Let R be a commutative ring, f(X)=Xd+∑i=1daiXd−i a monic polynomial of R[X], A the R-algebra R[X]/(f) and x the residue class of X in A. Then TrA/R(x)=−a1 and NA/R(x)=(−1)dad.
By the Cor. (IV, p. 11) the sequence (1,x,...,xd−1) is a basis of A ; further we have
x⋅1=x,x⋅x=x2,…,x⋅xd−2=xd−1,
x⋅xd−1=−ad⋅1−ad−1⋅x−…−a1⋅xd−1
The matrix which expresses the multiplication by x with respect to the basis (1,x,...,xd−1) of A is thus of the following form (we have taken d=5 to fix the ideas):
01000001000001000001−a−a−a3−a2−a1
The trace of this matrix is clearly −a; the determinant may be calculated by expanding by the first row, and we then find
(−1)d−1(−ad)=(−1)dad.
For the rest of this No. we denote by E an extension of finite degree of K and by x an element of E. We shall indicate how to calculate the norm and trace of x in various cases.
a) Case of a separable extension: suppose that E is separable of degree n over K, denote by Ω an algebraic closure of K and by σ1,…,σn the n distinct K-homomorphisms of E into R. By Formula (3) (V, p. 48) we have in R
TrE/K(x)=i=1∑nσi(x),NE/K(x)=i=1∏nσi(x).
b) Case of a p-radical extension: suppose that K is of characteristic p>0 and that the extension E is p-radical; there exists an integer e≥0 such that [E:K]=pe (V, p. 26, Prop. 4). If f is the height of x over K, the minimal polynomial of x over K is Xpf−xpf (V, p. 24, Prop. 1). By Prop. 3 we have NE/K(x)=(xpf)pe/pf, whence
NE/K(x)=xpe=x[E:K]
For the trace we find TrE/K(x)=−pe−fa, where a is the coefficient of Xpf−1 in the polynomial Xpf−xpf; in other words, we have
TrE/K(x)=pe⋅x=[E:K]x={x0if [E:K]=1if [E:K]>1.
c) General case: we can summarize the calculation of norm and trace in the following proposition:
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.