Bourbaki

Proposition 1

Tag 0CTV, permanent. Cite it as [Bourbaki, Tag 0CTV].

Commutative Algebra, chapter IX, § 1. Vecteurs de Witt

Let A be a filtered ring and (Jn)n∈Z(\mathbf{J}_n)_{n \in \mathbf{Z}} its filtration. Assume that one has J0=A\mathbf{J}_0 = \mathbf{A} and p.1A∈J1p.1_A \in \mathbf{J}_1. Let m and n be integers such that m⩾1m \geqslant 1 and n⩾0n \geqslant 0, and let a0,...,an,b0,...,bna_0, ..., a_n, b_0, ..., b_n be elements of A.

a) If one has ai≡bimod  Jma_i \equiv b_i \mod \mathbf{J}_m for 0⩽i⩽n0 \leqslant i \leqslant n, then one has

Φi(a0,...,ai)≡Φi(b0,...,bi)mod  Jm+ifor 0⩽i⩽n. \Phi_i(a_0, ..., a_i) \equiv \Phi_i(b_0, ..., b_i) \mod \mathbf{J}_{m+i} \quad \text{for } 0 \leqslant i \leqslant n .

b) Suppose that, for every integer k⩾1k \geqslant 1, and every x∈Ax \in \mathbf{A}, the relation p.x∈Jk+1p.x \in \mathbf{J}_{k+1} entails x∈Jkx \in \mathbf{J}_k. If one has Φi(a0,...,ai)≡Φi(b0,...,bi)mod  Jm+i\Phi_i(a_0, ..., a_i) \equiv \Phi_i(b_0, ..., b_i) \mod \mathbf{J}_{m+i} for 0⩽i⩽n0 \leqslant i \leqslant n, then one has ai≡bimod  Jma_i \equiv b_i \mod \mathbf{J}_m for 0⩽i⩽n0 \leqslant i \leqslant n.

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

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