Degree of extension $\mathbb{Q}(\sqrt{2},\sqrt{6})$ over $\mathbb{Q}(\sqrt{3})$ Find degree of extension$[\mathbb{Q}(\sqrt{2},\sqrt{6}):\mathbb{Q}(\sqrt{3})]$.
I think that $\sqrt{6}=\sqrt{2}\sqrt{3}$, then  a basis of vector space $\mathbb{Q}(\sqrt{2},\sqrt{6})$ over field $\mathbb{Q}(\sqrt{3})$ is $B=\{1,\sqrt{2}\}$. Please, can someone help me!
 A: Since $\mathbb{Q}(\sqrt 2, \sqrt 6) = \mathbb{Q}(\sqrt 2,\sqrt 3)$, $[\mathbb{Q}(\sqrt2,\sqrt 6):\mathbb{Q}(\sqrt 3)] = [\mathbb{Q}(\sqrt2,\sqrt 3):\mathbb{Q}(\sqrt 3)]$. It remains to show that $[\mathbb{Q}(\sqrt 2,\sqrt3):\mathbb{Q}(\sqrt3)] = 2$.
Since $\sqrt 2$ is zero of $x^2-2 \in \mathbb{Q}(\sqrt 3)[x]$, $[\mathbb{Q}(\sqrt 2,\sqrt3):\mathbb{Q}(\sqrt3)] \le 2$.
Next show that $\sqrt 2\notin \mathbb{Q}(\sqrt 3) = \{a+b\sqrt 3\mid a,b\in \mathbb{Q}\}$. Suppose $\sqrt 2 = a+b\sqrt 3$ for some $a,b\in \mathbb{Q}$. Square both sides to get some contradiction.
A: $\{1,\sqrt2\}$ is the basis of $Q(\sqrt2,\sqrt6)$ over $Q(\sqrt3)$. You have remarked that it generates $Q(\sqrt2,\sqrt6)$. You just have to show that it is free.
$a+b\sqrt2=0, a,b\in Q(\sqrt3)$ implies that $\sqrt2 =u+v\sqrt3, u,v\in Q$, the square of the last equations implies that $uv\sqrt3\in Q$, this implies that $u=0$ or $v=0$. If $v=0$, you have $\sqrt2\in Q$ impossible. If $u=0$, write $v={e\over f}$ where $e$ and $f$ are relatively prime, you have $2f^2=3e^2$, thus $2$ divides $e$ $e=2e'$ and $2f^2=12{e'}^2$, $f^2=6{e'}^2$, you deduce that $2$ divides $f$. Contradiction since it is assumed that $e$ and $f$ are relatively prime.
