The Stacks project

96.11 Restriction of scalars

Suppose $X \to Z \to B$ are morphisms of algebraic spaces over $S$. Given a scheme $T$ we can consider pairs $(a, b)$ where $a : T \to B$ is a morphism and $b : T \times _{a, B} Z \to X$ is a morphism over $Z$. Picture

96.11.0.1
\begin{equation} \label{criteria-equation-pairs} \vcenter { \xymatrix{ & X \ar[d] \\ T \times _{a, B} Z \ar[d] \ar[ru]^ b \ar[r] & Z \ar[d] \\ T \ar[r]^ a & B } } \end{equation}

In this situation we can define a functor

96.11.0.2
\begin{equation} \label{criteria-equation-restriction-of-scalars} \text{Res}_{Z/B}(X) : (\mathit{Sch}/S)^{opp} \longrightarrow \textit{Sets}, \quad T \longmapsto \{ (a, b)\text{ as above}\} \end{equation}

Sometimes we think of this as a functor defined on the category of schemes over $B$, in which case we drop $a$ from the notation.

Lemma 96.11.1. Let $S$ be a scheme. Let $X \to Z \to B$ be morphisms of algebraic spaces over $S$. Then

  1. $\text{Res}_{Z/B}(X)$ is a sheaf on $(\mathit{Sch}/S)_{fppf}$.

  2. If $T$ is an algebraic space over $S$, then there is a canonical bijection

    \[ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})}(T, \text{Res}_{Z/B}(X)) = \{ (a, b)\text{ as in }(05Y9)\} \]

Proof. Let $T$ be an algebraic space over $S$. Let $\{ T_ i \to T\} $ be an fppf covering of $T$ (as in Topologies on Spaces, Section 72.7). Suppose that $(a_ i, b_ i) \in \text{Res}_{Z/B}(X)(T_ i)$ such that $(a_ i, b_ i)|_{T_ i \times _ T T_ j} = (a_ j, b_ j)|_{T_ i \times _ T T_ j}$ for all $i, j$. Then by Descent on Spaces, Lemma 73.7.2 there exists a unique morphism $a : T \to B$ such that $a_ i$ is the composition of $T_ i \to T$ and $a$. Then $\{ T_ i \times _{a_ i, B} Z \to T \times _{a, B} Z\} $ is an fppf covering too and the same lemma implies there exists a unique morphism $b : T \times _{a, B} Z \to X$ such that $b_ i$ is the composition of $T_ i \times _{a_ i, B} Z \to T \times _{a, B} Z$ and $b$. Hence $(a, b) \in \text{Res}_{Z/B}(X)(T)$ restricts to $(a_ i, b_ i)$ over $T_ i$ for all $i$.

Note that the result of the preceding paragraph in particular implies (1).

Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say “pair” we mean a pair $(a, b)$ fitting into (96.11.0.1).

Let $v : T \to \text{Res}_{Z/B}(X)$ be a natural transformation. Choose a scheme $U$ and a surjective étale morphism $p : U \to T$. Then $v(p) \in \text{Res}_{Z/B}(X)(U)$ corresponds to a pair $(a_ U, b_ U)$ over $U$. Let $R = U \times _ T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_ U, b_ U)$ by $s$ and $t$ agree. Hence, since $\{ U \to T\} $ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, b)$ over $T$.

Conversely, let $(a, b)$ be a pair over $T$. Let $U \to T$, $R = U \times _ T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, b)|_ U$ gives rise to a transformation of functors $v : h_ U \to \text{Res}_{Z/B}(X)$ by the Yoneda lemma (Categories, Lemma 4.3.5). As the two pullbacks $s^*(a, b)|_ U$ and $t^*(a, b)|_ U$ are equal, we see that $v$ coequalizes the two maps $h_ t, h_ s : h_ R \to h_ U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma 64.9.1 and since $\text{Res}_{Z/B}(X)$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to \text{Res}_{Z/B}(X)$.

We omit the verification that the two constructions above are mutually inverse. $\square$

Of course the sheaf $\text{Res}_{Z/B}(X)$ comes with a natural transformation of functors $\text{Res}_{Z/B}(X) \to B$. We will use this without further mention in the following.

Lemma 96.11.2. Let $S$ be a scheme. Let $X \to Z \to B$ and $B' \to B$ be morphisms of algebraic spaces over $S$. Set $Z' = B' \times _ B Z$ and $X' = B' \times _ B X$. Then

\[ \text{Res}_{Z'/B'}(X') = B' \times _ B \text{Res}_{Z/B}(X) \]

in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})$.

Proof. The equality as functors follows immediately from the definitions. The equality as sheaves follows from this because both sides are sheaves according to Lemma 96.11.1 and the fact that a fibre product of sheaves is the same as the corresponding fibre product of pre-sheaves (i.e., functors). $\square$

Lemma 96.11.3. Let $S$ be a scheme. Let $X' \to X \to Z \to B$ be morphisms of algebraic spaces over $S$. Assume

  1. $X' \to X$ is étale, and

  2. $Z \to B$ is finite locally free.

Then $\text{Res}_{Z/B}(X') \to \text{Res}_{Z/B}(X)$ is representable by algebraic spaces and étale. If $X' \to X$ is also surjective, then $\text{Res}_{Z/B}(X') \to \text{Res}_{Z/B}(X)$ is surjective.

Proof. Let $U$ be a scheme and let $\xi = (a, b)$ be an element of $\text{Res}_{Z/B}(X)(U)$. We have to prove that the functor

\[ h_ U \times _{\xi , \text{Res}_{Z/B}(X)} \text{Res}_{Z/B}(X') \]

is representable by an algebraic space étale over $U$. Set $Z_ U = U \times _{a, B} Z$ and $W = Z_ U \times _{b, X} X'$. Then $W \to Z_ U \to U$ is as in Lemma 96.9.2 and the sheaf $F$ defined there is identified with the fibre product displayed above. Hence the first assertion of the lemma. The second assertion follows from this and Lemma 96.9.1 which guarantees that $F \to U$ is surjective in the situation above. $\square$

At this point we can use the lemmas above to prove that $\text{Res}_{Z/B}(X)$ is an algebraic space whenever $Z \to B$ is finite locally free in almost exactly the same way as in the proof that $\mathit{Mor}_ B(Z, X)$ is an algebraic spaces, see Proposition 96.10.4. Instead we will directly deduce this result from the following lemma and the fact that $\mathit{Mor}_ B(Z, X)$ is an algebraic space.

Lemma 96.11.4. Let $S$ be a scheme. Let $X \to Z \to B$ be morphisms of algebraic spaces over $S$. The following diagram

\[ \xymatrix{ \mathit{Mor}_ B(Z, X) \ar[r] & \mathit{Mor}_ B(Z, Z) \\ \text{Res}_{Z/B}(X) \ar[r] \ar[u] & B \ar[u]_{\text{id}_ Z} } \]

is a cartesian diagram of sheaves on $(\mathit{Sch}/S)_{fppf}$.

Proof. Omitted. Hint: Exercise in the functorial point of view in algebraic geometry. $\square$

Proposition 96.11.5. Let $S$ be a scheme. Let $X \to Z \to B$ be morphisms of algebraic spaces over $S$. If $Z \to B$ is finite locally free then $\text{Res}_{Z/B}(X)$ is an algebraic space.

Proof. By Proposition 96.10.4 the functors $\mathit{Mor}_ B(Z, X)$ and $\mathit{Mor}_ B(Z, Z)$ are algebraic spaces. Hence this follows from the cartesian diagram of Lemma 96.11.4 and the fact that fibre products of algebraic spaces exist and are given by the fibre product in the underlying category of sheaves of sets (see Spaces, Lemma 64.7.2). $\square$


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