
theorem split2:
for F being Field, E being FieldExtension of F
for n being non zero Nat
for a being Element of F
for b being Element of E st b = a holds X^(n,a) = X^(n,b)
proof
let F be Field, E be FieldExtension of F;
let n be non zero Nat, a be Element of F, b be Element of E;
assume AS: b = a;
H1: F is Subring of E by FIELD_4:def 1; then
H2: -b = -a by AS,FIELD_6:17;
now let i be Nat;
  per cases;
  suppose A: i = 0;
    hence X^(n,a).i = -b by H2,Lm10 .= X^(n,b).i by A,Lm10;
    end;
  suppose A: i = n;
    hence X^(n,a).i = 1.F by Lm10
                   .= 1.E by H1,C0SP1:def 3 .= X^(n,b).i by A,Lm10;
    end;
  suppose A: i <> 0 & i <> n;
    hence X^(n,a).i = 0.F by Lm11
                   .= 0.E by H1,C0SP1:def 3 .= X^(n,b).i by A,Lm11;
    end;
  end;
hence X^(n,a) = X^(n,b);
end;
