TITLE: Prime spectrum and going-down property
QUESTION [5 upvotes]: I want to show that
$f$ has the going-down property $\Leftrightarrow$ For any prime ideal $\mathfrak{q}$ of $B$, if $\mathfrak{p}=\mathfrak{q}^c$, then $f^{*}:\textrm{Spec}(B_{\mathfrak{q}}) \rightarrow \textrm{Spec}(A_{\mathfrak{p}})$ is surjective.
I have proved ($\Leftarrow$), but there's something wrong in ($\Rightarrow$).
pf of ($\Rightarrow$): First, I understood $ \textrm{Spec}(A_{\mathfrak{p}}) = \{\mathfrak{p}' \in \textrm{Spec}(A) | \mathfrak{p}' \subset \mathfrak{p} \}. $ Let $\mathfrak{p}' \subset \mathfrak{p}$. Then $f(\mathfrak{p}') \subset f(\mathfrak{p})$ are prime ideals in $f(A)$. From $f(\mathfrak{p})=f(f^{-1}(\mathfrak{q}))=\mathfrak{q} \cap f(A)$, since f has the going-down property, there exists $\mathfrak{q}' \subset \mathfrak{q}$ such that $\mathfrak{q}' \cap f(A) = f(\mathfrak{p}')$. Now $f^{*}(\mathfrak{q}')=f^{-1}(\mathfrak{q}')=f^{-1}(\mathfrak{q}' \cap f(A))=f^{-1}(f(\mathfrak{p}'))$
If $\mathfrak{p}' \supset \ker f$, then $f^{-1}(f(\mathfrak{p}'))=\mathfrak{p}'$, so the proof is done. But isn't it possible that $\mathfrak{p}'$ does not contain $\ker f$? 
But $f^{*}$ to be surjective, $\textrm{Spec}(A_{\mathfrak{p}})$ must consists of contracted ideal. Is the problem wrong?

REPLY [2 votes]: We actually don't need the argument with the kernel, since the contraction of any prime in $f(A)$ is a prime in $A$. (Then just use Gobi's argument). With a bit more detail:
By going-down property, we want to show that whenever we have $p'\subseteq p$ in Spec $f(A)$ and a $q\in $Spec $B$ such that $q\cap f(A)=p$, we can find some $q'\subseteq q$ such that $p'=q'\cap f(A)$. But $a'=f^{-1}(p')$ is some prime ideal in $A$, and is clearly contained in $a=f^{-1}(p)$, another prime in $A$. So from the surjectivity condition, since $a$ is the contraction of $q$, there is some prime in Spec$(A_p)$, that is, some $q'\subseteq q$ such that $f^*(q')=a'$. It is clear then that $q'\cap f(A)=p'$.