Bourbaki

Lemma 5

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

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

Let A be a ring, M an A-module and (Pn),(Qn)(P_n), (Q_n) two decreasing filtrations on M consisting of submodules. Suppose that one has Pn⊃QnP_n \supset Q_n and long⁡A(Pn/Qn)<+∞\operatorname{long}_A(P_n/Q_n) < +\infty for every n∈Zn \in \mathbf{Z} and that there exists an integer n1n_1 such that Qn1=MQ_{n_1} = M. In Z((T))\mathbf{Z}((T)), one has the inequalities

∑n∈Zlong⁡A((Pn+1∩Qn)/Qn+1).Tn≤∑n∈Zlong⁡A(Pn+1/Qn+1).Tn≤ \sum_{n \in \mathbf{Z}} \operatorname{long}_A((P_{n+1} \cap Q_n)/Q_{n+1}).T^n \leq \sum_{n \in \mathbf{Z}} \operatorname{long}_A(P_{n+1}/Q_{n+1}).T^n \leq
≤(1−T)−1∑n∈Zlong⁡A((Pn+1∩Qn)/Qn+1).Tn. \leq (1-T)^{-1} \sum_{n \in \mathbf{Z}} \operatorname{long}_A((P_{n+1} \cap Q_n)/Q_{n+1}).T^n .

It is a matter of proving that one has the inequalities

(15) long⁡((Pn+1∩Qn)/Qn+1)≤long⁡(Pn+1/Qn+1)\operatorname{long}((P_{n+1} \cap Q_n)/Q_{n+1}) \leq \operatorname{long}(P_{n+1}/Q_{n+1}), (16) long⁡(Pn+1/Qn+1)≤∑i≤nlong⁡((Pi+1∩Qi)/Qi+1)\operatorname{long}(P_{n+1}/Q_{n+1}) \leq \sum_{i \leq n} \operatorname{long}((P_{i+1} \cap Q_i)/Q_{i+1}).

The first is obvious. On the other hand, one has Pn+1∩Qi=Pn+1P_{n+1} \cap Q_i = P_{n+1} for i≤n1i \leq n_1 and Pn+1∩Qn+1=Qn+1P_{n+1} \cap Q_{n+1} = Q_{n+1}; whence one deduces the inequality

long⁡(Pn+1/Qn+1)≤∑i≤nlong⁡((Pn+1∩Qi)/(Pn+1∩Qi+1)). \operatorname{long}(P_{n+1}/Q_{n+1}) \leq \sum_{i \leq n} \operatorname{long}((P_{n+1} \cap Q_i)/(P_{n+1} \cap Q_{i+1})) .

But the A-module (Pn+1∩Qi)/(Pn+1∩Qi+1)(P_{n+1} \cap Q_i)/(P_{n+1} \cap Q_{i+1}) is isomorphic to a submodule of (Pi+1∩Qi)/Qi+1(P_{i+1} \cap Q_i)/Q_{i+1}, and inequality (16) follows.

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

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Cites

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