Support of a module

Remarks

If $M$ is finitely generated, then $\operatorname{Supp} M = V(\operatorname{Ann} M)$, the set of primes that contain the annihilator $\operatorname{Ann} M = \set{a \in R : aM = 0}$. In particular the support of a finitely generated module is closed in the Zariski topology. By the local property 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 $V(\operatorname{Ann}(\Q/\Z))$.