Galois Group of $x^6+3\in\mathbb F_7$ If I want to compute the Galois Group of $x^6+3\in\mathbb F_7$ I could use the theorem about automorphism groups of finite fields in conjunction with the fact that finite extensions of finite fields are normal and separable. So for a root $\alpha$ of $x^6+3$ we would simply obtain the splitting field to be $\mathbb F_7 (\alpha)\cong\mathbb F_{7^6}$ and thus $\text{Gal}(f)\cong\mathbb Z_{6}$. 
But for all this to work I need that $x^6+3$ is irreducible and I've ran out of ideas how to prove that. It has no roots and I couldn't find any factors because in $\mathbb F_7$ there are just too many possibilities.
 A: $x^6+3=(x^3-2)(x^3+2)$.
Note that $\Bbb F_7$ already has a primitive cube root of unity (i.e. $2$), so if $x^3 = 2$, then $x, 2x, 4x$ are distinct roots of $x^3-2$, and $-x, -2x, -4x$ are distinct roots of $x^3+2$.
Therefore $\Bbb F_7[X]/(X^3-2)$ is a field where $x^6+3$ splits.
The degree of extension is $3$, which is prime, so we see that it is the splitting field for $x^6+3$ since it has no root in the base field.
Therefore the Galois group is $C_3$, the cyclic group of $3$ elements.

Alternatively, if $x^3 = 2$, then $x$ would have order $9$, which divides $7^3 - 1 = 342$ and not $7^2 - 1 = 48$ or $7 - 1 = 6$, so it splits in $\Bbb F_{7^3}$, so the Galois group is $C_3$.
A: You checked that this polynomial has no linear factors. If it was reducible, then $x^6+3=g(x)f(x)$ and one of $g,f$ must be quadratic or cubic by degree considerations. You can then check all the irreducible quadratics/cubic over $\mathbb{F}_7$ and show they do not divide your polynomial.  
A: $x^6+3=(x^3+2)(x^3+5)$ over $\Bbb F_7$. Got it from here?
