
theorem tchar:
for R being preordered non degenerated Ring holds Char R = 0
proof
let R be preordered non degenerated Ring;
set P = the Preordering of R;
A: 1.R in P by ord3;
B: QS R c= P by ord2;
set n = Char R;
now assume AS: Char R <> 0;
  then C: n '*' 1.R = 0.R by RING_3:def 5;
  (n '*' 1.R) - 1.R = (n '*' 1.R) - (1 '*' 1.R) by RING_3:60;
  then F: (n-1) '*' 1.R  = - 1.R by C,RING_3:64;
  reconsider n1 = n-1 as Element of NAT by AS,INT_1:3;
  deffunc F(object) = 1.R;
  consider f being FinSequence such that
  D: len f = n1 & for k being Nat st k in dom f holds f.k = F(k)
     from FINSEQ_1:sch 2;
  now let o be object;
    assume o in rng f;
    then consider k being object such that
    E: k in dom f & o = f.k by FUNCT_1:def 3;
    f.k = 1.R by E,D;
    hence o in the carrier of R by E;
    end;
  then rng f c= the carrier of R;
  then reconsider f as FinSequence of R by FINSEQ_1:def 4;
  now let k be Element of NAT;
    assume 1 <= k & k <= n1;
    then E: k in dom f by D,FINSEQ_3:25;
    hence f/.k = f.k by PARTFUN1:def 6 .= 1.R by E,D;
    end;
  then E: Sum f = n1 * 1.R by D,POLYNOM8:4 .= -1.R by F,RING_3:def 2;
  now let k be Nat;
    assume G: k in dom f;
    1.R is square;
    hence ex a being Element of R st f.k = a^2 by G,D;
    end;
  then -1.R is sum_of_squares by E;
  then -1.R in QS R;
  then --1.R in -P by B;
  then 1.R in P /\ (-P) by A,XBOOLE_0:def 4;
  then 1.R in {0.R} by defppc;
  hence contradiction by TARSKI:def 1;
  end;
hence thesis;
end;
