The Stacks project

Proof. Let us denote $\epsilon = \text{id}_{small, {\acute{e}tale}, Zar}$ and $\mathcal{O}_{\acute{e}tale}$ and $\mathcal{O}_{Zar}$ the structure sheaves on $S_{\acute{e}tale}$ and $S_{Zar}$. We have to show that $\mathcal{O}_{\acute{e}tale}$ is a flat $\epsilon ^{-1}\mathcal{O}_{Zar}$-module. Recall that étale morphisms are open, see Morphisms, Lemma 29.36.13. It follows (from the construction of pullback on sheaves) that $\epsilon ^{-1}\mathcal{O}_{Zar}$ is the sheafification of the presheaf $\mathcal{O}'$ on $S_{\acute{e}tale}$ which sends an étale morphism $f : V \to S$ to $\mathcal{O}_ S(f(V))$. If both $V$ and $U = f(V) \subset S$ are affine, then $V \to U$ is an étale morphism of affines, hence corresponds to an étale ring map. Since étale ring maps are flat, we see that $\mathcal{O}_ S(U) = \mathcal{O}'(V) \to \mathcal{O}_{\acute{e}tale}(V) = \mathcal{O}_ V(V)$ is flat. Finally, for every étale morphism $f : V \to S$, i.e., object of $S_{\acute{e}tale}$, there is an affine open covering $V = \bigcup V_ i$ such that $f(V_ i)$ is an affine open in $S$ for all $i$¹. Thus the result by Modules on Sites, Lemma 18.28.4. $\square$

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 (2) hold.

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

The functor $\mathit{QCoh}(\mathcal{O}_ S) \to \textit{Mod}(S_{\acute{e}tale}, \mathcal{O})$, $\mathcal{F} \mapsto \mathcal{F}^ a$ 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 $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}^ a, \mathcal{G}^ a) = \mathcal{O}[2]$ 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

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

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

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. $\square$

Proof. The corresponding facts hold for quasi-coherent modules on the scheme $S$, see Schemes, Section 26.24. The proof will be to use Lemma 35.10.2 to transfer these truths to $S_{\acute{e}tale}$.

Proof of (1). Let $\mathcal{F}_ i$, $i \in I$ be a family of objects of $\mathit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$. Write $\mathcal{F}_ i = \mathcal{G}_ i^ a$ for some quasi-coherent modules $\mathcal{G}_ i$ on $S$. Then $\bigoplus \mathcal{F}_ i = (\bigoplus \mathcal{G}_ i)^ a$ by the lemma cited and we conclude.

Proof of (2). Let $\mathcal{I} \to \mathit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$, $i \mapsto \mathcal{F}_ i$ be a diagram. Write $\mathcal{F}_ i = \mathcal{G}_ i^ a$ so we get a diagram $\mathcal{I} \to \mathit{QCoh}(\mathcal{O}_ S)$. Then $\mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i = (\mathop{\mathrm{colim}}\nolimits \mathcal{G}_ i)^ a$ by the lemma cited and we conclude.

Proof of (3). Let $a : \mathcal{F} \to \mathcal{F}'$ be an arrow of $\mathit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$. Write $a = b^ a$ for some map $b : \mathcal{G} \to \mathcal{G}'$ of quasi-coherent modules on $S$. By the lemma cited we have $\mathop{\mathrm{Ker}}(a) = \mathop{\mathrm{Ker}}(b)^ a$ and $\mathop{\mathrm{Coker}}(a) = \mathop{\mathrm{Coker}}(b)^ a$ and we conclude.

Proof of (4). This follows from (3) except in the case when we know $\mathcal{F}_1$ and $\mathcal{F}_3$ are quasi-coherent. In this case write $\mathcal{F}_1 = \mathcal{G}_1^ a$ and $\mathcal{F}_3 = \mathcal{G}_3^ a$ with $\mathcal{G}_ i$ quasi-coherent on $S$. By Lemma 35.10.2 part (10) we conclude.

Proof of (5). Let $\mathcal{F}$ and $\mathcal{F}'$ be in $\mathit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$. Write $\mathcal{F} = \mathcal{G}^ a$ and $\mathcal{F}' = (\mathcal{G}')^ a$ with $\mathcal{G}$ and $\mathcal{G}'$ quasi-coherent on $S$. By the lemma cited we have $\mathcal{F} \otimes _\mathcal {O} \mathcal{F}' = (\mathcal{G} \otimes _{\mathcal{O}_ S} \mathcal{G}')^ a$ and we conclude.

Proof of (6). Let $\mathcal{F}$ and $\mathcal{G}$ be in $\mathit{QCoh}(S_{\acute{e}tale}, \mathcal{O})$ with $\mathcal{F}$ of finite presentation. Write $\mathcal{F} = \mathcal{H}^ a$ and $\mathcal{G} = (\mathcal{I})^ a$ with $\mathcal{H}$ and $\mathcal{I}$ quasi-coherent on $S$. By Lemma 35.8.10 we see that $\mathcal{H}$ is of finite presentation. By Lemma 35.10.2 part (8) we have $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G}) = (\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{H}, \mathcal{I}))^ a$ and we conclude. $\square$

Proof. The corresponding facts hold for quasi-coherent modules on the scheme $S$, see Schemes, Section 26.24. The proof will be to use Lemma 35.10.2 to transfer these truths to $(\mathit{Sch}/S)_\tau$.

Proof of (1). Let $\mathcal{F}_ i$, $i \in I$ be a family of objects of $\mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$. Write $\mathcal{F}_ i = \mathcal{G}_ i^ a$ for some quasi-coherent modules $\mathcal{G}_ i$ on $S$. Then $\bigoplus \mathcal{F}_ i = (\bigoplus \mathcal{G}_ i)^ a$ by the lemma cited and we conclude.

Proof of (2). Let $\mathcal{I} \to \mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$, $i \mapsto \mathcal{F}_ i$ be a diagram. Write $\mathcal{F}_ i = \mathcal{G}_ i^ a$ so we get a diagram $\mathcal{I} \to \mathit{QCoh}(\mathcal{O}_ S)$. Then $\mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i = (\mathop{\mathrm{colim}}\nolimits \mathcal{G}_ i)^ a$ by the lemma cited and we conclude.

Proof of (3). Let $a : \mathcal{F} \to \mathcal{F}'$ be an arrow of $\mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$. Write $a = b^ a$ for some map $b : \mathcal{G} \to \mathcal{G}'$ of quasi-coherent modules on $S$. By the lemma cited we have $\mathop{\mathrm{Coker}}(a) = \mathop{\mathrm{Coker}}(b)^ a$ (because a cokernel is a colimit) and we conclude.

Proof of (4). Write $\mathcal{F}_1 = \mathcal{G}_1^ a$ and $\mathcal{F}_3 = \mathcal{G}_3^ a$ with $\mathcal{G}_ i$ quasi-coherent on $S$. By Lemma 35.10.2 part (10) we conclude.

Proof of (5). Let $\mathcal{F}$ and $\mathcal{F}'$ be in $\mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$. Write $\mathcal{F} = \mathcal{G}^ a$ and $\mathcal{F}' = (\mathcal{G}')^ a$ with $\mathcal{G}$ and $\mathcal{G}'$ quasi-coherent on $S$. By the lemma cited we have $\mathcal{F} \otimes _\mathcal {O} \mathcal{F}' = (\mathcal{G} \otimes _{\mathcal{O}_ S} \mathcal{G}')^ a$ and we conclude.

Proof of (6). Write $\mathcal{F} = \mathcal{H}^ a$ for some quasi-coherent $\mathcal{O}_ S$-module. By Lemma 35.8.10 we see that $\mathcal{H}$ is finite locally free. The problem is Zariski local on $S$ (details omitted) hence we may assume $\mathcal{H} = \mathcal{O}_ S^{\oplus n}$ is finite free. Then $\mathcal{F} = \mathcal{O}^{\oplus n}$ and $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G}) = \mathcal{G}^{\oplus n}$ is quasi-coherent. $\square$

Proof. This lemma collects the results shown above in a slightly different manner. First of all, by Lemma 35.10.4 we already know the output of the construction in (1), (2), or (3) ends up in $\mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$. It remains to show in each case that the result is independent of the topology used. The key to this is that the equivalence $\mathit{QCoh}(\mathcal{O}_ S) \to \mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$, $\mathcal{F} \mapsto \mathcal{F}^ a$ of Proposition 35.8.9 is given by the same formula independent of the choice of the topology $\tau \in \{ Zariski, {\acute{e}tale}, fppf\}$.

Proof of (1). Let $\mathcal{I} \to \mathit{QCoh}((\mathit{Sch}/S)_{fppf}, \mathcal{O})$, $i \mapsto \mathcal{F}_ i$ be a diagram. Write $\mathcal{F}_ i = \mathcal{G}_ i^ a$ so we get a diagram $\mathcal{I} \to \mathit{QCoh}(\mathcal{O}_ S)$. Then $\mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i = (\mathop{\mathrm{colim}}\nolimits \mathcal{G}_ i)^ a$ in $\textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O})$ for $\tau \in \{ Zariski, {\acute{e}tale}, fppf\}$ by Lemma 35.10.2. This proves (1).

Proof of (2). Write $\mathcal{F} = \mathcal{H}^ a$ and $\mathcal{G} = (\mathcal{I})^ a$ with $\mathcal{H}$ and $\mathcal{I}$ quasi-coherent on $S$. Then $\mathcal{F} \otimes _\mathcal {O} \mathcal{G} = (\mathcal{H} \otimes _\mathcal {O} \mathcal{I})^ a$ in $\textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O})$ for $\tau \in \{ Zariski, {\acute{e}tale}, fppf\}$ by Lemma 35.10.2. This proves (2).

Proof of (3). Let $\mathcal{F}$ and $\mathcal{G}$ be in $\mathit{QCoh}((\mathit{Sch}/S)_{fppf}, \mathcal{O})$. Write $\mathcal{F} = \mathcal{H}^ a$ with $\mathcal{H}$ quasi-coherent on $S$. By Lemma 35.8.10 we have

This explains the parenthetical statement of part (3). Now, if these equivalent conditions hold, then $\mathcal{H}$ is finite locally free. The construction of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G})$ in Modules on Sites, Section 18.27 depends only on $\mathcal{F}$ and $\mathcal{G}$ as presheaves of modules (only whether the output $\mathop{\mathcal{H}\! \mathit{om}}\nolimits$ is a sheaf depends on whether $\mathcal{F}$ and $\mathcal{G}$ are sheaves). $\square$