Let A be a ring, M an A-module and (Pn),(Qn) two decreasing filtrations on M consisting of submodules. Suppose that one has Pn⊃Qn and longA(Pn/Qn)<+∞ for every n∈Z and that there exists an integer n1 such that Qn1=M. In Z((T)), one has the inequalities
n∈Z∑longA((Pn+1∩Qn)/Qn+1).Tn≤n∈Z∑longA(Pn+1/Qn+1).Tn≤
≤(1−T)−1n∈Z∑longA((Pn+1∩Qn)/Qn+1).Tn.
It is a matter of proving that one has the inequalities
(15) long((Pn+1∩Qn)/Qn+1)≤long(Pn+1/Qn+1), (16) long(Pn+1/Qn+1)≤∑i≤nlong((Pi+1∩Qi)/Qi+1).
The first is obvious. On the other hand, one has Pn+1∩Qi=Pn+1 for i≤n1 and Pn+1∩Qn+1=Qn+1; whence one deduces the inequality
long(Pn+1/Qn+1)≤i≤n∑long((Pn+1∩Qi)/(Pn+1∩Qi+1)).
But the A-module (Pn+1∩Qi)/(Pn+1∩Qi+1) is isomorphic to a submodule of (Pi+1∩Qi)/Qi+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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