Proof of local definition of Dedekind Domain without using unique factorization of ideals

Question

We say that an integral domain $A$ is a Dedekind domain if:

1. $A$ is Noetherian,
2. $A$ is integrally closed,
3. $\dim A = 1$ (in other words, every prime ideal is maximal).

I would like to show that assuming these three conditions, the localization of $A$ at a nonzero prime ideal $\mathfrak{p}$ is a DVR (local PID that is not a field), but I would like to do this $\textit{without}$ using the fact that being a Dedekind domain implies that ideals factor uniquely into prime ideals.

Since the localization of a Noetherian domain is Noetherian, and the localization at a prime ideal is a local ring, it suffices to show that the maximal ideal in $A_\mathfrak{p}$ is principal. The issue is that every proof I can find uses unique factorization of ideals at this point. The argument usually goes:

1. There exists $\pi \in \mathfrak{p} \setminus \mathfrak{p}^2$ (unique factorization implies $\mathfrak{p} \neq \mathfrak{p}^2$),
2. Every ideal factors uniquely into primes and localization commutes with products, so every ideal in $A_\mathfrak{p}$ is a power of the maximal ideal.
3. Thus $(\pi)_\mathfrak{p}$ strictly contains $\mathfrak{m}^2$ and is contained in $\mathfrak{m}$, so $(\pi)_\mathfrak{p} = \mathfrak{m}$ and the maximal ideal is principal.

Is there a way to get around invoking unique factorization of ideals in this argument?

$\textbf{Context}$: I would like to have a coherent mental schema including both that Dedekind domains have unique factorization of ideals and that rings of integers in Number fields have this property. My preferred method for proving that Dedekind domains have unique factorization of ideals into prime ideals is by setting up an isomorphism between the group of fractional ideals and the free abelian group generated by the prime ideals. This starts with the definition of a Dedekind domain that $A_\mathfrak{p}$ is a DVR for all $\mathfrak{p}$. When it comes to proving that the ring of integers is a Dedekind domain, however, I find it most insightful to think of the ring of integers as a lattice and prove the three properties listed above. I would then like to say that the ring of integers has unique factorization of ideals by saying that it has the three properties listed above, therefore every localization is a DVR, and therefore unique factorization holds, rather than arguing for unique factorization directly from the three properties.

Answer 1

Yes. You can find a proof of this in J. Milne's Algebraic Number Theory, Proposition $3.2$.