4.3. 代数空间

Recall that stack is a category fibered in groupoids where descent holds for all covering maps.

命题 4.3.0.1. If $\mathcal{F}\to \mathcal{C}$ is a stack, then for all $X\in \mathcal{C}$ and $x,y\in \mathcal{F}(X)$, we have $\mathrm{Isom}(x,y)$ is a sheaf on $\mathcal{C}/X$.

证明. If $Y^{\prime }\stackrel{f}{\to }Y$ is a covering, then consider$$Y^{\prime \prime }=Y^{\prime }{\times }_Y{Y}^{\prime }\underset{p_1}{\overset{p_2}{\Rightarrow }}{Y}^{\prime }\stackrel{f}{\to }Y$$in $\mathcal{C}/X$, i.e. we also have an arrow $Y\to X$ by default. Then we get $x,y\in X$ and hence we get a bunch of pullbacksNote here the double arrows of pullback is actually a square

Suppose $p_1^{\ast }\varphi =p_2^{\ast }\varphi$. We want $\varphi$ to descent to $g^{\ast }x\stackrel{\sim }{\to }{g}^{\ast }y$. By definition, $\varphi$ defines an isomorphism $(f^{\ast }{g}^{\ast }x,{\sigma }_{\mathrm{can}})\stackrel{\sim }{\to }(f^{\ast }{g}^{\ast }y,{\sigma }_{\mathrm{can}})$ in $\mathcal{F}(Y^{\prime }\to Y)$. However, we know $\mathcal{F}(Y)\underset{ϵ}{\overset{\sim }{\to }}\mathcal{F}(Y^{\prime }\to Y)$ as $\mathcal{F}$ is a stack, thus we are done as desired.

$\square$

命题 4.3.0.2. Letbe maps of stacks over $\mathcal{C}$. Let $\mathcal{F}:={\mathcal{F}}_1{\times }_{{\mathcal{F}}_3}{\mathcal{F}}_2$ be the fibered product of category fibered in groupoids. Then $\mathcal{F}$ is a stack.

证明. By definition, $({\mathcal{F}}_1{\times }_{{\mathcal{F}}_3}{\mathcal{F}}_2)(X)={\mathcal{F}}_1(X){\times }_{{\mathcal{F}}_3(X)}{\mathcal{F}}_2(X)$. Similarly, an easy check shows$$\mathcal{F}(\{X_i\to X\})={\mathcal{F}}_1(\{X_i\to X\}){\times }_{{\mathcal{F}}_3(\{X_i\to X\})}{\mathcal{F}}_2(\{X_i\to X\})$$

Each ${\mathcal{F}}_i(X)\stackrel{\sim }{\to }{\mathcal{F}}_i(\{X_i\to X\})$ and hence $\mathcal{F}$ is a stack.

$\square$

Just like for sheaves we have sheafification, for stacks we have “stack-fication”.

定理 4.3.0.3 (Theorem 4.6.5 in Martin). Let $\mathcal{F}$ be category fibered in groupoids over $\mathcal{C}$. Then there exists stack ${\mathcal{F}}^a/\mathcal{C}$ and $\mathcal{F}\to {\mathcal{F}}^a$ such that for all stacks $\mathcal{H}/\mathcal{C}$, we have ${\mathrm{HOM}}_{\mathcal{C}}({\mathcal{F}}^a,\mathcal{H})\stackrel{\sim }{\to }{\mathrm{HOM}}_{\mathcal{C}}(\mathcal{F},\mathcal{H})$.

Next, we are going to define algebraic spaces, but first, we give some ideas.

Idea: what is a scheme? A scheme is affine schemes glued in Zariski topology. Then algebraic space is affine schemes glued in etale topology.

Of course, now we are just begging for the question of what if fppf topology. It turns out, it is not so easy to answer this question. The answer is that a theorem of Artin, where he showed they are the same as algebraic spaces.

Let $S$ be a scheme, $\mathcal{C}=(\mathsf{Sch}/S{)}_{et}$.

定义 4.3.0.4. A morphism of sheaves $\mathcal{F}\to \mathcal{H}$ is representable by schemes if for all $T\to \mathcal{H}$ with $T=h_T$ scheme, the fibered product (as cat fibered in sets) $\mathcal{F}{\times }_{\mathcal{H}}T$ is a scheme.

So, when we say $\mathcal{F}\to \mathcal{H}$ is representable by schemes, we mean for all $T$ we get the following diagram

定义 4.3.0.5. Let $P$ be a property of morphisms of schemes. If for all diagramswe have:

1.	$f$ has $P$ implies $f^{\prime }$ has $P$, then we say $P$ is stable under base change.
2.	if $g$ is a Zar (et, sm, fppf) covering, then $f$ has $P$ iff $f^{\prime }$ has $P$, then we say $P$ is local on the base for the Zar (et, sm, fppf) topology.

Finally, we say $P$ is local on the source for the Zar (et, sm, fppf) topology, if for all diagramswith $\pi$ a Zar (et, sm, fppf) covering, $f$ has $P$ iff $g$ has $P$.

定义 4.3.0.6. If $\mathcal{F}\stackrel{f}{\to }\mathcal{H}$ is representable by schemes and $P$ is a property which is stable under base change and local on the base, then we say $\mathbf{f}$ has $\mathbf{P}$ iff for all $T\to \mathcal{H}$ with $T$ schemes, $\mathcal{F}{\times }_{\mathcal{H}}T\to T$ has $P$.

We remark that if $\mathcal{F}=h_X$ and $\mathcal{H}=h_Y$ then $\mathcal{F}\to \mathcal{H}$ has $P$ iff $X\to Y$ has $P$ because we can consider the identity $Y\to \mathcal{H}$ map and pull it back using this.

引理 4.3.0.7. Let $\mathcal{F}$ be a presheaf on $\mathsf{Sch}/S$. Then $\Delta :\mathcal{F}\to {\mathcal{F}}_S\mathcal{F}$ is representable by schemes if and only if for all $T\to \mathcal{F}$ is representable by schemes for all schemes $T$.

证明. ConsiderThen we get

Here $\mathcal{H}$ is $(x,t,t^{\prime })$ with $(x,x)=\Delta (x)=(f(t),f^{\prime }(t))$. So, $\mathcal{H}\cong T^{\prime }{\times }_{\mathcal{F}}T$. $\Delta$ representable by schemes, so $\mathcal{H}$ is scheme, so $T^{\prime }{\times }_{\mathcal{F}}T$ is scheme, as desired.

$\square$

定义 4.3.0.8. An algebraic space over $S$ is a sheaf $\mathcal{F}$ for the big etale topology on $\mathsf{Sch}/S$ such that:

1.	$\Delta :\mathcal{F}\to \mathcal{F}{\times }_S\mathcal{F}$ is representable by scheme.
2.	there exists scheme $U$ and etale covering $\pi :U\twoheadrightarrow F$

We note $(2)$ makes sense because $(1)$ implies $\pi$ is representable by schemes and etale surjections are property that are stable under base change and local on the base.

注 4.3.0.9. Schemes are algebraic spaces because we see $(1)$ is true as $\mathcal{F}=h_X$ is a scheme, and for $(2)$ we choose $\pi ={\mathrm{Id}}_X$. This is because $h_X$ is sheaf for fppf topology, so also it is also sheaf for etale topology.

定义 4.3.0.10. A morphism $\mathfrak{X}\to \mathfrak{Y}$ of stacks is representable if for all diagrams with $Y$ algebraic spacewe get $X$ is an algebraic space.

命题 4.3.0.11. A morphism of stacks $\mathfrak{X}\to \mathfrak{Y}$ is representable iff for all diagrams with $Y$ a schemewe get $X$ is algebraic space.

证明. If $Y$ is algebraic space then we know $X$ is a stack, which proves the forward direction. We show the converse. Hence, suppose we are given $Y\to \mathfrak{Y}$ where $Y$ is algebraic space, we want to show $\mathfrak{Z}$ is an algebraic space, where the stack $\mathfrak{Z}$ is defined by

$(1)$: we want to show $\mathfrak{Z}$ is a sheaf. Since $\mathfrak{Z}$ is a stack, so we just need to show $\mathfrak{Z}$ is fibered in sets. Now consider this diagramwhere we know $Y^{\prime }$ is a scheme, $Y$ an algebraic space, and $T$ a scheme. We have $x\underset{\varphi }{\overset{\sim }{\to }}y$ in $\mathfrak{Z}(T)$ and we want $\varphi$ to be the identity map. However, ${\mathfrak{Z}}^{\prime }$ is algebraic space by hypothesis, $T^{\prime }$ is a scheme by definition of $Y$ being algebraic space. We get $g^{\ast }\varphi :x^{\prime }\stackrel{\sim }{\to }{y}^{\prime }$ and ${\mathfrak{Z}}^{\prime }$ is algebraic space, so $g^{\ast }\varphi =\mathrm{Id}$. $g$ is etale covering and $\mathfrak{Z}$ is a stack, and $g^{\ast }\varphi =\mathrm{Id}$, hence $\varphi =\mathrm{Id}$.

$(2)$: we want to show $\mathfrak{Z}$ has etale covering by scheme. We get $\mathrm{sch}\twoheadrightarrow {\mathfrak{Z}}^{\prime }\twoheadrightarrow \mathfrak{Z}$ where ${\mathfrak{Z}}^{\prime }$ is algebraic space and all arrows are etale, hence we are done.

$(3)$: It remains to show ${\Delta }_{\mathfrak{Z}}$ is repable by schemes. We do a diagram again:where $T$ is a scheme. This gives$$\mathfrak{Z}{\times }_{\Delta ,\mathfrak{Z}{\times }_Y\mathfrak{Z},g}T={\mathfrak{Z}}_T{\times }_{\Delta ,{\mathfrak{Z}}_T{\times }_T{\mathfrak{Z}}_T,s}T=:W$$where $W$ is a scheme since ${\mathfrak{Z}}_T$ is an algebraic space. This concludes the proof.

$\square$