# Solution of x^p = 1 in a finite field of char p

• August 31st 2008, 06:30 AM
petersmith
Solution of x^p = 1 in a finite field of char p
I was wondering if x^p = 1 implies that x =1 in a finite field of characteristic p, or in other words
If x^p = 1 (mod p), does that imply that x = 1 (mod p) ?

If so why?
• August 31st 2008, 06:38 AM
Moo
Hello,
Quote:

Originally Posted by petersmith
I was wondering if x^p = 1 implies that x =1 in a finite field of characteristic p, or in other words
If x^p = 1 (mod p), does that imply that x = 1 (mod p) ?

If so why?

x cannot be =0, otherwise, it wouldn't be possible that x^p=1
Considering that x is an element of the field, $\neq 0$, we have by Fermat's little theorem :
$x^p \equiv x (\bmod{p})$
So if $x^p \equiv 1 (\bmod{p})$, then ...

Hmmm I think it killed the question... or I've made a mistake ^^'
• August 31st 2008, 07:25 AM
ThePerfectHacker
If $\text{char}(F)=p$ then $(x+y)^p = x^p+y^p$.
Thus if $x^p = 1 \implies x^p -1 = 0 \implies (x-1)^p = \implies x=1$.

Quote:

Originally Posted by Moo
Hmmm I think it killed the question... or I've made a mistake ^^'

What you did is correct if $F=\mathbb{F}_p$.
However, a finite field need not be a the integers modulo $p$.
Thus, it does not cover all the cases.

This is Mine 1:)4:):)th Post!!!