Bourbaki

Proposition 5

Tag 0CRG, permanent. Cite it as [Bourbaki, Tag 0CRG].

Commutative Algebra, chapter VIII, § 6. Dimension des anneaux gradués

Suppose H0H_0 local and artinian, HH generated by H0∪H1H_0 \cup H_1, H1H_1 a finitely generated H0H_0-module, MM nonzero and a finitely generated HH-module. Then MnM_n is an H0H_0-module of finite length for each nn, and there exists Q(T)∈Z[T,T−1]Q(T) \in \mathbf{Z}[T, T^{-1}] such that Q(1)>0Q(1) > 0 and that one has in the ring Z((T))\mathbf{Z}((T))

∑n∈Zlong⁡H0(Mn)⋅Tn=(1−T)−d⋅Q(T), \sum_{n \in \mathbf{Z}} \operatorname{long}_{H_0}(M_n) \cdot T^n = (1 - T)^{-d} \cdot Q(T),

where d=dim⁡H(M)d = \dim_H(M).

The ring S−1HS^{-1}H is local and noetherian (No. 1, remark 1), the S−1HS^{-1}H-module S−1MS^{-1}M is nonzero and finitely generated, and of dimension d=dim⁡H(M)d = \dim_H(M) (prop. 4). Moreover, S−1H≥1S^{-1}H_{\geq 1} is an ideal of definition of S−1HS^{-1}H (§ 3, No. 2, lemma 2), and S−1M≥nS^{-1}M_{\geq n} is an S−1H≥1S^{-1}H_{\geq 1}-good filtration on S−1MS^{-1}M (lemma 3). Finally, one has long⁡S−1HS−1M≥n/S−1M≥n+1=long⁡H0(Mn)\operatorname{long}_{S^{-1}H} S^{-1}M_{\geq n}/S^{-1}M_{\geq n+1} = \operatorname{long}_{H_0}(M_n) for all nn. It is enough therefore to apply Theorems 2 and 3 of § 4 (Nos. 3 and 4).

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

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