
theorem aXn:
for F1 being Field,
    F2 being F1-homomorphic Field
for h being Homomorphism of F1,F2
for a being Element of F1
for n being Nat holds (PolyHom h).((X-a)`^n)= (X-(h.a))`^n
proof
let F1 be Field, F2 be F1-homomorphic Field, h be Homomorphism of F1,F2;
let a be Element of F1, n be Nat;
defpred P[Nat] means (PolyHom h).((X-a)`^($1))= (X-(h.a))`^($1);
  (PolyHom h).((X-a)`^0)
     = (PolyHom h).(1_.(F1)) by POLYNOM5:15
    .= 1_.(F2) by FIELD_1:23
    .= (X-(h.a))`^0 by POLYNOM5:15; then
IA: P[0];
IS: now let k be Nat;
    assume IV: P[k];
    reconsider p = (X-a)`^k, q = X-a
        as Element of the carrier of Polynom-Ring F1 by POLYNOM3:def 10;
    A: (PolyHom h).p = (X-(h.a))`^k &
       (PolyHom h).q = X-(h.a) by IV,FIELD_13:25;
    (PolyHom h).((X-a)`^(k+1))
        = (PolyHom h).(((X-a)`^k) *' (X-a)) by POLYNOM5:19
       .= (PolyHom h).(p * q) by POLYNOM3:def 10
       .= (PolyHom h).p * (PolyHom h).q by FIELD_1:25
       .= ((X-(h.a))`^k) *' (X-(h.a)) by A,POLYNOM3:def 10
       .= (X-(h.a))`^(k+1) by POLYNOM5:19;
    hence P[k+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis;
end;
