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Suppose you have an exact sequence $1\to G_1\to G_2\to G_3\to 1$ of affine group schemes, over a field $k$. By this a mean that $G_2\to G_3$ is a quotient map (i.e., the map on coordinate algebras is injective, or faithfully flat), and the kernel is $G_1$. Moreover, assume $G_1$ is commutative (if necessary).

Is it true that the exact sequence is the pull-back of an exact sequence of the form $1\to G_1\to G_2'\to G_3'\to 1$, where $G_2'$ and $G_3'$ are algebraic quotients of $G_2$ and $G_3$ respectively?

On a related question: if $\overline k$ is an algebraic closure of $k$, then $Gal(\overline k/k)$ acts on each $G_i(\overline k)$ but not necessarily in a continous way if the $G_i$ are not algebraic. Do you still have some sort of Galois cohomology exact sequence?

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If $G_1$ isn't finite type then there cannot be $G'_2$ of finite type in which $G_1$ is a closed subscheme, so what are you trying to do in the first question? And even with algebraicity throughout, if $G_1$ is not $k$-smooth then Galois cohomology is absolutely the wrong notion to consider, so some explanation of the motivation for the 2nd question would be helpful. (There's always fpqc topology, etc., but saying what is useful to consider is much informed by knowing the context to which you plan to apply an answer.) –  user30180 Apr 18 '13 at 3:34
    
Well, I might have been too optimistic, but I don't need that much generality. In my situation, $G_1$ is algebraic, and $k$ is a number field. Also, I know that $H^1(L,(G_1)_L)$ vanishes for some finite extension $L/k$, and I want to construct a section of $(G_3)_L\to(G_2)_L$. –  unknown Apr 18 '13 at 18:42
    
Oops, I meant $(G_2)_L\to(G_3)_L$ –  unknown Apr 18 '13 at 18:43
    
Vanishing of such cohomology does not suffice to make a section. For $n > 1$, consider ${\rm{GL}}_n \rightarrow {\rm{PGL}}_n$. This has sections Zariski-locally on the base (even as schemes over $\mathbf{Z}$), but there is no section over the entire base, over any field. Indeed, if $s$ were a global section then WLOG $s(1)=1$ and composing $s$ with det would define a global unit on ${\rm{PGL}}_n$ with value $1$ at $1$. The only such unit is $1$ (since ${\rm{PGL}}_n$ is semisimple), so $s$ would be a section to ${\rm{SL}}_n \rightarrow {\rm{PGL}}_n$, contradiction via connectedness. –  user30180 Apr 19 '13 at 11:36
    
Thanks a lot for the detailed explanation. However I don't understand your connectedness argument, since both groups are connected. –  unknown Apr 19 '13 at 20:27

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