrt[3]{4}, ~ \zeta_3\sqrt[3]{2}, ~ \zeta_3\sqrt[3]{4}\}$. Now, and here is where my understanding may be incorrect, I believe that Gal$(\mathbb{Q}(\zeta_3, \sqrt[3]{2}) ~ / ~ \mathbb{Q})$ must be generated by two elements, namely:

\begin{gather*} \sigma: \sqrt[3]{2} \mapsto \zeta_3\sqrt[3]{2}, && \tau: \zeta_3 \mapsto \zeta_3^2 \end{gather*}

Where $\sigma, \tau$ fix all other elements. I understand intuitively why these functions must be defined this way, since the image of a root must still be a root of its original minimal polynomial. And from these definitions, (and given the order of the group), it is not hard to see that Gal$(\mathbb{Q}(\zeta_3, \sqrt[3]{2}) ~ / ~ \mathbb{Q}) \cong S_3$. Furthermore, the subgroups of Gal$(\mathbb{Q}(\zeta_3, \sqrt[3]{2})$ are precisely <$\sigma$>, <$\tau$>, <$\sigma\tau$>, and <$\sigma^2\tau$>.

The fundamental theorem of Galois theory tells us that these subgroups are in a 1-1 correspondence with the intermediate fields of $\mathbb{Q}(\zeta_3, \sqrt[3]{2})$, where the intermediate field associated with a given subgroup is the field by which all of its elements are jointly fixed by all the automorphisms in a given subgroup. Now, it is easy to see that the field $\mathbb{Q}(\zeta_3)$ corresponds to <$\sigma$>, since $\sigma$ fixes $\zeta_3$. Likewise, $\mathbb{Q}(\sqrt[3]{2})$ corresponds to <$\tau$>. What is less clear is what <$\sigma\tau$> and <$\sigma^2\tau$> correspond to.

Both $\sigma\tau$ and $\sigma^2\tau$ are order 2, so we just need to find which elements these automorphisms fix. When I evaluate these automorphisms on the basis elements, I find that $\sigma\tau$ fixes $\zeta_3\sqrt[3]{4}$, and $\sigma^2\tau$ fixes $\zeta_3\sqrt[3]{2}$. Does this mean the corresponding intermediate fields are $\mathbb{Q}(\zeta_3\sqrt[3]{4})$ and $\mathbb{Q}(\zeta_3\sqrt[3]{2})$?

It seems like I should be getting $\mathbb{Q}(\zeta_3\sqrt[3]{2})$, $\mathbb{Q}(\zeta_3^2\sqrt[3]{2})$ instead, and I cannot for the life of me figure out why. I also can see that $\mathbb{Q}(\zeta_3) = \mathbb{Q}(\zeta_3^2)$, and $\mathbb{Q}(\sqrt[3]{2}) = \mathbb{Q}(\sqrt[3]{4})$, but I'm not sure if this is helpful. If anyone can see a mistake in my analysis and help point out what I did wrong, I would appreciate it greatly.

Pancho_Squancho
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    "Where , fix all other elements..." This is not how it works. If $\sigma$ does what it does to $\sqrt[3]{2}$, what does it do to $\zeta_3\sqrt[3]{2}$? Unclear if this impacts your error about what $\sigma \tau$ does or not. – tkr Apr 19 '24 at 02:35
  • $\sigma\tau$ does not fix the element you claim it does, you should probably check your calculations again. After, you should check what $\sigma\tau$ does to the element $\zeta_3^2\sqrt{2}$ – Fotis Apr 19 '24 at 02:38
  • What tkr said. To specify an automorphism, you need to prescribe what it does to all the generating elements, here both $\zeta_3$ and $\root3\of2$. Then the rest of its properties follow from the definition of an automorphism. Anyway, you seem to need an extra exercise explaining why $\Bbb{Q}(\zeta\root3\of4)=\Bbb{Q}(\zeta^2\root3\of2)$? – Jyrki Lahtonen Apr 19 '24 at 05:05
  • @tkr, Maybe I didn’t put it the best way, but I really meant $\sigma$ fixes $\zeta_3$. So we still have $\sigma(\zeta_3\sqrt[3]{2}) = \zeta_3^2\sqrt[3]{2}$, for example. – Pancho_Squancho Apr 19 '24 at 14:53
  • @Fotis, is there a reason to see what $\zeta_3^2\sqrt[3]{2}$ maps to? It’s not an element of our basis, which is why I originally didn’t check. I get that $\sigma\tau$ maps the element to itself, since $\tau$ sends it to $\zeta_3^4\sqrt[3]{2} = \zeta_3\sqrt[3]{2}$, and $\sigma$ sends it back to $\zeta_3^2\sqrt[3]{2}$. – Pancho_Squancho Apr 19 '24 at 14:56
  • @Pancho_Squancho Right. You understand then. Then, you should look at what Jyrki Lahtonen wrote in their comment. I only made my comment because it was the earliest point at which I found confusion may be arising. – tkr Apr 19 '24 at 14:59
  • @tkr, Ok, cool, thank you for clarifying and checking my work! I’ll think about Jykri’s comment. – Pancho_Squancho Apr 19 '24 at 15:02

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