Proposition 4

Let $R, S$ be two rings, $E$ a right $R$-module and $F$ an $(R, S)$-bimodule. Suppose that $E$ is faithfully flat. Then, for $F$ to be a flat (resp. faithfully flat) $S$-module, it is necessary and sufficient that $E \otimes_R F$ be so.

(1) If $F$ is flat, $E \otimes_R F$ is flat ($§ 2$, no. 7, Proposition 8).

(2) Suppose that $E \otimes_R F$ is flat and let $\nu : N' \to N$ be an injective left $S$-module homomorphism. The homomorphism

is then injective ($§ 2$, no. 3, Proposition 1). It follows from Proposition 2 of no. 1 that $l_F \otimes \nu : F \otimes_S N' \to F \otimes_S N$ is injective; then $F$ is a flat $S$-module ($§ 2$, no. 3, Proposition 1).

(3) Suppose that $F$ is faithfully flat and let $N$ be a left $S$-module such that $E \otimes_R F \otimes_S N = 0$. Since $E$ is faithfully flat, this implies that $F \otimes_S N = 0$, whence $N = 0$ since $F$ is faithfully flat; this proves that $E \otimes_R F$ is faithfully flat.

(4) Suppose that $E \otimes_R F$ is faithfully flat and let $N$ be a left $S$-module such that $F \otimes_S N = 0$. Then $E \otimes_R F \otimes_S N = 0$, whence $N = 0$, which shows that $F$ is faithfully flat.