# Basis for $\mathbb Q (\sqrt2 , \sqrt3 )$ over $\mathbb Q$

List a basis for $\mathbb K =\mathbb Q (\sqrt2 , \sqrt3 )$ as a vector space over $\mathbb Q$.

I don't know how people come to the conclusion of a claimed basis. Like I am pretty sure that we just claim $\{1, \sqrt2 , \sqrt3, \sqrt6 \}$ is a basis and then prove that it is LI and it is a spanning set of $\mathbb K$ but how do you even think of this claim? That is my first question. I know that the dimension of $\mathbb K$ is $4$ so the basis should have $4$ elements but it doesn't really tell you which $4$ specifically.

Secondly, to prove that these are LI:

$1$ is LI to all the other elements clearly since the others are irrational and $1$ is rational. Not sure on how the other three are LI to each other though.

Also I am unsure on the spanning part.

• – Git Gud Feb 19 '16 at 22:00
• @GitGud I still don't understand.... please help – snowman Feb 19 '16 at 22:07

## 2 Answers

One can use the fact that $\mathbb{Q}(\sqrt2,\sqrt3)$=$(\mathbb{Q}(\sqrt2))(\sqrt3)$ the latter of which has elements of the form $a+b\sqrt3$ where $a,b\in \mathbb{Q}(\sqrt2)$ since $[(\mathbb{Q}(\sqrt2))(\sqrt3) : \mathbb Q(\sqrt2)]=2$. From here we observe that $a=a_1+a_2\sqrt2$ and $b=b_1+b_2\sqrt2$ for some rationals $a_i$ and $b_i$ since $[\mathbb Q(\sqrt2) : \mathbb Q]=2$ . Together substituting back into $a+b\sqrt3$ you can see that $\{1, \sqrt2 , \sqrt3, \sqrt6 \}$ is a spanning set. Since the dimension of $\mathbb{Q}(\sqrt2,\sqrt3)$ as a vector space over $\mathbb{Q}$ is $4$ we know that the spanning set is also a basis as desired.

• With your first line, are you saying that all elements have the form $a+b\sqrt3$? Because I thought all elements of $Q(\sqrt2,\sqrt3)$ had the form $a+b\sqrt2+c\sqrt3+d\sqrt6$... – snowman Feb 19 '16 at 22:29
• Thanks for the edit. So how can we be sure that this is a spanning set? I do see that you used general elements of the sets to construct the basis but the spanning part is a bit unclear... – snowman Feb 19 '16 at 22:35
• Please help, I am unsure on the spanning set part. – snowman Feb 20 '16 at 0:27
• I have shown that every $\alpha\in\mathbb{Q}(\sqrt2,\sqrt3)$ can be written as a linear combination of $\{1,\sqrt2,\sqrt3,\sqrt6\}$ over $\mathbb{Q}$ which means it is a spanning set. – Sri-Amirthan Theivendran Feb 20 '16 at 0:37

You can note that $\sqrt{3}$ has degree $2$ over $\mathbb{Q}(\sqrt{2})$, because $$(a+b\sqrt{2})^2=3$$ leads to $$a^2+2b^2=3,\qquad 2ab=0$$ which has no solution in the rational numbers. Therefore $\{1,\sqrt{3}\}$ is a basis of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$, which in turn has $\{1,\sqrt{2}\}$ as a basis over $\mathbb{Q}$. The standard proof of the dimension theorem says that $$\{1\cdot 1,1\cdot\sqrt{2},1\cdot\sqrt{3},\sqrt{2}\cdot\sqrt{3}\}$$ is a basis of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$.