Show that Q adjoin square roots of 2, 3 is a vector space of dimension 4 over Q

[Ragnarok7]

Let \(\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3})\) be the field generated by elements of the form \(\displaystyle a+b\sqrt{2}+c\sqrt{3}\), where \(\displaystyle a,b,c\in\mathbb{Q}\). Prove that \(\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3})\) is a vector space of dimension 4 over \(\displaystyle \mathbb{Q}\). Find a basis for \(\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3})\).

I suspect the basis is \(\displaystyle B=\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}\), but I am unsure how to show this. Clearly span\(\displaystyle (B)\supset\mathbb{Q}(\sqrt{2},\sqrt{3})\), and I think I can show the other way around as well, but how does one show linear independence? If we suppose

\(\displaystyle d_1\cdot1+d_2\cdot\sqrt{2}+d_3\cdot\sqrt{3}+d_4\cdot\sqrt{6}=0\)

where \(\displaystyle d_i\in\mathbb{Q}\), then how do we show that each \(\displaystyle d_i=0\)?

Thank you!

 

[mathbalarka]

Your suspicion is correct. Here's a walkthrough which works in general for finding basis :

$\Bbb Q(\sqrt{3}, \sqrt{2})$ sits over $\Bbb Q(\sqrt{2})$ which in turn sits over $\Bbb Q$.

Arbitrary elements of $\Bbb Q(\sqrt{2})$ are $a + b\sqrt{2}$. Thus, arbitrary elements of $\Bbb Q(\sqrt{2}, \sqrt{3}) = \Bbb Q(\sqrt{3})(\sqrt{2})$ are of the form $(a+b\sqrt{2}) + \sqrt{3}(c+d\sqrt{2}) = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$.

You just have to show linear independence now to prove that $\Bbb Q(\sqrt{2}, \sqrt{3})$ is spanned by $\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}$ over $\Bbb Q$. Can you show pairwise linear independence?