Tate modules of commutative group schemes over finite field

Question

Let $G$ be a finite type commutative group scheme over a finite field $k=\Bbb F_q$ with $\Gamma=Gal(\bar k/k)$ and $l$ be a prime such that $(l,q)=1$, we can also define the Tate module $T_lG =\varprojlim_nG[l^n](\bar k)$. Then we can also consider the natural map

$Hom_k(G_1,G_2) \otimes \Bbb Z_l \rightarrow Hom_\Gamma(T_lG_1,T_lG_2)$

for any two commutative group scheme over $k$, or more generally the map

$\text{Ext}^i_k(G_1,G_2) \otimes \Bbb Z_l \rightarrow \text{Ext}^i_\Gamma(T_lG_1,T_lG_2)$ for every natural number $i$ at least when the extension groups are well-defined (This question may be related).

My first question is: when is this map surjective, injective or an isomorphism? If $G_i$ are abelian varieties and $i=0$, it's well-known that this map is an isomorphism. If $G_2=\Bbb G_m$ and $G_1$ is an abelian variety, this is also true for $i=0$ as both sides are zero, and is true for $i=1$ and $\text{dim}G_1=1$ as $\text{Ext}^1(A,\Bbb G_m)$ is the dual of $A$ and one can compute both sides explicitly (see this master thesis). However, this map can fail to be injective for trivial reasons (For instance, right side could be zero for some finite abelian groups).

So my second question is: how to describe extension groups between commutative group schemes using linear algebra datas in a systematic way? For instance, how to describe extension groups of finite group schemes like $\alpha_p$, $\mu_p$?

The last question is about growth of $\#G[l^n]$ for commutative group schemes (in order to study the Tate module). Let $k$ be any field, $G$ be a commutative group scheme over a field $k$ (not necessarily finite type) and $p$ be a prime. Assume $G[p^n]$ is a finite group scheme for every $n$, does there always exist $C >0,h \geq 0$ such that $\#G[p^n]=Cp^{nh}$ for $n >> 0$? Here the order of a finite group scheme means the dimension of its global section ring. I know the result holds for etale group schemes and abelian varieties. For example, if $G$ is constant i.e an abstract abelian group and $\#G[p^n]$ is finite for every $n$, then we know $G[p^\infty]\cong (\Bbb Q_p/\Bbb Z_p)^h\oplus T$ for some $h \geq 0$ and finite $p$-group $T$ (this result is used when studying Tate-Shafarevich group of elliptic curves).

Edit: there are good theory for commutative linear algebraic groups over complex number, for example every connected one is a product of $\Bbb G_a$ and $\Bbb G_m$, see here. Here is some discussion for higher extension groups. I am more interested in higher extension groups or non-reduced case.

Answer 1

I'll try to answer your last question on the order of $\#G[\ell^n]$ for $(\ell,p) = 1$.

Assumption: $G$ a commutative connected group scheme locally of finite type over $k$.

Since you're interested in the order of the group scheme $G[\ell^n]$, it is harmless to pass to $k = \overline{k}$ (a connected group scheme locally of finite type over $k$ is geometrically connected). Then $G_{\text{red}}$ is a closed reduced subgroup scheme of $G$ with $G_{\text{red}}[\ell^n] = G[\ell^n]$ as the latter is \'{e}tale over $k$. Therefore we may assume that $G$ is reduced, a fortiori smooth. By Chevalley's Theorem, we have an exact sequence $$0 \to H \to G \to A \to 0$$ where $H$ is a closed linear algebraic subgroup and $A$ an abelian variety.

Lemma: For every $n > 0$, the sequence $$0 \to H[\ell^n] \to G[\ell^n] \to A[\ell^n] \to 0$$ is still exact.

Proof: We have to show that $H/\ell^n H = 0$. Since $(\ell,p) = 1$ the multiplication by $\ell^n$ map is \'{e}tale and therefore open. However a homomorphism of group schemes always has closed image, and so by connectedness of $H$ we conclude the result.

Since you know the answer to your last question in the case of an abelian variety $A$, I'll just treat the case of $H$. Now $k$ is algebraically closed, in particular perfect so we know that $$ H \cong D \times U$$ where $D$ is a diagonalizable group and $U$ is unipotent. Any unipotent $U$ group over a perfect field has a filtration with successive quotients isomorphic to $\mathbf{G}_a$. Therefore if $(\ell,p) = 1$ (as is the case here), $U[\ell^n] = 0$ and so $$H[\ell^n] \cong D[\ell^n].$$ In particular, we may assume that $H$ is connected diagonalizable and therefore isomorphic to $\mathbf{G}_m^q$ for some $q \in \mathbf{N}$ (there are no forms of $\mathbf{G}_m^q$ over an algebraically closed field). Hence $$\mathbf{G}_m^q[\ell^n] \cong (\mu_{\ell^n})^q$$ which has order $\ell^{nq}$.