# $Spec(R)$ Noetherian and going up theorem

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?

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as stated it is not correct. You probably want some finiteness condition. – Soarer Jun 3 '11 at 4:30
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 math.stackexchange.com/questions/57270/…) – Andrea Aug 24 '11 at 21:45

$\mathbb{Z}\subset\mathbb{Z}[x]$
but $(x+n)\cap\mathbb{Z}=0$
for all natural number $n$.