Let A be a filtered commutative ring whose filtration we denote by (an)n∈Z; let A′ be the ring of formal power series A[[X1,…,Xs]]. For all e=(e1,…,es)∈Ns, we write ∣e∣=∑i=1sei, Xe=∏i=1sXiei so that every element P∈A′ can be written uniquely P=∑e∈Nsαe,PXe where αe,P∈A. For all n∈Z, let an′ denote the set of P∈A′ such that αe,P∈an−∣e∣ for all e∈Ns; we show that an′ is an ideal of A′. Clearly an′ is an additive subgroup of A′; on the other hand, if P∈an′ and Q∈A′, then, for all e∈Ns, αe,PQ=∑e′+e′′=eαe′,Qαe′′,P; as the relation e′+e′′=e implies ∣e′′∣≤∣e∣, PQ∈an′. Moreover, if Q∈am′, then, for e′+e′′=e, αe′,Qαe′′,P∈am−∣e′∣an−∣e′′∣⊂am+n−∣e∣, which proves that (an′)n∈Z is a filtration compatible with the ring structure on A′ (for obviously 1∈a0′). When in future we speak of A′ as a filtered ring, we shall mean, unless otherwise stated, with the filtration (ah). Clearly ⋂n∈Zan′ is the set of formal power series all of whose coefficients belong to ⋂n∈Zan; then, if A is Hausdorff, so is A′. If a0=A, then a0′=A′.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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