1
$\begingroup$

Let $S \subseteq R$ be commutative rings with 1 and suppose $Spec(R)$ is a Noetherian topological space. How do we show that the number of each $T \in Spec(R)$ lying over $P \in Spec(S)$ is finite?

I guess the idea is to use the going up theorem. We let $T \in Spec(R)$ such that $T$ lies over $P$. I don't see how to relate $T$ with a closed set of the form $V(J)$ for some ideal $J$ and then how to produce a descending chain of closed subsets of $Spec(R)$ so that we can use the Noetherian condition to guarantee this chain stabilize?

  • 0
    In fact, if $S \subseteq R$ is a finite extension of rings (suppose that $R$ is can be generated by a set of $n$ elements as an $S$-module), then the number of primes of $R$ lying over a prime of $S$ is not greater than $n$. (See my answer http://math.stackexchange.com/questions/57270/is-the-number-of-prime-ideals-of-a-zero-dimensional-ring-stable-under-base-change/57277#57277)2011-08-24

1 Answers 1

3

It seems not correct.

$\mathbb{Z}\subset\mathbb{Z}[x]$

but $(x+n)\cap\mathbb{Z}=0$

for all natural number $n$.