Proposition 3

Let A, B be two commutative rings, $\rho : A \to B$ a ring homomorphism and E, E' two A-modules. There exists one and only one B-homomorphism

mapping $(1 \otimes x) \otimes (1 \otimes x')$ to $1 \otimes (x \otimes x')$ for $x \in E, x' \in E'$, and this homomorphism is bijective.

The left hand side of (5) may be written $(B \otimes_A E) \otimes_B (B \otimes_A E')$ and is identified with $(E \otimes_A B) \otimes_B (B \otimes_A E')$ since A and B are commutative; the latter product is identified successively with $E \otimes_A (B \otimes_B B) \otimes_A E'$, $E \otimes_A (B \otimes_A E')$, $E \otimes_A (E' \otimes_A B)$ and finally $(E \otimes_A E') \otimes_A B$, using the associativity of the tensor product (§ 3, no. 8, Proposition 8), Proposition 4, § 3, no. 4, and the commutativity of A and B. The desired isomorphism is the composition of the successive canonical isomorphisms.

Clearly if S is a generating system of E, the image of S under the canonical mapping $E \to E_{(B)}$ is a generating system of $E_{(B)}$; in particular, if E is a finitely generated A-module, $E_{(B)}$ is a finitely generated B-module.