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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 $q$.

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up vote 8 down vote accepted

We show that there is not a prime $Q\subset P$ such that $Q\cap R=q$.

Let us compute $p,q$. The prime $q=(X)\cap R=\{Xh\mid h(1,1)=0,h(X,Y)\in k[X,Y]\}$, the prime $p=P\cap R=\{g(X,Y)\mid g(0,0)=g(1,1)=0, g\in k[X,Y]\}$. Now it is clear $q\subset p$ and not equal $p$, since $X-Y\in p\setminus q$.

We are ready to show our statement. Suppose there is a prime ideal $Q\subset P$ such that $Q\cap R=q$, then $Q\neq P,0$. Thus $Q$ must be a principal ideal $(f)$ with an irreducible polynomial $f$ such that $f(1,1)=0$. But in this case, $(f)\cap R=\{fh\mid f(0,0)h(0,0)=0,h\in k[X,Y]\}$. We may find an irreducible polynomial $g\in k[X,Y]$ such that $g(1,1)=0$ but $(g)\neq (f)$, then $Xg(X,Y)\in q$ but $Xg(X,Y)\notin (f)$. We are done.

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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 .

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.

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I just want to say $q$ is not contained in $p$ in the setting of the op. Since $x(y-1)\in q$ but $x(y-1)\notin p=(x-1,y)\cap R$.. :) – wxu May 27 '12 at 12:02
+1, Nice answer! – wxu May 27 '12 at 12:08
Dear wxu, you are absolutely right! Obviously you were too polite to mention that in your own answer. The least I can do to thank you for this observation is upvote you: done! – Georges Elencwajg May 27 '12 at 12:08
However, I think your proof is more direct, and more succinct, only for one word "Going downing" may be a little not appropriate. :) – wxu May 27 '12 at 12:12
Dear wxu, exactly for that reason I was modifying my answer while you posted your comment and in the edit I urged users to read your post, which does give a counterexample to Going Down. By the way, an interesting consequence of your result is that $A$ is not flat over $R$, since going down holds for flat algebras: Matsumura, Commutative ring theory, Theorem 9.5. – Georges Elencwajg May 27 '12 at 12:27

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