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Let $S$ be a Dedekind scheme with rational functions field $K$. Consider an exact sequence $$ 0 \to G'_K \to G_K \to G''_K \to 0$$ of smooth $K$-group schemes of finite type. Assume that $G'_K$ and $G''_K$ admit Néron models $G'$ resp. $G''$ over $S$.

Question: Is there a way to extend the above sequence to an exact sequence $$ 0 \to G'_U \to G_U \to G''_U \to 0 $$ over some dense open $U \subset S$, where $G'_U = G' \times_S U$, $G''_U = G'' \times_S U$.

Ansatz: One may extend $G_K$ to $G_U$ over $U$ by using a limit process (EGA IV3, 8.2.2), and by taking a smaller $U$ one may assume, that $G'$ is smooth over $U$. Now using the Néron mapping property there is a unique map $G_U \to G''_U$ extending $G_K \to G''_K$, since $G''_U$ is also a Néron model of $G''_K$ over $U$. Now I am stuck.

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The question is motivated by the proof of the Proposition 7.5/1 in [BLR] Néron Models. –  finite Nov 2 '12 at 10:27
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2 Answers

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Let $H$ be the kernel of $G_U\to G''_U$. Then $H_K=G'_K$. Hence shrinking $U$ if necessary, $H$ becomes isomorphic to $G'_U$. Therefore we have an exact sequence of group schemes $$ 0\to G'_U \to G_U \to G''_U. $$ At the generic fiber, $G_K\to G''_K$ is surjective. The image of $G_U\to G''_U$ is a constructible subset (Chevalley's theorem) containing the generic fiber. Hence it contains an open subset $V$ containing $G''_K$. The image of $G''_U\setminus V$ in $U$ is constructible and doesn't contain the generic point of $U$, hence contained in a proper closed subset $Z\subset U$. Replacing $U$ with $U\setminus Z$, $G_U\to G''_U$ becomes surjective and $$ 0\to G'_U \to G_U \to G''_U\to 0$$ is exact.

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I know that the question is rather old, but I believe I should add this. As far as I understand the other answer, it is concerned with the scheme theoretic surjectivity of the last morphism. However, exactness for group schemes is defined differently: a group scheme can be regarded as representable sheaf over the fppf (or fpqc) topology and for fppf (or fpqc) sheaves there is a good notion of exactness. That's the one we use, cf. [SGA III$_\text{new}$] and [Poonen] Chapter 5.1.

Why should we use the fppf topology?

Short answer: we need an approachable definition which is still strong enough, i.e. we want to have the most general notion of exactness which is still useful.
Slightly longer answer: let us restrict ourselves to short exact sequences, then (as said below) we only need to worry about quotients. There are several possible notions, for example the categorical quotient, sheave quotient (in a certain topology), etc. The categorical quotient is certainly the most general quotient. Is it a good notion? For two practical reasons the answer is no. Firstly, how can we check that something is the categorical quotient? I don't know of any usable criterion. Secondly, the categorical quotient doesn't play nice with cohomology. Hence, it is probably hard to extract information. However, the last points hints at a way out. If we interpret our group schemes as representable sheaves in a topos and use its notion of exactness, then it is very compatible with cohomology. And there is also an easy way to check exactness (for short exact sequences), see below.
Indeed, [SGA III$_\text{new}$] uses the fpqc topos presumably for these reasons. We have to note two things: firstly, the fpqc topology has serious draw backs from a user point of view. For example, the fpqc cohomology has severe set-theoretical problems you cannot ignore as you are usually inclined to do. (key word: universes, catastrophic example: fpqc cohomology groups may depend on the universe) Secondly, one can show (see below) that for group schemes which are locally of finite presentation a short exact sequence is fqpc-exact if and only if it is fppf-exact. Hence, if we are not pushing the limits, the (well behaving) fppf topology is good enough.
The situation regarding cohomology groups is even a bit better than one would anticipate due to the following theorem:

Theorem [Milne, Thm. III.3.9] There is the following canonical isomorphism between étale and fppf cohomology for a smooth, quasi-projective, commutative group scheme $G$ over $X$: $$ H^i_\text{ét}(X, G) \cong H^i_\text{fppf}(X, G). $$

(The theorem still works if one replaces fppf with fpqc, I think.) The importance of the theorem lies in the fact that we can compute étale cohomology groups.

Furthermore, we have the following useful tool: over a field $k$, the functor $G \mapsto G(\bar{k})$ is exact. You can prove it either directly (section functor left exact, surjectivity: use def, any geometric point of the covering is a good candidate) or use the theorem above.

Remark The requirements on $G$ may seem like a lot, however for a group scheme over a field of char $0$ they aren't as can be seen as follows: let $G$ be a connected commutative group scheme of finite type over a field of characteristic zero. Then it is reduced hence smooth (!) by a theorem of Cartier. [SGA III$_\text{new}$] Exp. VI$_\text{B}$ Cor. 1.6.1. Now using a corollary of Chevalley’s Theorem ([Conrad] Cor. 1.2) we see that it is also quasi-projective.

Remark Chapter 4.3 (p.61) of [Moonen] is very interesting as it discusses the different quotients and their relations. (See also p.65.)

[EGA IV$_3$] Theorem 8.8.2 in the literature

The "spreading out" behaviour ([EGA IV$_3$] Theorem 8.8.2) is often cited, you can find it in [BLR] Lemma 1.2/5, in [GW] Section 10.3 (Schemes over inductive limits of rings) and in [Poonen] Theorem 3.2.1. A list of which properties extend is given in [GW] Appendix C: Permanence properties of morphism of schemes and in [Poonen] Appendix C: Properties under base extensions. (Well, they often just reference EGA IV$_3$.) For example, you find a reference to EGA IV$_3$ for the (scheme theoretic) surjectivity statement. (Be aware: both list aren't complete, e.g. [GW] misses surjectivity and [Poonen] misses flatness, hence both miss faithful flatness.)

How can we prove that exact sequences "spread out"/extend?

Let us assume we are in the general situation of the theorem, we don't need any extra assumptions. Still I keep the original notation of the question.

First a few things on kernels and cokernels: the (co-)kernel of $H \rightarrow G$ are the fppf-sheafifications of the naive (co-)kernel pre-sheaves $T \mapsto (co\text{-})ker(H(T) \rightarrow G(T))$. (This makes sense as these are the correct choices in the pre-sheaf category and sheafification is exact.) These objects are only sheaves, the interesting question now is, can these sheaves be represented by a scheme?

Representability of the kernel: let $f: H \rightarrow G$ be a homomorphism of group schemes over $S$ and let $e: G \rightarrow S$ be the identity section, then the kernel of $f$ is $H \times_{G,f,e} S$, i.e. the kernel is always representable. (See also [Poonen] Def. 5.1.11.)

Representability of the cokernel: that is in general a hard question. [SGA III$_\text{new}$] Exp. VI$_\text{B}$ Rem. 9.3 has some nice references. See also [SGA III$_\text{new}$] Exp. V Cor. 10.1.3. Note that for $S$ a Dedekind scheme one has the result referenced in Rem. 9.3 ([Anantharaman] Thm. 4.C) at hand which states that the quotient (more or less) always exists. (I believe this result can also be extracted from [BLR].)

Luckily we can bypass representability questions due to the following statement:
Short exact sequences: the sequence $0 \rightarrow G' \rightarrow G \stackrel{f}{\rightarrow} G'' \rightarrow 0$ is exact in the fppf (fpqc) topology if and only if $G'$ is the kernel of $f$ and $f$ is faithfully flat and locally of finite presentation (faithfully flat and quasi compact). [SGA III$_\text{new}$] Exp. IV Cor. 6.3.3

Behaviour under base change: the base change of the (co-)kernel is indeed the (co-)kernel of the base change as base change is exact. They are universal, see Remark 4.30 in Chapter 4 of [Moonen]. (To be precise: the fppf/fqpc-quotients are universal. The categorical quotient in general isn't universal.)

Finally we can state the proof:
Proof: We know that $G'$, $G$ and $G''$ extend $G'_K$, $G_K$ and $G''_K$ and the homomorphisms also extend. By shrinking $U$ we can also assume that we have a sequence over $U$. Let $X$ be the kernel of $G \rightarrow G''$, hence $X_K$ is isomorphic to $G'_K$ by the base change property. This isomorphism extends, i.e. $G'$ is the kernel of $G \rightarrow G''$ after possibly shrinking $U$. It remains to show that $G \rightarrow G''$ is faithfully flat and locally of finite presentation. The last bit is clear, and the first claim is also true after shrinking $U$ as flat as well as surjective "spread out". ([EGA IV$_3$]: surjective: 8.10.5 vi), flat: 11.2.6)


[Anantharaman] Anantharaman - Thesis: Schémas en groupes, espaces homogènes et espaces algébriques sur une base de dimension 1
[BLR] Bosch, S.; Lütkebohmert, W.; Raynaud, M. - Néron Models
[Conrad] Conrad, B. - A Modern Proof of Chevalley's Theorem on Algebraic Groups, PDF
[GW] Görtz, U.; Wedhorn, T. - Algebraic Geometry I
[Milne] Milne, J. - Etale cohomology, 1980
[Moonen] Moonen, B. - Draft for a book on abelian varieties, Web
[Poonen] Poonen - Rational points on varieties, PDF
[SGA III$_\text{new}$] TeXed and slightly expanded version of SGA III, URL

EDIT: in a previous version I overlooked that faithfully flat "spreads out" and used a more complicated argument based on [SGA III$_\text{new}$] Exp. VI$_\text{B}$ Lem. 10.6. Cantlog thankfully pointed out that one can considerably simplify the argument.
EDIT2: overall improvements

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+1 for the detailed explanations. However, in the other answer, the set-theoretical surjectivity $G_U\to G''_U$ implies the morphism is faithfully flat because the kernel is smooth over $U$. So the sequence is exact as fppf-sheaves, right ? –  Cantlog Feb 28 at 22:38
The context of the original question is [BLR] and they never say what they mean by an exact sequence so I assume it is the SGA III definition, i.e. $G_K \rightarrow G''_K$ is an fppf covering (maybe I should have added something like this above). user18119's argumentation however assumes that the homomorphism is set-theoretically surjective. –  y_z Mar 1 at 11:41
Sure, but fppf covering implies surjectivity. –  Cantlog Mar 2 at 16:17
Indeed, just ignore my other comment. Next time I should think before I post. Anyway, I do not see at the moment how you deduce from $G_U'/U$ is flat and the kernel that $G_U \rightarrow G_U''$ is also flat. (Though it looks like an easy exercise, can you enlighten me?) This statement more or less replaces Lem. 10.8/Prop. 10.6 (with probably a similar method). Still, how do you get the flatness of $G_U'/U$? (Not mentioned in my main post.) The trouble with flatness is that it doesn't "spread out". (Why?) [...] –  y_z Mar 4 at 13:19
Depending on the application our groups are often smooth and smoothness does "spread out" and also implies flatness. So, in addition to user18119's argument you need the smoothness of your group schemes and one more "spread out" argument as well as your argument. When I read her/his answer this wasn't transparent to me that's why I wrote my answer (which is by no means comprehensive). –  y_z Mar 4 at 13:20
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