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Find the Galois group of $x^3-2x+2$ over $\mathbb{Q}$ and over $\mathbb{Q}(i\sqrt{19})$

One of its roots is absurd: http://www.wolframalpha.com/input/?i=x%5E3-2x%2B2, so its not pratical to divide this polynomial by $x-root$ to find a second degree one that might me irreducible. If I somehow know that the real root is not rational, I still would need to know how the second degree polynomial looks like when we divide by $x-root$, so I know if its roots are real or complex.

What should I do???

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    $\begingroup$ Did you show that for a rational irreducible cubic its splitting field is $\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3)=\mathbb{Q}(\alpha_1,\sqrt{\Delta})$ ? Iff $\sqrt{\Delta} \not \in \mathbb{Q}$ then it is a degree $6$ Galois extension with Galois group $S_3$. $\endgroup$ – reuns Oct 1 '17 at 20:44
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Let $K$ be a splitting field for $g(X) = X^{3}-2X+2$ over $\mathbb{Q}$. Note that $g(X)$ is is irreducible over $\mathbb{Q}$ by Eisenstein at $2$, so $G := \mathrm{Gal}(K/\mathbb{Q})$ must be a transitive subgroup of $S_{3}$, hence $A_{3}$ or $S_{3}$. We can distinguish the two possibilities by computing the discriminant $\Delta$ of $g$; if $\Delta$ is a square in $\mathbb{Q}$, then $G \cong A_{3}$, and if not, then $G \cong S_{3}$. We compute

$$\Delta = -4(-2)^{3} - 27(2)^{2} = -76$$

which is not a square in $\mathbb{Q}$, so $G \cong S_{3}$. Now, $\mathbb{Q}(\sqrt{\Delta})$ is a subfield of $K$ (since $\sqrt{\Delta}$ is a product of differences of roots of $g$), so $\mathbb{Q}(\sqrt{\Delta}) = \mathbb{Q}(i\sqrt{19})$ is the subfield of $K$ corresponding to an index $2$ subgroup of $G$. There is only one subgroup of $G$ of index $2$, namely the cyclic subgroup generated by $(1, 2, 3)$, so $\mathrm{Gal}(K/\mathbb{Q}(i\sqrt{19})) \cong \mathbb{Z}/3\mathbb{Z}$.

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