How is $[Q(\sqrt2, \sqrt3 ) : Q(\sqrt2)]=2$? $\mathbb{Q}$ is the rationals. I know that $\sqrt3 \notin \mathbb{Q}(\sqrt2)$ but so what? The answer to this question seems to be based upon that. Really don't understand what that means in finding the minimal polynomial.

Suppose $\sqrt3\in Q(\sqrt2)$, then write $\sqrt3=a+b\sqrt2, a,b\in Q$, 
Remark that $a,b$ are not zero, if $a=0$, then $\sqrt2\sqrt3=b(\sqrt2)^2=2b=\sqrt6$, this is not true since $\sqrt6$ is not in $Q$. Suppose $b=0$, then $\sqrt3=a\in Q$ this is not true.
you obtain $3=(a+b\sqrt2)^2=a^2+2ab\sqrt2+2b^2$. This implies that $\sqrt2={{3-a^2-2b^2}\over{2ab}}\in Q$. This is not true, so $\sqrt3$ is not in $Q(\sqrt2)$.
Since the minimal polynomial of $\sqrt3\in Q(\sqrt2,\sqrt3)$ is $x^2-3$. You deduce that $[Q(\sqrt2,\sqrt3):Q(\sqrt2)]\leq 2$ thus is $2$ since $\sqrt3$ is not an element of $Q(\sqrt2)$.
