Bourbaki

Remark

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

Algebra, chapter IV, § 1. Polynomials

— Let x∈AIx \in A^I and let ff be the mapping u↦u(x)u \mapsto u(x) of A[X]A[X] into AA. Given an AA-module MM, we consider the homomorphism 1⊗f1 \otimes f of M[X]=M⊗AA[X]M[X] = M \otimes_A A[X] into M⊗AA=MM \otimes_A A = M. For each v∈M[X]v \in M[X] we have (1⊗f)(v)=v(x)(1 \otimes f)(v) = v(x). If v=∑ν∈N(I)eνXνv = \sum_{\nu \in \mathbf{N}^{(I)}} e_\nu X^\nu, then v(x)=∑ν∈N(I)xνeνv(x) = \sum_{\nu \in \mathbf{N}^{(I)}} x^\nu e_\nu.

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.