Corollary 1.10.5 Assume L/K is not separable. Then Tr_{L/K}=0.

proof: Let x in L with x_i the roots (in some algebraic closure of K, counting multiplicities) of its minimal polynomial P over K.

case (1) If x is SEPARABLE over K.

case (2) If x is not separable over K.

Example 1.10.6

(1) Let K be a field, x algebraic over K and P(X)=X^n+a_1X+...+a_n in K[X] its minimal polynomial, then

Tr_{K(x)/K}(x)=-a_1

N_{K(x)/K}(x)=(-1)^na_n

chi_{x,L/K}=P.

(2) If L/K is a SEPARABLE finite extension, K^bar an algebraic closure of K and Hom_{K-alg}(L,K^bar)={sigma_1, ... , sigma_d}, we have d=[L : K], and 

Tr_{L/K}(x)=Sum sigma_i(x)

N_{L/K}(x)=Sum sigma_i(x)

Corollary 1.10.7

Let A be an integrally closed domain, K = FracA, L/K a finite extension and B the integral closure of A in L. If b in B, then Tr_{L/K}(b), N_{L/K}(b) in A and chi_{b, L/K} in A[X] and we have b in B* if and only if N_{L/K}(x) in A*.