Bourbaki

Lemma 1

Tag 04D6, permanent. Cite it as [Bourbaki, Tag 04D6].

Commutative Algebra, chapter III, § 1. Finitely generated graded algebras

Let AA be a graded commutative ring such that A=A0[A1]A = A_0[A_1], MM a graded AA-module and (yλ)λ∈L(y_\lambda)_{\lambda \in L} a system of homogeneous generators of MM such that deg⁡(yλ)⩽n0\deg(y_\lambda) \leqslant n_0 for all λ∈L\lambda \in L. Then, for all n⩾n0n \geqslant n_0 and all k⩾0k \geqslant 0, Mn+k=Ak⋅MnM_{n+k} = A_k \cdot M_n.

Let n⩾n0n \geqslant n_0 and x∈Mnx \in M_n. Since the yλy_\lambda generate MM, there exists a family (aλ)λ∈L(a_\lambda)_{\lambda \in L} of elements of AA of finite support such that x=∑λaλyλx = \sum_\lambda a_\lambda y_\lambda; we may further suppose that each aλa_\lambda is homogeneous and of degree n+1−deg⁡(yλ)n + 1 - \deg(y_\lambda) (replacing it if need be by its homogeneous component of that degree). As

A=A0[A1]A = A_0[A_1] and deg⁡(aλ)>0\deg(a_\lambda) > 0, each aλa_\lambda is a sum of elements of the form bb′bb' where b∈A0,b′∈Ab \in A_0, b' \in A, whence x∈A1Mnx \in A_1 M_n. Then Mn+1=A1MnM_{n+1} = A_1 M_n, whence Mn+k=AkMnM_{n+k} = A_k M_n by induction on kk.

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 volume. Not translated.