Let us assume that we have $n$ polynomials in $n$ variables, $p\_1(\vec{x}), p\_2(\vec{x}),\ldots p\_n(\vec{x})$.
Jacobian, $J(p)$ of $\vec{p}$ is a matrix with $i,j$ entry $\frac{\partial p\_i}{\partial x\_j}$
 Assume that $p\_i$ are algebraically dependent, i.e., there exists polynomial $f$ such that $f(p\_1,\ldots, p\_n)=0$ then using chain rule of derivatives one can get that $J(p)\cdot (\partial f(p))=0$.


Over the field of zero characteristic the converse is also true. If Jacobian is not of the full rank then polynomials are algebraicly dependent. My question is if the stronger claim is true: 


Assume that $\vec{v}=(v\_1(x),\ldots, v\_n(x))$ is a vector of polynomials such that $J(p)v=0$, then there exists polynomial $f$, such that $(\partial f)(\vec{p})=\vec{v}$.



Jacobian and Algebraic independence
I think perhaps this problem is not appropriate here. At any rate, here is a counterexample: $n$ equals $2$, $p\_1(x,y) = x^2$, $p\_2(x,y) = x^2$ and $\vec{v}(x,y) = (x,-x)$. Plugging in $\vec{p}(x)$ into any polynomial results in a polynomial in $x^2$. So the entries of $\partial f(\vec{p})$ would be polynomials in $x^2$. However, $\vec{v}$ is not a polynomial in $x^2$.


