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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 $(\partial f)(\vec{p})=\vec{v}$.

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  • $\begingroup$ You seem to be multiplying matrices in the wrong order here. $\endgroup$
    – Will Sawin
    Commented Jul 18, 2012 at 18:06

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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 $x^2$.

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    $\begingroup$ More simply, you can take $p_1=p_2=0$. $\endgroup$
    – Will Sawin
    Commented Jul 18, 2012 at 18:04
  • $\begingroup$ @Will -- That is definitely simpler. $\endgroup$
    – Jason Starr
    Commented Jul 18, 2012 at 18:35
  • $\begingroup$ Thanks, you are right the I need to state the question in a different way. What I am interested in is when $v$ is an image of $\partial f(p)$. $\endgroup$
    – Klim Efremenko
    Commented Jul 19, 2012 at 6:08

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