$Gal(\mathbb{Q}(\sqrt[4]{2},i)/\mathbb{Q})$, choice of automorphisms I have an exercise that asks for me to find $Gal(\mathbb{Q}(\sqrt[4]{2},i)/\mathbb{Q})$. The irreducible polynomial is $(x^4-2)(x^2+1)$. Its known that the automorphisms will permute the roots of each of the irreducible polynomials. My book picks two permutations:
$$r(\sqrt[4]{2}) = i\sqrt[4]{2}, r(i) = i\\s(\sqrt[4]{2}) = \sqrt[4]{2}, s(i) = -i$$
But I know that $i\sqrt[4]{2}, -\sqrt[4]{2}, -i\sqrt[4]{2}$ are also roots of the polynomial. Why my book picks $\sqrt[4]{2}\to i\sqrt[4]{2}$ specifically? 
Its clear to me that I don't need to specify the other automorphisms, because if I have $\phi(\sqrt[4]{2})$ and $\phi(i)$, then $\phi(\pm i\sqrt[4]{2}) = \phi(\pm i)\phi(\sqrt[4]{2})$, right? So I only need to choose one from each irreducible polynomial
 A: I feel like the other answer misses an important point.
The map $\sqrt[4]{2}\to -\sqrt[4]{2}$ does not generate the whole permutation group, and therefore cannot be used. This is because this is a map of order $2$. Both of the other non-trivial maps have order 4, and you can choose either of them to be your generator. $i\sqrt[4]{2}$ was probably chosen over $-i\sqrt[4]{2}$ because it was perceived to be simpler, or even just less characters, by the author. Either choice would be correct.
Identifying which roots are appropriate generators is easy. You are looking at something of the form $x^n-m$, so the roots form a cyclic group of order $n$ (to prove this!). Lay them out in $\mathbb{C}$ and connect them to form regular $n$-gon. The permutations of the roots correspond to rotations of this $n$-gon. Rotating in such as way as to move each root over by one will always generate the whole group. In your case, this would be the two permutations I mentioned as working, depending on if the rotation is clockwise ($-i\sqrt[4]{2}$) or counter clockwise ($i\sqrt[4]{2}$). To find all the generating permutations simply fix a $g$ that generates the group and compute $\{g^r|gcd(r,n)=1\}$. There is no reason to choose one element of this set over the others when writing down the generators, it's up to the author's preference.
Note that the proceeding paragraph is only correct because you are examining a cyclic group. When the Galois group of a factor of the polynomial isn't the cyclic group, modifications to what I have said need to be made. However, you're still looking for the generators of the subgroup.
A: $\Bbb Q(\sqrt[4]2,i)$ is generated by two elements, so the automorphism is completely determined by the image of this two numbers.
Let $r$ be an automorphism of $\Bbb Q(\sqrt[4]2,i)$ that fixes $\Bbb Q$.
Then, $r(\sqrt[4]2)^4=r(\sqrt[4]2^4)=r(2)=2$, so $r(\sqrt[4]2)$ is one of the four roots of $x^4-2=0$.
$r(i)^2=r(i^2)=r(-1)=-1$, so $r(i) = \pm i$.
This gives us a group of order $8$.
One generator would fix $i$ and the other would change $\sqrt[4]2$. If they picked $r(\sqrt[4]2)=-\sqrt[4]2$ instead, this wouldn't generate the automorphism that fixes $i$ and takes $\sqrt[4]2$ to $i\sqrt[4]2$.
