TITLE: Is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ the same as $\mathbb{Q}(\sqrt{2},\sqrt{3})$?
QUESTION [9 upvotes]: Is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ the same as $\mathbb{Q}(\sqrt{2},\sqrt{3})$?
I mean, $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ can be viewed as: $\mathbb{Q}[\sqrt{2}][\sqrt{3}]$, as polynomials of $\sqrt{3}$ having coefficients of $\mathbb{Q}[\sqrt{2}]$?, and $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\mathbb{Q}, \sqrt{2}\,\sqrt{3}$. Since $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is a field we must have that $\mathbb{Q}(\sqrt{2},\sqrt{3}) \subseteq \mathbb{Q}[\sqrt{2},\sqrt{3}]$?, but since all the terms in $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ are made with sums and multiplications of $\mathbb{Q},\sqrt{2},\sqrt{3}$, we must have that $\mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}(\sqrt{2},\sqrt{3})$?
Can someone confirm that this is correct?
But do we then also have that:
$F[\alpha]=F(\alpha)$, always?
and 
$F[\alpha_1,\alpha_2]=F(\alpha_1,\alpha_2)$?

REPLY [3 votes]: More generally, if $A$ is a domain which is a finite-dimensional algebra over a field $F$, then $A$ is a field.
Indeed, take $a \in A$, $a\ne 0$, and consider $\phi: A \to A$ given by $\phi(x)=ax$. Then $\phi$ an $F$-linear transformation. Moreover, $\phi$ is injective because $A$ is a domain. Since $A$ is finite-dimensional algebra over $F$, $\phi$ injective implies $\phi$ surjective, and so $1$ is in the image of $\phi$ and $a$ has an inverse in $A$. This inverse can be found by expressing $\phi$ with respect to a basis of $A$ and solving a linear system.
$\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is clearly a domain because it is contained in $\mathbb{R}$.
$\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is also a finite-dimensional algebra over $\mathbb{Q}$. Indeed, $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is the set of all polynomial expressions in $\sqrt{2}$ and $\sqrt{3}$ with coefficients in $\mathbb{Q}$, and so a basis is $\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6} \}$ because all powers greater than $1$ can be reduced.