Basis for a Field Extension 
Let $E$ be the field extension $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ over $\mathbb{Q}(\sqrt{3})$. Find the degree and basis for $E$.

I think the degree is $2$, since $\sqrt{2} + \sqrt{3}$ still has degree $2$. But I'm having trouble finding the basis. For $\mathbb{Q}(\sqrt{3})$, we simply have the basis $B$ = {$1, \sqrt{3}$}. But, now we're adding $\sqrt{2}$ to it. Does this mean we can simply add $\sqrt{2}$ to the basis $B$? But that would have $3$ elements instead of $2$, which doesn't match the degree. How should I approach this problem? Thank you.
 A: $$r=\sqrt2+\sqrt3\implies r^2-2\sqrt3\,r+3=2\implies r\;\;\text{is a root of the polynomial}\;\;$$
$$f(x)=x^2-2\sqrt3\,x+1\in\Bbb Q(\sqrt3)[x]$$
and thus $\;[\Bbb Q(\sqrt2+\sqrt3):\Bbb Q(\sqrt3)]\le2\;$ . But this degree cannot be one, otherwise:
$$\sqrt2+\sqrt3\in\Bbb Q(\sqrt3)\implies\;\exists\,a,b\in\Bbb Q\;\;s.t.\;\;\sqrt2+\sqrt3=a+b\sqrt3\implies$$
$$\sqrt2-a=(b-1)\sqrt3\implies a^2+2-2a\sqrt2=3(b-1)^2\implies$$
$$\implies\sqrt2=\frac{a^2+2-3(b-1)^2}{2a}\in\Bbb Q$$
(also $\;a=0\;$ leads to a straighfroward contradiction).
A: For $\alpha=\sqrt{2}+\sqrt{3}$, you have.
$$(\alpha-\sqrt{3})^2- 2=0$$ Hence $\alpha$ is a root of the polynomial
$$P(x) = x^2 - 2 \sqrt{3}x +1.$$ $P$ belongs to $\mathbb Q(\sqrt{3})[X]$ and is irreducible. To prove it, prove that $\sqrt{2}$ cannot be written as $a+b\sqrt{3}$ with $a,b \in \mathbb Q$.
Hence the degree of $\mathbb Q(\sqrt{2}+\sqrt{3})$ over $\mathbb Q(\sqrt{3})$ is equal to $2$ and a basis is $(1,\sqrt{2}+\sqrt{3})$ and as $\sqrt{3} \in \mathbb Q(\sqrt{3})$ another basis is $(1,\sqrt{2})$.
