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I hope this is not too elementary!

Let $G$ be a algebraic reductive group over $\mathbb{C}$. The group $G(\mathbb{C}[[t]])$ can be given the structure of a pro algebraic group as follows.

Let $l\in \mathbb{N}$ and $J^l:=\mathbb{C}[[t]] /t^l\mathbb{C}[[t]])$. Then $G(J^l)$ is known to be an algebraic group and $G(\mathbb{C}[[t]])$ can be identified with $\varprojlim_l G(J^l)$.

So my question is:

Why is $G(J^l)$ an algebraic group.

I mean I can prove this with dumb calculations. However, I wonder if there is an nice argument to see this?

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    $\begingroup$ Your notation is unfortunate. I suspect you mean the Weil restriction $\operatorname{Res}^{J^\ell}_{J^1} G_{J^\ell}$ instead of the group $G(J^\ell)$ of $J^\ell$-rational points. Algebraicity follows from the facts that $J^1 \to J^\ell$ is finite and flat, and $G$ is affine of finite type. $\endgroup$
    – S. Carnahan
    Feb 18, 2014 at 10:56
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    $\begingroup$ You can replace "reductive" by "linear". Then a more elementary (than S. Carnahan's) answer could consist in taking $\text{GL}$ in place of $G$ closed in $\text{GL}$ and in seeing the fact directly, without "dumb calculations". $\endgroup$ Feb 18, 2014 at 11:45

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As hinted by S. Carnahan, this is a particular case of a general construction, known as Weil restriction, Greenberg functor, or arc space, depending on the context.

Let $X$ be a scheme of finite type over a field $k$. Then there is a scheme $\mathcal L(X)$, which is a projective limit of schemes $\mathcal L_m(X)$ of finite type over $k$, such that $\mathcal L(X)(F)=X(F[[t]])$ and $\mathcal L_m(X)(F)=X(F[[t]]/(t^m))$, for every $k$-algebra $F$.

Most of the construction is, in fact, relatively easy.

Begin with $X=\mathbf A^1$. Then, it suffices to take $\mathcal L_m(X)=\mathbf A^m$, the identification of $\mathcal L_m(X)(F)$ with $X(F[[t]]/(t^m))$ begin given by $(x_0,\dots,x_{m-1})\mapsto x_0+x_1t+\dots+x_{m-1}t^{m-1}$. Then $\mathcal L(X)=\mathbf A^\infty=\mathop{\mathrm {Spec}}(T_0,T_1,T_2,\dots)$.

This generalizes readily to $X=\mathbf A^n$ (take the $n$th power of the preceding schemes).

Now, if $X$ is a closed subscheme of $\mathbf A^n$, with ideal $I=(P,\dots)$, one can expand $P(x_0+x_1t+\dots+x_{m-1}t^{m-1})=P_0(x)+P_1(x)t+\dots+P_{m-1}(x)t^{m-1} \pmod {t^m}$ and $\mathcal L_m(X)$ is viewed as a closed subscheme of $\mathcal L_m(\mathbf A^n)$ by adding the equations $P_0=\dots=P_{m-1}=0$ for every polynomial $P\in I$ (or in a generating subset of $I$). In your particular case, where $X$ is an (affine) algebraic group, this is all you need.

If $U$ is an affine open subscheme of $X$, then $\mathcal L_m(U)$ identifies as an open affine subscheme of $\mathcal L_m(X)$. This will allow to define $\mathcal L_m(X)$ in general by gluing $\mathcal L_m(U)$, for affine open subschemes $U$ of $X$.

One then takes the limit $\mathcal L(X)=\varprojlim_m \mathcal L_m(X)$, which exists as a scheme, because the transition morphisms $\mathcal L_{m+1}(X)\to \mathcal L_m(X)$ are affine.

Finally, the formula $\mathcal L(X)(F)=X(F[[t]])$ is easy if $F$ is a field, or if $X$ is affine (this is all you need), and is relatively easy if $X$ is quasiprojective. The general case is due to B. Bhatt (private communication). His proof used techniques of derived algebraic geometry (a theorem of Lurie/Brandenburg-Chiravasitu) and existence of enough perfect complexes (Thomason-Trobaugh).

EDIT: I realize that I did not fully answer your question, which was why one gets a pro-algebraic group when one begins with an algebraic group. The point is that this construction is functorial and commutes with products. Consequently, if $X$ is a group scheme, then one gets morphisms $\mathcal L_m(X)\times_k \mathcal L_m(X)\to \mathcal L_m(X)$ and $\mathcal L(X)\times_k\mathcal L(X)\to\mathcal L(X)$ which give $\mathcal L_m(X)$ a structure of algebraic group, and $\mathcal L(X)$ a structure of group scheme, projective limit of algebraic groups.

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  • $\begingroup$ What is wrong with taking ${\mathcal O}(\text{GL}_n)\to{\mathcal O}(G)\to0$ and tensoring it by $J^l$ (in notation of OP), where ${\mathcal O}(G)$ stands for the hopf algebra of a linear group $G$? It remains to observe that ${\mathcal O}(\text{GL}_n)\otimes J_l$ coincides with the hopf algebra of the group of automorphisms of the free $J^l$-module of rank $n$. $\endgroup$ Feb 18, 2014 at 13:34

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