In [Jeffrey Lang, *A Jacobian identity in positive characteristic*, J. Commut. Algebra, Volume 7, Number 3 (2015), pp. 393--409](https://projecteuclid.org/euclid.jca/1450102162), the following result is proven:



> 
> **Theorem 1.** Let $p$ be a prime. Let $\mathbf{k}$ be a commutative $\mathbb{F}\_p$-algebra. Let $n$ be a nonnegative integer. Let $R$ be the polynomial ring $\mathbf{k}\left[x\_1, x\_2, \ldots, x\_n\right]$. Let $f\_1, f\_2, \ldots, f\_n$ be $n$ polynomials in $R$. Let $\nabla$ be the differential operator $\prod\limits\_{i=1}^n \left(\dfrac{\partial}{\partial x\_i}\right)^{p-1}$ on $R$. Let $M \in R^{n\times n}$ be the Jacobian matrix of $f\_1, f\_2, \ldots, f\_n$; this is the $n\times n$-matrix over $R$ whose $\left(i,j\right)$-th entry is $\dfrac{\partial f\_i}{\partial x\_j}$. Then,
>  \begin{align}
> & \sum\limits\_{\left(i\_1, i\_2, \ldots, i\_n\right) \in \left\{0,1,\ldots,p-1\right\}^n} f\_1^{i\_1} f\_2^{i\_2} \cdots f\_n^{i\_n} \nabla\left(f\_1^{p-1-i\_1} f\_2^{p-1-i\_2} \cdots f\_n^{p-1-i\_n} \right) \\
> & = \left(-1\right)^n \left(\det M\right)^{p-1} .
> \end{align}
> 
> 
> 


(I have taken the liberty to correct the typo in the paper where the sum ranged over $\left\{1,2,\ldots,p-1\right\}^n$ instead of $\left\{0,1,\ldots,p-1\right\}^n$. Note that later in the paper, in Proposition 1.5, there is a "$0 \leq j \leq n-1$" that should be a "$1 \leq j \leq n-1$"; this suggests blaming a miscommunication between author and editors about which $0$ to replace by a $1$.)


Theorem 1 generalizes Glynn's determinant formula (see Section 6 of [Hendrik Lenstra, *The unit theorem for finite-dimensional algebras*, arXiv:1703.07273v1](https://arxiv.org/abs/1703.07273v1)); indeed, it is easy to see that we can obtain the latter formula from Theorem 1 by setting $f\_j = \sum\limits\_{i=1}^n a\_{ij} x\_i$.



> 
> **Question.** Is there an elementary (e.g., combinatorial, inductive, or Hopf-algebraic) proof of Theorem 1?
> 
> 
> 


The proof in Lang's paper relies on [a different paper](https://www.jstage.jst.go.jp/article/kyotoms1969/26/3/26_3_453/_article/-char/ja), which in turn relies on some algebraic geometry. I am not sure what is actually used in the proof, but the whole construct seems rather indirect.


One approach that looks promising is to consider the $\mathbf{k}$-linear map $\widetilde{\nabla} : R \otimes R \to R$ (all tensors are over $\mathbf{k}$) that sends each pure tensor $a \otimes b$ to $a \nabla\left(b\right)$. Then, the left hand side of Theorem 1 is
\begin{equation}
\widetilde{\nabla}\left( \prod\_{i=1}^n \left(1 \otimes f\_i - f\_i \otimes 1\right)^{p-1} \right)
\end{equation}
(where the product is in $R \otimes R$). This is because of the classical fact that $\left(X-Y\right)^{p-1} = \sum\limits\_{i=0}^{p-1} X^i Y^{p-1-i}$ in $\mathbb{F}\_p\left[X,Y\right]$. But the question is whether $\widetilde{\nabla}$ has any good properties with respect to the Hopf algebra $R$.


Also, the operator $\nabla$ can be described rather explicitly on monomials: For any $n$ nonnegative integers $a\_1, a\_2, \ldots, a\_n$, we have
\begin{align}
& \nabla \left(x\_1^{a\_1} x\_2^{a\_2} \cdots x\_n^{a\_n}\right) \\
& = \begin{cases}
\left(-1\right)^n x\_1^{a\_1-\left(p-1\right)} x\_2^{a\_2-\left(p-1\right)} \cdots x\_n^{a\_n-\left(p-1\right)}, & \text{ if } a\_i \equiv p-1 \mod p \text{ for all } i ; \\
0, & \text{ otherwise }.
\end{cases}
\end{align}



Lang's Jacobian identity: slicker, elementary proof?
Awesome question! I haven't looked at Lang's paper yet, so I can't comment on whether this will be a different approach, but it is elementary. I will make use of Glynn's determinant formula at some point later on, so in order to keep this self contained I will start by giving a combinatorial proof of it.


**Lemma 1:** (Glynn's determinant formula) Suppose $A$ is a matrix with entries in a commutative $\mathbb F\_p$-algebra $T$. The coefficient of $x\_{1}^{p-1}x\_{2}^{p-1}\cdots x\_n^{p-1}$ in the expansion of $\prod\_{i=1}^{n}\left(\sum\_{j=1}^n a\_{ij}x\_j\right)^{p-1}$ is equal to $(\det A)^{p-1}$.


**Proof:** By the [Macmahon Master Theorem](https://en.wikipedia.org/wiki/MacMahon_Master_theorem) this coefficient is the same as the coefficient of $x\_{1}^{p-1}x\_{2}^{p-1}\cdots x\_n^{p-1}$ in the series $\det (I-AX)^{-1}\in T[[x\_1,x\_2,\cdots,x\_n]]$ where $X$ is the diagonal matrix with $x\_i$ on the diagonal. Notice that this coefficient is unchanged if we multiply by $\det(I-AX)^p$ since this doesn't affect monomials where all exponents are $<p$. However this leaves us with $\det(I-AX)^{p-1}$ which is a polynomial of total degree $(p-1)n$ so our coefficient comes from the top homogeneous part, and from here is easily seen to equal $\left(\det A\right)^{p-1}$.




---


Define the ideal $I=\langle \epsilon\_1^p-1,\dots,\epsilon\_n^p-1\rangle$ in $R\_1=R[\epsilon\_1,\dots,\epsilon\_n]=\mathbf{k}[x\_1,x\_2,\dots,x\_n,\epsilon\_1,\dots,\epsilon\_n]$. We can distinguish between two operators on $R\_1$:
$$\nabla\_{x}=\prod\limits\_{i=1}^n \left(\dfrac{\partial}{\partial x\_i}\right)^{p-1}, \nabla\_{\epsilon}=\prod\limits\_{i=1}^n \left(\dfrac{\partial}{\partial \epsilon\_i}\right)^{p-1}$$
and notice that since $\frac{\partial}{\partial \epsilon\_{i}}$ annihilate $\epsilon\_i^p-1$, these are well defined in $R\_1/I$ as well.


**Lemma 2:** We have the following relation in $R\_1/I$
$$\begin{align}
& \sum\limits\_{\left(i\_1, i\_2, \ldots, i\_n\right) \in \left\{0,1,\ldots,p-1\right\}^n} f\_1^{i\_1} f\_2^{i\_2} \cdots f\_n^{i\_n} \nabla\_x\left(f\_1^{p-1-i\_1} f\_2^{p-1-i\_2} \cdots f\_n^{p-1-i\_n} \right) \\
& = \frac{\nabla\_{\epsilon}\prod\_{i=1}^n(f\_i(\epsilon\_1 x\_1,\dots,\epsilon\_n x\_n)-f\_i(x\_1,\dots,x\_n))^{p-1}}{x\_1^{p-1}x\_2^{p-1}\cdots x\_n^{p-1}}
\end{align}$$


**Proof:** We will make use of the identity in $R\_1/I$ $$\nabla\_x F(x\_1,\cdots,x\_n)=\frac{\nabla\_{\epsilon} F(\epsilon\_1x\_1,\dots,\epsilon\_n x\_n)}{x\_1^{p-1}x\_2^{p-1}\cdots x\_n^{p-1}}$$
which is pretty much immediate from the explicit description of $\nabla$ given at the end of your question. By applying this identity to every instance of $\nabla\_x$ on the left hand side of our equation and then using the fact that
$$\sum\_{i=0}^{p-1} f\_j(x)^i f\_j(\epsilon x)^{p-1-i}=(f\_j(\epsilon x)-f\_j(x))^{p-1}$$
we arrive at the right hand side.




---


**Lemma 3:** We have
$$\prod\_{i=1}^n(f\_i(\epsilon\_1 x\_1,\dots,\epsilon\_n x\_n)-f\_i(x\_1,\dots,x\_n))^{p-1}=\prod\_{i=1}^n \left(\sum\_{j=1}^n (\epsilon\_j-1)x\_j\frac{\partial f\_i}{\partial x\_j}\right)^{p-1}$$


**Proof:** We can write 
$$f\_i(\epsilon\_1 x\_1,\dots,\epsilon\_n x\_n)-f\_i(x\_1,\dots,x\_n)=\sum\_{j=1}^n (\epsilon\_j-1)x\_j\frac{\partial f\_i}{\partial x\_j}+S\_i$$
when expanding in powers of the form $(\epsilon\_1-1)^{r\_1}(\epsilon\_2-1)^{r\_2}\cdots (\epsilon\_n-1)^{r\_n}$ with coefficients in $R$, where $S\_i$ consists of "higher order terms", i.e. terms where $r\_1+r\_2+\cdots+r\_n\geq 2$. Since we have $(\epsilon\_i-1)^p=0$ we notice that in the expansion of $$\prod\_{i=1}^n \left(S\_i+\sum\_{j=1}^n (\epsilon\_j-1)x\_j\frac{\partial f\_i}{\partial x\_j}\right)^{p-1}$$
any term which uses a monomial from any $S\_i$ will have degree $r\_1+r\_2+\cdots +r\_n> (p-1)n$ and will therefore vanish, since this implies that some $r\_i\geq p$.




---


We are now ready to complete the proof of Theorem 1 since
$$\nabla\_{\epsilon}\prod\_{i=1}^n \left(\sum\_{j=1}^n (\epsilon\_j-1)x\_j\frac{\partial f\_i}{\partial x\_j}\right)^{p-1}$$
is $(-1)^n$ times the coefficient of $\epsilon\_1^{p-1}\epsilon\_2^{p-1}\cdots \epsilon\_n^{p-1}$ in $\prod\_{i=1}^n \left(\sum\_{j=1}^n (\epsilon\_j-1)x\_j\frac{\partial f\_i}{\partial x\_j}\right)^{p-1}$. This last coefficient is of course the same as taking the coefficient of $(\epsilon\_1-1)^{p-1}\cdots (\epsilon\_n-1)^{p-1}$ in the same expression but expanded in monomials $(\epsilon\_1-1)^{r\_1}(\epsilon\_2-1)^{r\_2}\cdots (\epsilon\_n-1)^{r\_n}$. By Glynn's determinant this last coefficient is $x\_1^{p-1}x\_2^{p-1}\cdots x\_n^{p-1}\left(\det M\right)^{p-1}$ and this completes our proof.


**Remark:** This proof, and all the lemmas, adapt easily (essentially just change every occurrence of $(p-1)$ to $(p^m-1)$) to give the following generalization of Lang's identity: For any $m\geq 1$, let's denote by $\nabla^{(m)}$ the linear operator that sends monomials of the form $x\_1^{a\_1}x\_2^{a\_2}\cdots x\_n^{a\_n}$ to $x\_1^{a\_1-p^m+1}x\_2^{a\_2-p^m+1}\cdots x\_n^{a\_n-p^m+1}$ when $a\_i\cong -1\pmod{p^m}$ for all $1\le i\le n$, and to zero otherwise. The following identity holds:
$$\begin{align}
& \sum\limits\_{\left(i\_1, i\_2, \ldots, i\_n\right) \in \left\{0,1,\ldots,p^m-1\right\}^n} f\_1^{i\_1} f\_2^{i\_2} \cdots f\_n^{i\_n} \nabla^{(m)}\left(f\_1^{p^m-1-i\_1} f\_2^{p^m-1-i\_2} \cdots f\_n^{p^m-1-i\_n} \right)= \left(\det M\right)^{p^m-1} .
\end{align}$$


