4
$\begingroup$

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?

Thanks in advance.

$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

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 $D_8$.

$\endgroup$
17
  • $\begingroup$ 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. $\endgroup$
    – Mathlearner
    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? $\endgroup$
    – Mathlearner
    Jan 28, 2015 at 0:14
  • 2
    $\begingroup$ 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
  • 5
    $\begingroup$ 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 $\mathbb Q(\sqrt{2+\sqrt{15}})$ $\endgroup$
    – Mathlearner
    Jan 28, 2015 at 12:11
4
+25
$\begingroup$

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($L/\mathbb{Q}$) $=$ $<\sigma , \tau> \cong$ $<(1234),(24)>=D_8$.

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

$\endgroup$
3
  • 1
    $\begingroup$ 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 $\endgroup$
    – Mathlearner
    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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.