## 36.26 Cohomology and base change, V

In Section 36.22 we saw a base change theorem holds when the morphisms are tor independent. Even in the affine case there cannot be a base change theorem without such a condition, see More on Algebra, Section 15.60. In this section we analyze when one can get a base change result “one complex at a time”.

To make this work, suppose we have a commutative diagram

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

of schemes (usually we will assume it is cartesian). Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. For a point $x' \in X'$ set $x = g'(x') \in X$, $s' = f'(x') \in S'$ and $s = f(x) = g(s')$. Then we can consider the maps

\[ K_ x \otimes _{\mathcal{O}_{S, s}}^\mathbf {L} \mathcal{O}_{S', s'} \to K_ x \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X', x'} \to K'_{x'} \]

where the first arrow is More on Algebra, Equation (15.60.0.1) and the second comes from $(L(g')^*K)_{x'} = K_ x \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X', x'}$ and the given map $L(g')^*K \to K'$. For each $i \in \mathbf{Z}$ we obtain a $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S', s'}$-module structure on $H^ i(K_ x \otimes _{\mathcal{O}_{S, s}}^\mathbf {L} \mathcal{O}_{S', s'})$. Putting everything together we obtain canonical maps

36.26.0.1
\begin{equation} \label{perfect-equation-bc} H^ i(K_ x \otimes _{\mathcal{O}_{S, s}}^\mathbf {L} \mathcal{O}_{S', s'}) \otimes _{(\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S', s'})} \mathcal{O}_{X', x'} \longrightarrow H^ i(K'_{x'}) \end{equation}

of $\mathcal{O}_{X', x'}$-modules.

Lemma 36.26.1. Let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

be a cartesian diagram of schemes. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. The following are equivalent

for any $x' \in X'$ and $i \in \mathbf{Z}$ the map (36.26.0.1) is an isomorphism,

for $U \subset X$, $V' \subset S'$ affine open both mapping into the affine open $V \subset S$ with $U' = V' \times _ V U$ the composition

\[ R\Gamma (U, K) \otimes _{\mathcal{O}_ S(U)}^\mathbf {L} \mathcal{O}_{S'}(V') \to R\Gamma (U, K) \otimes _{\mathcal{O}_ X(U)}^\mathbf {L} \mathcal{O}_{X'}(U') \to R\Gamma (U', K') \]

is an isomorphism in $D(\mathcal{O}_{S'}(V'))$, and

there is a set $I$ of quadruples $U_ i, V_ i', V_ i, U_ i'$, $i \in I$ as in (2) with $X' = \bigcup U'_ i$.

**Proof.**
The second arrow in (2) comes from the equality

\[ R\Gamma (U, K) \otimes _{\mathcal{O}_ X(U)}^\mathbf {L} \mathcal{O}_{X'}(U') = R\Gamma (U', L(g')^*K) \]

of Lemma 36.3.8 and the given arrow $L(g')^*K \to K'$. The first arrow of (2) is More on Algebra, Equation (15.60.0.1). It is clear that (2) implies (3). Observe that (1) is local on $X'$. Therefore it suffices to show that if $X$, $S$, $S'$, $X'$ are affine, then (1) is equivalent to the condition that

\[ R\Gamma (X, K) \otimes _{\mathcal{O}_ S(S)}^\mathbf {L} \mathcal{O}_{S'}(S') \to R\Gamma (X, K) \otimes _{\mathcal{O}_ X(X)}^\mathbf {L} \mathcal{O}_{X'}(X') \to R\Gamma (X', K') \]

is an isomorphism in $D(\mathcal{O}_{S'}(S'))$. Say $S = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$, $S' = \mathop{\mathrm{Spec}}(R')$, $X' = \mathop{\mathrm{Spec}}(A')$, $K$ corresponds to the complex $M^\bullet $ of $A$-modules, and $K'$ corresponds to the complex $N^\bullet $ of $A'$-modules. Note that $A' = A \otimes _ R R'$. The condition above is that the composition

\[ M^\bullet \otimes _ R^\mathbf {L} R' \to M^\bullet \otimes _ A^\mathbf {L} A' \to N^\bullet \]

is an isomorphism in $D(R')$. Equivalently, it is that for all $i \in \mathbf{Z}$ the map

\[ H^ i(M^\bullet \otimes _ R^\mathbf {L} R') \to H^ i(M^\bullet \otimes _ A^\mathbf {L} A') \to H^ i(N^\bullet ) \]

is an isomorphism. Observe that this is a map of $A \otimes _ R R'$-modules, i.e., of $A'$-modules. On the other hand, (1) is the requirement that for compatible primes $\mathfrak q' \subset A'$, $\mathfrak q \subset A$, $\mathfrak p' \subset R'$, $\mathfrak p \subset R$ the composition

\[ H^ i(M^\bullet _\mathfrak q \otimes _{R_\mathfrak p}^\mathbf {L} R'_{\mathfrak p'}) \otimes _{(A_\mathfrak q \otimes _{R_\mathfrak p} R'_{\mathfrak p'})} A'_{\mathfrak q'} \to H^ i(M^\bullet _{\mathfrak q} \otimes _{A_\mathfrak q}^\mathbf {L} A'_{\mathfrak q'}) \to H^ i(N^\bullet _{\mathfrak q'}) \]

is an isomorphism. Since

\[ H^ i(M^\bullet _\mathfrak q \otimes _{R_\mathfrak p}^\mathbf {L} R'_{\mathfrak p'}) \otimes _{(A_\mathfrak q \otimes _{R_\mathfrak p} R'_{\mathfrak p'})} A'_{\mathfrak q'} = H^ i(M^\bullet \otimes _ R^\mathbf {L} R') \otimes _{A'} A'_{\mathfrak q'} \]

is the localization at $\mathfrak q'$, we see that these two conditions are equivalent by Algebra, Lemma 10.23.1.
$\square$

Lemma 36.26.2. Let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

be a cartesian diagram of schemes. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. If

the equivalent conditions of Lemma 36.26.1 hold, and

$f$ is quasi-compact and quasi-separated,

then the composition $Lg^*Rf_*K \to Rf'_*L(g')^*K \to Rf'_*K'$ is an isomorphism.

**Proof.**
We could prove this using the same method as in the proof of Lemma 36.22.5 but instead we will prove it using the induction principle and relative Mayer-Vietoris.

To check the map 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 $X$ is a quasi-compact and quasi-separated scheme. We will use the induction principle of Cohomology of Schemes, Lemma 30.4.1 to prove that for any quasi-compact open $U \subset X$ the similarly constructed map $Lg^*R(U \to S)_*K|_ U \to R(U' \to S')_*K'|_{U'}$ is an isomorphism. Here $U' = (g')^{-1}(U)$.

If $U \subset X$ is an affine open, then we find that the result is true by assumption, see Lemma 36.26.1 part (2) and the translation into algebra afforded to us by Lemmas 36.3.5 and 36.3.8.

The induction step. Suppose that $X = U \cup V$ is an open covering with $U$, $V$, $U \cap V$ quasi-compact such that the result holds for $U$, $V$, and $U \cap V$. Denote $a = f|_ U$, $b = f|_ V$ and $c = f|_{U \cap V}$. Let $a' : U' \to S'$, $b' : V' \to S'$ and $c' : U' \cap V' \to S'$ be the base changes of $a$, $b$, and $c$. Using the distinguished triangles from relative Mayer-Vietoris (Cohomology, Lemma 20.33.5) we obtain a commutative diagram

\[ \xymatrix{ Lg^*Rf_*K \ar[r] \ar[d] & Rf'_* K' \ar[d] \\ Lg^*Ra_* K|_ U \oplus Lg^*Rb_* K|_ V \ar[r] \ar[d] & Ra'_* K'|_{U'} \oplus Rb'_* K'|_{V'} \ar[d] \\ Lg^*Rc_* K|_{U \cap V} \ar[r] \ar[d] & Rc'_* K'|_{U' \cap V'} \ar[d] \\ Lg^*Rf_* K[1] \ar[r] & Rf'_* K'[1] } \]

Since the 2nd and 3rd horizontal arrows are isomorphisms so is the first (Derived Categories, Lemma 13.4.3) and the proof of the lemma is finished.
$\square$

Lemma 36.26.3. Let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

be a cartesian diagram of schemes. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. If the equivalent conditions of Lemma 36.26.1 hold, then

for $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ the equivalent conditions of Lemma 36.26.1 hold for $L(g')^*(E \otimes ^\mathbf {L} K) \to L(g')^*E \otimes ^\mathbf {L} K'$,

if $E$ in $D(\mathcal{O}_ X)$ is perfect the equivalent conditions of Lemma 36.26.1 hold for $L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K')$, and

if $K$ is bounded below and $E$ in $D(\mathcal{O}_ X)$ pseudo-coherent the equivalent conditions of Lemma 36.26.1 hold for $L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K')$.

**Proof.**
The statement makes sense as the complexes involved have quasi-coherent cohomology sheaves by Lemmas 36.3.8, 36.3.9, and 36.10.8 and Cohomology, Lemmas 20.44.3 and 20.46.6. Having said this, we can check the maps (36.26.0.1) are isomorphisms in case (1) by computing the source and target of (36.26.0.1) using the transitive property of tensor product, see More on Algebra, Lemma 15.58.17. The map in (2) and (3) is the composition

\[ L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, L(g')^*K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K') \]

where the first arrow is Cohomology, Remark 20.39.13 and the second arrow comes from the given map $L(g')^*K \to K'$. To prove the maps (36.26.0.1) are isomorphisms one represents $E_ x$ by a bounded complex of finite projective $\mathcal{O}_{X. x}$-modules in case (2) or by a bounded above complex of finite free modules in case (3) and computes the source and target of the arrow. Some details omitted.
$\square$

Lemma 36.26.4. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $\mathcal{G}^\bullet $ be a bounded above complex of quasi-coherent $\mathcal{O}_ X$-modules flat over $S$. Then formation of

\[ Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet ) \]

commutes with arbitrary base change (see proof for precise statement).

**Proof.**
The statement means the following. Let $g : S' \to S$ be a morphism of schemes and consider the base change diagram

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

in other words $X' = S' \times _ S X$. The lemma asserts that

\[ Lg^*Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet ) \longrightarrow Rf'_*\left( L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{X'}} (g')^*\mathcal{G}^\bullet \right) \]

is an isomorphism. Observe that on the right hand side we do **not** use the derived pullback on $\mathcal{G}^\bullet $. To prove this, we apply Lemmas 36.26.2 and 36.26.3 to see that it suffices to prove the canonical map

\[ L(g')^*\mathcal{G}^\bullet \to (g')^*\mathcal{G}^\bullet \]

satisfies the equivalent conditions of Lemma 36.26.1. This follows by checking the condition on stalks, where it immediately follows from the fact that $\mathcal{G}^\bullet _ x \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S', s'}$ computes the derived tensor product by our assumptions on the complex $\mathcal{G}^\bullet $.
$\square$

Lemma 36.26.5. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. Let $E$ be an object of $D(\mathcal{O}_ X)$. Let $\mathcal{G}^\bullet $ be a complex of quasi-coherent $\mathcal{O}_ X$-modules. If

$E$ is perfect, $\mathcal{G}^\bullet $ is a bounded above, and $\mathcal{G}^ n$ is flat over $S$, or

$E$ is pseudo-coherent, $\mathcal{G}^\bullet $ is bounded, and $\mathcal{G}^ n$ is flat over $S$,

then formation of

\[ Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet ) \]

commutes with arbitrary base change (see proof for precise statement).

**Proof.**
The statement means the following. Let $g : S' \to S$ be a morphism of schemes and consider the base change diagram

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

in other words $X' = S' \times _ S X$. The lemma asserts that

\[ Lg^*Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet ) \longrightarrow R(f')_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, (g')^*\mathcal{G}^\bullet ) \]

is an isomorphism. Observe that on the right hand side we do **not** use the derived pullback on $\mathcal{G}^\bullet $. To prove this, we apply Lemmas 36.26.2 and 36.26.3 to see that it suffices to prove the canonical map

\[ L(g')^*\mathcal{G}^\bullet \to (g')^*\mathcal{G}^\bullet \]

satisfies the equivalent conditions of Lemma 36.26.1. This was shown in the proof of Lemma 36.26.4.
$\square$

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