Linear independency implies linear independence of the gradients Consider a set of linear independent and homogeneous polynomials $f_i:\mathbb{R}^n \rightarrow \mathbb{R}$ $i =1,\cdots,m$ and $m \leq n $. I wonder if the Linear independency of $\\{f_i\\}$ implies the linear independence of its gradients $\\{\nabla f_i\\}$ evaluated at some point in $\mathbb{R}^n$?. Thanks in advance.

This is of course false if we allow the polynomials to be constant, because the gradient of a constant is zero and thus linear dependent. But if we assume that all polynomials have positive degree, it is true as the following proof shows:

In characteristic zero, the gradient of a polynomial is zero if and only if the polynomial is constant.

Assume we have a linear combination $\sum a_i \
abla f_i=0$, then the polynomial $f := \sum a_if_i$ satisfies $\
abla f=0$, i.e. $f$ is constant. But as all $f_i$ have no constant term by assumption, we have that $f$ has no constant term either. This shows $f=0$, i.e. all $a_i=0$ by the linear independence of the $f_i$.