Consider $L=\mathbb{Q}(\sqrt[4]{5},i)$. Let $\tau_2 \in Gal(L/\mathbb{Q}(\sqrt[4]{5})$ be such that $\tau_2(\sqrt[4]{5})=\sqrt[4]{5}$ and $\tau_2(i)=-i$ and let $\tau_1\in Gal(L/\mathbb{Q}(i))$ be such that $\tau_1(\sqrt[4]{5})=i\sqrt[4]{5}$ and $\tau_1(i)=i$
How do I show that $\mathbb{Q}(\sqrt[4]{5},i)^{\{id, \tau_2\tau_1\}}$ = $\mathbb{Q}(\sqrt[4]{5}(1+i))$
My idea is to show that $|L:\mathbb{Q}|=8$ and then use then consider the basis $1,\theta, \theta^2,\theta^3,i,i\theta,i\theta^2,i\theta^3$ then use $\tau_2\tau_1$ on rational linear combinations of the above basis and compare coefficients. However, this seems tedious to do. I was wondering, is there an easier method?
Furthermore, what would be the intuition behind the equality? Because what I have written is written as in the solution, but I do not know a priori why one would think of $\sqrt[4]{5}(1+i)$.
Let $F$ be the fixed field of $\{1,\tau_1\circ\tau_2\}$. Then $L/F$ is a Galois extension dimension 2. Clearly, $F\supseteq\Bbb Q(\sqrt[4]5(1+i))$ and the minimal polynomial of $\sqrt[4]5(1+i)$ over $\Bbb Q$ is $x^4+20$ which is irreducible over $\Bbb Q$ by Eiseinstein's criterion (for $p=5$). $$\underbrace{\Bbb Q\xrightarrow 4\Bbb Q(\sqrt[4]5(1+i))\to F\xrightarrow 2 L}_8$$ Thus $[\Bbb Q(\sqrt[4]5(1+i)):\Bbb Q]=4$, hence $F=\Bbb Q(\sqrt[4]5(1+i))$.