Bourbaki

Lemma 3

Tag 06LG, permanent. Cite it as [Bourbaki, Tag 06LG].

Algebra, V, p. 50, § 8. Norms and traces

Let R be a commutative ring, f(X)=Xd+∑i=1daiXd−if(X) = X^d + \sum_{i=1}^d a_i X^{d-i} a monic polynomial of R[X]\mathbf{R}[X], A the R-algebra R[X]/(f)\mathbf{R}[X]/(f) and x the residue class of X in A. Then TrA/R(x)=−a1\text{Tr}_{A/R}(x) = -a_1 and NA/R(x)=(−1)dad\text{N}_{A/R}(x) = (-1)^d a_d.

By the Cor. (IV, p. 11) the sequence (1,x,...,xd−1)(1, x, ..., x^{d-1}) is a basis of A ; further we have

x⋅1=x,x⋅x=x2,…,x⋅xd−2=xd−1, x \cdot 1 = x , \quad x \cdot x = x^2 , \ldots , x \cdot x^{d-2} = x^{d-1} ,
x⋅xd−1=−ad⋅1−ad−1⋅x−…−a1⋅xd−1 x \cdot x^{d-1} = -a_d \cdot 1 - a_{d-1} \cdot x - \ldots - a_1 \cdot x^{d-1}

The matrix which expresses the multiplication by x with respect to the basis (1,x,...,xd−1)(1, x, ..., x^{d-1}) of AA is thus of the following form (we have taken d=5d = 5 to fix the ideas):

(0000−a1000−a0100−a30010−a20001−a1) \begin{pmatrix} 0 & 0 & 0 & 0 & -a \\ 1 & 0 & 0 & 0 & -a \\ 0 & 1 & 0 & 0 & -a_3 \\ 0 & 0 & 1 & 0 & -a_2 \\ 0 & 0 & 0 & 1 & -a_1 \end{pmatrix}

The trace of this matrix is clearly −a-a; the determinant may be calculated by expanding by the first row, and we then find

(−1)d−1(−ad)=(−1)dad. (-1)^{d-1} (-a_d) = (-1)^d a_d .

For the rest of this No. we denote by E an extension of finite degree of K and by xx an element of E. We shall indicate how to calculate the norm and trace of xx in various cases.

a) Case of a separable extension: suppose that E is separable of degree nn over K, denote by Ω\Omega an algebraic closure of K and by σ1,…,σn\sigma_1, \ldots, \sigma_n the nn distinct KK-homomorphisms of E into R. By Formula (3) (VV, p. 48) we have in R

TrE/K(x)=∑i=1nσi(x),NE/K(x)=∏i=1nσi(x). \text{Tr}_{E/K}(x) = \sum_{i=1}^n \sigma_i(x) , \quad N_{E/K}(x) = \prod_{i=1}^n \sigma_i(x) .

b) Case of a p-radical extension: suppose that K is of characteristic p>0p > 0 and that the extension E is p-radical; there exists an integer e≥0e \geq 0 such that [E:K]=pe[E:K] = p^e (VV, p. 26, Prop. 4). If ff is the height of xx over K, the minimal polynomial of xx over K is Xpf−xpfX^{p^f} - x^{p^f} (VV, p. 24, Prop. 1). By Prop. 3 we have NE/K(x)=(xpf)pe/pfN_{E/K}(x) = (x^{p^f})^{p^e/p^f}, whence

NE/K(x)=xpe=x[E:K] N_{E/K}(x) = x^{p^e} = x^{[E:K]}

For the trace we find TrE/K(x)=−pe−fa\text{Tr}_{E/K}(x) = -p^{e-f} a, where aa is the coefficient of Xpf−1X^{p^f-1} in the polynomial Xpf−xpfX^{p^f} - x^{p^f}; in other words, we have

TrE/K(x)=pe⋅x=[E:K]x={xif [E:K]=10if [E:K]>1. \text{Tr}_{E/K}(x) = p^e \cdot x = [E:K] x = \begin{cases} x & \text{if } [E:K] = 1 \\ 0 & \text{if } [E:K] > 1 . \end{cases}

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.

Cited by

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Cites

Transcribed from the printed edition, 2003, Springer. Not translated.