- #76

member 587159

I understand my mistake: I missed the [itex]\alpha \in K[/itex] requirement. We don't have [itex]\alpha\in\mathbb Q(\beta)[/itex], hence argument that followed is gibberish.

Put [itex]\alpha := \sqrt{1-\sqrt{3}}[/itex] and [itex]\beta := \sqrt{1+\sqrt{3}}[/itex]. The polynomial [itex]x^4-2x^2 -2[/itex] is irreducible by Eisenstein, hence minimal polynomial of [itex]\beta[/itex], which yields [itex][\mathbb Q(\beta),\mathbb Q] = 4[/itex] with a basis of [itex]\{1,\beta,\beta ^2,\beta ^3\}[/itex]. Observe that [itex]\alpha^2 = 2-\beta ^2\in\mathbb Q(\beta)[/itex] so it suffices to have [itex]\{1,\alpha\}[/itex] to generate [itex]\mathbb Q(\alpha,\beta) / \mathbb Q(\beta)[/itex] and they are linearly independent over [itex]\mathbb Q(\beta)[/itex]. Thus

[tex]

[\mathbb Q(\alpha,\beta),\mathbb Q] = 8.

[/tex]

Per definition, the Galois group is the group of automorphisms on [itex]\mathbb Q(\alpha,\beta)[/itex] that pointwise fix [itex]\mathbb Q[/itex]. If [itex]\tau[/itex] is such an automorphism and [itex]\alpha[/itex] is a root, then [itex]\tau (\alpha)[/itex] is also a root. An automorphism is therefore determined by how it acts on the roots. This has to be (isomorphic to) a subgroup of [itex]S_4[/itex], one of order [itex]8[/itex] to be exact.

One also notices that the polynomial is even so the roots come in pairs: [itex]\alpha, -\alpha[/itex] and [itex]\beta,-\beta[/itex].

I'm really looking for the exact isomorphism structure of the Galois group. There are only 5 groups of order 8 (up to isomorphism). Which one is it?