Field Extensions - Algebraic Extensions

Question

I have a question, How can I prove that $[\mathbb{Q}({\sqrt{2},\sqrt{5},\sqrt{10}}):\mathbb{Q}]= 4$?.

Thank you.

Answer 1

Hint: $\mathbb{Q}(\sqrt{2},\sqrt{5},\sqrt{10}) = \mathbb{Q}(\sqrt{2},\sqrt{5})$

Answer 2

First, notice that $\mathbb{Q}({\sqrt{2},\sqrt{5},\sqrt{10}})=\mathbb{Q}({\sqrt{2},\sqrt{5}})$ as $\sqrt {10}$ is in that field. Second, it is possible (for example with Eisenstein's criterion) to show that $x^2-2$ and $x^2-5$ are irreducible. The only thing that's left then would be showing that they cannot be acquired from each other which is a result of the fact they are relatively prime, and you can show this by showing that the set $\{1,\sqrt 2,\sqrt 5,\sqrt {10}\}$ is linearly independent over $\mathbb Q$ (which is almost directly from the fact that the two prime roots are irrational).

Answer 3

$\sqrt{10} = \sqrt{2} \sqrt{5}$, so $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{10}) = \mathbb{Q}(\sqrt{2}, \sqrt{5})$. $\sqrt{2} \not\in \mathbb{Q}$, and $(\sqrt{2})^2 - 2 = 0$, so $\mathrm{deg}_{\mathbb{Q}}(\sqrt{2}) = 2 = |\mathbb{Q}(\sqrt{2}) : \mathbb{Q}|$. $\sqrt{5} \not\in \mathbb{Q}$, and $(\sqrt{5})^2 - 5 = 0$, so $\mathrm{deg}_{\mathbb{Q}}(\sqrt{5}) = 2 = |\mathbb{Q}(\sqrt{5}) : \mathbb{Q}|$. Therefore, $|\mathbb{Q}(\sqrt{2}, \sqrt{5}) : \mathbb{Q}| \in \{2,4\}$. Perhaps you know $\sqrt{5} \not \in \mathbb{Q}(\sqrt{2})$. Otherwise, suppose $\sqrt{5} \in \mathbb{Q}(\sqrt{2})$, then${}^{*}$ $\exists b,c \in \mathbb{Q}$ such that either $b \sqrt{2} = c \sqrt{5}$, so $\sqrt{5/2} \in \mathbb{Q}$, a contradiction, or $1 + b \sqrt{2} = c \sqrt{5}$, $1 + 2 \sqrt{2} b + 2 b^2 = 5 c^2$, so $\sqrt{2} \in \mathbb{Q}$, also a contradiction.

${}^{*}$: $a + b \sqrt{2} = c \sqrt{5}$. If $a = 0$, $\frac{b}{c} = \sqrt{5/2}$. Alternatively, $a \neq 0$, so $1 + \frac{b}{a} \sqrt{2} = \frac{c}{a} \sqrt{5}$.