Can a non-PID noetherian integral domain have all PID localizations? If a noetherian integral domain R has all its localizations by maximal ideals being PIDs, does it imply that R itself is a PID?
Thank you for your help. 
 A: Short Version: No, an example would be the ring $R=\mathbb{Z}[\sqrt{5}i]$. I'll explain in the long version, why this example was chosen and why it does not satisfy your criteria.
Motivation: The idea is to first give a name to a Noetherian domain $A$ of Krull dimension one (if localization at every maximal is a PID, then the dimension is necessarily one), such that localization at every prime (not just maximal) ideal is a PID. This, as we will see, is a Dedekind domain. It is also known that if a Dedekind domain is a UFD, this it is a PID. So I have translated (a stronger version of) your question into:

Q: Does there exists a Dedekind domain that is not a UFD? A: Yes!

Long Version:
If you consult Atiyah & MacDonald, chapter 9 (Valuation Rings), you will see the following statements:

Proposition 9.2: Let $A$ be a Noetherian local  domain of [Krull] dimension one, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$
its residue field. Then the following are equivalent:

*

*$A$ is a discrete valuation ring;

*$A$ is integrally closed;

*$\mathfrak{m}$ is a principal ideal;

*$\dim \mathfrak{m}/\mathfrak{m^2}=1$;

*Every non-zero ideal is a power of $\mathfrak{m}$.

*There exists $x\in A$ such that every non-zero ideal is of the form $(x^k)$, $k\geq 0$.


In particular, if a Noetherian local domain of dimension one $A$ is a PID then $A$ is a DVR (by $3\to 1$) and conversely, if $A$ is a DVR then $A$ is a PID (by $1\to 6$).
Now we define a Dedekind domain:

Proposition 9.3: Let $A$ be a Noetherian domain of [Krull] dimension one. Then the following are equivalent:

*

*$A$ is integrally closed;

*Every primary ideal in $A$ is a prime power;

*Every local ring $A_\mathfrak{p}$ ($\mathfrak{p}\neq 0$) is a DVR;

A ring satisfying any (so all) of these conditions is called a Dedekind domain.

A very famous example of Dedekind domains are the following group (for a proof see here, proposition 13.5)

Let $\mathbb{Q}\subset K$ be a finite field extension [This is called a number field], and let $R$ be the integral closure of $\mathbb{Z}$ in $K$. Then $R$ is a Dedekind domain.

This is how the example in the short version is made. I'll leave it to you to show $\mathbb{Z}[\sqrt{5}i]$ is not a UFD.
