Lie algebra of an algebraic group I want to understand the construction of the lie algebra of a group functor following Milne's notes (https://www.jmilne.org/math/CourseNotes/LAG.pdf). My goal is to understand better the ind-group scheme $R \mapsto \mathrm{Aut}_c(R[[x]])$.
Some notation:
Let $k$ be a field, $\mathrm{Alg}_k$ the category of unital associative commutative $k-$algebras, a group functor over $k$ is a functor $G: \mathrm{Alg}_k \to \mathrm{Grp}$. Let $R$ be a $k-$algebra, denote $D_R = R[\varepsilon]/(\varepsilon^2)$ the ring of dual number over $R$ and $\pi_R: D(R) \to R$ the map
$$\pi_R(a+\varepsilon b) = a$$
$G(\pi_R): G(D_R) \to D(R)$ is a group morphism and we can take
$$\mathfrak{g}(R) = \ker\, G(\pi_R) \quad \subseteq G(D_R)$$
This subgroup have a natural action of $R$: let $r \in R$, define $\mu_r: D_R \to D_R$ by
$$\mu_r(a+ \varepsilon b) = a + \varepsilon rb$$
this is a morphism of $R-$algebras satisfying $\mu_r \pi_R = \pi_R$ so
$$G(\mu_R) \, \mathfrak{g}(R) \subseteq \mathfrak{g}(R)$$
and
$\mu_{rr'} = \mu_r \mu_{r'}$, $\mu_{1} = \mathrm{id}$ and $\mu_0 = \iota_R \pi_R$ then we really have a action of $R$ on $\mathfrak{g}(R)$.


The question:
Now I am trying to prove that $\mathfrak{g}(R)$ is a abelian subgroup of $G(D_R)$ so it will be a $R-$module.
A minimal example is $G = \mathrm{GL_n}$. We have
$$\mathfrak{g}(R) = \{\mathbb{I} + \varepsilon A \, |\, A \in \mathrm{Mat}_n(R)\}$$
and
$$(\mathbb{I} + \varepsilon A)(\mathbb{I} + \varepsilon B) = \mathbb{I} + \varepsilon (A+B)$$
so $\mathfrak{g}(R)$ is really a abelian normal subgroup. I don't know how make it work to a general functor $G$. Milne's notes do it to an affine groups (i.e., $G$ is representable), but I am really interested in non-affine case.


Another question:
Is there a relation of this construction (on non affine case) with derivations? I would like to see a connection with the Lie algebra of a Lie group.
 A: In general, this sounds to be false. More generally, let $F: \mathrm{Alg}_k \to \mathrm{Set}$ be a functor and $p \in F(R)$, we could define $T_pF = F(\pi_R)^{-1}(p)$ and as before $(F(\mu_r)) (T_pF) \subseteq T_pF$, nothing there uses de group structure of the functor.
Now the question is: how can we define the sum on $T_pF$?
In general we can't. Suppose that $F$ preserves the following pullback diagram

Let $\sigma: R[\varepsilon_1, \varepsilon_1]/(\varepsilon_1^2,\varepsilon_2^2) \to D_R$ be the $R-$morphism generated by $\varepsilon_1, \varepsilon_2 \mapsto \varepsilon, \varepsilon$, define

One could prove that this operation is commutative and so on, so that $T_pF$ inherit a $R-$module structure, the so called tangent space of $F$ at $p \in F(R)$.
Now, returning to the original question: let $G: \mathrm{Alg}_k \to \mathrm{Grp}$ be a (covariant) functor and suppose that $G$ preserves that pullback, is   $\mathfrak{g}(R) = T_e G = \ker (G(\pi_R))$ an abelian normal subgroup of $G(D_R)$?
The answers is positive and it is a direct application of Eckmann-Hilton argument: let $a, b, c, d \in \mathfrak{g}(R)$,
$$
(a+b)\cdot(c+d)
=G(\sigma) \phi (a,b) \, \cdot \, G(\sigma) \phi (c,d)\\
=G(\sigma) \phi ((a,b) \cdot (c,d))\\
=G(\sigma) \phi (a \cdot c, b \cdot d)\\
= (a \cdot c) + (b \cdot d)
$$
where I used that $\phi$ is not just a (natural) bijection, but a isomorphism of groups. So both operations are equal and commutative.
