Lost on Galois Group over $\mathbb{Q}(i)$ Question.

Question

The goal is to find the Galois Group of $t^8-i$ over $\mathbb{Q}(i)$ where $i=\sqrt{-1}$.

I've found the roots of the polynomial because I know I'll need them and they are

$\pm i^\frac{1}{8}$,$\pm i^\frac{5}{8}$$\pm i^\frac{9}{8}$$\pm i^\frac{13}{8}$.

I need to show that the given polynomial $t^8-i$ is irreducible over $\mathbb{Q}(i)$ right? Is it enough to show that no product of any two roots and no single root is in $\mathbb{Q}(i)$ like I would for a polynomial over $\mathbb{Q}$?

Since I have a root that looks like $i^\frac{1}{8}$ am I right in thinking that the splitting field is then $\mathbb{Q}(i^\frac{1}{8},i)$ and thus the extension is order 8 over $\mathbb{Q}(i)$?.

Also the elements of the Galois Group fix $\mathbb{Q}(i)$ and send a root of $t^8-i$ to another root of $t^8-i$ it seems kind of easy to see that there is something that can send $i^\frac{1}{8} \to -i^\frac{1}{8}$ but how could I find if there is something that performs a more complicated swap like $i^\frac{1}{8} \to -i^\frac{13}{8}$ is brute force verifying that the has the correct properties the only way?

If someone can point me in the right direction it would be most appreciated.

Update: Am I right in thinking that the Galois Group is isomorphic to the Quaternion Group? I think I've identified 3 elements that seem generate different order 4 subgroups and an element that generates an order 2 subgroup which seems to rule out all other groups of order 8.

Answer 1

Well, an eighth root of $i$ is a $32$-nd root of unity, right? And I think you can see that by adjoining one such root of $i$, you get all $32$-nd roots of $1$. So you should think about the field of these roots of $1$, both over $\mathbb Q$ and over $\mathbb Q(i)$. This is conceptually much simpler, I’m sure you’ll agree.

And let me add: if you want to do explicit computations, you might want to work with $\zeta=\exp(\pi i/16)$, clearly a primitive $32$-nd root of unity, and its powers.

Answer 2

If you're allowed to use this generalization of the Rational Root theorem, it becomes really easy.

Let $R$ be a UFD, $Q$ be its fraction field, and let $p(x) \in R[x]$. If $r/s \in Q$ (lowest terms) is a root, then $r$ divides the constant term, and $s$ divides the leading term.

The proof for this is as straightforward as it was for the case $R = \mathbb{Z}$. Therefore, $p(x) = x^8 - i$ could only have the roots $\pm 1$, $\pm i$. But those aren't roots, so it's irreducible.

EDIT: I forgot to do the other half of the question.

Your guess for the splitting field and its degree is right. If we adjoin $\zeta_{32} = \exp \frac{2 \pi i}{32}$, we get all the roots, so $(\mathbb{Q}(i))(\zeta_{32})$ is the splitting field of $p$. The minimal polynomial for $\zeta_{32}$ has degree $8$, so so does our field extension.

I don't actually know Galois theory (taking it this term though), so I can't much help with determining the Galois group. I feel like the image of any of the four "primitive" roots determines the rest of the automorphism, though I can't prove it.