Bourbaki

Proposition 8

Tag 06F9, permanent. Cite it as [Bourbaki, Tag 06F9].

Algebra, chapter IV, § 6. Symmetric functions

— Suppose that f and g are monic of degrees p and q respectively. With the above notation, the A-module F is free with basis (xiyj)0≤i<p, 0≤j<q(x^i y^j)_{0 \leq i < p,\ 0 \leq j < q} and we have

res(f,g)=NF/A(x−y). \text{res}(f, g) = N_{F/A}(x - y).

Put E=A[X]/(f)E = A[X]/(f) and E′=A[Y]/(g)E' = A[Y]/(g). By II, p. 253, Cor. 1, the homomorphism σ\sigma of E@E′E @ E' into F derived from the canonical homomorphism A[X]⊗A[Y]→A[X,Y]A[X] \otimes A[Y] \to A[X, Y] is bijective; this proves the assertion about the basis of F. We now identify E with its image in F under σ\sigma. Then the E-algebra homomorphism of E[Y]/(g(Y))E[Y]/(g(Y)) into F which maps Y to y is an isomorphism.

By the transitivity of the norm (III, p. 546) we have

NF/A(x−y)=NE/A(NF/E(x−y)). N_{F/A}(x - y) = N_{E/A}(N_{F/E}(x - y)).

By Prop. 7 (IV, p. 77), NF/E(x−y)N_{F/E}(x - y) is the resultant of the polynomials g(Y) and x − Y in E[Y], hence equal to g(x) (IV, p. 78, Cor. 3). By formula (34) and Prop. 7 (IV, p. 77) we thus have

NF/A(x−y)=NE/A(g(x))=res(f,g) N_{F/A}(x - y) = N_{E/A}(g(x)) = \text{res}(f, g)

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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