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Let $\alpha=\sqrt{2+\sqrt{3}}$.

Let $L=\mathbb{Q}(\alpha)$.

Show that $L$ is a normal extension of $\mathbb{Q}$.

I know that we need to prove that the Galois group for this extension is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z_2}$ (which i dont know how to prove ). An extension is normal if and only if the subgroup of automorphisms which fix each intermediate field is a normal subgroup of the Galois group. When this group is abelian, every subgroup is normal and thus every extension is Galois.

Thanks

moona
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1 Answers1

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We know already that $\mathbb{Q}(\sqrt{2+\sqrt{3}})=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is a biquadratic field, because $2^2-3=1$ is a square:

Proving $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}\iff a^{2}-b$ is a square

These biquadratic fields are Galois over $\mathbb{Q}$, because they are separable (the field has characteristic zero) and normal (the roots of $(x^2-m)(x^2-n)$ are again in the field).

Edit: Here is actually a duplicate:

Determine the Galois group of $\mathbb{Q}(\sqrt{a+b\sqrt{d}})$

We see that the extension is normal, if $a^2-db^2$ is a square. For $(a,b,d)=(2,3,1)$ we obtain our case.

Dietrich Burde
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