Let $k$ be a field of characteristic zero and $A$ be a $k$-algebra. A derivation on $A$ is a $k$-linear map $D: A \to A$ such that $D(ab)=aD(b)+bD(a), \forall a,b \in A$. A derivation is called locally nilpotent if for every $a\in A$, $\exists n_a\in \mathbb N$ such that $D^{n_a} (a)=0$, where $D^n$ means $D$ composed with itself $n$-times.

My question is : Given a finitely generated $k$-algebra $A$ which is also an integral domain, does there exist a locally nilpotent derivation on $A[X,Y]$ whose kernel is $A$ ?

  • $\begingroup$ It is immediate that if this is possible, then the same is true with $A$ replaced with its fraction field. For $A$ a field, easy to see that it is not possible. $\endgroup$ – Mohan May 22 '18 at 13:20
  • $\begingroup$ @Mohan: why is it not true when $A=k(a_1,...,a_n)$ is a field for some $a_1,...,a_n \in A$ ? $\endgroup$ – user111524 May 22 '18 at 16:12
  • $\begingroup$ Your question was not about fields, nor was my answer. $\endgroup$ – Mohan May 22 '18 at 17:10
  • $\begingroup$ @Mohan: You did say " For $A$ a field, it is not true ... " please help me understand what I am missing ... $\endgroup$ – user111524 May 22 '18 at 17:25
  • $\begingroup$ I said that if $A$ is a field (which is enough for your case) and $D\in LND(A[X,Y])$, then kernel of $D\neq A$. $\endgroup$ – Mohan May 22 '18 at 17:28

I will write it as an answer for your last comment. Assume $A$ is a field and $ker D=A$ as in the question. Then $D\neq 0$ and so $D(f)\neq 0$ for some $f\in R=A[X,Y]$. Since $D\in LND$, $D^k(f)\neq 0, D^{k+1}(f)=0$ for some $k\geq 1$. Let $s=D^{k-1}(f)$. Then $D(s)\neq 0, D^2(s)=0$, so if $ker D=A$, $D(s)\in A$ and not zero. Replacing $s$ by $D(s)^{-1}s$, we may assume $D(s)=1$. Then, $R=A[s]$, by the usual Taylor expansion techniques as follows:

Define degree of $f\in R$ to be the largest integer such that $D^k(f)\neq 0$. We show by induction on the degree that $f\in A[s]$. If $\deg f=0$, then $D(f)=0$ and thus $f\in A$. Assume proved for smaller degrees. Let $\deg f=k>0$. Then, $D^{k+1}(f)=0$ and thus $D^k(f)=a\in A$. Now, let $g=f-as^k/k!$. Then, $D^k(g)=a-aD^k(s^k/k!)=a-a=0$, so $\deg g<k$ and thus, by induction $g\in A[s]$. Then clearly $f\in A[s]$.

But transcendence degree of $A[s]$ over $A$ is (at most) one and that of $R$ is two, a contradiction.

  • $\begingroup$ So indeed you prove that if $D \in LND (A^{[n]})$, and $\ker D=A$, then $n\le 1$, right ? $\endgroup$ – user111524 May 22 '18 at 21:15

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