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 $g'_*\mathcal{O}_{X'} = f^*g_*\mathcal{O}_{S'}$. 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).
$\square$

## Comments (3)

Comment #2600 by Kiran Kedlaya on

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