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I'm working on a problem that asks to determine the Galois group of the polynomial $(x^3-2)(x^2-5)$ over $\mathbb{Q}$. I know by a Theorem in Hungerford that the Galois group of $X^3-2$ is isomorphic to $S_3$. Additionally, since the splitting field of $x^2-5$ is $\mathbb{Q}(\sqrt{5})$, it's easy to show the automorphism group group over $\mathbb{Q}$ (i.e. the Galois group) is just $\mathbb{Z}_2$.

What I'm guessing is that since $x^2-5$ doesn't split over $\mathbb{Q}(2^{1/3}, \omega)$ (where $\omega$ is a complex cube root of unity) and $x^3-2$ doesn't split over $\mathbb{Q}(\sqrt{5})$, the Galois group is just $S_3 \times \mathbb{Z}_2$, but I'm not sure if that's the right reasoning (or even the right answer). Am I supposed to try and reinterpret it as a subgroup of $S_5$ and mess around with cycles?

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We have the translation theorem:

Given two extensions $L/K$, $M/K$ - with the first being galois - we have that $LM/M$ is a galois extension with $$Gal(LM/M) \cong Gal(L/M\cap L).$$

In the case $M \cap L=K$, we get $$Gal(LM/M) \cong Gal(L/K).$$

If $M/K$ is also galois, we can do the same after changing roles of $L$ and $M$ and get $$Gal(LM/L) \cong Gal(M/K).$$

Note that both subgroups $Gal(LM/M),Gal(LM/L) \subset Gal(LM/K)$ are normal and their intersection is trivial, since an automorphism in the intersection fixes $L$ and $M$, hence all of $LM$.

Two normal subgroups with trivial intersection is precisely the criteria for $Gal(LM/M)Gal(LM/L) \cong Gal(LM/M) \times Gal(LM/L)$, hence we have

$$Gal(L/K) \times Gal(M/K) \cong Gal(LM/M) \times Gal(LM/L) \cong Gal(LM/M)Gal(LM/L) \subset Gal(LM/K)$$

The groups to the very left and very right have the same number of elements, hence the inclusion must be an equality:

$$Gal(L/K) \times Gal(M/K) \cong Gal(LM/K)$$

So you are left to proof $\mathbb Q(\sqrt[3]{2}, \omega) \cap \mathbb Q(\sqrt 5) = \mathbb Q$, which you might have already done by saying "What I'm guessing is that since $x^2-5$ doesn't split over $\mathbb{Q}(2^{1/3},\omega)$"

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