
theorem ll:
for F being Field
for a being Element of F, p being Polynomial of F
for E being FieldExtension of F
for b being Element of E, q being Polynomial of E
st a = b & p = q holds a * p = b * q
proof
let F be Field, a be Element of F, p be Polynomial of F;
let E be FieldExtension of F, b be Element of E, q be Polynomial of E;
H: F is Subring of E by FIELD_4:def 1;
assume AS: a = b & p = q;
now let o be object;
  assume o in NAT;
  then reconsider i = o as Element of NAT;
  (a * p).i = a * (p.i) by POLYNOM5:def 4
           .= b * (q.i) by AS,H,FIELD_6:16
           .= (b * q).i by POLYNOM5:def 4;
  hence (a * p).o = (b * q).o;
  end;
hence thesis;
end;
