Why is $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$? I need to use the so called Ring Tower Theorem to show that $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$, but I'm quite confused with the notation and some concepts. First of all, my book says that if $a$ in some extension $E$ of $\mathbb{Q}$ is algebraic over $\mathbb{Q}$ then the elements of $\mathbb{Q}(a)$ can be written in the form $p+qa$, with $p,q\in\mathbb{Q}$. Hence we know that, since $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(\sqrt2+\sqrt3)$, elements of $\mathbb{Q}(\sqrt2,\sqrt3)$ can be written in the form $p+(\sqrt2+\sqrt3)q$. But doesn't this imply that the the dimension of the basis for $\mathbb{Q}(\sqrt2,\sqrt3)$ is 2, like it is for $\mathbb{Q}(\sqrt2)$?

The Tower Theorem says that if $E$ is a finite extension field of the field $G$ and $G$ is a finite extension of the field $F$, then $E$ is a finite extension of the $F$, and $[E:F]=[E:G][G:F]$.

If I try to apply this theorem to the case in question, I would do as follows:
$$[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)][\mathbb{Q}(\sqrt2):\mathbb{Q}] $$
So I know that $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$, but how do I find $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)]$?

Let $x=\sqrt 2+\sqrt 3$. Then $x-\sqrt 3=\sqrt 2$. So $x^2-2\sqrt 3x+1=0$. But this is not the minimum degree polynomial over $\mathbb{Q}$ since $-2\sqrt 3 \notin\mathbb{Q}$. Thus with $x^2+1=2\sqrt 3x$ squaring both sides gives $x^4-10x^2+1=0$. Hence $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$.
