Let $A$ and $B$ be abelian schemes over a base scheme $S$. There is the $\underline{\mathrm{Hom}}(A,B)$ functor $T \mapsto \mathrm{Hom}(A \times T, B \times T)$, where $\mathrm{Hom}$ means homomorphisms of group schemes.

Is it true that $\underline{\mathrm{Hom}}(A,B)$ is representable by a scheme of locally finite presentation over $S$?

(By an abelian scheme over $S$, I mean a smooth proper group scheme over $S$, with geometrically integral fibers.)

I am ok with assuming that $S$ is locally Noetherian if necessary.


If the abelian schemes $A$ and $B$ are moreover projective, one can make use of representability of the Hilbert scheme. I am not sure what to do if $A$ and $B$ are just proper, and not necessarily projective.

In the answer to this question

Representability of Hom of two finite flat group schemes

R. van Dobben de Bruyn mentions there is still an argument to show the representability of $\underline{\mathrm{Hom}}(A,B)$ by a scheme, but I have not yet found the argument.

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    $\begingroup$ Maybe one can use the algebraic space result linked in the linked answer and check some criterion for the algebraic space to be a scheme? In reasonable situations it's a disjoint union of closed subschemes of $S$, so one should have a lot of clearance to apply these criteria. $\endgroup$
    – Will Sawin
    Apr 5, 2022 at 20:49

1 Answer 1


Thanks to Will Sawin for the suggestion. That works, here are the details:

Write $\underline{\mathrm{Mor}}(A,B)$ for the functor describing morphisms of schemes (as opposed to $\underline{\mathrm{Hom}}(A,B)$ for morphisms of group schemes).

By [Tag 0D1C] we know that $\underline{\mathrm{Mor}}(A,B)$ is representable by an algebraic space, locally of finite presentation over $S$, and admits an open immersion into the Hilbert functor $\mathrm{Hilb}_{A \times_S B/S}$ [Tag 0D1B] (which is also an algebraic space). This Hilbert functor is separated over $S$ [Tag 0DM7], so $\underline{\mathrm{Mor}}(A,B)$ is also separated over $S$. Taking a fiber product as in the prevously linked post [https://mathoverflow.net/questions/314723/representability-of-hom-of-two-finite-flat-group-schemes] shows that $\underline{\mathrm{Hom}}(A,B)$ is representable by an algebraic space, which is separated and locally of finite presentation over $S$.

By the criterion in [https://mathoverflow.net/questions/4573/when-is-an-algebraic-space-a-scheme] (more precisely, the reference to Théorème A.2 in Champs Algébriques by Laumon and Moret-Bailly which says that a separated locally quasi-finite morphism of algebraic spaces is represented by schemes), it is enough to show that $\underline{\mathrm{Hom}}(A,B)$ is locally quasi-finite over $S$. For this, we reduce to the case where $S = \mathrm{Spec}~ k$ for a field $k$. In this case, one knows that $\underline{\mathrm{Hom}}(A,B)$ is in fact étale over $\mathrm{Spec}~k$ (rigidity for morphisms of abelian schemes), hence locally quasi-finite.

  • $\begingroup$ Welcome new contributor. The Hom functor of two Abelian varieties over Spec of a field is often not quasi-finite. Consider composition with "multiplication by $n$" isogenies (which gives a non-quasi-compact subgroup scheme). Even worse things can happen in families. $\endgroup$ Apr 5, 2022 at 22:40
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    $\begingroup$ Actually I wrote some version of this in my dissertation (§4.1) as I couldn't find a reference (but that doesn't mean that a reference doesn't exist!), exactly along the lines of this answer. DOI: 10.7916/D89K5TB0, or my website for a version with better typesetting. $\endgroup$ Apr 5, 2022 at 22:48
  • $\begingroup$ @JasonStarr Agreed, for example if $A = B$. In the argument though, I only need locally quasi-finite rather than quasi-finite. And thanks Remy for the reference. $\endgroup$
    – 351910953
    Apr 6, 2022 at 1:42

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