# quotient of an algebraic group by the connected component containing identity

Suppose $G$ is a finite flat group over scheme $S$, let $G^0$ be the connected component containing identity. Is it true that the quotient sheaf $G^0 \G$ is always representable by a group scheme over $S$? If so, is it always flat over $S$?

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I'm not so sure. Perhaps it depends on what you mean by $G^0$. Consider for example $\mu_p$ over $\mathbf Z_p$. The special fibre of this is connected and the generic fibre is not. So do you want $G^0$ to be the entire special fibre and just the identity in the generic fibre? This I guess is a finite non-flat group scheme. I don't think the quotient will exist in this case---will it? I certainly can't believe that the quotient can be flat, in any case. –  Kevin Buzzard Nov 23 '10 at 21:29
[The reason I ask is that when $S$ is a DVR sometimes $G^0$ is used to mean something else, at least in the $p$-divisible group case] –  Kevin Buzzard Nov 23 '10 at 21:37
If $S$ is spec of henselian local ring (e.g., complete local noeth) then $G^0$ means the unique open and closed subscheme containing the identity section, but (as Kevin notes) its formation doesn't commute with non-local base change (e.g., passage to generic fiber over complete dvr). It's a subgp scheme. In general if $G$ is a finite loc. free comm. gp scheme over a scheme $S$ and $H$ is a finite loc. free closed subgp scheme then the fppf quotient sheaf $G/H$ is always reptble. Nice exercise to prove the dual map $D(G) \rightarrow D(H)$ is fppf, and that dual of its kernel does the job. –  BCnrd Nov 23 '10 at 22:10
As a separate example, it is proved in Messing's thesis that if $G \rightarrow S$ is a finite loc. free commutative group scheme of $p$-power order with $p$ locally nilpotent on $S$ and if the fibers $G_s$ have identity component whose order is locally constant in $s \in S$ then there exists a unique finite loc. free closed $S$-subgroup $G'$ in $G$ such that $G'_s = G_s$ for all $s \in S$; often this $G'$ is denoted as $G^0$. An interesting example is an $\mathbf{F}_p$-scheme $S$ and $G = A[p^n]$ for $A \rightarrow S$ an abelian scheme whose fibers have constant $p$-rank (e.g., all ordinary). –  BCnrd Nov 23 '10 at 22:17
I see, so the naive definition of the unique open and closed subscheme containing the identity section for $G^0$ is not interesting over a general base scheme because it's not going to be radical. But over an artinian scheme it's still going to work! That makes a lot of sense, thanks Kevin and Brian! –  Taisong Jing Nov 23 '10 at 22:38