Prove that $\displaystyle x^{p^n}-x+1$ is irreducible over $\displaystyle \mathbb{F}_p$ only when $\displaystyle n=1$ or $\displaystyle n=p=2$.
if $\displaystyle n=p^km,$ where $\displaystyle m > 1, \ k \geq 0, \ \gcd(p,m)=1,$ then it's fairly easy to see that $\displaystyle x^{p^{p^k}} - x + m^{-1}$ will divide $\displaystyle x^{p^n}-x+1.$ so we may assume that $\displaystyle n=p^k, \ k \geq 0.$
Last edited by NonCommAlg; Dec 9th 2008 at 11:08 PM.
if $\displaystyle n=p^km,$ where $\displaystyle m > 1, \ k \geq 0, \ \gcd(p,m)=1,$ then it's fairly easy to see that $\displaystyle x^{p^{p^k}} - x + m^{-1}$ will divide $\displaystyle x^{p^n}-x+1.$ so we may assume that $\displaystyle n=p^k, \ k \geq 0.$
Why does $\displaystyle x^{p^{p^k}} - x + m^{-1}$ divide $\displaystyle x^{p^n}-x+1.$?
Please...give me some more detail explanation.