# fppf-extension of algebraic groups is an algebraic group

The problem is the following:

Let $$k$$ be a field and let $$N,G,Q: (\operatorname{Sch}/k)_{\operatorname{fppf}} \longrightarrow \operatorname{Grps}$$ be fppf-sheaves from the big fppf-site of $$\operatorname{Sch}/k$$ to the category of groups. Assume that we have an exact sequence of fppf-sheaves $$e\longrightarrow N \longrightarrow G \longrightarrow Q \longrightarrow e$$ and that $$N$$ and $$Q$$ are both algebraic group schemes. Then $$G$$ is representable by an algebraic group scheme.

I know that after some fppf base change $$Q'\longrightarrow Q$$ we have an equality $$G\times_Q Q' \cong N\times_k Q'$$ and so it becomes representable. Moreover, if $$N$$ is affine I can use fppf-descent obtaining the desired scheme. Nevertheless I am not sure how to deal with the general case.

This is an exercise of Milne's book on Algebraic Groups (Ex 5-10).

Here is a possible road to a solution (I am fairly sure that Milne had something more elementary in mind). Algebraic spaces satisfy fppf descent; hence $$G$$ is a group object is in the category of algebraic spaces of finite type over $$k$$. But it is well known that a group object is in the category of finitely generated algebraic spaces over a field is in fact a group scheme. This seems to be "well known", although I have never seen a complete proof. Over an algebraically closed field this follows immediately from the result of Grothendieck that every algebraic space contains a dense open subscheme. In the general case this requires a little more care.