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The identity model of a smooth group with abelian generic fibre

Statement

Assume AC and DC as inherited from the supplied algebra and scheme results. Let $R$ be a discrete valuation ring with fraction field $K$ and residue field $k$ , and let $G$ be a smooth separated finite-type $R$ -group scheme (Group schemes over a base scheme) whose generic fibre $G_{K}$ is an abelian variety (Abelian varieties over a field). Then $G^{0} = G_{K} \cup (G_{k})^{0}$ is an open $R$ -subgroup scheme of $G$ , smooth, separated and of finite type over $R$ , with geometrically connected fibres. On each geometric fibre of $G$ , the orbits of $G^{0}$ are exactly the connected components of that fibre.

Facts & Assumptions

Given: AC and DC, a DVR $R$ with fraction field $K$ and residue field $k$ , a smooth separated finite-type $R$ -group scheme $G$ with abelian generic fibre $G_{K}$ , and the identity component $(G_{k})^{0}$ of the special fibre.

[F1]

$G_{K}$ is an abelian variety, hence connected; $G_{k}$ is a smooth finite-type $k$ -group scheme with finitely many connected components, the identity component $(G_{k})^{0}$ being open and a subgroup scheme (Abelian varieties over a field, Group schemes over a base scheme).

[F2]

A connected smooth finite-type group scheme over a field is geometrically connected, and a smooth connected such group is geometrically integral; these statements propagate through field extension (Connected finite-type groups are geometrically connected, assuming AC).

[F3]

A scheme which is smooth over a discrete valuation ring is flat over it, and a closed subscheme of a scheme over a DVR whose generic and special fibres are both empty is empty; a morphism of $R$ -schemes whose restriction to the generic fibre and to the special fibre both factor through an open subscheme factors through it (Locally Noetherian and Noetherian schemes, Group schemes over a base scheme).

Proof

technique · direct: the subscheme is open by construction, closed under the group operations fibrewise, and its orbits are computed by translation

1.1F1F2givenconstruct

The set $G^{0} = G_{K} \cup (G_{k})^{0}$ is open in $G$ : by [F1], $G_{k}$ has finitely many connected components, so $G_{k} ∖ (G_{k})^{0}$ is closed in $G_{k}$ . Since the special fibre $G_{k}$ is closed in $G$ , this complement is closed in $G$ . Its complement in $G$ is exactly $G_{K} \cup (G_{k})^{0}$ , which is therefore open. It is an open subscheme, hence smooth, separated and of finite type over $R$ . Its generic fibre is the connected abelian variety $G_{K}$ and its special fibre is $(G_{k})^{0}$ , connected; by [F2] the special fibre is geometrically connected and the generic fibre is geometrically connected, so the fibres over the two points of $Spec R$ are geometrically connected.

2.1F1F3step 1.1algebra

The multiplication, inverse and unit of $G$ restrict to $G^{0}$ : the multiplication $m_{G}$ maps $G_{K} \times_{K} G_{K}$ into $G_{K}$ and $(G_{k})^{0} \times_{k} (G_{k})^{0}$ into $(G_{k})^{0}$ because $(G_{k})^{0}$ is a subgroup scheme; hence the preimage $m_{G}^{- 1} (G^{0}) \subseteq G \times_{R} G$ is an open subscheme containing both $G_{K} \times_{K} G_{K}$ and $(G_{k})^{0} \times_{k} (G_{k})^{0}$ . The complement of this preimage inside the open subscheme $G^{0} \times_{R} G^{0}$ is closed with empty generic and special fibres, hence empty by [F3]; therefore $m_{G}$ restricts to $G^{0} \times_{R} G^{0} \to G^{0}$ . The same argument with the inverse and the unit section (whose value at the closed point lies in $(G_{k})^{0}$ ) shows that $G^{0}$ is an $R$ -subgroup scheme of $G$ .

2.2F1F2step 1.1algebra

Let $\overset{ˉ}{s}$ be a geometric point of $Spec R$ . If $\overset{ˉ}{s}$ lies over the generic point, $G_{\overset{ˉ}{s}} = (G_{K})_{\overset{ˉ}{s}}$ is connected by [F2]; if $\overset{ˉ}{s}$ lies over the closed point, $G_{\overset{ˉ}{s}} = (G_{k})_{\overset{ˉ}{s}}$ and $(G^{0})_{\overset{ˉ}{s}} = ((G_{k})^{0})_{\overset{ˉ}{s}} = (G_{\overset{ˉ}{s}})^{0}$ because the identity component of a smooth group scheme over a field is geometrically connected by [F2]. Thus $G_{\overset{ˉ}{s}}^{0}$ is the identity component of the smooth group $G_{\overset{ˉ}{s}}$ .

3.1F1step 2.2algebra∎

On a geometric fibre $G_{\overset{ˉ}{s}}$ , translation by a point $x$ is an isomorphism $G_{\overset{ˉ}{s}} \to G_{\overset{ˉ}{s}}$ carrying the identity component onto the connected component of $x$ ; since $G_{\overset{ˉ}{s}}^{0}$ is the identity component by step 2.2, the orbit of $x$ under $G_{\overset{ˉ}{s}}^{0}$ is exactly the connected component of $x$ . Hence the orbits of $G^{0}$ on each geometric fibre are the connected components, as claimed.

Depends on

Used by

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Bosch, Lutkebohmert, Raynaud, Neron Models (1990), 6.4 (identity components over a DVR) (standard reference, not scraped)