Show that $\lvert \mathbb{Q}(\sqrt2 , \sqrt{1+i}) : \mathbb{Q} \rvert = 8$

Question

This is the problem $2.2$ of the book Fields and Galois Theory by Falko Lorenz. It says: "show that $\lvert \mathbb{Q}(\sqrt2 , \sqrt{1+i}) : \mathbb{Q} \rvert = 8$. Hint: For $w=\sqrt{1+i}$ we have $w\bar{w}=\sqrt2$, so $\mathbb{Q}(\sqrt2,w)=\mathbb{Q}(i,w,\bar{w})$."

I really don't know how to deal with this problem. Any Idea would be appreciated.

Answer 1

Use multiplication of degrees $$\lvert \mathbb{Q}(\sqrt2 , \sqrt{1+i}) : \mathbb{Q} \rvert=[\Bbb Q(\sqrt{1+i}):\Bbb Q(\sqrt2)]\cdot[\Bbb Q(\sqrt2):\Bbb Q]$$

Let $x=\sqrt{1+i}$ and find its minimal polynomial over $\Bbb Q(\sqrt2)$ $$x^2=1+i\notin\Bbb Q(\sqrt2)[x]$$ because $1+i\notin\Bbb Q(\sqrt2)$ but $(x^2-1)^2=-1\Rightarrow x^4-2x^2+2\in\Bbb Q[x]\subset\Bbb Q(\sqrt2)$ then the degree $[\Bbb Q(\sqrt{1+i}):\Bbb Q(\sqrt2)]=4$ hence the proposed degree 8.

Answer 2

Use the fact that $\sqrt{a}\in\mathbb{K}(\sqrt{b})$ iff $ab$ is a perfect square in $K$.

Now you have $\mathbb{Q}\subseteq \mathbb{Q}(i)\subseteq \mathbb{Q}(i,w)\subseteq \mathbb{Q}(i,w,\overline{w})$

Answer 3

Note that $ \mathbf Q(\sqrt{2}, \sqrt{1+i}) = \mathbf Q(\zeta_{16}) $ and $ \varphi(16) = 8 $....