## 36.7 The coherator

Let $X$ be a scheme. The coherator is a functor

$Q_ X : \textit{Mod}(\mathcal{O}_ X) \longrightarrow \mathit{QCoh}(\mathcal{O}_ X)$

which is right adjoint to the inclusion functor $\mathit{QCoh}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X)$. It exists for any scheme $X$ and moreover the adjunction mapping $Q_ X(\mathcal{F}) \to \mathcal{F}$ is an isomorphism for every quasi-coherent module $\mathcal{F}$, see Properties, Proposition 28.23.4. Since $Q_ X$ is left exact (as a right adjoint) we can consider its right derived extension

$RQ_ X : D(\mathcal{O}_ X) \longrightarrow D(\mathit{QCoh}(\mathcal{O}_ X)).$

Since $Q_ X$ is right adjoint to the inclusion functor $\mathit{QCoh}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X)$ we see that $RQ_ X$ is right adjoint to the canonical functor $D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathcal{O}_ X)$ by Derived Categories, Lemma 13.30.3.

In this section we will study the functor $RQ_ X$. In Section 36.21 we will study the (closely related) right adjoint to the inclusion functor $D_\mathit{QCoh}(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$ (when it exists).

Lemma 36.7.1. Let $f : X \to Y$ be an affine morphism of schemes. Then $f_*$ defines a derived functor $f_* : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$. This functor has the property that

$\xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[d]_{f_*} \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[d]^{Rf_*} \\ D(\mathit{QCoh}(\mathcal{O}_ Y)) \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ Y) }$

commutes.

Proof. The functor $f_* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ is exact, see Cohomology of Schemes, Lemma 30.2.3. Hence $f_*$ defines a derived functor $f_* : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$ by simply applying $f_*$ to any representative complex, see Derived Categories, Lemma 13.16.9. The diagram commutes by Lemma 36.5.1. $\square$

Lemma 36.7.2. Let $f : X \to Y$ be a morphism of schemes. Assume $f$ is quasi-compact, quasi-separated, and flat. Then, denoting

$\Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$

the right derived functor of $f_* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ we have $RQ_ Y \circ Rf_* = \Phi \circ RQ_ X$.

Proof. We will prove this by showing that $RQ_ Y \circ Rf_*$ and $\Phi \circ RQ_ X$ are right adjoint to the same functor $D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathcal{O}_ X)$.

Since $f$ is quasi-compact and quasi-separated, we see that $f_*$ preserves quasi-coherence, see Schemes, Lemma 26.24.1. Recall that $\mathit{QCoh}(\mathcal{O}_ X)$ is a Grothendieck abelian category (Properties, Proposition 28.23.4). Hence any $K$ in $D(\mathit{QCoh}(\mathcal{O}_ X))$ can be represented by a K-injective complex $\mathcal{I}^\bullet$ of $\mathit{QCoh}(\mathcal{O}_ X)$, see Injectives, Theorem 19.12.6. Then we can define $\Phi (K) = f_*\mathcal{I}^\bullet$.

Since $f$ is flat, the functor $f^*$ is exact. Hence $f^*$ defines $f^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$ and also $f^* : D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$. The functor $f^* = Lf^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$ is left adjoint to $Rf_* : D(\mathcal{O}_ X) \to D(\mathcal{O}_ Y)$, see Cohomology, Lemma 20.28.1. Similarly, the functor $f^* : D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$ is left adjoint to $\Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$ by Derived Categories, Lemma 13.30.3.

Let $A$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ Y))$ and $E$ an object of $D(\mathcal{O}_ X)$. Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ Y))}(A, RQ_ Y(Rf_*E)) & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(A, Rf_*E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(f^*A, E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ X))}(f^*A, RQ_ X(E)) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ Y))}(A, \Phi (RQ_ X(E))) \end{align*}

This implies what we want. $\square$

Lemma 36.7.3. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. Then

1. $Q_ X : \textit{Mod}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ X)$ is the functor which sends $\mathcal{F}$ to the quasi-coherent $\mathcal{O}_ X$-module associated to the $A$-module $\Gamma (X, \mathcal{F})$,

2. $RQ_ X : D(\mathcal{O}_ X) \to D(\mathit{QCoh}(\mathcal{O}_ X))$ is the functor which sends $E$ to the complex of quasi-coherent $\mathcal{O}_ X$-modules associated to the object $R\Gamma (X, E)$ of $D(A)$,

3. restricted to $D_\mathit{QCoh}(\mathcal{O}_ X)$ the functor $RQ_ X$ defines a quasi-inverse to (36.3.0.1).

Proof. The functor $Q_ X$ is the functor

$\mathcal{F} \mapsto \widetilde{\Gamma (X, \mathcal{F})}$

by Schemes, Lemma 26.7.1. This immediately implies (1) and (2). The third assertion follows from (the proof of) Lemma 36.3.5. $\square$

At this point we are ready to prove a criterion for when the functor $D(\mathit{QCoh}(\mathcal{O}_ X)) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ is an equivalence.

Lemma 36.7.4. Let $X$ be a quasi-compact and quasi-separated scheme. Suppose that for every affine open $U \subset X$ the right derived functor

$\Phi : D(\mathit{QCoh}(\mathcal{O}_ U)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$

of the left exact functor $j_* : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_ X)$ fits into a commutative diagram

$\xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ U)) \ar[d]_\Phi \ar[r]_{i_ U} & D_\mathit{QCoh}(\mathcal{O}_ U) \ar[d]^{Rj_*} \\ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[r]^{i_ X} & D_\mathit{QCoh}(\mathcal{O}_ X) }$

Then the functor (36.3.0.1)

$D(\mathit{QCoh}(\mathcal{O}_ X)) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_ X)$

is an equivalence with quasi-inverse given by $RQ_ X$.

Proof. Let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $A$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ X))$. We have to show that the adjunction maps

$RQ_ X(i_ X(A)) \to A \quad \text{and}\quad E \to i_ X(RQ_ X(E))$

are isomorphisms. Consider the hypothesis $H_ n$: the adjunction maps above are isomorphisms whenever $E$ and $i_ X(A)$ are supported (Definition 36.6.1) on a closed subset of $X$ which is contained in the union of $n$ affine opens of $X$. We will prove $H_ n$ by induction on $n$.

Base case: $n = 0$. In this case $E = 0$, hence the map $E \to i_ X(RQ_ X(E))$ is an isomorphism. Similarly $i_ X(A) = 0$. Thus the cohomology sheaves of $i_ X(A)$ are zero. Since the inclusion functor $\mathit{QCoh}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X)$ is fully faithful and exact, we conclude that the cohomology objects of $A$ are zero, i.e., $A = 0$ and $RQ_ X(i_ X(A)) \to A$ is an isomorphism as well.

Induction step. Suppose that $E$ and $i_ X(A)$ are supported on a closed subset $T$ of $X$ contained in $U_1 \cup \ldots \cup U_ n$ with $U_ i \subset X$ affine open. Set $U = U_ n$. Consider the distinguished triangles

$A \to \Phi (A|_ U) \to A' \to A \quad \text{and}\quad E \to Rj_*(E|_ U) \to E' \to E$

where $\Phi$ is as in the statement of the lemma. Note that $E \to Rj_*(E|_ U)$ is a quasi-isomorphism over $U = U_ n$. Since $i_ X \circ \Phi = Rj_* \circ i_ U$ by assumption and since $i_ X(A)|_ U = i_ U(A|_ U)$ we see that $i_ X(A) \to i_ X(\Phi (A|_ U))$ is a quasi-isomorphism over $U$. Hence $i_ X(A')$ and $E'$ are supported on the closed subset $T \setminus U$ of $X$ which is contained in $U_1 \cup \ldots \cup U_{n - 1}$. By induction hypothesis the statement is true for $A'$ and $E'$. By Derived Categories, Lemma 13.4.3 it suffices to prove the maps

$RQ_ X(i_ X(\Phi (A|_ U))) \to \Phi (A|_ U) \quad \text{and}\quad Rj_*(E|_ U) \to i_ X(RQ_ X(Rj_*E|_ U))$

are isomorphisms. By assumption and by Lemma 36.7.2 (the inclusion morphism $j : U \to X$ is flat, quasi-compact, and quasi-separated) we have

$RQ_ X(i_ X(\Phi (A|_ U))) = RQ_ X(Rj_*(i_ U(A|_ U))) = \Phi (RQ_ U(i_ U(A|_ U)))$

and

$i_ X(RQ_ X(Rj_*(E|_ U))) = i_ X(\Phi (RQ_ U(E|_ U))) = Rj_*(i_ U(RQ_ U(E|_ U)))$

Finally, the maps

$RQ_ U(i_ U(A|_ U)) \to A|_ U \quad \text{and}\quad E|_ U \to i_ U(RQ_ U(E|_ U))$

are isomorphisms by Lemma 36.7.3. The result follows. $\square$

Proposition 36.7.5. Let $X$ be a quasi-compact scheme with affine diagonal. Then the functor (36.3.0.1)

$D(\mathit{QCoh}(\mathcal{O}_ X)) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_ X)$

is an equivalence with quasi-inverse given by $RQ_ X$.

Proof. Let $U \subset X$ be an affine open. Then the morphism $U \to X$ is affine by Morphisms, Lemma 29.11.11. Thus the assumption of Lemma 36.7.4 holds by Lemma 36.7.1 and we win. $\square$

Lemma 36.7.6. Let $f : X \to Y$ be a morphism of schemes. Assume $X$ and $Y$ are quasi-compact and have affine diagonal. Then, denoting

$\Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$

the right derived functor of $f_* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ the diagram

$\xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[d]_\Phi \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[d]^{Rf_*} \\ D(\mathit{QCoh}(\mathcal{O}_ Y)) \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ Y) }$

is commutative.

Proof. Observe that the horizontal arrows in the diagram are equivalences of categories by Proposition 36.7.5. Hence we can identify these categories (and similarly for other quasi-compact schemes with affine diagonal). The statement of the lemma is that the canonical map $\Phi (K) \to Rf_*(K)$ is an isomorphism for all $K$ in $D(\mathit{QCoh}(\mathcal{O}_ X))$. Note that if $K_1 \to K_2 \to K_3 \to K_1$ is a distinguished triangle in $D(\mathit{QCoh}(\mathcal{O}_ X))$ and the statement is true for two-out-of-three, then it is true for the third.

Let $U \subset X$ be an affine open. Since the diagonal of $X$ is affine, the inclusion morphism $j : U \to X$ is affine (Morphisms, Lemma 29.11.11). Similarly, the composition $g = f \circ j : U \to Y$ is affine. Let $\mathcal{I}^\bullet$ be a K-injective complex in $\mathit{QCoh}(\mathcal{O}_ U)$. Since $j_* : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_ X)$ has an exact left adjoint $j^* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ U)$ we see that $j_*\mathcal{I}^\bullet$ is a K-injective complex in $\mathit{QCoh}(\mathcal{O}_ X)$, see Derived Categories, Lemma 13.31.9. It follows that

$\Phi (j_*\mathcal{I}^\bullet ) = f_*j_*\mathcal{I}^\bullet = g_*\mathcal{I}^\bullet$

By Lemma 36.7.1 we see that $j_*\mathcal{I}^\bullet$ represents $Rj_*\mathcal{I}^\bullet$ and $g_*\mathcal{I}^\bullet$ represents $Rg_*\mathcal{I}^\bullet$. On the other hand, we have $Rf_* \circ Rj_* = Rg_*$. Hence $f_*j_*\mathcal{I}^\bullet$ represents $Rf_*(j_*\mathcal{I}^\bullet )$. We conclude that the lemma is true for any complex of the form $j_*\mathcal{G}^\bullet$ with $\mathcal{G}^\bullet$ a complex of quasi-coherent modules on $U$. (Note that if $\mathcal{G}^\bullet \to \mathcal{I}^\bullet$ is a quasi-isomorphism, then $j_*\mathcal{G}^\bullet \to j_*\mathcal{I}^\bullet$ is a quasi-isomorphism as well since $j_*$ is an exact functor on quasi-coherent modules.)

Let $\mathcal{F}^\bullet$ be a complex of quasi-coherent $\mathcal{O}_ X$-modules. Let $T \subset X$ be a closed subset such that the support of $\mathcal{F}^ p$ is contained in $T$ for all $p$. We will use induction on the minimal number $n$ of affine opens $U_1, \ldots , U_ n$ such that $T \subset U_1 \cup \ldots \cup U_ n$. The base case $n = 0$ is trivial. If $n \geq 1$, then set $U = U_1$ and denote $j : U \to X$ the open immersion as above. We consider the map of complexes $c : \mathcal{F}^\bullet \to j_*j^*\mathcal{F}^\bullet$. We obtain two short exact sequences of complexes:

$0 \to \mathop{\mathrm{Ker}}(c) \to \mathcal{F}^\bullet \to \mathop{\mathrm{Im}}(c) \to 0$

and

$0 \to \mathop{\mathrm{Im}}(c) \to j_*j^*\mathcal{F}^\bullet \to \mathop{\mathrm{Coker}}(c) \to 0$

The complexes $\mathop{\mathrm{Ker}}(c)$ and $\mathop{\mathrm{Coker}}(c)$ are supported on $T \setminus U \subset U_2 \cup \ldots \cup U_ n$ and the result holds for them by induction. The result holds for $j_*j^*\mathcal{F}^\bullet$ by the discussion in the preceding paragraph. We conclude by looking at the distinguished triangles associated to the short exact sequences and using the initial remark of the proof. $\square$

Remark 36.7.7 (Warning). Let $X$ be a quasi-compact scheme with affine diagonal. Even though we know that $D(\mathit{QCoh}(\mathcal{O}_ X)) = D_\mathit{QCoh}(\mathcal{O}_ X)$ by Proposition 36.7.5 strange things can happen and it is easy to make mistakes with this material. One pitfall is to carelessly assume that this equality means derived functors are the same. For example, suppose we have a quasi-compact open $U \subset X$. Then we can consider the higher right derived functors

$R^ i(\mathit{QCoh})\Gamma (U, -) : \mathit{QCoh}(\mathcal{O}_ X) \to \textit{Ab}$

of the left exact functor $\Gamma (U, -)$. Since this is a universal $\delta$-functor, and since the functors $H^ i(U, -)$ (defined for all abelian sheaves on $X$) restricted to $\mathit{QCoh}(\mathcal{O}_ X)$ form a $\delta$-functor, we obtain canonical tranformations

$t^ i : R^ i(\mathit{QCoh})\Gamma (U, -) \to H^ i(U, -).$

These transformations aren't in general isomorphisms even if $X = \mathop{\mathrm{Spec}}(A)$ is affine! Namely, we have $R^1(\mathit{QCoh})\Gamma (U, \widetilde{I}) = 0$ if $I$ an injective $A$-module by construction of right derived functors and the equivalence of $\mathit{QCoh}(\mathcal{O}_ X)$ and $\text{Mod}_ A$. But Examples, Lemma 109.46.2 shows there exists $A$, $I$, and $U$ such that $H^1(U, \widetilde{I}) \not= 0$.

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