At the moment I have some difficulties in understanding the quotient of group schemes and so exact sequences. I am aware that precise answers would be difficult to be given without speaking of sheaves over categories, that is why I am just looking for the idea behind. In particular I have the following questions.
If $k$ is a field, $F$ a field extending $k$, and
$$1\longrightarrow H\longrightarrow G\longrightarrow K\longrightarrow 1$$
is an exact sequence of group schemes over $k$. Is it true that then
$$1\longrightarrow H(F)\longrightarrow G(F)\longrightarrow K(F)\longrightarrow 1$$ is an exact sequence of groups?
What is the relation between $G/H$ and $K$?
And what can we say about $(G/H)(F)$, $G(F)$ and $H(F)$? I mean, is there something like $(G/H)(F)=G(F)/H(F)$ or $(G/H)(F)\simeq G(F)/H(F)$ holding?
Thank you.
The exactness of $1\to H\to G\to K$ means $H\to G$ induces an isomorphism from $H$ to the kernel of $G\to K$ which is the fibere product $G\times_K \{1_K\}$. The exactness of $G\to K\to 1$ means $G\to K$ is faithfully flat. This is equivalent to $G\to K$ is surjective when $G, K$ are smooth over $k$.
For any field extension $F/k$, $$ 1\to H(F)\to G(F) \to K(F)$$ is exact because $(G\times_K \{1_K\})(F)=G(F)\times_{K(F)} \{ 1_K\}$ by the universal product of fiber product. This exactness can also be seen directly.
If moreover $F$ is algebraically closed, $G(F) \to K(F)\to 1$ is also exact at least if your group schemes are of finite type over $k$. This is because $G\to K$ is a surjective morphism.
However, for a general extension $F/k$, $G(F) \to K(F)\to 1$ is not exact in general. If $F$ is perfect, we can use Galois cohomology to bound the cokernel: $$ 1\to H(F)\to G(F) \to K(F) \to H^1(F, H(F^{alg})). $$ To give an example example of non-exactness at right, fix an odd integer $n\ge 3$ and consider the Kummer exact sequence for the multiplicative group $\mathbb G_m$ over $k=\mathbb Q$: $$ 1\to \mu_n \to \mathbb G_m \xrightarrow{()^n} \mathbb G_m \to 1 $$ where $\mu_n$ is the group of the $n$-th roots of unit and the second map is the power $n$ map. When you take rational points over $\mathbb Q$, you get $$ 1\to 1 \to \mathbb Q^* \xrightarrow{()^n} \mathbb Q^*$$ where the last map the power $n$ map. It is of course not surjective.
Finally, $G/H=K$ by definition, and $G(F)/H(F)$ is a subgroup of $K(F)$, not equal to $K(F)$ in general by the above example.