# Is Q(√2,√3)=Q(√2+√3)\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})?

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ?

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?

$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.
Now note that hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence
$\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3})$.