How to prove $\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\gamma]$? 
Possible Duplicate:
Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? 

On Page  379, Algebra, Artin(1991)

Let $\mathbb{Q}[\alpha,\beta]$ denote the smallest subring of $\mathbb{C}$ containing $\mathbb{Q}$, $\alpha$=$\sqrt{2}$,  $\beta$=$\sqrt{3}$, and let $\gamma=\alpha+\beta$. Probe that $\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\gamma]$.

To be precise, how to show $\alpha, \beta \in \mathbb{Q}[\gamma]$?It seems to me two subrings have different dimensions, how can they equal?
 A: $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.
Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 -  3} = \sqrt{3} - \sqrt{2}$$ 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}) $.
As pointed out by YACP, if $a$ is algebraic over $K$ then $K[a]=K(a)$, see e.g. here.
A: You can check that $(\sqrt 2 + \sqrt 3)^3 = 11\sqrt 2 + 9\sqrt 3$, which is conveniently not a multiple of $\sqrt 2 + \sqrt 3$.
So you can obtain $\sqrt 2$ by a suitable linear combination of $(\sqrt 2 + \sqrt 3)^3$ and $\sqrt 2 + \sqrt 3$ with rational coefficients.
In general, if $K$ is a field and $x$ is algebraic over $K$, $K[x] = K(x)$ : $K[x]$ is a finite dimensional $K$-vector space, and if you pick $y \in K[x]$, multiplication by $y$ is a $K$-linear endomorphism of $K[x]$. It is invertible if and only if it is injective, which is equivalent to $y$ being nonzero.
A: More directly, without reference to theory of algebraic extensions: $(\sqrt 2 + \sqrt 3)^3 = 11 \sqrt 2 + 9 \sqrt 3 \in \mathbb Q[\gamma]$. And $(\sqrt 2 + \sqrt 3)^3 - 11(\sqrt 2 + \sqrt 3) = -2 \sqrt 3$, so $\sqrt 3 \in \mathbb Q[\gamma]$. Similary $\sqrt 2 \in \mathbb Q[\gamma]$.
