Bourbaki

Proposition 4

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

Algebra, chapter IV, § 6. Symmetric functions

*Let B be a commutative ring, ρ\rho a homomorphism of A into B and ξ1,...,5\xi_1, ..., 5, elements of B. Suppose that the relation ρf(X)=∏i=1n(X−ξi)^\rho f(X) = \prod_{i=1}^n (X - \xi_i) holds in B[X]. Then there exists one and only one ring homomorphism u:Ef→Bu : E_f \to B such that ρ(a)=u(a.1)\rho(a) = u(a . 1) for all a∈Aa \in A and u(xi)=ξiu(x_i) = \xi_i for 1≤i≤n1 \leq i \leq n.

We consider B as associative, commutative and unital A-algebra by means of ρ\rho. Then the relation ρf(X)=∏i=1n(X−ξi)^\rho f(X) = \prod_{i=1}^n (X - \xi_i) may also be written in the form

∑i1<...<ikξi1...ξik=(−1)kak.1(1≤k≤n) \sum_{i_1 < ... < i_k} \xi_{i_1} ... \xi_{i_k} = (-1)^k a_k . 1 \quad (1 \leq k \leq n)

in B. Since the relations (24) define a presentation of EfE_f, Prop. 4 follows.

Prop. 4 justifies the name « universal decomposition algebra of ff » for EfE_f. The relation f(X)=∏i=1n(X−xi,f)f(X) = \prod_{i=1}^n (X - x_{i,f}) is called the « universal decomposition of ff ». Let σ∈Sn\sigma \in S_n be a permutation ; since f(X)=∏i=1n(X−xσ(i),f)f(X) = \prod_{i=1}^n (X - x_{\sigma(i),f}), there exists an automorphism tσt_\sigma of the A-algebra EfE_f characterized by tσ(xi,f)=xσ(i),ft_\sigma(x_{i,f}) = x_{\sigma(i),f} for 1≤i≤n1 \leq i \leq n. We have tστ=tσ∘tτt_{\sigma \tau} = t_\sigma \circ t_\tau for σ,τ\sigma, \tau in SnS_n, and hence obtain an action of the group G\mathscr{G}, an the A-algebra EfE_f.

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

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Cites

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