Let X⊂R^n and let R_n denote the polynomial ring in n variables. For a set G⊂R_n, write G^deg(g)-1 for the relevant degree-restricted subset, and suppose that G generates the vanishing ideal I(X). Assume that G is degree-restriction compatible. Gradient criterion for redundancy. For some g∈ G, one has g∈langle G^deg(g)-1rangle if and only if, for every boldsymbolx∈ X, there exist coefficients alpha_g^prime,boldsymbolx∈R such that nabla g(boldsymbolx)=∑_g^prime∈ G^deg(g)-1alpha_g^prime,boldsymbolxnabla g^prime(boldsymbolx). This gives a numerical, noise-tolerant and monomial-agnostic criterion for detecting redundant basis polynomials, avoiding Gröbner-basis computations; the conjecture's resolution is not established in the supplied text.

Gradient criterion for redundant generators of vanishing ideals

Let $X\subset\mathbb{R}^n$ and let $\mathcal{R}_n$ denote the polynomial ring in $n$ variables. For a set $G\subset\mathcal{R}_n$ , write $G^{\mathrm{deg}(g)-1}$ for the relevant degree-restricted subset, and suppose that $G$ generates the vanishing ideal $\mathcal{I}(X)$ . Assume that $G$ is degree-restriction compatible. Gradient criterion for redundancy. For some $g\in G$ , one has

g\in\langle G^{\mathrm{deg}(g)-1}\rangle

if and only if, for every $\boldsymbol{x}\in X$ , there exist coefficients $\alpha_{g^{\prime},\boldsymbol{x}}\in\mathbb{R}$ such that

\nabla g(\boldsymbol{x})=\sum_{g^{\prime}\in G^{\mathrm{deg}(g)-1}}\alpha_{g^{\prime},\boldsymbol{x}}\nabla g^{\prime}(\boldsymbol{x}).

This gives a numerical, noise-tolerant and monomial-agnostic criterion for detecting redundant basis polynomials, avoiding Gröbner-basis computations; the conjecture's resolution is not established in the supplied text.

Progress summary

Solved

A reader-written construction claims the conjecture is false by giving a degree-compatible generating set whose gradients pass the test although one generator is not redundant.

The conjecture asserts that, for a degree-compatible generating set $G$ of a vanishing ideal, membership $g\in\langle G^{\mathrm{deg}(g)-1}\rangle$ is equivalent to pointwise expressibility of $\nabla g$ through lower-degree generator gradients. Kera and Hasegawa state the claim as an unproved conjecture.

Known results

The membership-to-gradient implication is readily proved.
The converse is stated as Conjecture 1 and explicitly left unproved.
Conjecture 9.1 restates the criterion under degree-restriction compatibility and uses it heuristically for numerical redundancy removal.
The associated least-squares test for approximate data is an application, not a proof.

Posted attempt

A construction with $X=\{0,1,2\}^2\setminus\{(2,2)\}$ and generators $f=x(x-1)(x-2)$ , $h=y(y-1)(y-2)$ , and $g=x(x-1)y(y-1)$ claims all degree-compatibility hypotheses hold, the gradients of $f$ and $h$ span every required gradient at points of $X$ , but $g\notin(f,h)$ . It therefore claims a complete counterexample; the argument has not been independently verified.

Current status (as of August 2026): the forward implication is established, while the converse is challenged by an unverified claimed counterexample and is not settled.

Sources

Sources & referencesView supporting material

Primary source

Hiroshi Kera and Yoshihiko Hasegawa, “Monomial-agnostic computation of vanishing ideals”, arXiv:2101.00243 (2023).

1 week ago

Counterexample satisfying all degree-compatibility hypotheses.

Let

D=\{0,1,2\},\qquad X=D^2\setminus\{(2,2)\}\subset\mathbb R^2,

and define

f=x(x-1)(x-2),\qquad h=y(y-1)(y-2),\qquad g=x(x-1)y(y-1).

Take $G=\{f,h,g\}$ .

The Chinese remainder theorem identifies

\mathbb R[x,y]/(f,h) \cong\prod_{(a,b)\in D^2}\mathbb R

by evaluation on the nine-point grid. Under this identification, $g$ vanishes at all eight points of $X$ and equals $4$ at $(2,2)$ . Its image therefore generates the coordinate factor supported at the omitted point. Hence

I(X)=(f,h,g).

The required degree compatibility also holds. If $u\in I(X)$ has total degree at most $3$ , divide by the monic univariate polynomials $f,h$ . Its remainder $R$ has degree at most $2$ in each variable and total degree at most $3$ , while still vanishing on $X$ . Uniqueness of interpolation on $D^2$ implies

R(x,y)=\frac{R(2,2)}4\,x(x-1)y(y-1).

Since the right-hand side has total degree $4$ unless its coefficient is zero, $R=0$ . Thus

I(X)_{\le2}=\{0\},\qquad I(X)_{\le3}=\operatorname{span}_{\mathbb R}\{f,h\}.

For degrees at least $4$ , all three generators are available and generate $I(X)$ . Consequently every $u\in I(X)$ belongs to the ideal generated by elements of $G$ of degree at most $\deg u$ , exactly as required.

Now $\deg g=4$ , so the lower-degree generators are precisely $f,h$ . Write $F(t)=t(t-1)(t-2)$ . At every $(a,b)\in X$ ,

\nabla f(a,b)=(F'(a),0),\qquad \nabla h(a,b)=(0,F'(b)).

Since

F'(0)=2,\qquad F'(1)=-1,\qquad F'(2)=2,

these gradients form a basis of $\mathbb R^2$ . Therefore

\nabla g(a,b) =\frac{\partial_xg(a,b)}{F'(a)}\nabla f(a,b) +\frac{\partial_yg(a,b)}{F'(b)}\nabla h(a,b)

at every point of $X$ , fulfilling the proposed gradient criterion.

Nevertheless $g\notin(f,h)$ : otherwise evaluating at $(2,2)$ would give $g(2,2)=0$ , whereas $g(2,2)=4$ . Hence the pointwise gradient condition does not imply redundancy, and the conjecture is false.

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Shivam Patel · 2 days ago