
theorem fresh3:
for p being Prime, a being Element of Z/p holds a|^p = a
proof
let p be Prime, a be Element of Z/p;
set F = Z/p; set M = MultGroup F;
H: the carrier of F = (the carrier of M) \/ {0.F} by UNIROOTS:15;
Char F = p by RING_3:def 6; then
consider n being non zero Nat such that A: card F = p|^n by FIELD_15:92;
I: now assume J: n <> 1;
   n >= 0 + 1 by INT_1:7;
   then n > 1 by J,XXREAL_0:1;
   then p|^n > p|^1 by NAT_6:2;
   hence contradiction by A,fresh3a;
   end;
per cases;
suppose a <> 0.F;
  then not a in {0.F} by TARSKI:def 1;
  then reconsider b = a as Element of M by H,XBOOLE_0:def 3;
  card M = p - 1 by I,A,UNIROOTS:18; then
  b|^(p - 1) = 1_M by GR_CY_1:9; then
  (1_M) * b
       = b|^(p - 1) * b|^1 by GROUP_1:26
      .= b|^(p - 1 + 1) by GROUP_1:33
      .= a|^p by FIELD_15:10;
  hence a|^p = a by GROUP_1:def 4;
  end;
suppose a = 0.F;
  hence a|^p = a;
  end;
end;
