This is an exercise from Rotman, introduction to homological algebra.
A right $R$-module $B$ is called faithfully flat if :
1) $B$ is flat
2) If $X$ is a left $R$-module and $B \otimes_R X =0 $ then $X = 0$
Prove that a right $R$-module $B$ is faithfully flat $\Leftrightarrow$ $B$ is flat and $B \otimes_R R/I \neq 0$ for every proper left ideal $I$.
My attempt:
The direction $\Longrightarrow$ is clear.
For the other direction, suppose $X \neq 0$ and $B \otimes_R X =0$. Then $$I = \text{Ann}(X)$$ is a proper ideal. Maybe it is useful, then ?
Hint: let $x\in X$, $x\ne 0$; the sequence $$ 0\to Rx\to X\to X/Rx\to 0 $$ is exact, so also $$ 0\to B\otimes_RRx\to B\otimes_RX\to B\otimes_R(X/Rx)\to 0 $$ is exact. Let $I=\operatorname{Ann}_R(x)$; then $Rx\cong R/I$.