All prime ideals are invertible $\Rightarrow$ Dedekind domain

Question

$\newcommand{\p}{\mathfrak{p}}$ $\newcommand{\a}{\mathfrak{a}}$ 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 $\eqref{a}$.

Answer 1

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.

Answer 2

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.