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?
Hello,
x cannot be =0, otherwise, it wouldn't be possible that x^p=1
Considering that x is an element of the field, $\displaystyle \neq 0$, we have by Fermat's little theorem :
$\displaystyle x^p \equiv x (\bmod{p})$
So if $\displaystyle x^p \equiv 1 (\bmod{p})$, then ...
Hmmm I think it killed the question... or I've made a mistake ^^'
If $\displaystyle \text{char}(F)=p$ then $\displaystyle (x+y)^p = x^p+y^p$.
Thus if $\displaystyle x^p = 1 \implies x^p -1 = 0 \implies (x-1)^p = \implies x=1$.
What you did is correct if $\displaystyle F=\mathbb{F}_p$.Originally Posted by Moo
However, a finite field need not be a the integers modulo $\displaystyle p$.
Thus, it does not cover all the cases.
This is Mine 14th Post!!!