Show that $\frac{1+i}{\sqrt{2}}$ is in the splitting field of $x^8 - 3$ over $\mathbb{Q}$

Question

I want to show $\frac{1+i}{\sqrt{2}}$ is in the splitting field of $x^8 - 3$ over $\mathbb{Q}$. Now I know $\frac{1+i}{\sqrt{2}}$ = $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$. Since $\alpha_0 = i\sqrt[8]{3}$ and $\alpha_1 = \sqrt[8]{3}$ are both roots for $x^8 - 3$, I can write $i = \alpha_0\alpha_1^{-1}$ in terms of elements in the splitting field. However I am not sure how to write $\sqrt{2}$ in terms of the elements in the splitting field, and I am not even sure whether it's possible as the 8 powers of $\sqrt[8]{3}$ seems to be linearly independent with $\sqrt{2}$?

Answer 1

Notice that $$ \left(\frac{1+i}{\sqrt{2}}\right)^8=1 $$ and the splitting field of $x^8-3$ over $\mathbb{Q}$ must contain a primitive $8$-th root of unity.