
theorem Fpp1:
for p being Prime
for F being finite p-characteristic Field
for a being Element of F ex b being Element of F st b|^p = a
proof
let p be Prime, F be finite p-characteristic Field, a be Element of F;
set M = MultGroup F;
Char F = p by RING_3:def 6; then
consider n being non zero Nat such that A: card F = p|^n by CF;
reconsider n1 = n - 1 as Element of NAT by INT_1:3;
H: the carrier of F = (the carrier of M) \/ {0.F} by UNIROOTS:15;
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|^n - 1 by A,UNIROOTS:18; then
  b|^(p|^n - 1) = 1_M by GR_CY_1:9; then
  (1_M) * b
       = b|^(p|^n - 1) * b|^1 by GROUP_1:26
      .= b|^(p|^n - 1 + 1) by GROUP_1:33
      .= a|^(p|^(n1+1)) by MG
      .= a|^(p|^n1 * p) by NEWTON:6
      .= (a|^(p|^n1))|^p by BINOM:11; then
  a = (a|^(p|^n1))|^p by GROUP_1:def 4;
  hence ex b being Element of F st b|^p = a;
  end;
suppose C: a = 0.F;
  (0.F)|^p = 0.F;
  hence ex b being Element of F st b|^p = a by C;
  end;
end;
