Library / Algebra / Chapter IV / § 6
Corollary 3
Tag 06F8, permanent. Cite it as [Bourbaki, Tag 06F8].
Algebra, chapter IV, § 6. Symmetric functions
— For every λ ∈ A \lambda \in A λ ∈ A
res p , 1 ( f , λ − X ) = f ( λ ) , res 1 , q ( X − λ , g ) = g ( λ ) .
\operatorname{res}_{p,1}(f, \lambda - X) = f(\lambda), \quad \operatorname{res}_{1,q}(X - \lambda, g) = g(\lambda).
res p , 1 ( f , λ − X ) = f ( λ ) , res 1 , q ( X − λ , g ) = g ( λ ) . When
f ( X ) = X − λ f(X) = X - \lambda f ( X ) = X − λ , the algebra E is equal to A and we have
x = A x = A x = A ; the second formula (32) now follows from Prop. 7 (IV, p. 77). By remarks 1 and 3 (IV, p. 76, 77) we conclude
res p , 1 ( f , λ − X ) = ( − 1 ) p res 1 , p ( λ − X , f ) = ( − 1 ) p + p res 1 , p ( X − A , f ) = f ( A ) .
\operatorname{res}_{p,1}(f, \lambda - X) = (-1)^p \operatorname{res}_{1,p}(\lambda - X, f) = (-1)^{p+p} \operatorname{res}_{1,p}(X - A, f) = f(A).
res p , 1 ( f , λ − X ) = ( − 1 ) p res 1 , p ( λ − X , f ) = ( − 1 ) p + p res 1 , p ( X − A , f ) = f ( A ) . Suppose now that f and g are monic. We denote by F the A-algebra A [ X , Y ] / ( f ( X ) , g ( Y ) ) A[X, Y]/(f(X), g(Y)) A [ X , Y ] / ( f ( X ) , g ( Y ))
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.