Bourbaki

Lemma 1

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

Algebra, chapter II, § 7. Vector spaces

Let (a_i)_{i \in I} be a free family of elements of E; if b \in E does not belong to the subspace F generated by (a_i), the subset of E consisting of the a_i and b is free.

Suppose that there were a relation μb+∑iλiai=0\mu b + \sum_i \lambda_i a_i = 0 with μ∈K\mu \in K and λi∈K\lambda_i \in K for all i∈Ii \in I, the family (λi)(\lambda_i) having finite support; if μ≠0\mu \neq 0, it would follow that b=−∑i(μ−1λi)aib = -\sum_i (\mu^{-1} \lambda_i) a_i and hence b∈Fb \in F contrary to the hypothesis; hence μ=0\mu = 0 and the relation becomes ∑iλiai=0\sum_i \lambda_i a_i = 0, which implies λi=0\lambda_i = 0 for all i∈Ii \in I by hypothesis; whence the lemma.

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

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Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.