
theorem hirr:
for F being Field,
    E being FieldExtension of F
for p being Element of the carrier of Polynom-Ring F,
    q being Element of the carrier of Polynom-Ring E
st q = p holds q is irreducible implies p is irreducible
proof
let F be Field, E be FieldExtension of F;
let p be Element of the carrier of Polynom-Ring F,
    q be Element of the carrier of Polynom-Ring E;
assume AS1: q = p;
assume AS: q is irreducible;
now assume p is reducible; then
  per cases by RING_4:41;
  suppose p = 0_.(F); then
    q = 0_.(E) by AS1,FIELD_4:12;
    hence q is reducible;
    end;
  suppose p is Unit of Polynom-Ring F; then
    deg p = 0 by RING_4:37; then
    deg q = 0 by AS1,FIELD_4:20;
    hence q is reducible by RING_4:37;
    end;
  suppose ex p1 being Element of the carrier of Polynom-Ring F
          st p1 divides p & 1 <= deg p1 & deg p1 < deg p; then
    consider p1 being Element of the carrier of Polynom-Ring F such that
    A: p1 divides p & 1 <= deg p1 & deg p1 < deg p;
    consider r being Polynomial of F such that
    B: p1 *' r = p by A,RING_4:1;
    p1 is Polynomial of E & r is Polynomial of E by FIELD_4:8; then
    reconsider q1 = p1, s = r as Element of the carrier of Polynom-Ring E
      by POLYNOM3:def 10;
    q1 *' s = q & deg q1 = deg p1 & deg q = deg p
      by B,AS1,FIELD_4:17,FIELD_4:20;
    hence q is reducible by A,RING_4:1,RING_4:41;
    end;
  end;
hence p is irreducible by AS;
end;
