Lemma 51.5.4. Let $A$ be a Noetherian ring. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. The functor $D^+(\text{Mod}_{A, T}) \to D^+_ T(A)$ is an equivalence.

**Proof.**
Let $M$ be an object of $\text{Mod}_{A, T}$. Choose an embedding $M \to J$ into an injective $A$-module. By Dualizing Complexes, Proposition 47.5.9 the module $J$ is a direct sum of injective hulls of residue fields. Let $E$ be an injective hull of the residue field of $\mathfrak p$. Since $E$ is $\mathfrak p$-power torsion we see that $H^0_ T(E) = 0$ if $\mathfrak p \not\in T$ and $H^0_ T(E) = E$ if $\mathfrak p \in T$. Thus $H^0_ T(J)$ is injective as a direct sum of injective hulls (by the proposition) and we have an embedding $M \to H^0_ T(J)$. Thus every object $M$ of $\text{Mod}_{A, T}$ has an injective resolution $M \to J^\bullet $ with $J^ n$ also in $\text{Mod}_{A, T}$. It follows that $RH^0_ T(M) = M$.

Next, suppose that $K \in D_ T^+(A)$. Then the spectral sequence

(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 $RH^0_ T(K) \to K$ is an isomorphism. Thus the functor $D^+(\text{Mod}_{A, T}) \to D^+_ T(A)$ is an equivalence with quasi-inverse given by $RH^0_ T$. $\square$

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