In the field Fp, which has characteristic p, is it true that for any element a in Fp, we have a^p=a?

Answer 1

Yes it is, look up Fermats little theorem.

It states that a^p = a mod p for all a and for any prime p.

The proof is as follows:

examine the list {a,2*a,3*a,4*a...(p-1)*a}
each entry in the list is distinct, becuase we can divide in Fp (it is a field!) (proof: if na = ma mod p then n = m mod p)

Furthermore since we have a list of p-1 elements, none of which are 0, we have a bijective mapping:
{a,2*a,3*a,4*a...(p-1)*a} <-> {1,2,3,4,5...p-1}
now simply multiple both sets of lists

on the left we have
a*2*a*3*a*4*a*...*(p-1)*a = (p-1)!*a^(p-1)
on the right we have 1*2*3*4*...*p-1 = (p-1)!
and these are equal mod p:

(p-1)!a^(p-1) = (p-1)! mod p

since (p-1)! is none zero, cancel it to be left with:

a^(p-1) = 1 mod p
and multiply by a...
a^p = a mod p

QED

Answer 2

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