
15.79 Torsion modules

In this section “torsion modules” will refer to modules supported on a given closed subset $V(I)$ of an affine scheme $\mathop{\mathrm{Spec}}(R)$. This is different, but analogous to, the notion of a torsion module over a domain (Definition 15.22.1).

Definition 15.79.1. Let $R$ be a ring. Let $M$ be an $R$-module.

1. Let $I \subset R$ be an ideal. We say $M$ is an $I$-power torsion module if for every $m \in M$ there exists an $n > 0$ such that $I^ n m = 0$.

2. Let $f \in R$. We say $M$ is an $f$-power torsion module if for each $m \in M$, there exists an $n > 0$ such that $f^ n m = 0$.

Thus an $f$-power torsion module is the same thing as a $I$-power torsion module for $I = (f)$. We will use the notation

$M[I^ n] = \{ m \in M \mid I^ nm = 0\}$

and

$M[I^\infty ] = \bigcup M[I^ n]$

for an $R$-module $M$. Thus $M$ is $I$-power torsion if and only if $M = M[I^\infty ]$ if and only if $M = \bigcup M[I^ n]$.

Lemma 15.79.2. Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $I$-power torsion module. Then $M$ admits a resolution

$\ldots \to K_2 \to K_1 \to K_0 \to M \to 0$

with each $K_ i$ a direct sum of copies of $R/I^ n$ for $n$ variable.

Proof. There is a canonical surjection

$\oplus _{m \in M} R/I^{n_ m} \to M \to 0$

where $n_ m$ is the smallest positive integer such that $I^{n_ m} \cdot m = 0$. The kernel of the preceding surjection is also an $I$-power torsion module. Proceeding inductively, we construct the desired resolution of $M$. $\square$

Lemma 15.79.3. Let $R$ be a ring. Let $I$ be an ideal of $R$. For any $R$-module $M$ set $M[I^ n] = \{ m \in M \mid I^ nm = 0\}$. If $I$ is finitely generated then the following are equivalent

1. $M[I] = 0$,

2. $M[I^ n] = 0$ for all $n \geq 1$, and

3. if $I = (f_1, \ldots , f_ t)$, then the map $M \to \bigoplus M_{f_ i}$ is injective.

Proof. This follows from Algebra, Lemma 10.23.4. $\square$

Lemma 15.79.4. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$.

1. For any $R$-module $M$ we have $(M/M[I^\infty ])[I] = 0$.

2. An extension of $I$-power torsion modules is $I$-power torsion.

Proof. Let $m \in M$. If $m$ maps to an element of $(M/M[I^\infty ])[I]$ then $Im \subset M[I^\infty ]$. Write $I = (f_1, \ldots , f_ t)$. Then we see that $f_ i m \in M[I^\infty ]$, i.e., $I^{n_ i}f_ i m = 0$ for some $n_ i > 0$. Thus we see that $I^ Nm = 0$ with $N = \sum n_ i + 2$. Hence $m$ maps to zero in $(M/M[I^\infty ])$ which proves the first statement of the lemma.

For the second, suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of modules with $M'$ and $M''$ both $I$-power torsion modules. Then $M[I^\infty ] \supset M'$ and hence $M/M[I^\infty ]$ is a quotient of $M''$ and therefore $I$-power torsion. Combined with the first statement and Lemma 15.79.3 this implies that it is zero $\square$

Lemma 15.79.5. Let $I$ be a finitely generated ideal of a ring $R$. The $I$-power torsion modules form a Serre subcategory of the abelian category $\text{Mod}_ R$, see Homology, Definition 12.9.1.

Proof. It is clear that a submodule and a quotient module of an $I$-power torsion module is $I$-power torsion. Moreover, the extension of two $I$-power torsion modules is $I$-power torsion by Lemma 15.79.4. Hence the statement of the lemma by Homology, Lemma 12.9.2. $\square$

Lemma 15.79.6. 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. $\square$

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