Lemma 15.88.6. Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal. The subcategory $I^\infty \text{-torsion} \subset \text{Mod}_ R$ depends only on the closed subset $Z = V(I) \subset \mathop{\mathrm{Spec}}(R)$. In fact, an $R$-module $M$ is $I$-power torsion if and only if its support is contained in $Z$.

Proof. Let $M$ be an $R$-module. Let $x \in M$. If $x \in M[I^\infty ]$, then $x$ maps to zero in $M_ f$ for all $f \in I$. Hence $x$ maps to zero in $M_\mathfrak p$ for all $\mathfrak p \not\supset I$. Conversely, if $x$ maps to zero in $M_\mathfrak p$ for all $\mathfrak p \not\supset I$, then $x$ maps to zero in $M_ f$ for all $f \in I$. Hence if $I = (f_1, \ldots , f_ r)$, then $f_ i^{n_ i}x = 0$ for some $n_ i \geq 1$. It follows that $x \in M[I^{\sum n_ i}]$. Thus $M[I^\infty ]$ is the kernel of $M \to \prod _{\mathfrak p \not\in Z} M_\mathfrak p$. The second statement of the lemma follows and it implies the first. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).