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I'm trying to compute de Galois group of $p(x) = x^4-2x^2+2$, and I seem to have arrived to a contradiction, which of course means that there is some mistake that I'm not able to find.

The first thing I did was to compute the roots:

$\alpha_1 = \sqrt[4]2 \space e^{\pi i/8}$

$\alpha_2 = -\sqrt[4]2 \space e^{\pi i/8}$

$\alpha_3 = \sqrt[4]2 \space e^{-\pi i/8}$

$\alpha_4 = -\sqrt[4]2 \space e^{-\pi i/8}$

From the expressions, one can see that:

$\alpha_3 = \alpha_1^{15} \space \frac{\sqrt 2}{16}$

So that $\sqrt 2$ is in the splitting field, and with $\sqrt 2$ and $\alpha_1$ it's possible to generate the other roots. Therefore, the splitting field must be:

$\mathbb{Q}(\sqrt 2,\alpha_1)$

We need to see if the degree is equal to 4 or 8. If the degree of $\alpha_1$ over $\mathbb{Q}(\sqrt 2)$ is 4, the total degree will be 8. And this is the case, because $p(x)$ splits in $\mathbb{Q}(\sqrt 2)[x]$ if and only if there are $\alpha_i$ and $\alpha_j$ so that $(x-\alpha_i)(x-\alpha_j)$ have coefficients in $\mathbb{Q}(\sqrt 2)$, which applying Cardano's formulas requires $\alpha_i + \alpha_j \in \mathbb{Q}(\sqrt 2)$ and $\alpha_i \alpha_j \in \mathbb{Q}(\sqrt 2)$. It's easy to rule out all combinations of $\alpha_i$ and $\alpha_j$ (perhaps easier if one considers the expression $\alpha_i = \pm\sqrt[4]{2} (\frac{1}{2}\sqrt{2+\sqrt2} \pm \frac{i}{2}\sqrt{2-\sqrt2})$), and to prove that it's not possible, so that the degree of the extension is 8.

In this point, the Galois group $G$ can be seen as a subgroup of $S_4$ (because the group acts permuting the four roots) of order 8, so that, according to the structure of subgroups of $S_4$ (see https://groupprops.subwiki.org/wiki/Symmetric_group:S4), the Galois group is isomorphic to $D_8$.

However, I wanted a more direct proof, using the expression of the splitting field, which allows us to consider the group as acting on $\alpha_1$ (sending it to either $\alpha_1$, $\alpha_2$, $\alpha_3$ or $\alpha_4$) and $\sqrt 2$ (sending it to $\sqrt 2$ or $-\sqrt 2$). Using this, the Galois automorphisms have to be:

\begin{array}{|c|c|c|} \hline & \sigma(\alpha_1) & \sigma(\sqrt 2) \\ \hline \sigma_1 & \alpha_1 & \sqrt 2\\ \hline \sigma_2 & \alpha_2 &\sqrt 2\\ \hline \sigma_3 & \alpha_3 &\sqrt 2\\ \hline \sigma_4 & \alpha_4 &\sqrt 2\\ \hline \sigma_5 & \alpha_1 &-\sqrt 2\\ \hline \sigma_6 & \alpha_2 &-\sqrt 2\\ \hline \sigma_7 & \alpha_3 &-\sqrt 2\\ \hline \sigma_8 & \alpha_4 &-\sqrt 2\\ \hline \end{array}

Moreover, we can consider these two subgroups:

$N = Gal(\mathbb{Q}(\sqrt 2,\alpha_1)/\mathbb{Q}(\sqrt 2)) = \{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}$

$H = Gal(\mathbb{Q}(\sqrt 2,\alpha_1)/\mathbb{Q}(\alpha_1)) = \{\sigma_1,\sigma_5\}$

$N$ is normal, because the field extension $\mathbb{Q}(\sqrt 2)/\mathbb{Q}$ is a normal extension.

And $N\cap H = \{\sigma_1\}$, so that $N H= G$ and $G = N \rtimes H$

It seems that we are arriving to the same conclusion, but we need to prove that $N \simeq \mathbb{Z}_4$ in order to end. And this is the point that fails, because all elements of $N$ are of order $1$ or $2$:

$\sigma_2^2(\alpha_1) = \sigma_2(\alpha_2) = \sigma_2(-\alpha_1) = -\alpha_2 = \alpha_1$

$\sigma_3^2(\alpha_1) = \sigma_3(\alpha_3) = \sigma_3(\alpha_1^{15} \space \frac{\sqrt 2}{16}) = \alpha_3^{15} \space \frac{\sqrt 2}{16} = \alpha_1$

$\sigma_4^2(\alpha_1) = \sigma_4(\alpha_4) = \sigma_4(-\alpha_1^{15} \space \frac{\sqrt 2}{16}) = -\alpha_4^{15} \space \frac{\sqrt 2}{16} = \alpha_1$

So that $N \simeq \mathbb{Z}_2 \times \mathbb{Z}_2$, and $G \simeq (\mathbb{Z}_2 \times \mathbb{Z}_2)\rtimes \mathbb{Z}_2$

Please, can you tell me where is the mistake??

Thanks a lot in advance!

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    $\begingroup$ Compare with this question, for $x^4+ax+b$. $\endgroup$ – Dietrich Burde Nov 26 '17 at 15:27
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    $\begingroup$ there is no mistake except you thinking that $N$ has to be $\Bbb Z_4$ $\endgroup$ – mercio Nov 26 '17 at 16:14
  • $\begingroup$ Thanks a lot!! Yes, that was my mistake: I did not know that the group $D_8$ is also isomorfic to a semidirect product of the Klein group and the 2-cyclic group. $\endgroup$ – Jaime Arboleda Castilla Nov 26 '17 at 16:44

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