Bourbaki

Corollary

Tag 04I7, permanent. Cite it as [Bourbaki, Tag 04I7].

Commutative Algebra, chapter III, § 4. Lifting in complete rings

Let qq be a closed ideal of AA contained in mm. Then the relation x≡x′(modqxn)x \equiv x' \pmod{q^{x^n}} is equivalent to f(x)≡f(x′)(modqxn)f(x) \equiv f(x') \pmod{q^{x^n}}.

For every formal power series f∈A[[X1,…,Xn]]f \in A[[X_1, \ldots, X_n]],

f(X1,…,Xn)−f(Y1,…,Yn)=∑i=1n(Xi−Yi)hi(X1,…,Xn,Y1,…,Yn) f(X_1, \ldots, X_n) - f(Y_1, \ldots, Y_n) = \sum_{i=1}^n (X_i - Y_i) h_i(X_1, \ldots, X_n, Y_1, \ldots, Y_n)

where the hih_i belong to A[[X1,…,Xn,Y1,…,Yn]]A[[X_1, \ldots, X_n, Y_1, \ldots, Y_n]] (Algebra, Chapter IV, § 5, no. 8, Proposition 9); it follows immediately that the relation x≡x′(modqxn)x \equiv x' \pmod{q^{x^n}} implies f(x)≡f(x′)(modqxn)f(x) \equiv f(x') \pmod{q^{x^n}}. The converse is obtained by replacing ff by its "inverse" gg.

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.