Bourbaki

Proposition 4

Tag 069V, permanent. Cite it as [Bourbaki, Tag 069V].

Algebra, chapter IV, § 1. Polynomials

Let E be an associative, commutative and unital A-algebra, x=(xi)i∈Ix = (x_i)_{i \in I} a family of elements of E, u an element of A[(Xi)i∈I]A[(X_i)_{i \in I}] and y=u(x)y = u(x). Then for every derivation D of E into an E-module we have

Dy=∑i∈I(Diu)(x)⋅Dxi. Dy = \sum_{i \in I} (D_i u)(x) \cdot Dx_i .

It suffices to prove the proposition when uu is a monomial, and in that case it follows from III, p. 558, Prop. 6.

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

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Transcribed from the printed edition, 2003, Springer. Not translated.