Bourbaki

Example 1

Tag 0CO4, permanent. Cite it as [Bourbaki, Tag 0CO4].

Commutative Algebra, chapter VIII, § 4. Séries de Hilbert-Samuel

Suppose H0H_0 artinian and take M=HM = H. Then, with the preceding notations, we have R=0R = 0 and S=H′S = H', hence by (5), QH=QH′Q_H = Q_{H'}; since we have QH0=long⁡(H0)Q_{H_0} = \operatorname{long}(H_0), we infer by induction that QH=long⁡(H0)Q_H = \operatorname{long}(H_0), that is,

PH=long⁡(H0)⋅∏i∈I(1−Tdi)−1. P_H = \operatorname{long}(H_0) \cdot \prod_{i \in I} (1 - T^{d_i})^{-1} .

Now suppose H endowed with the usual grading, for which di=1d_i = 1 for all i∈Ii \in I, and put r=Card(I)r = \mathrm{Card}(I); we have PM=QM(T).(1−T)−rP_M = Q_M(T).(1 - T)^{-r}. Put cM=QM(1)c_M = Q_M(1). Then, by formula (2) of No. 1, we have:

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

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.