# Tate modules of commutative group schemes over finite field

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.

• I know how to answer the last question but not any of the others. – Will Sawin Apr 21 '18 at 18:28
• This seems like a question that is too broad to really be answered in one go. Let me attempt to say some words on 2. You can assume that each $G_i$ is built, via extensions, from building block group schemes. These fall, essentially, into several classes: abelian varieties, tori, smooth unipotent groups, finite etale group schemes, and finite connected group schemes. To try and uniformly describe extensions for all of these seems unreasonable, and so I think you need to consider them on a case-by-case basis. Let met just mention the specific examples you pointed out of $\alpha_p$ and $\mu_p$. – SomeGuy Apr 30 '18 at 11:30
• These fall into the category of finite connected group schemes. Such objects are tailor built to be handled by Dieudonne theory. So, for example, to compute $\text{Ext}^1(\alpha_p,\alpha_p)$ one really computes $\text{Ext}^1(D(\alpha_p),D(\alpha_p))$ where $D(\alpha_p)$ is the Dieudonne module of $\alpha_p$ which, here, just looks like $\mathbb{F}_p$ with the Frobenius operator 0. So, you're essentially trying to find extensions of $\mathbb{F}_p$ by itself that have operators $F$ and $V$ that restriction $0$ on $D(\alpha_p)$ and map to $0$ on $D(\alpha_p)$ and satisfy $FV=VF=0$. I leave it to – SomeGuy Apr 30 '18 at 11:36
• you to figure out what such things look like. In particular, as this example shows for some of the types of group schemes mentioned above $T_\ell$ is a horrible choice since it is essentially tailor made to pick up on etale data of $G$. If $G$ has no etale data (e.g. it's a connected finite group scheme) is essentially useless. Dieudonne theory is meant to fill the whole there (i.e. in the case of finite connected group schemes). – SomeGuy Apr 30 '18 at 11:38
• @SomeGuy Thank you, I also want to know whether higher extension group vanishes. – zzy Apr 30 '18 at 12:26

## 1 Answer

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 $\ell^{nq}$.

• Thank you for your kind help. Now I am more interseted in removing your assumption of being finite type or prime to the characterestic. – zzy Apr 29 '18 at 20:39