$\mathbb Q(\sqrt2 + \sqrt3)$ is what set? I thought $$\mathbb Q(\sqrt2 + \sqrt3) = \left\{a + b\left(\sqrt2+\sqrt3\right) | a,b \in \mathbb Q\right\}$$ I appears I am mistaken. But what set is it?
 A: $\mathbb{Q}(\sqrt 2 + \sqrt3)$ is the smallest field containing both $\mathbb{Q}$ and $\sqrt{2}+\sqrt{3}$. Since the minimal polynomial $x^4-10x^2+1$ of $\sqrt{2}+\sqrt{3}$ is of degree $4$ this is a field extension of degree $4$, which turns out to be equal to $$\mathbb{Q}(\sqrt{2},\sqrt{3}) = \{a + b\sqrt2 + c \sqrt3 + d\sqrt6 : a,b,c,d \in \mathbb{Q} \}.$$
A: The notation $\,\Bbb Q(\alpha)\,$ denotes the smallest ring containing $\,\Bbb Q$ and $\,\alpha.\,$ The set you give is not a ring since it is not closed under multiplication, e.g. it does not contain $(\sqrt 2 + \sqrt 3)^2 = 5 + 2\sqrt 6.\,$ Note that any ring $R$ containing $\,\Bbb Q\,$ and $\,\sqrt 2 +\sqrt 3\,$  also contains $\,\sqrt 2 -\sqrt 3,\,$ so also $\sqrt 2$ and $\sqrt 3$ by the argument below. So any ring containing $\,\Bbb Q\,$ and $\,\sqrt2 + \sqrt 3\,$ also contains $\,\sqrt2,\, \sqrt 3,\, \sqrt 6.\,$ Thus it contains the set $\, S = \{ a + b\sqrt 2 + c\sqrt 3 + d\sqrt 6\ :\ a,b,c,d\in\Bbb Q\},\,$ which one easily checks is closed under multiplication (and addition) so forms a ring. Conversely, $\,S\,$ contains $\,\Bbb Q\,$ and $\,\sqrt 2 +\sqrt 3,\,$ so it is the sought smallest such ring.

Notice that $\rm\ F = \mathbb Q(\sqrt{a} + \sqrt{b})\ $ contains the $\,\Bbb Q$-linearly independent $\rm\ \sqrt{a} - \sqrt{b}\ $ since 
$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{\ a\,-\,b}{\sqrt{a}+\sqrt{b}}\ \in\ F = \mathbb Q(\sqrt{a}+\sqrt{b}) $$
To be explicit, notice that $\rm\ u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in F\ $ so solving the linear system for the roots yields $\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (u-v)/2,\ $ both of which are clearly $\rm\,\in F,\:$ since $\rm\:u,\:v\in F\:$ and $\rm\:2\ne 0\:$ in $\rm\:F,\:$ so $\rm\:1/2\:\in F.\:$ This works over any field where $\rm\:2\ne 0\:,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field. Further discussion is here.
