65.16 Change of base scheme
In this section we briefly discuss what happens when we change base schemes. The upshot is that given a morphism $S \to S'$ of base schemes, any algebraic space over $S$ can be viewed as an algebraic space over $S'$. And, given an algebraic space $F'$ over $S'$ there is a base change $F'_ S$ which is an algebraic space over $S$. We explain only what happens in case $S \to S'$ is a morphism of the big fppf site under consideration, if only $S$ or $S'$ is contained in the big site, then one first enlarges the big site as in Section 65.15.
Lemma 65.16.1. Suppose given a big site $\mathit{Sch}_{fppf}$. Let $g : S \to S'$ be morphism of $\mathit{Sch}_{fppf}$. Let $j : (\mathit{Sch}/S)_{fppf} \to (\mathit{Sch}/S')_{fppf}$ be the corresponding localization functor. Let $F$ be a sheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Then
for a scheme $T'$ over $S'$ we have $j_!F(T'/S') = \coprod \nolimits _{\varphi : T' \to S} F(T' \xrightarrow {\varphi } S),$
if $F$ is representable by a scheme $X \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, then $j_!F$ is representable by $j(X)$ which is $X$ viewed as a scheme over $S'$, and
if $F$ is an algebraic space over $S$, then $j_!F$ is an algebraic space over $S'$, and if $F = U/R$ is a presentation, then $j_!F = j(U)/j(R)$ is a presentation.
Let $F'$ be a sheaf of sets on $(\mathit{Sch}/S')_{fppf}$. Then
for a scheme $T$ over $S$ we have $j^{-1}F'(T/S) = F'(T/S')$,
if $F'$ is representable by a scheme $X' \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S')_{fppf})$, then $j^{-1}F'$ is representable, namely by $X'_ S = S \times _{S'} X'$, and
if $F'$ is an algebraic space, then $j^{-1}F'$ is an algebraic space, and if $F' = U'/R'$ is a presentation, then $j^{-1}F' = U'_ S/R'_ S$ is a presentation.
Proof.
The functors $j_!$, $j_*$ and $j^{-1}$ are defined in Sites, Lemma 7.25.8 where it is also shown that $j = j_{S/S'}$ is the localization of $(\mathit{Sch}/S')_{fppf}$ at the object $S/S'$. Hence all of the material on localization functors is available for $j$. The formula in (1) is Sites, Lemma 7.27.1. By definition $j_!$ is the left adjoint to restriction $j^{-1}$, hence $j_!$ is right exact. By Sites, Lemma 7.25.5 it also commutes with fibre products and equalizers. By Sites, Lemma 7.25.3 we see that $j_!h_ X = h_{j(X)}$ hence (2) holds. If $F$ is an algebraic space over $S$, then we can write $F = U/R$ (Lemma 65.9.1) and we get
\[ j_!F = j(U)/j(R) \]
because $j_!$ being right exact commutes with coequalizers, and moreover $j(R) = j(U) \times _{j_!F} j(U)$ as $j_!$ commutes with fibre products. Since the morphisms $j(s), j(t) : j(R) \to j(U)$ are simply the morphisms $s, t : R \to U$ (but viewed as morphisms of schemes over $S'$), they are still étale. Thus $(j(U), j(R), s, t)$ is an étale equivalence relation. Hence by Theorem 65.10.5 we conclude that $j_!F$ is an algebraic space.
Proof of (4), (5), and (6). The description of $j^{-1}$ is in Sites, Section 7.25. The restriction of the representable sheaf associated to $X'/S'$ is the representable sheaf associated to $X'_ S = S \times _{S'} Y'$ by Sites, Lemma 7.27.2. The restriction functor $j^{-1}$ is exact, hence $j^{-1}F' = U'_ S/R'_ S$. Again by exactness the sheaf $R'_ S$ is still an equivalence relation on $U'_ S$. Finally the two maps $R'_ S \to U'_ S$ are étale as base changes of the étale morphisms $R' \to U'$. Hence $j^{-1}F' = U'_ S/R'_ S$ is an algebraic space by Theorem 65.10.5 and we win.
$\square$
Note how the presentation $j_!F = j(U)/j(R)$ is just the presentation of $F$ but viewed as a presentation by schemes over $S'$. Hence the following definition makes sense.
Definition 65.16.2. Let $\mathit{Sch}_{fppf}$ be a big fppf site. Let $S \to S'$ be a morphism of this site.
If $F'$ is an algebraic space over $S'$, then the base change of $F'$ to $S$ is the algebraic space $j^{-1}F'$ described in Lemma 65.16.1. We denote it $F'_ S$.
If $F$ is an algebraic space over $S$, then $F$ viewed as an algebraic space over $S'$ is the algebraic space $j_!F$ over $S'$ described in Lemma 65.16.1. We often simply denote this $F$; if not then we will write $j_!F$.
The algebraic space $j_!F$ comes equipped with a canonical morphism $j_!F \to S$ of algebraic spaces over $S'$. This is true simply because the sheaf $j_!F$ maps to $h_ S$ (see for example the explicit description in Lemma 65.16.1). In fact, in Sites, Lemma 7.25.4 we have seen that the category of sheaves on $(\mathit{Sch}/S)_{fppf}$ is equivalent to the category of pairs $(\mathcal{F}', \mathcal{F}' \to h_ S)$ consisting of a sheaf on $(\mathit{Sch}/S')_{fppf}$ and a map of sheaves $\mathcal{F}' \to h_ S$. The equivalence assigns to the sheaf $\mathcal{F}$ the pair $(j_!\mathcal{F}, j_!\mathcal{F} \to h_ S)$. This, combined with the above, leads to the following result for categories of algebraic spaces.
Lemma 65.16.3. Let $\mathit{Sch}_{fppf}$ be a big fppf site. Let $S \to S'$ be a morphism of this site. The construction above give an equivalence of categories
\[ \left\{ \begin{matrix} \text{category of algebraic}
\\ \text{spaces over }S
\end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{category of pairs }(F', F' \to S)\text{ consisting}
\\ \text{of an algebraic space }F'\text{ over }S'\text{ and a}
\\ \text{morphism }F' \to S\text{ of algebraic spaces over }S'
\end{matrix} \right\} \]
Proof.
Let $F$ be an algebraic space over $S$. The functor from left to right assigns the pair $(j_!F, j_!F \to S)$ to $F$ which is an object of the right hand side by Lemma 65.16.1. Since this defines an equivalence of categories of sheaves by Sites, Lemma 7.25.4 to finish the proof it suffices to show: if $F$ is a sheaf and $j_!F$ is an algebraic space, then $F$ is an algebraic space. To do this, write $j_!F = U'/R'$ as in Lemma 65.9.1 with $U', R' \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S')_{fppf})$. Then the compositions $U' \to j_!F \to S$ and $R' \to j_!F \to S$ are morphisms of schemes over $S'$. Denote $U, R$ the corresponding objects of $(\mathit{Sch}/S)_{fppf}$. The two morphisms $R' \to U'$ are morphisms over $S$ and hence correspond to morphisms $R \to U$. Since these are simply the same morphisms (but viewed over $S$) we see that we get an étale equivalence relation over $S$. As $j_!$ defines an equivalence of categories of sheaves (see reference above) we see that $F = U/R$ and by Theorem 65.10.5 we see that $F$ is an algebraic space.
$\square$
The following lemma is a slight rephrasing of the above.
Lemma 65.16.4. Let $\mathit{Sch}_{fppf}$ be a big fppf site. Let $S \to S'$ be a morphism of this site. Let $F'$ be a sheaf on $(\mathit{Sch}/S')_{fppf}$. The following are equivalent:
The restriction $F'|_{(\mathit{Sch}/S)_{fppf}}$ is an algebraic space over $S$, and
the sheaf $h_ S \times F'$ is an algebraic space over $S'$.
Proof.
The restriction and the product match under the equivalence of categories of Sites, Lemma 7.25.4 so that Lemma 65.16.3 above gives the result.
$\square$
We finish this section with a lemma on a compatibility.
Lemma 65.16.5. Let $\mathit{Sch}_{fppf}$ be a big fppf site. Let $S \to S'$ be a morphism of this site. Let $F$ be an algebraic space over $S$. Let $T$ be a scheme over $S$ and let $f : T \to F$ be a morphism over $S$. Let $f' : T' \to F'$ be the morphism over $S'$ we get from $f$ by applying the equivalence of categories described in Lemma 65.16.3. For any property $\mathcal{P}$ as in Definition 65.5.1 we have $\mathcal{P}(f') \Leftrightarrow \mathcal{P}(f)$.
Proof.
Suppose that $U$ is a scheme over $S$, and $U \to F$ is a surjective étale morphism. Denote $U'$ the scheme $U$ viewed as a scheme over $S'$. In Lemma 65.16.1 we have seen that $U' \to F'$ is surjective étale. Since
\[ j(T \times _{f, F} U) = T' \times _{f', F'} U' \]
the morphism of schemes $T \times _{f, F} U \to U$ is identified with the morphism of schemes $T' \times _{f', F'} U' \to U'$. It is the same morphism, just viewed over different base schemes. Hence the lemma follows from Lemma 65.11.4.
$\square$
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