Is $x^6-3$ irreducible over $\mathbb{F}_7$? I know that $\mathbb{F}_7=\mathbb{Z}_7$, and the all possible solutions of $x^6-1=0$ over $\mathbb{Z}_7$ are 1~6, so if we let the root of equation $x^6-3$ as $t $ then the solutions of $x^6-3=0$ is $t$~$6t$, which shows that all the solutions are not in $\mathbb{Z}_7$. However I can't say that $x^6-3$ is irreducible by this fact(since there could exist 2 or more degree polynomial divides $x^6-3$.) Is it irreducible over $\mathbb{Z_7}$? Why?

Since you can get $7 | x^6-1$ for all x by Fermat Little Theorem and $(x^6 -1 , x^6 -3) \leq 2$; hence, 7 doesn't divide $ x^6 -3$. Hence, it is irreducible. 
