Bourbaki

Proposition 7

Tag 06C8, permanent. Cite it as [Bourbaki, Tag 06C8].

Algebra, chapter IV, § 4. Formal power series

— Let (yi)i∈I(y_i)_{i \in I}, be a family of elements of E. When I is infinite, we assume that yiy_i tends to 0 along the filter of complements of finite subsets of I. There exists then a unique continuous A-derivation D of A[[I]]A[[I]] into the A[[I]]A[[I]]*-module* E such that D(Xi)=yiD(X_i) = y_i for all i∈Ii \in I. We have

D(u)=∑i∈I(Diu)(x)⋅yi(u∈A[[I]]). D(u) = \sum_{i \in I} (D_i u)(x) \cdot y_i \quad (u \in A[[I]]) .

Since 0 admits in E a fundamental system of neighbourhoods consisting of ideals, the family ((Diu)(x)⋅yi)i∈I((D_i u)(x) \cdot y_i)_{i \in I} is summable in E for all u∈A[[I]]u \in A[[I]] (Gen. Top., III, p. 263, Cor. 2). Formula (16) thus defines an A-linear mapping D:A[[I]]→ED : A[[I]] \to E. We leave the reader to verify that D is a continuous derivation.

Let D1D_1 be a continuous A-derivation of A[[I]]A[[I]] into E, such that D1(Xi)=yiD_1(X_i) = y_i for all i∈Ii \in I. The kernel of the continuous derivation D−D1D - D_1 is a closed subalgebra B of A[[I]]\mathbf{A}[[\mathbf{I}]] containing 1 and the indeterminates XiX_i. Since the polynomial algebra A[(Xi)i∈I]\mathbf{A}[(X_i)_{i \in I}] is dense in A[[I]]\mathbf{A}[[\mathbf{I}]], we have B=A[[I]]B = \mathbf{A}[[\mathbf{I}]] and so D1=DD_1 = D.

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 edition, 2003, Springer. Not translated.