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I need to show that the extension $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$, and compute its Galois group.

I am learning Galois theory by myself and got stuck in this exercise. I know the fundamental theorem of Galois theory. Any help would be useful

Thanks

Gottfried
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    Have you tried computing the minimal polynomial for $\sqrt{2+\sqrt{2}}$? – Jeff Apr 01 '15 at 01:53
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    What are all the conjugates of $\sqrt{2+\sqrt{2}}$? – lhf Apr 01 '15 at 02:23
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    The idea that the previous two commenters were leading you towards is this: if $f(X)$ is the minimal polynomial for your given irrationality $\alpha$, you want to show that your field is already the splitting field of $f$. This means showing that the other roots of $f$ are already in $\Bbb Q(\alpha)$. – Lubin Apr 01 '15 at 04:12
  • Related: https://math.stackexchange.com/questions/1241221, https://math.stackexchange.com/questions/575171 – Watson Dec 26 '16 at 10:17
  • Related: https://math.stackexchange.com/questions/1367383 – Watson Dec 26 '16 at 12:56

1 Answers1

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Let $\alpha = \sqrt{2 + \sqrt{2}}$.

An extension is Galois if it is separable and normal. Both of these properties can be resolved by considering the minimal polynomial $p$ of $\alpha$ (i.e. $p(x) = 0$).

Now $(\alpha^2 - 2)^2 - 2 = 0$ so the minimal polynomial is $x^4 - 4x^2 + 2$.

The extension is Galois since this polynomial doesn't have repeated roots. That is easy to verify by computing the gcd of $p$ with its derivative $p'$.

We can show the extension is normal by showing that $p$ splits in the field $\mathbb Q(\alpha)$. The four roots of $p$ are $\pm \sqrt{2 + \pm \sqrt{2}}$ so if we can construct $\alpha' = \sqrt{2 + \sqrt{2}}$ from $\alpha$ we would have $p(x) = (x-\alpha)(x+\alpha)(x-\alpha')(x+\alpha')$.

Observe that:

  • $\alpha^2 - 2 = \sqrt{2}$
  • $1/\alpha = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}\sqrt{2 - \sqrt{2}}} = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2^2 - 2}} = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2}}$

so $\alpha' = (\alpha^2 - 2)/\alpha$.

This proves that the extension is Galois.

Now to compute the Galois group. The extension can be taken in two, degree two, steps: first adjoin $\sqrt{2}$ (a root of $x^2-\sqrt{2}$) to $\mathbb Q$ then a root of $x^2-(2+\sqrt{2})$. $$[\mathbb Q : \mathbb Q(\sqrt{2+\sqrt{2}})] = [\mathbb Q : \mathbb Q(\sqrt{2})] [ \mathbb Q(\sqrt{2}) : \mathbb Q(\sqrt{2+\sqrt{2}})] = 4.$$ so by Galois theory the group too should have order 4: The group must be $C_4$ or $C_2 \times C_2$.

Next we should look at the automorphisms in detail to determine which group type this is. From Galois theory we know the group acts transitively on the roots, i.e. there is an automorphism $\sigma$ that maps $\alpha$ to any of the other roots.

Suppose $\sigma (\alpha) = \alpha'$, then $\sigma (-\alpha) = -\alpha'$ and using the relation defining $\alpha'$ compute that $\sigma (\alpha') = -\alpha$. So $\sigma$ has order for, generating $C_4$.

Yulia Alexandr
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Brennan.Tobias
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