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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?

Robin
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Mathlearner
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2 Answers2

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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 $D_8$.

Community
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Sary
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  • 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. – Mathlearner Jan 28 '15 at 00:12
  • Yes. What's its order ? – Sary Jan 28 '15 at 00:13
  • its order is 2 :) but there are more elements of the galois group, right? – Mathlearner Jan 28 '15 at 00:14
  • Yes my last comment was perhaps irrelevant. Anyway it may be easy now to guess if ${\mathbf Q}(\sqrt{2+\sqrt{15}})$ generates the whole extension. Note that if that's true, then you can express all the other roots in terms of that one, by taking combinations with rational coefficients. – Sary Jan 28 '15 at 00:34
  • Yeah. And thats my problem. I already tried to express $\sqrt{2-\sqrt{15}}$ as $\sqrt{2+\sqrt{15}}$ by squaring, mutplicate them etc. to see the connection between them. But I obtained values which are "irrelevant". So I guess they are "independent". But maybe I am missing something? So they could still be a connection between those roots. Is there a possibility who I can check it for sure? – Mathlearner Jan 28 '15 at 07:51
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    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 ? – Sary Jan 28 '15 at 12:04
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    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 $\mathbb Q(\sqrt{2+\sqrt{15}})$ – Mathlearner Jan 28 '15 at 12:11
  • Hello. One more question: Just in case..The degree of the extension is $8$, correct? – Mathlearner Jan 31 '15 at 15:28
  • That is wondering me a bit. Let $x_{1,2}=\pm\sqrt{2+\sqrt{15}}$ and $x_{3,4}=\pm\sqrt{2-\sqrt{15}}$. We know that the splitting field is $L=\mathbb Q(x_1,x_3)$. Hence $[L:\mathbb Q]=[L:\mathbb Q(x_1)]\cdot [\mathbb Q(x_1):\mathbb Q]$. But the first factor is 2, since $\sqrt{15}\in \mathbb Q(x_1)$, then the minimal polynomial of $x_3$ is $x^2+\sqrt{15}-2$. The second factor is 4. Because the polynomial $x^4-4x^2-11$ is irreducible over $\mathbb Q$. (I think thats right,...well I hope so). I showed this by showing irreducibility over $\mathbb Z$.. where is the mistake? – Mathlearner Jan 31 '15 at 16:34
  • I withdraw my comment ! You are perfectly right. Sorry. There are indeed 8 generators (among which $\sqrt{11}$). Did you manage to compute the Galois group ? – Sary Jan 31 '15 at 19:14
  • You mean degree 8? Because there are just 2 generators. Or did I a mistake? :/. And I was just able to find 6 elements of the galois group. – Mathlearner Jan 31 '15 at 19:25
  • I should have been more precise : 8 generators as a vector space, but only 2 as a field extension. – Sary Jan 31 '15 at 19:27
  • Alright :). I was just able to find 6 automorphisms, but maybe I found the mistake. I know that $\mathbb Q$-automorphisms permute roots of the polynomial $f$. So, for example any $\mathbb Q$-automorphism permutes the roots of the minimal polynomial of $x_1$,$x_2$ etc. But the minimal polynomial of $x_3$ over $\mathbb Q(x_1)$ has just two roots. So I thought, the image of $x_3$ under an automorphism only can have two different images, but this is wrong, right? Since THIS minimal polynomial is irrelevant. Important is the minimal polynomial in $\mathbb Q$, correct? Which has degree 4. – Mathlearner Jan 31 '15 at 19:33
  • Hence there are more automorphisms than I thought. Hopefully you understand what I mean and where I think my mistake is.. my english is not perfect :) – Mathlearner Jan 31 '15 at 19:34
  • Hm, something is wrong, can you help me to determine all the automorphisms? – Mathlearner Jan 31 '15 at 19:50
  • That's right. But you can use the fundamental theorem of Galois theory, which tells you (for instance) that ${\rm Aut}({\bf Q}(x_1, x_3)/{\bf Q}(x_1))$ is a subgroup of the whole Galois group. – Sary Jan 31 '15 at 19:50
  • Let us continue this discussion in chat. – Mathlearner Jan 31 '15 at 19:50
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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

Robin
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Arch
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    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). – ahulpke Feb 01 '15 at 16:38
  • Hi. Where I can found this in Artins Algebra? Did not found it on p.494 – Mathlearner Feb 02 '15 at 15:17
  • Hi. Do you have 2nd edition? I combined the methods of Example 16.9.2 a) and b). – Arch Feb 02 '15 at 15:36