0
$\begingroup$

This question already has an answer here:

The splitting field of $x^4 - 2x^2 - 6$ is $\mathbb{Q}(\alpha, \sqrt{-6}) = \mathbb{Q}(\alpha, \beta)$ where $\alpha = \sqrt{1+\sqrt{7}}$, $\beta = \sqrt{1-\sqrt{7}}$. From the first representation, $x^4 - 2x^2 - 6$ being irreducible (say, Eisenstein for $2$) and $\sqrt{-6}$ being non-real it follows that the extension (and hence the Galois group) has order $8$.

Now one can sneakily show that the group is $D_8$ as follows -- it is not Abelian, as then by FTGT any intermediate extension must be Galois, whereas $\mathbb{Q}(\alpha) : \mathbb{Q}$ is not. On the other hand, $(\alpha \to -\alpha)$ (and fix the rest) and $(\sqrt{-6} \to -\sqrt{-6})$ (and fix the rest) are two distinct morphisms of order $2$, hence the group cannot be quaternion. So it must be $D_8$.

But what about a set of generators? I think that $(\sqrt{-6} \to -\sqrt{-6})$ can be the reflection one but I can't think of a suitable rotation one.

Any help appreciated!

$\endgroup$

marked as duplicate by Jyrki Lahtonen abstract-algebra May 25 at 12:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Great question. $\endgroup$ – Kaj Hansen May 25 at 11:01
  • 2
    $\begingroup$ An other way to see that the Galois group must be $D_8$ is this. Since the polynomial is irreducible, the Galois group acts transitively on the roots. Furthermore it is a subgroup of $S_4$, and the only transitive subgroup of $S_4$ with order 8 is $D_8$. $\endgroup$ – rae306 May 25 at 11:23
  • $\begingroup$ Yet another way to see it is $D_8$ is to observe that $D_8$ is the only group of order $8$ with a non-normal subgroup. $\endgroup$ – EpsilonDelta May 25 at 11:30
2
$\begingroup$

The roots of $X^4-2X^2-6$ are $\alpha_1=\sqrt{1+\sqrt{7}},\alpha_2=\sqrt{1-\sqrt{7}}$, $\alpha_3=-\alpha_1$ and $\alpha_4=-\alpha_2$.

You have already determined that the Galois group is isomorphic to the dihedral group $D_4=\langle\rho,\sigma\mid \rho^4=1,\sigma^2=1,\sigma\rho\sigma^{-1}=\rho^{-1}\rangle$.

To give explicit generators, we seek an element $\rho$ of order $4$ and and element $\sigma$ of order $2$ that fulfill the relation $\sigma\rho\sigma^{-1}=\rho^{-1}.$

An element $f$ of the Galois group can send $\alpha_1$ to any of the four other roots, then there are two possibilites left for $f(\alpha_2)$. (Since $f(\alpha_1)$ and $f(\alpha_3)=f(-\alpha_1)=-f(\alpha_1)$ are already determined.)

Take $\rho:\begin{cases}\alpha_1 \longmapsto \alpha_2 \\ \alpha_2\longmapsto -\alpha_1\end{cases}$ and $\sigma:\begin{cases}\alpha_1 \longmapsto \alpha_2 \\ \alpha_2\longmapsto \alpha_1\end{cases}$. Verify as an exercise that these do the job.

$\endgroup$
  • $\begingroup$ And a great answer. Really satisfying thread. $\endgroup$ – Kaj Hansen May 25 at 11:37
  • $\begingroup$ Yeah, that's actually my problem - I was also thinking about the $\rho$ map. Why is it well-defined, i.e. how are we sure there is such an automorphism? Sorry if the question is dumb. $\endgroup$ – DesmondMiles May 25 at 11:38
  • 1
    $\begingroup$ @DesmondMiles, we are sort of meeting in the middle: on one hand, we know $\text{Gal}(f) \cong D_4$, and so this element of order $4$ we're looking for must exist. On the other side, we know an element of $\text{Gal}(f)$ is determined uniquely by its action on the roots of $f$. And so we reason as above about where roots of $f$ would have to go to be of order $4$. Such an automorphism must exist, and what it could be is then forced by the logic. $\endgroup$ – Kaj Hansen May 25 at 11:52
  • $\begingroup$ That's cool :) But what if I didn't know at first that the group is $D_8$ and I just tried to compute it via finding a presentation (generators + relations)? $\endgroup$ – DesmondMiles May 25 at 11:57

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