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Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$

I've been struggling with some of these Galois group questions.

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Welcome to math.stackexchange! What have you tried? – talmid Jun 13 '12 at 5:16

Let $f=(x^3-1)(x^2-5)\in\mathbb{Q}[x]$. Note that $$\begin{align}f&=(x-1)(x^2+x+1)(x^2-5)\\ &=(x-1)(x-\zeta_3)(x-\zeta_3^2)(x-\sqrt{5})(x+\sqrt{5})\end{align}$$ where $\zeta_3=\frac{-1+\sqrt{3}}{2}$ is a primitive cube root of unity. Thus, the splitting field of $f$ over $\mathbb{Q}$ is $$K=\mathbb{Q}(\zeta_3,\sqrt{5})=\{a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5}\mid a,b,c,d\in\mathbb{Q}\}.$$

Suppose that $\phi:K\to K$ is an automorphism of $K$ fixing $\mathbb{Q}$. Then $$\begin{align}\phi(a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5})&=\phi(a)+\phi(b)\phi(\zeta_3)+\phi(c)\phi(\sqrt{5})+\phi(d)\phi(\zeta_3)\phi(\sqrt{5}) \\ &=a+b\phi(\zeta_3)+c\phi(\sqrt{5})+d\phi(\zeta_3)\phi(\sqrt{5})\end{align}$$ so that $\phi$ is determined entirely by the values of $\phi(\zeta_3)$ and $\phi(\sqrt{5})$; that is, if $\psi$ is an automorphism of $K$ such that $\psi(\zeta_3)=\phi(\zeta_3)$ and $\psi(\sqrt{5})=\phi(\sqrt{5})$, then in fact $\psi=\phi$.

Because $\zeta_3^2+\zeta_3+1=0$ and $(\sqrt{5})^2-5=0$, we must have that $$0=\phi(0)=\phi(\zeta_3^2+\zeta_3+1)=\phi(\zeta_3)^2+\phi(\zeta_3)+1$$ and $$0=\phi(0)=\phi((\sqrt{5})^2-5)=\phi(\sqrt{5})^2-5$$ so that $\phi(\zeta_3)$ must be one of the two roots of $x^2+x+1$, namely $\zeta_3$ and $\zeta_3^2$, and $\phi(\sqrt{5})$ must be one of the two roots of $x^2-5$, namely $\sqrt{5}$ and $-\sqrt{5}$.

Thus, there are at most four elements of $\operatorname{Gal}(K/\mathbb{Q})$ (there are two choices for where $\zeta_3$ goes, and two choices for where $\sqrt{5}$ goes): $$\begin{align} \operatorname{id}_K(\zeta_3)&=\zeta_3 & \operatorname{id}_K(\sqrt{5})&=\sqrt{5}\\ \phi(\zeta_3) &= \zeta_3^2 & \phi(\sqrt{5})&=\sqrt{5}\\ \rho(\zeta_3) &= \zeta_3 & \rho(\sqrt{5})&=-\sqrt{5}\\ \sigma(\zeta_3) &= \zeta_3^2 & \sigma(\sqrt{5})&=-\sqrt{5}\\ \end{align}$$ Check that all four do in fact define valid automorphisms, and see what the group structure on $\operatorname{Gal}(K/\mathbb{Q})$ is by composing them and seeing what comes up. For example, we have that $\phi\circ \rho=\sigma$, because $$(\phi\circ\rho)(\zeta_3)=\phi(\rho(\zeta_3))=\phi(\zeta_3)=\zeta_3^2=\sigma(\zeta_3)$$ $$(\phi\circ\rho)(\sqrt{5})=\phi(\rho(\sqrt{5}))=\phi(-\sqrt{5})=-\phi(\sqrt{5})=-\sqrt{5}=\sigma(\sqrt{5})$$ Are there any elements of order 4?

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The splitting field of $(x^3-1)(x^2-5) = (x-1)(x^2+x+1)(x^2-5)$ is the compositum of the splitting fields of $x^2+x+1$ and of $x^2-5$; it is not hard to check that it is of degree $4$ over $\mathbb{Q}$. Hence, the Galois group must be either cyclic of order $4$, or isomorphic to the Klein $4$-group.

Now, which is it? By the Fundamental Theorem of Galois Theory, the subgroups of the Galois group correspond to intermediate fields of the extension. We can distinguish the cyclic group of order $4$ from the Klein $4$-group in many ways, but one of them is that the former has a unique subgroup of index $2$, whereas the latter has three distinct subgroups of index $2$. A subgroup of index $2$ corresponds to an intermediate extension of degree $2$.

So... does the splitting field contain a unique quadratic extension of $\mathbb{Q}$? Or does it contain more than one?

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