What is the "strongest" non-local property of a ring/module that is true of all localizations at maximal ideals? Given a commutative ring $A$ we say that a property P is local if 

$A$ has P $\leftrightarrow$ $A_{p}$ has P for all prime ideals $p$ of $A$

It is usually the case that this requirement is equivalent to $A_{m}$ having P for all maximal ideals $m$ of $A$. I was wondering which (if any) are the strongest/most interesting local properties $P$ of a commutative ring that do not satisfy the second equivalence. Similarly, I would like to know the strongest/most interesting non-local properties P that are true at all localizations at $p$.
That is to say, what are the most interesting properties P of $A$ such that:

(1) $A_{p}$ has P for all prime ideals $p$ of $A$ but P is NOT local

or

(2) P is local BUT it is NOT true that if $A_m$ has P for all maximal ideals $m$ of $A$ then $A$ has P.

EDIT: After comments and answers received have edited (and expanded) the question. Hope it is clear and unambiguous now.
 A: On the class of noetherian rings, the property "having finite Krull dimension" holds for every local ring, hence is equivalent for $A_m$ at all $m$ maximal, or for $A_p$ at all $p$. However the property is not local since there are noetherian rings of infinite Krull dimension (Nagata).
If you want the property to be defined over all commutative rings, just build-in noetherianity by changing P into: "is non-noetherian or of finite Krull dimension". ;-)
As to the final such property, it is probably P="being local". Indeed, it is a non-local property but it holds for all $A_m$, or equivalently for all $A_p$. At this stage I'm wondering whether I understood the question right. :)))
A: Let P be the property of "being an integral domain". Then


1: If $A$ is an integral domain, then $A_p$ is an integral domain for every prime ideal $p\subseteq A$.


On the other hand.


2:
    Let $A=A_1\oplus A_2$ be a direct sum of two integral domains. Then it is obviously not an integral domain, although $A_p$ is an integral domain for every prime ideal $p\subseteq A$.
    So P is not local.


A: If I understand correctly (1), the following properties are not local 


*

*being a PID (including fields): take a non principal Dedekind domain;

*being Noetherian (consider an infinite product of $\mathbb F_2$);

*being Artinian (same example as above). 

A: A simple example is obtained by taking $P$ to mean "has positive dimension".
Every  local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all  maximal ideals (i.e. just  at $ \mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta=Frac(A)$ has dimension zero, being a field.
Edit In order to address Chuck's comment, let me emphasize that the answer above is very easily adapted to non-local rings.
For example any finitely generated domain $A$ of positive dimension $d$ over a field has property $P$ when localized at a  maximal ideal $\frak m$ but not at the zero ideal.
More precisely, $dim A_{(0)}=0$ and $dimA_{\frak m}=d$ for any maximal ideal ${\frak m}$ : this  equidimensionality result follows from Noether's normalization theorem.
This shows that if property $P_d$ is " has dimension $d$ ", then $P_d$ holds for $A_{\frak m}$ if ${\frak m}$ is maximal and does not hold for $A_{\frak p}$ if the prime $\frak p$ is not maximal.
