For a finite field $F$ of order $n$, all elements are roots of $x^n - x$

Question

I need to prove two things at $F[X]$ but don't know how, Ill glad to get help...

$F$ is a finite field.
$|F|=n$. We look at $p(x)=x^n-x\in F[X]$
1. How we can show that every $c\in F$ is a root of $p(x)$ (Lagrange Theorem).
2. $p(x)=\prod_{c\in F}(x-c)$

Thank you!

Answer 1

Hints:

(1) The multiplicative group $F^\times$ is a finite group of order $n - 1$. This gives $x^{n-1} = 1$ for all $x \in F - \{0\}$ by Lagrange's theorem.

(2) In the polynomial ring over a field $F$, if $c \in F$ is a root of $x^n - x$, then $x - c$ divides $x^n - x$.

Answer 2

1) You gave yourself a hint. ;) Apply Lagrange's theorem in the group of units of F.

2) Recall that if $a$ is a root of $p(x)$, then $p(x) = (x-a)q(x)$ for some $q(x)$ whose degree is one less than the degree of $p(x)$.

Answer 3

Use the hint, Lagranges' Theorem to dispatche $(1)\ \ $ (that's quite a big hint!)

By $(1)$ and the Factor Theorem, $\,x^n\!-x$ is disvisible by the nonassociate primes $\, x- c,\, $ all $\,c\in F,\,$ therefore it is divisible by their lcm = product. Finally compare degree and lead coefficient.