Bourbaki

Proposition 3

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

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

Let A be a filtered ring, (An) its filtration, B the subring ⋃n∈ZAn\bigcup_{n \in \mathbf{Z}} A_n of A, E a filtered B-module, (E_n) its filtration and F the sub-B-module ⋃n∈ZEn\bigcup_{n \in \mathbf{Z}} E_n of E. Then the mapping (a,x)↦ax(a, x) \mapsto ax from B×FB \times F to F is continuous.*

Let a,∈B,x0∈Fa, \in B, x_0 \in F; there exists by hypothesis integers r,sr, s such that a0∈Ara_0 \in A_r and x0∈Esx_0 \in E_s. The relation

ax−a0x0=(a−a0)x0+a0(x−x0)+(a−a0)(x−x0) ax - a_0x_0 = (a - a_0)x_0 + a_0(x - x_0) + (a - a_0)(x - x_0)

shows that if a−a0∈Aia - a_0 \in A_i and x−x0∈Ejx - x_0 \in E_j, ax−a0x0ax - a_0x_0 belongs to

Ei+s+Ej+r+Ei+j. E_{i+s} + E_{j+r} + E_{i+j}.

Then, given an integer nn, ax−a0x0∈Enax - a_0x_0 \in E_n provided i≥n−s,j≥n−ri \geq n - s, j \geq n - r and i+j≥ni + j \geq n, that is so long as ii and jj are sufficiently large.

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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Transcribed from the printed volume. Not translated.