TITLE: A counterexample to the going down theorem
QUESTION [9 upvotes]: 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 $q$.

REPLY [3 votes]: I'll show that the existence of   $Q\in Spec(A)$  satisfying  $Q\subsetneq P$ and $q= Q\cap R$ leads to a contradiction.    
We have  $X\cdot (X-1)\in q $ , so   $X\cdot (X-1)\in Q$.
Hence we  have $(X-1)\in Q$  (since  $X\notin Q$ because   $X\notin P$).
But this  forces $Q=(X-1)A$, since $(X-1)A\subset Q\subsetneq P=(X-1,Y)$.
But then $(X-1)Y\in Q\cap R \setminus q$  : contradiction . 
Edit
As wxu remarks in his comment, $q$ as defined by eltonjohn is not included in $p$. 
The above answer remains correct (I am sure of that, because else wxu would have noticed!),  but it is not a counterexample to Going Down.
I advise users to read  wxu's post: he modified eltonjohn's question precisely in order to give such a counterexample.