I'm trying to find all the subfields of the splitting field of $x^4-2$. It's $\mathbb{Q}(\sqrt[4]{2}, i)$. I already figured out the Galois group is isomorphic to $D_4$ and I found the correspondent fixed fields for all subgroups but $\langle sr^1\rangle$ and $\langle sr^3\rangle$. I know the fixed field for $\langle sr,sr^3\rangle$ is $\mathbb{Q}(\sqrt2i)$ and so the two I'm missing should be in between $\mathbb{Q}(\sqrt[4]{2}, i)$ and $\mathbb{Q}(\sqrt2i)$, but none of them is $\mathbb{Q}(\sqrt2, i)$, for that is the fixed field of $\langle r^2\rangle$. I just can't figure out what to try with.
Mind that I'm calling the complex conjugation $s$ and $r$ to the element satisfying $r(\sqrt[4]{2})=\sqrt[4]{2}i$ and $r(i)=i$.
Besides the particular problem, any kind of advice or method to proceed with in order to find subfields corresponding to Galois subgroups is welcome.
Best Answer
So thanks to @AndrewHubery I found the answer. As both $sr$ and $sr^3$ have order two, any element of the form $x+sr(x)$ will remain fixed by $sr$, and likewise, any $x+sr^3(x)$ will remain fixed by $sr^3$. Choosing an $x$ such that the obtained fixed element extends $\mathbb{Q}(\sqrt2i)$ we have the solution. $x=\sqrt[4]{2}$ does the trick and hence $\mathbb{Q}(\sqrt[4]{2}(1+i))$ is $sr$'s fixed field and $\mathbb{Q}(\sqrt[4]{2}(1-i))$ si $sr^3$'s.