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

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

• You're along the right lines for a), can you think what possible automorphisms there could be? How many can you have? For b), can you think of some roots, possibly (or not) in a splitting field? – B. Mehta May 4 '18 at 23:17
• For a? I think 2 because you have the identity and the map that sends to the negative – user558377 May 4 '18 at 23:18
• Is sending to the negative an automorphism of $\mathbb{Q}(i)$ that fixes $\mathbb{Q}$? – B. Mehta May 4 '18 at 23:18
• For b, the roots are the same? -1 and 1 are in the field and again $-i, i$ are not. So $\mathbb{F}_5(i)$ would be a splitting field and if its separable I'm done? – user558377 May 4 '18 at 23:20
• @B.Mehta I suppose not – user558377 May 4 '18 at 23:20

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 $p \equiv 1 \pmod{4}$.

• Do you know a nice theorem we can appeal to in order to show $f(x)$ splits in $\mathbb{F}_{2017}$? – User0112358 May 4 '18 at 23:56
• Hint: the special case of quadratic reciprocity for $\left(\frac{-1}{p}\right)$. – Tob Ernack May 4 '18 at 23:58
• It does not get much nicer than quadratic reciprocity! – User0112358 May 5 '18 at 0:02