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Let $f(x)= 2x^6-10x^2-5$. Then we know that by Eisenstein's criterion, $f$ is irreducible. Also, we know that there are 2 real roots and 4 complex roots.

From here, I'm not sure how to determine the Galois group for $f$.

Any help will be appreciated!

Toasted_Brain
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  • Hint: Consider the relationship to the Galois group for $2x^3-10x+5.$ – Thomas Andrews Apr 06 '23 at 00:44
  • I know that the Galois group for $2x^3-10x+5$ is $S_3$, since the discriminant is not a perfect square. I still don't see how that relates to the given polynomial $f$... – Toasted_Brain Apr 06 '23 at 01:03
  • If $\alpha$ is a real root and $\beta$ a complex root then the splitting field is $\mathbb{Q} (\alpha, \beta, \overline{\beta}) $ and it appears the degree of this splitting field is $48$ (however I need to confirm this with more calculations). – Paramanand Singh Apr 06 '23 at 02:05
  • If $g (x)=2x^3-10x+5$ then $g(x^2)$ is your original polynomial. – Thomas Andrews Apr 06 '23 at 03:03
  • How can it be $S_6$ since the root pairs ${\alpha,-\alpha}$ have to go to another root pair ${\beta,-\beta}?$ If you know $\alpha\to\beta,$ then $-\alpha\to-\beta.$ – Thomas Andrews Apr 06 '23 at 13:10
  • @ThomasAndrews Would it be $S_3$ then? – Toasted_Brain Apr 06 '23 at 20:51
  • No, because you can send $\alpha\to-\alpha.$ – Thomas Andrews Apr 06 '23 at 21:08
  • @ThomasAndrews Since $\alpha \rightarrow -\alpha$... will it be $\mathbb{Z}/2\mathbb{Z}$? Thank you!! – Toasted_Brain Apr 07 '23 at 02:46
  • No, there will be three pairs $\alpha_i,-\alpha_i$ where the $\alpha_i^2$ are the roots of the cubic. You have to combine the $S_3$ part and more than one $\mathbb Z/2\mathbb Z.$ It is some work. I'm not even sure what the correct answer is off the top of my head. – Thomas Andrews Apr 07 '23 at 13:00
  • What Thomas Andrews's comment is leading to is that the Galois group will be a subgroup of the wreath product $C_2\wr S_3$. In other words, the semidirect product $(C_2\times C_2\times C_3)\rtimes S_3$ with $S_3$ acting on the threefold Cartesian product by permuting the components. If the splitting field has degree $48$, then that settles the matter because this group has order $48$. It migh be a proper subgroup, if there are dependencies among the square roots of the roots of the cubic. I don't see any right away, but cannot confirm the degree either. – Jyrki Lahtonen Apr 07 '23 at 17:23
  • Ok. A bit of testing revealed that modulo $p=29$ the polynomial has four roots in $\Bbb{Z}_{29}$ and one irreducible quadratic factor. Dedekind's theorem then tells us that the Galois group has an element $\sigma$ that acts on the six roots as a $2$-cycle. This can only be if $\sigma$ interchanges a pair $\pm\alpha_1$, $\alpha_1$ a root of the cubic. Because the cubic has full $S_3$ as its Galois group, some conjugates of $\sigma$ do the same to other pairs of roots $\pm\alpha_2$ and $\pm\alpha_3$. Therefore all of the $C_2\times C_2\times C_2$ is in there. – Jyrki Lahtonen Apr 07 '23 at 17:35
  • Don't know if we can see this without resorting to Dedekind. I haven't really tried since Dedekind is my go to -technique nowadays :-( – Jyrki Lahtonen Apr 07 '23 at 17:38

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