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darij grinberg
darij grinberg
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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)=\Sigma_{n=0}^\infty\frac{D^na}{n!}(\frac{-\epsilon}{D\epsilon})^n$$\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 $m\geq 0$?

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 $\Phi(a)=\Sigma_{n=0}^\infty\frac{D^na}{n!}(\frac{-\epsilon}{D\epsilon})^n$. Why can this sum be written in the form $\frac{b}{(D\epsilon)^m}$ for some $b\in\mathrm{ker}D$ and $m\geq 0$?

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 $m\geq 0$?

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

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 $\Phi(a)=\Sigma_{n=0}^\infty\frac{D^na}{n!}(\frac{-\epsilon}{D\epsilon})^n$. Why can this sum be written in the form $\frac{b}{(D\epsilon)^m}$ for some $b\in\mathrm{ker}D$ and $m\geq 0$?