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?

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- Aug 31st 2008, 06:30 AMpetersmithSolution 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? - Aug 31st 2008, 06:38 AMMoo
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 ^^' - Aug 31st 2008, 07:25 AMThePerfectHacker
If $\displaystyle \text{char}(F)=p$ then $\displaystyle (x+y)^p = x^p+y^p$.

Thus if $\displaystyle x^p = 1 \implies x^p -1 = 0 \implies (x-1)^p = \implies x=1$.

Quote:

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.

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