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

Solution 1:

Hint $\ $ Note $\,\ \sqrt{3}\,\not\in \Bbb Z[\alpha]\ $ for $\ \alpha\, =\, \sqrt 3 +\! \sqrt{2}\ $ of degree $\,4\,$ over $\Bbb Q,\,$ else $$\!\!\!\! \begin{eqnarray} \alpha\,(2\sqrt 3-\!\alpha)&=&\phantom{._{I^{I^I}}}\!\!\!\!\!\!\!\!\!\! (\sqrt 3+\!\sqrt 2)(\sqrt 3-\!\sqrt 2)\, =\, \color{#0a0}{1}\\ \Rightarrow\ \ \alpha\sqrt 3\, =\, \dfrac{\alpha^2}{\color{#c00}2}\!&+&\!\dfrac{\color{#0a0}{1}}2\,\in\,\color{}{\Bbb Z}[\alpha]\, =\,\color{}{\Bbb Z}\!+\!\alpha\Bbb Z\!+\!\color{}{\alpha^2{\color{#c00}{\Bbb Z}}}\!+\!\alpha^3\Bbb Z \,\ \Rightarrow\ \dfrac{1}{\color{#c00}2} \in \color{#c00}{\Bbb Z}\end{eqnarray}$$

Solution 2:

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 $\mathbb Q$.