irreducible

**Stiger** · December 9th 2008, 06:46 PM

Prove that $x^{p^n}-x+1$ is irreducible over $\mathbb{F}_p$ only when $n=1$ or $n=p=2$ .

**NonCommAlg** · December 9th 2008, 10:49 PM

Originally Posted by Stiger

Prove that $x^{p^n}-x+1$ is irreducible over $\mathbb{F}_p$ only when $n=1$ or $n=p=2$ .

if $n=p^km,$ where $m > 1, \ k \geq 0, \ \gcd(p,m)=1,$ then it's fairly easy to see that $x^{p^{p^k}} - x + m^{-1}$ will divide $x^{p^n}-x+1.$ so we may assume that $n=p^k, \ k \geq 0.$

**Stiger** · December 11th 2008, 06:27 AM

Originally Posted by NonCommAlg

if $n=p^km,$ where $m > 1, \ k \geq 0, \ \gcd(p,m)=1,$ then it's fairly easy to see that $x^{p^{p^k}} - x + m^{-1}$ will divide $x^{p^n}-x+1.$ so we may assume that $n=p^k, \ k \geq 0.$

Why does $x^{p^{p^k}} - x + m^{-1}$ divide $x^{p^n}-x+1.$ ?

Please...give me some more detail explanation.

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