Bourbaki

Example 2

Tag 04DY, permanent. Cite it as [Bourbaki, Tag 04DY].

Commutative Algebra, chapter III, § 2. General results on filtered rings and modules

Let AA be a filtered commutative ring whose filtration we denote by (an)n∈Z(a_n)_{n \in \mathbf{Z}}; let A′A' be the ring of formal power series A[[X1,…,Xs]]A[[X_1, \ldots, X_s]]. For all e=(e1,…,es)∈Nse = (e_1, \ldots, e_s) \in \mathbf{N}^s, we write ∣e∣=∑i=1sei|e| = \sum_{i=1}^s e_i, Xe=∏i=1sXieiX^e = \prod_{i=1}^s X_i^{e_i} so that every element P∈A′P \in A' can be written uniquely P=∑e∈Nsαe,PXeP = \sum_{e \in \mathbf{N}^s} \alpha_{e,P} X^e where αe,P∈A\alpha_{e,P} \in A. For all n∈Zn \in \mathbf{Z}, let an′a'_n denote the set of P∈A′P \in A' such that αe,P∈an−∣e∣\alpha_{e,P} \in a_{n-|e|} for all e∈Nse \in \mathbf{N}^s; we show that an′a'_n is an ideal of A′A'. Clearly an′a'_n is an additive subgroup of A′A'; on the other hand, if P∈an′P \in a'_n and Q∈A′Q \in A', then, for all e∈Nse \in \mathbf{N}^s, αe,PQ=∑e′+e′′=eαe′,Qαe′′,P\alpha_{e,PQ} = \sum_{e'+e''=e} \alpha_{e',Q}\alpha_{e'',P}; as the relation e′+e′′=ee' + e'' = e implies ∣e′′∣≤∣e∣|e''| \leq |e|, PQ∈an′PQ \in a'_n. Moreover, if Q∈am′Q \in a'_m, then, for e′+e′′=ee' + e'' = e, αe′,Qαe′′,P∈am−∣e′∣an−∣e′′∣⊂am+n−∣e∣\alpha_{e',Q}\alpha_{e'',P} \in a_{m-|e'|}a_{n-|e''|} \subset a_{m+n-|e|}, which proves that (an′)n∈Z(a'_n)_{n \in \mathbf{Z}} is a filtration compatible with the ring structure on A′A' (for obviously 1∈a0′1 \in a'_0). When in future we speak of A′A' as a filtered ring, we shall mean, unless otherwise stated, with the filtration (ah). Clearly ⋂n∈Zan′\bigcap_{n \in \mathbf{Z}} a'_n is the set of formal power series all of whose coefficients belong to ⋂n∈Zan\bigcap_{n \in \mathbf{Z}} a_n; then, if AA is Hausdorff, so is A′A'. If a0=Aa_0 = A, then a0′=A′a'_0 = A'.

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

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Transcribed from the printed volume. Not translated.