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 ⊗ y = 0 in E ⊗A F implies x = 0 or y = 0.
Let (aλ) be a basis of E, (bμ) a basis of F and let x=∑λaλξλ,y=∑μημbμ; then x⊗y=∑λ,μ((aλξλημ)⊗bμ and the relation x⊗y=0 implies ξλημ=0 for every ordered pairs of indices (λ,μ) (Corollary 1). Hence, if x=0, that is ξλ=0 for at least one λ, it follows that ημ=0 for all μ, whence y=0.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.