I'm stuck on this exercise ( from Bosch ) :
Let $\phi :R \to R' $ a flat ring morphism. Show that $\phi$ is faithfully flat if and only if the associated map $Spec(R') \to Spec(R)$ , $\mathfrak{p}' \mapsto \mathfrak{p}' \cap R$ is surjective.
Definitions: An $R$-module $M$ is faithfully flat when $M$ is flat and $$M \otimes_R N = 0 \Rightarrow N = 0$$ $\phi :R \to R' $ flat ring morphism means that $R'$ is flat as $R$-module via $\phi$.
Similarly for $\phi$ faithfully flat.
How to proceed ?
Assume that $\mathrm{Spec}(R') \to \mathrm{Spec}(R)$ is surjective. It suffices to prove that $\mathfrak{p} R' \neq R'$ for any prime ideal $\mathfrak{p}$ of $R$. But there is some prime ideal $\mathfrak{p'}$ of $R'$ restricting to $\mathfrak{p}$. Hence, $\mathfrak{p} R'$ is contained in $\mathfrak{p'}$.
Now assume that $R \to R'$ is faithfully flat and let $\mathfrak{p}$ be a prime ideal of $R$. Then $R' \otimes_R Q(R/\mathfrak{p}) \neq 0$, hence this ring has a prime ideal. It restricts to a prime ideal $\mathfrak{p}'$ of $R'$ which restricts to $\mathfrak{p}$.