I will appreciate any enlightenment on the following which must be an exercise in a certain textbook. (I don't recognize where it comes from.) I understand that the going down property does not hold since $R$ is not integrally closed (in fact, it is not a UFD), but I have no idea how to show that $q$ is such a counterexample.
Let $k$ be a field, $A = k[X, Y]$ be a polynomial ring, $R = \lbrace f \in A \colon f(0, 0) = f (1, 1) \rbrace \subset A$ be a subring. Define $q = (X)\cap R$, $p = (X - 1, Y - 1) \cap R$, $P = (X - 1, Y - 1)$. Show that there is no $Q \in \operatorname{Spec} A$, $Q\subset P$ that goes down to $q$.