Suppose H0 artinian and take M=H. Then, with the preceding notations, we have R=0 and S=H′, hence by (5), QH=QH′; since we have QH0=long(H0), we infer by induction that QH=long(H0), that is,
PH=long(H0)⋅i∈I∏(1−Tdi)−1.
Now suppose H endowed with the usual grading, for which di=1 for all i∈I, and put r=Card(I); we have PM=QM(T).(1−T)−r. Put cM=QM(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.