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I've found the Galois group of $x^3-2$, isomorphic to $\mathbf{S}_3$. It has 6 subgroups (including the trivial subgroup and the group itself), and thus by the Galois Correspondence there should be 6 intermediate fields $\mathbf{E}$, where $\mathbb{Q} \le \mathbf{E} \le \mathbf{K}$, with each a fixed field of the corresponding subgroups of $\mathrm{G}(\mathbf{K}/\mathbb{Q})$.

I've found what I think are the correct 6 intermediate fields: $\mathbb{Q}, \mathbb{Q}(i\sqrt{3}), \mathbb{Q}(\alpha_1), \mathbb{Q}(\alpha_2), \mathbb{Q}(\alpha_3)$, and $\mathbf{K}$, where $$\alpha_1 = \sqrt[3]{2}$$ $$\alpha_2 = \sqrt[3]{2}\frac{-1 + i\sqrt{3}}{2}$$ $$\alpha_3 = \sqrt[3]{2}\frac{-1 - i\sqrt{3}}{2}$$ are the three zeros of $x^3-2$.

My problem comes here: how do we know these are the only intermediate fields? I know the Galois Correspondence says they are, but (for a specific example) what about fields like $\mathbb{Q}(\alpha_1^2)$, $\mathbb{Q}(\alpha_2^2)$, and $\mathbb{Q}(\alpha_3^2)$? These are obviously subfields of $\mathbf{K}$, but they are not (explicitly) included in my above list of the 6 intermediate fields. So I suppose my question is, either: 1) Did I get the wrong intermediate fields in the above listing, or 2) Are the $\mathbb{Q}(\alpha_i^2)$ isomorphic to some of the fields in my listing? Or, I could have the wrong idea about what the Galois Correspondence is saying? Because otherwise it seems like these $\mathbb{Q}(\alpha_i^2)$ are intermediate fields not taken into account by Galois.

Thanks a lot, and P.S. this is my first question so apologies for any formatting issues or otherwise.

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The Galois Correspondence between subfields of the splitting field of a polynomial and subgroups of the Galois group is a bijection. Thus the theory tells you that the six intermediate fields that you found must be all of them.

Each of the fields you are wondering about are actually one of the six you listed above. For example, $\mathbb{Q}(\alpha_1^2) = \mathbb{Q}(\alpha_1)$ since $(\alpha_1^2)^2 = 2\alpha_1 \implies \alpha_1 \in \mathbb{Q}(\alpha_1^2)$, while the other inclusion is clear.

I'll leave it to you to find out which of the six fields $\mathbb{Q}(\alpha_2^2)$ and $\mathbb{Q}(\alpha_3^2)$ are. Perhaps consider writing them in exponential form, e.g. $\alpha_2 = 2^{\frac{1}{3}}e^{\frac{2\pi i}{3}}$

This is my first answer so any critiques are appreciated, thanks!

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  • $\begingroup$ looks fine to me. $\endgroup$ – DanielWainfleet Sep 3 '15 at 4:22
  • $\begingroup$ My critique: excellent answer, and efficient. Keep making contributions as good! $\endgroup$ – Lubin Sep 4 '15 at 0:25
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Hint: $(\alpha_i^2)^2 = \alpha_i^4 = 2 \alpha_i$.

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  • $\begingroup$ Oh wow - thanks so much, I got it now. A quick question on the same topic: is there a good way to prove that $\mathbb{Q}(\alpha_2) \neq \mathbb{Q}(\alpha_3)$? And similarly, is $\alpha_1 \le \mathbb{Q}(\alpha_2)$? $\endgroup$ – Chris Sep 3 '15 at 4:34

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