Jacobian criterion for algebraic independence over a perfect field in positive characteristics It is well known that the Jacobian criterion for algebraic independence does not hold in general for fields of positive characteristics. However, the following partial statement seems promising:

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*Consider the polynomial algebra $\mathbb{F}[x_1,\cdots,x_n]$, where $\mathbb{F}$ be a perfect field of characteristic $p>0$, i.e., for all $a\in\mathbb{F}$, the equation $x^p=a$ has a solution. (For instance, $\mathbb{F}=\mathbb{F}_p$.) Let $\varphi_1,\cdots,\varphi_m\in\mathbb{F}[x_1,\cdots,x_n], m\leq n$ be polynomials such that the Jacobian matrix $(\partial\varphi_j/\partial x_i)$ is of rank $m$. Then $\varphi_1,\cdots,\varphi_m$ are algebraically independent.

The argument is as follows. Suppose $\varphi_1,\cdots,\varphi_m$ are algebraically dependent. Let $f(y_1,\cdots,y_m)$
be the nontrivial polynomial of the lowest degree such that we have
$$f(\varphi_1,\cdots,\varphi_m)=0.$$
Since the Jocobian matrix is of full rank, we have $\partial f/\partial\varphi_i=0$ for all $i$. Therefore, we have
$$f(\varphi_1,\cdots,\varphi_m)=g(\varphi_1^p,\cdots,\varphi_m^p)$$
for some polynomial
$$g(z_1,\cdots,z_m)=\Sigma_{i_1,\cdots,i_m}a_{i_1,\cdots i_m}z_1^{i_1}\cdots z_m^{i_m}.$$
Let $b_{i_1,\cdots,i_m}\in\mathbb{F}$ satisfy $b_{i_1,\cdots,i_m}^p=a_{i_1,\cdots,i_m}$, and let
$$\bar{g}(w_1,\cdots w_m)=\Sigma_{i_1,\cdots,i_m}b_{i_1,\cdots i_m}w_1^{i_1}\cdots w_m^{i_m}\neq0.$$
Then we have
$$0=f(\varphi_1,\cdots,\varphi_m)=g(\varphi_1^p,\cdots,\varphi_m^p)=\Sigma_{i_1,\cdots,i_m}(b_{i_1,\cdots i_m}\varphi_1^{i_1}\cdots \varphi_m^{i_m})^p=\bar{g}(\varphi_1,\cdots,\varphi_m)^p.$$
Therefore, $\bar{g}(\varphi_1,\cdots,\varphi_m)=0$ is a nontrivial polynomial relation for $\varphi_1,\cdots,\varphi_m$, and the polynomial $\bar{g}$ has degree lower than that of $f$, a contradiction. Therefore, $\varphi_1,\cdots,\varphi_m$ are algebraically independent.
My questions are:  Am I missing anything in the above argument? If not, is it discussed anywhere in the literature?
 A: The question is raised in comments about whether full rank Jacobian still implies algebraic independence in characteristic $p$ when the group field is imperfect. The answer is yes. Here is a reference, and a short proof.
Reference Let $k$ be a field of characteristic $p$, let $K = k(x_1, \ldots, x_n)$ and let $F = k(f_1, \ldots, f_n) \subseteq K$. Then $dx_1$, $dx_2$, ..., $dx_n$ is a basis for $\Omega_{K/k}$ as a $K$-vector space. We have $df_j = \sum \tfrac{\partial f_j}{\partial x_i} d x_i$, so the Jacobian matrix expresses the $df_j$ as $K$-linear combinations of the basis $d x_i$, and thus the Jacobian matrix has full rank if and only if the $df_j$ are a $K$-basis for $\Omega_{K/k}$.
Theorem 16.14 in Eisenbud, Commutative Algebra with a View toward Algebraic Geometry tells us that $df_j$ are a $K$-basis for $\Omega_{K/k}$ if and only the $df_i$ form a separating transcendence basis for $K/k$ (meaning that the $f_j$ are a transcendence basis for $F$ and $K/F$ is separable). In particular, it implies that the $f_j$ are a transcendence basis for the field $F$ that they generate, and thus they are linearly independent.
Quick Proof Suppose for contradiction that the Jacobian matrix is invertible and we have an algebraic relation $g(f_1, f_2, \ldots, f_n) = 0$ for some polynomial $g(z_1, z_2, \ldots, z_n)$. Take $g$ of minimal total degree.
Now, applying the chain rule to the equation $g(f_1, f_2, \ldots, f_n) = 0$, we get
$\begin{bmatrix} \tfrac{\partial g}{\partial z_j} \end{bmatrix}\begin{bmatrix} \tfrac{\partial f_j}{\partial x_i} \end{bmatrix}=0$. Since we assumed that the Jacobian $\begin{bmatrix} \tfrac{\partial f_j}{\partial x_i} \end{bmatrix}$ is invertible, this means that $\tfrac{\partial g}{\partial z_1}=\tfrac{\partial g}{\partial z_2}=\cdots = \tfrac{\partial g}{\partial z_n}=0$. This means that $g$ is a polynomial in $z_1^p$, $z_2^p$, ..., $z_n^p$, say $\sum g_{i_1 i_2 \cdots i_n} z_1^{pi_1} z_2^{p i_2} \cdots z_n^{p i_n}$. However, we don't know that the coefficients $ g_{i_1 i_2 \cdots i_n}$ are $p$-th powers. Presumably, this is where the commenters got stuck.
A little trick can unstick us. Consider $k$ as a $k^p$ vector space, so $k^p$ is a one-dimensional subspace of $k$. Choose a $k^p$-linear retraction $\lambda: k \to k^p$. (By a retraction, I mean that $\lambda(a^p) = a^p$ for $a^p \in k^p$.) Define $\lambda : k[x_1, \ldots, x_n] \to k^p[x_1, \ldots, x_n]$ by acting on the coefficients. Now, apply the map $\lambda$ to the equation $\sum g_{i_1 i_2 \cdots i_n} f_1^{p i_1} f_2^{p i_2} \cdots f_n^{p i_n} = 0$. Each of the polynomials $f_j^{p i_j}$ is already in $k^p[x_1, \ldots, x_n]$, so we get $\sum \lambda(g_{i_1 i_2 \cdots i_n}) f_1^{p i_1} f_2^{p i_2} \cdots f_n^{p i_n} = 0$. Moreover, replacing $g$ by $ag$ for some $a \in k$, we may assume that some coefficient of $g$ is equal to $1$, and hence the $\lambda(g_{i_1 i_2 \cdots i_n})$ are not all $0$.
So we get a new algebraic relation $\sum \lambda(g_{i_1 i_2 \cdots i_n}) f_1^{p i_1} f_2^{p i_2} \cdots f_n^{p i_n} = 0$ between the $f_j$'s. But now all the coefficients $\lambda(g_{i_1 i_2 \cdots i_n})$ are in $k^p$, so we may take a $p$-th root and get an algebraic relation $\sum \lambda(g_{i_1 i_2 \cdots i_n})^{1/p} f_1^{i_1} f_2^{i_2} \cdots f_n^{i_n} = 0$ of lower degree, a contradiction.
