
theorem CF:
for F being Field holds
F is finite iff ex n being non zero Nat st card F = (Char F)|^n
proof
let F1 be Field;
now assume F1 is finite; then
  reconsider F = F1 as finite Field;
  set p = Char F;
  H: F is p-characteristic by RING_3:def 6;
  set M = multMagma(#the carrier of F,the addF of F#);
  M is Group-like associative
    proof
    ex e being Element of M st
    for h being Element of M holds h * e = h & e * h = h &
      ex g being Element of M st h * g = e & g * h = e
      proof
      reconsider z = 0.F as Element of M;
      take z;
      now let h be Element of M;
        reconsider j = h as Element of F;
        j + 0.F = j & 0.F + j = j;
        hence h * z = h & z * h = h;
        thus ex g being Element of M st h * g = z & g * h = z
          proof
          reconsider g = -j as Element of M;
          take g;
          thus h * g = j + (-j) .= z by RLVECT_1:5;
          thus g * h = (-j) + j .= z by RLVECT_1:5;
          end;
        end;
      hence thesis;
      end;
    hence M is Group-like by GROUP_1:def 2;
    now let x,y,z be Element of M;
      reconsider a = x, b = y, c = z as Element of F;
      thus (x * y) * z
         = (a + b) + c
        .= a + (b + c) by RLVECT_1:def 3
        .= x * (y * z);
      end;
    hence M is associative by GROUP_1:def 3;
    end; then
  reconsider M as finite Group;
  H1: 1_M = 0.F
      proof
      reconsider z = 0.F as Element of M;
      now let h be Element of M;
        reconsider j = h as Element of F;
        j + 0.F = j & 0.F + j = j;
        hence h * z = h & z * h = h;
        end;
      hence thesis by GROUP_1:def 4;
      end;
  H2: now let a be Element of F, b be Element of M, n be non zero Nat;
      assume AS: a = b;
      defpred P[Nat] means
        for a being Element of F, b being Element of M
        holds a = b implies b |^ ($1) = ($1) * a;
      IA: P[1]
          proof
          now let a be Element of F, b be Element of M;
            assume a = b;
            hence b |^ 1 = a by GROUP_1:26 .= 1 * a by BINOM:13;
            end;
          hence thesis;
          end;
      IS: now let k be Nat;
          assume k >= 1;
          assume B: P[k];
          now let a be Element of F, b be Element of M;
            assume C: a = b; then
            D: b|^k = k * a by B;
            thus b |^ (k+1)
             = b|^k * b by GROUP_1:34
            .= (k * a) + (1 * a) by D,C,BINOM:13
            .= (k+1) * a by BINOM:15;
            end;
          hence P[k+1];
          end;
      I: for k being Nat st k >= 1 holds P[k] from NAT_1:sch 8(IA,IS);
      n >= 0 + 1 by INT_1:7;
      hence b |^ n = n * a by I,AS;
      end;
  reconsider e = 1.F as Element of M;
  ord e = p
      proof
      B0: p is Element of NAT by ORDINAL1:def 12;
      B1: e |^ p = p * 1.F by H2
                .= p '*' 1.F by RING_3:def 2
                .= 1_M by H1,RING_3:def 5; then
      B2: not e is being_of_order_0 by GROUP_1:def 10;
      now let m be Nat;
        assume B4: e |^ m = 1_M & m <> 0;
        m '*' 1.F = m * 1.F by RING_3:def 2 .= 0.F by B4,H2,H1;
        hence p <= m by B4,RING_3:def 5;
        end;
      hence thesis by B0,B1,B2,GROUP_1:def 11;
      end; then
  A: p divides (card M) by GR_CY_1:8;
  B: now let q be Prime;
     assume E: p <> q & q divides (card M); then
     consider b being Element of M such that C: ord b = q by GROUP_10:11;
     reconsider a = b as Element of F;
     q > 1 by INT_2:def 4; then
     D: a is non zero by H1,C,GROUP_1:42;
     q * a = b |^ q by H2 .= 0.F by C,GROUP_1:41,H1;
     then p divides q & p > 1 by H,D,Lm2a,INT_2:def 4;
     hence contradiction by E,INT_2:def 4;
     end;
  ex n being Nat st card F = p|^n
     proof
     consider k,m being Nat such that
     C: card F = m * p|^k & not p divides m by CF1;
     m <= 1 implies m = 0 or ... or m = 1; then
     per cases;
     suppose m = 0 or m = 1;
       hence ex n being Nat st card F = p|^n by C;
       end;
     suppose m > 1;
       then m >= 1 + 1 by INT_1:7;
       then ex q being Element of NAT st q is prime & q divides m by INT_2:31;
       hence ex n being Nat st card F = p|^n by B,C,INT_2:2;
       end;
     end;
  then consider n being Nat such that F: card F = p|^n;
  now assume n is zero; then
    card F = 1 & p > 1 by F,NEWTON:4,INT_2:def 4;
    hence contradiction by A,NAT_D:7;
    end;
  hence ex n being non zero Nat st card F1 = (Char F1)|^n by F;
  end;
hence thesis;
end;
