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What I mean by the question is for example the following statement:

Let $R$ be a ring and suppose that for each prime ideal $\mathfrak{p}$ the local ring $R_{\mathfrak{p}}$ has no nilpotent elements. Then $R$ itself also has no nilpotent elements.

So we call having nilpotent elements a local property. What are other local properties or what are non-local properties? For example being an integral domain is not a local property.

Constantin K
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    I think this question is very good and potential answers can be very helpful. Someone voted to close because it is too broad - Maybe it would be a good idea to make this a community wiki? – Jesko Hüttenhain Jun 14 '17 at 09:56

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$\newcommand{\p}{\mathfrak p} \newcommand{\m}{\mathfrak m}$I took these properties from Atiyah-Macdonald. The first key local property is : if $M$ is a $A$-module, then there is an equivalence between :

  • $M=0$
  • For all prime $\p$, $M_{\p} = 0$
  • For all maximal ideals $\m$, $M_{\m} = 0$

We have similar equivalence for the following properties :

  • Injectivity of a morphism of $A$-module $f : M \to N$.
  • Flatness of a $A$-module $M$.
  • $M$ is torsion-free.
  • $A$ is integrally closed.
  • $M$ is an invertible fractional ideal.

After, some properties are by definition local, for example being smooth or normal. Sometimes, properties can be checked locally but you have to ask different conditions. For example, $A$ is absolutely flat if and only if for all maximal ideals $\m$, $A_{\m}$ is a field.

  • What about being finitely generated? Is that a local property? – Olivier Bégassat Nov 17 '19 at 13:47
  • Being finitely generated is not a local property over arbitrary rings. For any infinite field $k$, we could take $\oplus_{a\in k} k[x]/(x-a)$ as a $k[x]$-module. This is not finitely generated, but is locally finitely generated. – desiigner Nov 03 '22 at 00:48
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Jesko Hüttenhain
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