Lemma 47.12.1. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $R\Gamma _ Z$ be as in Lemma 47.9.1. Let ${\ }^\wedge $ denote derived completion as in More on Algebra, Lemma 15.84.10. For an object $K$ in $D(A)$ we have

in $D(A)$.

Let $A$ be a ring and let $I$ be a finitely generated ideal. In this case we can consider the derived category $D_{I^\infty \text{-torsion}}(A)$ of complexes with $I$-power torsion cohomology modules (Section 47.9) and the derived category $D_{comp}(A, I)$ of derived complete complexes (More on Algebra, Section 15.84). In this section we show these categories are equivalent. A more general statement can be found in [Dwyer-Greenlees].

Lemma 47.12.1. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $R\Gamma _ Z$ be as in Lemma 47.9.1. Let ${\ }^\wedge $ denote derived completion as in More on Algebra, Lemma 15.84.10. For an object $K$ in $D(A)$ we have

\[ R\Gamma _ Z(K^\wedge ) = R\Gamma _ Z(K) \quad \text{and}\quad (R\Gamma _ Z(K))^\wedge = K^\wedge \]

in $D(A)$.

**Proof.**
Choose $f_1, \ldots , f_ r \in A$ generating $I$. Recall that

\[ K^\wedge = R\mathop{\mathrm{Hom}}\nolimits _ A\left((A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}), K\right) \]

by More on Algebra, Lemma 15.84.10. Hence the cone $C = \text{Cone}(K \to K^\wedge )$ is given by

\[ R\mathop{\mathrm{Hom}}\nolimits _ A\left((\prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}), K\right) \]

which can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(A_{f_{i_0} \ldots f_{i_ p}}, K), \quad p > 0 \]

These complexes vanish on applying $R\Gamma _ Z$, see Lemma 47.9.4. Applying $R\Gamma _ Z$ to the distinguished triangle $K \to K^\wedge \to C \to K[1]$ we see that the first formula of the lemma is correct.

Recall that

\[ R\Gamma _ Z(K) = K \otimes ^\mathbf {L} (A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}) \]

by Lemma 47.9.1. Hence the cone $C = \text{Cone}(R\Gamma _ Z(K) \to K)$ can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to

\[ K \otimes _ A A_{f_{i_0} \ldots f_{i_ p}}, \quad p > 0 \]

These complexes vanish on applying ${\ }^\wedge $, see More on Algebra, Lemma 15.84.11. Applying derived completion to the distinguished triangle $R\Gamma _ Z(K) \to K \to C \to R\Gamma _ Z(K)[1]$ we see that the second formula of the lemma is correct. $\square$

The following result is a special case of a very general phenomenon concerning admissible subcategories of a triangulated category.

Proposition 47.12.2. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. The functors $R\Gamma _ Z$ and ${\ }^\wedge $ define quasi-inverse equivalences of categories

\[ D_{I^\infty \text{-torsion}}(A) \leftrightarrow D_{comp}(A, I) \]

**Proof.**
Follows immediately from Lemma 47.12.1.
$\square$

The following addendum of the proposition above makes the correspondence on morphisms more precise.

Lemma 47.12.3. With notation as in Lemma 47.12.1. For objects $K, L$ in $D(A)$ there is a canonical isomorphism

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , L^\wedge ) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\Gamma _ Z(K), R\Gamma _ Z(L)) \]

in $D(A)$.

**Proof.**
Say $I = (f_1, \ldots , f_ r)$. Denote $C = (A \to \prod A_{f_ i} \to \ldots \to A_{f_1 \ldots f_ r})$ the alternating Čech complex. Then derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ (More on Algebra, Lemma 15.84.10) and local cohomology by $C \otimes ^\mathbf {L} -$ (Lemma 47.9.1). Combining the isomorphism

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(K \otimes ^\mathbf {L} C, L \otimes ^\mathbf {L} C) = R\mathop{\mathrm{Hom}}\nolimits _ A(K, R\mathop{\mathrm{Hom}}\nolimits _ A(C, L \otimes ^\mathbf {L} C)) \]

(More on Algebra, Lemma 15.69.1) and the map

\[ L \to R\mathop{\mathrm{Hom}}\nolimits _ A(C, L \otimes ^\mathbf {L} C) \]

(More on Algebra, Lemma 15.69.6) we obtain a map

\[ \gamma : R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(K \otimes ^\mathbf {L} C, L \otimes ^\mathbf {L} C) \]

On the other hand, the right hand side is derived complete as it is equal to

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(C, R\mathop{\mathrm{Hom}}\nolimits _ A(K, L \otimes ^\mathbf {L} C)). \]

Thus $\gamma $ factors through the derived completion of $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ by the universal property of derived completion. However, the derived completion goes inside the $R\mathop{\mathrm{Hom}}\nolimits _ A$ by More on Algebra, Lemma 15.84.12 and we obtain the desired map.

To show that the map of the lemma is an isomorphism we may assume that $K$ and $L$ are derived complete, i.e., $K = K^\wedge $ and $L = L^\wedge $. In this case we are looking at the map

\[ \gamma : R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(R\Gamma _ Z(K), R\Gamma _ Z(L)) \]

By Proposition 47.12.2 we know that the cohomology groups of the left and the right hand side coincide. In other words, we have to check that the map $\gamma $ sends a morphism $\alpha : K \to L$ in $D(A)$ to the morphism $R\Gamma _ Z(\alpha ) : R\Gamma _ Z(K) \to R\Gamma _ Z(L)$. We omit the verification (hint: note that $R\Gamma _ Z(\alpha )$ is just the map $\alpha \otimes \text{id}_ C : K \otimes ^\mathbf {L} C \to L \otimes ^\mathbf {L} C$ which is almost the same as the construction of the map in More on Algebra, Lemma 15.69.6). $\square$

Lemma 47.12.4. Let $I$ and $J$ be ideals in a Noetherian ring $A$. Let $M$ be a finite $A$-module. Set $Z =V(J)$. Consider the derived $I$-adic completion $R\Gamma _ Z(M)^\wedge $ of local cohomology. Then

we have $R\Gamma _ Z(M)^\wedge = R\mathop{\mathrm{lim}}\nolimits R\Gamma _ Z(M/I^ nM)$, and

there are short exact sequences

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}_ Z(M/I^ nM) \to H^ i(R\Gamma _ Z(M)^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^ i_ Z(M/I^ nM) \to 0 \]

In particular $R\Gamma _ Z(M)^\wedge $ has vanishing cohomology in negative degrees.

**Proof.**
Suppose that $J = (g_1, \ldots , g_ m)$. Then $R\Gamma _ Z(M)$ is computed by the complex

\[ M \to \prod M_{g_{j_0}} \to \prod M_{g_{j_0}g_{j_1}} \to \ldots \to M_{g_1g_2\ldots g_ m} \]

by Lemma 47.9.1. By More on Algebra, Lemma 15.86.6 the derived $I$-adic completion of this complex is given by the complex

\[ \mathop{\mathrm{lim}}\nolimits M/I^ nM \to \prod \mathop{\mathrm{lim}}\nolimits (M/I^ nM)_{g_{j_0}} \to \ldots \to \mathop{\mathrm{lim}}\nolimits (M/I^ nM)_{g_1g_2\ldots g_ m} \]

of usual completions. Since $R\Gamma _ Z(M/I^ nM)$ is computed by the complex $ M/I^ nM \to \prod (M/I^ nM)_{g_{j_0}} \to \ldots \to (M/I^ nM)_{g_1g_2\ldots g_ m}$ and since the transition maps between these complexes are surjective, we conclude that (1) holds by More on Algebra, Lemma 15.80.1. Part (2) then follows from More on Algebra, Lemma 15.80.4. $\square$

Lemma 47.12.5. With notation and hypotheses as in Lemma 47.12.4 assume $A$ is $I$-adically complete. Then

\[ H^0(R\Gamma _ Z(M)^\wedge ) = \mathop{\mathrm{colim}}\nolimits H^0_{V(J')}(M) \]

where the filtered colimit is over $J' \subset J$ such that $V(J') \cap V(I) = V(J) \cap V(I)$.

**Proof.**
Since $M$ is a finite $A$-module, we have that $M$ is $I$-adically complete. The proof of Lemma 47.12.4 shows that

\[ H^0(R\Gamma _ Z(M)^\wedge ) = \mathop{\mathrm{Ker}}(M^\wedge \to \prod M_{g_ j}^\wedge ) = \mathop{\mathrm{Ker}}(M \to \prod M_{g_ j}^\wedge ) \]

where on the right hand side we have usual $I$-adic completion. The kernel $K_ j$ of $M_{g_ j} \to M_{g_ j}^\wedge $ is $\bigcap I^ n M_{g_ j}$. By Algebra, Lemma 10.50.5 for every $\mathfrak p \in V(IA_{g_ j})$ we find an $f \in A_{g_ j}$, $f \not\in \mathfrak p$ such that $(K_ j)_ f = 0$.

Let $s \in H^0(R\Gamma _ Z(M)^\wedge )$. By the above we may think of $s$ as an element of $M$. The support $Z'$ of $s$ intersected with $D(g_ j)$ is disjoint from $D(g_ j) \cap V(I)$ by the arguments above. Thus $Z'$ is a closed subset of $\mathop{\mathrm{Spec}}(A)$ with $Z' \cap V(I) \subset V(J)$. Then $Z' \cup V(J) = V(J')$ for some ideal $J' \subset J$ with $V(J') \cap V(I) \subset V(J)$ and we have $s \in H^0_{V(J')}(M)$. Conversely, any $s \in H^0_{V(J')}(M)$ with $J' \subset J$ and $V(J') \cap V(I) \subset V(J)$ maps to zero in $M_{g_ j}^\wedge $ for all $j$. This proves the lemma. $\square$

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.

## Comments (3)

Comment #4190 by Amnon Yekutieli on

Comment #4197 by Amnon Yekutieli on

Comment #4384 by Johan on