The following lemma tells us that the functor $R\Gamma _ Z$ is related to cohomology with supports.
Lemma 51.2.1. Let $A$ be a ring and let $I$ be a finitely generated ideal. Set $Z = V(I) \subset X = \mathop{\mathrm{Spec}}(A)$. For $K \in D(A)$ corresponding to $\widetilde{K} \in D_\mathit{QCoh}(\mathcal{O}_ X)$ via Derived Categories of Schemes, Lemma 36.3.5 there is a functorial isomorphism where on the left we have Dualizing Complexes, Equation (47.9.0.1) and on the right we have the functor of Cohomology, Section 20.34.
\[ R\Gamma _ Z(K) = R\Gamma _ Z(X, \widetilde{K}) \]
Proof. By Cohomology, Lemma 20.34.5 there exists a distinguished triangle
where $U = X \setminus Z$. We know that $R\Gamma (X, \widetilde{K}) = K$ by Derived Categories of Schemes, Lemma 36.3.5. Say $I = (f_1, \ldots , f_ r)$. Then we obtain a finite affine open covering $\mathcal{U} : U = D(f_1) \cup \ldots \cup D(f_ r)$. By Derived Categories of Schemes, Lemma 36.9.4 the alternating Čech complex $\text{Tot}(\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \widetilde{K^\bullet }))$ computes $R\Gamma (U, \widetilde{K})$ where $K^\bullet $ is any complex of $A$-modules representing $K$. Working through the definitions we find
It is clear that $K^\bullet = R\Gamma (X, \widetilde{K^\bullet }) \to R\Gamma (U, \widetilde{K}^\bullet )$ is induced by the diagonal map from $A$ into $\prod A_{f_ i}$. Hence we conclude that
By Dualizing Complexes, Lemma 47.9.1 this complex computes $R\Gamma _ Z(K)$ and we see the lemma holds. $\square$
Lemma 51.2.2. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Set $X = \mathop{\mathrm{Spec}}(A)$, $Z = V(I)$, $U = X \setminus Z$, and $j : U \to X$ the inclusion morphism. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ U$-module. Then
there exists an $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$,
given $M$ there is an exact sequence
and isomorphisms $H^ p(U, \mathcal{F}) = H^{p + 1}_ Z(M)$ for $p \geq 1$,
we may take $M = H^0(U, \mathcal{F})$ in which case we have $H^0_ Z(M) = H^1_ Z(M) = 0$.
\[ 0 \to H^0_ Z(M) \to M \to H^0(U, \mathcal{F}) \to H^1_ Z(M) \to 0 \]
Proof. The existence of $M$ follows from Properties, Lemma 28.22.1 and the fact that quasi-coherent sheaves on $X$ correspond to $A$-modules (Schemes, Lemma 26.7.5). Then we look at the distinguished triangle
of Cohomology, Lemma 20.34.5. Since $X$ is affine we have $R\Gamma (X, \widetilde{M}) = M$ by Cohomology of Schemes, Lemma 30.2.2. By our choice of $M$ we have $\mathcal{F} = \widetilde{M}|_ U$ and hence this produces an exact sequence
and isomorphisms $H^ p(U, \mathcal{F}) = H^{p + 1}_ Z(X, \widetilde{M})$ for $p \geq 1$. By Lemma 51.2.1 we have $H^ i_ Z(M) = H^ i_ Z(X, \widetilde{M})$ for all $i$. Thus (1) and (2) do hold. Finally, setting $M' = H^0(U, \mathcal{F})$ we see that the kernel and cokernel of $M \to M'$ are $I$-power torsion. Therefore $\widetilde{M}|_ U \to \widetilde{M'}|_ U$ is an isomorphism and we can indeed use $M'$ as predicted in (3). It goes without saying that we obtain zero for both $H^0_ Z(M')$ and $H^0_ Z(M')$. $\square$
Lemma 51.2.3. Let $I, J \subset A$ be finitely generated ideals of a ring $A$. If $M$ is an $I$-power torsion module, then the canonical map is an isomorphism for all $i$.
\[ H^ i_{V(I) \cap V(J)}(M) \to H^ i_{V(J)}(M) \]
Proof. Use the spectral sequence of Dualizing Complexes, Lemma 47.9.6 to reduce to the statement $R\Gamma _ I(M) = M$ which is immediate from the construction of local cohomology in Dualizing Complexes, Section 47.9. $\square$
Lemma 51.2.4. Let $S \subset A$ be a multiplicative set of a ring $A$. Let $M$ be an $A$-module with $S^{-1}M = 0$. Then $\mathop{\mathrm{colim}}\nolimits _{f \in S} H^0_{V(f)}(M) = M$ and $\mathop{\mathrm{colim}}\nolimits _{f \in S} H^1_{V(f)}(M) = 0$.
Proof. The statement on $H^0$ follows directly from the definitions. To see the statement on $H^1$ observe that $R\Gamma _{V(f)}$ and $H^1_{V(f)}$ commute with colimits. Hence we may assume $M$ is annihilated by some $f \in S$. Then $H^1_{V(ff')}(M) = 0$ for all $f' \in S$ (for example by Lemma 51.2.3). $\square$
Lemma 51.2.5. Let $I \subset A$ be a finitely generated ideal of a ring $A$. Let $\mathfrak p$ be a prime ideal. Let $M$ be an $A$-module. Let $i \geq 0$ be an integer and consider the map Then
$\mathop{\mathrm{Im}}(\Psi )$ is the set of elements which map to zero in $H^ i_{V(I)}(M)_\mathfrak p$,
if $H^{i - 1}_{V(I)}(M)_\mathfrak p = 0$, then $\Psi $ is injective,
if $H^{i - 1}_{V(I)}(M)_\mathfrak p = H^ i_{V(I)}(M)_\mathfrak p = 0$, then $\Psi $ is an isomorphism.
\[ \Psi : \mathop{\mathrm{colim}}\nolimits _{f \in A, f \not\in \mathfrak p} H^ i_{V((I, f))}(M) \longrightarrow H^ i_{V(I)}(M) \]
Proof. For $f \in A$, $f \not\in \mathfrak p$ the spectral sequence of Dualizing Complexes, Lemma 47.9.6 degenerates to give short exact sequences
This proves (1) and part (2) follows from this and Lemma 51.2.4. Part (3) is a formal consequence. $\square$
Lemma 51.2.6. Let $I \subset I' \subset A$ be finitely generated ideals of a Noetherian ring $A$. Let $M$ be an $A$-module. Let $i \geq 0$ be an integer. Consider the map The following are true:
if $H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for all $\mathfrak p \in V(I) \setminus V(I')$, then $\Psi $ is surjective,
if $H^{i - 1}_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for all $\mathfrak p \in V(I) \setminus V(I')$, then $\Psi $ is injective,
if $H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = H^{i - 1}_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for all $\mathfrak p \in V(I) \setminus V(I')$, then $\Psi $ is an isomorphism.
\[ \Psi : H^ i_{V(I')}(M) \to H^ i_{V(I)}(M) \]
Proof. Proof of (1). Let $\xi \in H^ i_{V(I)}(M)$. Since $A$ is Noetherian, there exists a largest ideal $I \subset I'' \subset I'$ such that $\xi $ is the image of some $\xi '' \in H^ i_{V(I'')}(M)$. If $V(I'') = V(I')$, then we are done. If not, choose a generic point $\mathfrak p \in V(I'')$ not in $V(I')$. Then we have $H^ i_{V(I'')}(M)_\mathfrak p = H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ by assumption. By Lemma 51.2.5 we can increase $I''$ which contradicts maximality.
Proof of (2). Let $\xi ' \in H^ i_{V(I')}(M)$ be in the kernel of $\Psi $. Since $A$ is Noetherian, there exists a largest ideal $I \subset I'' \subset I'$ such that $\xi '$ maps to zero in $H^ i_{V(I'')}(M)$. If $V(I'') = V(I')$, then we are done. If not, then choose a generic point $\mathfrak p \in V(I'')$ not in $V(I')$. Then we have $H^{i - 1}_{V(I'')}(M)_\mathfrak p = H^{i - 1}_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ by assumption. By Lemma 51.2.5 we can increase $I''$ which contradicts maximality.
Part (3) is formal from parts (1) and (2). $\square$
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