Let $G/S$ be a group scheme and $H \leq G$ an open subgroup scheme. Is $H \subseteq G$ closed? I want to apply this to $G^0 \leq G$ (see SGA 3, VI_B, Théorème 3.10) for $G$ commutative.

(*) If $S = \mathrm{Spec}(K)$, this is proven in http://jmilne.org/math/CourseNotes/iAG200.pdf Proposition 1.27 by the usual argument: the complement is the disjoint open union of the cosets of $H$.

If $G/H$ is representable by a group scheme and everything is separated, apply Exercise 1(ii) of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf to $\pi: G \to G/H$. But I want to prove this without having to assume $G/H$ being representable. It seems that if $H \subseteq G$ is closed, $G/H$ is representable (under certain conditions) ...

(This is a cross-post from https://math.stackexchange.com/questions/1670288/open-subgroup-scheme-closed)

Edit: Assume $S$ the spectrum of a DVR. Then, by (*), $G^0_\eta \subseteq G_\eta$ and $G^0_s \subseteq G_s$ are closed. Now, if everything is separated and flat, does it follow that $G^0 \subseteq G$ is closed?


Nope, because the affine line with zero doubled seen as a group scheme over the affine line is a counter example with subgroup being $G^0$. A more interesting (separated) example is to take an elliptic curve degenerating to a banana curve and taking the connected component of the N'eron model.

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    $\begingroup$ Among the interesting examples near the end of Raynaud’s thesis (Springer LNM 119) is a smooth affine group $G$ over $\mathbf{A}^2_k$ (${\rm{char}}(k)=0$) whose open relative identity component $G^0$ is shown to be not affine (even worse than not closed!): see (iii) in section 3 of Chapter VII. This is also optimal for being quasi-affine but not affine because any flat separated group scheme of finite type over a discrete valuation ring with affine generic fiber is affine: see Prop. 3.1 of the paper “On quasi-reductive group schemes” by G. Prasad and J-K Yu for Raynaud's proof of this fact. $\endgroup$ – nfdc23 Feb 25 '16 at 14:38

For $k=\mathbb{C}$ (or any field of characteristic $\ne2$) let $G$ be the subgroup scheme of $SL(2)\times\mathbb{A}^1$ consisting of all matrices of the form $\begin{pmatrix}x&sy\\y&x\end{pmatrix}$. The defining equation is $x^2-sy^2=1$ which shows that $G$ is an irreducible variety. For $s\ne0$ the fiber is a torus, hence connected. The fiber over $s=0$ is $\mathbf{G}_a\times\{\pm1\}$, hence disconnected. Thus $G^0$ is $G$ minus $\{s=0,x=-1\}$, hence open but not closed.


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