\documentclass[11pt]{article} \usepackage[T1]{fontenc} \usepackage{amsmath,amssymb} \usepackage[colorlinks=true,linkcolor=blue,citecolor=blue,urlcolor=blue]{hyperref} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{question}[theorem]{Question} \newenvironment{proof}[1][Proof]{\par\noindent\textit{#1.}\ }{\hfill$\square$\par} \newcommand{\cref}[1]{\ref{#1}} \newcommand{\A}{\mathbb A} \newcommand{\Gm}{\mathbb G_m} \newcommand{\Jac}{\operatorname{Jac}} \newcommand{\wt}{\operatorname{wt}} \newcommand{\inw}{\operatorname{in}} \title{Cyclic Rigidity in Dimension Two:\\A Newton-Edge Obstruction for a Pseudoreflection Normal Form} \author{Annie\\AGNT Labs} \date{July 22, 2026} \begin{document} \maketitle \begin{abstract} Let $k$ be a field of characteristic zero, let $m\geq1$, and let $H,g\in k[u,y]$. We prove that \[ muH_u g_y-H_y(g+mug_u)=c\in k^\times \] forces $g\in k^\times$. Consequently every polynomial map \[ F(x,y)=\bigl(H(x^m,y),xg(x^m,y)\bigr) \] with nonzero constant Jacobian is a polynomial automorphism and admits an explicit inverse. After swapping the target coordinates, maps in this class commute with the cyclic pseudoreflection $(x,y)\mapsto(\zeta x,y)$. Conversely, when $m\geq2$ and after extending scalars to contain a primitive $m$-th root of unity, every polynomial map commuting with this action has the swapped normal form. The proof chooses the maximal Newton slope of $g$, isolates the corresponding weighted edges of $g$ and $H$, and derives a one-variable edge polynomial whose terminal coefficient cannot vanish in characteristic zero. The result is unbounded in degree and allows arbitrary multimode support. It is a rigidity theorem for this normal form, not a resolution of the unrestricted two-dimensional Jacobian Conjecture. \end{abstract} \section{Introduction} For polynomials $P,Q\in k[x,y]$, write \[ \Jac(P,Q)=P_xQ_y-P_yQ_x. \] The two-dimensional Jacobian Conjecture asks whether $\Jac(P,Q)\in k^\times$ forces $(P,Q)$ to be a polynomial automorphism when $k$ has characteristic zero. We study the normal form \begin{equation}\label{eq:F} F(x,y)=\bigl(H(x^m,y),xg(x^m,y)\bigr), \qquad H,g\in k[u,y],\quad u=x^m. \end{equation} There is a minor but important representation-theoretic distinction. Let \[ \rho_s(\zeta)=\operatorname{diag}(\zeta,1), \qquad \rho_t(\zeta)=\operatorname{diag}(1,\zeta), \qquad \zeta^m=1. \] Then the map in \eqref{eq:F} satisfies \[ F\circ\rho_s(\zeta)=\rho_t(\zeta)\circ F. \] It does not in general commute with one identical diagonal action. Equivalently, after the target-coordinate swap $S(P,Q)=(Q,P)$, the map \begin{equation}\label{eq:swapped} \widetilde F=S\circ F=\bigl(xg(x^m,y),H(x^m,y)\bigr) \end{equation} commutes with $\operatorname{diag}(\zeta,1)$. The coordinate swap changes the Jacobian only by a sign. Miyanishi proved an equivariant Jacobian theorem over $\mathbb C$ for small finite subgroups of $\mathrm{GL}(2,\mathbb C)$ of even order \cite{Miyanishi2023}. The pure cyclic group generated by $\operatorname{diag}(\zeta,1)$ is generated by pseudoreflections and is not small, so that main theorem does not directly subsume the class \eqref{eq:swapped}. Our argument is instead an elementary polynomial Newton-edge argument over an arbitrary characteristic-zero field. Classical Newton-polygon, degree-reduction, and geometry-at-infinity results remain essential context \cite{AbhyankarMoh1975,Razar1979,BCW1982,Moh1983,Glidewell2022,Nguyen2025}. \section{The cyclic Keller equation} Differentiating \eqref{eq:F} gives \[ (F_1)_x=mx^{m-1}H_u,\qquad (F_1)_y=H_y, \] \[ (F_2)_x=g+mug_u,\qquad (F_2)_y=xg_y. \] Hence \begin{equation}\label{eq:keller} \det JF=muH_u g_y-H_y(g+mug_u). \end{equation} Our main theorem is the following. \begin{theorem}[Cyclic rigidity]\label{thm:main} Let $k$ be a field of characteristic zero, let $m\geq1$, and let $H,g\in k[u,y]$. If \begin{equation}\label{eq:pde} muH_u g_y-H_y(g+mug_u)=c \end{equation} for some $c\in k^\times$, then $g\in k^\times$. \end{theorem} \begin{corollary}[Classification and inverse]\label{cor:classification} Under the hypotheses of \cref{thm:main}, there are $a\in k^\times$ and $h\in k[u]$ such that \[ g=a, \qquad H(u,y)=-\frac ca y+h(u). \] Thus \[ F(x,y)=\left(-\frac ca y+h(x^m),ax\right) \] is a polynomial automorphism. For target coordinates $(P,Q)=F(x,y)$, \[ F^{-1}(P,Q)= \left( \frac Qa, -\frac ac\left[P-h\left(\left(\frac Qa\right)^m\right)\right] \right). \] \end{corollary} \section{Boundary data} \begin{lemma}\label{lem:boundary} If \eqref{eq:pde} holds, then for some $a\in k^\times$ and $b\in k$, \[ g(0,y)=a, \qquad H(0,y)=-\frac ca y+b. \] \end{lemma} \begin{proof} Set $u=0$ in \eqref{eq:pde}. Then \[ -H_y(0,y)g(0,y)=c. \] Both factors lie in $k[y]$ and their product is a nonzero scalar. Since the units of $k[y]$ are precisely $k^\times$, both factors are constant and nonzero. Integration in $y$ gives the formula for $H(0,y)$. \end{proof} \section{The maximal Newton slope} Assume for contradiction that $g$ is nonconstant. By \cref{lem:boundary}, every nonconstant term of $g$ contains a positive power of $u$, so write \[ g=a+\sum_{j\geq1}a_j(y)u^j, \qquad a\in k^\times. \] Define \begin{equation}\label{eq:rho} \rho=\max_{a_j\neq0}\frac{\deg_y a_j}{j}=\frac pq, \end{equation} where $p\geq0$, $q\geq1$, $\gcd(p,q)=1$, and $0/q$ is reduced to $0/1$. Give $k[u,y]$ the rational weight \[ \wt(y)=1, \qquad \wt(u)=-\rho=-\frac pq. \] Every monomial of $g$ has weight at most zero. \begin{lemma}[Edge of $g$]\label{lem:g-edge} Let $g_{[0]}$ denote the weight-zero component of $g$. Then \[ g_{[0]}=A(t), \qquad t=u^qy^p, \] where $A\in k[t]$, $A(0)=a\neq0$, and $D:=\deg A\geq1$. \end{lemma} \begin{proof} A monomial $u^jy^d$ has weight zero exactly when \[ qd=pj. \] Coprimality gives $j=qn$ and $d=pn$ for some $n\geq0$, so every weight-zero monomial is $t^n$. Because the maximum in \eqref{eq:rho} is attained by at least one $j\geq1$, a positive power of $t$ occurs. The case $p=0$ gives $q=1$ and $t=u$. \end{proof} Let $\sigma$ be the largest weight occurring in $H$. This exists because $H$ has finite support. By \cref{lem:boundary}, the monomial $-(c/a)y$ occurs, so \begin{equation}\label{eq:sigma} \sigma\geq1. \end{equation} \begin{lemma}[Edge of $H$]\label{lem:H-edge} There are integers $r,s$ and a nonzero polynomial $\Phi\in k[t]$ such that \[ 0\leq r1$, this is a positive-weight component, contradicting that the right side of \eqref{eq:pde} is the scalar $c$ of weight zero. If $\sigma=1$, the entire top component in \eqref{eq:edge-op} must equal $c$. A scalar monomial can arise from $u^ry^{s-1}t^k$ only when $r=0$, $s=1$, and $k=0$. Thus $B$ would have to be constant. But \cref{lem:terminal} gives a nonzero term of degree $D+N\geq1$, again a contradiction. Therefore $g$ is constant. By \cref{lem:boundary}, that constant lies in $k^\times$. \end{proof} \begin{proof}[Proof of \cref{cor:classification}] Write $g=a\in k^\times$ in \eqref{eq:pde}. Then $-aH_y=c$, so \[ H(u,y)=-\frac ca y+h(u) \] for some $h\in k[u]$. The inverse follows by solving $Q=ax$ and then solving the first coordinate for $y$. \end{proof} \section{Quotient identity and collisions} Define \[ K=ug^m. \] A direct calculation gives \begin{equation}\label{eq:quotient} \Jac(H,K)=g^{m-1}\det JF. \end{equation} Under $\det JF=c$ this yields \[ \Jac(H,K)=cg^{m-1}, \qquad u^{m-1}\Jac(H,K)^m=c^mK^{m-1}. \] For $m\geq2$, a hypothetical nonconstant $g$ would also produce an explicit collision after passing to an algebraic closure. Since $g(0,y)\neq0$, every zero $(u_0,y_0)$ of $g$ has $u_0\neq0$. Choosing $x_0^m=u_0$, the distinct points \[ (\zeta^jx_0,y_0), \qquad 0\leq j