Bourbaki

Proposition 6

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

Algebra, chapter IV, § 6. Symmetric functions

— Let f∈A[X]f \in A[X] be a monic polynomial of degree n, and let P be a symmetric polynomial in X1,...,XnX_1, ..., X_n with coefficients in A. Then there exists precisely one element a of A with the following property: (FS) For any ring homomorphism ρ:A→B\rho : A \to B and decomposition 0f(X)=∏i=1n(X−ξi)^0f(X) = \prod_{i=1}^n (X - \xi_i) in B[X] we have ρ(a)=P(ξ1,...,ξn)\rho(a) = P(\xi_1, ..., \xi_n).

Write f=Xn+∑k=1nakXn−kf = X^n + \sum_{k=1}^n a_k X^{n-k}, then by Th. 1, IV, p. 62 there exists a polynomial Π\Pi in n indeterminates with coefficients in A, such that P=Π(s1,...,sn)P = \Pi(s_1, ..., s_n). Put a=Π(−a1,a2,...,(−1)nan)a = \Pi(-a_1, a_2, ..., (-1)^n a_n). Under the hypothesis (FS) we have

sk(ξ1,...,ξn)=(−1)kρ(ak) s_k(\xi_1, ..., \xi_n) = (-1)^k \rho(a_k)
whence
ρ(a)=Π(−ρ(a1),ρ(a2),...,(−1)nρ(an))=Π(s1(ξ1,...,ξn),...,sn(ξ1,...,ξn))=P(ξ1,...,ξn). \begin{align*} \rho(a) &= \Pi(-\rho(a_1), \rho(a_2), ..., (-1)^n \rho(a_n)) \\ &= \Pi(s_1(\xi_1, ..., \xi_n), ..., s_n(\xi_1, ..., \xi_n)) \\ &= P(\xi_1, ..., \xi_n) . \end{align*}

This proves the existence of an element a satisfying (FS). The uniqueness of a follows from the Cor. to Prop. 5, because we have a⋅1=P(x1,f,...,xn,f)a \cdot 1 = P(x_{1,f}, ..., x_{n,f}) in the universal decomposition algebra EfE_f.

With the notation of Prop. 6 one sometimes writes a=P∗(f)a = P^*(f). Here are some examples.

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.