# Showing $\mathbb{Q}(\sqrt[4]{2},i)=\mathbb{Q}(\sqrt[4]{2}+i)$ using the Galois orbit of $\sqrt[4]{2} + i$

The following is a problem from I. Martin Isaac's Algebra. Let $E=\mathbb{Q}(\sqrt[4]{2}+i)$. I am trying to show $\mathbb{Q}(\sqrt[4]{2},i)=E$ with the following hint:

Find at least five different elements in the orbit of $i+\sqrt[4]{2}$ under $\text{Gal}(E/\mathbb{Q})$.

I have solved the problem in the usual manner i.e. by showing $\mathbb{Q}(\sqrt[4]{2}+i)\subseteq \mathbb{Q}(\sqrt[4]{2},i)$ and $\mathbb{Q}(\sqrt[4]{2},i)\subseteq \mathbb{Q}(\sqrt[4]{2}+i)$ with a few calculations.

My question is the following:

What is the theoretical framework behind Isaacs' hint?

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Are you sure that you don't want to use the Galois group of $\mathbf Q(\sqrt[4]2, i)$ over $\mathbf Q$? – Dylan Moreland Jul 12 '12 at 16:10

Let $K/k$ be a Galois extension with group $G$. Let $\alpha$ be an element of $K$ with minimal polynomial $f$ over $k$. Then $G$ acts transitively on the set of roots of $f$ in $K$, which all have multiplicity $1$ because our extension is separable. Hence $$[k(\alpha) : k] = \deg f = \#(G\alpha) \quad \text{divides} \quad \#(G) = [K : k].$$ Is clear to you how one might find five conjugates in this case, by the way?
It's easy to see that $\mathbb{Q}(\sqrt[4]{2} + i) \subseteq \mathbb{Q}(\sqrt[4]{2}, i)$. Additionally, $\mathbb{Q}(\sqrt[4]{2}, i)$ has degree $8$ over $\mathbb{Q}$. So the degree of $\mathbb{Q}(\sqrt[4]{2} + i)$ over $\mathbb{Q}$ must be a divisor of $8$. If you can show that it has degree at least $5$, you know that it has to be $8$ and so the fields are equal.