# Jacobian of n linearly independent forms in n variables

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)?

• $f_i$ must be functionally independed – Leox Apr 23 '15 at 8:50
• 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? – Hans Apr 23 '15 at 8:58
• It means that there is no any function $F[y_1,y_2,\ldots, y_n]$ such that holds $F[f_1,f_2,\ldots, f_n] \neq 0.$ – Leox Apr 23 '15 at 9:00
• I see (probably $F \neq 0$). Is functional independence equivalent to the statement in the question? – Hans Apr 23 '15 at 9:03
• yes, that are equivalent statements – Leox Apr 23 '15 at 9:04

## 2 Answers

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 $x_3$.

$f_i$ must be functionally independent.

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