Let \(A\) be a noetherian ring, and let \(\mfa\) be an ideal of \(A\).
Show that \(\Gamma_{\mfa}(\wait)\) (II, Ex. 5.6) is a left-exact functor from the category of \(A\)-modules to itself. We denote its right derived functors, calculated in \(\Mod(A)\), by \(H_{\mfa}^i(\wait)\).
Now let \(X=\Spec A\), \(Y=V(\mfa)\). Show that for any \(A\)-module \(M\), \[ H_{\mfa}^i(M)=H_Y^i(X, \tilde{M}), \] where \(H_Y^i(X, \wait)\) denotes cohomology with supports in \(Y\) (Ex. 2.3).
For any \(i\ge0\), show that \(\Gamma_{\mfa}(H_{\mfa}^i(M))=H_{\mfa}^i(M)\).
Solution
For an \(A\)-module \(M\), the ideal-torsion submodule is \[ \Gamma_{\mfa}(M)=\{m\in M:\mfa^n m=0\text{ for some integer }n\ge1\}. \] No finite-generation hypothesis is placed on \(M\).
1. The assignment \(M\mapsto\Gamma_{\mfa}(M)\) is an additive left exact functor, proving (a).
If two elements are killed by powers \(\mfa^r\) and \(\mfa^s\), their sum is killed by \(\mfa^{\max(r,s)}\). Multiplying an element by a scalar in \(A\) preserves any annihilating power. Thus \(\Gamma_{\mfa}(M)\) is a submodule. An \(A\)-linear homomorphism sends an element killed by \(\mfa^n\) to one killed by the same power, so the construction is functorial and additive.
For an exact sequence \(0\to M'\xrightarrow{u}M\xrightarrow{v}M''\), the induced map on torsion submodules is injective at the left. If \(m\in\Gamma_{\mfa}(M)\) lies in \(\ker v\), write \(m=u(m')\). For a power \(\mfa^n\) killing \(m\), injectivity of \(u\) gives \(\mfa^n m'=0\). Hence \(m'\) lies in \(\Gamma_{\mfa}(M')\), proving exactness at the middle term.
2. Under \(M\cong\Gamma(X,\widetilde M)\), one has a natural equality\[ \Gamma_{\mfa}(M)=\Gamma_Y(X,\widetilde M). \]
The support of the section corresponding to \(m\) is \(V(\Ann_A(m))\), by P-AGH256SUPPORT, part (a). It is contained in \(V(\mfa)\) exactly when \(\mfa\subseteq\sqrt{\Ann_A(m)}\). If \(\mfa^n m=0\), this containment holds.
Conversely, choose generators \(a_1,\ldots,a_r\) of \(\mfa\) and positive integers \(n_j\) with \(a_j^{n_j}m=0\). Every monomial of total degree \(N=1+\sum_j(n_j-1)\) in the generators is divisible by one of these annihilating powers. Thus \(\mfa^N m=0\). For \(\mfa=0\) the assertion holds directly, since all elements are killed by its first power and \(Y=X\). This proves equality, and its construction commutes with all module homomorphisms.
3. For every \(i\ge0\) there is a natural \(A\)-linear isomorphism\[ \boxed{H_{\mfa}^i(M)\cong H_Y^i(X,\widetilde M),} \]proving (b).
Choose an injective resolution \(M\to I^\bullet\) in the category of \(A\)-modules. The associated-sheaf functor on \(\Spec A\) is exact, as can be checked by localization at every prime, so \(\widetilde M\to\widetilde{I^\bullet}\) is an exact resolution. Since \(A\) is noetherian, each \(\widetilde{I^q}\) is flasque [@Har10a, Proposition III.3.4]. Flasque sheaves are acyclic for sections supported in a closed subset, by P-AGH323SUPPORTS, part (c). Consequently this resolution computes the supported cohomology of \(\widetilde M\). It is not necessary that the associated sheaves be injective as abelian sheaves.
Step 2 identifies, term by term and compatibly with the differentials, the complexes \[ \Gamma_{\mfa}(I^\bullet)\cong\Gamma_Y(X,\widetilde{I^\bullet}). \] The cohomology of the first is \(H_{\mfa}^i(M)\) by definition and that of the second is \(H_Y^i(X,\widetilde M)\) by the preceding acyclicity. This gives the required isomorphism. All maps preserve scalar multiplication, and comparison of injective resolutions makes the isomorphism natural in \(M\) [@Har10a, Chapter III, §1].
4. Every element of \(H_{\mfa}^i(M)\) is killed by some power of \(\mfa\), proving (c).
3.A cohomology class has a cocycle representative \(z\in\Gamma_{\mfa}(I^i)\).By definition, some power \(\mfa^n\) kills \(z\) and therefore also kills its cohomology class.Thus every element belongs to \(\Gamma_{\mfa}(H_{\mfa}^i(M))\), proving equality with the whole module.The exponent may depend on the class; a single exponent annihilating the entire cohomology module is not required.
5. Q.E.D.
Step 1 proves (a), steps <1>2–<1>3 prove (b), and step 4 proves (c).