Assume that the algebraically independent polynomials $f\_1,\ldots, f\_n\in\mathbb{C}[x\_1,\ldots, x\_n]$ are such that the Jacobian matrix $\text{Jac}\_{x\_1,\ldots, x\_n}^{f\_1,\ldots, f\_n}\in\mathbb{C}\setminus\{0\}$.


Is it true that every derivation $D$ of algebra $\mathbb{C}[f\_1,\ldots, f\_n]$ can be continued to the derivation $U$ of $\mathbb{C}[x\_1,\ldots, x\_n]$, in other words does there exists a derivation $U$ such that $U|\_{\mathbb{C}[f\_1,\ldots, f\_n]}= D$?



Continuations of derivations of Jacobian subring
Let $R=\mathbb{C}[f\_1,\ldots,f\_n]\subset\mathbb{C}[x\_1,\ldots, x\_n]=A$. A derivation $U$ of $A$ is completely determined by $U(x\_i)$, since, then $U(P(x\_1,\ldots,x\_n))=\sum\frac{\partial P}{\partial x\_i} U(x\_i)$ for any $P\in A$. In particular, one has $U(f\_i)=\sum \frac{\partial f\_i}{\partial x\_j}U(x\_j)$. If $U\_{|R}=D$, then $U(f\_i)=D(f\_i)$. Now using the invertibility of the Jacobian, one ca calculate $U(x\_i)$ in terms of $D(f\_i)$ and $\frac{\partial f\_j}{\partial x\_i}$ and then you have such a $U$ by the first sentence.


