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Let $S$ be some base scheme, $H$ a finite flat group scheme over $S$, and $\alpha: \mu_p \to H$ a homomorphism of group schemes ($p$ a prime). Is the kernel of $\alpha$ necessarily flat over $S$?

(I know that kernels of general homomorphisms of FFGS $G \to H$ need not be flat, but I don't know of a counterexample when $G = \mu_p$.)

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This holds for any homomorphism $f: G\to H$ with $G$ of multiplicative type and of finite type, and $H$ separated and finitely presented. Here I assume that by "finite flat" you mean "finite locally free".

Reference: SGA3, IX, Thm 6.8.

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If I'm not mistaken, the kernel of $\alpha$ is necessarily flat, and indeed either equal to $\mu_p$ or $0$ if $S$ is connected. The key is that $\mu_p$ is of multiplicative type, and we have the following result (Corollary B.3.3 in Conrad's notes on reductive group schemes):

Let $S$ be a scheme and $H$ an $S$-group scheme of multiplicative type. Then any fppf closed subgroup of $H$ is also of multiplicative type (hence flat).

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    $\begingroup$ I don't follow: if the subgroup is fppf (= faithfully flat and finitely presented), it's in particular flat, so isn't this circular? Or is "fppf closed" something different from being closed and fppf over the base? $\endgroup$
    – David Loeffler
    Commented Sep 4, 2018 at 19:22
  • $\begingroup$ Sorry, you are right, this does not answer your question. $\endgroup$
    – Simon Pepin Lehalleur
    Commented Sep 4, 2018 at 19:49

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