Bourbaki

Proposition 19

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

Algebra, chapter IV, § 5. Symmetric tensors and polynomial mappings

— Let M be a free A-module, N an A-module, G an infinite additive subgroup of A, and suppose that the homotheties of N defined by the nonzero elements of G are injective. Then Pol(M, N) is the direct sum of the Polq^q(M, N).

Let f0,f1,…,fnf_0, f_1, \ldots, f_n be such that fi∈Poli(M,N)f_i \in \mathrm{Pol}^i(M, N) and suppose that we have the relation f0+…+fn=0f_0 + \ldots + f_n = 0. Let x∈Mx \in M, then for all λ∈G\lambda \in G we have

0=∑i=0nfi(λx)=∑i=0nλifi(x). 0 = \sum_{i=0}^n f_i(\lambda x) = \sum_{i=0}^n \lambda^i f_i(x).
By Lemma 2, applied to the polynomial ∑i=0nfi(x)Xi\sum_{i=0}^n f_i(x) X^i we have
f0(x)=⋯=fn(x)=0 f_0(x) = \cdots = f_n(x) = 0

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.