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The Stacks project

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 15.94.1, 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 15.73.2. 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


Comments (1)

Comment #10009 by local cohomology on

Even though the canonical morphism is not an equivalence, the functor should be right? The latter functor depends only on as a topological space?


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