so i came up with a proof that..well.. Let L/K be a field extension and we have defined the norm and trace of an element in L, call it a, to be the determinant (resp. trace) of the linear transformation L -> L given by x->ax. Now it's well known that the determinant and trace are the product/sum of the eigenvalues of a linear transformation. An eigenvalue here would be an element c in K such that for some nonzero x in L we have f(x)=cx where f(x)=ax which would mean we'd have cx=ax. So c=a, how is this possible if a is NOT in K? i'm clearly missing something very obvious here because this is nonsense!