Let $\alpha$ be a root of $x^4+1$.. so we conclude that all roots of $x^4 + 1$ is primitive 8th root of unity. And the generator is $e^\frac{(2πi)}{8}$. Which is equal to $\cos(2\pi/8) + i \sin(2\pi/8)$ and equal to $\frac{1+ i }{\sqrt{2}}$
And all other roots is generating by it. So my my doubt is
< Then splitting field should be $\mathbb{Q}( \frac{1+ i }{\sqrt{2}}$
). . But it's not , we solve it further and conclude that 1 and i must be there and splitting field must be $\mathbb{Q}( \sqrt2 , i )$ . But in my opinions splitting field can be $\mathbb{Q}( \frac{1+ i }{\sqrt{2}}$ )
too because it's also contains all roots and can even generate $ \sqrt{2}$ and $i$>
Can please anyone help me to solve this puzzle for me and why we choose splitting field to $\mathbb{Q}( \sqrt2 , $i$ )$ .
Best Answer
Since $\frac{1+i}{\sqrt 2} \in \mathbb{Q}(\sqrt 2, i)$ by closure, it is true that $\mathbb{Q}(\frac{1+i}{\sqrt 2}) \subseteq \mathbb{Q}(\sqrt 2,i)$ .
Now since $i = \big(\frac{1+i}{\sqrt 2}) ^2$ , $i \in \mathbb{Q}(\frac{1+i}{\sqrt 2} )$ then so does $1+i$ and by closure $\sqrt 2 \in\mathbb{Q}(\frac{1+i}{\sqrt 2} ) $
By definition $\mathbb{Q}(\sqrt 2, i)$ is the smallest subfield to contain $\sqrt 2$ and $i$, thus $\mathbb{Q}(\frac{1+i}{\sqrt 2} ) = \mathbb{Q}(\sqrt 2, i)$