Lemma 110.71.1. Let $W$ be a two dimensional regular integral Noetherian scheme with function field $K$. Let $G \to W$ be an abelian scheme. Then the map $H^1_{fppf}(W, G) \to H^1_{fppf}(\mathop{\mathrm{Spec}}(K), G)$ is injective.
110.71 An example of a non-algebraic Hom-stack
Let $\mathcal{Y}, \mathcal{Z}$ be algebraic stacks over a scheme $S$. The Hom-stack $\underline{\mathop{\mathrm{Mor}}\nolimits }_ S(\mathcal{Y}, \mathcal{Z})$ is the stack in groupoids over $S$ whose category of sections over a scheme $T$ is given by the category
\[ \mathop{\mathrm{Mor}}\nolimits _ T(\mathcal{Y} \times _ S T, \mathcal{Z} \times _ S T) \]
whose objects are $1$-morphisms and whose morphisms are $2$-morphisms. We omit the proof this is indeed a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$ (insert future reference here). Of course, in general the Hom-stack will not be algebraic. In this section we give an example where it is not true and where $\mathcal{Y}$ is representable by a proper flat scheme over $S$ and $\mathcal{Z}$ is smooth and proper over $S$.
Let $k$ be an algebraically closed field which is not the algebraic closure of a finite field. Let $S = \mathop{\mathrm{Spec}}(k[[t]])$ and $S_ n = \mathop{\mathrm{Spec}}(k[t]/(t^ n)) \subset S$. Let $f : X \to S$ be a map satisfying the following
$f$ is projective and flat, and its fibres are geometrically connected curves,
the fibre $X_0 = X \times _ S S_0$ is a nodal curve with smooth irreducible components whose dual graph has a loop consisting of rational curves,
$X$ is a regular scheme.
To make such a surface $X$ we can take for example
\[ X\quad :\quad T_0T_1T_2 - t(T_0^3 + T_1^3 + T_2^3) = 0 \]
in $\mathbf{P}^2_{k[[t]]}$. Let $A_0$ be a non-zero abelian variety over $k$ for example an elliptic curve. Let $A = A_0 \times _{\mathop{\mathrm{Spec}}(k)} S$ be the constant abelian scheme over $S$ associated to $A_0$. We will show that the stack $\mathcal{X} = \underline{\mathop{\mathrm{Mor}}\nolimits }_ S(X, [S/A]))$ is not algebraic.
Recall that $[S/A]$ is on the one hand the quotient stack of $A$ acting trivially on $S$ and on the other hand equal to the stack classifying fppf $A$-torsors, see Examples of Stacks, Proposition 95.15.3. Observe that $[S/A] = [\mathop{\mathrm{Spec}}(k)/A_0] \times _{\mathop{\mathrm{Spec}}(k)} S$. This allows us to describe the fibre category over a scheme $T$ as follows
\begin{align*} \mathcal{X}_ T & = \underline{\mathop{\mathrm{Mor}}\nolimits }_ S(X, [S/A])_ T \\ & = \mathop{\mathrm{Mor}}\nolimits _ T(X \times _ S T, [S/A] \times _ S T) \\ & = \mathop{\mathrm{Mor}}\nolimits _ S(X \times _ S T, [S/A]) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathrm{Spec}}(k)}(X \times _ S T, [\mathop{\mathrm{Spec}}(k)/A_0]) \end{align*}
for any $S$-scheme $T$. In other words, the groupoid $\mathcal{X}_ T$ is the groupoid of fppf $A_0$-torsors on $X \times _ S T$. Before we discuss why $\mathcal{X}$ is not an algebraic stack, we need a few lemmas.
Sketch of proof. Let $P \to W$ be an fppf $G$-torsor which is trivial in the generic point. Then we have a morphism $\mathop{\mathrm{Spec}}(K) \to P$ over $W$ and we can take its scheme theoretic image $Z \subset P$. Since $P \to W$ is proper (as a torsor for a proper group algebraic space over $W$) we see that $Z \to W$ is a proper birational morphism. By Spaces over Fields, Lemma 72.3.2 the morphism $Z \to W$ is finite away from finitely many closed points of $W$. By (insert future reference on resolving indeterminacies of morphisms by blowing quadratic transformations for surfaces) the irreducible components of the geometric fibres of $Z \to W$ are rational curves. By More on Groupoids in Spaces, Lemma 79.11.3 there are no nonconstant morphisms from rational curves to group schemes or torsors over such. Hence $Z \to W$ is finite, whence $Z$ is a scheme and $Z \to W$ is an isomorphism by Morphisms, Lemma 29.54.8. In other words, the torsor $P$ is trivial. $\square$
Lemma 110.71.2. Let $G$ be a smooth commutative group algebraic space over a field $K$. Then $H^1_{fppf}(\mathop{\mathrm{Spec}}(K), G)$ is torsion.
Proof. Every $G$-torsor $P$ over $\mathop{\mathrm{Spec}}(K)$ is smooth over $K$ as a form of $G$. Hence $P$ has a point over a finite separable extension $L/K$. Say $[L : K] = n$. Let $[n](P)$ denote the $G$-torsor whose class is $n$ times the class of $P$ in $H^1_{fppf}(\mathop{\mathrm{Spec}}(K), G)$. There is a canonical morphism
\[ P \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} P \to [n](P) \]
of algebraic spaces over $K$. This morphism is symmetric as $G$ is abelian. Hence it factors through the quotient
\[ (P \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} P)/S_ n \]
On the other hand, the morphism $\mathop{\mathrm{Spec}}(L) \to P$ defines a morphism
\[ (\mathop{\mathrm{Spec}}(L) \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(L))/S_ n \longrightarrow (P \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} P)/S_ n \]
and the reader can verify that the scheme on the left has a $K$-rational point. Thus we see that $[n](P)$ is the trivial torsor. $\square$
To prove $\mathcal{X} = \underline{\mathop{\mathrm{Mor}}\nolimits }_ S(X, [S/A])$ is not an algebraic stack, by Artin's Axioms, Lemma 98.9.5, it is enough to show the following.
Lemma 110.71.3. The canonical map $\mathcal{X}(S) \to \mathop{\mathrm{lim}}\nolimits \mathcal{X}(S_ n)$ is not essentially surjective.
Sketch of proof. Unwinding definitions, it is enough to check that $H^1(X, A_0) \to \mathop{\mathrm{lim}}\nolimits H^1(X_ n, A_0)$ is not surjective. As $X$ is regular and projective, by Lemmas 110.71.2 and 110.71.1 each $A_0$-torsor over $X$ is torsion. In particular, the group $H^1(X, A_0)$ is torsion. It is thus enough to show: (a) the group $H^1(X_0, A_0)$ is non-torsion, and (b) the maps $H^1(X_{n + 1}, A_0) \to H^1(X_ n, A_0)$ are surjective for all $n$.
Ad (a). One constructs a nontorsion $A_0$-torsor $P_0$ on the nodal curve $X_0$ by glueing trivial $A_0$-torsors on each component of $X_0$ using non-torsion points on $A_0$ as the isomorphisms over the nodes. More precisely, let $x \in X_0$ be a node which occurs in a loop consisting of rational curves. Let $X'_0 \to X_0$ be the normalization of $X_0$ in $X_0 \setminus \{ x\} $. Let $x', x'' \in X'_0$ be the two points mapping to $x_0$. Then we take $A_0 \times _{\mathop{\mathrm{Spec}}(k)} X'_0$ and we identify $A_0 \times {x'}$ with $A_0 \times \{ x''\} $ using translation $A_0 \to A_0$ by a nontorsion point $a_0 \in A_0(k)$ (there is such a nontorsion point as $k$ is algebraically closed and not the algebraic closure of a finite field – this is actually not trivial to prove). One can show that the glueing is an algebraic space (in fact one can show it is a scheme) and that it is an nontorsion $A_0$-torsor over $X_0$. The reason that it is nontorsion is that if $[n](P_0)$ has a section, then that section produces a morphism $s : X'_0 \to A_0$ such that $[n](a_0) = s(x') - s(x'')$ in the group law on $A_0(k)$. However, since the irreducible components of the loop are rational to section $s$ is constant on them ( More on Groupoids in Spaces, Lemma 79.11.3). Hence $s(x') = s(x'')$ and we obtain a contradiction.
Ad (b). Deformation theory shows that the obstruction to deforming an $A_0$-torsor $P_ n \to X_ n$ to an $A_0$-torsor $P_{n + 1} \to X_{n + 1}$ lies in $H^2(X_0, \omega )$ for a suitable vector bundle $\omega $ on $X_0$. The latter vanishes as $X_0$ is a curve, proving the claim. $\square$
Proposition 110.71.4. The stack $\mathcal{X} = \underline{\mathop{\mathrm{Mor}}\nolimits }_ S(X, [S/A])$ is not algebraic.
Proof. See discussion above. $\square$
Remark 110.71.5. Proposition 110.71.4 contradicts [Theorem 1.1, AokiHomStacks]. The problem is the non-effectivity of formal objects for $\underline{\mathop{\mathrm{Mor}}\nolimits }_ S(X, [S/A])$. The same problem is mentioned in the Erratum [AokiHomStacksErr] to [AokiHomStacks]. Unfortunately, the Erratum goes on to assert that $\underline{\mathop{\mathrm{Mor}}\nolimits }_ S(\mathcal{Y}, \mathcal{Z})$ is algebraic if $\mathcal{Z}$ is separated, which also contradicts Proposition 110.71.4 as $[S/A]$ is separated.
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