Local property

Builds on Commutative ring Module over a ring Prime ideal Localization of a module Localization at a prime ideal

Definition

Let $P$ be a property of modules over commutative rings. The property $P$ is local if, for every commutative ring $R$ and every $R$ -module $M$ , the following are equivalent:

$M$ has $P$ as an $R$ -module;
for every prime ideal $\mathfrak p$ of $R$ , the localized module $M_{\mathfrak p}$ has $P$ as an $R_{\mathfrak p}$ -module.

A property of commutative rings is local in the same way, with $R$ in place of $M$ and the localizations $R_{\mathfrak p}$ in place of $M_{\mathfrak p}$ . A property of module homomorphisms $\varphi \colon M \to N$ is local in the same way, with the localized maps $\varphi_{\mathfrak p} \colon M_{\mathfrak p} \to N_{\mathfrak p}$ .

Remarks

Example: the property " $M = 0$ " is local. If $M \ne 0$ , pick $m \ne 0$ in $M$ ; its annihilator is a proper ideal, so it lies in a maximal ideal $\mathfrak m$ . Then $m/1 \ne 0$ in $M_{\mathfrak m}$ , since $um \ne 0$ for all $u \notin \mathfrak m$ . The converse direction is clear. It follows that injectivity and surjectivity of a module map, exactness of a sequence, and flatness are local.

In each case it is enough to check the maximal ideals instead of all prime ideals.

Non-example: being an integral domain is not a local property of rings. For a field $k$ , the ring $R = k \times k$ has the two prime ideals $k \times 0$ and $0 \times k$ , and both localizations are isomorphic to $k$ , a domain. But $(1, 0) \cdot (0, 1) = 0$ in $R$ , so $R$ is not a domain. Being Noetherian is also not local; see localization of a Noetherian ring is Noetherian.

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Tag	`FD3T25A1DX`
URI	`m2://FD3T25A1DX`
Kind	definition
Writing	W0 stub
MSC	13B30, 13C99
Defines	Local property
Uses	Commutative ring, Module over a ring, Prime ideal, Localization of a module, Localization at a prime ideal
Changed	2 November 2026

Local property

Definition

Remarks

Reading path

Further reading

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