# Showing linear independence of power basis of $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$.

I have been told that $$\mathbb{Q}(\sqrt{3})$$ as a vector space over $$\mathbb{Q}$$ has dimension $$4$$. If $$\alpha = \sqrt{3}$$, then I am guessing a basis is $$1, \alpha, \alpha^2, \alpha^3$$. I can see that these elements span all of $$\mathbb{Q}(\alpha)$$, but I am not sure how to show that they are linearly independent.

Now, I think I can show that any two of the elements are linearly independent, but I know this doesn't imply that all four are linearly independent.

• $\alpha$ is a root of $p(x)=x^4-3$. Have you covered enough theory to show that $p(x)$ is irreducible over $\Bbb{Q}$? If so, then you are basically done, because a linear dependency relation here would mean that $\alpha$ is also a zero of a lower degree polynomial (and an irreducible polynomial with $\alpha$ as a zero is easily seen to be the lowest degree one). Mar 5 '19 at 18:39
• @JyrkiLahtonen: I kinda see this ahead, but we aren't there yet. Is there a more direct way to do it? Mar 5 '19 at 18:42
• I think any proof that $[\Bbb{Q}(\sqrt{3}):\Bbb{Q}]=4$ will directly or indirectly involve proving that $X^4-3$ is the minimal polynomial of $\sqrt{3}$ over $\Bbb{Q}$, and hence proving that $X^4-3$ is irreducible over $\Bbb{Q}$. Mar 5 '19 at 18:47

If the set $$\{1,\alpha,\alpha^2,\alpha^3\}$$ is linearly dependent over $$\Bbb{Q}$$, then there exist $$x_0,x_1,x_2,x_3\in\Bbb{Q}$$ not all zero such that $$x_0+x_1\alpha+x_2\alpha^2+x_3\alpha^3=0.$$ This means $$\alpha$$ is a root of the polynomial $$f:=x_3X^3+x_2X^2+x_1X+x_0\in\Bbb{Q}[X]$$. Because $$\alpha$$ is also a root of the polynomial $$g:=X^4-3\in\Bbb{Q}[X]$$, it is also a root of $$\gcd(f,g)$$. Now it suffices to show that $$X^4-3$$ is irreducible in $$\Bbb{Q}[X]$$ to reach a contradiction.