2

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.

i.a.m
  • 2,376

1 Answers1

2

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 $0$.

Martin Brandenburg
  • 163,620
  • Thank you for your answer, I have a question, since $S$ is a finite extention can I write $S=R[x_1,...,x_n]$? can I use the following proof? https://docs.google.com/viewer?a=v&q=cache:sfivbscEByEJ:spot.colorado.edu/~kearnes/F09/HW/ca7p2.pdf+&hl=en&gl=ca&pid=bl&srcid=ADGEESgM9rT6YzLpeUoptz3_NqFeCHSLKt9rJJxbf5sq4yU9Toe9PwUpbvOAmwI6bMztpvasrRR5Y8rckVtf5Dibj1lv-SHprI183Vsi-E22E24ABctqBbu2z1N0ZPg6G9K5cWf02idG&sig=AHIEtbSStrj9PxT8CIlZgOfv2AHqkC86AQ , and how can I show that the integral closure of $\Bbb Z$ in $\Bbb C$ has infinite primes over $0$. Thank you. – i.a.m Mar 27 '13 at 03:42
  • @iam: This is also a proof, yes. They use integral homomorphims of finite type - but they coincide with finite homomorphisms. – Martin Brandenburg Mar 27 '13 at 10:24