Bourbaki

Proposition 20

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

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

— If the A-module M is free and finitely generated, then 00 is an isomorphism of graded algebras.

Let (ei)i∈I(e_i)_{i \in I} be a basis of MM and (ei∗)i∈I(e_i^*)_{i \in I} the dual basis of M∗M^*. For ν∈NI\nu \in \mathbf{N}^I put

eν=∏i∈Iγνi(ei)∈TS(M) e_\nu = \prod_{i \in I} \gamma_{\nu_i}(e_i) \in \mathbf{TS}(M)

By Prop. 4 (IV, p. 47) the family (eν)ν,NI(e_\nu)_\nu,_{\mathbf{N}^I} is a basis of TS(M)TS(M); let (eν∗)(e_\nu^*) be the basis of TS(M)∗gr\mathbf{TS}(M)^{*gr} dual to (eν)(e_\nu). In the light of III, p. 505, Th. 1 it is enough to show that for any ν∈NI\nu \in \mathbf{N}^I we have

eν∗=∏i∈I(ei∗)νi, e_\nu^* = \prod_{i \in I} (e_i^*)^{\nu_i},

or also that for ρ,σ∈NI\rho, \sigma \in \mathbf{N}^I we have eρ∗⋅eσ∗=eρ+σ∗e_\rho^* \cdot e_\sigma^* = e_{\rho + \sigma}^*; but this last assertion follows from IV, p. 50, Formula (12).

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.