Test irreducibility of polynomial over finite field I am reading a text, where it says that $X^5-X-1$ is irreducible modulo 3. I am not sure how I can know that. Could someone help? 
By the way, is there some practical trick to judge whether a polynomial irreducible over a finite field in general? Because when I calculate the galois group I find this information is very important, so I am very curious. Thanks!
 A: Hint $\:\!\color{#0a0}f\:\!$ has no roots $\in\Bbb F_3$  so no linear factors. If it splits it has an irreducible $\rm\color{#90f}{quadratic}$ factor $\color{#90f}g.\,$ In  $\,\Bbb F_9\! = \Bbb F_3[x]/\color{#90f}g\!:\: $ $\underbrace{x^8\! = 1}_{\!\!\!\!\!\!\!\!\!\large \Rightarrow\,\ \color{#c00}{{x^4}\ =\ \pm1}}\,$ so $\,0\! =\underbrace{\color{#c00}{x^4}\color{0a0a}x\!-\!x\!-\!1}_{\large \color{#0a0}{\deg f\ \le\ 1}}\, $ so $\,\color{#90f}g\mid\color{#0a0} f\,$ in $\,\Bbb F_3[x],\,$ contra $\,\color{#90f}{\deg g \!=\! 2},\,\color{#0a0}{\deg f\!\le\! 1}$.
Remark $ $ We used $\,0\neq f\in \Bbb F_9\Rightarrow f^{\large \color{#c00}8}\! = 1\,$ (for $\,f =x),\,$ an analogue of Fermat's little Theorem, which is true because $\,\Bbb  F_9$ has multiplicative group $\,\Bbb F_9^*$ of size $\,\color{#c00}8 = 9\color{#0a0}{-1}\,$ ($\rm non\color{#0a0}{zero}$ elements are invertible in a field), so Lagrange's Theorem $\Rightarrow f^{\large \color{#c00}8}\! = 1\,$ for all $\,f\neq 0.$
Above is essentially a special case of a general  polynomial irreducibility test over finite fields - which is an an efficient analog of the impractical Pocklington-Lehmer integer primality test.
