The Stacks project

36.22 Cohomology and base change, IV

This section continues the discussion of Cohomology of Schemes, Section 30.22. First, we have a very general version of the projection formula for quasi-compact and quasi-separated morphisms of schemes and complexes with quasi-coherent cohomology sheaves.

Lemma 36.22.1. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of schemes. For $E$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and $K$ in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ the map

\[ Rf_*(E) \otimes _{\mathcal{O}_ Y}^\mathbf {L} K \longrightarrow Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*K) \]

defined in Cohomology, Equation (20.51.2.1) is an isomorphism.

Proof. To check the map is an isomorphism we may work locally on $Y$. Hence we reduce to the case that $Y$ is affine.

Suppose that $K = \bigoplus K_ i$ is a direct sum of some complexes $K_ i \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. If the statement holds for each $K_ i$, then it holds for $K$. Namely, the functors $Lf^*$ and $\otimes ^\mathbf {L}$ preserve direct sums by construction and $Rf_*$ commutes with direct sums (for complexes with quasi-coherent cohomology sheaves) by Lemma 36.4.5. Moreover, suppose that $K \to L \to M \to K[1]$ is a distinguished triangle in $D_\mathit{QCoh}(Y)$. Then if the statement of the lemma holds for two of $K, L, M$, then it holds for the third (as the functors involved are exact functors of triangulated categories).

Assume $Y$ affine, say $Y = \mathop{\mathrm{Spec}}(A)$. The functor $\widetilde{\ } : D(A) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ is an equivalence (Lemma 36.3.5). Let $T$ be the property for $K \in D(A)$ that the statement of the lemma holds for $\widetilde{K}$. The discussion above and More on Algebra, Remark 15.58.13 shows that it suffices to prove $T$ holds for $A[k]$. This finishes the proof, as the statement of the lemma is clear for shifts of the structure sheaf. $\square$

Definition 36.22.2. Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. We say $X$ and $Y$ are Tor independent over $S$ if for every $x \in X$ and $y \in Y$ mapping to the same point $s \in S$ the rings $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ are Tor independent over $\mathcal{O}_{S, s}$ (see More on Algebra, Definition 15.60.1).

Lemma 36.22.3. Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes. The following are equivalent

  1. $X$ and $Y$ are tor independent over $S$, and

  2. for every affine opens $U \subset X$, $V \subset Y$, $W \subset S$ with $f(U) \subset W$ and $g(V) \subset W$ the rings $\mathcal{O}_ X(U)$ and $\mathcal{O}_ Y(V)$ are tor independent over $\mathcal{O}_ S(W)$.

  3. there exists an affine open overing $S = \bigcup W_ i$ and for each $i$ affine open coverings $f^{-1}(W_ i) = \bigcup U_{ij}$ and $g^{-1}(W_ i) = \bigcup V_{ik}$ such that the rings $\mathcal{O}_ X(U_{ij})$ and $\mathcal{O}_ Y(V_{ik})$ are tor independent over $\mathcal{O}_ S(W_ i)$ for all $i, j, k$.

Proof. Omitted. Hint: use More on Algebra, Lemma 15.60.6. $\square$

Lemma 36.22.4. Let $X \to S$ and $Y \to S$ be morphisms of schemes. Let $S' \to S$ be a morphism of schemes and denote $X' = X \times _ S S'$ and $Y' = Y \times _ S S'$. If $X$ and $Y$ are tor independent over $S$ and $S' \to S$ is flat, then $X'$ and $Y'$ are tor independent over $S'$.

Proof. Omitted. Hint: use Lemma 36.22.3 and on affine opens use More on Algebra, Lemma 15.60.4. $\square$

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 15.60.6), 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). $\square$

The following lemma will be used in the chapter on dualizing complexes.

Lemma 36.22.6. Consider a cartesian square

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

of quasi-compact and quasi-separated schemes. Assume $g$ and $f$ Tor independent and $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$ affine. For $M, K \in D(\mathcal{O}_ X)$ the canonical map

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(M, K) \otimes ^\mathbf {L}_ R R' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*K) \]

in $D(R')$ is an isomorphism in the following two cases

  1. $M \in D(\mathcal{O}_ X)$ is perfect and $K \in D_\mathit{QCoh}(X)$, or

  2. $M \in D(\mathcal{O}_ X)$ is pseudo-coherent, $K \in D_\mathit{QCoh}^+(X)$, and $R'$ has finite tor dimension over $R$.

Proof. There is a canonical map $R\mathop{\mathrm{Hom}}\nolimits _ X(M, K) \to R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*K)$ in $D(\Gamma (X, \mathcal{O}_ X))$ of global hom complexes, see Cohomology, Section 20.41. Restricting scalars we can view this as a map in $D(R)$. Then we can use the adjointness of restriction and $- \otimes _ R^\mathbf {L} R'$ to get the displayed map of the lemma. Having defined the map it suffices to prove it is an isomorphism in the derived category of abelian groups.

The right hand side is equal to

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(M, R(g')_*L(g')^*K) = R\mathop{\mathrm{Hom}}\nolimits _ X(M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

by Lemma 36.5.3. In both cases the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K)$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.10.8. There is a natural map

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K) \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

which is an isomorphism in both cases by Lemma 36.10.9. To see that this lemma applies in case (2) we note that $g'_*\mathcal{O}_{X'} = Rg'_*\mathcal{O}_{X'} = Lf^*g_*\mathcal{O}_ X$ the second equality by Lemma 36.22.5. Using Lemma 36.10.4 and Cohomology, Lemma 20.45.4 we conclude that $g'_*\mathcal{O}_{X'}$ has finite Tor dimension. Hence, in both cases by replacing $K$ by $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K)$ we reduce to proving

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

is an isomorphism. Note that the left hand side is equal to $R\Gamma (X', L(g')^*K)$ by Lemma 36.5.3. Hence the result follows from Lemma 36.22.5. $\square$

Remark 36.22.7. With notation as in Lemma 36.22.6. The diagram

\[ \xymatrix{ R\mathop{\mathrm{Hom}}\nolimits _ X(M, Rg'_*L) \otimes _ R^\mathbf {L} R' \ar[r] \ar[d]_\mu & R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*Rg'_*L) \ar[d]^ a \\ R\mathop{\mathrm{Hom}}\nolimits _ X(M, R(g')_*L) \ar@{=}[r] & R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L) } \]

is commutative where the top horizontal arrow is the map from the lemma, $\mu $ is the multiplication map, and $a$ comes from the adjunction map $L(g')^*Rg'_*L \to L$. The multiplication map is the adjunction map $K' \otimes _ R^\mathbf {L} R' \to K'$ for any $K' \in D(R')$.

Lemma 36.22.8. Consider a cartesian square of schemes

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

Assume $g$ and $f$ Tor independent.

  1. If $E \in D(\mathcal{O}_ X)$ has tor amplitude in $[a, b]$ as a complex of $f^{-1}\mathcal{O}_ S$-modules, then $L(g')^*E$ has tor amplitude in $[a, b]$ as a complex of $f^{-1}\mathcal{O}_{S'}$-modules.

  2. If $\mathcal{G}$ is an $\mathcal{O}_ X$-module flat over $S$, then $L(g')^*\mathcal{G} = (g')^*\mathcal{G}$.

Proof. We can compute tor dimension at stalks, see Cohomology, Lemma 20.45.5. If $x' \in X'$ with image $x \in X$, then

\[ (L(g')^*E)_{x'} = E_ x \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X', x'} \]

Let $s' \in S'$ and $s \in S$ be the image of $x'$ and $x$. Since $X$ and $S'$ are tor independent over $S$, we can apply More on Algebra, Lemma 15.60.2 to see that the right hand side of the displayed formula is equal to $E_ x \otimes _{\mathcal{O}_{S, s}}^\mathbf {L} \mathcal{O}_{S', s'}$ in $D(\mathcal{O}_{S', s'})$. Thus (1) follows from More on Algebra, Lemma 15.65.13. To see (2) observe that flatness of $\mathcal{G}$ is equivalent to the condition that $\mathcal{G}[0]$ has tor amplitude in $[0, 0]$. Applying (1) we conclude. $\square$

Lemma 36.22.9. Consider a cartesian diagram of schemes

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

where $i$ is a closed immersion. If $Z$ and $X'$ are tor independent over $X$, then $Ri'_* \circ Lg^* = Lf^* \circ Ri_*$ as functors $D(\mathcal{O}_ Z) \to D(\mathcal{O}_{X'})$.

Proof. Note that the lemma is supposed to hold for all $K \in D(\mathcal{O}_ Z)$. Observe that $i_*$ and $i'_*$ are exact functors and hence $Ri_*$ and $Ri'_*$ are computed by applying $i_*$ and $i'_*$ to any representatives. Thus the base change map

\[ Lf^*(Ri_*(K)) \longrightarrow Ri'_*(Lg^*(K)) \]

on stalks at a point $z' \in Z'$ with image $z \in Z$ is given by

\[ K_ z \otimes _{\mathcal{O}_{X, z}}^\mathbf {L} \mathcal{O}_{X', z'} \longrightarrow K_ z \otimes _{\mathcal{O}_{Z, z}}^\mathbf {L} \mathcal{O}_{Z', z'} \]

This map is an isomorphism by More on Algebra, Lemma 15.60.2 and the assumed tor independence. $\square$


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