# A stupid question on norm and trace of fields

1. Jul 31, 2010

### sidm

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!

2. Jul 31, 2010

That "the determinant and trace are the product/sum of the eigenvalues of a linear transformation" is true only when the underlying field is algebraically closed (and when you count multiplicity). You have just proven that your linear transformation has no eigenvalues in K if a is not in K. (If K is algebraically closed, then L = K since L/K is a finite extension.)

Last edited: Aug 1, 2010