# Galois group of $x^4 - 2x^2 - 6$ - generators [duplicate]

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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!

## marked as duplicate by Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 25 at 12:39

• Great question. – Kaj Hansen May 25 at 11:01
• 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 $D_8$. – rae306 May 25 at 11:23
• 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. – EpsilonDelta May 25 at 11:30

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.

• And a great answer. Really satisfying thread. – Kaj Hansen May 25 at 11:37
• 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. – DesmondMiles May 25 at 11:38
• @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. – Kaj Hansen May 25 at 11:52
• 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)? – DesmondMiles May 25 at 11:57