
theorem lem21:
for F being Field
for p being Element of the carrier of Polynom-Ring F
st deg p is odd
ex q being non constant monic Element of the carrier of Polynom-Ring F
st q divides p & q is irreducible & deg q is odd
proof
let F be Field, p1 be Element of the carrier of Polynom-Ring F;
assume AS: deg p1 is odd;
per cases;
suppose p1 is zero; then
  A: (X-(1.F)) *' 0_.(F) = p1 by UPROOTS:def 5;
  deg(X-(1.F)) = 2 * 0 + 1 by FIELD_5:def 1;
  hence thesis by A,RING_4:1;
  end;
suppose p1 is non zero; then
reconsider p = p1 as non zero Element of the carrier of Polynom-Ring F;
defpred Q[Nat] means
  for p be non zero Element of the carrier of Polynom-Ring F
  st deg p = 2 * ($1) + 1
  ex q being non constant monic Element of the carrier of Polynom-Ring F
  st q divides p & q is irreducible & deg q is odd;
IS: now let k be Nat;
    assume AS1: for n being Nat st n < k holds Q[n];
    per cases;
    suppose AS2: k = 0;
      now let p be non zero Element of the carrier of Polynom-Ring F;
        set q = NormPolynomial p;
        p *' 1_.(F) = p; then
        A: q divides p by RING_4:25,RING_4:1;
        C: 1 = 2 * 0 + 1;
        assume deg p = 2 * 0 + 1; then
        B: deg q = 1 by lemdeg; then
        q is linear by FIELD_5:def 1;
        hence ex q being non constant monic
                              Element of the carrier of Polynom-Ring F
              st q divides p & q is irreducible & deg q is odd by A,B,C;
      end;
      hence Q[k] by AS2;
      end;
    suppose k > 0;
      now let p be non zero Element of the carrier of Polynom-Ring F;
        assume AS3: deg p = 2 * k + 1; then
        p is non constant by RING_4:def 4; then
        consider q being non constant monic
                     Element of the carrier of Polynom-Ring F such that
        A: q divides p & q is irreducible by lem22;
        per cases;
        suppose deg q is odd;
          hence ex q being non constant monic
                              Element of the carrier of Polynom-Ring F
                st q divides p & q is irreducible & deg q is odd by A;
          end;
        suppose B: deg q is even;
          consider r being Polynomial of F such that
          C: q *' r = p by A,RING_4:1;
          q <> 0_.(F) & r <> 0_.(F) by C; then
          I: deg q + deg r = deg p by C,HURWITZ:23; then
          deg r is odd by AS3,B; then
          consider i being Integer such that
          E: deg r = 2 * i + 1 by ABIAN:1;
          now assume E2: i < 0;
            r is non zero by C;
            hence contradiction by E,E2,INT_1:7;
            end; then
          reconsider i as Element of NAT by INT_1:3;
          reconsider r as Element of the carrier of Polynom-Ring F
            by POLYNOM3:def 10;
          reconsider r as non zero Element of the carrier of Polynom-Ring F
            by C;
          deg r <= deg p & deg r <> deg p
                 by C,I,RING_4:def 4,RING_4:1,RING_5:13; then
          2 * i + 1 < 2 * k + 1 by AS3,E,XXREAL_0:1; then
          2 * i + 1 - 1 < 2 * k + 1 - 1 by XREAL_1:9; then
          2 * i - 2 * k < 2 * k - 2 * k by XREAL_1:9; then
          2 * (i - k) < 0; then
          i - k < 0; then
          i - k + k < 0 + k by XREAL_1:6; then
          consider s being non constant monic
                       Element of the carrier of Polynom-Ring F such that
          F: s divides r & s is irreducible & deg s is odd by E,AS1;
          consider t being Polynomial of F such that
          H: s *' t = r by F,RING_4:1;
          s *' (t *' q) = p by H,C,POLYNOM3:33;
          hence ex q being non constant monic
                              Element of the carrier of Polynom-Ring F
                st q divides p & q is irreducible & deg q is odd by F,RING_4:1;
          end;
        end;
      hence Q[k];
      end;
    end;
I: for k being Nat holds Q[k] from NAT_1:sch 4(IS);
consider i being Integer such that H: deg p = 2 * i + 1 by AS,ABIAN:1;
i >= 0 by H,INT_1:7; then
reconsider i as Element of NAT by INT_1:3;
deg p = 2 * i + 1 by H;
hence thesis by I;
end;
end;
