# The Stacks Project

## Tag 08IB

Lemma 35.21.3. 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 $E \in D_\textit{QCoh}(\mathcal{O}_X)$ we have $Rf'_*L(g')^*E = Lg^*Rf_*E$.

Proof. For any object $E$ of $D(\mathcal{O}_X)$ we can use Cohomology, Remark 20.29.3 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 35.5.1 (and use Lemmas 35.3.8, 35.3.9, and 35.4.1 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 28.11.8). By Lemma 35.5.2 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 35.21.1 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 $X = \mathop{\rm Spec}(A)$, $S = \mathop{\rm Spec}(R)$ and $S' = \mathop{\rm Spec}(R')$. Our assumption implies that $A$ and $R'$ are Tor independent over $R$ (More on Algebra, Lemma 15.57.5), 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 35.3.8). $\square$

The code snippet corresponding to this tag is a part of the file perfect.tex and is located in lines 4240–4256 (see updates for more information).

\begin{lemma}
\label{lemma-compare-base-change}
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
$E \in D_\QCoh(\mathcal{O}_X)$ we have
$Rf'_*L(g')^*E = Lg^*Rf_*E$.
\end{lemma}

\begin{proof}
For any object $E$ of $D(\mathcal{O}_X)$ we can use
Cohomology, Remark \ref{cohomology-remark-base-change} 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 \ref{lemma-affine-morphism}
(and use Lemmas \ref{lemma-quasi-coherence-pullback},
\ref{lemma-quasi-coherence-tensor-product}, and
\ref{lemma-quasi-coherence-direct-image}
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 \ref{morphisms-lemma-base-change-affine}).
By Lemma \ref{lemma-affine-morphism-pull-push} 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 \ref{lemma-cohomology-base-change} 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 $X = \Spec(A)$, $S = \Spec(R)$ and $S' = \Spec(R')$. Our assumption
implies that $A$ and $R'$ are Tor independent over $R$
(More on Algebra, Lemma \ref{more-algebra-lemma-tor-independent}), 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 \ref{lemma-quasi-coherence-pullback}).
\end{proof}

Comment #2600 by Kiran Kedlaya on June 5, 2017 a 9:50 pm UTC

In the first displayed equation in the proof, there is an occurrence of $Rh_*$ which should be $Rg'_*$.

Comment #2625 by Johan (site) on July 7, 2017 a 12:32 pm UTC

Thanks, fixed here.

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