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 $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$.
 A: Hint $\ $ It is an immediate consequence of the following more general lemma, which is the basis of a general result on linear independence of square roots due to Besicovitch (see below).
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $  all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K$
Proof $\ \ $  Let  $\rm\ L = K(\sqrt{b}).\,$  $\rm\,  [L:K] = 2\:$  via  $\rm\:\sqrt{b}  \not\in K,\,$ so it is suffices to prove $\rm\: [L(\sqrt{a}):L] = 2.\:$ It fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b}).\, $ Then $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K.\:$ But that is impossible since squaring yields $\ \rm\color{#c00}{(1)}:\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ contra hypotheses, as follows  
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $\,\color{#c00}{(1)}\,$ for $\rm\sqrt{b}\:,\:$ using  $\rm\:2 \ne 0$  
$\rm\qquad\qquad  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r \in K$ 
$\rm\qquad\qquad  r = 0\ \ \Rightarrow\ \  \sqrt{a\:b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\ \sqrt{b},\: $ times $\rm\:\sqrt{b}\quad\,$ QED

Remark $\ $ Using the above as the inductive step one easily proves the following 
Theorem $\ $  Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$  square roots  $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts  $\rm\ a,\:b,\:\ldots \in  Q\:.\:$
If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive 
adjunction  $\rm\ Q(\sqrt{a}),\  Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n.\:$  Hence the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q.$
A: 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,  $\mathbb{Q}(\sqrt{2},\sqrt{3})\neq \mathbb{Q}(\sqrt{6})$.
