Going down theorem with modification. Going-down Thm: Let $A\subseteq B$ be an integral extension. Assume that $B$ is an integral domain and that $A$ is integrally closed. Then going down holds for the above extension.
Question1: Can we remove the hypothesis that $B$ is a domain and replace it with $B$ is reduced.
Question2: if not, can someone give me a counterexample?
Thanks.
 A: I find it easier to think geometrically, so I'll argue that way. Suppose that $B$ is reduced and Noetherian, so that Spec $B$ is the union of finitely many irreducible components Spec $B_i$, each of which is reduced and irreducible, i.e. integral.   Suppose also that each component of Spec $B$ dominates Spec $A$.  (In ring-theoretic terms, suppose that each of the induced maps $A \to B_i$ is injective.)
Each of the maps $f_i:$ Spec $B_i \to$ Spec $A$ is also finite, because it is the composite of the closed immersion Spec $B_i \hookrightarrow $ Spec $B$ and the map $f:$ Spec $B \to $ Spec $A$, which is finite by assumption.  Thus going down holds for each of the maps Spec $B_i \to$ Spec $A$.  
Now any point $x$ of Spec $B$ belongs to one of the Spec $B_i$, and so given $y' \in $ Spec $A$ generalizing $y = f_i(x)$, going down gives $x' \in$ Spec $B_i$ generalizing $x$ such that $y' = f_i(x')$.  Now think of $y$ just as an element of Spec $B$, and recall that $f_i$ is nothing but the restriction of $f$ to Spec $B_i$.  We thus see that $f(y') = x'$, and so going down holds for the map $f$.
(You can easily convert this argument into pure commutative algebra: the point is that $B$ embeds into the product $\prod_i B_i$ of the domains $B_i$, and so any prime ideal of $B$ is pulled back from a prime ideal of one of the $B_i$.  so going down for the $B_i$ implies going down for $B$.  I leave the details to you.)

Note that it is crucial to assume that each Spec $B_i$ dominates Spec $A$, not just that Spec $B$ dominates Spec $A$.  Otherwise we could just take Spec $B$ to be the disjoint union of Spec $B_1$ dominating Spec $A$, and some Spec $B_2$ which does not dominate Spec $A$; since going down does not hold for the map Spec $B_2 \to $ Spec $A$, it won't hold for the disoint union.  This is what happens in wxu's counterexample: it is the disjoint union of a line and a plane mapping to a plane.  
A: I think it is not right. 
Let $A=k[x,y]$, $B=k[x,y]\times k[x]$. The map $A\to B$ is sending $f(x,y)$ to $(f(x,y),f(x,0))$. So $A\to B$ is injective and $B$ is integral over $A$. 
Let $\mathfrak{p}_1=(y)\subset A$ and $\mathfrak{q}_1=k[x,y]\times (0)\subset B$, then $\mathfrak{q}_1\cap A=\mathfrak{p}_1$. Let $\mathfrak{p}_2=(0)\subset A$. Then there is no prime $\mathfrak{q}_2\subset \mathfrak{q}_1$ of $B$ such that $\mathfrak{q}_2\cap A=\mathfrak{p}_2$.
EDIT: In this example, it is clear $B$ is integral over $A$. For $(a,0)\in B$, $(a,0)^2-(a,a)(a,0)=0$. For $(0,b)\in B$, $(0,b)^2-(b,b)(0,b)=0$.
