Lemma 51.5.7. Let $A \to B$ be a flat homomorphism of Noetherian rings. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $T' \subset \mathop{\mathrm{Spec}}(B)$ be the inverse image of $T$. Then the canonical map
is an isomorphism for $K \in D^+(A)$. If $A$ and $B$ have finite dimension, then this is true for $K \in D(A)$.
Proof. From the map $R\Gamma _ T(K) \to K$ we get a map $R\Gamma _ T(K) \otimes _ A^\mathbf {L} B \to K \otimes _ A^\mathbf {L} B$. The cohomology modules of $R\Gamma _ T(K) \otimes _ A^\mathbf {L} B$ are supported on $T'$ and hence we get the arrow of the lemma. This arrow is an isomorphism if $T$ is a closed subset of $\mathop{\mathrm{Spec}}(A)$ by Dualizing Complexes, Lemma 47.9.3. Recall that $H^ i_ T(K)$ is the colimit of $H^ i_ Z(K)$ where $Z$ runs over the (directed set of) closed subsets of $T$, see Lemma 51.5.3. Correspondingly $H^ i_{T'}(K \otimes _ A^\mathbf {L} B) = \mathop{\mathrm{colim}}\nolimits H^ i_{Z'}(K \otimes _ A^\mathbf {L} B)$ where $Z'$ is the inverse image of $Z$. Thus the result because $\otimes _ A B$ commutes with filtered colimits and there are no higher Tors. $\square$
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