What is the Galois group of $(x^3-2)(x^3-3)$ over $\mathbb{Q}$?

Question

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?

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

Answer 1

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 $\sqrt[3]{\alpha}(\sqrt[3]{\beta})^2\in F$.

Answer 2

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.