Given a locally nilpotent derivation over a field of characteristic 0 and a local slice, how is the ring homomorphism below defined?

Question

Let $K$ be a field of characteristic 0 and $A$ a $K$-domain. Let $D:A\longrightarrow A$ be a locally nilpotent K-derivation, that is, $D(k)=0$ for all $k\in K$, $D(ab)=(Da)b+a(Db)$ for all $a,b\in A$, and given $a\in A$ there exists an integer $n\geq 0$ s.t. $D^na=0$. Suppose $\epsilon\in A$ satisfies $D\epsilon\not=0$, $D^2\epsilon=0$. Then there exists a ring homomorphism $\Phi:A\longrightarrow(\mathrm{ker}D)_{D\epsilon}$ by $\Phi(a)=\sum_{n=0}^\infty\frac{D^na}{n!}(\frac{-\epsilon}{D\epsilon})^n$. Why can this sum be written in the form $\frac{b}{(D\epsilon)^m}$ for some $b\in\mathrm{ker}D$ and $m\geq 0$?

Answer 1

Let $m$ be maximal such that $D^ma\neq 0$. Then, $$\Phi(a)=\frac{b}{(D\epsilon)^m}$$ where $$ b=\sum_{k=0}^m\frac{(-1)^k}{k!}(D^ka)(\epsilon)^k(D\epsilon)^{m-k}. $$ Now you need to show that $D(b)=0$. This is a straightforward exercise. Check the cases $m=1,2$ and you'll see what is going on.