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Calculate the Galois Group $G$ of $K$ over $F$ when $F=\mathbb{Q}$ and $K=\mathbb{Q}\big(i,\sqrt2,\sqrt3 \big)$.

My thoughts are as follows:

By the Tower Lemma, we can see that $|\text{Gal}(K/F)|=8$, since $K$ is a degree $8$ extension over $F$.

Now, $\phi \in\text{Gal}(K/F)$ where $\phi$ represents complex conjugation. This satisfies $\phi ^2 =\text{Id}$. Similarly, the map which switches $\sqrt2$ and $\sqrt3$ around is also an automoprhism fixing $\mathbb{Q}$, say $\tau$, satisfying $\tau ^ 2=\text{Id}$.

How can I find the other elements of the Galois Group? Have I missed an easier method?

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  • $\begingroup$ Switching $\sqrt 2$ and $\sqrt 3$ doesn't give you an automorphism fixing $\Bbb Q$. You want $\sqrt 2 \mapsto - \sqrt 2$. $\endgroup$ Apr 21, 2014 at 17:59
  • $\begingroup$ Ah, of course, because $2 \mapsto 3$. Thanks for pointing that out. $\endgroup$
    – Evariste
    Apr 21, 2014 at 18:02

2 Answers 2

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It's just $(\mathbb{Z}/2\mathbb{Z})^3$.

To see this, indeed complex conjugation is an automorphism of order 2. However, so are the automorphisms sending $\sqrt{2}\mapsto -\sqrt{2}$, and $\sqrt{3}\mapsto-\sqrt{3}$ (and fixing the rest). Hence, you have 3 elements of order 2, which commute and generate your group (check this!). These three automorphisms then give you a nice isomorphism to $(\mathbb{Z}/2\mathbb{Z})^3$.

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  • $\begingroup$ Thanks a lot, I've realise thanks to @Ayman Hourieh that my $\tau$ is not in fact a homomorphism. Thanks for filling in the missing pieces for me. $\endgroup$
    – Evariste
    Apr 21, 2014 at 18:03
  • $\begingroup$ How do we know that complex conjugation is an automorphism? $\endgroup$
    – thinker
    Apr 8, 2016 at 21:03
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Though your question has already been answered, I think it's important to emphasize exactly why these roots can only go to their negatives (perhaps for readers in the future).

Take an arbitrary $\mathbb{Q}$-automorphism $\phi$. What if $\phi(\sqrt{2}) = \sqrt{3}$? Well, we get the following contradiction:

$$\phi(2) = \phi((\sqrt{2})^2) = \phi(\sqrt{2}) \phi(\sqrt{2})=\sqrt{3}\sqrt{3} = 3$$

And this is no good because $\phi(2) = 2$ by virtue of $\phi$ fixing $\mathbb{Q}$.

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  • $\begingroup$ Thanks for that - I'd picked up on this in response to one of the comments earlier, but your exposition adds clarity. $\endgroup$
    – Evariste
    Apr 21, 2014 at 19:25

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