
theorem split0:
for p being Prime
for F being p-characteristic Field
for a,b being Element of F st b|^p = a holds X^(p,a) = (X-b)|^p
proof
let p be Prime, R be p-characteristic Field, a,b be Element of R;
assume AS: b|^p = a;
per cases;
suppose C: p = 2; then
  D: (X-b)|^p
   = <%-b,1.R%>`^2 by RING_5:10
  .= <%-b,1.R%> *' <%-b,1.R%> by POLYNOM5:17
  .= <%(-b)*(-b),(-b)*(1.R)+(1.R)*(-b),(1.R)*(1.R)%> by FIELD_9:24
  .= <%(-b)*(-b),b*(1.R)+(1.R)*(-b),1.R%> by C,t2
  .= <%(-b)*(-b),0.R,1.R%> by RLVECT_1:5
  .= <%b*b,0.R,1.R%> by VECTSP_1:10
  .= <%b|^2,0.R,1.R%> by RING_5:3
  .= <%-a,0.R,1.R%> by AS,C,t2;
  now let j be Nat;
    j <= 2 implies j = 0 or ... or j = 2; then
    per cases;
    suppose E: j = 0;
      hence ((X-b)`^p).j = -a by D,FIELD_9:16 .= (X^(p,a)).j by E,Lm10;
      end;
    suppose E: j = 2;
      hence ((X-b)`^p).j = 1.R by D,FIELD_9:16 .= (X^(p,a)).j by E,C,Lm10;
      end;
    suppose E: j = 1;
      hence ((X-b)`^p).j = 0.R by D,FIELD_9:16 .= (X^(p,a)).j by E,C,Lm11;
      end;
    suppose E: j > 2;
      then j >= 2 + 1 by INT_1:7;
      hence ((X-b)`^p).j = 0.R by D,FIELD_9:16 .= (X^(p,a)).j by E,C,Lm11;
      end;
    end;
  hence X^(p,a) = (X-b)`^p .= (X-b)|^p;
  end;
suppose H0: p <> 2;
  p <= 2 implies p = 0 or ... or p = 2; then
  H: p is 2_greater by H0,INT_2:def 4,NAT_6:def 1;
  set q = X^(p,a);
  reconsider q1 = <%0.R,1.R%>, q2 = <%-b%>
              as Element of the carrier of Polynom-Ring R by POLYNOM3:def 10;
  set r = q1`^p + q2`^p;
  q2 = <%-b,0.R%> by POLYNOM5:43; then
  A: q1 + q2 = <%0.R,1.R%> + <%-b,0.R%> by POLYNOM3:def 10
            .= <%(0.R)+(-b),(1.R)+(0.R)%> by NIVEN:30
            .= X-b by RING_5:10;
  B: (q1 + q2)|^p = q1|^p + q2|^p by fresh;
  now let j be Nat;
  D0: r.j = (q1`^p).j + (q2`^p).j by NORMSP_1:def 2;
  H1: j is Element of NAT by ORDINAL1:def 12;
  per cases;
  suppose j in Support q; then
    q.j <> 0.R by POLYNOM1:def 3; then
    per cases by Lm11;
    suppose D: j = p; then
      D4: j >= 1 by INT_2:def 4;
      D1: (<%-b%>`^p).j = <%power(-b,p)%>.j by POLYNOM5:36
                       .= 0.R by H1,D4,POLYNOM5:32;
      D5: q1 = rpoly(1,-0.R) by RING_5:10;
      D3: deg(<%0.R,1.R%>`^p) = p * (deg <%0.R,1.R%>) by t1
                             .= p * 1 by D5,HURWITZ:27;
      thus q.j = 1.R by D,Lm10
              .= LC(<%0.R,1.R%>`^p) + (<%-b%>`^p).j by D1,D5,RATFUNC1:def 7
              .= r.j by D0,D3,D,FIELD_6:2;
      end;
    suppose D: j = 0; then
      D1: (<%0.R,1.R%>`^p).j = (<%0.R,1.R%>.0)|^p by t3
                            .= (0.R)|^p by POLYNOM5:38
                            .= 0.R;
      thus q.j = -(b|^p) by D,AS,Lm10
              .= (-b)|^p by H,todd
              .= <%power(-b,p)%>.j by D,POLYNOM5:32
              .= r.j by D0,D1,POLYNOM5:36;
      end;
    end;
  suppose D: not j in Support q; then
    D8: q.j = 0.R by H1,POLYNOM1:def 4;
    per cases;
    suppose D1: j = 0; then
      D2: -a = 0.R by D8,Lm10;
      D3: (<%-b%>`^p).0 = (<%-b%>.0)|^p by t3
                       .= (-b)|^p by POLYNOM5:32
                       .= 0.R by AS,D2,H,todd;
      (<%0.R,1.R%>`^p).0 = (<%0.R,1.R%>.0)|^p by t3
                        .= (0.R)|^p by POLYNOM5:38
                        .= 0.R;
      hence q.j = r.j by D0,D,D1,D3,POLYNOM1:def 4;
      end;
    suppose j = p;
      hence q.j = r.j by D8,Lm10;
      end;
    suppose D2: j <> 0 & j <> p; then
      D4: j >= 1 + 0 by INT_1:7;
      (<%-b%>`^p).j = <%power(-b,p)%>.j by POLYNOM5:36
                   .= 0.R by D4,H1,POLYNOM5:32;
      hence q.j = r.j by D8,D0,D2,t4;
      end;
    end;
  end;
  hence X^(p,a) = r .= (X-b)|^p by B,A,POLYNOM3:def 10;
  end;
end;
