Describe $\operatorname{Gal}_F(f)$

Question

Describe $\operatorname{Gal}_F(f)$ up to isomorphism for $f=x^4 - 1$

a) $F = \mathbb{Q}$

b) $F = \mathbb{F}_5$

c) $F = \mathbb{F}_{2017}$

I don't know if I am approaching this the right way. I know that f has roots $-1, 1, -i, i$ so in the case of $F = \mathbb{Q}$ the splitting field would be $\mathbb{Q}(i)$ so the Galois extension is just $\mathbb{Q}(i)$ since $\mathbb{Q}$ has characteristic 0 and is therefore separable. In the case of $\mathbb{F}_5$, the characteristic is nonzero so I don't know what to do. I guess we should use the substitution of $x \rightarrow cx + d, cd \in F$ since its the only thing I see in my notes that doesn't require a field of chracteristic 0 or $f$ to be a quadratic or cubic polynomial, but I don't see how

Answer 1

Summarising what was discussed in chat:

1. Two of the roots of $f$ are in $\mathbb{Q}$.
2. The splitting field of $f$ over $\mathbb{Q}$ is $\mathbb{Q}(i)$, since $i$ and $-i$ solve $f$, and this is all the roots.
3. The automorphisms of $\mathbb{Q}(i)$ fixing $\mathbb{Q}$ are the identity and complex conjugation.
4. We have $[\mathbb{Q}(i):Q]=2$, so the size of the Galois group is at most $2$, so we can find the Galois group...
5. For $\mathbb{F}_5$, $2$ and $3$ solve $f$, as well as $1$ and $4$, so $f$ splits over $\mathbb{F}_5$.
6. So the splitting field is the base field, and we can find the Galois group.

As discussed in the comments below and in chat, $x^2 + 1$ splits over $\mathbb{F}_p$ for a prime $p$ if and only if $p \equiv 1 \pmod{4}$.