Galois theory : Intermediate fields of $L = \mathbb{Q} ( \sqrt 2 , \sqrt 3 , \sqrt 5)$

Question

I want to get all the intermediate fields of $L=\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$.

Please help me and give me advice.

Answer 1

The Galois Group of Your Extenson is $\left(\mathbb{Z} / 2\mathbb{Z}\right)^3$.

Now use the Galois Correspondence to link every subgroup of $\left(\mathbb{Z} / 2\mathbb{Z}\right)^3$ to a subfields of $L$.

Note that these are all the subextension of $L$ (anf that are all Galois Extensions)

Answer 2

Let $L=\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)$. Then $L/\mathbb Q$ has degree 8. By the way, element of Galois group has to through permute roots of $x^2-2$, permute roots of $x^2-3$ and roots of $x^2-5$. Then all element has order $2$, and thus $$\text{Gal}(L/\mathbb Q)=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z.$$

Find the subgroup of $\text{Gal}(L/\mathbb Q)$ and conclude using the Galois correspondance theorem.

Answer 3

The Galois group of the extension is $G=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, now:

$1)$ Show that besides trivial subgroups, $G$ has $6$ subgroups of order $2$, and $6$ subgroups of order $4$.

$2)$ Conclude that these are all the intermediate fields of $L/\mathbb{Q}$.

$$\mathbb{Q},$$$$\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{5}),\mathbb{Q}(\sqrt{6}),\mathbb{Q}(\sqrt{10}),\mathbb{Q}(\sqrt{15}),$$$$\mathbb{Q}(\sqrt{2},\sqrt{3}),\mathbb{Q}(\sqrt{2},\sqrt{5}),\mathbb{Q}(\sqrt{3},\sqrt{5}),\mathbb{Q}(\sqrt{5},\sqrt{6}),\mathbb{Q}(\sqrt{10},\sqrt{3}),\mathbb{Q}(\sqrt{15},\sqrt{2}),$$$$\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}).$$