$\newcommand{\p}{\mathfrak{p}}$ $\newcommand{\tp}{\tilde{\mathfrak{p}}}$ $\newcommand{\tA}{\tilde{A}}$
I have a question about Neukirch, Algebraic Number Theory, page 92. The problem is to show the following.

Let $A$ be a one-dimensional Noetherian domain, and $\tA$ be its normalization. Let $\p$ be a prime ideal of $A$, and $\p\tA = \tp_1^{e_1} \dots \tp_r^{e_r}$ be a prime decomposition.
Then, $\p$ is regular (i.e. $A_\p$ is a DVR) $\Leftrightarrow$ $r=1$, $e_1=1$ and $f_1 = (\tA/\tp_1 : A/\p) = 1$.

For ($\Rightarrow$), I proved $r = 1$ by $A_\p =\tilde{A_\p} = \tA_{\tp_i}$, but I can't prove $e_1 = f_1 = 1$.

I have no idea about ($\Leftarrow$).

  • $\begingroup$ Neukirch says "regular point", since "regular prime" has another meaning. $\endgroup$ – Heinrich Nov 22 '15 at 21:32
  • $\begingroup$ @Heinrich At p79 in Section 12, he uses "regular prime ideal" for the same meaning. $\endgroup$ – nohm Nov 23 '15 at 12:23

"$\Rightarrow$" Let $S=A\setminus\mathfrak p$. Then $S^{-1}\tilde A=\tilde A_{\mathfrak p}=A_{\mathfrak p}$. If prime ideal $\tilde{\mathfrak p}\subset\tilde A$ lies over $\mathfrak p$ then it survives in $A_{\mathfrak p}$, so there is only one with this property. Then $\mathfrak p\tilde A=\tilde{\mathfrak p}^e$, so $S^{-1}(\mathfrak p\tilde A)=S^{-1}\tilde{\mathfrak p}^e$. This entails $\mathfrak p(S^{-1}\tilde A)=(S^{-1}\tilde{\mathfrak p})^e$, that is, $\mathfrak pA_{\mathfrak p}=(\mathfrak pA_{\mathfrak p})^e$ and thus we get $e=1$. It follows $\mathfrak p\tilde A=\tilde{\mathfrak p}$. Now it is obvious that $f_1=1$.

"$\Leftarrow$" From $f_1=1$ we get $\tilde A=A+\mathfrak p\tilde A$. Now localize at $\mathfrak p$ and get $\tilde A_{\mathfrak p}=A_{\mathfrak p}+\mathfrak p\tilde A_{\mathfrak p}$. At this point I have to invoke a condition which appears in Neukirch on page 78:

(*) $\tilde A$ is a finitely generated $A$-module.

Then $\tilde A_{\mathfrak p}$ is a finitely generated $A_{\mathfrak p}$-module and by Nakayama lemma we get $\tilde A_{\mathfrak p}=A_{\mathfrak p}$, so $A_{\mathfrak p}$ is integrally closed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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