fppf-extension of algebraic groups is an algebraic group

Question

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).

Answer 1

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.