2.2. 概形是 fppf 层 (仍然在翻译中)

Next, we will show if $X$ is a scheme, then $h_X$ is sheaf for the fppf topology.

We expand the definitions to see what this means. We see all $\{Z_i\to Z{\}}_{i\in I}$ we get $Z_i\to Z$ flat locally of finite presentation and $\coprod Z_i\to Z$ surjective. We also get exact sequence $$h_X(Z){\prod }_i{h}_X(Z_i){\prod }_{i,j}{h}_X(Z_i{\times }_Z{Z}_j)$$Then, $h_X$ sheaf for fppf topology means we get$$Z_{ij}Z{Z}_i{Z}_{j}X{\alpha }_{ji}{\alpha }_{ij}\exists !$$then there exists unique arrow from $Z$ to $X$.

A special case of this is that, $Z=\bigcup Z_i$ open cover and $f:Z\to X$ is equivalent to $f_i:Z_i\to X$ such that $f_i{|}_{Z_{ij}}=f_j{|}_{Z_{ij}}$.

The big theorem we are working towards at this point is that, if $X$ is a scheme, then $h_X$ is a sheaf for the fppf site on all schemes, i.e. the fppf site on $\mathrm{Spec}\mathbb{Z}$. In particular, if $X\to Y$ then $h_X$ is a sheaf on the fppf site on $Sch/Y$.

例 2.2.0.1. Let $L/K$ be a Galois field extension with Galois group $G$. Then $\mathrm{Spec}L\to \mathrm{Spec}K$ is etale and hence fppf. We showed for any sheaf $\mathcal{F}$, we see $\mathcal{F}(K)=\mathcal{F}(L{)}^L$. In particular, taking $\mathcal{F}=h_X$, we see a morphism $\mathrm{Spec}K\to X$ is the same as a $G$-invariant morphism $\mathrm{Spec}L\to X$.

Since Zariski and etale covers are examples of fppf covers, the big theorem also says $h_X$ is a sheaf for the (big) etale and Zariski topologies. We will start prove the theorem.

命题 2.2.0.2. If $A\to B$ is faithful flat and $M$ is $A$-module, then$$MM{\otimes }_{A}BM{\otimes }_{A}B{\otimes }_{A}B{p}_1{p}_2$$is exact. Here the maps are given by $m\mapsto m\otimes 1$ and $p_1:m\otimes b\mapsto m\otimes 1\otimes b$ and $p_2:m\otimes b\mapsto m\otimes b\otimes 1$.

证明. Exactness is equivalent to exactness of $0\to M\stackrel{f}{\to }M\otimes B\stackrel{p_1-p_2}{\to }M\otimes B\otimes B$. Thus it is enough to show (since $A\to B$ is faithfully flat) that$$MM\otimes B\otimes BM\otimes B\otimes B\otimes B{f}^{\prime }:=f\otimes 1p_1^{\prime }{p}_2^{\prime }$$where$$p_1^{\prime }(m\otimes b\otimes b^{\prime })=m\otimes b\otimes 1\otimes b^{\prime }$$$$p_2^{\prime }(m\otimes b\otimes b^{\prime })=m\otimes 1\otimes b\otimes b^{\prime }$$and $f^{\prime }(m\otimes b)=m\otimes 1\otimes b$. The point of doing this is that now we get a section, i.e. we have$$M\otimes BM\otimes B\otimes BM\otimes B\otimes B\otimes B\gamma \tau$$where$$\tau (m\otimes b\otimes b^{\prime }\otimes b^{\prime \prime })=m\otimes b\otimes b^{\prime }{b}^{\prime \prime }$$$$\gamma (m\otimes b\otimes b^{\prime })=m\otimes b{b}^{\prime }$$In particular, we get $\tau p_1^{\prime }=\mathrm{Id}$ and $\tau p_2^{\prime }=f^{\prime }\gamma$. Indeed,$$\tau p_1^{\prime }(m\otimes b\otimes b^{\prime })=\tau (m\otimes b\otimes 1\otimes b^{\prime })=m\otimes b\otimes 1b^{\prime }=m\otimes b\otimes b^{\prime }$$$$\tau p_2^{\prime }(m\otimes b\otimes b^{\prime })=\tau (m\otimes 1\otimes b\otimes b^{\prime })=m\otimes 1\otimes b{b}^{\prime }$$where we note$$f^{\prime }\gamma (m\otimes b\otimes b^{\prime })=f^{\prime }(m\otimes b{b}^{\prime })=m\otimes 1\otimes b{b}^{\prime }$$One can also check $\gamma f^{\prime }=\mathrm{Id}$. Hence, we indeed get a section which implies $f^{\prime }$ is injective as desired (which proves exactness on the left).

Next, we check exactness on the middle. Suppose $\alpha \in M\otimes B\otimes B$, then we have $p_1^{\prime }(\alpha )=p_2^{\prime }(\alpha )$. In particular, we get$$\tau p_1^{\prime }(\alpha )=\tau p_2^{\prime }(\alpha )=f^{\prime }(\gamma (\alpha ))=\alpha$$This concludes the exactness on the middle as well.

$\square$

推论 2.2.0.3. If $U,V,X$ are affine schemes. If $V\twoheadrightarrow U$ is fppf cover, then we get exact sequence$$h_X(U)h_X(V)h_X(V{\times }_{U}V)$$

证明. Say $U=\mathrm{Spec}A,V=\mathrm{Spec}B$ and $X=\mathrm{Spec}R$. Then t ake $M=A$ in previous proposition, we get$$ABB{\otimes }_{A}B\iota$$is exact. In particular, $\iota$ is injective. We want that, when we take $\mathrm{Hom}(R,-)$ to the above sequence, we get exact sequence, i.e. we want to show the following sequence is exact$$\mathrm{Hom}(R,A)\mathrm{Hom}(R,B)\mathrm{Hom}(R,B{\otimes }_{A}B)$$

Exact on the left: Suppose we have $R\underset{\beta }{\overset{\alpha }{\Rightarrow }}A\stackrel{\iota }{\to }B$ with $\iota \alpha =\iota \beta$ and $\iota$ injective. However, this implies $\alpha =\beta$ as $\iota$ is injective, as desired.

Exact on the middle: Say $f:R\to B$ such that for all $v\in R$, $f(r)\otimes 1=1\otimes f(r)$. By previous proposition, we know$$f(r)\in A\subseteq B$$Hence $f$ factors through $A$, which proves our claim.

$\square$

引理 2.2.0.4. Let $\mathcal{F}:{\mathsf{Sch}}^{opp}\to \mathsf{Sets}$ be a big Zariski sheaf. Then $\mathcal{F}$ is a sheaf for fppf topology iff for all $V\twoheadrightarrow U$ fppf, we get exact sequence$$\mathcal{F}(U)\mathcal{F}(V)\mathcal{F}(V{\times }_{U}V)$$

证明. Let $\{U_i\to U\}\in \mathrm{Cov}(U)$. Let $V=\coprod U_i$. Then we get a sequence$$\mathcal{F}(U){\prod }_i\mathcal{F}(U_i){\prod }_{i,j}\mathcal{F}(U_i{\times }_U{U}_j)$$However, we can complete the above diagram to$$\mathcal{F}(U){\prod }_i\mathcal{F}(U_i){\prod }_{ij}\mathcal{F}(U_i{\times }_U{U}_j)\mathcal{F}(U)\mathcal{F}(V)\mathcal{F}(V{\times }_{U}V)=\sim \sim$$The vertical maps are isomorphisms because $U_i$ form open (Zariski) cover of $V$, and $U_i\cap U_j=\varnothing$ in $V$. So apply the Zariski sheaf axiom exact sequence and we conclude the bottom is exact.

$\square$

引理 2.2.0.5. Let $\mathcal{F}:{\mathsf{Sch}}^{opp}\to \mathsf{Sets}$ be a presheaf. Assume $\mathcal{F}$ is a sheaf for the big Zariski topology. Then $\mathcal{F}$ is fppf sheaf iff for all $V\twoheadrightarrow U$ fppf, $V,U$ affine, we have$$\mathcal{F}(U)\mathcal{F}(V)\mathcal{F}(V{\times }_{U}V)$$is exact.

证明. By previous lemma, it is enough to check sheaf axioms on $V\twoheadrightarrow U$ not necessarily affine but singleton covers.

Exactness on the left: We need to show $\mathcal{F}(U)$ injects into $\mathcal{F}(V)$. Let $U=\bigcup U_i$ be open affine cover and $f:V\twoheadrightarrow U$ be fppf. Then $f^{-1}(U_i)={\bigcup }_j{V}_{ij}$ be an open affine cover. Hence $V={\bigcup }_{ij}{V}_{ij}$. We get the following diagram$$\mathcal{F}(U)\mathcal{F}(V){\prod }_i\mathcal{F}(U_i){\prod }_{i,j}\mathcal{F}(V_{ij})$$Thus, to prove the exactness on the left, it is enough to show the bottom arrow is injective. Since $f$ is flat locally of finite presentation, we see $f$ is open. Thus $f(V_{ij})$ are open in $U_i$. Also, $f$ is surjective as its fppf, we see$$U_i=f(f^{-1}(U_i))=\bigcup _{j}f(V_{ij})$$where the each $f(V_{ij})$ are open. Since $U_i$ is affine, it is quasi-compact, we see we can take finite subcover $U_i={\bigcup }_{k=1}^{l}f(V_{i{j}_k})$. In particular, we see the map$$\coprod _{k=1}^l{V}_{i{j}_k}\twoheadrightarrow U$$is fppf because$$V_{ij}\subseteq f^{-1}(U_i)U_{i}VUfppf\subseteq \subseteq fppf$$where the inclusion $V_{ij}\subseteq f^{-1}(U_i)$ is flat. Thus we see since ${\coprod }_{k=1}^l{V}_{i{j}_k}$ and $U_i$ are both affine, so by assumption, we get$$\mathcal{F}(U_i){\prod }_{k=1}^l\mathcal{F}(V_{i{j}_k}){\prod }_j\mathcal{F}(V_{ij})\subseteq$$which shows exactness on the left.

Exactness on the middle: Suppose $V\twoheadrightarrow U$ is fppf with $V,U$ not necessarily affine. We will do step by step.

Step 1: we show we may assume $U$ affine. Let $U=\bigcup U_i$ is affine open cover. Let $V_i=f^{-1}(U_i)$. Then we get$$\mathcal{F}(U)\mathcal{F}(V)\mathcal{F}(V{\times }_{U}V){\prod }_i\mathcal{F}(U_i){\prod }_i\mathcal{F}(V_i){\prod }_i\mathcal{F}(V_i{\times }_{U_i}{V}_i){\prod }_{i,j}\mathcal{F}(U_i\cap U_j){\prod }_{i,j}\mathcal{F}(V_i\cap V_j)abc$$We get $a,b$ injective since $\mathcal{F}$ is big Zariski sheaf, the $c$ is injective since $V_i\cap V_j\twoheadrightarrow U_i\cap U_j$ is fppf and we apply exactness on the left shown above. A diagram chase shows exactness in the middle as desired (if we can show the $U$ affine case).

The diagram chase is roughly as follows: start with the top middle bullet ${\bullet }_1$, we want to ask if there exists ${\bullet }_{?}$ in $\mathcal{F}(U)$ that maps to ${\bullet }_1$:$${\bullet }_{?}\in \mathcal{F}(U){\bullet }_1\in \mathcal{F}(V){\bullet }_2={\bullet }_3\in \mathcal{F}(V{\times }_{U}V)?$$There is not much we can do at this point, thus we send ${\bullet }_1$ to the bottom via $b$ and get$${\bullet }_{?}\in \mathcal{F}(U){\bullet }_1\in \mathcal{F}(V){\bullet }_2={\bullet }_3\in \mathcal{F}(V{\times }_{U}V){\bullet }_b?$$We want to show ${\bullet }_b$ comes from $\prod \mathcal{F}(U_i)$. Thus we get$${\bullet }_{?}\in \mathcal{F}(U){\bullet }_1\in \mathcal{F}(V){\bullet }_2={\bullet }_3\in \mathcal{F}(V{\times }_{U}V){\bullet }_a{\bullet }_b{\bullet }_{2b}={\bullet }_{3b}??$$However, note the middle row is exact by assumption, we indeed get$${\bullet }_{?}\in \mathcal{F}(U){\bullet }_1\in \mathcal{F}(V){\bullet }_2={\bullet }_3\in \mathcal{F}(V{\times }_{U}V){\bullet }_a{\bullet }_b{\bullet }_{2b}={\bullet }_{3b}??$$Viz, we have ${\bullet }_b$ comes from ${\bullet }_a$ and it remains to show ${\bullet }_a$ comes from the injection $?$ arrow from $\mathcal{F}(U)$. To that end, we note the left vertical line is exact, hence to show ${\bullet }_a$ lives in the image of $\mathcal{F}(U)$, we just need to show ${\bullet }_a$ has the same image in ${\prod }_{i,j}\mathcal{F}(U_i\cap U_j)$. To show that, we map ${\bullet }_a$ forward via the two different maps, and get ${\bullet }_{a2}$ and ${\bullet }_{a3}$, i.e. we get$${\bullet }_{?}\in \mathcal{F}(U){\bullet }_1\in \mathcal{F}(V){\bullet }_2={\bullet }_3\in \mathcal{F}(V{\times }_{U}V){\bullet }_a{\bullet }_b{\bullet }_{2b}={\bullet }_{3b}{\bullet }_{a2},{\bullet }_{a3}??$$where at the bottom, we must have ${\bullet }_{a2}$ and ${\bullet }_{a3}$ map to the same element because the middle row $\mathcal{F}(V)\to {\prod }_i\mathcal{F}(V_i)\Rightarrow {\prod }_{i,j}\mathcal{F}(V_i\cap V_j)$ is exact and the image of ${\bullet }_{a2}$ and ${\bullet }_{a3}$ must equal the image of ${\bullet }_1$. Hence, this forces ${\bullet }_{a2}={\bullet }_{a3}$ which forces ${\bullet }_a$ to come from ${\bullet }_{?}$ and hence shows ${\bullet }_1$ indeed comes from ${\bullet }_{?}$ as desired.

Step 2: we show we can assume $V$ is quasi-compact. We showed last time there exists $V=\bigcup V_j$ open cover by quasi-compacts such that $V_j\twoheadrightarrow U$ fppf. Consider the restriction map$$x\in \mathrm{Eq}(\mathcal{F}(U)\mathcal{F}(V{\times }_{U}V))x_j\in \mathrm{Eq}(\mathcal{F}(V_j)\mathcal{F}(V_j{\times }_U{V}_j))$$Our goal is to show $x$ comes from $\mathcal{F}(U)$.

Assume quasi-compact case. Since $V_j\twoheadrightarrow U$ is fppd, we get the following sequence$$\mathcal{F}(U)\mathcal{F}(V_j)\mathcal{F}(V_j{\times }_U{V}_j)$$is exact. Thus there exists unique $y_j\in \mathcal{F}(U)$ mapping to $x_j$. We claim $y_j$ is independent of $j$. Indeed, consider the diagram$$V_i{\times }_U{V}_j{V}_j{V}_{i}VU\subseteq \subseteq$$where we used fppf maps $V_j\to U$ and $V_i\to U$ to get the fibered product. Now we apply $\mathcal{F}$ to the whole diagram. First, we get the following injections$$\mathcal{F}(V_i{\times }_U{V}_j)\mathcal{F}(V_j)\mathcal{F}(V_i)\mathcal{F}(V)\mathcal{F}(U)$$Now let’s chase elements:$$\mathcal{F}(V_i{\times }_U{V}_j)x_j\in \mathcal{F}(V_j)x_i\in \mathcal{F}(V_i)x\in \mathcal{F}(V)\mathcal{F}(U)y_j\mapsto x_j{y}_i\mapsto x_i$$However, since $x$ maps to $x_i$ and $x_j$, we know $x_i$ and $x_j$ must map to the same thing in $\mathcal{F}(V_i{\times }_U{V}_j)$. However, $V_i{\times }_U{V}_j\to V_i\to U$ is fppf cover, thus the two arrows $\mathcal{F}(U)\Rightarrow \mathcal{F}(V_i{\times }_U{V}_j)$ are injective. Thus, we must have $y_i=y_j$, hence we can denote this as $y=y_i=y_j$. Moreover we have $y\mapsto x$. This is exactly what we wanted, and hence this finishes step $2$.

Step 3: finish the proof. We may assume $V\twoheadrightarrow U$ fppf with $V$ quasi-compact and $U$ affine. Let $V=\bigcup V_j$ be a finite affine cover. In particular, since the union is finite, we see $\coprod V_j$ is affine and hence $\coprod V_j\twoheadrightarrow U$ is fppf.

Thus we get$$\mathcal{F}(U)\mathcal{F}(V)\mathcal{F}(V{\times }_{U}V)\mathcal{F}(U)\mathcal{F}({\coprod }_j{V}_j)\mathcal{F}({\coprod }_j{V}_j{\times }_U{\coprod }_j{V}_j)=$$The vertical arrows are injective since $\coprod V_j\twoheadrightarrow V$ is fppf cover. The bottom row is exact because ${\coprod }_j{V}_j\twoheadrightarrow U$ is fppf and both of them are affine and we are assuming the affine case holds. Hence the top row is exact. This concludes the proof.

$\square$

推论 2.2.0.6. If $X$ is affine, then $h_X$ is fppf sheaf.

证明. We proved sheaf axiom for $V\twoheadrightarrow U$ where $V,U$ are affine and the arrow is fppf. Also, it is easy to check $h_X$ is big Zariski sheaf. Hence $h_X$ is fppf sheaf.

$\square$

Last time, we proved a big lemma says if $\mathcal{F}$ is sheaf for big Zariski topology, then $\mathcal{F}$ is sheaf for fppf topology iff for affine singleton covers we get the exact sequence.

The next step is to show we can move from affine $X$ to any $X$.

定理 2.2.0.7. Let $X$ be a scheme, then $h_X$ is a fppf sheaf.

证明. Let $X={\bigcup }_i{X}_i$ be open affine cover. Let $V\twoheadrightarrow U$ be fppf with $U,V$ affine. We just need to show$$\mathcal{F}(U)\mathcal{F}(V)\mathcal{F}(V{\times }_{U}V)$$is exact.

Exact on the left: say we have$$VUXtfg$$such that $ft=gt$. We want $f=g$. Set-theoretically, we know $f=g$ as $t$ is surjection. Thus we just need to show this equality is scheme-theoretically. Since $X=\bigcup X_i$ and $f=g$ as set maps, we see $f^{-1}(X_i)=g^{-1}(X_i)$. Thus let’s define $U_i:=f^{-1}(X_i)=g^{-1}(X_i)$. Now take the diagram and restrict to $U_i$ with $V_i:=t^{-1}(U_i)$, we get$$V_i{U}_i{X}_{i}t{|}_{V_i}fppff{|}_{U_i}g{|}_{U_i}$$In particular, we get $f{|}_{U_i}\circ t{|}_{V_i}=g{|}_{U_i}\circ t{|}_{V_i}$. However, $h_{X_i}$ is a sheaf as $X_i$ is affine, thus we see $f{|}_{U_i}=g{|}_{U_i}$ scheme-theoretically. However, this holds for all $U_i$ and hence we see $f=g$ scheme-theoretically globally.

Exact on the middle: say we have$$V{\times }_{U}VVUX{p}_1{p}_{2}tf\exists h$$with $f{p}_1=f{p}_2$. We want to show the dash arrow exists, i.e. we want to show there exists $h$. Let $|T|$ be the underlying topological space of any scheme $T$. Then we see we get the same diagram for topological spaces$$|V{\times }_{U}V||V||U||X||p_1||p_2||t||f|\exists h$$However, in this case, $|h|$ exists because of the following claim.

Claim:$$|V{\times }_{U}V||V||U||p_1||p_2||t|$$is a coequalizer of topological spaces.

Suppose this claim holds, then $h$ exists topologically , and so we can talk about subschemes $U_i:=h^{-1}(X_i)\subseteq U$, $V_i:=f^{-1}(X_i)\subseteq V$. Then, we get$$V_i{U}_i{X}_{i}fppft{|}_{V_i}f{|}_{V_i}\exists !h_i$$where the existence of $h_i$ is by affine case. Moreover, we see $h_i{|}_{U_i\cap U_j}=h_j{|}_{U_i\cap U_j}$ because we can cover $X_i\cap X_j$ by affine open and $h_i$ restrict to open affine agrees with $h_j$ restricts to open affine by the uniqueness statement in the affine case. Since $h_i{|}_{U_i\cap U_j}=h_j{|}_{U_i\cap U_j}$, the $h_i$’s glue to a map $h:U\to X$ scheme-theoretically. Thus, if we can prove the above claim, we are done. We are going to prove it in small steps.

Next, we claim that for fppf $t:V\twoheadrightarrow U$,:

1.	there exists natural surjection $|V{\times }_U\times V|\twoheadrightarrow |V|{\times }_{|U|}|V|$
2.	$R\subseteq |U|$ open iff $|t{|}^{-1}(R)\subseteq |V|$ is open.
3.	$|V{\times }_{U}V||V||U||p_1||p_2||t|$ is coequalizer in category of topological spaces.

$(1)$: Take $x,x^{\prime }\in |V|$ with the same image $\overline{x}\in |U|$. Then we see we get$$\mathrm{Spec}\kappa (x)\mathrm{Spec}\kappa (x^{\prime })V\mathrm{Spec}\kappa (\overline{x})Ut$$What we do next is to take fibered products on the right hand side (of the residue fields) and we get$$\mathrm{Spec}\kappa (x){\times }_{\mathrm{Spec}(\overline{x})}\mathrm{Spec}\kappa (x^{\prime })V{\times }_{U}V$$where we see the right hand side is $\mathrm{Spec}(\kappa (x){\otimes }_{\kappa (\overline{x})}\kappa (x^{\prime }))$ and we just choose any point of this. This yields a point of $|V{\times }_{U}V|$ mapping to $(x,x^{\prime })\in |V|{\times }_{|U|}|V|$.

$(2)$: $R\subseteq |U|$ is open then since $t$ is continuous we get $|t{|}^{-1}(R)$ is open. Conversely, $t$ is fppf implies it is surjective, thus $R=t(t^{-1}(R))$ and hence its open as $t^{-1}(R)$ is open and $t$ is open map.

$(3)$: we need to show the diagram is a coequalizer diagram. In this part, we will drop the bars, and just move to the category of topological spaces. What we want is that for any $W$ topological space, we want$$V{\times }_{U}VVUW{p}_1{p}_{2}tf\exists !h$$If $h$ exists, then it is unique as $t$ is surjective, i.e. $u\in |U|$, then choose $v\in |V|$ such that $t(v)=u$, then $h(u)=f(v)$.

Thus we just need to show if $v,v^{\prime }\in |V|$ and $t(v)=t(v^{\prime })$, then $f(v)=f(v^{\prime })$. However, since $v,v^{\prime }\in |V|$ with $t(v)=t(v^{\prime })$, this means $(v,v^{\prime })\in |V|{\times }_{|U|}|V|$. Earlier, we showed there is surjection$$q:|V{\times }_{U}V|\twoheadrightarrow |V|{\times }_{|U|}|V|$$Let $v^{\prime \prime }\in |V{\times }_{U}V|$ so $q(v^{\prime \prime })=(v,v^{\prime })$. Then, we see we know $f{p}_1=f{p}_2$ by assumption, thus we see $f{p}_1(v^{\prime \prime })=f{p}_2(v^{\prime \prime })$ but by definition $f{p}_1(v^{\prime \prime })=f(v)$ and $f{p}_2(v^{\prime \prime })=f(v)$.

At this point, we have defined $h$ as set map, and we need to show $h$ is continuous. Thus take $W^{\prime }\subseteq W$ be open, then we see $h^{-1}(W^{\prime })\subseteq U$ is open iff $t^{-1}{h}^{-1}(W^{\prime })$ is open in $|V|$ by part $(2)$. However,$$t^{-1}{h}^{-1}(W^{\prime })=f^{-1}(W^{\prime })$$where $f$ is continuous, hence $h$ is continuous as desired.

$\square$