# 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

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 $A_3$.
• 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 $\rho\mapsto\rho^{7^m}$, $0\le m\le5$.