# Linear independence of $\{1,\sqrt2,\sqrt3,\sqrt6 \}$ over $\mathbb Q$ [duplicate]

We want to prove that $a+b\sqrt2+c\sqrt3 +d\sqrt6 =0$ where $a,b,c,d \in \mathbb Q$ such that $a=b=c=d=0$.

Can we prove that

1. there exists no linear combination of $1, \sqrt3, \sqrt6$ that gives $\sqrt2$

2. there exists no linear combination of $1, \sqrt2, \sqrt6$ that gives $\sqrt3$

3. there exists no linear combination of $1, \sqrt3, \sqrt2$ that gives $\sqrt6$

4. there exists no linear combination of $\sqrt2, \sqrt3, \sqrt6$ that gives $1$

Which would mean they are all linearly independent?

So for the first one, I would do suppose there does exist $a,b,c \in \mathbb Q$ such that $\sqrt2=a+b\sqrt3 + c\sqrt6$ but we can minus $a$ from both sides and then square both sides, then we would get after rearranging: $$\sqrt2 = \frac{2+a^2 -3b^2 - 6c^2}{2a+6bc}$$ which can't be true so there is no linear combination.

And then do similar for the rest of the 3 parts I named?

This question is very linked to Basis for $\mathbb Q (\sqrt2 , \sqrt3 )$ over $\mathbb Q$ but I decided to make a new thread because asking this is very lengthy as you can see.

• Note that what you really get is not that $\sqrt2 = \frac{2+a^2 -3b^2 - 6c^2}{2a+6bc}$, but rather that $$(2a+6bc)\sqrt 2 = 2+a^2 -3b^2 - 6c^2$$which means that we must have $2a+6bc = 0$ and $2+a^2 -3b^2 - 6c^2 = 0$. Then you need to derive a contradiction from there to really prove that $\sqrt 2$ cannot be written as a linear combination of the others. That being said, I would probably go directly from $a+b\sqrt2+c\sqrt3 +d\sqrt6 =0$ and show that $a = b = c = d = 0$ rather than splitting up into four different cases. I think that's shorter. – Arthur Feb 20 '16 at 12:33
• In principle: yes. If the elements are not linearly dependent then at least one of the four statements is not true. – drhab Feb 20 '16 at 12:36
• @Arthur What was wrong with the way I wrote it? I was trying to say that the RHS is rational and the LHS is not which makes it a contradiction. Also can you show me how to do the shorter way please. I was trying it for ages. – snowman Feb 20 '16 at 12:40
• The general fact is that if $a_1,\ldots,a_n,$ are pairwise coprime and squarefree, then $1,\sqrt{a_1},\ldots,\sqrt{a_n}$ are linearly independent. As far as I can recall, the simplest way to prove it is by proving first that the automorphism group of ${\bf Q}[\sqrt{a_1},\ldots,\sqrt{a_n}]$ is of order $2^n$. – tomasz Feb 20 '16 at 16:48

Suppose you have $$a+b\sqrt2+c\sqrt3 +d\sqrt6 =0$$ for rational coefficients. Then $$(a+b\sqrt2)+ \sqrt3 (c +d\sqrt2) =0.$$ If $c +d\sqrt2 \ne 0$, we have that $$\sqrt{3} = - \frac{a+b\sqrt2}{c +d\sqrt2} = \frac{(a + b \sqrt{2})(c - d \sqrt{2})}{c^2 - 2 d^2} = \frac{a c - 2 b d}{c^2 - 2 d^2} + \frac{b c -a d }{c^2 - 2 d^2} \cdot \sqrt{2} = u + v \sqrt{2}$$ for some rational $u, v$. Prove that this is impossible. (Just square both sides and distinguish a couple of cases.)

If $c +d\sqrt2 = 0$, then $c = d = 0$, as $\sqrt{2}$ is irrational. Therefore $a+b\sqrt2 = 0$, so that also $a = b = 0$.

Suppose $a+b\sqrt2+c\sqrt3 +d\sqrt6 =0$.

We may assume that $a,b,c\in \mathbb Z$ and are coprime.

Multiplying the equation $a+b\sqrt2+c\sqrt3 +d\sqrt6 =0$ by $\sqrt2$, $\sqrt3$, $\sqrt6$, we get $$\pmatrix{ a & b & c & d \\ 2b & a & 2d & c \\ 3c & 3d & a & b \\ 6d & 3c & 2b & a} \pmatrix{ 1 \\ \sqrt2 \\ \sqrt3 \\ \sqrt6} = 0$$ The matrix is thus singular and its determinant must be zero: $$a^4-4a^2b^2-6a^2c^2-12a^2d^2+48abcd+4b^4-12b^2c^2-24b^2d^2+9c^4-36c^2d^2+36d^4=0$$ Hopefully, this will imply that $a,b,c,d$ are not coprime. I'll check the details later.

• Motivated by math.stackexchange.com/a/829083/589. – lhf Feb 20 '16 at 12:58
• You have transformed an easy problem into a hard one! – TonyK Feb 20 '16 at 13:16
• @TonyK, quite so. I thought it'd be easier to see that they are not coprime. – lhf Feb 20 '16 at 13:23