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

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 $$(x^3 - 2)(x^2 + 3)$$.

• I believe the splitting field is $\mathbb{Q}[\sqrt[3]{2},i \sqrt{3}]$ which is properly contained in $\mathbb{Q}[\sqrt[3]{2}, \sqrt{3}, i]$. – Matthias Klupsch Nov 16 '18 at 15:17
• The splitting field of $x^3-2$ includes a primitive cube root of unity, $\omega$. I think the splitting field is $\Bbb Q(\sqrt[3]2,\omega,i\sqrt3)$. – Arthur Nov 16 '18 at 15:19
• @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. – Math enthusiast Nov 16 '18 at 15:49

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.