<s>[INST] Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? 
If $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}$ and $\mathbf{Q}(\sqrt{6})= \{a+b\sqrt{6} | a,b \in \mathbf{Q}\}$
Assume an element in $\mathbf{Q}(\sqrt{6})$ , then obviously it is also in $\mathbf{Q}(\sqrt{2},\sqrt{3})$. 
Assume an element in $\mathbf{Q}(\sqrt{2},\sqrt{3})$ , because we can write: $\sqrt{6} = \sqrt{3} \cdot \sqrt{2} $ and $\sqrt{6} = \sqrt{2} \cdot \sqrt{3}$ the sum of $b\sqrt{3} + c\sqrt{2}$ is equal to a fraction of $\sqrt{6}$. 
So $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$.
 [/INST] Suppose $\sqrt{2}\in \mathbb{Q}(\sqrt{6})$. Then there are rational numbers $a,b\in\mathbb{Q}$ such that $a+b\sqrt{6}=\sqrt{2}$. If $a=0$ but $b\neq 0$, then $\sqrt{3}=1/b\in\mathbb{Q}$. That's impossible, as we know that $\sqrt{3}$ is irrational. Similarly, if $b=0$ and $a\neq 0$, then $\sqrt{2}=a$, and again we reach a contradiction. Otherwise, assume $ab\neq 0$. Thus,
$$2 = a^2+6b^2+2ab\sqrt{6}.$$
In particular, 
$$\sqrt{6} = \frac{2-a^2-6b^2}{2ab},$$
so $\sqrt{6}\in \mathbb{Q}$. This is absurd, so we have reached a contradiction. Our original assumption $\sqrt{2}\in \mathbb{Q}(\sqrt{6})$ was false, and $\mathbb{Q}(\sqrt{2})\not\subseteq \mathbb{Q}(\sqrt{6})$. In particular,  $\mathbb{Q}(\sqrt{2},\sqrt{3})\neq \mathbb{Q}(\sqrt{6})$.
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