Homogeneous-polynomial extension of the generalized relation

Let ${\bf x}^m=\{x_1,\ldots,x_m\}$ and let ${\bf 1}^m$ denote the $m$ -tuple all of whose entries are $1$ . Let $U_n({\bf x}^m,{\bf y}^m)$ be the function defined from the symmetric polynomials $F_n$ and the rows of ${\bf S}({\bf x}^m,{\bf y}^m)$ . Homogeneous-polynomial extension. For $n\geq m$ , specializing ${\bf y}^m={\bf 1}^m$ gives a homogeneous symmetric polynomial $Z_{n-m}({\bf x}^m)$ of degree $n-m$ :

U_n({\bf x}^m,{\bf 1}^m)=Z_{n-m}({\bf x}^m),\qquad n\geq m.

The claim extends the preceding relation beyond $n=m-1$ and asserts polynomiality after the specialization; the source gives no evidence resolving it.

Progress summary

Solved

A 2025 preprint claims the extension is valid in full generality, and a posted complete-proof attempt is unverified.

The problem asks whether specializing the generalized relation at ${\bf y}^m={\bf 1}^m$ always produces a homogeneous symmetric polynomial for $n\ge m$ . Boris Y. Rubinstein's 2025 preprint claims a broader result: the underlying relations hold for every homogeneous symmetric polynomial.

October 2025 claimed generalization

Rubinstein claims that the relations are valid for arbitrary homogeneous symmetric polynomials, which would imply the stated extension for the specified family $F_n$ . The retrieved record contains no independent verification.

Posted attempt

A complete proof attempt argues that all possible poles at $x_i=0$ and $x_i=x_j$ cancel by symmetry, leaving a homogeneous symmetric polynomial of degree $n-m$ . This attempt has not been independently verified.

Current status (as of August 2026): A 2025 preprint claims the extension, but no independent verification is recorded, so the claim remains unsettled.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Boris Y. Rubinstein, “A New Class of Relations for Homogeneous Symmetric Polynomials”, arXiv:2510.25749 (2025).

homogeneous-polynomials symmetric-polynomials

1 week ago

The asserted extension holds for every symmetric homogeneous polynomial, not only for the specified Bell-polynomial family.

Let $S$ be any symmetric homogeneous polynomial of degree $n$ in $m$ variables, and specialize $y_1=\cdots=y_m=1$ . Put

(s_i)_j= \begin{cases} x_i,&j=i,\\ x_j-x_i,&j\ne i. \end{cases}

The rational expression in the conjecture becomes

U(x,\mathbf1) = \frac{S(x)}{\prod_{r=1}^m x_r} - \sum_{i=1}^m \frac{S(s_i)} {x_i\prod_{j\ne i}(x_j-x_i)}.

Its only possible polar factors are the distinct irreducible linear forms

x_i,\qquad x_j-x_i.

When $x_i=0$ , the vector $s_i$ agrees with $x$ with its $i$ -th coordinate set to zero. Consequently the residue of the $i$ -th summand is

\frac{S(x_1,\ldots,x_{i-1},0,x_{i+1},\ldots,x_m)} {\prod_{j\ne i}x_j},

which exactly cancels the residue of $S(x)/\prod_r x_r$ .

When $x_i=x_j$ , the vectors $s_i$ and $s_j$ differ only by interchanging coordinates $i$ and $j$ . Symmetry gives

S(s_i)=S(s_j).

All other denominator factors coincide, whereas the singular factors satisfy

x_j-x_i=-(x_i-x_j).

Thus the residues of the $i$ -th and $j$ -th summands cancel.

Every possible pole is therefore removable, proving

U(x,\mathbf1)\in\mathbb Q[x_1,\ldots,x_m].

Each summand is homogeneous of degree $n-m$ , so the resulting polynomial is homogeneous of degree $n-m$ , or zero. Finally, every permutation of the $x$ -coordinates permutes the summands, while symmetry of $S$ preserves their numerators. Hence $U(x,\mathbf1)$ is symmetric.

Taking $S=F_n$ and $n\ge m$ gives exactly the conjectured homogeneous symmetric polynomial $Z_{n-m}$ .

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Shivam Patel · 2 days ago