Bourbaki

Proposition 5

Tag 04I3, permanent. Cite it as [Bourbaki, Tag 04I3].

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

Let A\mathbf{A} be a commutative ring and f=(f1,…,fn)\mathbf{f} = (f_1, \ldots, f_n) a system without constant term of nn series in A[[X1,…,Xn]]\mathbf{A}[[X_1, \ldots, X_n]]. Suppose that Jf(0)J_f(0) is invertible in A\mathbf{A}. Then there exists a system without constant term g=(g1,…,gn)\mathbf{g} = (g_1, \ldots, g_n) of nn series in A[[X1,…,Xn]]\mathbf{A}[[X_1, \ldots, X_n]] such that

(16)(16)
f∘g=1n.f \circ g = \mathbf{1}_n.

This system is unique and

(17) g∘f=11g \circ f = 1_1.

The existence and uniqueness of gg follows from Algebra, Chapter IV, § 5, no. 9, Proposition 10, applied to the nn formal power series

fi(Y1,…,Yn)−Xi(1≤i≤n). f_i(Y_1, \ldots, Y_n) - X_i \quad (1 \leq i \leq n).

It follows from (15) and (16) that Jf(0)Jg(0)=1J_f(0)J_g(0) = 1 and hence Jg(0)J_g(0) is also invertible. We conclude that there exists a system h=(h1,…,hn)h = (h_1, \ldots, h_n) of nn series without constant term in A[[X1,…,Xn]]A[[X_1, \ldots, X_n]] such that g∘h=11g \circ h = 1_1; from this relation and (16) it then follows, with the aid of (10), that

h=1n∘h=(f∘g)∘h=f∘(g∘h)=f∘1n=f. h = 1_n \circ h = (f \circ g) \circ h = f \circ (g \circ h) = f \circ 1_n = f.

Proposition 5 and formulae (10) and (15) show that under the law of composition (f,g)↦f∘g(f, g) \mapsto f \circ g the set of systems f=(f1,…,fn)f = (f_1, \ldots, f_n) of nn series without constant term in A[[X1,…,Xn]]A[[X_1, \ldots, X_n]] for which Jf(0)J_f(0) is invertible in AA, is a group.

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

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