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

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

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

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

of algebraic spaces over $K$. This morphism is symmetric as $G$ is abelian. Hence it factors through the quotient

On the other hand, the morphism $\mathop{\mathrm{Spec}}(L) \to P$ defines a morphism

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.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)