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I know that there is an answer in splitting field of $(x^3-2)(x^3-3)$ over $\mathbb Q$. I know the splitting field is $L=Q(ζ3,\sqrt[3]2,\sqrt[3]3)$. But I'm quite confused with [L:Q]=18. How can I get it?

And what are the subgroups of this Galois group?

Now I know how to get [L:Q]=18. But what is the Galois group? Is it C3⋊S3 or C3xS3?

Lewei He
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  • The answer in the linked question shows why the index of the field extension is $18$. What part of the answer are you confused with? – Rushabh Mehta Jun 18 '20 at 15:14
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    A way of looking at the Galois group, implicit in the linked thread, is that it is a subgroup of $S_6$ with one copy of $S_3$ acting on ${1,2,3}$ and another acting on ${4,5,6}$ together with the further constraint that the permutations on the two halves should have the same parity - either both even or both odd. You should go through the process of checking what the automorphisms do the six roots to see this. – Jyrki Lahtonen Jun 19 '20 at 18:52
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    As an abstract group it comes out as the semidirect produt $(C_3\times C_3)\rtimes C_2$ with the generator of $C_2$ acting by inverting everything. – Jyrki Lahtonen Jun 19 '20 at 18:53

2 Answers2

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We have the following elements

  • $\zeta_3$ is a quadratic
  • $\alpha = \sqrt[3]{2}$
  • $\beta = \sqrt[3]{3}$

and the following field extension degrees

  • $[\mathbb Q(\zeta_3) : \mathbb Q] = 2$ (Galois)
  • $[\mathbb Q(\zeta_3, \alpha) : \mathbb Q] = 6$ (Galois)
  • $[\mathbb Q(\zeta_3, \beta) : \mathbb Q] = 6$ (Galois)
  • $[\mathbb Q(\zeta_3, \alpha) : \mathbb Q(\zeta_3)] = 3$ (Galois)
  • $[\mathbb Q(\zeta_3, \beta) : \mathbb Q(\zeta_3)] = 3$ (Galois)

Now we would like to find the degree of the composite $$\mathbb Q(\zeta_3, \alpha)\mathbb Q(\zeta_3, \beta) = \mathbb Q(\zeta_3, \alpha, \beta)$$

define the following

  • $K = \mathbb Q(\zeta_3, \alpha)$
  • $K' = \mathbb Q(\zeta_3, \beta)$
  • $F = \mathbb Q(\zeta_3)$

We use the following lemmas (from Dummit and Foote) about composite Galois extensions:

Proposition 21: Suppose $K/F$ and $K'/F$ are Galois extensions, then $K \cap K'/F$ and $KK'/F$ are Galois extensions. (+ information about the Galois group which we will not use here).

Corollary 22: If $K \cap K'/F = F$ then $$[KK' : F] = [K : F][K' : F]$$ and the Galois group is $$\operatorname{Gal}(K/F) \times \operatorname{Gal}(K'/F).$$

We just need to show now that $$\mathbb Q(\zeta_3, \alpha) \cap \mathbb Q(\zeta_3, \beta) = \mathbb Q(\zeta_3)$$ and this holds because $\alpha$ and $\beta$ are real.

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    Can I use Q(ζ3,α)∩Q(β)=Q to prove Q(ζ3,α,β) not equal to Q(ζ3,α) Then [Q(ζ3,α,β):Q(ζ3,α)]=3, thus [Q(ζ3,α,β):Q]=3*6=18? – Lewei He Jun 18 '20 at 17:34
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    Something wrong with your writing of Proposition 19. It is false as stated. Consider $F=\Bbb{Q}$, $K=\Bbb{Q}(\sqrt2)$, $F'=\Bbb{Q}(\root3\of2)$. Then $KF'=\Bbb{Q}(\root6\of2)$, but that is not a Galois extension of $F$. I suspect the claim is that $KF'/F'$ is Galois. At least that would be a true statement :-) – Jyrki Lahtonen Jun 19 '20 at 07:01
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    Also the logic in your last paragraph is not watertight. From $\zeta_3\notin\Bbb{Q}(\beta)$ it does not follow that $$\Bbb{Q}(\zeta_3,\alpha)\cap \Bbb{Q}(\beta)=\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta).$$ A counterexample of this would be $\alpha=\zeta_3\root3\of2$, $\beta=\root3\of2$. In that case $\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)$ is trivial but $\Bbb{Q}(\beta)\subset\Bbb{Q}(\alpha,\zeta_3)$. – Jyrki Lahtonen Jun 19 '20 at 07:06
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    The conclusion is true, but (using your $\alpha$ and $\beta$) I would go, for example, via the route: $\Bbb{Q}(\beta)\subset\Bbb{R}$, $\Bbb{Q}(\zeta_3,\alpha)\cap\Bbb{R}=\Bbb{Q}(\alpha)$. Therefore $$\Bbb{Q}(\zeta_3,\alpha)\cap \Bbb{Q}(\beta)=\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta).$$ – Jyrki Lahtonen Jun 19 '20 at 07:09
  • @JyrkiLahtonen That is so much nicer, thank you! I have corrected the F' thing. –  Jun 19 '20 at 07:55
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    Thank you! I got [L:Q]=18. But what is the Galois group of this? – Lewei He Jun 19 '20 at 18:19
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Let $F=\mathbf{Q}(\zeta_3)$. It follows from the lemma below (whose proof is easy using Galois theory) that $F(\sqrt[3]{2})=F(\sqrt[3]{2})$ if and only if $\sqrt[3]{2}\sqrt[3]{3}=\sqrt[3]{6}$ or $\sqrt[3]{2}(\sqrt[3]{3})^2=\sqrt[3]{18}$ is in $F$, neither of which is the case. Hence $F(\sqrt[3]{2},\sqrt[3]{3})$ is a degree $18$ extension of $\mathbf{Q}$.

Lemma. Let $F$ be a field of characteristic $\neq 3$ such that $\zeta_3\in F$. Let $\alpha,\beta\in F$ be two elements such that $\sqrt[3]{\alpha},\sqrt[3]{\beta}\not\in F$. Then $F(\sqrt[3]{\alpha})=F(\sqrt[3]{\alpha})$ if and only if $\sqrt[3]{\alpha}\sqrt[3]{\beta}\in F$ or $\sqrt[3]{\alpha}(\sqrt[3]{\beta})^2\in F$.

rae306
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