I know faithful and flat module is not always faithfully flat. For example $\mathbb Q \otimes_{\mathbb Z} \mathbb Z/2\mathbb Z=0$ implies $\mathbb Q$ is not faithfully flat as $\mathbb Z$-module unless it is faithful and flat as $\mathbb Z$-module.
But is the converse true?
I want to know the proof of
faithfully flat module is faithful and flat（flatness is obvious）
Consider the annihilator $I$ of a flat module $M$ and the commutative diagram with exact rows $$\require{AMScd} \begin{CD} 0 @>>> I\otimes_R M @>>> R\otimes_R M @>>> R/I\otimes_R M @>>> 0 \\ @. @VfVV @VgVV @VhVV \\ 0 @>>> IM @>>> M @>>> M/IM @>>> 0 \end{CD} $$ The obvious morphisms $g$ and $h$ are isomorphisms even if $M$ is not flat; if $M$ is flat, also $f$ is an isomorphism. Since $IM=0$ by definition, we have $I\otimes_RM=0$ and, if $M$ is also faithfully flat, this implies $I=0$.