What is the degree of the field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$?
I know the formula
$$[\mathbb{Q}(\sqrt{2},\sqrt{3})]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$$ Since $x^2-2$ is the minimal polynomial in $\mathbb{Q}$ that has $2$ as root, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, but I am stuck with the second one..
Suppose $x^2-3$ is reducible over $\Bbb Q(\sqrt 2)$.
Since it is irreducible over $\Bbb Q$ this would mean it has two roots in $\Bbb Q(\sqrt 2) \setminus \Bbb Q$. Since the polynomial is rational they would have to be conjugate to each other.
So there are rationals $a,b$ such that $x^2-3 = (x-(a+b\sqrt 2))(x-(a-b\sqrt 2))$. From there you get $a=0$ and $2b^2 = 3$, or $(2b)^2 = 6$
But since $6$ is not the square of a rational, this is impossible.
So $x^2-3$ is irreducible over $\Bbb Q(\sqrt 2)$, and $[\Bbb Q(\sqrt 2,\sqrt 3) : \Bbb Q(\sqrt 2)] = 2$