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Let $k$ be a perfect field of characteristic $p>0$. Let $K$ be a finite, totally ramified extension of $K_0:=\mathrm{Frac}\ W(k)$ and let $\mathcal{O}_K$ be the ring of integers of $K$. All group schemes in this question will be assumed to be commutative. If $\mathscr{V}$ is a finite flat group scheme over $\mathcal{O}_K$, we say that $\mathscr{V}$ is $\mathit{maximal}$ if for any other finite flat (commutative) group scheme $\mathscr{W}$ over $\mathcal{O}_K$ which has the same generic fibre as $\mathscr{V}$, there is a morphism $\mathscr{V}\to\mathscr{W}$ which restricts to the identity on the generic fibre. By a theorem of Raynaud, we can replace any finite flat group scheme over $\mathcal{O}_K$ by a maximal one without changing the generic fibre.

Now the question: Let $\mathscr{V}$ be a finite flat group scheme as before and suppose that $\mathscr{V}$ is maximal. Is it true that the connected component $\mathscr{V}^0$ of $\mathscr{V}$ is maximal as well?

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  • $\begingroup$ In general, $\mathscr{V}^0$ doesn't make sense. $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 17, 2015 at 20:09
  • $\begingroup$ @LaurentMoret-Bailly: By $\mathscr{V}^0$ I believe XYZ means the connected component of the identity section; this is a good notion over a henselian local base, as you know. In what way does it "not make sense" for the setting of the question posed? $\endgroup$
    – nfdc23
    Commented Nov 17, 2015 at 22:11
  • $\begingroup$ @nfdc23: OK, so doesn't have connected fibers in general. In SGA, the notation $G^0$ for a group scheme $G$ is used for the "scheme of identity components of fibers" whenever that exists. $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 18, 2015 at 7:42
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    $\begingroup$ @LaurentMoret-Bailly: Yes, I agree with you. In the setting over a finite flat group scheme over a henselian local ring I believe the (abuse of) notation $G^0$ to mean the literal connected component of the identity section rather than the SGA3 notion is often used as a convention, such as going back to Tate's paper on p-divisible groups. $\endgroup$
    – nfdc23
    Commented Nov 20, 2015 at 0:31

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Yes, and the same holds with $\mathscr{V}^0$ replaced by any finite flat $O_K$-subgroup $\mathscr{V}'$ of $\mathscr{V}$ (and with $O_K$ replaced by any discrete valuation ring of generic characteristic 0). Consider a map $f:\mathscr{V}' \rightarrow \mathscr{H}$ between finite flat $O_K$-group schemes that is an isomorphism on generic fibers. We want to show that $f$ is an isomorphism.

Form the pushout of the exact sequence $$1 \rightarrow \mathscr{V}' \rightarrow \mathscr{V} \rightarrow \mathscr{V}'' \rightarrow 1$$ against $f$ to get a short exact sequence of finite flat $O_K$-group schemes $$1 \rightarrow \mathscr{H} \rightarrow \mathscr{G} \rightarrow \mathscr{V}'' \rightarrow 1.$$ There is an evident commutative diagram from the first of these to the second inducing $f$ along the left and the identity map for $\mathscr{V}''$ on the right, so the map in the middle $\mathscr{V} \rightarrow \mathscr{G}$ is an isomorphism on generic fibers (as the outer maps have that property). By maximality of $\mathscr{V}$ it follows that this middle map is an isomorphism, so the same then holds for $f$ by the snake lemma or whatever for fppf abelian sheaves.

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  • $\begingroup$ Doesn't this show that finite flat $\mathcal{O}_K$-subgroups of minimal finite flat group schemes are minimal (with the notion of minimality defined exactly as in the question, only with the arrows reversed)? I think in order to deduce maximality of $\mathscr{V}'$ one would have to consider a morphism $\mathscr{H}\to\mathscr{V}'$ rather than the other way round, in which case one cannot take pullbacks or pushouts anymore. $\endgroup$
    – XYZ
    Commented Nov 18, 2015 at 10:51
  • $\begingroup$ I was interpreting "maximal" in the sense of containments at the generic-fiber level among finite flat group schemes over the dvr, rather than among coordinate rings; i.e., I thought by "maximal" you meant what you seem to have regarded as "minimal". So indeed I have done the opposite of what you wanted. I haven't looked at Raynaud's paper in a long time, so I didn't remember offhand which way he chose for his conventions. Sorry for misunderstanding. $\endgroup$
    – nfdc23
    Commented Nov 20, 2015 at 0:34

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