# 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 ${Z_{i}→Z}_{i∈I}$ we get $Z_{i}→Z$ flat locally of finite presentation and $∐Z_{i}→Z$ surjective. We also get exact sequence $h_{X}(Z)∏_{i}h_{X}(Z_{i})∏_{i,j}h_{X}(Z_{i}×_{Z}Z_{j})$Then, $h_{X}$ sheaf for fppf topology means we get$Z_{ij}ZZ_{i}Z_{j}Xα_{ji}α_{ij}∃!$then there exists unique arrow from $Z$ to $X$.

A special case of this is that, $Z=⋃Z_{i}$ open cover and $f:Z→X$ is equivalent to $f_{i}:Z_{i}→X$ 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 $SpecZ$. In particular, if $X→Y$ 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 $SpecL→SpecK$ is etale and hence fppf. We showed for any sheaf $F$, we see $F(K)=F(L)_{L}$. In particular, taking $F=h_{X}$, we see a morphism $SpecK→X$ is the same as a $G$-invariant morphism $SpecL→X$.

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 $A→B$ is faithful flat and $M$ is $A$-module, then$MM⊗_{A}BM⊗_{A}B⊗_{A}Bp_{1}p_{2}$is exact. Here the maps are given by $m↦m⊗1$ and $p_{1}:m⊗b↦m⊗1⊗b$ and $p_{2}:m⊗b↦m⊗b⊗1$.

**证明.** Exactness is equivalent to exactness of $0→Mf M⊗Bp_{1}−p_{2} M⊗B⊗B$. Thus it is enough to show (since $A→B$ is faithfully flat) that$MM⊗B⊗BM⊗B⊗B⊗Bf_{′}:=f⊗1p_{1}p_{2}$where$p_{1}(m⊗b⊗b_{′})=m⊗b⊗1⊗b_{′}$$p_{2}(m⊗b⊗b_{′})=m⊗1⊗b⊗b_{′}$and $f_{′}(m⊗b)=m⊗1⊗b$. The point of doing this is that now we get a section, i.e. we have$M⊗BM⊗B⊗BM⊗B⊗B⊗Bγτ$where$τ(m⊗b⊗b_{′}⊗b_{′′})=m⊗b⊗b_{′}b_{′′}$$γ(m⊗b⊗b_{′})=m⊗bb_{′}$In particular, we get $τp_{1}=Id$ and $τp_{2}=f_{′}γ$. Indeed,$τp_{1}(m⊗b⊗b_{′})=τ(m⊗b⊗1⊗b_{′})=m⊗b⊗1b_{′}=m⊗b⊗b_{′}$$τp_{2}(m⊗b⊗b_{′})=τ(m⊗1⊗b⊗b_{′})=m⊗1⊗bb_{′}$where we note$f_{′}γ(m⊗b⊗b_{′})=f_{′}(m⊗bb_{′})=m⊗1⊗bb_{′}$One can also check $γf_{′}=Id$. Hence, we indeed get a section which implies $f_{′}$ is injective as desired (which proves exactness on the left).

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推论 2.2.0.3. If $U,V,X$ are affine schemes. If $V↠U$ is fppf cover, then we get exact sequence$h_{X}(U)h_{X}(V)h_{X}(V×_{U}V)$

**证明.** Say $U=SpecA,V=SpecB$ and $X=SpecR$. Then t ake $M=A$ in previous proposition, we get$ABB⊗_{A}Bι$is exact. In particular, $ι$ is injective. We want that, when we take $Hom(R,−)$ to the above sequence, we get exact sequence, i.e. we want to show the following sequence is exact$Hom(R,A)Hom(R,B)Hom(R,B⊗_{A}B)$

Exact on the left: Suppose we have $Rαβ Aι B$ with $ια=ιβ$ and $ι$ injective. However, this implies $α=β$ as $ι$ is injective, as desired.

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引理 2.2.0.4. Let $F:(Sch)_{opp}→(Sets)$ be a big Zariski sheaf. Then $F$ is a sheaf for fppf topology iff for all $V↠U$ fppf, we get exact sequence$F(U)F(V)F(V×_{U}V)$

**证明.**Let ${U_{i}→U}∈Cov(U)$. Let $V=∐U_{i}$. Then we get a sequence$F(U)∏_{i}F(U_{i})∏_{i,j}F(U_{i}×_{U}U_{j})$However, we can complete the above diagram to$F(U)∏_{i}F(U_{i})∏_{ij}F(U_{i}×_{U}U_{j})F(U)F(V)F(V×_{U}V)=∼∼$The vertical maps are isomorphisms because $U_{i}$ form open (Zariski) cover of $V$, and $U_{i}∩U_{j}=∅$ in $V$. So apply the Zariski sheaf axiom exact sequence and we conclude the bottom is exact.

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引理 2.2.0.5. Let $F:(Sch)_{opp}→(Sets)$ be a presheaf. Assume $F$ is a sheaf for the big Zariski topology. Then $F$ is fppf sheaf iff for all $V↠U$ fppf, $V,U$ affine, we have$F(U)F(V)F(V×_{U}V)$is exact.

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

Exactness on the left: We need to show $F(U)$ injects into $F(V)$. Let $U=⋃U_{i}$ be open affine cover and $f:V↠U$ be fppf. Then $f_{−1}(U_{i})=⋃_{j}V_{ij}$ be an open affine cover. Hence $V=⋃_{ij}V_{ij}$. We get the following diagram$F(U)F(V)∏_{i}F(U_{i})∏_{i,j}F(V_{ij})$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$U_{i}=f(f_{−1}(U_{i}))=j⋃ f(V_{ij})$where the each $f(V_{ij})$ are open. Since $U_{i}$ is affine, it is quasi-compact, we see we can take finite subcover $U_{i}=⋃_{k=1}f(V_{ij_{k}})$. In particular, we see the map$k=1∐l V_{ij_{k}}↠U$is fppf because$V_{ij}⊆f_{−1}(U_{i})U_{i}VUfppf⊆⊆fppf$where the inclusion $V_{ij}⊆f_{−1}(U_{i})$ is flat. Thus we see since $∐_{k=1}V_{ij_{k}}$ and $U_{i}$ are both affine, so by assumption, we get$F(U_{i})∏_{k=1}F(V_{ij_{k}})∏_{j}F(V_{ij})⊆$which shows exactness on the left.

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

Step 1: we show we may assume $U$ affine. Let $U=⋃U_{i}$ is affine open cover. Let $V_{i}=f_{−1}(U_{i})$. Then we get$F(U)F(V)F(V×_{U}V)∏_{i}F(U_{i})∏_{i}F(V_{i})∏_{i}F(V_{i}×_{U_{i}}V_{i})∏_{i,j}F(U_{i}∩U_{j})∏_{i,j}F(V_{i}∩V_{j})abc$We get $a,b$ injective since $F$ is big Zariski sheaf, the $c$ is injective since $V_{i}∩V_{j}↠U_{i}∩U_{j}$ 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 $∙_{1}$, we want to ask if there exists $∙_{?}$ in $F(U)$ that maps to $∙_{1}$:$∙_{?}∈F(U)∙_{1}∈F(V)∙_{2}=∙_{3}∈F(V×_{U}V)?$There is not much we can do at this point, thus we send $∙_{1}$ to the bottom via $b$ and get$∙_{?}∈F(U)∙_{1}∈F(V)∙_{2}=∙_{3}∈F(V×_{U}V)∙_{b}?$We want to show $∙_{b}$ comes from $∏F(U_{i})$. Thus we get$∙_{?}∈F(U)∙_{1}∈F(V)∙_{2}=∙_{3}∈F(V×_{U}V)∙_{a}∙_{b}∙_{2b}=∙_{3b}??$However, note the middle row is exact by assumption, we indeed get$∙_{?}∈F(U)∙_{1}∈F(V)∙_{2}=∙_{3}∈F(V×_{U}V)∙_{a}∙_{b}∙_{2b}=∙_{3b}??$Viz, we have $∙_{b}$ comes from $∙_{a}$ and it remains to show $∙_{a}$ comes from the injection $?$ arrow from $F(U)$. To that end, we note the left vertical line is exact, hence to show $∙_{a}$ lives in the image of $F(U)$, we just need to show $∙_{a}$ has the same image in $∏_{i,j}F(U_{i}∩U_{j})$. To show that, we map $∙_{a}$ forward via the two different maps, and get $∙_{a2}$ and $∙_{a3}$, i.e. we get$∙_{?}∈F(U)∙_{1}∈F(V)∙_{2}=∙_{3}∈F(V×_{U}V)∙_{a}∙_{b}∙_{2b}=∙_{3b}∙_{a2},∙_{a3}??$where at the bottom, we must have $∙_{a2}$ and $∙_{a3}$ map to the same element because the middle row $F(V)→∏_{i}F(V_{i})⇒∏_{i,j}F(V_{i}∩V_{j})$ is exact and the image of $∙_{a2}$ and $∙_{a3}$ must equal the image of $∙_{1}$. Hence, this forces $∙_{a2}=∙_{a3}$ which forces $∙_{a}$ to come from $∙_{?}$ and hence shows $∙_{1}$ indeed comes from $∙_{?}$ as desired.

Step 2: we show we can assume $V$ is quasi-compact. We showed last time there exists $V=⋃V_{j}$ open cover by quasi-compacts such that $V_{j}↠U$ fppf. Consider the restriction map$x∈Eq(F(U)F(V×_{U}V))x_{j}∈Eq(F(V_{j})F(V_{j}×_{U}V_{j}))$Our goal is to show $x$ comes from $F(U)$.

Assume quasi-compact case. Since $V_{j}↠U$ is fppd, we get the following sequence$F(U)F(V_{j})F(V_{j}×_{U}V_{j})$is exact. Thus there exists unique $y_{j}∈F(U)$ mapping to $x_{j}$. We claim $y_{j}$ is independent of $j$. Indeed, consider the diagram$V_{i}×_{U}V_{j}V_{j}V_{i}VU⊆⊆$where we used fppf maps $V_{j}→U$ and $V_{i}→U$ to get the fibered product. Now we apply $F$ to the whole diagram. First, we get the following injections$F(V_{i}×_{U}V_{j})F(V_{j})F(V_{i})F(V)F(U)$Now let’s chase elements:$F(V_{i}×_{U}V_{j})x_{j}∈F(V_{j})x_{i}∈F(V_{i})x∈F(V)F(U)y_{j}↦x_{j}y_{i}↦x_{i}$However, since $x$ maps to $x_{i}$ and $x_{j}$, we know $x_{i}$ and $x_{j}$ must map to the same thing in $F(V_{i}×_{U}V_{j})$. However, $V_{i}×_{U}V_{j}→V_{i}→U$ is fppf cover, thus the two arrows $F(U)⇒F(V_{i}×_{U}V_{j})$ are injective. Thus, we must have $y_{i}=y_{j}$, hence we can denote this as $y=y_{i}=y_{j}$. Moreover we have $y↦x$. This is exactly what we wanted, and hence this finishes step $2$.

Step 3: finish the proof. We may assume $V↠U$ fppf with $V$ quasi-compact and $U$ affine. Let $V=⋃V_{j}$ be a finite affine cover. In particular, since the union is finite, we see $∐V_{j}$ is affine and hence $∐V_{j}↠U$ is fppf.

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推论 2.2.0.6. If $X$ is affine, then $h_{X}$ is fppf sheaf.

**证明.**We proved sheaf axiom for $V↠U$ 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.

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Last time, we proved a big lemma says if $F$ is sheaf for big Zariski topology, then $F$ 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 $X=⋃_{i}X_{i}$ be open affine cover. Let $V↠U$ be fppf with $U,V$ affine. We just need to show$F(U)F(V)F(V×_{U}V)$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 $X=⋃X_{i}$ 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$V_{i}U_{i}X_{i}t∣_{V}fppff∣_{U}g∣_{U}$In particular, we get $f∣_{U_{i}}∘t∣_{V_{i}}=g∣_{U_{i}}∘t∣_{V_{i}}$. 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$V×_{U}VVUXp_{1}p_{2}tf∃h$with $fp_{1}=fp_{2}$. 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$∣V×_{U}V∣∣V∣∣U∣∣X∣∣p_{1}∣∣p_{2}∣∣t∣∣f∣∃h$However, in this case, $∣h∣$ exists because of the following claim.

Claim:$∣V×_{U}V∣∣V∣∣U∣∣p_{1}∣∣p_{2}∣∣t∣$is a coequalizer of topological spaces.

Suppose this claim holds, then $h$ exists topologically , and so we can talk about subschemes $U_{i}:=h_{−1}(X_{i})⊆U$, $V_{i}:=f_{−1}(X_{i})⊆V$. Then, we get$V_{i}U_{i}X_{i}fppft∣_{V}f∣_{V}∃!h_{i}$where the existence of $h_{i}$ is by affine case. Moreover, we see $h_{i}∣_{U_{i}∩U_{j}}=h_{j}∣_{U_{i}∩U_{j}}$ because we can cover $X_{i}∩X_{j}$ 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 $h_{i}∣_{U_{i}∩U_{j}}=h_{j}∣_{U_{i}∩U_{j}}$, the $h_{i}$’s glue to a map $h:U→X$ 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 $t:V↠U$,:

1. | there exists natural surjection $∣V×_{U}×V∣↠∣V∣×_{∣U∣}∣V∣$ |

2. | $R⊆∣U∣$ open iff $∣t∣_{−1}(R)⊆∣V∣$ is open. |

3. | $∣V×_{U}V∣∣V∣∣U∣∣p_{1}∣∣p_{2}∣∣t∣$ is coequalizer in category of topological spaces. |

$(1)$: Take $x,x_{′}∈∣V∣$ with the same image $x∈∣U∣$. Then we see we get$Specκ(x)Specκ(x_{′})VSpecκ(x)Ut$What we do next is to take fibered products on the right hand side (of the residue fields) and we get$Specκ(x)×_{Spec(x)}Specκ(x_{′})V×_{U}V$where we see the right hand side is $Spec(κ(x)⊗_{κ(x)}κ(x_{′}))$ and we just choose any point of this. This yields a point of $∣V×_{U}V∣$ mapping to $(x,x_{′})∈∣V∣×_{∣U∣}∣V∣$.

$(2)$: $R⊆∣U∣$ 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$V×_{U}VVUWp_{1}p_{2}tf∃!h$If $h$ exists, then it is unique as $t$ is surjective, i.e. $u∈∣U∣$, then choose $v∈∣V∣$ such that $t(v)=u$, then $h(u)=f(v)$.

Thus we just need to show if $v,v_{′}∈∣V∣$ and $t(v)=t(v_{′})$, then $f(v)=f(v_{′})$. However, since $v,v_{′}∈∣V∣$ with $t(v)=t(v_{′})$, this means $(v,v_{′})∈∣V∣×_{∣U∣}∣V∣$. Earlier, we showed there is surjection$q:∣V×_{U}V∣↠∣V∣×_{∣U∣}∣V∣$Let $v_{′′}∈∣V×_{U}V∣$ so $q(v_{′′})=(v,v_{′})$. Then, we see we know $fp_{1}=fp_{2}$ by assumption, thus we see $fp_{1}(v_{′′})=fp_{2}(v_{′′})$ but by definition $fp_{1}(v_{′′})=f(v)$ and $fp_{2}(v_{′′})=f(v)$.

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