I am reconsidering an argument previously I thought obvious but now I feel I do not understand. Let $G$ be a finite flat group scheme over a finite field $k$. $G^\vee$ the Cartier dual of it. Let $TG$ denote its Lie algebra. $\mathbb{G}\_a$ the additive group scheme. Then we have


$$
TG\cong \mathrm{Hom}(G^\vee, \mathbb{G}\_a)
$$


Previously I am thinking $G\cong \mathrm{Hom}(G^\vee,\mathbb{G}\_m)$ then take the tangent space we have the argument. But it seems I can not prove it strictly. Any help?



Tangent space of a finite flat group scheme
This is correct. Here's another to see it (assume for simplicity we are working over a field $k$). Let $R\_\epsilon : = k[\epsilon]/\epsilon^2$ and let $S\_\epsilon:= Spec R\_\epsilon.$ This is a scheme with a single $k$-point, $s\in S$ and a one-dimensional tangent space. Now recall that the tangent space can be interpreted as a mapping space (in the category of schemes): $$TG = \{f\in Hom(S\_\epsilon, G)\mid f(s) = e\}.$$


On the other hand,
$$Hom(G^\vee, \mathbb{G}\_a) : = Hom\_{Hopf}(k[x], O\_G^\vee) \cong Hom\_{Hopf}(O\_G, k[x]^\vee),$$ where Hom is taken in the category of Hopf algebras. Now $k[x]$ has basis $x^n$ and $k[x]^\vee$ has dual elements $x^\vee\_i$ (this is not quite a basis, unless we define the dual more carefully, but that isn't relevant here). Now you can check directly that $$(x\_1^\vee)^2 = 0,$$ so given $f\in \hom(G^\vee, \mathbb{G}\_a)$, the function $f^\vee: O\_G\to k[x]^\vee$ can be restricted to the subring $k[x\_1^\vee]\cong R\_\epsilon$, giving a tangent vector (in $Hom(Spec R\_\epsilon, G)$, of schemes). Conversely, say you have a map $S\_\epsilon\to G$ with the closed point going to $e\in G$. Dually, you have a map of co-rings $k\cdot 1^\vee\oplus k\cdot\epsilon^\vee\to O\_G^\vee,$ taking $1^\vee$ to the identity. Now remembering the ring structure on $O\_G^\vee,$ this extends to a map from the polynomial ring $k[\epsilon^\vee]\to O\_G^\vee,$ and setting $x = \epsilon^\vee,$ you can check that this construction is converse to the previous one (in particular, the resulting map is a map of Hopf algebras). 


There is a fancy way to generalize this idea in terms of adjoint functors, as follows. Let $Ring\_\*$ be the category of *augmented rings*, dual to the category of affine schemes with a basepoint (here rings and Hopf algebras are assumed to be commutative and co-commutative), and let $coRing\_\*$ be the category of *coaugmented corings* (with all operations and axioms reversed). 


Then the forgetful functor $$\mathtt{forg}:Hopf\to coRing\_\*$$ (which ignores multiplication but remembers comultiplication and the coaugmetnation — equivalently, unit) has a left adjoint, $$\mathtt{free}:coRing\_\*\to Hopf.$$ This is actually very easy to see: any coaugmented coring $R$ can be canonically written as $k\cdot 1 \oplus R/k,$ and the free Hopf algebra on $k\oplus R/k$ is simply $k[R/k],$ with comultiplication induced from that on $R$. In particular, plugging in the co-augmented co-ring corresponding to $S\_\epsilon,$ which is $R\_\epsilon^\vee$, we can compute $\mathtt{free}R\_\epsilon^\vee \cong O\_{\mathbb{G}\_a}$ as Hopf algebras, and the statement you want follows from the universal property.


