How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.
I want to know the detail about the proof. Thanks in advance.
Actually I know any two of them and three of them are linearly independent.
 A: Observe that $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} = 0$ if and only if $(a + b\sqrt{2}) + (c + d\sqrt{2})\sqrt{3} = 0$. Hence it is enough to show that $\sqrt{3}$ and $1$ are independent over $\mathbb{Q}(\sqrt{2})$, but this is equivalent to showing that $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, or that $\sqrt{3}$ cannot be written as $a + b\sqrt{2}$. This is obvious (square both sides and play with the result).
A: I have no doubt that many slick solutions will be given here. I'll try to post an elementary one; which uses perhaps the most straightforward approach I can imagine. (The only things that are needed are some algebraic manipulation and basic properties of rational numbers; such as that the only solution of $x^2=6y^2$ in $\mathbb Q$ are $x=y=0$.)

This is equivalent to showing that if
$$a+b\sqrt6=c\sqrt3+d\sqrt2$$
then $a=b=c=d=0$.
By squaring both sides of the above equation we get
$$
\begin{align*}
a^2+6b^2+2ab\sqrt6&=3c^2+2d^2+2cd\sqrt6\\
2(ab-cd)\sqrt6=3c^2+2d^2-a^2-6b^2
\end{align*}
$$
Since $a,b,c,d\in\mathbb Q$, this implies
$$
\begin{align*}
ab-cd&=0\\
3c^2+2d^2-a^2-6b^2&=0
\end{align*}
$$
which is the same as
$$
\begin{align*}
ab&=cd\\
3c^2+2d^2&=a^2+6b^2
\end{align*}
$$
Suppose that $b\ne0$, $c\ne0$. (I'll leave the solution of these cases to the reader.) Then we can rewrite the first equation as $\frac ac = \frac db = x$, where $x\in\mathbb Q$. Now the second equation becomes
$$
\begin{align*}
3c^2+2x^2b^2&=x^2c^2+6b^2\\
x^2(2b^2-c^2)&=3(2b^2-c^2)\\
(x^2-3)(2b^2-c^2)&=0
\end{align*}$$
This implies that $x^2=3$ or $2b^2=c^2$. None of them has non-zero solutions in rational numbers.

Alternatively, we could solve the case $b=0$ first. Once we're looking only at $b\ne0$, we can assume w.l.o.g. that $b=1$. (After dividing both sides with $b$.)
So we're looking at $a+\sqrt6=c\sqrt2+d\sqrt2$. And squaring, as above, gives us now $a=cd$. Thus we get
\begin{align*}
cd-c\sqrt2-d\sqrt3+\sqrt6&=0\\
(c-\sqrt3)(d-\sqrt2)&=0
\end{align*}
which means that $c=\sqrt3$ or $d=\sqrt2$. In either case, it is the contradiction with the assumption $c,d\in\mathbb Q$.
A: HINT $\ $ Specialize the Lemma below to $\rm\ K = \mathbb Q,\ a,b\ =\ 2,3\:. $
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  $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\:  [L:K] = 2\:$  via  $\rm\:\sqrt{b}  \not\in K\:,\:$  so it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ which contradicts hypotheses as follows:  
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using  $\rm\:2 \ne 0$  
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r \in K$ 
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\  r = 0\ \ \Rightarrow\ \  \sqrt{a\:b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED
REMARK $\ $ By induction, the lemma easily generalizes to algebraic extensions generated by adjoining $\rm\:n\:$ square-roots, see my post here on Besicovic's Theorem, which includes references to generalizations by Mordell and Siegel. These results are elementary special cases of the Galois theory of Kummer extensions.
