Lemma 47.10.1. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal.

1. the adjunction $R\Gamma _ I(K) \to K$ is an isomorphism for $K \in D_{I^\infty \text{-torsion}}(A)$,

2. the functor (47.8.3.1) $D(I^\infty \text{-torsion}) \to D_{I^\infty \text{-torsion}}(A)$ is an equivalence,

3. the transformation of functors (47.10.0.1) is an isomorphism, in other words $R\Gamma _ I(K) = R\Gamma _ Z(K)$ for $K \in D(A)$.

Proof. A formal argument, which we omit, shows that it suffices to prove (1).

Let $M$ be an $I$-power torsion $A$-module. Choose an embedding $M \to J$ into an injective $A$-module. Then $J[I^\infty ]$ is an injective $A$-module, see Lemma 47.3.9, and we obtain an embedding $M \to J[I^\infty ]$. Thus every $I$-power torsion module has an injective resolution $M \to J^\bullet$ with $J^ n$ also $I$-power torsion. It follows that $R\Gamma _ I(M) = M$ (this is not a triviality and this is not true in general if $A$ is not Noetherian). Next, suppose that $K \in D_{I^\infty \text{-torsion}}^+(A)$. Then the spectral sequence

$R^ q\Gamma _ I(H^ p(K)) \Rightarrow R^{p + q}\Gamma _ I(K)$

(Derived Categories, Lemma 13.21.3) converges and above we have seen that only the terms with $q = 0$ are nonzero. Thus we see that $R\Gamma _ I(K) \to K$ is an isomorphism.

Suppose $K$ is an arbitrary object of $D_{I^\infty \text{-torsion}}(A)$. We have

$R^ q\Gamma _ I(K) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ q_ A(A/I^ n, K)$

by Lemma 47.8.2. Choose $f_1, \ldots , f_ r \in A$ generating $I$. Let $K_ n^\bullet = K(A, f_1^ n, \ldots , f_ r^ n)$ be the Koszul complex with terms in degrees $-r, \ldots , 0$. Since the pro-objects $\{ A/I^ n\}$ and $\{ K_ n^\bullet \}$ in $D(A)$ are the same by More on Algebra, Lemma 15.87.1, we see that

$R^ q\Gamma _ I(K) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ q_ A(K_ n^\bullet , K)$

Pick any complex $K^\bullet$ of $A$-modules representing $K$. Since $K_ n^\bullet$ is a finite complex of finite free modules we see that

$\mathop{\mathrm{Ext}}\nolimits ^ q_ A(K_ n, K) = H^ q(\text{Tot}((K_ n^\bullet )^\vee \otimes _ A K^\bullet ))$

where $(K_ n^\bullet )^\vee$ is the dual of the complex $K_ n^\bullet$. See More on Algebra, Lemma 15.69.2. As $(K_ n^\bullet )^\vee$ is a complex of finite free $A$-modules sitting in degrees $0, \ldots , r$ we see that the terms of the complex $\text{Tot}((K_ n^\bullet )^\vee \otimes _ A K^\bullet )$ are the same as the terms of the complex $\text{Tot}((K_ n^\bullet )^\vee \otimes _ A \tau _{\geq q - r - 2} K^\bullet )$ in degrees $q - 1$ and higher. Hence we see that

$\mathop{\mathrm{Ext}}\nolimits ^ q_ A(K_ n, K) = \text{Ext}^ q_ A(K_ n, \tau _{\geq q - r - 2}K)$

for all $n$. It follows that

$R^ q\Gamma _ I(K) = R^ q\Gamma _ I(\tau _{\geq q - r - 2}K) = H^ q(\tau _{\geq q - r - 2}K) = H^ q(K)$

Thus we see that the map $R\Gamma _ I(K) \to K$ is an isomorphism. $\square$

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