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 $D(I^\infty \text{-torsion}) \to D(A)$.
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 15.88.6. 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 15.88.2). 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).
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 in $D(A)$ and as modules for all $q \in \mathbf{Z}$.
\[ R\Gamma _ I(K) = \text{hocolim}\ R\mathop{\mathrm{Hom}}\nolimits _ A(A/I^ n, K) \]
\[ R^ q\Gamma _ I(K) = \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Ext}}\nolimits _ A^ q(A/I^ n, K) \]
Proof. Let $J^\bullet $ be a K-injective complex representing $K$. Then
\[ R\Gamma _ I(K) = J^\bullet [I^\infty ] = \mathop{\mathrm{colim}}\nolimits J^\bullet [I^ n] = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(A/I^ n, J^\bullet ) \]
where the first equality is the definition of $R\Gamma _ I(K)$. By Derived Categories, Lemma 13.33.7 we obtain the first displayed equality in the statement of the lemma. The second displayed equality in the statement of the lemma then follows 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 15.88.5 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.
$\mathop{\mathrm{Hom}}\nolimits _{D(A)}(M, N) = \mathop{\mathrm{Hom}}\nolimits _{D({I^\infty \text{-torsion}})}(M, N)$,
$\mathop{\mathrm{Ext}}\nolimits ^1_{D(A)}(M, N) = \mathop{\mathrm{Ext}}\nolimits ^1_{D({I^\infty \text{-torsion}})}(M, N)$,
$\mathop{\mathrm{Ext}}\nolimits ^2_{D({I^\infty \text{-torsion}})}(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^2_{D(A)}(M, N)$ is not surjective in general,
(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 $A[f]$. $\square$
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