4.1. 范畴叠

定义 4.1.0.1. A stack on a site $\mathcal{C}$ is a category fibered in groupoids $p:\mathcal{F}\to \mathcal{C}$, such that descent data is effective for covering maps, i.e. if $\{X_i\to Y{\}}_i\in \mathrm{Cov}(Y)$, then $\mathcal{F}(Y)\stackrel{ϵ}{\to }\mathcal{F}(\{X_i\to Y\})$ is an equivalence between categories.

In short, stacks are just category fibered in groupoids where descent holds.

注 4.1.0.2. This is what we really call categorical stack, as we haven’t done any actual geometry yet. Fibered categories are analogous to presheaves (they are presheaves when fibered in sets). So stacks are presheaves where sheaf axiom holds.

What this means is that, for example, take fppf topology on schemes and choose any sheaf $\mathcal{F}$. Then we say $\mathcal{F}$ is “geometric” if $\mathcal{F}=h_X$ for some scheme $X$. Those $\mathcal{F}$ are of course example of stacks.

Hence, in general, we want “representable stacks” (the technical term is algebraic stacks, or Artin stack) instead of arbitrary stacks.

Our next goal is to get a feeling about cat stacks, and then we try to find what would be a nice notion for algebraic stacks.

Well, I lied. Here is the punch line for what algebraic stacks are.

注 4.1.0.3 (Spoiler Alert). The idea for algebraic stacks is that, if $\mathcal{F}$ is a stack over schemes. To import/involve geometry, we require that there exists $X$ and arrow $X=h_X\to \mathcal{F}$, so that $h_X\twoheadrightarrow \mathcal{F}$ is a “smooth cover”. This makes no sense, as we don’t know what smooth covers are.

Thus, what should smooth cover $X\twoheadrightarrow \mathcal{F}$ be? Well, it should have the property that, for any scheme $T$, if we take fibered productthen the arrow $X{\times }_{\mathcal{F}}T\to T$ is a smooth cover. Well, this helps a little bit, as now over base becomes a scheme $T$. But, what is $X{\times }_{\mathcal{F}}T$? We don’t know, hence we just insists that it should be nice, i.e. it should be a scheme by definition (this is actually not the full definition, i.e. the actual def is $X{\times }_{\mathcal{F}}T$ should be algebraic space).

In other word, algebraic stacks are stacks over scheme that we get a smooth cover $X\twoheadrightarrow \mathcal{F}$, where smooth cover means when we pullback along scheme $T$ we always get $X{\times }_{\mathcal{F}}T$ be a scheme and $X{\times }_{\mathcal{F}}T\twoheadrightarrow T$ is a smooth cover of schemes.

Next, we consider an example of effective descent morphism.

命题 4.1.0.4. If $f:X\to Y$ admits a section $s:Y\to X$ then descent data is effective for $f$.

Note this holds for any category.

证明. We have $\mathcal{F}(Y)\stackrel{ϵ}{\to }\mathcal{F}(X\stackrel{f}{\to }Y)$ by $F\mapsto (f^{\ast }F,{\sigma }_{\mathrm{can}})$. Thus we define $\eta :\mathcal{F}(X\to Y)\to \mathcal{F}(Y)$ by$$(E,\sigma )\mapsto s^{\ast }E$$where we recall $E\in \mathcal{F}(X)$ and $\sigma :p_1^{\ast }E\stackrel{\sim }{\to }{p}_2^{\ast }E$. Let’s check this is what we wanted. Indeed, we see $\eta ϵ(F)=\eta (f^{\ast }F,{\sigma }_{\mathrm{can}})=s^{\ast }{f}^{\ast }F\cong F$.

Now we need to go the other way, i.e. we start with $(E,\sigma )$. We get the diagramThen, we getThis commutes because $f\circ \mathrm{Id}=f$ and $fsf=f$. Hence the dotted arrow $h$ in the above exists.

Then we get $\sigma :p_1^{\ast }E\stackrel{\sim }{\to }{p}_2^{\ast }E$ and hence we get diagramHere $h^{\ast }{p}_2^{\ast }\cong (p_{2}h{)}^{\ast }\cong (sf{)}^{\ast }\cong f^{\ast }{s}^{\ast }$ hence $h^{\ast }{p}_2^{\ast }E\cong f^{\ast }{s}^{\ast }E$ and similarly $h^{\ast }{p}_1^{\ast }E\cong E={\mathrm{Id}}^{\ast }E$. This gives an arrow $E\to f^{\ast }{s}^{\ast }(E,\sigma )$. Next, by pullback the cocycle diagram along the morphism $({\mathrm{Id}}_{X{\times }_{Y}X},sf{p}_2):X{\times }_{Y}X\to (X{\times }_{Y}X){\times }_{Y}X$, we get this map is compatible with $\sigma$. Hence this yields an isomorphism $(E,\sigma )\stackrel{\sim }{\to }ϵ\eta (E,\sigma )$.

$\square$

This is a silly example, but it is actually very useful, as we can frequently reduce to the case where we have sections.

Now we consider descent for sheaves. Let $\mathcal{C}$ be a site where finite limits exist. Then take $f:X\to Y$ in $\mathcal{C}$, we get maps between their topoi (here we use $\widetilde{\mathcal{C}/X}$ to denote topos). Here we get $(f^{\ast }\mathcal{F})(W\to X)=\mathcal{F}(W\to X\stackrel{f}{\to }Y)$. So, $f^{\ast }{g}^{\ast }=(gf{)}^{\ast }$ are equivalent.

Let $p:\mathsf{Sh}\to \mathcal{C}$ be the following category. The objects are $(X,E)$ where $X\in \mathcal{C}$ and $E\in \widetilde{\mathcal{C}/X}$. The morphisms from $(X,E)$ to $(Y,F)$ are given by $X\stackrel{f}{\to }Y$ plus $E\to f^{\ast }F$. The projection is $p(X,E)=X$. We note this is not fibered in groupoids.

定理 4.1.0.5. If $f:X\to Y$ is a covering in $\mathcal{C}$, then $f$ is effective descent morphism for $\mathsf{Sh}$.

证明. Consider $\mathsf{Sh}(X\to Y)$. This has objects $(E,\sigma )$ where $E\in \widetilde{\mathcal{C}/X}$ and $\sigma :p_1^{\ast }E\stackrel{\sim }{\to }{p}_2^{\ast }E$ in $\widetilde{\mathcal{C}/(X{\times }_{Y}X)}$ that satisfies cocycle condition. Forwe want to construct an inverse functor $\eta :\mathsf{Sh}(X\to Y)\to \mathsf{Sh}(Y)$.

In this case, we get diagramThis is not a commutative diagram, thus we want to take equalizer. That is, we want to take $\mathrm{Eq}(f_{\ast }E\underset{{\sigma }^{-1}{p}_2^{\ast }}{\overset{p_1^{\ast }}{\Rightarrow }}{g}_{\ast }{p}_1^{\ast }E)=:\eta ((E,\sigma ))$ as our definition. Here we abused notations. In particular, we write $p_1^{\ast }$ to mean the arrow $f_{\ast }E\to f_{\ast }(p_1{)}_{\ast }{p}_1^{\ast }E$ given by apply $f_{\ast }$ to the adjunction map $E\to (p_1{)}_{\ast }{p}_1^{\ast }E$ then take the reverse arrow of the arrow $g_{\ast }{p}_1^{\ast }E\to f_{\ast }(p_1{)}_{\ast }{p}_1^{\ast }E$. Similarly $p_2^{\ast }$ is the arrow $f_{\ast }E\to g_{\ast }{p}_2^{\ast }E$ obtained from the above diagram. Also, that ${\sigma }^{-1}{p}_2^{\ast }$ is also abuse of notations, as what we really meant is $(g_{\ast }{\sigma }^{-1})p_2^{\ast }$ as in the above diagram.

Then, we claim $\mathrm{Id}\stackrel{\sim }{\to }\eta \circ ϵ$. Indeed, if $F\in \mathsf{Sh}(Y)$, then $ϵ(F)=(f^{\ast }F,{\sigma }_{\mathrm{can}})$. Then $\eta ϵ(F)=\eta (f^{\ast }F,{\sigma }_{\mathrm{can}})=\mathrm{Eq}(f_{\ast }{f}^{\ast }F\underset{{\sigma }_{\mathrm{can}}^{-1}{p}_2^{\ast }}{\overset{p_1^{\ast }}{\Rightarrow }}{g}_{\ast }{p}_1^{\ast }{f}^{\ast }F)$. Now note $p_1^{\ast }{f}^{\ast }F=g^{\ast }F$ and hence we just want to show that $F$ is an equalizer of the arrow $f_{\ast }{f}^{\ast }F\Rightarrow g_{\ast }{g}^{\ast }F$, then it will conclude $\eta ϵ(F)=F$, where we note there is a natural map $F\to f_{\ast }{f}^{\ast }F$, i.e. we want to show $F\to f_{\ast }{f}^{\ast }F\Rightarrow g_{\ast }{g}^{\ast }F$ is an equalizer diagram.

To do this, we prove it on $Z$-valued points, i.e. we take arbitrary $Z\to Y$, and we show it holds when we apply $F$ to $Z$. LetThen, we see we getThat is exactly by def of sheaf an equalizer. Hence we see $\eta \circ ϵ\cong \mathrm{Id}$ by Yoneda.

Next, we let $(E,\sigma )$ be given and set $F:=\eta (E,\sigma )$. Then $F\to f_{\ast }E$ by construction and so we get $f^{\ast }F\to f^{\ast }{f}_{\ast }E\to E$, i.e. we get canonical map $\rho (E,\sigma ):(f^{\ast }F,{\sigma }_{\mathrm{can}})=ϵ\eta ((E,\sigma ))\to (E,\sigma )$. We want to show $\rho (E,\sigma )$ is isomorphism.

To show $\rho (E,\sigma )$ is isomorphism, it is okay to do that locally on $Y$ (left as exercise). To say do this locally, we mean that ifthen $g^{\ast }$ on topoi is exact, so $g^{\ast }\circ \mathrm{equalizer}=\mathrm{equalizer}\circ g^{\ast }$. In other word, we get commutative diagramThen $(g^{\prime }{)}^{\ast }\rho (E,\sigma )=\rho ((g^{\prime }{)}^{\ast }E,(g^{\prime }{)}^{\ast }\sigma )$. In particular, we see ${\rho }^{\prime }$ (this is the image of $\rho (E,\sigma )$ in $\mathsf{Sh}(X^{\prime }\to Y^{\prime })$ in the above diagram) is isomorphism implies $\rho (E,\sigma )$ is isomorphism (this claim is left as exercise).

Now here is the trick: since we can take any cover $g$, we choose $g=f$. Now we getand $s^{\prime }$ is a section via the diagonal map, i.e. $x\mapsto (x,x)\in X{\times }_{Y}X$. Hence by the last theorem, we are done.

$\square$

Last time, we defined fibered category of sheaves $\mathsf{Sh}\to \mathcal{C}$ over $\mathcal{C}$. We showed $\mathsf{Sh}$ is a “stack not fibered in groupoids”, i.e. we have descent for covering in $\mathcal{C}$.

As a corollary of the theorem we proved above, we get

命题 4.1.0.6. Let $X,Y,S$ be schemes with diagram:where $X^{\prime }=X{\times }_S{S}^{\prime },X^{\prime \prime }=X{\times }_S{S}^{\prime \prime }$, and similarly for $Y^{\prime },Y^{\prime \prime }$. THe $f^{\prime }:X^{\prime }\to Y^{\prime }$ over $S^{\prime }$ is such that $p_1^{\ast }{f}^{\prime }=p_2^{\ast }{f}^{\prime }$. Then there exists unique $f:X\to Y$ over $S$ so that $g^{\ast }f=f^{\prime }$.

证明. We showed a big theorem: $h_X,h_Y$ are fppf sheaves. Thus $f^{\prime }$ yields $h_{X^{\prime }}\to h_{Y^{\prime }}$. In particular, $p_1^{\ast }{f}^{\prime }=p_2^{\ast }{f}^{\prime }$ means this extends to $(h_{X^{\prime }},{\sigma }_{\mathrm{can}})\to (h_{Y^{\prime }},{\sigma }_{\mathrm{can}})$ in $\mathsf{Sh}(S^{\prime }\twoheadrightarrow S)$. By big theorem last time, we see $\mathsf{Sh}(S)\stackrel{\sim }{\to }\mathsf{Sh}(S^{\prime }\twoheadrightarrow S)$ and hence this arrow $(h_{X^{\prime }},{\sigma }_{\mathrm{can}})\to (h_{Y^{\prime }},{\sigma }_{\mathrm{can}})$ correspond to an arrow $h_X\to h_Y$. Now Yoneda lemma tells us we have the desired arrow $f:X\to Y$.

$\square$

Next, we talk about variant of descent for sheaves. Let $\mathcal{O}$ be a sheaf of rings on a site $\mathcal{C}$. For all $X\in \mathcal{C}$, let ${\mathcal{O}}_X\in \widetilde{\mathcal{C}/X}$ be defined by ${\mathcal{O}}_X(Y\to X):=\mathcal{O}(Y)$. Then for all $f:X\to Y$, we get a map $(\widetilde{\mathcal{C}/X},{\mathcal{O}}_X)\to (\widetilde{\mathcal{C}/Y},{\mathcal{O}}_Y)$ map of “ringed topoi”, i.e. map of topoi plus ${\mathcal{O}}_Y\to f_{\ast }{\mathcal{O}}_X$.

We will show this is almost a stack.

For $X\in \mathcal{C}$, let ${\mathrm{Mod}}_X$ be the category of ${\mathcal{O}}_X$-modules in $\widetilde{\mathcal{C}/X}$. Then for all $f:Y\to X$, we get $f^{\ast }:{\mathrm{Mod}}_X\to {\mathrm{Mod}}_Y$ by $(f^{\ast }M)(Z\to Y):=M(Z\to Y\to X)$.

Now we define fibered category $\mathrm{MOD}\stackrel{p}{\to }\mathcal{C}$ as follows: it has object $(E,X)$ where $X\in \mathcal{C}$, $E\in {\mathrm{Mod}}_X$. The morphisms are $(F,Y)\to (E,X)$ is $f:Y\to X$ in $\mathcal{C}$ and $ϵ:F\to f^{\ast }E$ in ${\mathrm{Mod}}_Y$.

定理 4.1.0.7. For all $f:Y\to X$ covers in $\mathcal{C}$, ${\mathrm{Mod}}_X\stackrel{\sim }{\to }\mathrm{MOD}(Y\to X)$.

Now we have defined modules, the next topic is of course quasi-coherent sheaves.

Let $S$ be a scheme, $\mathcal{C}=(\mathsf{Sch}/S{)}_{fppf}$ be the fppf site associated with $\mathsf{Sch}/S$.

Now $\mathcal{O}$ be the presheaf of rings on $\mathcal{C}$ defined as: $\mathcal{O}(T\to S):=\Gamma ({\mathcal{O}}_T)={\mathrm{Hom}}_S(T,{\mathbb{A}}_S^1)$. This is just the global sections, i.e. it is the sheaf represented by ${\mathbb{A}}_S^1$. In other word, $\mathcal{O}=h_{{\mathbb{A}}_S^1}$ and hence we see this is a fppf sheaf as $h_{{\mathbb{A}}_S^1}$ is fppf sheaf.

Next, we want to figure out what’s a reasonable notion of quasi-coherent for fppf topology.

For any scheme $S$, let $\mathsf{Qcoh}(S)$ be the category of quasi-coherent ${\mathcal{O}}_S$-modules in Zariski topology, i.e. for $S_{Zar}$.

Given $\mathcal{F}\in \mathsf{Qcoh}(S)$ we get ${\mathcal{F}}_{big}$, a presheaf of $\mathcal{O}$-modules on $\mathcal{C}=(\mathsf{Sch}/S{)}_{fppf}$, defined as follows$${\mathcal{F}}_{big}(T\stackrel{f}{\to }S):=(f^{\ast }\mathcal{F})(T)$$Note this depends on choice of pullback.

引理 4.1.0.8. ${\mathcal{F}}_{big}$ is an fppf sheaf.

证明. Recall from awhile ago, to prove ${\mathcal{F}}_{big}$ is an fppf sheaf, we just need to check:

1.	$\forall T\to S$, ${\mathcal{F}}_{big}{|}_{T_{Zar}}$ is a sheaf, and
2.	sheaf condition on fppf arrow $\mathrm{Spec}B\twoheadrightarrow \mathrm{Spec}A$.

To see $(1)$, we see it is enough to show ${\mathcal{F}}_{big}{|}_{T_{Zar}}$ is sheaf for small Zariski site. However, we see this is clearly a sheaf, because$${\mathcal{F}}_{big}{|}_{T_{Zar}}=f^{\ast }\mathcal{F}\in \mathsf{Qcoh}(T)$$by definition. So it is indeed a sheaf.

For $(2)$, let $f:\mathrm{Spec}A\to S$ and $f^{\ast }\mathcal{F}$ be $M$ an $A$-module. We need to checkWe showed this when showing schemes are fppf sheaves.

$\square$

So, $\mathcal{F}\in \mathsf{Qcoh}(S)$ yields ${\mathcal{F}}_{big}\in {\widetilde{(\mathsf{Sch}/S)}}_{fppf}$. Conversely, given $\mathcal{H}\in {\widetilde{(\mathsf{Sch}/S)}}_{fppf}$ sheaf of $\mathcal{O}$-mods, and $T\stackrel{f}{\to }S$, we get ${\mathcal{H}}_T\in \mathsf{Qcoh}(T)$ defined by$${\mathcal{H}}_T(U\subseteq T)=\mathcal{H}(U)$$By construction, $\mathcal{F}\in \mathsf{Qcoh}(S)$ is given by $({\mathcal{F}}_{big}{)}_S=\mathcal{F}$.

命题 4.1.0.9. Let $\mathcal{F}\in \mathsf{Qcoh}(S)$, $\mathcal{G}$ a fppf sheaf of $\mathcal{O}$-mods. Then$${\mathrm{Hom}}_{\mathcal{O}}({\mathcal{F}}_{big},\mathcal{G})\stackrel{\sim }{\to }{\mathrm{Hom}}_{{\mathcal{O}}_S}(\mathcal{F},{\mathcal{G}}_S)$$given by $\alpha \mapsto \alpha {|}_{S_{Zar}}$.

证明. Exercise: it is enough to check Zariski local on $S$, so we can assume $S$ is affine. Then we get$${\mathcal{F}}_2:={\mathcal{O}}_S^J\to {\mathcal{F}}_1:={\mathcal{O}}_S^T\to \mathcal{F}\to 0$$Then $f^{\ast }$ is right exact, so ${\mathcal{F}}_{2,big}\to {\mathcal{F}}_{1,big}\to {\mathcal{F}}_{big}\to 0$. As a result, we getSo, we can assume $\mathcal{F}={\mathcal{O}}_S^I$. Therefore, we may assume $\mathcal{F}={\mathcal{O}}_S$, i.e. $|I|=1$. Now, observe elements in ${\mathrm{Hom}}_{\mathcal{O}}(\mathcal{O},\mathcal{G})$ means that we have compatible maps ${\varphi }_T:\mathcal{O}(T\to S)\to \mathcal{G}(T\to S)$. In particular, this means for any $f$ with diagramwe get $f:(T\to S)\to (S\to S)$. Thus we get diagramso ${\varphi }_T$ is determined by ${\varphi }_S$. Thus the map ${\mathrm{Hom}}_{\mathcal{O}}(\mathcal{O},\mathcal{G})\stackrel{\eta }{\to }{\mathrm{Hom}}_{{\mathcal{O}}_S}({\mathcal{O}}_S,{\mathcal{G}}_S)$ is isomorphism and hence ${\varphi }_S$ iff $\xi \in \mathcal{G}(S)$ then ${\varphi }_T(1)=f^{\ast }\xi$.

$\square$

定义 4.1.0.10. A big quasi-coherent sheaf on $\mathbf{S}$ of $\mathcal{O}$-mods is $\mathcal{F}$ on $S_{fppf}$ such that:

1.	$\forall T\to S$, ${\mathcal{F}}_T:=\mathcal{F}{|}_{T_{Zar}}\in \mathsf{Qcoh}(T)$
2.	$\forall T\stackrel{f}{\to }S$, ${\mathcal{F}}_T\to f^{\ast }{\mathcal{F}}_S$ is an isomorphism.

命题 4.1.0.11. THere is an equivalent of categories$$\begin{array}{c}\mathsf{Qcoh}(S_{Zar})\stackrel{\sim }{\to }\mathsf{Qcoh}(S_{fppf})\\ \mathcal{F}\mapsto {\mathcal{F}}_{big}\\ {\mathcal{G}}_S\text{leftmapsto}\mathcal{G}\end{array}$$

We get fibered category $\mathsf{Qcoh}\to \mathcal{C}$.

定理 4.1.0.12. If we have fppf $f:Y\twoheadrightarrow X$ then $\mathsf{Qcoh}(X)\stackrel{\sim }{\to }\mathsf{Qcoh}(Y\to X)$.

证明. We only show the local case and the full proof can be found in the book.

Martin reduces to the case where $f:Y\to X$ is qcqs (quasi-compact, quasi-separated). In this case, pushforward of quasi-coherent sheaf is quasi-coherent, i.e. $f(\mathrm{qcoh})=\mathrm{qcoh}$.

Then, we note$$\mathsf{Sh}(Y\stackrel{f}{\to }X)\underset{\eta }{\overset{\sim }{\to }}\mathsf{Sh}(X)$$$$(\mathcal{F},\sigma )\mapsto \text{Equalizer of }{f}_{\ast }^{\prime }S$$Since $f_{\ast }(\mathrm{qcoh})=\mathrm{qcoh}$, and equalizer of qcoh is qcoh, we see $\eta$ sends $\mathsf{Qcoh}(Y\to X)$ to $\mathsf{Qcoh}(X)$.

$\square$

Now we consider something that we already know, but in the new language.

We consider descent for closed subschemes:

命题 4.1.0.13. Suppose we have fppf coverThen the set of closed $W\subseteq Y$ is equivalent to the set of closed $Z\subseteq X$ such that $p_1^{\ast }Z=p_2^{\ast }Z$ given by the map $W\mapsto f^{-1}(W)$.

证明. We note closed $W\subseteq Y$ is the same as quasi-coherent sheaf of ideals ${\mathcal{I}}_W\subseteq {\mathcal{O}}_Y$. Because $f,p_1,p_2$ are flat, pullback of ideal is an ideal. The result follows from descent of quasi-coherent sheaves applied to ideal sheaves.

$\square$

In a very similar manner, we get descent for affine maps.

Let $\mathsf{Aff}\to \mathsf{Sch}$ be the fibered category with objects $(X^{\prime }\stackrel{g}{\to }X,X)$ with $g$ affine map.

命题 4.1.0.14. If we have fppf cover $S^{\prime }\twoheadrightarrow S$, then ${\mathsf{Aff}}_S\stackrel{\sim }{\to }\mathsf{Aff}(S^{\prime }\to S)$.

证明. Say $X\to S$ is affine, this is the same as $X=\mathrm{Spec}\mathcal{A}$ where $\mathcal{A}$ is qcoh sheaf of ${\mathcal{O}}_S$-algebra. Then we have descent for quasi-coherent sheaves, and we want to make sure it is still ${\mathcal{O}}_S$-algebra.

That is, how do we know if ${\mathcal{A}}^{\prime }=f^{\ast }\mathcal{A}$ and ${\mathcal{A}}^{\prime }$ a ${\mathcal{O}}_{S^{\prime }}$-algebra, then $\mathcal{A}$ is ${\mathcal{O}}_S$-algebra?

Being an algebra means we have $m:\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ with commutativity diagrams. But this is a map and hence $m^{\prime }:{\mathcal{A}}^{\prime }\times {\mathcal{A}}^{\prime }\to {\mathcal{A}}^{\prime }$ descents to get $m$ and diagrams can be checked locally.

$\square$