TITLE: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
QUESTION [82 upvotes]: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ? 

$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$$ 
$$\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace  $$
So if an element is in $\mathbf Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbf{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.
How to conclude from there?

REPLY [2 votes]: Another solution:
Clearly it has to be a subfield of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$. Thus its degree has to be one of these numbers: $1,2,4$. We just have to rule out it is of degree $2$. If it is, then there is a minimal polynomial of degree two s.t $\sqrt{2}+\sqrt{3}$ is its root. However, if there was such polynomial, say $x^2+bx+c$ then $5+c+2\sqrt{6}+b(\sqrt{3}+\sqrt{2})=0$ which is impossible as $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$ and the equation implies it is linearly dependent!