MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the proof, the author consider the normalization $\tilde{A}$ of $A$ and show $\tilde{A}/t \tilde{A}$ is a integral domain. He showed that the localizations at points of Spec $A$ are domains, but we know a non-domain ring can have integral localizations. How should I understand the proof? Thanks a lot.

share|cite|improve this question
Thank for Schwede's edit. – MZWang Feb 5 '12 at 10:30
I don't understand your question: $A$ is supposed to be a local noetherian domain -- so he doesn't have to work to prove it is a domain! – Julien Puydt Aug 26 '12 at 14:47
Julien, the problem is to show $\tilde{A}/t\tilde{A}$ is a domain. – MZWang Aug 30 '12 at 3:32

Retain all notation of the lemma as stated in Hartshorne. Pose $S=A-\mathfrak p$. Since normalization commutes with localization, the integral closure of $A_\mathfrak p$ is $S^{-1}\tilde A$. But $A_\mathfrak p$ is a DVR so in fact $S^{-1}\tilde A=A_{\mathfrak p}$. Therefore $\tilde A$ has precisely one prime $\mathfrak p'$ lying over $\mathfrak p$, and $A_\mathfrak p=\tilde A_{\mathfrak p'}$. Note that as $\tilde A$ is normal it satisfies condition (S2) and therefore $\tilde A/t\tilde A$ has no embedded associated primes. To conclude that $\mathfrak p'$ is the only associated prime of $\tilde A/t\tilde A$, it would suffice to show that every minimal associated prime of $\tilde A/t\tilde A$ contracts to $\mathfrak p$, since we already know that only $\mathfrak p'$ contracts to $\mathfrak p$, and the set of associated primes of $\tilde A/t\tilde A$ is nonempty, as $\tilde A$ is finite over $A$, hence noetherian, since $A$ is japanese.

Let $\mathfrak r\in\operatorname{Spec}\tilde A$ be a minimal associated prime of $\tilde A/t\tilde A$. As $A$ is a localization of an affine ring, it is universally catenary. By incomparability (i.e. all fibers of $A\rightarrow\tilde A$ are of dimension zero), a fortiori $\mathfrak r$ is maximal among those primes meeting $A$ in $A\cap\mathfrak r$; since $A$ is japanese, we can apply Theorem 13.8 of Eisenbud (a variant of Nagata's altitude formula) to conclude $$\operatorname{ht} \mathfrak r=\operatorname{ht} \mathfrak r\cap A+\dim K(A)\otimes\tilde A,$$ but of course $K(A)\otimes\tilde A=K(A)$, so we conclude that $\operatorname{ht} \mathfrak r=\operatorname{ht}\mathfrak r\cap A$. But now of course $\operatorname{ht}\mathfrak r=1$ by Krull's hauptidealsatz, and as $t\in\mathfrak r\cap A$ and $\operatorname{ht}\mathfrak r\cap A=1$, we have that $\mathfrak r\cap A$ is minimal over $t$, again by the hauptidealsatz. But by hypothesis, the only minimal associated prime of $tA$ is $\mathfrak p$. Hence $\mathfrak r\cap A=\mathfrak p$.

We now have enough to infer that $t\tilde A$ is a $\mathfrak p'$-primary ideal of $\tilde A$. Let $S(t\tilde A)$ denote the saturation of the ideal $t\tilde A$ with respect to $\tilde A-\mathfrak p'$ (the contraction in $\tilde A$ of $t\tilde A_{\mathfrak p'}$). As $S\cap\mathfrak p'=\emptyset$, it follows from standard facts about the behavior of primary ideals under saturation (Atiyah-Macdonald Prop. 4.8) that $S(t\tilde A)=t\tilde A$. But $t\tilde A_{\mathfrak p'}=tA_{\mathfrak p}$, and $t$ generates the maximal ideal of $A_\mathfrak p$ by hypothesis; hence $t\tilde A=S(t\tilde A)$ is the contraction of a prime and so we finally see that $\tilde A/t\tilde A$ is a domain. $\blacksquare$

(This proof is adapted from the one in EGA IV 2 (5.12.7).)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.