How do I show that $\mathbb{Q}(\sqrt[4]{7},\sqrt{-1})$ is Galois? How do I show that $\mathbb{Q}(\sqrt[4]{7},\sqrt{-1})$ is Galois?
At first I thought it was the splitting field of $x^4-7$, but I was only able to prove that it was a subfield of the splitting field. Any ideas?
I'm trying to find all the intermediate fields in terms of their generators, but I don't understand how. I am trying to imitate Dummit and Foote on this. I am looking at the subgroups of the Galois group in terms of $\tau$ where $\tau$ is the automorphism that takes $\sqrt[4]{7}$ to itself and $i$ to $-i$, and $\sigma$ that takes $\sqrt[4]{7}$ to $i\sqrt[4]{7}$ and $i$ to itself. How do I, for example, find the subfield corresponding to $\{1, \tau\sigma^3\}$? I know I am supposed to find four elements of the galois group that $\tau\sigma^3$ fixes, but so far i can only find $-\sqrt[4]{7}^3$.

The polynomial $f(x)=x^4-7$ factors as $(x-7^{1/4})(x+7^{1/4})(x-i7^{1/4})(x+i7^{1/4})$, and all these irreducible factors are distinct. Hence, $x^4-7$ is separable. Moreover, the field $L= \mathbb{Q}(7^{1/4},i)$ contains all its roots and is the minimal field where $x^4-7$ factors completely in. Hence, it is the minimal splitting field of $f(x)$. So the extension $K \subset L$ is both normal and separable and hence, is Galois.
