Bourbaki

Corollary 4

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

Algebra, chapter II, § 3. Tensor products

Let A be a ring with no divisor of zero, E a free right A-module and F a free left A-module. Then the relation x ⊗\otimes y = 0 in E ⊗A\otimes_A F implies x = 0 or y = 0.

Let (aλ)(a_\lambda) be a basis of E, (bμ)(b_\mu) a basis of F and let x=∑λaλξλ,y=∑μημbμx = \sum_\lambda a_\lambda \xi_\lambda, y = \sum_\mu \eta_\mu b_\mu; then x⊗y=∑λ,μ((aλξλημ)⊗bμx \otimes y = \sum_{\lambda, \mu} ((a_\lambda \xi_\lambda \eta_\mu) \otimes b_\mu and the relation x⊗y=0x \otimes y = 0 implies ξλημ=0\xi_\lambda \eta_\mu = 0 for every ordered pairs of indices (λ,μ)(\lambda, \mu) (Corollary 1). Hence, if x≠0x \neq 0, that is ξλ≠0\xi_\lambda \neq 0 for at least one λ\lambda, it follows that ημ=0\eta_\mu = 0 for all μ\mu, whence y=0y = 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.