Lang's Jacobian identity: slicker, elementary proof?

In Jeffrey Lang, A Jacobian identity in positive characteristic, J. Commut. Algebra, Volume 7, Number 3 (2015), pp. 393--409, 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); 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, 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 $$\widetilde{\nabla}\left( \prod_{i=1}^n \left(1 \otimes f_i - f_i \otimes 1\right)^{p-1} \right)$$ (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}

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 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}

• Also, in Lemma 1, you probably want the entries to be in a commutative $\mathbf{k}$-algebra, not in $\mathbf{k}$, right? Commented Aug 27, 2018 at 7:42
• @darijgrinberg That's true, but fortunately MMT has bijective/symbolic proofs that will work in any generality. I will have to make the edit. Commented Aug 27, 2018 at 7:45
• I don't quite see the first identity in the proof of Lemma 2. Is $F$ supposed to be a polynomial over $\mathbb{k}$ ? Should the ideal $I$ perhaps be generated by the $\epsilon^{p-1}-1$, not the $\epsilon^p - 1$? Commented Aug 27, 2018 at 7:50
• @darijgrinberg Yes, $F$ is any arbitrary polynomial. The image of $\nabla$ consists of monomials whose exponents are all zero mod p which is why we need $\epsilon ^p=1$. Commented Aug 27, 2018 at 7:53
• Wow, this is a beautiful argument, if long and somewhat hard to read due to the terseness. I think it would make a nice paper! Jeffrey Lang's proof is completely different, and there is novelty in yours. (Even your proof of Glynn's determinant formula is new to my knowledge!) Commented Aug 27, 2018 at 8:58