The Stacks project

47.8 Deriving torsion

Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal (if $I$ is not finitely generated perhaps a different definition should be used). Let $Z = V(I) \subset \mathop{\mathrm{Spec}}(A)$. Recall that the category $I^\infty \text{-torsion}$ of $I$-power torsion modules only depends on the closed subset $Z$ and not on the choice of the finitely generated ideal $I$ such that $Z = V(I)$, see More on Algebra, Lemma 15.82.6. In this section we will consider the functor

\[ H^0_{I} : \text{Mod}_ A \longrightarrow I^\infty \text{-torsion},\quad M \longmapsto M[I^\infty ] = \bigcup M[I^ n] \]

which sends $M$ to the submodule of $I$-power torsion.

Let $A$ be a ring and let $I$ be a finitely generated ideal. Note that $I^\infty \text{-torsion}$ is a Grothendieck abelian category (direct sums exist, filtered colimits are exact, and $\bigoplus A/I^ n$ is a generator by More on Algebra, Lemma 15.82.2). Hence the derived category $D(I^\infty \text{-torsion})$ exists, see Injectives, Remark 19.13.3. Our functor $H^0_ I$ is left exact and has a derived extension which we will denote

\[ R\Gamma _ I : D(A) \longrightarrow D(I^\infty \text{-torsion}). \]

Warning: this functor does not deserve the name local cohomology unless the ring $A$ is Noetherian. The functors $H^0_ I$, $R\Gamma _ I$, and the satellites $H^ p_ I$ only depend on the closed subset $Z \subset \mathop{\mathrm{Spec}}(A)$ and not on the choice of the finitely generated ideal $I$ such that $V(I) = Z$. However, we insist on using the subscript $I$ for the functors above as the notation $R\Gamma _ Z$ is going to be used for a different functor, see (47.9.0.1), which agrees with the functor $R\Gamma _ I$ only (as far as we know) in case $A$ is Noetherian (see Lemma 47.10.1).

Lemma 47.8.1. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. The functor $R\Gamma _ I$ is right adjoint to the functor $D(I^\infty \text{-torsion}) \to D(A)$.

Proof. This follows from the fact that taking $I$-power torsion submodules is the right adjoint to the inclusion functor $I^\infty \text{-torsion} \to \text{Mod}_ A$. See Derived Categories, Lemma 13.30.3. $\square$

Lemma 47.8.2. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. For any object $K$ of $D(A)$ we have

\[ R\Gamma _ I(K) = \text{hocolim}\ R\mathop{\mathrm{Hom}}\nolimits _ A(A/I^ n, K) \]

in $D(A)$ and

\[ R^ q\Gamma _ I(K) = \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Ext}}\nolimits _ A^ q(A/I^ n, K) \]

as modules for all $q \in \mathbf{Z}$.

Proof. Let $J^\bullet $ be a K-injective complex representing $K$. Then

\[ R\Gamma _ I(K) = J^\bullet [I^\infty ] = \mathop{\mathrm{colim}}\nolimits J^\bullet [I^ n] = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(A/I^ n, J^\bullet ) \]

The first equality is the definition. By Derived Categories, Lemma 13.33.7 we obtain the second equality. The third equality is clear because $H^ q(\mathop{\mathrm{Hom}}\nolimits _ A(A/I^ n, J^\bullet )) = \mathop{\mathrm{Ext}}\nolimits ^ q_ A(A/I^ n, K)$ and because filtered colimits are exact in the category of abelian groups. $\square$

Lemma 47.8.3. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet $ be a complex of $A$-modules such that $f : K^\bullet \to K^\bullet $ is an isomorphism for some $f \in I$, i.e., $K^\bullet $ is a complex of $A_ f$-modules. Then $R\Gamma _ I(K^\bullet ) = 0$.

Proof. Namely, in this case the cohomology modules of $R\Gamma _ I(K^\bullet )$ are both $f$-power torsion and $f$ acts by automorphisms. Hence the cohomology modules are zero and hence the object is zero. $\square$

Let $A$ be a ring and $I \subset A$ a finitely generated ideal. By More on Algebra, Lemma 15.82.5 the category of $I$-power torsion modules is a Serre subcategory of the category of all $A$-modules, hence there is a functor

47.8.3.1
\begin{equation} \label{dualizing-equation-compare-torsion} D(I^\infty \text{-torsion}) \to D_{I^\infty \text{-torsion}}(A) \end{equation}

see Derived Categories, Section 13.17.

Lemma 47.8.4. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $M$ and $N$ be $I$-power torsion modules.

  1. $\mathop{\mathrm{Hom}}\nolimits _{D(A)}(M, N) = \mathop{\mathrm{Hom}}\nolimits _{D({I^\infty \text{-torsion}})}(M, N)$,

  2. $\mathop{\mathrm{Ext}}\nolimits ^1_{D(A)}(M, N) = \mathop{\mathrm{Ext}}\nolimits ^1_{D({I^\infty \text{-torsion}})}(M, N)$,

  3. $\mathop{\mathrm{Ext}}\nolimits ^2_{D({I^\infty \text{-torsion}})}(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^2_{D(A)}(M, N)$ is not surjective in general,

  4. (47.8.3.1) is not an equivalence in general.

Proof. Parts (1) and (2) follow immediately from the fact that $I$-power torsion forms a Serre subcategory of $\text{Mod}_ A$. Part (4) follows from part (3).

For part (3) let $A$ be a ring with an element $f \in A$ such that $A[f]$ contains a nonzero element $x$ annihilated by $f$ and $A$ contains elements $x_ n$ with $f^ nx_ n = x$. Such a ring $A$ exists because we can take

\[ A = \mathbf{Z}[f, x, x_ n]/(fx, f^ nx_ n - x) \]

Given $A$ set $I = (f)$. Then the exact sequence

\[ 0 \to A[f] \to A \xrightarrow {f} A \to A/fA \to 0 \]

defines an element in $\mathop{\mathrm{Ext}}\nolimits ^2_ A(A/fA, A[f])$. We claim this element does not come from an element of $\mathop{\mathrm{Ext}}\nolimits ^2_{D(f^\infty \text{-torsion})}(A/fA, A[f])$. Namely, if it did, then there would be an exact sequence

\[ 0 \to A[f] \to M \to N \to A/fA \to 0 \]

where $M$ and $N$ are $f$-power torsion modules defining the same $2$ extension class. Since $A \to A$ is a complex of free modules and since the $2$ extension classes are the same we would be able to find a map

\[ \xymatrix{ 0 \ar[r] & A[f] \ar[r] \ar[d] & A \ar[r] \ar[d]_\varphi & A \ar[r] \ar[d]_\psi & A/fA \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A[f] \ar[r] & M \ar[r] & N \ar[r] & A/fA \ar[r] & 0 } \]

(some details omitted). Then we could replace $M$ by the image of $\varphi $ and $N$ by the image of $\psi $. Then $M$ would be a cyclic module, hence $f^ n M = 0$ for some $n$. Considering $\varphi (x_{n + 1})$ we get a contradiction with the fact that $f^{n + 1}x_ n = x$ is nonzero in $A[f]$. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BJA. Beware of the difference between the letter 'O' and the digit '0'.