
theorem Fpp:
for p being Prime
for F being p-characteristic Field holds F is perfect iff F == F|^p
proof
let p be Prime, F be p-characteristic Field;
A: now assume A1: not F == F|^p;
   set a = the Element of (the carrier of F) \ (the carrier of F|^p);
   A2: the carrier of F|^p c= the carrier of F by EC_PF_1:def 1;
       now assume B: the carrier of F|^p = the carrier of F;
       then the addF of F|^p = (the addF of F)||(the carrier of F) &
            the multF of F|^p = (the multF of F)||(the carrier of F) &
            1.(F|^p) = 1.F & 0.(F|^p) = 0.F by deffp;
       hence contradiction by B,A1;
       end; then
   the carrier of F|^p c< the carrier of F by A2,XBOOLE_0:def 8; then
   A3: (the carrier of F) \ (the carrier of F|^p) <> {} by XBOOLE_1:105; then
   reconsider a as Element of F by XBOOLE_0:def 5;
   not a in F|^p by A3,XBOOLE_0:def 5;
   then X^(p,a) is irreducible inseparable by contr;
   hence F is non perfect;
   end;
now assume A0: F == F|^p;
   now assume F is non perfect; then
     consider q being irreducible
                          Element of the carrier of Polynom-Ring F such that
     A1: q is inseparable;
     consider E being FieldExtension of F such that
     A2: q splits_in E &
         ex a being Element of E st a is_a_root_of q,E & multiplicity(q,a) <> 1
         by A1,ThSep0;
     consider a being Element of E such that
     A3: a is_a_root_of q,E & multiplicity(q,a) <> 1 by A2;
     multiplicity(q,a) >= 1 by A3,mulzero; then
     multiplicity(q,a) > 1 by A3,XXREAL_0:1; then
     A4: (Deriv F).q = 0_.(F) by A2,lemsep1;
     now let i be Nat;
       assume i in Support q;
       hence p divides i by A4,DP0;
       q.i in the carrier of F|^p by A0; then
       q.i in the set of all a|^p where a is Element of F by deffp;
       hence ex a being Element of F st a|^p = q.i;
       end; then
     consider r being Element of the carrier of Polynom-Ring F such that
     A5: r|^p = q by DP1;
     reconsider r as non constant Element of the carrier of Polynom-Ring F
         by A5;
     H: r|^((p-1)+1) = (r|^(p-1)) * r|^1 by BINOM:10
                    .= (r|^(p-1)) * r by BINOM:8
                    .= (r`^(p-1)) *' r by POLYNOM3:def 10;
     A6: 0 + 1 <= deg r by INT_1:7,RING_4:def 4;
     r <> 0_.(F) & r`^(p-1) <> 0_.(F); then
     K: deg q = deg(r`^(p-1)) + deg r by A5,H,HURWITZ:23
             .= (p-1) * (deg r) + deg r by t1
             .= p * (deg r);
     now assume deg r >= deg q;
       then deg q >= p * (deg q) by K,XREAL_1:64;
       then (deg q) / (deg q) >= (p * (deg q)) / (deg q) by XREAL_1:64;
       then L: 1 >= (p * (deg q)) / (deg q) by K,A6,XCMPLX_1:60;
       (p * (deg q)) / (deg q) = p * ((deg q) / (deg q))
                              .= p * 1 by K,A6,XCMPLX_1:60;
       hence contradiction by L,INT_2:def 4;
       end;
     hence contradiction by H,A5,A6,RING_4:1,RING_4:40;
     end;
   hence F is perfect;
   end;
hence thesis by A;
end;
