Denote by $P_n$ the space of all polynomials in $n$ variables, with coefficients in a field $\mathbb F$.
A collection of polynomials $(f_1,\cdots,f_m)=\vec f\in P_n\!^m$ is called algebraically dependent if there is some polynomial $g\neq0$ in $P_m$ such that $g\circ\vec f=0$ in $P_n$.
According to the Jacobian criterion, $(f_1,\cdots,f_m)$ is algebraically dependent if and only if $(\vec\nabla f_1,\cdots,\vec\nabla f_m)$ is linearly dependent over $P_n$. But this equivalence generally requires $\mathbb F$ to have characteristic $0$.
With characteristic $p\neq0$, is the "only if" part still true?
See this thesis by Sinhababu (section 2.5), also this MO question.
Seeking a contradiction, suppose $(f_1,\cdots,f_m)$ is algebraically dependent but $(\vec\nabla f_1,\cdots,\vec\nabla f_m)$ is linearly independent. Let $g\neq0$ be a polynomial of minimal degree such that $g\circ\vec f=0$. Differentiating by the chain rule,
$$0=\vec\nabla(g\circ\vec f)=\sum_{k=1}^m\Big((\partial_k g)\circ\vec f\Big)\vec\nabla f_k,$$
so, by linear independence, each coefficient must vanish:
$$0=(\partial_k g)\circ\vec f.$$
If any $\partial_k g\in P_m$ is non-zero, since its degree is lower than that of $g$, this would contradict minimality of $g$. Hence
$$0=\partial_k g$$
for each $k$; that is, $\vec\nabla g=0$.
If $\mathbb F$ has characteristic $0$, this implies that $g=c\in\mathbb F$ is constant; but then $0\neq g=c=g\circ\vec f=0$.
Now suppose $\mathbb F$ has characteristic $p\neq0$. Vanishing derivative $\vec\nabla g=0$ is equivalent to $g=h\circ(x_1^p,\cdots,x_m^p)$ for some polynomial $h\in P_m$, where $x_1,\cdots,x_m$ are the variables. (I was trying to use notation which doesn't explicitly refer to these variables....)
If, further, $\mathbb F$ is perfect, meaning that the Frobenius function $\varphi:x\mapsto x^p$ for $x\in\mathbb F$ is bijective, then $h\circ(x_1^p,\cdots,x_m^p)$ can be written as $(\varphi^{-1}h)^p$, where $\varphi^{-1}$ applies only to the coefficients of $h$, but the $p$th power applies to the whole polynomial. Thus
$$0=g\circ\vec f=(\varphi^{-1}h)^p\circ\vec f=\Big((\varphi^{-1}h)\circ\vec f\Big)^p.$$
But then $(\varphi^{-1}h)\circ\vec f=0$ contradicts minimality of $g$, as the degree of $\varphi^{-1}h$ is the same as the degree of $h$ which is $\tfrac1p$ of the degree of $g$ (and $g$ is not constant).
How can we proceed when $\mathbb F$ is not perfect (for example, an infinite field of rational expressions over a finite field)?
Or is the statement not true? At least it is true for $m=1$: if a single polynomial $f$ is algebraically dependent, then it's constant (else $g\circ f$ would have positive degree), so its derivative is $\vec\nabla f=0$, which is linearly dependent.
