# Galois group of a splitting field of a polynomial over $\mathbb{F}_7$

Question: Find the Galois group of the splitting field $(x^2-3)(x^3+x+1)$ over $\mathbb{F}_7$.

I know the splitting field is $K:=\mathbb{F}_7(\sqrt{3},\alpha_1)$, where $\alpha_1$ is one of the roots of the polynomial $x^3+x+1$. I know that the possible automorphisms of K fixing F must have the mappings $\sqrt{3} \mapsto \pm\sqrt{3}$ and $\alpha_1 \mapsto \{\alpha_1,\alpha_2,\alpha_3\}$ where $\alpha_1,\alpha_2,\alpha_3$ are the distinct roots of $x^3+x+1$. But when I wanted to write out all the automorphsism explicitly, I have some trouble. Any help will be appreciated

## 2 Answers

Note that the Galois group is some subgroup of the direct product of the Galois groups of each factor considered individually. Since the splitting field of $x^2 - 3$ over $\Bbb{F}_7$ has degree two, the splitting field of $x^3 +x+1$ has degree three, and the degrees are coprime the splitting field of their product has degree 6. The direct product of the Galois groups of the factors, $\Bbb{Z}_2 \times \Bbb{Z}_3$, has order 6, and the Galois group of $K$ is a 6 element subgroup of this so it must be the whole group. If you want it explicitly, a generator is the permutation $\sigma$ sending $\sqrt{3}$ to its negative and sending $\alpha _1 \to \alpha _2 \to \alpha_3 \to \alpha _1$. This is necessarily an automorphism, because the Galois group acts on the $\alpha _i$ as the alternating group $A_3$.

• "Note that the Galois group is some subgroup of the direct product of the Galois groups of each factor considered individually." May I know where is the proof of this? I am new to galois theory so I am not very well-versed with the subject thank you! – Noob4398 Oct 9 '16 at 10:48
• Sure. The important thing here is that both of the extensions $\Bbb{Q}[\sqrt{3}]$ and $\Bbb{Q}[\alpha _1]$ are normal, so any automorphism of $K/\Bbb{Q}$ must restrict to an automorphism of each of those subfields. This immediately implies the statement. Whenever you find the Galois group of an reducible polynomial it's a subgroup of the direct product of the Galois groups of the irreducible factors. – Vik78 Oct 9 '16 at 16:21
• If you have $K = \Bbb{Q}[\alpha _1, ..., \alpha _n]$ normal over $\Bbb{Q}$, there is no isomorphism from $K$ to any other field besides $K$ that fixes $\Bbb{Q}$, since any isomorphism of this type is defined by its action on the $\alpha _i$ and must send the $\alpha _i$ to other roots of their minimal polynomials, which are all in $K$. – Vik78 Oct 9 '16 at 16:26

You might as well make good use of the fact that the Galois group of an extension of finite fields is cyclic, generated by $x\mapsto x^q$, where $q$ is the cardinality of the smaller field. I think it’s clear that $\sqrt3+\alpha_1$ generates the whole degree-six extension $K$, and we can call this quantity $\rho$, so that everything in $K$ can be written uniquely in form $\sum_0^5n_i\rho^i$ with all $n_i\in\Bbb F_7$. The minimal polynomial for $\rho$ over $\Bbb F_7$ seems to be $2 -X + 2X^3 + X^6$. At any rate, your six automorphisms are $\rho\mapsto\rho^{7^m}$, $0\le m\le5$.