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In SGA 3 XI proposition 3.12 (b) (https://webusers.imj-prg.fr/~patrick.polo/SGA3/Expo11.pdf)

It is shown that:

If $G$ is an affine group scheme over a base $S$ and $H$ is a finite group scheme over $S$ of multiplicative type then $\underline{Hom}_{S-grp} (H,G)$ is representable.

Question: If we assume that $G$ is finite over $S$ is it possible that $\underline{Hom}_{S-grp} (H,G)$ will be a proper $S$ scheme?

A perhaps related is result of Abramovich, Olsson and Vistoli is that if G and H are both finite, flat linearly reductive group schemes then $\underline{Hom}_{S-grp} (H,G)$ is finite over $S$. This is Lemma 6.4 of https://arxiv.org/pdf/0801.3040.pdf.

Thanks!

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    $\begingroup$ Consider the case that $H$ is $\mu_p$ over $S=\text{Spec}\ \mathbb{Z}_p$ and $G$ is the finite ‘etale group scheme that is constant with fiber $\mathbb{Z}/p\mathbb{Z}$. $\endgroup$
    – Jason Starr
    Jan 7, 2022 at 16:13
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    $\begingroup$ I believe you can also get examples in equal characteristic. For a DVR over a field of characteristic $p$, there is a finite flat group scheme $G$ with generic fiber $\mu_p$ and special fiber $\alpha_p$. For $H = \mu_p$, it seems that the scheme $Hom(H,G)$ has special fiber a reduced point, and the generic fiber has $p$ points, so it can't possible be proper. In the positive direction, if $H$ is smooth then it is true that $Hom(H, G)$ is proper. $\endgroup$
    – afh
    Jan 7, 2022 at 16:57
  • $\begingroup$ @afh In the case I had in mind I think that $H$ will actually be smooth so the last sentence you wrote could also be useful to me. Do you mind sketching how that proof would go? $\endgroup$
    – Anette
    Jan 7, 2022 at 17:21
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    $\begingroup$ @Anette Sure, properness will actually be true for any finite etale $H$ (not necessarily of multiplicative type). First, since $H$ is essentially free over $S$ and $G$ is separated, it can be shown that $Hom_{gr}(H,G)$ is a closed subfunctor of the functor of all morphisms of schemes $Hom(H,G)$ (this can also be proven by hand in this special case). It suffices to check that $Hom(H,G)$ is represented by a finite $S$-scheme. This can be checked etale locally on $S$ by descent. After passing to an etale cover of $S$, we assume that the finite etale $H$ is a disjoint union of $n$ copies of $S$. $\endgroup$
    – afh
    Jan 7, 2022 at 17:42
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    $\begingroup$ In that case we have $Hom(H,G) = G^n$, which is finite over $S$ (because $G$ is finite over $S$). $\endgroup$
    – afh
    Jan 7, 2022 at 17:43

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