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).
2026-03-27 10:09:05.1774606145
Nonzero subdeterminants conjecture: has anybody seen this anywhere?
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