# Galois group of $(x^3-2)(x^5-1)$ over $\mathbb{Q}$.

I am studying for my Galois theory final for tomorrow (and I'm really getting burned out), I need help with the following question:

Galois group $$G$$ of $$f(x)=(x^3-2)(x^5-1)$$ over $$\mathbb{Q}$$.

Let $$\alpha = \sqrt[3]{2}$$ and let $$\omega_3$$ be the primitive cubed root of unity, and $$\omega_5$$ the primitive $$5$$th root of unity. The splitting field for $$f$$ is $$E=\mathbb{Q}(\alpha,\omega_3, \omega_5)$$. We know that $$E:\mathbb{Q}$$ is separable since $$\text{char}\mathbb{Q}=0$$ and that it is normal since it is a splitting field. Hence $$E:\mathbb{Q}$$ is Galois. So $$|G|=|E:\mathbb{Q}|$$. It is easy for me to show that $$\text{Gal}((x^3-2)/\mathbb{Q}) \cong S_3$$ and that $$\text{Gal}((x^5-1)/\mathbb{Q}) \cong C_4$$.

But how can I determine $$\text{Gal}(f)$$? How can I find the size of the extension $$|E:\mathbb{Q}|$$? I know it is divisible by $$6$$ and $$4$$.

Edit: I see people mentioning the result about cartesian products. I have not seen this result. Given that this is a past exam question I would be interested in a proof that does not use the result

• Do you know the degrees of the cyclotomic extensions? If you know that $[\Bbb{Q}(\zeta_{15}):\Bbb{Q}]=8$, and that the extension is abelian, then you are basically done. May 26, 2019 at 18:46
• The splitting fields of $x^3-2$ and $x^5-1$ are linearly disjoint, so $G$ is the product of their Galois groups. May 26, 2019 at 18:46
• @JyrkiLahtonen I know both of those facts. But why is that the case? May 26, 2019 at 18:47
• @LordSharktheUnknown hm, never saw that results. The past papers I'm doing are from a different lecturer so seems like it was not covered this year. May 26, 2019 at 18:48
• The extension generated by both $\omega_3$ and $\omega_5$ is the same extension you get by adjoining $\omega_{15}$. Do you see why? Then $8$ is coprime to $3$. That is enough to give you the degree. I first thought that you need to use the fact that the cyclotomic extension is abelian to deduce that $\root3\of2\notin\Bbb{Q}(\omega_{15})$, but that was an error. May 26, 2019 at 18:50

The Galois group of $$f=gh$$ is equal to the cartesian product of the Galois group of $$g$$ and the Galois group of $$h$$ iff the splitting field of $$g$$ over $$\mathbb{Q}$$ is disjoint with the splitting field of $$h$$ over $$\mathbb{Q}$$. I.e. you have to determine whether $$\mathbb{Q}(\omega_3,\sqrt[3]{2})\cap\mathbb{Q}(\omega_5)=\mathbb{Q}$$. If shown, you have that Gal$$(\mathbb{Q}(\omega_3,\sqrt[3]{2},\omega_5)/\mathbb{Q})\cong S_3\times C_4$$.