
theorem thirr:
for F being Field,
    p being Element of the carrier of Polynom-Ring F
st deg p = 1 holds p is irreducible
proof
let R be Field, p be Element of the carrier of Polynom-Ring R;
set K = Polynom-Ring R;
reconsider x = p as Element of K;
assume AS: deg p = 1;
then p <> 0_.(R) by HURWITZ:20; then
A: x <> 0.K by POLYNOM3:def 10;
B: not x is Unit of K by AS,T88;
now let a be Element of K;
  assume a divides x;
  then consider b being Element of K such that H1: a * b = x;
  reconsider q = a, r = b as Element of the carrier of K;
  q *' r = p by H1,POLYNOM3:def 10;
  then H2: q <> 0_.(R) & r <> 0_.(R) by AS,HURWITZ:20,POLYNOM3:34;
  then H4: deg q + deg r = deg(q *' r) by HURWITZ:23
                        .= 1 by AS,H1,POLYNOM3:def 10;
  per cases;
  suppose C0: deg q = 0;
    then q is constant; then consider u being Element of R such that
    C1: q = u|R by T11;
    u is non zero by C1,C0,T11a;
    hence a is Unit of K or a is_associated_to x by Th90,C1;
    end;
  suppose deg q <> 0;
    then deg q > 0 by H2,T8b;
    then deg q + deg r > 0 + deg r by XREAL_1:6;
    then r is constant by H4,NAT_1:14;
    then consider u being Element of R such that C1: r = u|R by T11;
    u is non zero by C1,T6,H2;
    then r is Unit of K by Th90,C1;
    hence a is Unit of K or a is_associated_to x by H1,GCD_1:18;
    end;
  end;
hence thesis by A,B,RING_2:def 9;
end;
