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I already posted this question on Mathematics StackExchange. A user there suggested that I rather post it on mathoverflow, since it is a research question. So here it is.

Let $m\geq2$, $n\geq1$ be natural numbers. Let $M$ be the set of all lists $(k_1,\ldots,k_n)$ of natural numbers such that $k_1+\cdots+k_n=m$, and let $M'$ be the set of all lists $(k'_1,\ldots,k'_n)$ of natural numbers such that $k'_1+\cdots+k'_n=m-1$. For $x=(x_1,\ldots,x_n)$, where the $x_i$ are formal variables, and for any list $\lambda=(l_1,\ldots,l_n)$ of natural numbers, we write $x^\lambda:=x_1^{l_1}\cdots x_n^{l_n}$. Let $$ f = \sum_{\mu\in M} c_\mu\, x^\mu~, $$ where the $c_\mu$, $\mu\in M$, are formal variables. The coefficients of partial derivatives $$ \frac{\partial f}{\partial x_i} = \sum_{\mu'\in M'}c'_{i,\,\mu'}\,x^{\mu'}~, \qquad\quad i=1,\,\ldots,\,n~, $$ are integer multiples of the coefficients of $f$. We form the $n\!\times\!\binom{n+m-2}{m-1}$-matrix $$ \Gamma = \bigl[\,c'_{i,\,\mu'}\,\big|\,1\leq i\leq n,\,\mu'\in M'\,\bigr]~. $$ Conjecture. The determinant of every $n\!\times\!n$-submatrix of the matrix $\Gamma$ is a nonzero polynomial, with integer coefficients, of the formal variables $c_\mu$, $\mu\in M$. Equivalently: for every choice of $n$ columns of the matrix $\Gamma$ there exist integer values for the coefficients $c_\mu$, $\mu\in M$, which make the chosen columns linearly independent (over integers/rational numbers).

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  • $\begingroup$ In other words, you want to prove that the vectors $\left(c_{\nu+e_1}, c_{\nu+e_2}, \ldots, c_{\nu+e_n}\right)$ for any $n$ distinct values of $\nu \in M^{\prime}$ (where $e_1,e_2,\ldots,e_n$ is the standard basis of the $\mathbb{Z}$-module $\mathbb{Z}^n$, and we regard $\nu$ as a vector in this module) are linearly independent over the polynomial ring in the variables $c_\mu$. Right? $\endgroup$
    – darij grinberg
    Commented Jun 4, 2018 at 16:41
  • $\begingroup$ Okay, let me try directly attacking your Conjecture. Consider the $n\times n$-submatrix of $\Gamma$ obtained by taking columns $\nu_1, \nu_2, \ldots, \nu_n$. WLOG assume that $\nu_1 > \nu_2 > \cdots > \nu_n$ in lexicographic order. We regard elements of $M$ and of $M^\prime$ as vectors in the $\mathbb{Z}$-module $\mathbb{Z}^n$, and we let $\left(e_1,e_2,\ldots,e_n\right)$ be the basis of this $\mathbb{Z}$-module. Then I believe that the monomial $c_{\nu_1 + e_1} c_{\nu_2 + e_2} \cdots c_{\nu_n + e_n}$ appears only once in the determinant, ... $\endgroup$
    – darij grinberg
    Commented Jun 4, 2018 at 16:45
  • $\begingroup$ ... namely as the result of multiplying through the main diagonal. All other monomials are smaller in the lexicographic order of monomials (where the variables $c_\mu$ themselves are ordered by the lexicographic order of the $\mu \in M$). So this monomial isn't cancelled by anything, and thus is nonzero. Note that I'm assuming the ground field to have characteristic $0$ here; otherwise I don't know. $\endgroup$
    – darij grinberg
    Commented Jun 4, 2018 at 16:46
  • $\begingroup$ @darij grinberg - This is a promising way to attack the problem, though your reasoning at the start is incorrect. Note that the characteristic is 0 since the ground ring is the polynomial ring in the variables $c_\mu$, $\mu\in M$, with integer coefficients. $\endgroup$
    – chizhek
    Commented Jun 6, 2018 at 10:03
  • $\begingroup$ Oops, yes, I forgot to mention the coefficients (though I thought of them, as they're the reason for the characteristic-$0$ assumption). Thanks for clearing it up. You never said "integer coefficients", I believe, so I figured I should answer as generally as I could. $\endgroup$
    – darij grinberg
    Commented Jun 6, 2018 at 13:09

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