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$\newcommand{\p}{\mathfrak{p}}$ $\newcommand{\a}{\mathfrak{a}}$ Let $A$ be a one-dimensional Noetherian domain. I am thinking about this claim:

If all prime ideals of $A$ are invertible, then $A$ is a Dedekind domain.

Is that true?

I tried to show a unique factorization into prime ideals. Indeed, an integral domain is a Dedekind domain, iff all nonzero ideals admit a unique prime factorization.

To show this, I'm going to show following two lemmas. If it is proved, we can prove a unique factorization by the same way as a Dedekind domain (cf. Neukirch, Algebraic Number Theory).

Let $\a \neq 0$ be an ideal of $A$. I proved the first lemma:

There exists prime ideals $\p_1, \dots, \p_r$ such that $$ \a \supseteq \p_1\dots \p_r $$

by the same way as a Dedekind domain, because it is not used that $A$ is integrally closed.

But I can't show the second lemma:

$$\a\p^{-1} \supsetneq \a.$$

I found that to show the second lemma, it is sufficient to show that $$ \bigcap_{n=1}^\infty \p^n = 0. \label{a}\tag{1}$$ We assume that it is proved. If $\a\p^{-1} = \a$, $\a = \a\p$. By induction, we get $\a = \a\p^n$ for $\forall n > 0$. We use $\eqref{a}$, then we get $\a = 0$. It is a contradiction.

However, I can't prove $\eqref{a}$.

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  • $\begingroup$ How do you define invertible ideal? $\endgroup$ – oxeimon Nov 21 '15 at 7:40
  • $\begingroup$ @oxeimon $\mathfrak{a}\mathfrak{a}^{-1} = 1$ for $\mathfrak{a}^{-1} = \{x \in \mathrm{Frac}(A) \mid x\mathfrak{a} \subset A \}$. $\endgroup$ – nohm Nov 21 '15 at 7:53
  • $\begingroup$ Though it is a different way from my thought, what I want to show is proved in linked question. $\endgroup$ – nohm Nov 23 '15 at 13:49
  • $\begingroup$ Closely related: math.stackexchange.com/questions/607305/… $\endgroup$ – user26857 Nov 26 '15 at 9:42
  • $\begingroup$ @nohm Ask proving (1) as a different question. The present question is more or less settled by my answer. $\endgroup$ – user26857 Nov 26 '15 at 11:49
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If $\mathfrak a\mathfrak p^{-1}=\mathfrak a$, then $\mathfrak a\mathfrak p=\mathfrak a$. Moreover $(\mathfrak aA_{\mathfrak p})(\mathfrak pA_{\mathfrak p})=\mathfrak aA_{\mathfrak p}$. By Nakayama lemma we get $\mathfrak aA_{\mathfrak p}=0$, so $\mathfrak a=0$ which I suppose you don't want.

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  • $\begingroup$ Why Nakayama's lemma can be applied? As $\mathbb{Z}$-module? $\endgroup$ – nohm Nov 23 '15 at 13:55
  • $\begingroup$ @nohm Can you see a maximal ideal in my answer? And what about a finitely generated module? $\endgroup$ – user26857 Nov 23 '15 at 13:57
  • $\begingroup$ I see that $\mathfrak{p}A_\mathfrak{p}$ is a maximal ideal of $A_\mathfrak{p}$. I can't see a finitely generated module. $\endgroup$ – nohm Nov 23 '15 at 14:07
  • $\begingroup$ @nohm What about $\mathfrak aA_{\mathfrak p}$ (as an $A_{\mathfrak p}$-module)? (Since $\mathfrak a$ is finitely generated as an $A$-module, then...) $\endgroup$ – user26857 Nov 23 '15 at 14:11
  • $\begingroup$ I see $\mathfrak{a}A_\mathfrak{p}$ is a finitely generated module, but I think Statement 3 of this link can't be applied. $\endgroup$ – nohm Nov 23 '15 at 14:18
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The proof I know for the second lemma also assumes that $A$ is integrally closed: since $1\in \mathfrak{p}^{-1}$ we have $\mathfrak{a}\subseteq \mathfrak{ap}^{-1}$. So we have to show that the inclusion is strict. Assume that $\mathfrak{a}= \mathfrak{ap}^{-1}$. Let $a\in \mathfrak{p}^{-1}$. Then $a$ is integral over $A$ iff there is a finitely-generated $A$-module $N$ with $Na\subseteq N$. But $\mathfrak{a}$ is such an $A$-module, so that $a$ is indeed integral. Since $A$ is integrally closed, we must have $a\in A$, i.e., $\mathfrak{p}^{-1}\subseteq \mathfrak{a}$. Since $\mathfrak{p}$ is maximal, and $A$ is a Noetherian domain of dimension $1$, we obtain $A\subsetneq \mathfrak{p}^{-1}\subseteq \mathfrak{a}$, a contradiction.

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  • $\begingroup$ Should the last containment say $A \subseteq \mathfrak{p}$ instead? $\endgroup$ – Barry Smith Nov 21 '15 at 17:54
  • $\begingroup$ No, the results is that a maximal Ideal $M$ in a Dedekind ring $A$ satisfies $A\subsetneq M^{-1}$. $\endgroup$ – Dietrich Burde Nov 21 '15 at 18:37
  • $\begingroup$ Ah, your additional edit clarifies. With your original wording, I thought that was supposed to be the contradiction. It makes sense now. Thanks. $\endgroup$ – Barry Smith Nov 21 '15 at 20:03
  • $\begingroup$ @DietrichBurde This proof seems to be using that A is a Dedekind domain, but we only assume that A is a one-dimensional Noetherian domain and all prime ideals are invertible. $\endgroup$ – nohm Nov 22 '15 at 1:38

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