Bourbaki

Remark

Tag 05LJ, permanent. Cite it as [Bourbaki, Tag 05LJ].

Algebra, chapter II, § 1. Modules

(1) By Definition 10, if A≠{0}A \neq \{0\} and (al)l∈I(a_l)_{l \in I} is a free family, no element aκa_\kappa can be equal to a linear combination of the ala_l of index l≠κl \neq \kappa. But conversely, a family (al)(a_l) satisfying this condition is not necessarily a free family. For example, let A be an integral domain and a,ba, b two distinct non-zero elements; in A, considered as an A-module, aa and bb form a related system, since (−b)a+ab=0(-b)a + ab = 0. But in general there does not exist an element x∈Ax \in A such that a=xba = xb of b=xab = xa (cf. however § 7, no. 1, Remark).

An element xx of a module E is called free if {x}\{x\} is a free subset, that is if the relation αx=0\alpha x = 0 implies α=0\alpha = 0. Every element of a free subset is free and in particular 0 cannot belong to any free subset when A≠{0}A \neq \{0\}.

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, 1998, Springer. Not translated.