Lemma 35.10.2. Let $S$ be a scheme. Let $\tau \in \{ Zariski, \linebreak {\acute{e}tale}, \linebreak smooth, \linebreak syntomic, \linebreak fppf\}$. The functors

$\mathit{QCoh}(\mathcal{O}_ S) \longrightarrow \textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O}) \quad \text{and}\quad \mathit{QCoh}(\mathcal{O}_ S) \longrightarrow \textit{Mod}(S_\tau , \mathcal{O})$

defined by the rule $\mathcal{F} \mapsto \mathcal{F}^ a$ seen in Proposition 35.8.9 are

1. fully faithful,

2. compatible with direct sums,

3. compatible with colimits,

4. right exact,

5. exact as a functor $\mathit{QCoh}(\mathcal{O}_ S) \to \textit{Mod}(S_{\acute{e}tale}, \mathcal{O})$,

6. not exact as a functor $\mathit{QCoh}(\mathcal{O}_ S) \to \textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O})$ in general,

7. given two quasi-coherent $\mathcal{O}_ S$-modules $\mathcal{F}$, $\mathcal{G}$ we have $(\mathcal{F} \otimes _{\mathcal{O}_ S} \mathcal{G})^ a = \mathcal{F}^ a \otimes _\mathcal {O} \mathcal{G}^ a$,

8. if $\tau = {\acute{e}tale}$ or $\tau = Zariski$, given two quasi-coherent $\mathcal{O}_ S$-modules $\mathcal{F}$, $\mathcal{G}$ such that $\mathcal{F}$ is of finite presentation we have $(\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, \mathcal{G}))^ a = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}^ a, \mathcal{G}^ a)$ in $\textit{Mod}(S_\tau , \mathcal{O})$,

9. given two quasi-coherent $\mathcal{O}_ S$-modules $\mathcal{F}$, $\mathcal{G}$ we do not have $(\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, \mathcal{G}))^ a = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}^ a, \mathcal{G}^ a)$ in $\textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O})$ in general even if $\mathcal{F}$ is of finite presentation, and

10. given a short exact sequence $0 \to \mathcal{F}_1^ a \to \mathcal{E} \to \mathcal{F}_2^ a \to 0$ of $\mathcal{O}$-modules then $\mathcal{E}$ is quasi-coherent1, i.e., $\mathcal{E}$ is in the essential image of the functor.

Proof. Part (1) we saw in Proposition 35.8.9.

We have seen in Schemes, Section 26.24 that a colimit of quasi-coherent sheaves on a scheme is a quasi-coherent sheaf. Moreover, in Remark 35.8.6 we saw that $\mathcal{F} \mapsto \mathcal{F}^ a$ is the pullback functor for a morphism of ringed sites, hence commutes with all colimits, see Modules on Sites, Lemma 18.14.3. Thus (3) and its special case (3) hold.

This also shows that the functor is right exact (i.e., commutes with finite colimits), hence (4).

The functor $\mathit{QCoh}(\mathcal{O}_ S) \to \mathit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$, $\mathcal{F} \mapsto \mathcal{F}^ a$ is left exact because an étale morphism is flat, see Morphisms, Lemma 29.36.12. This proves (5).

To see (6), suppose that $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Then $2 : \mathcal{O}_ S \to \mathcal{O}_ S$ is injective but the associated map of $\mathcal{O}$-modules on $(\mathit{Sch}/S)_\tau$ isn't injective because $2 : \mathbf{F}_2 \to \mathbf{F}_2$ isn't injective and $\mathop{\mathrm{Spec}}(\mathbf{F}_2)$ is an object of $(\mathit{Sch}/S)_\tau$.

Part (7) holds because, as mentioned above, the functor $\mathcal{F} \mapsto \mathcal{F}^ a$ is the pullback functor for a morphism of ringed sites and such commute with tensor products by Modules on Sites, Lemma 18.26.2.

Part (8) is obvious if $\tau = Zariski$ because the category of $\mathcal{O}$-modules on $S_{Zar}$ is the same as the category of $\mathcal{O}_ S$-modules on the topological space $S$. If $\tau = {\acute{e}tale}$ then (8) holds because, as mentioned above, the functor $\mathcal{F} \mapsto \mathcal{F}^ a$ is the pullback functor for the flat morphism of ringed sites $(S_{\acute{e}tale}, \mathcal{O}) \to (S_{Zar}, \mathcal{O}_ S)$, see Lemma 35.10.1. Pullback by flat morphisms of ringed sites commutes with taking internal hom out of a finitely presented module by Modules on Sites, Lemma 18.31.4.

To see (9), suppose that $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Let $\mathcal{F} = \mathop{\mathrm{Coker}}(2 : \mathcal{O}_ S \to \mathcal{O}_ S)$ and $\mathcal{G} = \mathcal{O}_ S$. Then $\mathcal{F}^ a = \mathop{\mathrm{Coker}}(2 : \mathcal{O} \to \mathcal{O})$ and $\mathcal{G}^ a = \mathcal{O}$. Hence $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}^ a, \mathcal{G}^ a) = \mathcal{O}$ is equal to the $2$-torsion in $\mathcal{O}$, which is not zero, see proof of (6). On the other hand, the module $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, \mathcal{G})$ is zero.

Proof of (10). Let $0 \to \mathcal{F}_1^ a \to \mathcal{E} \to \mathcal{F}_2^ a \to 0$ be a short exact sequence of $\mathcal{O}$-modules with $\mathcal{F}_1$ and $\mathcal{F}_2$ quasi-coherent on $S$. Consider the restriction

$0 \to \mathcal{F}_1 \to \mathcal{E}|_{S_{Zar}} \to \mathcal{F}_2$

to $S_{Zar}$. By Proposition 35.9.3 we see that on any affine $U \subset S$ we have $H^1(U, \mathcal{F}_1^ a) = H^1(U, \mathcal{F}_1) = 0$. Hence the sequence above is also exact on the right. By Schemes, Section 26.24 we conclude that $\mathcal{F} = \mathcal{E}|_{S_{Zar}}$ is quasi-coherent. Thus we obtain a commutative diagram

$\xymatrix{ & \mathcal{F}_1^ a \ar[r] \ar[d] & \mathcal{F}^ a \ar[r] \ar[d] & \mathcal{F}_2^ a \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_1^ a \ar[r] & \mathcal{E} \ar[r] & \mathcal{F}_2^ a \ar[r] & 0 }$

To finish the proof it suffices to show that the top row is also right exact. To do this, denote once more $U = \mathop{\mathrm{Spec}}(A) \subset S$ an affine open of $S$. We have seen above that $0 \to \mathcal{F}_1(U) \to \mathcal{E}(U) \to \mathcal{F}_2(U) \to 0$ is exact. For any affine scheme $V/U$, $V = \mathop{\mathrm{Spec}}(B)$ the map $\mathcal{F}_1^ a(V) \to \mathcal{E}(V)$ is injective. We have $\mathcal{F}_1^ a(V) = \mathcal{F}_1(U) \otimes _ A B$ by definition. The injection $\mathcal{F}_1^ a(V) \to \mathcal{E}(V)$ factors as

$\mathcal{F}_1(U) \otimes _ A B \to \mathcal{E}(U) \otimes _ A B \to \mathcal{E}(U)$

Considering $A$-algebras $B$ of the form $B = A \oplus M$ we see that $\mathcal{F}_1(U) \to \mathcal{E}(U)$ is universally injective (see Algebra, Definition 10.82.1). Since $\mathcal{E}(U) = \mathcal{F}(U)$ we conclude that $\mathcal{F}_1 \to \mathcal{F}$ remains injective after any base change, or equivalently that $\mathcal{F}_1^ a \to \mathcal{F}^ a$ is injective. $\square$

 Warning: This is misleading. See part (6).

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