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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 $-\sqrt[4]{7}^3$.

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    Gerry, so sorry! I didn't know there was such a thing. I will check the meta-section and accept ASAP. This ite is wonderful.2011-12-04

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The splitting field of your polynomial is generated by the $a=\sqrt 7$, $-a$, $ia$ and $-ia$, its roots. It therefore contains $a$ and $i=\frac{ia}{a}$. Obviously, the field generated by $a$ and $i$ also contains the four roots.

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    You should :) ${}$2011-12-04