# 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.

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I think the more natural and common definition is: $P$ is local if it can be tested locally with respect to the Zariski topology. This means that $A$ has $P$ iff there is a partition of unity $f_1,...,f_n$ such that each $A_{f_i}$ has $P$. Anyway, your question is interesting. So you're asking for a property such that "$A_m$ has P for all maximal ideals $m$ iff $A$ has P", but it is not true for prime ideals? – Martin Brandenburg Oct 23 '11 at 7:47
@Martin Yes, i.e. there is a prime ideal for which P doesn't hold at $A_p$ even though it holds for all maximal $m$ at $A_m$ – Chuck Oct 23 '11 at 14:50
A standard example is the ideal class group of a Dedekind domain $R$, which measures the failure of freeness for projective rank one modules over $R$. Locally the group is trivial. – Tommaso Centeleghe Oct 23 '11 at 21:35
@Martin: this is a pretty widely used notion of a property being local. The idea is that if your property is open and local in this sense, then it is local in your sense. A lot of the properties for which this interpretation of local is used is a singularity condition, such as being regular, Cohen-Macaulay, Gorenstein, $S_n$, etc. are inherently open, so the two interpretations of local agrees for them. – Sándor Kovács Oct 23 '11 at 23:44
Dear Chuck, I think your formulation is slightly ambiguous. My interpretation is that you want a property that holds for the local rings of $A$ at maximal ideals, but does not hold for some (all?) localizations at non-maximal primes. Since the other two answers interpret your question as finding a property that holds at *all primes but not true for the ring itself, it might be useful if you merged your two requirements in grey boxes into a statement that we can all agree upon. Thanks in advance and sorry for giving you some work: your interesting question deserves a crystal-clear formulation. – Georges Elencwajg Oct 24 '11 at 8:38

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.

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Thanks for this Georges - I considered local rings too just to make sure that there is such a P as the one I am asking about. I would like to know about more interesting non-local P's with the maximal property though - the question was about the most 'interesting/strongest (i.e. local-flavoured)' non-local P for which this holds. I actually can't think of any (non-trivial ones) for a non-local ring, which is why I asked the question. – Chuck Oct 23 '11 at 17:11
Dear Chuck, I have added a class of non-local examples in an edit. Needless to say, however, I will never write an answer claiming that it contains "the most interesting/strongest " property, example,... in any subject whatsoever. – Georges Elencwajg Oct 23 '11 at 18:38
That was stupid of me, apologies – Chuck Oct 24 '11 at 0:35
A more general version of this would be "having dimension at least $d$" for some positive $d$. – Sándor Kovács Oct 24 '11 at 2:38
Dear Chuck, no need to apologize: it is perfectly legitimate for you to require the most interesting and strongest answer! It's another matter with the answerer... – Georges Elencwajg Oct 24 '11 at 8:00

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. :)))

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Grüezi, Paul ! – Georges Elencwajg Oct 24 '11 at 8:27
+1 P="being local" – Reimundo Heluani Oct 24 '11 at 10:43
No, "being local" is not nearly as strong as some other such properties, for example "being isomorphic to $\mathbb C$". – Tom Goodwillie Oct 24 '11 at 14:28
Right! I update strongest into "final". – Paul Balmer Oct 25 '11 at 0:11

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.

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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).
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