Proposition 6

Let $A$ be a commutative ring, $B$ an algebra over $A$ and $\rho : a \mapsto a . 1$ the canonical homomorphism of $A$ to $B$. Suppose that $B$ is a faithfully flat $A$-module. Then, for an $A$-module $E$ to be flat (resp. faithfully flat), it is necessary and sufficient that the right $B$-module $E_{(B)} = E \otimes_A B$ be flat (resp. faithfully flat).

(1) If $E$ is flat (resp. faithfully flat), $E_{(B)}$ is flat (resp. faithfully flat) by § 2, no. 7, Corollary 2 to Proposition 8 (resp. by Proposition 5).

(2) Suppose that $E_{(B)}$ is flat and let $\nu : N' \to N$ be an injective $A$-module homomorphism. By § 2, no. 7, Corollary 3, the $A$-module $E \otimes_A B$ is flat, hence the homomorphism $1, \otimes 1_B \otimes \nu : E \otimes_A B \otimes_A N' \to E \otimes_A B \otimes_A N$ is injective. As the right and left $A$-module structures on $B$ coincide, this homomorphism is identified with

As $B$ is a faithfully flat $A$-module, it follows that $1_E \otimes \nu : E \otimes_A N' \to E \otimes_A N$ is injective (no. 1, Proposition 2), which shows that $E$ is flat.

(3) Suppose finally that $E_{(B)}$ is faithfully flat. First of all $E$ is flat by (2). Also let $N$ be an $A$-module such that $E \otimes_A N = 0$. Then $E \otimes_A N \otimes_A B = 0$, whence, since the right and left $A$-module structures on $B$ coincide, $E \otimes_A B \otimes_A N = 0$, which may also be written $(E \otimes_A B) \otimes_B (B \otimes_A N) = 0$. As $E_{(B)}$ is a faithfully flat $B$-module, this implies that $B \otimes_A N = 0$ (no. 1, Proposition 1), whence $N = 0$ since $B$ is a faithfully flat $A$-module (no. 1, Proposition 1).