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Let $k$ be a field of characteristic zero and let $f_1, \ldots, f_n \in k[x_1, \ldots, x_n]_d$ be linearly independent forms of degree $d$ in $n$ variables.

Is there a nice algebraic argument for proving that the determinant of the Jacobian matrix $(\frac{\partial{f_i}}{\partial x_j})_{1\leq i,j \leq n}$ is not identically zero (if that statement is correct)?

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  • $\begingroup$ $f_i$ must be functionally independed $\endgroup$ – Leox Apr 23 '15 at 8:50
  • $\begingroup$ I'm not quite sure what you mean by "functionally independent". Since $k$ has infinitely many elements the forms $f_1, \ldots, f_n$ are linearly independent as functions from $k^n$ to $k$. Could you specify that? $\endgroup$ – Hans Apr 23 '15 at 8:58
  • $\begingroup$ It means that there is no any function $F[y_1,y_2,\ldots, y_n]$ such that holds $F[f_1,f_2,\ldots, f_n] \neq 0.$ $\endgroup$ – Leox Apr 23 '15 at 9:00
  • $\begingroup$ I see (probably $F \neq 0$). Is functional independence equivalent to the statement in the question? $\endgroup$ – Hans Apr 23 '15 at 9:03
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    $\begingroup$ yes, that are equivalent statements $\endgroup$ – Leox Apr 23 '15 at 9:04
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The statement is incorrect as attested by the case $n=3, d=2$ and the linearly independent polynomials $$f_1=x_1^2,\quad f_2=x_1x_2,\quad f_3=x_2^2 $$ The jacobian determinant is identically zero.
Indeed its third column is zero, because the $f_i$'s do not not depend on $x_3$.

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$f_i$ must be functionally independent.

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  • $\begingroup$ Dear Leox: could you explain what you mean, given that the statement in the question is false. $\endgroup$ – Georges Elencwajg Apr 24 '15 at 9:41
  • $\begingroup$ I mean a general situation not the statement. $\endgroup$ – Leox Apr 24 '15 at 9:46

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