Bourbaki

Lemma 2

Tag 04D7, permanent. Cite it as [Bourbaki, Tag 04D7].

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] and let S=⨁i≥0SiS = \bigoplus_{i \geq 0} S_i, be a graded commutative AA-algebra with positive degrees, which is a finitely generated AA-module. Then there exists an integer n0≥0n_0 \geq 0 such that:

(i) For n≥n0n \geq n_0 and k≥0k \geq 0, Sn+k=Sk⋅SnS_{n+k} = S_k \cdot S_n. (ii) For d≥n0d \geq n_0, S(d)=S0[Sd]S^{(d)} = S_0[S_d].

By Lemma 1 there exists an integer n0≥0n_0 \geq 0 such that, for n≥n0n \geq n_0 and k≥0k \geq 0, Sn+k=AkSnS_{n+k} = A_k S_n, whence a fortiori Sn+k=SkSnS_{n+k} = S_k S_n, which establishes (i). Then, for d≥n0d \geq n_0 and m>0m > 0, Smd=(Sd)mS_{m d} = (S_d)^m as follows by induction on mm applying (i); this establishes (ii).

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

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