Lemma 36.22.5. Let g : S' \to S be a morphism of schemes. Let f : X \to S be quasi-compact and quasi-separated. Consider the base change diagram
\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }
If X and S' are Tor independent over S, then for all E \in D_\mathit{QCoh}(\mathcal{O}_ X) we have Rf'_*L(g')^*E = Lg^*Rf_*E.
Proof.
For any object E of D(\mathcal{O}_ X) we can use Cohomology, Remark 20.28.3 to get a canonical base change map Lg^*Rf_*E \to Rf'_*L(g')^*E. To check this is an isomorphism we may work locally on S'. Hence we may assume g : S' \to S is a morphism of affine schemes. In particular, g is affine and it suffices to show that
Rg_*Lg^*Rf_*E \to Rg_*Rf'_*L(g')^*E = Rf_*(Rg'_* L(g')^* E)
is an isomorphism, see Lemma 36.5.2 (and use Lemmas 36.3.8, 36.3.9, and 36.4.1 to see that the objects Rf'_*L(g')^*E and Lg^*Rf_*E have quasi-coherent cohomology sheaves). Note that g' is affine as well (Morphisms, Lemma 29.11.8). By Lemma 36.5.3 the map becomes a map
Rf_*E \otimes _{\mathcal{O}_ S}^\mathbf {L} g_*\mathcal{O}_{S'} \longrightarrow Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'})
Observe that g'_*\mathcal{O}_{X'} = f^*g_*\mathcal{O}_{S'} (by affine base change, see Cohomology of Schemes, Lemma 30.5.1). Thus by Lemma 36.22.1 it suffices to prove that Lf^*g_*\mathcal{O}_{S'} = f^*g_*\mathcal{O}_{S'}. This follows from our assumption that X and S' are Tor independent over S. Namely, to check it we may work locally on X, hence we may also assume X is affine. Say X = \mathop{\mathrm{Spec}}(A), S = \mathop{\mathrm{Spec}}(R) and S' = \mathop{\mathrm{Spec}}(R'). Our assumption implies that A and R' are Tor independent over R (More on Algebra, Lemma 15.61.6), i.e., \text{Tor}_ i^ R(A, R') = 0 for i > 0. In other words A \otimes _ R^\mathbf {L} R' = A \otimes _ R R' which exactly means that Lf^*g_*\mathcal{O}_{S'} = f^*g_*\mathcal{O}_{S'} (use Lemma 36.3.8).
\square
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