Why is $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] \le 4$

Question

I am working on the following exercise:

Show that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $4$ over $\mathbb{Q}$ by showing that $1,\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent.

I have shown that $1,\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent, but if I am not mistaken this only shows that the degree of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ is at least $4$, i.e. $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] \ge 4$. I do not see how this should also prove that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] \le 4$. Could you explain that to me?

Answer 1

That it has at most $4$ follows from the fact that $p(x)=(x^2-2)(x^3-3)$ has degree $4$ and splits over $\Bbb Q(\sqrt{2},\sqrt{3})$. So that's already obvious. (By standard results the minimal polynomial of this extension, whose degree equals $[\Bbb Q(\sqrt{2},\sqrt{3}): \Bbb Q]$, must thus be a divisor of $p$ and so $2$ or $4$).

Answer 2

Hint : $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]\cdot[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$.