# Galois group of splitting field of $x^3-5$ over $\mathbb F _7$

Honestly, I'm not even sure where to start. I think I understand how to find the Galois group of a field extension with $$\textrm{char}\mathbb F=0$$ but for some reason I'm confused when it comes to finite fields.

1. I want to say that the roots of $$x^3-5$$ are $$\sqrt 5, -\sqrt 5, \sqrt5 \zeta_3$$ and so the splitting field is $$\mathbb F_7 (\sqrt5, \zeta_3)$$ which has degree $$3$$, but I also think that I'm not supposed to think of $$\mathbb F_7$$ as if it's contained in $$\mathbb C$$ but in some other algebraic closed field.
2. Even after I know what the splitting field is, how do I determine what automorphisms are there in the Galois group? Each automorphism sends a root of $$x^3-5$$ to some other root, so that's why I need to find exactly what's the splitting field? Is it enough to determine what are the elements of the Galois group just by showing what are the possibilities of matching between roots of the polynomial?

I know there are similar questions here, but what I'm really looking for is an explanation such that each step is justified with the appropriate theorem/claim it's based on, so I would highly appericiate it if you could explain in each step why what you're saying is true.

• In $\mathbb{F}_5$, $5=0$. Hence you just need the splitting field of $x^3$ over $\mathbb{F}_5$, which is $\mathbb{F}_5$. I advise thinking about reducing a polynomial first, instead of finding the root (ex $5^{1/3}$) and thinking about what happens mod 5. Commented Jun 10 at 13:21
• @KaiWang Oh no! I've copied the question incorrectly, it's supposed to be $\mathbb F_7$. I'll amend it. Commented Jun 10 at 13:26
• In $\Bbb{F}_7$ you have $2^3=4^3=1$, so if $\alpha$ is one root (in some extension field) then $2\alpha$ and $4\alpha$ are the others. RHspqr made the same observation (+1). Commented Jun 10 at 16:38
• Here's a similar question with 3 quite different answers: math.stackexchange.com/q/2127987 . I want to point out that, even if they may be less familiar, the structure of finite extensions of finite fields is much simpler than for fields of characteristic $0$. Given an irreducible polynomial $f$ of degree $n$ over $\mathbb{F}_q$, the field obtained by adjoining one root of $f$ will always be the splitting field of $f$, with Galois group $C_n$, the cyclic group of order $n$, generated by the Frobenius map $x \mapsto x^q$. See here. Commented Jun 10 at 20:58

You can generalize the idea for characteristic $$0$$ fields. However, for $$\mathbb{F}_7$$, you can verify that $$x^3-1$$ splits over $$\mathbb{F}_7$$ already, so the field extension $$\mathbb{F}_7(t)$$ with $$t^3 =5$$ suffices to be the splitting field.