Is it obvious that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)$? Is it obvious that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)$ ? If not how can I show it ?
 A: The first inclusion is obvious because $\mathbb{Q}(\sqrt{3},\sqrt{5})$
 is closed under addition. So $\mathbb{Q}(\sqrt{3}  +\sqrt{5}) \subseteq \mathbb{Q}(\sqrt{3}  ,\sqrt{5})$. 
To second inclusion observe that
$$(\sqrt{3}  +\sqrt{5})(\sqrt{3}  -\sqrt{5}) = 3 - 5 = -2 \Rightarrow \sqrt{5}  -\sqrt{3} = \frac{-2}{\sqrt{5}  +\sqrt{3}}$$
then $\sqrt{5}  -\sqrt{3} \in \mathbb{Q}(\sqrt{3}  +\sqrt{5})$. Now use sum and subtraction to find that $\sqrt{3}, \sqrt{5} \in \mathbb{Q}(\sqrt{3}  +\sqrt{5})$. 
Can you take it from here? 
A: It is obvious if the facts below are obvious:


*

*$\mathbb Q(\sqrt 3,\sqrt 5)$ has degree 4

*$\mathbb Q(\sqrt 3+\sqrt 5)$ has degree 4

*$\mathbb Q(\sqrt 3+\sqrt 5) \subseteq \mathbb Q(\sqrt 3,\sqrt 5)$
Of these, only the last one is really obvious.
A: $(\sqrt3+\sqrt 5)^2\in \mathbb Q(\sqrt 3+\sqrt  5)$
$\implies 8+2\sqrt3.\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt  5)$
$\implies \sqrt3.\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt  5)$
$\implies (\sqrt3+\sqrt 5).\sqrt3.\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt  5)$
$\implies 3\sqrt 5+5\sqrt3\in \mathbb Q(\sqrt 3+\sqrt  5)$
$3(\sqrt3+\sqrt 5)\in \mathbb Q(\sqrt 3+\sqrt  5)$
$\implies 2\sqrt 3\in \mathbb Q(\sqrt 3+\sqrt  5)$
$\implies \sqrt 3\in \mathbb Q(\sqrt 3+\sqrt  5)$
similarly $\sqrt 5\in \mathbb Q(\sqrt 3+\sqrt  5)$
Thus $\mathbb Q(\sqrt 3+\sqrt  5)\subseteq \mathbb Q(\sqrt3,\sqrt5)$
The other part is obvious
