Let $k$ be an algebraically closed field and $R$ a noetherian (as ring) $k$-algebra. we denote by $N$ the ideal of nilpotent elements. assume we know that $R = k + N$.
I want to figure out that the structure of unit group of $R$ is $$R^*=k^*(1+N)$$
partial solution: the "$k^*(1+N) \subset R^*$" inclusion is simple: $k \to R$ is injective, therefore $k^*$ is embedded in $R^*$ as multiplicative subgroup.
let $n \in N$ then there exist a minimal natural $k \in \mathbb{N}$ with $n^k=0$. then the element $r := \sum_{0 \le j \le k-1} (-n)^j$ inverts $1+n=1-(-n)$. this implies $(1+N) \subset R^*$.
therefore I need only to show the intractable "$R^* \subset k^*(1+N) $" inclusion. does anybody see the argument?
Suppose $x\in R^*$. Because $x\in R=k+N$, we know we can write $x=a+n$ for $a\in k$ and $n\in N$. If we can show $a\in k^*$, i.e. $a\neq0$, then we will have $x=a(1+a^{-1}n)$ and since $a^{-1}n\in N$ we will have $x$ in the desired form.
Now you just need to notice that if $a=0$ then $x\in N$, but then $x$ cannot possibly be a unit in $R$ (why?)