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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!

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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.

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  • $\begingroup$ I don't think this is what the OP asked. He asks for a property (X) such that if $R_P$ has $(X)$ for all primes $P$ then $R$ has $(X)$, but if $R_M$ has $(X)$ for all maximal $M$ then $R$ does not have $(X)$. $\endgroup$
    – user26857
    Oct 1, 2015 at 14:22
  • $\begingroup$ @user26857 That is exactly what I was asking for. $\endgroup$
    – Alan
    Oct 2, 2015 at 1:19
  • $\begingroup$ I do not know why we even have to discuss about this: the OP asked for a local property than cannot be tested on maximal ideals only. I gave a proof, that such a property does not exist. So we are done here. $\endgroup$
    – MooS
    Oct 6, 2015 at 6:21
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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 $M$.

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  • $\begingroup$ I can't see the connection between this answer and the OP's question. $\endgroup$
    – user26857
    Oct 3, 2015 at 20:39
  • $\begingroup$ Certainly. As I wrote, it is an answer only to the question as it was understood by you in your comment to MooS' answer. Interpreting the OP's answer as MooS (and I) did, MooS answer is correct. $\endgroup$
    – SlavaM
    Oct 3, 2015 at 21:12
  • $\begingroup$ I also can't see the connection of this answer to the way I interpreted the OP's question (and he conceded with). $\endgroup$
    – user26857
    Oct 3, 2015 at 21:16
  • $\begingroup$ ?? My example $X$= "being a local ring of dimension 1" answers your question. $\endgroup$
    – SlavaM
    Oct 3, 2015 at 21:21
  • $\begingroup$ The point is that your interpretation of the question is so unnatural, that you can choose examples so trivial as the one I chose (that is, simply there is no ring that satisfies the property $X$ for all localizations at all prime ideals, but there are rings that satisfy it for all maximal ideals). $\endgroup$
    – SlavaM
    Oct 3, 2015 at 21:25

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