Ring homomorphism between commutative rings whose kernel has only nilpotent elements

Question

Let $f: A\to B$ be a ring homomorphism of commutative rings such that $\ker f$ is contained in the nilradical. Hence the induced continuous map $f^* : \operatorname{Spec} B \to \operatorname{Spec} A$ has dense image in $ \operatorname{Spec} A$. My question is : how to show that for every minimal prime ideal $P$ in $A$, there is a prime ideal $Q$ in $B$ such that $f^{-1}(Q)=P$ ?

Answer 1

Hint:

If you localise at a minimal prime ideal $\mathfrak p$, the local ring $A_{\mathfrak p}$ has a single prime ideal, and the ring $B_{\mathfrak p}$ will necessarily have all its prime ideals above $\,\mathfrak pA_{\mathfrak p}$, unless $\,B_{\mathfrak p}=\{0\}$.

Can you prove this is not the case?