Do "enveloping algebras" for algebraic groups exist?

Question

Let $K$ be a field, for simplicity algebraically closed of characteristic $0$. Let $G$ be a reductive group over $K$.

Def. A finite-dimensional $K$-algebra $R$ is an enveloping algebra for $G$, if $R^{\times} \cong G$ as algebraic groups over $K$.

[Since $R$ is a finite-dimensional $K$-vector space, being invertible in $R$ is an algebraic condition, so I denote by $R^{\times}$ the algebraic group of units of $R$.]

Question: What is known about the existence of an enveloping algebra for general reductive groups?

For example the matrix algebra $M_n$ is an enveloping algebra for $\mathrm{GL}_n$. As of 2020-02-22 I don't know of any other examples. We know, that if an enveloping algebra exists, then $G$ is not simple.

Application. Let $R$ be an enveloping algebra for $G$. Then $$ \mathrm{Hom}_{K\text{-}\mathrm{Rings}}(K[\Gamma], R) = \mathrm{Hom}_{\mathrm{Groups}}(\Gamma, G(K))$$ for all groups $\Gamma$.

A kind of converse to this statement is also true, so precisely the enveloping algebras for $G$ can be used to linearize representations.

Answer 1

Enveloping algebras in this sense don't always exist. In particular, no simple algebraic group has an enveloping algebra.

Let $A$ be a finite-dimensional $k$-Algebra and $J(A)$ its radical. $J(A)$ is Zariski-closed in $A$ and hence $1+J(A)$ is Zariski-closed in $A^\times$. Now as a variety, $1+J(A)$ is isomorphic to $J(A)$ which is connected, as $J(A)$ is a sub vector space of $A$.

We have a short exact sequence $1 \to 1+J(A) \to A^\times \to (A/J(A))^\times \to 1$

This shows that if we assume that $A^\times$ is a simple algebraic group, then $1+J(A)=1$ and $A^\times \cong (A/J(A))^\times$. By Artin-Wedderburn $A/J(A)^\times$ is a product of copies of $\mathrm{GL}_n$ for varying $n$ which is never simple.

Answer 2

Thanks to Lukas' answer I think I could figure out a complete classification of reductive groups, that admit enveloping algebras.

Let $G$ be a reductive group over an arbitrary field $K$ of arbitrary characteristic and $A$ a finite-dimensional $K$-algebra, such that $G \cong A^{\times}$.

Lukas has seen, that $(1+J(A))$ is a connected closed normal subgroup of $G$. The groups $(1+J(A)^n)/(1+J(A)^{n+1})$ are abelian and since $J(A)$ is nilpotent, $(1+J(A))$ is solvable. This together with connectedness shows, that $(1+J(A))$ is contained in the radical $\mathrm{rad}(G)$ (the identity component of the maximal normal solvable closed subgroup of $G$). Again, since $J(A)$ is nilpotent, $(1+J(A))$ consists of unipotent elements and hence is contained in the unipotent radical $U(G)$, but since $G$ is reductive $U(G)=1$. We conlude, that $J(A)=0$ and $A$ is semisimple. As above by Artin-Wedderburn $A$ is isomorphic to a product of matrix algebras over divison rings $D$ and in particular $G$ is a product of $\mathrm{GL}_n(D)$'s.