Extension of a short exact sequence of group schemes

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

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. [Poonen] Chapter 5.1 and [SGA III$_\text{new}$].

Why should we use the fppf topology?

Well, that's the wrong question. We want to have the most general notion of exactness which is still useful. That's why the fpqc topology is used in [SGA III$_\text{new}$]. However, from a user stand-point it is not exactly practical (substantial set theoretic trouble, fpqc is a bit scary). The next best option is the fppf topology. And indeed it is useful, for example we can compute cohomology groups via the following

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 $X$: $$H^i_\text{ét}(X, G) \cong H^i_\text{fppf}(X, G).$$

Furthermore, we have 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 much, 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$_\text{new}$] Exposé VI$_\text{B}$ Corollaire 1.6.1. Now using a corollary of Chevalley’s Theorem ([Conrad] Corollary 1.2) we see that it is also quasi-projective.

Remark Chapter 4.3 (p.61) of [Moonen] is also interesting as it discusses a bit of the philosophy.

Remark Except for the set theoretical problems, everything above still seems to be valid for the fpqc topology.

[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.

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 (except the assumptions we import when we use certain existence theorems for the quotient, see below.) Still I keep the original notation of the question.

First a few things on kernels and cokernels: The kernel and the cokernel of $H \rightarrow G$ are the fppf-sheafifications of the naive (co-)kernel pre-sheaves $T \mapsto ker(H(T) \rightarrow G(T))$ resp. $T \mapsto G(T)/H(T)$. (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?

Existence of a kernel: Let $f: H \rightarrow G$ be a homomorphism of group schemes over $S$ and let $e: G \rightarrow S$ 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.)

Existence of a cokernel: That is a hard question. [SGA III$_\text{new}$] Exposé VI$_\text{B}$ Remarque 9.3 has some nice references. However, we are in a special situation as we know that the quotient of $G_K / G'_K \cong G''_K$ is representable, hence $G / G'$ is also representable (after possibly shrinking $U$) when we assume that $G'$ is flat by [SGA III$_\text{new}$] Exposé VI$_\text{B}$ Proposition 10.6.
Several remarks are needed: Proposition 10.6 originally states the representability of the fpqc-cokernel, however, in our situation the fpqc-representability condition coincides with the fppf-representability condition: Proposition 10.6 is proved using Lemma 10.8 which indeed proves a long the way the equivalence. There are also some alternative proofs possible: One could also use [SGA III$_\text{new}$] Exposé V Corollaire 10.1.3. For $S$ a Dedekind scheme we could also use the result referenced in Remarque 9.3 ([Anantharaman] Theorem 4.C) to state that the quotient always exists if $G'$ is flat. (I believe this result can also be extracted from [BLR].)

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].)

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. 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 $U$. Thus the sequence is left exact. Now we do the same for the quotient: Let $Y$ be the cokernel of $G' \rightarrow G$. Note that it exists by the discussion above. Again, after base change $Y_K$ is isomorphic to $G''_K$. The isomorphism extends and hence $Y$ is isomorphic to $G''$ over a possibly smaller $U$, i.e. the sequence is exact.

EDIT: As Cantlog points out one can also reason as follows (and hence bypass the whole representability discussion): Assume the left exactness of the sequence over $U$, i.e. apply the first part of the argument above. We are left to show that $G \rightarrow G''$ is fppf surjective, i.e. an fppf covering. (This is more or less the content of Lemma 10.8.) An fppf covering morphism is a faithfully flat morphism (locally of finite presentation) and this property "spreads out". (in [EGA IV$_3$]: surjective: 8.10.5 vi), flat: 11.2.6)

Epilogue

In [BLR] Proposition 7.5/1 they say they "apply a limit argument" to get the result. In the end this is true, but one needs the really nice characterisation of the quotient as the diagram seen in [SGA III$_\text{new}$] Exposé VI$_\text{B}$ Lemma 10.8 which is the heart of the surjectivity statement, i.e. the whole proof.

References

[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 (you can find the PDF online)
[GW] Görtz, U.; Wedhorn, T. - Algebraic Geometry I
[Milne] Milne, J. - Etale cohomology, 1980
[Moonen] Moonen, B. - Draft for a book on abelian varieties (you can find the PDF on his website)
[Poonen] Poonen - Rational points on varieties, PDF
[SGA III$_\text{new}$] TeXed and slightly expanded version of SGA III, URL

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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