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Task: Determine the Galois group of $x^4-4x^2-11$ over $\mathbb{Q}$.

The roots are obviously $\pm\sqrt{2\pm\sqrt{15}}$.

But I have problems in checking whether just $\sqrt{2+\sqrt{15}}$ is generating the extension or the other root is also needed.

Any ideas?

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2 Answers 2

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Hint: Are the roots real ? If not, how are they located ? Can you provide non-trivial elements of the Galois group ? Remember that it must send a root to a root.


Actually a similar question was answered before. Here the Galois group is a subgroup of the dihedral group $D_8$ (the symmetry group of the square) because the roots come in opposite pairs. Since it contains a transposition (the complex conjugation) and is transitive, it is the whole of $D_8$.

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  • $\begingroup$ Hi. $\pm\sqrt{2-\sqrt{15}}$ are complex roots. What you mean with "how are they located"?. One non-trivial element of the galois group is the conjugation, which swaps both complex roots. $\endgroup$ Jan 28, 2015 at 0:12
  • $\begingroup$ Yes. What's its order ? $\endgroup$
    – Sary
    Jan 28, 2015 at 0:13
  • $\begingroup$ its order is 2 :) but there are more elements of the galois group, right? $\endgroup$ Jan 28, 2015 at 0:14
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    $\begingroup$ It's better to write $i\sqrt{\sqrt{15}-2}$. Where are located the elements of ${\bf Q}(\sqrt{2+\sqrt{15}})$ in the complex plane ? $\endgroup$
    – Sary
    Jan 28, 2015 at 12:04
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    $\begingroup$ Ah, I know now what you want to say. The field $\mathbb Q(\sqrt{2+\sqrt{15}})$ is a subset of the reals, but the other root is a complex number, hence cant be in $\mathbb Q(\sqrt{2+\sqrt{15}})$ $\endgroup$ Jan 28, 2015 at 12:11
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I rewrote this proof after pointed out by ahulpke:

Let $L = \mathbb{Q}(\sqrt{2+\sqrt{15}}, \sqrt{2-\sqrt{15}})$.

$\sqrt{2+\sqrt{15}} \cdot \sqrt{2-\sqrt{15}}=\sqrt{-11} \not \in \mathbb{Q}(\sqrt{2+\sqrt{15}})$, so that |Gal($L / \mathbb{Q}$)| = $8$.

First, there exists $\tau \in$ Gal($L/\mathbb{Q}(\sqrt{2+\sqrt{15}})$) s.t. $\tau(\sqrt{-11}) = -\sqrt{-11}$, since $x^2+11$ is the minimal polynomial for $\sqrt{-11}$ over $\mathbb{Q}(\sqrt{2+\sqrt{15}})$, then $\tau^2 = id$.

We can take $\sigma \in $ Gal($L/\mathbb{Q}$) s.t. $\sigma(\sqrt{2+\sqrt{15}}) \mapsto \sqrt{2-\sqrt{15}} $.

Then, $\sigma(\sqrt{2+\sqrt{15}} \cdot \sqrt{2-\sqrt{15}}) = \sigma(\sqrt{-11})\mapsto \pm \sqrt{-11}$, and we can choose $\sigma(\sqrt{-11}) \mapsto -\sqrt{-11}$, by interchanging $\sigma $ with $\sigma \tau$, if necessary.

Then $ \sqrt{2-\sqrt{15}} \cdot \sigma(\sqrt{2-\sqrt{15}}) = - \sqrt{2+\sqrt{15}} \cdot \sqrt{2-\sqrt{15}}$, so that $\sigma(\sqrt{2-\sqrt{15}}) = - \sqrt{2+\sqrt{15}}$.

Constructing an isomorphism, Gal($L/\mathbb{Q}$) $=$ $\langle \sigma , \tau \rangle \cong$ $\langle (1234),(24) \rangle=D_8$.

This is basically the same approach as Artin's Algebra p.494

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    $\begingroup$ No, $\sqrt{15}$ and $\sqrt{-11}$ do not span the splitting field, but a normal subfield of index $2$ (corresponding to the derived subgroup of the Galois group, which is the dihedral group of order 8). $\endgroup$
    – ahulpke
    Feb 1, 2015 at 16:38
  • $\begingroup$ Hi. Where I can found this in Artins Algebra? Did not found it on p.494 $\endgroup$ Feb 2, 2015 at 15:17
  • $\begingroup$ Hi. Do you have 2nd edition? I combined the methods of Example 16.9.2 a) and b). $\endgroup$
    – Arch
    Feb 2, 2015 at 15:36

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