Which is the basis of this field extension?

Question

I know that: $\Bbb Q(\zeta_8)=\Bbb Q(i,\sqrt{2})$ .

Is the basis of $\Bbb Q(i,\sqrt{2})$ over $\Bbb Q\,$ $\Rightarrow$ $\{1,i,\sqrt{2}\}$ ? Or is: $\,\{1,i,\sqrt{2},\sqrt{2}i\}$ ? And why?

Can I write: $\,a+bi+c\sqrt{2}\,$ $\,(a,b,c\in{\Bbb Q})\,$ as such generic element of $\Bbb Q(i,\sqrt{2})\,$ or should I write instead: $\,a+bi+c\sqrt{2}+d\sqrt{2}i\,$ .

Thanks in advance.

Answer 1

You know that $[\Bbb Q(\zeta_n):\Bbb Q]=\varphi(n)$. For $n=8$ Euler's totient function is $\varphi(8)=4$. So the basis has $4$ elements, and not $3$ elements. The basis is $\{ 1,\zeta,\zeta^2,\zeta^3\}$, for example, with $\zeta:=\zeta_8$. We have $i=\zeta^2$ and $\sqrt{2}=\zeta+\zeta^{-1}$.