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I know this question was asked and answered before here, but I try do by myself and I had a different result. I would like to know if I'm wrong of if the answer of the previous topic is wrong.

I know that the splitting field of $(x^3 - 2)(x^2 + 3)$ over $\mathbb{Q}$ is $\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]$. I observed that

$[\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}] = \text{degree} ( \text{irr} (\sqrt[3]{2}, \mathbb{Q}) ) = \text{degree} ( x^3 - 2 ) = 3,$

$[\mathbb{Q}[\sqrt[3]{2}, i]:\mathbb{Q}[\sqrt[3]{2}]] = \text{degree} ( \text{irr} (i, \mathbb{Q}[\sqrt[3]{2}]) ) = \text{degree} ( x^2 + 1 ) = 2$

and

$[\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]:\mathbb{Q}[\sqrt[3]{2}, i]] = \text{degree} ( \text{irr} (\sqrt{3}, \mathbb{Q}[\sqrt[3]{2}, i]) ) = \text{degree} ( x^2 - 3 ) = 2.$

By tower law,

$[\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]:\mathbb{Q}] = [\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]:\mathbb{Q}[\sqrt[3]{2}, i]] \cdot [\mathbb{Q}[\sqrt[3]{2}, i]:\mathbb{Q}[\sqrt[3]{2}]] \cdot [\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}] = 2 \cdot 2 \cdot 3 = 12,$

then the Galois group of $(x^3 - 2)(x^2 + 3)$ over $\mathbb{Q}$ has order 12. I tried find the automorphisms of $\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]$ which fix $\mathbb{Q}$, but I just found these:

$\alpha: i \mapsto -i, \sqrt{3} \mapsto \sqrt{3}, \sqrt[3]{2} \mapsto \sqrt[3]{2}$,

$\beta: i \mapsto i, \sqrt{3} \mapsto -\sqrt{3}, \sqrt[3]{2} \mapsto \sqrt[3]{2}$,

$\gamma: i \mapsto i, \sqrt{3} \mapsto \sqrt{3}, \sqrt[3]{2} \mapsto -\sqrt[3]{2}$.

The other automorphisms are the identity, $\alpha \beta$, $\alpha \gamma$, $\beta \gamma$ and $\alpha \beta \gamma$. Since $\alpha, \beta$ and $\gamma$ has order $2$, I have that $\alpha \beta$, $\alpha \gamma$ and $\beta \gamma$ has order $2$, therefore $\alpha \beta = \beta \alpha$, $\alpha \gamma = \gamma \alpha$ and $\beta \gamma = \gamma \beta$, then I found $8$ elements on Galois group of $(x^3 - 2)(x^2 + 3)$.

I would like to know where I'm going wrong, if the answer in the previously topic are correct and, if it is not, I would like to know how to proceed in order to compute the Galois group of $(x^3 - 2)(x^2 + 3)$.

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    $\begingroup$ I believe the splitting field is $\mathbb{Q}[\sqrt[3]{2},i \sqrt{3}]$ which is properly contained in $\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]$. $\endgroup$ – Matthias Klupsch Nov 16 '18 at 15:17
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    $\begingroup$ The splitting field of $x^3-2$ includes a primitive cube root of unity, $\omega$. I think the splitting field is $\Bbb Q(\sqrt[3]2,\omega,i\sqrt3)$. $\endgroup$ – Arthur Nov 16 '18 at 15:19
  • $\begingroup$ @Arthur, but $\omega = e^{\frac{2\pi i}{3}} = \cos (\frac{2\pi}{3}) +i \sin(\frac{2\pi}{3}) = - \cos (\frac{\pi}{3}) + i \sin(\frac{\pi}{3}) = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \in \mathbb{Q}[\sqrt[3]{2},i\sqrt{3}]$. Maybe the spliiting is $\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}]$. I 'll try compute the Galois group again. $\endgroup$ – Math enthusiast Nov 16 '18 at 15:49
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The splitting field is as Matthias said it was. The splitting field cannot be $\mathbb{Q}[2^{\frac{1}{3}},\sqrt{3}]$ because this is strictly contained in $\mathbb{R}$ and $x^3-2$ has imaginary roots. First, it is well known that $x^3-2$ has Galois group $S_3$, to see this let $\zeta_3$ be a third root of unity and consider the automorphisms of complex conjugation and $ 2^{\frac{1}{3}} \to \zeta_3 2^{\frac{1}{3}}$. So there should be at least $6$ elements in the Galois group of $(x^3-2)(x^2+3)$. Now, you should be able to see that $\mathbb{Q}[2^{\frac{1}{3}},\sqrt{-3}]$ is a degree 6 extension, and that $x^2+3$ splits in this field. And indeed, it contains $\zeta_3$ by your argument in the comments so $x^3-2$ also splits in this field. Hope this helps.

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