Local properties that require prime, not just maximal

Question

So I'm taking a course using Atiyah and MacDonald's Introduction to Commutative Algebra, and we just got to local properties, which were defined as a property of a ring $R$ that holds if and only if it holds for $R_P$ for every prime ideal $P\subset R$. The authors then give several examples of local properties, but all of them turn out that there's a third, weaker condition in "the following are equivalent" statements, that the property hold at the localization at every maximal ideal $M$.

So I asked the obvious question: Are there any local properties that require the stronger condition of being true at the localization of every prime?

I.e., is there a ring and a property such that the property IS local, but just being true at every maximal ideal is not sufficient?

The professor not having an example off hand, I figured I'd ask here!

Answer 1

If we have a property $X$ such that for all rings $R$ the statements $$1) ~ R \text { has property } X$$ and $$2) ~ R_P \text { has property } X \text{ for all primes } P$$

are equivalent, then so is

$$3) ~R_M \text { has property } X \text{ for all maximal ideals } M.$$

This is trivial: We only have to show: If the third holds, then the second holds.

Assume $R_M$ has $X$ for all $M$. Let $P$ be a prime. We have $R_P=(R_M)_P$ for some maximal ideal $M$ containg $P$. The latter has $X$, since the implication "$1) \Rightarrow 2)$" is true.

Answer 2

This should be a comment to MooS' answer, but I have not enough reputation at present to write comments in this site. I understood the question as MooS did, that is, that a local property $X$ is defined as one that satisfy:

for all rings $R$, $ ~ R \text { has property } X$ if and only if $ ~ R_P \text { has property } X \text{ for all primes } P$.

The definition of a local property, as suggested, as one that only satisfy

for all rings $R$, $ ~ R \text { has property } X$ if $ ~ R_P \text { has property } X \text{ for all primes } P$

is very unnatural. A stupid example will show this (and also answer the question): $X$ = "being a local ring of dimension 1" would be a local property (in this second sense), simply because there is no ring with all its localizations of dimension 1. And if you take a non-local ring of equidimension 1, you have a local ring $R$ that does not satisfy $X$ but $R_M$ does it for all maximal ideals $M$.