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I want to find the Galois group over $\mathbb{Q}$ of $(x^3-2)(x^2-2).$


I already know that the Galois group over $\mathbb{Q}$ of $f(x) = x^3-2$ is isomorphic to $S_3$ and there are four intermediate fields between the splitting field of $f$ and $\mathbb{Q}$: $\mathbb{Q}(\sqrt[3]{2}),\,\mathbb{Q}(\omega),\,\mathbb{Q}(\omega\sqrt[3]{2}),\,\mathbb{Q}(\omega^2\sqrt[3]{2}),$ where $\omega$ is a primitive $3^\mathrm{rd}$ root of unity. It's easy to verify that $\sqrt{2}$ does not belong in one of these intermediate fields, so by the fundamental theorem, $\mathbb{Q}(\sqrt{2})\not\subset\mathbb{Q}(\sqrt[3]{2},\omega).$ Now we have that $$[\mathbb{Q}(\sqrt[3]{2},\omega,\sqrt{2}) : \mathbb{Q}]=12.$$ This is where I'm stuck. I don't know how to conclude what the Galois group of the polynomial is, other than it is some subgroup of $S_6$ of order $12$.

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    $\begingroup$ First, show that your extension field is a splitting field of $x^6-2$ over $\mathbb{Q}$. Then see this: math.stackexchange.com/questions/1575214/… $\endgroup$ Aug 5, 2016 at 22:50
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    $\begingroup$ Actually you should be able to realize this as a subgroup of $S_5$ because your polynomial only has five zeros. Anyway, the imaginary part of the eighth root of unity found a nice way of doing it. Also, this question is quite similar to this. It doesn't seem to matter, which square root you adjoin in addition to the cube roots (ok, $\sqrt{-3}$ would be an exception). $\endgroup$ Aug 5, 2016 at 23:11

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Well, you know what the generating automorphisms of the Galois group are, namely $\sigma, \tau, \rho$ which act as follows on the generators of the splitting field: $$\sigma(\sqrt[3]{2}) = \omega\sqrt[3]{2}; \sigma(\omega) = \omega; \sigma(\sqrt{2}) = \sqrt{2}$$ $$\tau(\sqrt[3]{2}) = \sqrt[3]{2}; \tau(\omega) = \omega^{2}; \tau(\sqrt{2}) = \sqrt{2}$$ $$\rho(\sqrt[3]{2}) = \sqrt[3]{2}; \rho(\omega) = \omega; \rho(\sqrt{2}) = -\sqrt{2}$$ Now, we compute $$(\sigma\rho)(\sqrt[3]{2}) = \omega\sqrt[3]{2}; (\sigma\rho)(\omega) = \omega; (\sigma\rho)(\sqrt{2}) = -\sqrt{2}$$ It is easy to see that $\sigma\rho$ has order $6$ (why?). Further, $(\sigma\rho)^{-1} = \rho^{-1}\sigma^{-1} = \rho\sigma^{2}$. Finally, note that $\sigma$ and $\rho$ commute, as well as $\tau$ and $\rho$, and $\sigma, \tau$ obey the relations $\sigma\tau = \tau\sigma^{2}$. Now, your group $G$ cannot be abelian, and must contain $S_{3}$. A good candidate for $G$ is thus $D_{6}$, which is nonabelian of order $12$ and contains an isomorphic copy of $D_{3} \cong S_{3}$. Let $\beta = \sigma\rho$. Then $$\beta\tau = \sigma\rho\tau = \sigma\tau\rho = \tau\sigma^{2}\rho = \tau\rho\sigma^{2} = \tau\beta^{-1}$$ Furthermore, $\beta^{4} = \sigma$, $\beta^{3} = \rho$, so $G$ has the presentation $\langle \beta, \tau \mid \beta^{6} = \tau^{2} = e, \beta\tau = \tau\beta^{-1}\rangle \cong D_{6}$.

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  • $\begingroup$ In order to relate this answer to the other, let me add that $D_6\simeq S_3\times\mathbb{Z}_2$. $\endgroup$
    – user26857
    May 28, 2023 at 22:18
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Let $L_3$ denote the splitting field of $x^3 - 2$ and $L_2$ denote the splitting field of $x^2-2.$ Now, we compute the Galois groups of the two polynomials. The cubic has Galois group $S_3$ since we can find intermediate fields of degree 2 and 3 by adjoining the third root of unity and the cubed root of 2 respectively. Now, the quadratic has Galois group $\mathbb{Z}_2$ since neither root is rational. We now ask what the intersection of $L_2$ and $L_3$ are. By the Galois Correspondence, $L_3$ has only one intermediate field of degree 2. Namely, $\mathbb{Q}(\sqrt{-3})$. The test for determining whether two extensions of the form $\mathbb{Q}(\sqrt{a})$ are equal is to determine if $ab$ is a square in $\mathbb{Q}$. In our case, $a=-3$ and $b=2$. Then, $ab<0$, so it can't possibly be a square. Thus, the intersection of $L_2$ and $L_3$ is $\mathbb{Q}$. Thus, if we let $K$ denote the splitting field of $f(x)$, then $$Gal(K/\mathbb{Q}) \cong Gal(L_2/\mathbb{Q}) \oplus Gal(L_3/\mathbb{Q}) \cong S_3 \oplus\mathbb{Z}_2.$$ To describe the automorphisms explicitly, let 1,2,3 be the roots of the cubic and 4,5 be the roots of the quadratic. Then, we embed $S_3 \oplus\mathbb{Z}_2$ into $S_5$ via $(\sigma,1)$ mapping to $\sigma(45)$ and $(\sigma,0)$ mapping to $\sigma$. This describes the action of the roots.

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