Necessity of an assumption in the Going-down Theorem In the Going-down Theorem (Theorem 5.16 p. 64) in Introduction to Commutative Algebra by Atiyah and MacDonald, we have an integral inclusion $A\subset B$ of domains, $A$ being integrally closed. The assumption that $B$ is a domain is used in the proof, and I suspect that this assumption is necessary, but I haven't been able to prove it.
Here is a way of stating the question:
Let $A\subset B$ be an integral inclusion of (commutative) rings, $A$ being an integrally closed domain. Let $\mathfrak m$ be a maximal ideal of $B$, let $\mathfrak p$ be the contraction of $\mathfrak m$ in $A$, and let $\mathfrak p'$ be a prime ideal of $A$ contained in $\mathfrak p$:
$$
\begin{matrix}
B&\supset&\mathfrak m\\ 
|&&|\\ 
A&\supset&\mathfrak p&\supset&\mathfrak p'
\end{matrix}
$$
Is $\mathfrak p'$ the contraction of a prime contained in $\mathfrak m$?
My suspicion is that the answer is "not necessarily".
Such an answer would show the necessity of the assumption that $B$ is a domain.
[The symbol $\subset$ is used as in Bourbaki, i.e. $X\subset X$ for all set $X$.]
 A: Here's a counterexample that shows you can't completely drop the assumption of $B$ being an integral domain.
Let $k$ be a field, and consider the commutative rings $\tilde{A}=k[x_1,x_2]$ and $B=k[x_1]\times k[x_1,x_2]$, where $A\subseteq B$ is the image of $\tilde{A}$ under the injection $f(x_1,x_2)\mapsto(f(x_1,0),f(x_1,x_2))$. 
First note that $B$ is not an integral domain since $(x_1,0)\cdot(0,x_1)=0.$ Now consider an arbitrary element $(f(x_1),0)\in B$ and note that
$$ (f(x_1),0)^2-(f(x_1),f(x_1))\cdot(f(x_1),0)=0,\qquad(f(x_1),f(x_1))\in A. $$
So $(f(x_1),0)$ is integral over $A$. Similarly, for $(0,f(x_1,x_2))\in B$ we see that 
$$ (0,f(x_1,x_2))^2 - (f(x_1,0),f(x_1,x_2))\cdot(0,f(x_1,x_2))=0,\qquad(f(x_1,0),f(x_1,x_2))\in A. $$
So $(0,f(x_1,x_2))$ is integral over $A$. Since the set of elements which are integral over $A$ form a subring of $B$, and since elements of these two forms generate $B$, we see that $B$ is integral over $B$.
Finally, we show that the going-down theorem does not hold in this case. Consider the prime ideal $\mathfrak{p}_1=((0,x_2))\subseteq A$. Note that $\mathfrak{q}_1=(0)\times k[x_1,x_2]$ is a prime ideal in $B$ such that $\mathfrak{q}_1\cap A=\mathfrak{p}_1$. Let $\mathfrak{p}_2=((0,0))\subseteq A$. We claim there is no prime ideal $\mathfrak{q}_2\subseteq\mathfrak{q}_1$ such that $\mathfrak{q}_2\cap A=\mathfrak{p}_2$. This is clear since the prime ideals of $B$ are of the form $\mathfrak{p}\times k[x_1,x_2]$ and $k[x_1]\times\mathfrak{q}$ where $\mathfrak{p}$ is a prime ideal of $k[x_1]$ and $\mathfrak{q}$ is a prime ideal of $k[x_1,x_2]$.
