The Stacks project

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

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

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. $\square$

Comments (2)

Comment #6521 by Mingchen on

You mention three equalities in the proof, but there are only two in the lemma.

Comment #6578 by on

Yes, very confusing. I've tried to clear it up. Changes are here.

There are also:

  • 4 comment(s) on Section 47.8: Deriving torsion

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