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!


2 Answers 2


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.

  • $\begingroup$ 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

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 $M$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.