4
$\begingroup$

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.

  • 0
    @wxu Now can you say a line or two about how you got to this counterexample?2012-06-09

2 Answers 2

1

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 $(0,b)\in B$, $(0,b)^2-(b,b)(0,b)=0$.

  • 0
    I am sorry in the previous example because it is not an integral extension. Now I think I fix it.2012-06-09
4

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.

  • 0
    @wxu: Dear wxu, Yes, you are right; I forgot to add that every component of Spec $B$ should dominate Spec $A$. Regards,2012-06-09