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Let $L = \mathbb{Q}(\sqrt{2}, \sqrt{3},\sqrt{5})$.

(a) Given a basis for $L/\mathbb{Q}$.

(b) List the elements of $Gal(L/\mathbb{Q})$ and show that $|Gal(L/\mathbb{Q})| = [L/\mathbb{Q}]$. Which known group is isomorphic to $Gal(L/\mathbb{Q})$?

(c) For each of the following fields $K_{i}$, determine the subgroup $H_{i}$ of $Gal(L/\mathbb{Q})$ corresponding to $K_{i}$ by the Galois correspondence: $$K_{1} = \mathbb{Q}(\sqrt{10}),\quad K_{2} = \mathbb{Q}(\sqrt{6} + \sqrt{15}),\quad K_{3} = \mathbb{Q}(\sqrt{2} + \sqrt{5}),\quad K_{4} = \mathbb{Q}(\sqrt{30}).$$

I already showed the items (a) and (b). If I'm correct, in the item (b), the elements of $Gal(L/\mathbb{Q})$ are the automorphisms that permutes $\sqrt{2},\sqrt{3},\sqrt{5}$ and their inverse. But, I don't know how to use it for find the subgroups correspondents. Can anybody help me?

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    $\begingroup$ Take $K_1=\mathbb{Q}(\sqrt{10})$ for example. That has degree 2 over $\Bbb{Q}$, so it must be fixed by a subgroup of order 4. Now obviously the automorphism that sends $\sqrt{3}$ to $-\sqrt{3}$ and fixes $\sqrt{2}$ and $\sqrt{5}$ fixes $K_1$. Also the automorphism which sends $\sqrt{2}$ to $-\sqrt{2}$ and $\sqrt{5}$ to $-\sqrt{5}$ fixes $K_1$. The product of those 2 automorphisms plus the identity constitutes the needed subgroup of ordr 4. $K_2$ has degree $4$ over $\Bbb{Q}$, so there should be a subgroup of order 2 which fixes $K_2$. Which one? $\endgroup$ – sharding4 Jun 17 '18 at 2:25
  • $\begingroup$ @sharding4, about $K_{1}$. For the second automorphism, I can define $\sqrt{3}$ to $-\sqrt{3}$? And, The subgroups would be $\lbrace id, \tau, \sigma, \tau\sigma \rbrace$ where $\tau, \sigma$ are the automorphism that you define, respectively? This groups is ciclic, right? $\endgroup$ – Corrêa Jun 17 '18 at 2:56
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    $\begingroup$ Not cyclic. $Gal(L/\Bbb{Q})$ is elementary ableian, i.e. a product of cyclic groups of order 2, and so are all its subgroups. You need 3 generators for $Gal(L/\Bbb{Q})$. Say $\rho, \sigma, \tau$ with $\rho: \sqrt{2} \mapsto -\sqrt{2}, \sigma: \sqrt{3} \mapsto -\sqrt{3}, \tau: \sqrt{5} \mapsto -\sqrt{5}$ Using that notation $K_1$ is fixed by $\{id, \sigma, \rho \tau, \rho \sigma \tau \}$ $\endgroup$ – sharding4 Jun 17 '18 at 3:11
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As an example, consider $K_2$. An element of the Galois group fixes $K_2$ pointwise iff it fixes $\alpha=\sqrt 6+\sqrt{15}$. The images of $\alpha$ under the Galois group are the four numbers $\pm\sqrt6\pm\sqrt{15}$. So the subgroup fixing $\alpha$ has order $8/4=2$, and consists of the identity and one more element $\sigma$. Bearing in mind that for each of the square roots $\sqrt2$, $\sqrt3$ and $\sqrt5$, $\sigma$ either fixes it or negates it, the $\sigma$ in question must have $\sqrt2\mapsto-\sqrt2$, $\sqrt3\mapsto-\sqrt3$ and $\sqrt5\mapsto-\sqrt5$.

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