Prime ideals and ring extensions Let $R\subset S$ be a finite extension and $P\in \text{Spec}(R)$. I'm trying to prove that$$\left \{Q\in\text{Spec}(S)\mid Q\cap R=P\right \}$$ is a finite set. Is this also true for any integral extension $R\subset S$?
I tried the following: let $P$ be a prime ideal of $R$, and consider the localization at $P$, $R_P$. Then $R_P$ is a local ring, with maximal ideal $PR_P$ (here is where I'm not sure if what I'm doing is correct or not), then the number of primes lying over $P$ and primes lying over $PR_P$ should be the same, and since $PR_P$ is maximal, then the number of prime lying over it must be finite, and hence the result. So is it true that  then the number of primes lying over $P$ and primes lying over $PR_P$ should be the same, and how to show this, and how can I show that primes lying over $PR_P$ are finite.For the second part of the question, I don't know if there is a counter example. I'm I on the correct path here?  Thank you for your help.
 A: If $R \to S$ is an arbitrary homomorphism and $P$ is a prime ideal of $R$, then there is a bijection between the prime ideals of $S$ lying over $P$ and the prime ideals of $S \otimes_R \mathrm{Quot}(R/P)$ (for the proof, just use the isomorphism to $(R \setminus P)^{-1} (S/PS)$ and the well-known descriptions of prime ideals in localizations and quotients, or look it up in any book on commutative algebra).
If $R \to S$ is finite, then also $\mathrm{Quot}(R/P) \to S \otimes_R \mathrm{Quot}(R/P)$ is finite. Thereby we reduce to the case of fields and have to prove: If $R$ is a field and $R \to S$ is a finite extension (automatically injective when wlog $S \neq 0$), then $S$ has only finitely many prime ideals. Well, actually $S$ is artinian, since it is a finite-dimensional vector space over $R$. Artinian rings have only finitely many prime ideals (Atiyah-Macdonald, 8.1 + 8.3).
For integral extension it fails. For example, the integral closure of $\mathbb{Z}$ in $\mathbb{C}$ has lots of prime ideals lying over $0$.
