---
tag: B0223N93XS
kind: definition
name: Support of a module
msc: [13C05, 13B30]
defines: FETMR0M0M8
uses: [0NZ8JA25WK, 00MHHZ9SNE, 68E0F8N94E, FB45BHHA2J, HB4XDKJTW8]
sources:
  - url: https://en.wikipedia.org/wiki/Support_of_a_module
---

## Definition

Let $R$ be a [[commutative ring|0310REJJY8]] and let $M$ be an $R$-[[module|0NZ8JA25WK]].
For a [[prime ideal|00MHHZ9SNE]] $\mathfrak p$, let $M_{\mathfrak p}$ be the [[localization|68E0F8N94E]] of $M$ at $\mathfrak p$.
The *support* of $M$ is the following subset of the [[spectrum|FB45BHHA2J]] $\operatorname{Spec} R$:
$$\operatorname{Supp} M = \set{\mathfrak p \in \operatorname{Spec} R : M_{\mathfrak p} \ne 0}.$$

## Remarks

If $M$ is [[finitely generated|05QY1VEBYD]], then $\operatorname{Supp} M = V(\operatorname{Ann} M)$, the set of primes that contain the [[annihilator|HB4XDKJTW8]] $\operatorname{Ann} M = \set{a \in R : aM = 0}$.
In particular the support of a finitely generated module is closed in the [[Zariski topology|8SQK7RTB63]].
By the [[local property|3S58E70Z7Z]] of vanishing, $\operatorname{Supp} M = \emptyset$ if and only if $M = 0$.

Example: let $R = \Z$ and $M = \Z/6\Z$.
For a prime $p \ne 2, 3$, the integer $6$ lies outside $p\Z$, and $m/s = 6m/(6s) = 0$, so $M_{(p)} = 0$.
Also $M_{(0)} = 0$ since $6 \notin (0)$.
For $p = 2$, the element $1/1$ is nonzero in $M_{(2)}$, since $u \cdot 1 \ne 0$ in $\Z/6\Z$ for every odd $u$.
The same holds for $p = 3$, so $\operatorname{Supp} M = \set{2\Z, 3\Z} = V(6\Z)$.

Non-example of the formula with the annihilator: the $\Z$-module $\Q/\Z$ is not finitely generated, and $\operatorname{Ann}(\Q/\Z) = 0$, so $V(\operatorname{Ann}) = \operatorname{Spec} \Z$.
But $(\Q/\Z)_{(0)} = 0$, since every element is killed by a nonzero integer.
So $(0) \notin \operatorname{Supp}(\Q/\Z)$, and the support is not $V(\operatorname{Ann}(\Q/\Z))$.