Prove that $\Bbb Z[\sqrt{2}, \sqrt{3}]$ does not equal $\Bbb Z[\sqrt{2} + \sqrt{3}]$. 
Prove that $\Bbb Z[\sqrt{2}, \sqrt{3}]$ does not equal $\Bbb Z[\sqrt{2} + \sqrt{3}]$.

This is homework. I want to prove that these are different sets. The first set is the smallest ring containing the integers and the two radicals separately. The second is the smallest ring containing the integers and the sum of the two radicals. Unfortunately I cannot find any similar proofs to learn from. I've been trying to show that the second set does not contain the square roof of $6$, for instance, or the square root of $3$. We haven't looked at what the second set might contain. We've only looked at adjoining a single radical and not the sum of radicals on class.  If someone could solve a similar example or give me a crucial insight, that would be great. 
 A: Hint $\ $ Note $\,\ \sqrt{3}\,\not\in \Bbb Z[\alpha]\ $ for  $\ \alpha\, =\, \sqrt 3 +\! \sqrt{2}\ $  of degree $\,4\,$ over $\Bbb Q,\,$ else
$$\!\!\!\! \begin{eqnarray}
\alpha\,(2\sqrt 3-\!\alpha)&=&\phantom{._{I^{I^I}}}\!\!\!\!\!\!\!\!\!\! (\sqrt 3+\!\sqrt 2)(\sqrt 3-\!\sqrt 2)\, =\, \color{#0a0}{1}\\
\Rightarrow\ \  \alpha\sqrt 3\,  =\, \dfrac{\alpha^2}{\color{#c00}2}\!&+&\!\dfrac{\color{#0a0}{1}}2\,\in\,\color{}{\Bbb Z}[\alpha]\, =\,\color{}{\Bbb Z}\!+\!\alpha\Bbb Z\!+\!\color{}{\alpha^2{\color{#c00}{\Bbb Z}}}\!+\!\alpha^3\Bbb Z \,\ \Rightarrow\ \dfrac{1}{\color{#c00}2} \in \color{#c00}{\Bbb Z}\end{eqnarray}$$
A: Not the most elegant solution, but note that
$$(\sqrt2+\sqrt3)^{2n+1}=a_n\sqrt2+b_n\sqrt3,$$
where $a_0=b_0=1$ and
$$(\sqrt2+\sqrt3)^2(a_n\sqrt2+b_n\sqrt3)=(5+\sqrt6)(a_n\sqrt2+b_n\sqrt3)=(5a_n+6b_n)\sqrt2+(5b_n+4a_n)\sqrt3,$$
so
$$2\mid a_{n+1}-b_{n+1}\iff 2\mid(5a_n+6b_n)-(5b_n+4a_n)=a_n-b_n+2b_n.$$
Therefore the difference $a_n-b_n$ is even for all $n$. Even powers $(\sqrt2+\sqrt3)^{2n}$ contain just $\sqrt6$. But the elements of $\mathbb Z[\sqrt2+\sqrt3]$ have the form
$$\sum_{k=0}^d z_k(\sqrt2+\sqrt3)^k.$$
This means the difference of multiples of $\sqrt2$ and $\sqrt3$ always has to be even, which contradicts $\sqrt2$ or $\sqrt3$ being elements of this ring.
I obviously assume we already know $\sqrt2, \sqrt3$ and $\sqrt6$ are linearly independent over $\mathbb Q$.
