Group Schemes and rational points

Question

Suppose $0\to F \to G \to H \to 0$ is a short exact sequence of group schemes over a field $k$. Then $0 \to F(k) \to G(k) \to H(k) \to 0$ is not exact ($G(k) \to H(k)$ is not surjective). However, is it true if we look at the $k^{sep}$-points or the $k^{alg}$-points instead of the $k$-points ($k^{sep}$ resp. $k^{alg}$ a separable resp. algebraic closure of $k$)? Is there a reference for this fact?

Answer 1

Let us assume that

a) You define a short exact sequence of group schemes to mean that the corresponding sequence of fppf group sheaves is exact

and

b) Your group schemes are of finite type over $k$.

Then one has that

$$0\to F(\overline{k})\to G(\overline{k})\to H(\overline{k})\to 0$$

is exact, and

$$0\to F(k^\mathrm{sep})\to G(k^\mathrm{sep})\to H(k^\mathrm{sep})\to 0$$

is exact if $F$ is smooth over $k$.

The point is that one has an exact sequence

$$0\to F(L)\to G(L)\to H(L)\to H^1_\text{fppf}(\mathrm{Spec}(L),F)$$

for any extension $L$ of $k$.

Now, if $L$ is algebraically closed then $H^1_\text{fppf}(\mathrm{Spec}(L),F)$ is trivial. Indeed, note that $F$ is separated (e.g. see [1, Proposition 1.22]) and thus any $F$-torsor $X$ is representable (e.g. see [2, Theorem 6.5.10(ii)]). But, since $X$ is of finite type over $L$, and $L$ is algebraically closed, we know that $X(L)\ne \varnothing$ and so $X$ is trivial (e.g. see [2, Proposition 6.5.3]).

If, in addition, $F$ is smooth, we need to show that $H^1_\text{fppf}(\mathrm{Spec}(L),F)$ is trivial when $L$ is separably closed. But, by the same argument as above any $F$-torsor $X$ is representable. But, since $F$ is smooth we know that $X$ is also smooth, and thus $X(L)\ne \varnothing$ (e.g. see [2, Proposition 3.5.70]) and thus $X$ is trivial once again.

[1] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[2] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..