Group schemes decomposition Given an abelian group scheme of finite type $(G,+)$ over $\mathbb{F}$ connected, and given two connected closed subgroup schemes of finite type $G$ over $\mathbb{F}$ connected $H$, $N$ of $G$. ($\mathbb{F}$ is the algebraic closure of $\mathbb{F}_p$)
Supose we have $G=H+N$.  Is the the following is true:
$G(R)=H(R)+N(R)$
For all ring $R$.
 A: No, this is false. The issue is that one has a surjective map of group schemes
$$H\times N\to G$$
(given by multiplication) but if we denote by $\mu$ the kernel of this map (which is just $H\cap N$) we a priori only get an exact sequence
$$1\to \mu(R)\to H(R)\times N(R)\to G(R)\to H^1_\mathrm{fppf}(R,\mu)\to H^1_\text{fppf}(R,H\times N)\to H^1_\text{fppf}(R,G)\to\cdots$$
and thus there is no reason that the map
$$H(R)\times N(R)\to G(R)$$
is surjective.
Here's an example of this phenomenon. Consider

*

*$G=\mathbb{G}_{m,\mathbb{F}}^n$.


*$H$ the subgroup of $\mathbb{G}_{m,\mathbb{F}}^n$ whose $R$-points is given by
$$\{(x_1,\ldots,x_n)\in (R^\times)^n:x_1\cdots x_n=1\}$$


*$N$ is the subgroup of $\mathbb{G}_{m,\mathbb{F}}^n$ whose $R$-points is given by
$$\{(x_1,\ldots,x_n)\in (R^\times)^n:x_1=\cdots=x_n\}$$
Then, the map
$$H\times N\to G$$
is evidently surjective since this can be checked on $\mathbb{F}$-points, for which the claim is clear.
That said, note that the kernel of this multiplication map is $\mu_n$. Thus, we have an exact sequence
$$1\to\mu_n(R)\to H(R)\times N(R)\to G(R)\to H^1_\text{fppf}(R,\mu_n)\to H^1_\text{fppf}(R,N\times H)$$
But, let $L:=\mathbb{F}(x)$. Then, by Hilbert's theorem 90 we have that
$$H^1_\text{fppf}(L,N\times H)=H^1_\text{et}(L,N\times H)=0$$
and so we see that
$$H(L)\times N(L)\to G(L)$$
is surjective if and only if $H^1_\text{fppf}(L,\mu_n)=0$. But, assume that $p\nmid n$. Then,
$$H^1_\text{fppf}(L,\mu_n)=H^1_\text{et}(L,\mu_n)=L^\times/(L^\times)^n$$
which is evidently non-zero if $n>1$.
