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 15.88.1. Let $R$ be a ring. Let $M$ be an $R$-module.
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 $I^ n m = 0$.
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 an $I$-power torsion module for $I = (f)$. We will use the notation
and
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 15.88.2. 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 with each $K_ i$ a direct sum of copies of $R/I^ n$ for $n$ variable.
\[ \ldots \to K_2 \to K_1 \to K_0 \to M \to 0 \]
Proof. There is a canonical surjection
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 15.88.3. 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
$M[I] = 0$,
$M[I^ n] = 0$ for all $n \geq 1$, and
if $I = (f_1, \ldots , f_ t)$, then the map $M \to \bigoplus M_{f_ i}$ is injective.
Proof. This follows from Algebra, Lemma 10.24.4. $\square$
Lemma 15.88.4. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$.
For any $R$-module $M$ we have $(M/M[I^\infty ])[I] = 0$.
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 15.88.3 this implies that it is zero. $\square$
Lemma 15.88.5. 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 12.10.1.
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 15.88.4. Hence the statement of the lemma by Homology, Lemma 12.10.2. $\square$
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. $\square$
The next lemma should probably go somewhere else.
Lemma 15.88.7. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $K$ be an object of $D(R)$ such that $K \otimes _ R^\mathbf {L} R/I = 0$ in $D(R)$. Then
$K \otimes _ R^\mathbf {L} R/I^ n = 0$ for all $n \geq 1$,
$K \otimes _ R^\mathbf {L} N = 0$ for any $I$-power torsion $R$-module $N$,
$K \otimes _ R^\mathbf {L} M = 0$ for any $M \in D^ b(R)$ whose cohomology modules are $I$-power torsion.
Proof. Proof of (2). We can write $N = \bigcup N[I^ n]$. We have $K \otimes _ R^\mathbf {L} N = \text{hocolim}_ n K \otimes _ R^\mathbf {L} N[I^ n]$ as tensor products commute with colimits (details omitted; hint: represent $K$ by a K-flat complex and compute directly). Hence we may assume $N$ is annihilated by $I^ n$. Consider the $R$-algebra $R' = R/I^ n \oplus N$ where $N$ is an ideal of square zero. It suffices to show that $K' = K \otimes _ R^\mathbf {L} R'$ is $0$ in $D(R')$. We have a surjection $R' \to R/I$ of $R$-algebras whose kernel $J$ is nilpotent (any product of $n$ elements in the kernel is zero). We have
by Lemma 15.60.5. Hence by Lemma 15.78.4 we find that $K'$ is a perfect complex of $R'$-modules. In particular $K'$ is bounded above and if $H^ b(K')$ is the right-most nonvanishing cohomology module (if it exists), then $H^ b(K')$ is a finite $R'$-module (use Lemmas 15.74.2 and 15.64.3) with $H^ b(K') \otimes _{R'} R'/J = H^ b(K')/JH^ b(K') = 0$ (because $K' \otimes _{R'}^\mathbf {L} R'/J = 0$). By Nakayama's lemma (Algebra, Lemma 10.20.1) we find $H^ b(K') = 0$, i.e., $K' = 0$ as desired.
Part (1) follows trivially from part (2). Part (3) follows from part (2), induction on the number of nonzero cohomology modules of $M$, and the distinguished triangles of truncation from Derived Categories, Remark 13.12.4. Details omitted. $\square$
Lemma 15.88.8. Let $R \to R'$ be a ring map. Let $I \subset R$ be an ideal such that $R/I^ n \to R'/I^ nR'$ is an isomorphism for $n > 0$. For any $I$-power torsion $R$-module $M$ the map $M \to M \otimes _ R R'$ is an isomorphism. For example, if $I$ is finitely generated and $R^\wedge $ is the completion of $R$ with respect to $I$, then we have $M \cong M \otimes _ R R^\wedge $.
Proof. If $M$ is annihilated by $I^ n$, then
If $M$ is $I$-power torsion, then $M = \bigcup M[I^ n]$. Since tensor products commute with direct limits (Algebra, Lemma 10.12.9), we obtain the desired isomorphism. The last statement is a special case of the first statement by Algebra, Lemma 10.96.3. $\square$
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