# On the two-dimensional Jacobian conjecture: Magnus’ formula revisited, I

William E. Hurst  Department of Mathematics, University of Alabama,
Tuscaloosa, AL 35487, U.S.A<EMAIL_ADDRESS>,  Kyungyong Lee
Department of Mathematics, University of Alabama, Tuscaloosa, AL 35487, U.S.A.
and Korea Institute for Advanced Study, Seoul 02455, Republic of Korea
<EMAIL_ADDRESS><EMAIL_ADDRESS>,  Li Li  Department of Mathematics and
Statistics, Oakland University, Rochester, MI 48309, U.S.A.
<EMAIL_ADDRESS>and  George D. Nasr  Department of Mathematics, University
of Oregon, Eugene, OR 97403, U.S.A<EMAIL_ADDRESS>To the memory of
Shreeram Shankar Abhyankar

###### Abstract.

Let $K$ be an algebraically closed field of characteristic 0. When the
Jacobian $({\partial f}/{\partial x})({\partial g}/{\partial y})-({\partial
g}/{\partial x})({\partial f}/{\partial y})$ is a constant for $f,g\in
K[x,y]$, Magnus’ formula from [23] describes the relations between the
homogeneous degree pieces $f_{i}$’s and $g_{i}$’s. We show a more general
version of Magnus’ formula and prove a special case of the two-dimensional
Jacobian conjecture as its application.

This paper grew out of an undergraduate research project for WH, which had
been initiated by the Randall Research Scholars Program at the University of
Alabama. KL was supported by the University of Alabama, Korea Institute for
Advanced Study, and the NSF grant DMS 2042786. GDN is supported by the NSF FRG
grant, Grant Number DMS-2053243.

##  1\. introduction

The Jacobian conjecture, raised by Keller [20], has been studied by many
mathematicians: a partial list of related results includes [1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 16, 17, 18, 19, 21, 22, 25, 26, 27, 28, 29, 30, 31,
32]. A survey is given in [14, 15]. In this paper we exclusively deal with the
plane case. Hence whenever we write the Jacobian conjecture, we mean the two-
dimensional Jacobian conjecture.

Let $K$ be an algebraically closed field of characteristic 0, and let
$\mathcal{R}=K[x,y]$.

Jacobian conjecture. _Let $f(x,y),g(x,y)\in\mathcal{R}$. Consider the
polynomial map _$\pi:\mathcal{R}\longrightarrow\mathcal{R}$_given by_
$\pi(x)=f(x,y)$ and $\pi(y)=g(x,y)$. _If the Jacobian of the map_

$\det\begin{pmatrix}{\partial f}/{\partial x}&{\partial g}/{\partial x}\\\
{\partial f}/{\partial y}&{\partial g}/{\partial y}\end{pmatrix}$

_is a non-zero constant, then the map is bijective._

For simplicity, let $[f,g]:=\det\begin{pmatrix}{\partial f}/{\partial
x}&{\partial g}/{\partial x}\\\ {\partial f}/{\partial y}&{\partial
g}/{\partial y}\end{pmatrix}\in\mathcal{R}$ for any pair of polynomials
$f,g\in\mathcal{R}$. Similarly $[f,g]$ is defined for
$f,g\in\mathcal{R}[[t]]=K[[t]][x,y]$.

A useful tool to study this conjecture is the Newton polygon. One source for
this is [10], but we redefine it here. Let

$f=\displaystyle\sum_{i,j\geq 0}f_{ij}x^{i}y^{j}$

be a polynomial in $\mathcal{R}$. The support of $f$ is defined as

$\operatorname{supp}(f)=\\{(i,j)\mid f_{ij}\neq
0\\}\subset\mathbb{Z}^{2}\subset\mathbb{R}^{2}.$

The Newton polygon for $f$, which we denote $N(f)$, is defined to be the
convex hull of $\operatorname{supp}(f)$ 111In [10], it was defined as the
convex hull of $\operatorname{supp}(f)\cup\\{(0,0)\\}$ in $\mathbb{R}^{2}$,
which is different from our definition. in $\mathbb{R}^{2}$. Note that
$N(f)\subset\mathbb{R}_{\geq 0}^{2}$. See Figure 1 for an example of a Newton
polygon. The support and Newton polygon of a Laurent polynomial in $K[x^{\pm
1},y^{\pm 1}]$ are similarly defined.

$x$$y$ Figure 1. $N(f)$ for
$f=y+7xy^{3}+\sqrt{7}x^{3}y^{4}-4x^{5}y^{3}+2x^{5}y^{2}-{1\over 2}x^{3}+xy+1$.

Throughout this paper whenever we consider pairs of polynomials $f,g$ with
$[f,g]\in K$, we only consider such pairs for which both $N(f)$ and $N(g)$
contain $(1,0),(0,1)$, and $(0,0)$. As long as $\deg(f)$ and $\deg(g)$ are
positive, it is always possible to obtain such a pair by adding a generic
constant to $f$ and $g$ and applying some linear change of variables $x$ and
$y$, which does not change $[f,g]$.

It is known that the following conjecture implies the Jacobian conjecture.
(For instance, see [15, Theorem 10.2.23].)

######  Conjecture A.

Let $a,b\in\mathbb{Z}_{>0}$ be relatively prime. Suppose that
$F,G\in\mathcal{R}$ satisfy the following:

_(1)_ $[F,G]\in K$;

_(2)_ $\\{(1,0),(0,1),(0,0)\\}\subset N(F)\cap N(G)$ and $N(F)$ is similar to
$N(G)$ with the origin as center of similarity and with ratio
$\deg(F):\deg(G)=a:b$; and

_(3)_ $\min(a,b)\geq 2$.

Then $[F,G]=0$.

Let $W=\\{(u,v)\in\mathbb{Z}^{2}\,:\,u>0\text{ or }v>0,\text{ and
}\gcd(|u|,|v|)=1\\}$. An element $w=(u,v)\in W$ is called a _direction_. To
each such a direction we consider its $w$-grading on $\mathcal{R}$ by defining
the $w$-degree of the monomial $x^{i}y^{j}$ as $n=ui+vj$. Define
$\mathcal{R}_{n}\subset\mathcal{R}$ to be the $K$-subspace generated by
monomials of $w$-degree $n$. Then
$\mathcal{R}=\oplus_{n\in\mathbb{Z}}\mathcal{R}_{n}$. A non-zero element $P$
of $\mathcal{R}_{n}$ is called a $w$-homogeneous element of $\mathcal{R}$; the
integer $n$ is called the $w$-degree of $P$ and is denoted $w$-$\deg(P)$. The
element of the highest $w$-degree in the homogeneous decomposition of a non-
zero polynomial $P$ is called its $w$-leading form and is denoted by $P_{+}$.
The $w$-degree of $P$ is by definition $w$-$\deg(P_{+})$.

In [23, Theorem 1], Magnus produced a formula which inspired much of the work
for this paper. His formula was published almost 70 years ago but has not been
used in almost any paper but [24]. Even in [24], only small piece of
information from the formula was utilized. The first main result in our paper
is a more general version of Magnus’ formula, as given in Theorem 1.1. In what
follows, the binomial coefficient ${A\choose B}$ is defined by

${A\choose B}:={A(A-1)\cdots(A-B+1)\over B!}$

for any real number $A$ and any nonnegative integer $B$.

######  Theorem 1.1.

Suppose $[F,G]\in K$. For any direction $w=(u,v)\in W$, let
$d=w\text{-}\deg(F_{+})$ and $e=w\text{-}\deg(G_{+})$. Write the
$w$-homogeneous degree decompositions $F=\sum_{i\leq d}F_{i}$ and
$G=\sum_{i\leq e}G_{i}$. Then there exists a unique 222Note that there is some
ambiguity of the notation $F_{d}^{1/r}$, since it is unique up to an $r$-th
root of unity. We fix a choice of $F_{d}^{1/r}$. Then the fractional power
$F_{d}^{c/r}:=(F_{d}^{1/r})^{c}$ is nonambiguous for any integer $c$. sequence
of constants $c_{0},c_{1},...,c_{d+e-u-v-1}\in K$ such that $c_{0}\neq 0$ and

(1.1)
$G_{e-\mu}=\sum_{\gamma=0}^{\mu}c_{\gamma}\sum{(e-\gamma)/d\choose\sum_{\alpha\leq
d-1}\nu_{\gamma,\alpha}}\frac{\left(\sum_{\alpha\leq
d-1}\nu_{\gamma,\alpha}\right)!}{\prod_{\alpha\leq
d-1}\nu_{\gamma,\alpha}!}F_{d}^{(e-\gamma)/d-\sum_{\alpha\leq
d-1}\nu_{\gamma,\alpha}}\prod_{\alpha\leq
d-1}F_{\alpha}^{\nu_{\gamma,\alpha}}$

for every integer $\mu\in\\{0,1,...,d+e-u-v-1\\}$, where the inner sum is to
run over all combinations of non-negative integers $\nu_{\gamma,\alpha}$
satisfying $\sum_{\alpha\leq d-1}(d-\alpha)\nu_{\gamma,\alpha}=\mu-\gamma$.
Furthermore, $c_{\gamma}=0$ if $r(e-\gamma)/d\notin\mathbb{Z}$, where
$r\in\mathbb{Z}_{>0}$ is the largest integer such that $F_{d}^{1/r}\in
K[x,y]$.

In a series of forthcoming papers, we will make progress toward Conjecture A,
using Theorem 1.1. The purpose of this paper is to write out a proof of
Theorem 1.1333In [23], a detailed proof was not given. Moreover, the original
statement in [23] was written only for $w=(1,1)$, and did not contain the
statement that starts with ”Furthermore”, which will play a pivotal role in a
series of our papers including this one. and illustrate how useful this
theorem is.

Let $F$ and $G$ satisfy the assumptions (1) and (2) in Conjecture A. Then
$[F,G]\in K$ implies $F_{+}^{1/a}\in K[x,y]$ for any direction $w\in W$ (for
instance, see [2, 4, 29]). Let $\mathcal{T}$ be the set of polynomials $f\in
K[x,y]$ such that $N(f)$ contains exactly two distinct lattice points, i.e.,
$N(f)$ is a line segment containing no lattice points other than its
endpoints.

######  Definition 1.2.

We say that the pair $(F,G)$ has _generic boundaries_ if it satisfies (1) and
(2) in Conjecture A, and the polynomial $F_{+}^{1/a}$ is not divisible by the
square of any polynomial in $\mathcal{T}$ for any direction $w\in W$. 444Note
that the latter condition is equivalent to that $F_{+}^{1/a}$ does not have as
a divisor any square of a non-monomial polynomial, since $K$ is algebraically
closed.

The following is the second main result in this paper.

######  Theorem 1.3.

If $(F,G)$ has generic boundaries with $a=2$, then Conjecture A is true. More
precisely, we have

$F={P}^{2}+u_{0}$

for some ${P}\in K[x,y]$ and some $u_{0}\in K$. In particular, $[F,G]=0$.

For a real number $r\in\mathbb{R}$ and a subset $S\subseteq\mathbb{R}^{2}$,
denote $rS:=\\{rs\ :s\in S\\}\subseteq\mathbb{R}^{2}$.

######  Corollary 1.4.

Suppose that each edge of $\frac{1}{a}N(F)$ contains either the origin or no
lattice points other than its endpoints. If $a=2$ then Conjecture A is true.

##  2\. Magnus’ Formula Revisited

The goal of this section is to prove Theorem 1.1 and present Proposition 2.4,
a useful application of this Theorem. We start by reinterpreting the equality
(1.1) in Theorem 1.1 as follows. Recall that $\mathcal{R}=K[x,y]$, where $K$
is an algebraically closed field of characteristic 0. For any
$\widetilde{F}\in\mathcal{R}[[t]]$, denote

$[\widetilde{F}]_{t^{i}}=\textrm{ the coefficient of $t^{i}$ in
$\widetilde{F}$},$

which is a polynomial in $x$ and $y$.

Recall the generalized multinomial theorem in the formal power series ring
$K[[x_{1},\dots,x_{n}]]$: for $A\in\mathbb{Q}$,

$(1+x_{1}+\cdots+x_{n})^{A}=\sum_{v_{1},\dots,v_{n}\in\mathbb{Z}_{\geq
0}}\frac{A(A-1)\cdots(A-\sum_{i=1}^{n}v_{i})}{\prod_{i=1}^{n}v_{i}!}x_{1}^{v_{1}}\cdots
x_{n}^{v_{n}}.$

Consider a variation of this. For $x_{1},\dots,x_{n}\in\mathcal{R}$, we have
the following expansion in the ring $\mathcal{R}[[t]]$:

$(1+x_{1}t+\cdots+x_{n}t^{n})^{A}=\sum_{v_{1},\dots,v_{n}\in\mathbb{Z}_{\geq
0}}\frac{A(A-1)\cdots(A-\sum_{i=1}^{n}v_{i})}{\prod_{i=1}^{n}v_{i}!}x_{1}^{v_{1}}\cdots
x_{n}^{v_{n}}t^{v_{1}+2v_{2}+\cdots+nv_{n}}.$

In general, for $x_{0},\dots,x_{n}\in\mathcal{R}$ and for $A=a/b$ where
$a\in\mathbb{Z}$, $b\in\mathbb{Z}_{>0}$, we have the following identity in the
ring $\mathcal{R}[x_{0}^{\pm 1/b}][[t]]$ (where we fix a choice of
$x_{0}^{1/b}$):

(2.1)
$(x_{0}+x_{1}t+\cdots+x_{n}t^{n})^{A}=\sum_{v_{1},\dots,v_{n}\in\mathbb{Z}_{\geq
0}}\frac{A\cdots(A-\sum_{i=1}^{n}v_{i})}{\prod_{i=1}^{n}v_{i}!}x_{0}^{A-\sum_{i=1}^{n}v_{i}}x_{1}^{v_{1}}\cdots
x_{n}^{v_{n}}t^{v_{1}+\cdots+nv_{n}}.$

######  Lemma 2.1.

The equality (1.1) can be rewritten as the following equality in
$\mathcal{R}[F_{d}^{\pm 1/d}]$:

(2.2)
$G_{e-\mu}=\sum_{\gamma=0}^{\mu}c_{\gamma}\bigg{[}\bigg{(}F_{d}+F_{d-1}t+F_{d-2}t^{2}+\cdots\bigg{)}^{(e-\gamma)/d}\bigg{]}_{t^{\mu-\gamma}}.$

###### Proof.

Let $A=(e-\gamma)/d$ and $s=\sum_{\alpha\leq d-1}\nu_{\gamma,\alpha}$. Then

${(e-\gamma)/d\choose\sum_{\alpha\leq
d-1}\nu_{\gamma,\alpha}}\frac{\left(\sum_{\alpha\leq
d-1}\nu_{\gamma,\alpha}\right)!}{\prod_{\alpha\leq
d-1}\nu_{\gamma,\alpha}!}=\frac{A(A-1)\cdots(A-s+1)}{s!}\frac{s!}{\prod_{\alpha\leq
d-1}\nu_{\gamma,\alpha}!}=\frac{A\cdots(A-s+1)}{\prod_{\alpha\leq
d-1}\nu_{\gamma,\alpha}!}.$

So the right side of (1.1), without the constraint $\sum_{\alpha\leq
d-1}(d-\alpha)\nu_{\gamma,\alpha}=\mu-\gamma$, is

$\sum_{\gamma=0}^{\mu}c_{\gamma}\bigg{(}F_{d}+F_{d-1}t+F_{d-2}t^{2}+\cdots\bigg{)}^{(e-\gamma)/d},$

thanks to (2.1). Then, note that the constraint $\sum_{\alpha\leq
d-1}(d-\alpha)\nu_{\gamma,\alpha}=\mu-\gamma$ is equivalent to the restriction
to degree $t^{\mu-\gamma}$. Thus the right side of (1.1) is equal to the right
side of (2.2). ∎

We will prove the following statement, which is equivalent to Theorem 1.1.

######  Theorem 2.2.

Suppose $[F,G]\in K$. For any direction $w=(u,v)\in W$, let
$d=w\text{-}\deg(F_{+})$ and $e=w\text{-}\deg(G_{+})$. Assume $d>0$. Write the
$w$-homogeneous decompositions $F=\sum_{i\leq d}F_{i}$ and $G=\sum_{i\leq
e}G_{i}$. Define

$\widetilde{F}=F_{d}+F_{d-1}t+\cdots\quad\text{and}\quad\widetilde{G}=G_{e}+G_{e-1}t+\cdots.$

Let $r\in\mathbb{Z}_{>0}$ be the largest integer such that $F_{d}^{1/r}\in
K[x,y]$. Then there exists a unique sequence of constants
$c_{0},c_{1},\dots,c_{d+e-u-v-1}\in K$ such that $c_{0}\neq 0$ and

(2.3)
$G_{e-\mu}=\sum_{\gamma=0}^{\mu}c_{\gamma}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{\mu-\gamma}}$

for every integer $\mu\in\\{0,1,...,d+e-u-v-1\\}$. Moreover, $c_{\gamma}=0$ if
$r(e-\gamma)/d\notin\mathbb{Z}$.

Note that the last condition implies that every nonzero summand appearing on
the right side of (2.3) is in $\mathcal{R}[F_{d}^{-1/r}]$, so must be a
rational function.

In order to prove Theorem 2.2, we need the following the lemma (cf. [29,
Propositions 1,2], [4, Lemma 22], [23, p258]) and for the readers’ convenience
we reproduce the proof here.

######  Lemma 2.3.

Let $w=(u,v)\in W$. Let $R$ be any polynomial ring over $K$, $f\in R$ be a
$w$-homogeneous polynomial of degree $d_{f}>0$, and $g$ be a nonzero
$w$-homogeneous function of degree $d_{g}\in\mathbb{Z}$ in the fractional
field of $R$ such that the Jacobian $[g,f]=0$. Define $r\in\mathbb{Z}_{>0}$ to
be the largest integer such that $h=f^{1/r}$ is a polynomial. Then there
exists a unique $c\in K\setminus\\{0\\}$ so that $g=c\cdot h^{s}$, where
$s=rd_{g}/d_{f}$ is an integer.

###### Proof.

If $g\in K$, the statement is trivial where $s=0$, $c=g$. So for the rest of
the proof we assume that $g$ is not a constant. By Euler’s Lemma,
$uxf_{x}+vyf_{y}=d_{f}f$, $uxg_{x}+vyg_{y}=d_{g}g$. Then

$\begin{bmatrix}d_{f}f\\\ d_{g}g\end{bmatrix}=\begin{bmatrix}f_{x}&f_{y}\\\
g_{x}&g_{y}\end{bmatrix}\begin{bmatrix}ux\\\ vy\end{bmatrix},$

so

$\begin{bmatrix}d_{f}fg_{y}-d_{g}gf_{y}\\\
-d_{f}fg_{x}+d_{g}gf_{x}\end{bmatrix}=\begin{bmatrix}g_{y}&-f_{y}\\\
-g_{x}&f_{x}\end{bmatrix}\begin{bmatrix}d_{f}f\\\
d_{g}g\end{bmatrix}=\begin{bmatrix}g_{y}&-f_{y}\\\
-g_{x}&f_{x}\end{bmatrix}\begin{bmatrix}f_{x}&f_{y}\\\
g_{x}&g_{y}\end{bmatrix}\begin{bmatrix}ux\\\
vy\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}\begin{bmatrix}ux\\\
vy\end{bmatrix}=\begin{bmatrix}0\\\ 0\end{bmatrix}$

Thus
$(g^{d_{f}}/f^{d_{g}})_{x}=d_{f}g^{d_{f}-1}g_{x}f^{-d_{g}}-d_{g}f^{-d_{g}-1}f_{x}g^{d_{f}}=g^{d_{f}-1}f^{-d_{g}-1}(d_{f}fg_{x}-d_{g}gf_{x})=0$,
and similarly $(g^{d_{f}}/f^{d_{g}})_{y}=0$. So
$g^{d_{f}}/f^{d_{g}}=c^{\prime}$ for some $c^{\prime}\in K\setminus\\{0\\}$.

Let $a_{1},a_{2}\in K\setminus\\{0\\}$, $p_{1},\dots,p_{n}$ be distinct
irreducible polynomials, $r_{1},\dots,r_{n}\in\mathbb{Z}_{\geq 0}$,
$s_{1},\dots,s_{n}\in\mathbb{Z}$, such that we have the prime factoriaztion
$f=a_{1}p_{1}^{r_{1}}\cdots p_{n}^{r_{n}}$, $g=a_{2}p_{1}^{s_{1}}\cdots
p_{n}^{s_{n}}$. Then $d_{f}s_{i}=d_{g}r_{i}$ for $1\leq i\leq n$, and $r={\rm
gcd}(r_{1},\dots,r_{n})$. Let $s^{\prime}={\rm gcd}(s_{1},\dots,s_{n})>0$. We
have $d_{f}s^{\prime}={\rm gcd}(d_{f}s_{1},\dots,d_{f}s_{n})={\rm
gcd}(d_{g}r_{1},\dots,d_{g}r_{n})=|d_{g}|r$. So $s=\pm s^{\prime}$ is an
integer, $r_{i}:s_{i}=d_{f}:d_{g}=r:s$. So the exponent of $p_{i}$ in the
prime factorization of $c=g/h^{s}$ is $s_{i}-s(r_{i}/r)=0$; thus $c$ is a
constant. The uniqueness of $c$ follows from the previous sentences. ∎

###### Proof of Theorem 2.2.

We proceed by induction on $\mu$. The base case of $\mu=0$ is
$G_{e}=c_{0}F_{d}^{e/d}$, which follows from Lemma 2.3. Note that $c_{0}\neq
0$, because otherwise $G_{e}=c_{0}F_{d}^{e/d}=0$ which contradicts the
assumption that $e=w\text{-}\deg(G_{+})$.

For the inductive step, assume $\mu>0$. By inductive assumption,
$c_{0},\dots,c_{\mu-1}$ are uniquely determined. The assumption $[F,G]\in K$
implies that each positive $w$-degree component of $[F,G]$ is 0. Note that if
the $w$-degrees of homogeneous rational functions $f$ and $g$ are $i$ and $j$
respectively, then the $w$-degree of $[f,g]$ is $i+j-u-v$. On the other hand,
the component in $[F,G]$ of $w\text{-degree}=(d+e-u-v-\mu)$ is just
$\big{[}[\widetilde{F},\widetilde{G}]\big{]}_{t^{\mu}}$. Since
$\mu<{d+e-u-v}$, we have

$0=\big{[}[\widetilde{F},\widetilde{G}]\big{]}_{t^{\mu}}=\big{[}[\sum_{i\geq
0}F_{d-i}t^{i},\sum_{j\geq
0}G_{e-j}t^{j}]\big{]}_{t^{\mu}}=\sum_{\stackrel{{\scriptstyle i,j\geq
0}}{{i+j=\mu}}}[F_{d-i},G_{e-j}],$

therefore

(2.4) $[G_{e-\mu},F_{d}]=-[F_{d},G_{e-\mu}]=\sum_{\stackrel{{\scriptstyle
i>0,j\geq 0}}{{i+j=\mu}}}[F_{d-i},G_{e-j}].$

Define

$H=G_{e-\mu}-\sum_{\gamma=0}^{\mu-1}c_{\gamma}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{\mu-\gamma}}$

Note that (2.3) holds if and only if $H=c_{\mu}F_{d}^{\frac{e-\mu}{d}}\
(=c_{\mu}[\widetilde{F}^{\frac{e-\mu}{d}}]_{t^{0}})$.

If $H=0$, then the equation $H=c_{\mu}F_{d}^{\frac{e-\mu}{d}}$ holds exactly
when $c_{\mu}=0$, so the choice of $c_{\mu}$ is unique.

Now assume $H\neq 0$. It is a homogeneous rational function in
$\mathcal{R}[F_{d}^{-1/r}]$ of $w\text{-deg}=(e-\mu)$ by the inductive
hypothesis. We claim that $[H,F_{d}]=0$. Indeed,

$\displaystyle[H,F_{d}]$
$\displaystyle=[G_{e-\mu},F_{d}]-[\sum_{\gamma=0}^{\mu-1}c_{\gamma}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{\mu-\gamma}},F_{d}]\quad\text{
(by the definition of $H$)}$ $\displaystyle=\sum_{\stackrel{{\scriptstyle
i>0,j\geq
0}}{{i+j=\mu}}}[F_{d-i},G_{e-j}]-\sum_{\gamma=0}^{\mu-1}[c_{\gamma}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{\mu-\gamma}},F_{d}]\quad\text{
(by \eqref{eq:GF=FG})}$
$\displaystyle=\sum_{j=0}^{\mu-1}[F_{d-\mu+j},\sum_{\gamma=0}^{j}c_{\gamma}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{j-\gamma}}]-\sum_{\gamma=0}^{\mu-1}[c_{\gamma}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{\mu-\gamma}},F_{d}]\quad\text{
(by the inductive hypothesis)}$
$\displaystyle=\sum_{\gamma=0}^{\mu-1}c_{\gamma}\Big{(}\sum_{j=\gamma}^{\mu-1}\big{[}F_{d-\mu+j},[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{j-\gamma}}\big{]}-\big{[}[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{\mu-\gamma}},F_{d}\big{]}\Big{)}$
$\displaystyle=\sum_{\gamma=0}^{\mu-1}c_{\gamma}\Big{(}\sum_{j=\gamma}^{\mu}\big{[}F_{d-\mu+j},[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{j-\gamma}}\big{]}\Big{)}=\sum_{\gamma=0}^{\mu-1}c_{\gamma}\Big{(}\sum_{j=\gamma}^{\mu}\big{[}[\widetilde{F}]_{t^{\mu-j}},[\widetilde{F}^{\frac{e-\gamma}{d}}]_{t^{j-\gamma}}\big{]}\Big{)}$
$\displaystyle=\sum_{\gamma=0}^{\mu-1}c_{\gamma}\Big{(}\big{[}[\widetilde{F},\widetilde{F}^{\frac{e-\gamma}{d}}]\big{]}_{t^{\mu-\gamma}}\Big{)}=0\quad\text{
(because $[\widetilde{F},\widetilde{F}^{\frac{e-\gamma}{d}}]=0$)}$

This equality together with Lemma 2.3 (with $g=H,f=F_{d},h=F_{d}^{1/r}$)
implies that there is a unique element $c_{\mu}\in K\setminus\\{0\\}$ such
that

$H=c_{\mu}h^{s}=c_{\mu}F_{d}^{\deg H/\deg
F_{d}}=c_{\mu}F_{d}^{\frac{e-\mu}{d}},$

where $s=r(e-\mu)/d\in\mathbb{Z}$. In other words, $c_{\mu}=0$ if
$r(e-\mu)/d\notin\mathbb{Z}$. ∎

For any real numbers $r_{1}\leq r_{2}$, we use the usual notation for a closed
interval $[r_{1},r_{2}]:=\\{x\in\mathbb{R}\ :\ r_{1}\leq x\leq r_{2}\\}$, and
introduce the notation
$[r_{1},r_{2}]_{\mathbb{Z}}:=[r_{1},r_{2}]\cap\mathbb{Z}$. For a line segment
$\overline{AB}\subset\mathbb{R}^{2}$ whose endpoints are both in
$\mathbb{Z}^{2}\subset\mathbb{R}^{2}$, we define the length ${\rm
len}(\overline{AB})\in\mathbb{Z}_{\geq 0}$ to be one less than the number of
lattice points on $\overline{AB}$. For any direction $w=(u,v)\in W$ and for
any $w$-homogeneous Laurent polynomial $h\in K[x^{\pm 1},y^{\pm 1}]$, we
define ${\rm len}(h)$ to be the length of $N(h)$; that is, if
$h=a_{0}x^{b}y^{c}+a_{1}x^{b+v}y^{c-u}+a_{2}x^{b+2v}y^{c-2u}+\cdots+a_{l}x^{b+lv}y^{c-lu}$
with $a_{0}\neq 0$ and $a_{l}\neq 0$, then ${\rm len}(h)=l$.

In the following statement, for each polynomial $F,P$ and $R$, we fix
$w=(u,v)\in W$ and write the $w$-homogeneous degree decompositions
$F=\sum_{i}F_{i}$, $P=\sum_{i}P_{i}$, and $R=\sum_{i}R_{i}$.

######  Proposition 2.4.

Let $k\in\mathbb{Z}_{>0}$ and assume that $(F,G)$ has generic boundaries with
$a=2$ and $b=2k+1$. Denote $d:=w\emph{-deg}(F_{+})$ and
$e:=w\emph{-deg}(G_{+})$. Let $P\in\mathcal{R}$ such that
$w\emph{-deg}(P_{+})=m=d/2$ and $F_{d}=P_{m}^{2}$. Let $R=F-P^{2}$ and
$h\in[1,2m-1]_{\mathbb{Z}}$. If $R_{d-\ell}=0$ for all $\ell<h$, then
$P_{m}^{-1}R_{d-h}\in K[x^{\pm 1},y^{\pm 1}]$.

###### Proof.

We assume $P_{m}$ is not a monomial (thus $F_{d}$ is not a monomial) since the
statement is trivial otherwise. Let

$\displaystyle\widetilde{F}=F_{d}+F_{d-1}t+F_{d-2}t^{2}+\cdots\in\mathcal{R}[[t]],$
$\displaystyle Q=P_{m}+P_{m-1}t+\cdots+P_{m-h}t^{h}\in\mathcal{R}[[t]],$
$\displaystyle\text{and }T=\widetilde{F}-Q^{2}.$

For each positive integer $z$, let $\mathcal{O}(t^{z})$ denote an element of
the form $\sum_{i\geq z}f_{i}t^{i}$ in $\mathcal{R}[[t]]$, where each $f_{i}$
is a polynomial in $\mathcal{R}$ which we do not have to care about. Since
$R_{d-\ell}=0$ for $\ell<h$, we get

$T=\left(F_{d-h}-\sum_{j=0}^{h}P_{m-j}P_{m-h+j}\right)t^{h}+\mathcal{O}(t^{h+1})=R_{d-h}t^{h}+\mathcal{O}(t^{h+1}).$

We will apply Theorem 2.2 to the case of $\mu=h(k+1)$. For that purpose, we
need to check that $h(k+1)\leq d+e-u-v-1$. Observe:

$\displaystyle d+e-u-v-1-h(k+1)$ $\displaystyle\geq
2m+(2k+1)m-u-v-1-(2m-1)(k+1)$ $\displaystyle=m-u-v+k\geq
m-u-v+1=m+1-(w\text{-deg}(xy))\geq 0,$

where the last inequality holds because $m-(w\text{-deg}(xy))\geq 0$ if the
lattice point $(1,1)$ is contained in $N(P)$, or $w\text{-deg}(xy)=m+1$
otherwise. Indeed, if $(1,1)$ is not contained in $N(P)$, then $N(P)$ is
either the triangle with vertices $(0,0),(c,0),(0,1)$ or the triangle with
vertices $(0,0),(1,0),(0,c)$ for some $c\in\mathbb{Z}_{>0}$. Without loss of
generality we assume the former. Since $P_{m}$ is not a monomial, both the
points $(c,0)$ and $(0,1)$ must lie in the support of $P_{m}$. So $w=(1,c)$,
$m=c$, and thus $w\text{-deg}(xy)=1+c=m+1$.

Since $(F,G)$ has generic boundaries, Theorem 2.2 gives the following:

$\displaystyle G_{e-h(k+1)}$ $\displaystyle=\sum_{r=0}^{\lfloor
h(k+1)/m\rfloor}c_{rm}\bigg{[}\widetilde{F}^{k+1/2-r/2}\bigg{]}_{t^{h(k+1)-rm}}$
$\displaystyle=\sum_{r=0}^{\lfloor
h(k+1)/m\rfloor}c_{rm}\bigg{[}(Q^{2}+T)^{k+1/2-r/2}\bigg{]}_{t^{h(k+1)-rm}}$
$\displaystyle=\sum_{r=0}^{\lfloor
h(k+1)/m\rfloor}c_{rm}\bigg{[}\sum_{i=0}^{\infty}\binom{k+1/2-r/2}{i}Q^{2k+1-r-2i}T^{i}\bigg{]}_{t^{h(k+1)-rm}}.$

Note that $T=\mathcal{O}(t^{h})$, which implies $T^{i}=\mathcal{O}(t^{hi})$.
Then it is enough to look at $r$ and $i$ such that $hi\leq h(k+1)-rm$, or
equivalently $i\leq k+1-rm/h$. Then the exponent of $Q$ satisfies

$2k+1-r-2i\geq 2k+1-r-2(k+1-rm/h)=2rm/h-r-1\geq 2rm/d-r-1=r-r-1=-1.$

Here the first “$\geq$” becomes “$=$” only when $i=k+1-rm/h$, and the second
“$\geq$” becomes “$=$” only when $r=0$. Since the exponent of $Q$ is an
integer, it is always nonnegative except when “$r=0$ and $i=k+1$” (in which
case the exponent is $-1$). Since $G_{e-h(k+1)}$ and
$[Q^{2k+1-r-2i}T^{i}]_{t^{h(k+1)-rm}}$ are polynomials in $K[x,y]$ whenever
$2k+1-r-2i\geq 0$, the following must also be a polynomial in $K[x,y]$:

$\displaystyle c_{0}\binom{k+1/2}{k+1}[Q^{-1}T^{k+1}]_{t^{h(k+1)}}$
$\displaystyle=c_{0}\binom{k+1/2}{k+1}\bigg{[}(P_{m}+\mathcal{O}(t))^{-1}\Big{(}R_{d-h}t^{h}+\mathcal{O}(t^{h+1})\Big{)}^{k+1}\bigg{]}_{t^{h(k+1)}}$
$\displaystyle=c_{0}\binom{k+1/2}{k+1}P_{m}^{-1}(R_{d-h})^{k+1}.$

Since $c_{0}\neq 0$ and $\binom{k+1/2}{k+1}\neq 0$, we get that
$P_{m}^{-1}(R_{d-h})^{k+1}$ is a polynomial. Since $(F,G)$ has generic
boundaries, the polynomial $P_{m}$ is not divisible by the square of any
polynomial in $\mathcal{T}$. This implies $P_{m}^{-1}R_{d-h}\in K[x^{\pm
1},y^{\pm 1}]$, because $K$ is algebraically closed. ∎

######  Corollary 2.5.

The same hypotheses as above. If $\text{len}(R_{d-h})<\text{len}(P_{m})$ then
$R_{d-h}=0$.

###### Proof.

If $R_{d-h}\neq 0$ then $\text{len}(R_{d-h})\geq\text{len}(P_{m})$, because
$P_{m}^{-1}R_{d-h}\in K[x^{\pm 1},y^{\pm 1}]$. ∎

The following lemma is elementary but makes Proposition 2.4 useful. Let
$F\in\mathcal{R}$ be a polynomial with a nonzero constant term such that all
vertices of $N(F)$ are in $(2\mathbb{Z})^{2}$, and let $C$ be any vertex of
$N(F)$ other than the point of origin $O$. Let
$\mathcal{N}^{\prime}=\frac{1}{2}N(F)$ which is defined at the end of Section
1, and
$\mathcal{N}^{\prime\prime}=\mathcal{N}^{\prime}+\frac{1}{2}\overrightarrow{OC}$.
(For example, see the polygons shown in Figure 4.)

######  Lemma 2.6.

Let $F=\sum_{i,j}\lambda_{ij}x^{i}y^{j}$. There exists a polynomial
$P=\sum_{(i,j)\in\mathcal{N}^{\prime}}p_{ij}x^{i}y^{j}$, unique up to a sign,
such that $\emph{supp}(F-P^{2})\cap\mathcal{N}^{\prime\prime}=\emptyset$.

###### Proof.

First consider the case that $N(F)$ is a rectangle
$[0,2m^{\prime}]\times[0,2m]$ for $m^{\prime},m\in\mathbb{Z}_{>0}$. In
particular, $\lambda_{2m^{\prime},2m}\neq 0$. See Figure 2. Let
$C=(2m^{\prime},2m)$. The required property gives a system of
$(m+1)(m^{\prime}+1)$ quadratic equations with $(m+1)(m^{\prime}+1)$ variables
$p_{ij}$. We can solve $p_{ij}$ recursively, in the following “graded lex
order”:
$p_{m^{\prime},m}>p_{m^{\prime},m-1}>p_{m^{\prime}-1,m}>p_{m^{\prime},m-2}>p_{m^{\prime}-1,m-1}>p_{m^{\prime}-2,m}>\cdots$.
Namely, first use $\lambda_{2m^{\prime},2m}-p_{m^{\prime},m}^{2}=0$ to
determine $p_{m^{\prime},m}\neq 0$ up to a sign; next use
$\lambda_{2m^{\prime},2m-1}-2p_{m^{\prime},m}p_{m^{\prime},m-1}=0$ to uniquely
determine $p_{m^{\prime},m-1}$, etc.

Even if $N(F)$ is arbitrary, we can still solve $p_{ij}$ recursively with
respect to an appropriate order in the same way as follows. Let
$L_{c}=\\{(x,y)\ |\alpha x+\beta y=c\\}$ (for some $c>0$) be a line with
irrational slope that passes $C$ and intersect with $N(F)$ only at $C$. Then
$N(F)$ lies in the half plane $\alpha x+\beta y\leq c$. Arrange points
$\\{z_{i}=(x_{i},y_{i})\\}_{1\leq i\leq n}$ in $\mathcal{N}^{\prime}$ such
that $\alpha x_{1}+\beta y_{1}>\alpha x_{2}+\beta y_{2}>\cdots>\alpha
x_{n}+\beta y_{n}$. Then $z_{1}=\frac{1}{2}C$. For Denote ${\bf
x}^{z_{i}}=x^{x_{i}}y^{y_{i}}$. Then in our new notation,
$P=\sum_{z_{i}}p_{z_{i}}{\bf x}^{z_{i}}$. We claim that we can solve
$p_{z_{1}},p_{z_{2}},\dots,p_{z_{n}}$ recursively. First use
$\lambda_{2z_{1}}-p_{z_{1}}^{2}=0$ to determine $p_{z_{1}}$ up to a sign. If
$p_{z_{1}},\dots,p_{z_{k-1}}$ are determined, then using the equaltion

$\lambda_{z_{1}+z_{k}}=\sum_{z_{i}+z_{j}=z_{1}+z_{k}}p_{z_{i}}p_{z_{j}}=2p_{z_{1}}p_{z_{k}}+\sum_{z_{i}+z_{j}=z_{1}+z_{k},1<i,j<k}p_{z_{i}}p_{z_{j}}$

we can uniquely determine $p_{z_{k}}$. Since
$\mathcal{N}^{\prime\prime}=\\{z_{1}+z_{k}\ |\ 1\leq k\leq n\\}$, we have
found a unique $P$ (up to a sign) such that
$\text{supp}(F-P^{2})\cap\mathcal{N}^{\prime\prime}=\emptyset$. ∎

##  3\. Proof of Theorem 1.3 for the case where $N(F)$ is a rectangle

Let $F=\sum_{i,j}\lambda_{ij}x^{i}y^{j}$. In this section we prove Theorem
1.3, assuming that $N(F)$ is a rectangle $[0,2m^{\prime}]\times[0,2m]$ for
$m^{\prime},m\in\mathbb{Z}_{>0}$. In particular, $\lambda_{2m^{\prime},2m}\neq
0$. See Figure 2.

$O$$(0,2m)$$(2m^{\prime},2m)$$(2m^{\prime},0)$$(m^{\prime},2m)$$(2m^{\prime},m)$$(0,m)$$(m^{\prime},m)$$(m^{\prime},0)$$x$$y$
Figure 2. The case where $N(F)$ is a rectangle

Let $P$ be a polynomial given by Lemma 2.6, and let $R=F-P^{2}$. Write the
$w$-homogeneous degree decomposition $R=\sum_{i}R_{i}$ for $w=(0,1)\in W$, and
the $w^{\prime}$-homogeneous degree decomposition $R=\sum_{i}R^{\prime}_{i}$
for $w^{\prime}=(1,0)\in W$.

######  Proposition 3.1.

Suppose that $(F,G)$ has generic boundaries with $a=2$. Let
$d^{\prime}=2m^{\prime}$, $d=2m$ and $e^{\prime}=(2k+1)m^{\prime}$,
$e=(2k+1)m$ for some positive integer $k$. Then we have $R_{d-h}=0$ for
$h\in[0,d-1]_{\mathbb{Z}}$, and $R^{\prime}_{d^{\prime}-h}=0$ for
$h\in[0,d^{\prime}-1]_{\mathbb{Z}}$.

###### Proof.

We will use induction on $h$. The base case of $h=0$ is well known (for
instance, see [2, 4, 29]), and also follows from Theorem 1.1 for $\mu=0$.

Let $h\in[1,m]_{\mathbb{Z}}$, and assume the inductive hypothesis that
$R_{d-\ell}=0$ for $\ell<h$. Lemma 2.6 implies that
$\text{len}(R_{d-h})<m^{\prime}=\text{len}(P_{+})$. Hence $R_{d-h}=0$ by
Corollary 2.5.

Applying the same argument, we also get $R^{\prime}_{d^{\prime}-h}=0$ for
$h\in[0,m^{\prime}]_{\mathbb{Z}}$. Then the support of $R$ is contained in
$[0,m^{\prime}-1]\times[0,m-1]$, which in turn implies that $R_{d-h}=0$ for
$h\in[m+1,d-1]_{\mathbb{Z}}$, and $R^{\prime}_{d^{\prime}-h}=0$ for
$h\in[m^{\prime}+1,d^{\prime}-1]_{\mathbb{Z}}$. ∎

In light of this proposition, we can show the following.

######  Corollary 3.2.

Suppose that $(F,G)$ has generic boundaries with $a=2$. If $N(F)$ is a
rectangle, then $F=P^{2}+u_{0}$ for some constant $u_{0}\in K$. In particular,
$[F,G]=0$.

###### Proof.

Proposition 3.1 implies that the support of $R$ is either the origin or empty,
so $R$ is equal to a constant, say $u_{0}\in K$. That is, $F=P^{2}+u_{0}$.
This gives $[F,G]=[P^{2},G]=2P[P,G]\in K$. So $P\in K$, which gives $[F,G]=0$.
∎

######  Example 3.3.

Assume that $a=2,b=3$, $N(F)$ is the $2\times 2$ square, $N(G)$ is the
$3\times 3$ square, and $F$ takes the form

$F=\lambda_{2,2}x^{2}y^{2}+\lambda_{2,1}x^{2}y+\lambda_{1,2}xy^{2}+\lambda_{1,1}xy+\lambda_{2,0}x^{2}+\lambda_{0,2}y^{2}+\lambda_{1,0}x+\lambda_{0,1}y+\lambda_{0,0},$

where $\lambda_{i,j}\in K$ and $\lambda_{2,2}\lambda_{2,0}\lambda_{0,2}\neq
0$. See Figure 3.

$P$$O$$(0,2)$$(2,2)$$(2,0)$$(1,2)$$(2,1)$$(0,1)$$(1,1)$$(1,0)$$x$$y$ Figure 3.
The Newton polygons associated with $F$ and $P$, with $P$ in bold

By Lemma 2.6, there exists a polynomial
$P=P_{1,1}xy+P_{1,0}x+P_{0,1}y+P_{0,0}$ such that $\text{supp}(F-P^{2})$ is
contained in $\\{(0,2),(0,1),(0,0),(1,0),(2,0)\\}$. We immediately deduce
that:

$\displaystyle\lambda_{2,2}$ $\displaystyle=P_{1,1}^{2},$
$\displaystyle\lambda_{2,1}$ $\displaystyle=2P_{1,1}P_{1,0},$
$\displaystyle\lambda_{1,2}$ $\displaystyle=2P_{1,1}P_{0,1},$
$\displaystyle\lambda_{1,1}$ $\displaystyle=2P_{1,0}P_{0,1}+2P_{1,1}P_{0,0}.$

Suppose that $[F,G]\in K$. Applying (1.1) to $w=(0,1)$ and $\mu=0$, we have

$\lambda_{2,2}x^{2}y^{2}+\lambda_{1,2}xy^{2}+\lambda_{0,2}y^{2}=P_{1,1}^{2}x^{2}y^{2}+2P_{1,1}P_{0,1}xy^{2}+\lambda_{0,2}y^{2}=(P_{1,1}^{\prime}xy+P_{0,1}^{\prime}y)^{2}$

for some constants $P_{1,1}^{\prime},P_{0,1}^{\prime}\in K$, which implies
$\lambda_{0,2}=P_{0,1}^{2}$ and
$(P_{1,1}xy+P_{0,1}y)^{2}=(P_{1,1}^{\prime}xy+P_{0,1}^{\prime}y)^{2}$.

Applying (1.1) to $w=(0,1)$ and $\mu=2$, we get that
$\lambda_{2,1}x^{2}y+\lambda_{1,1}xy+\lambda_{0,1}y$ is divisible by
$P_{1,1}^{\prime}xy+P_{0,1}^{\prime}y$, hence divisible by
$P_{1,1}xy+P_{0,1}y$. This means that

$\displaystyle\lambda_{2,1}x^{2}y+\lambda_{1,1}xy+\lambda_{0,1}y$
$\displaystyle=2P_{1,1}P_{1,0}x^{2}y+(2P_{1,0}P_{0,1}+2P_{1,1}P_{0,0})xy+\lambda_{0,1}y$
$\displaystyle=2(P_{1,1}xy+P_{0,1}y)(P_{1,0}^{\prime}x+P_{0,0}^{\prime})$

for some constants $P_{1,0}^{\prime},P_{0,0}^{\prime}\in K$, which implies
$\lambda_{0,1}=2P_{0,1}P_{0,0}$.

Similarly, applying Magnus’ formula to $w=(1,0)$, we also have
$\lambda_{2,0}=P_{1,0}^{2}$ and $\lambda_{1,0}=2P_{1,0}P_{0,0}$. From here, it
follows that $F-P^{2}$ is a constant.

##  4\. Construction of broken lines and proof of Theorem 1.3

In this section, we prove Theorem 1.3. Besides from using Proposition 2.4,
Corollary 2.5 and Lemma 2.6, the rest is a purely combinatorial analysis on
subsets of $\mathbb{R}^{2}$ and $\mathbb{Z}^{2}$. Suppose that $(F,G)$ has
generic boundaries with $a=2$.

Denote the point of origin by $O=A_{0}=B_{0}$. Now we extend the result from
Section 3 to the general case that $N(F)$ is of arbitrary shape. We say that a
vertex $C=(c_{x},c_{y})\in N(F)$ is _northeastern_ if
$(v_{x}-c_{x},v_{y}-c_{y})\not\in\mathbb{Z}_{\geq 0}^{2}$ for any other vertex
$V=(v_{x},v_{y})\in N(F)$. Observe that a northeastern vertex exists.

Let $N(F)=OA_{1}\cdots A_{\alpha-1}CB_{\beta-1}\cdots B_{1}$ where $C$ is a
northeastern vertex of $N(F)$. We denote $A_{\alpha}=B_{\beta}=C$,
$C^{\prime}=\frac{1}{2}C$. Since $(1,0),(0,1)\in N(F)$, we must have that
$A_{1}$ lies on the $y$-axis and $B_{1}$ lies on the $x$-axis. Without loss of
generality, assume $\alpha>1$ (but we allow $\beta=1$).

Let $\mathcal{N}^{\prime}=\frac{1}{2}N(F)$ and
$\mathcal{N}^{\prime\prime}=\mathcal{N}^{\prime}+\frac{1}{2}\overrightarrow{OC}$
be the polygons shown in Figure 4.

###  4.1. Construction of parallelograms associated with $N(F)$

For $0\leq i\leq j\leq\alpha$, define
$A_{ij}=\frac{1}{2}(A_{i}+A_{j})\in\mathbb{R}^{2}$. In particular,
$A_{i,i}=A_{i}$.

We define the parallelogram $\mathcal{P}_{ij}$ ($1\leq i<j\leq\alpha$) by its
four vertices $A_{i-1,j-1}$, $A_{i-1,j}$, $A_{i,j-1}$, $A_{i,j}$. For
convenience, we call the line segments $\overline{A_{i-1,j-1}A_{i,j-1}}$,
$\overline{A_{i,j-1}A_{ij}}$, $\overline{A_{ij}A_{i-1,j}}$,
$\overline{A_{i-1,j}A_{i-1,j-1}}$ the west, north, east, south edges of
$\mathcal{P}_{ij}$, respectively. Similarly we also define the parallelogram
$\mathcal{P}^{\prime}_{ij}$ ($1\leq i<j\leq\beta$). See Figure 4.

$O$$A_{1}$$A_{2}$$A_{3}$$C=A_{4}=B_{3}$$C^{\prime}=A_{04}$$B_{2}$$B_{1}$$\mathcal{N}^{\prime\prime}$$\mathcal{N}^{\prime}$$\mathcal{P}_{12}$$\mathcal{P}_{13}$$\mathcal{P}_{14}$$\mathcal{P}_{23}$$\mathcal{P}_{24}$$\mathcal{P}_{34}$$\mathcal{P}^{\prime}_{12}$$\mathcal{P}^{\prime}_{13}$$\mathcal{P}^{\prime}_{23}$$x$$y$$A_{01}$$A_{02}$$A_{03}$$A_{14}$$A_{24}$$A_{34}$$O$$A_{1}$$A_{2}$$A_{3}$$C$$B_{2}$$B_{1}$$x$$y$
Figure 4. The case where $N(F)$ is arbitrary, $\alpha=4,\beta=3$. Left:
regions $\mathcal{N}^{\prime}$, $\mathcal{N}^{\prime\prime}$,
$\mathcal{P}_{ij}$; Right: various broken lines $T_{D}$ where the red dots are
the various positions of the point $D$

One can verify that
$\bullet$ $\mathcal{P}_{ij}$ is indeed a parallelogram,
$\bullet$ the lengths of the edges of $\mathcal{P}_{ij}$ are equal to
$\frac{1}{2}A_{i-1}A_{i}$ and $\frac{1}{2}A_{j-1}A_{j}$,
$\bullet$ $\mathcal{P}_{i,j}$ shares edges with $\mathcal{P}_{i,j\pm 1}$ and
$\mathcal{P}_{i\pm 1,j}$ (if the latter are defined).

We make the following claim.

######  Lemma 4.1.

(a) The union of the parallelograms $\mathcal{P}_{ij}$ is the following
(closed and non-convex) polygon:

$\mathcal{P}:=A_{01}A_{02}\cdots A_{0\alpha}A_{1\alpha}A_{2\alpha}\cdots
A_{\alpha-1,\alpha}A_{\alpha-1}A_{\alpha-2}\cdots A_{1}$

(b) These parallelograms do not overlap with each other. More precisely,
$\mathcal{P}_{ij}\cap\mathcal{P}_{i,j+1}=\overline{A_{i-1,j}A_{ij}}$,
$\mathcal{P}_{ij}\cap\mathcal{P}_{i+1,j}=\overline{A_{i,j-1}A_{ij}}$,
$\mathcal{P}_{ij}\cap\mathcal{P}_{i+1,j+1}=A_{ij}$,
$\mathcal{P}_{ij}\cap\mathcal{P}_{i+1,j-1}=A_{i,j-1}$, and
$\mathcal{P}_{ij}\cap\mathcal{P}_{i^{\prime},j^{\prime}}=\emptyset$ if
$|i-i^{\prime}|>1$ or $|j-j^{\prime}|>1$.

###### Proof.

(a) Given any point $r\in\mathcal{P}$, we assert that $r$ is in some
$\mathcal{P}_{ij}$.

Let ${\bf v}_{i}$ be the vector $\frac{1}{2}\overrightarrow{A_{i-1}A_{i}}$ for
$1\leq i\leq\alpha$. Then $A_{ij}=O+\sum_{k=1}^{i}{\bf
v}_{k}+\sum_{k=1}^{j}{\bf v}_{k}$, and the parallelogram
$\mathcal{P}_{ij}=\\{A_{i-1,j-1}+s{\bf v}_{i}+t{\bf v}_{j}|0\leq s,t\leq
1\\}$. The assumption that $C$ is a northeastern vertex of $N(F)$ has the
following consequences:
$\mathcal{N}^{\prime}\cap\mathcal{N}^{\prime\prime}=\\{C^{\prime}\\}$, and
${\bf v}_{1},\dots,{\bf v}_{\alpha}$ are in clockwise order and (strictly) in
the same half plane $y+\lambda x>0$ for a sufficiently large constant
$\lambda\gg 0$.

One can see that $\mathcal{P}=L_{1}+L_{2}$, the Minkowski sum of the following
two broken lines

$L_{1}:=OA_{01}A_{02}\cdots A_{0,\alpha-1}\quad\text{ and }\quad
L_{2}:=A_{01}A_{02}\cdots A_{0,\alpha-1}A_{0,\alpha}.$

It can be visualized as follows: as a point $p$ moves along $L_{1}$, the
broken line $p+L_{2}$ sweeps out the region $\mathcal{P}$. As a consequence,
any point $r\in\mathcal{P}$ is the sum of a point $p\in L_{1}$ and a point
$q\in L_{2}$. Without loss of generality we may assume that $p$ lies in
$\overline{A_{0,i-1}A_{0,i}}$ and $q$ lies in $\overline{A_{0,j-1}A_{0,j}}$
for some $i\leq j$. Then $p=O+\sum_{k=1}^{i-1}{\bf v}_{k}+s{\bf v}_{i}$ and
$q=O+\sum_{k=1}^{j-1}{\bf v}_{k}+t{\bf v}_{j}$. We consider three cases:

$O$$A_{1}$$A_{2}$$A_{3}$$C$$B_{2}$$B_{1}$$x$$y$ Figure 5. The red broken lines
are $p+L_{2}$ whose left endpoint $p$ is blue and right endpoint is red.

If $i<j$, then $r=p+q=A_{i-1,j-1}+s{\bf v}_{i}+t{\bf v}_{j}$ lies in
$\mathcal{P}_{i,j}$.

If $i=j$ and $s+t\leq 1$, then $r=p+q=A_{i-1,i-1}+(s+t){\bf
v}_{i}=A_{i-2,i-1}+1{\bf v}_{i-1}+(s+t){\bf v}_{i}$ lies in
$\mathcal{P}_{i-1,i}$.

If $i=j$ and $s+t>1$, then $r=p+q=A_{i-1,i-1}+(s+t){\bf
v}_{i}=A_{i-1,i}+(s+t-1){\bf v}_{i}+0{\bf v}_{i+1}$ lies in
$\mathcal{P}_{i,i+1}$.

So we have proved that in all cases, the point $r$ lies in some parallelogram
$\mathcal{P}_{ij}$. This proves (a).

(b) Assume that there exists a point
$r\in\mathcal{P}_{ij}\cap\mathcal{P}_{i^{\prime}j^{\prime}}$ where $i<j$ and
$i^{\prime}<j^{\prime}$ and $i\leq i^{\prime}$. Then

$r=O+\sum_{k=1}^{i-1}{\bf v}_{k}+\sum_{k=1}^{j-1}{\bf v}_{k}+s{\bf
v}_{i}+t{\bf v}_{j}=O+\sum_{k=1}^{i^{\prime}-1}{\bf
v}_{k}+\sum_{k=1}^{j^{\prime}-1}{\bf v}_{k}+s^{\prime}{\bf
v}_{i^{\prime}}+t^{\prime}{\bf v}_{j^{\prime}}$

for some $s,t,s^{\prime},t^{\prime}\in[0,1]$.

If $i^{\prime}=i$ and $j^{\prime}>j$, then $s{\bf v}_{i}+t{\bf
v}_{j}=\sum_{k=j}^{j^{\prime}-1}{\bf v}_{k}+s^{\prime}{\bf
v}_{i}+t^{\prime}{\bf v}_{j^{\prime}}$, hence $(s-s^{\prime}){\bf
v}_{i}=(1-t){\bf v}_{j}+\sum_{k=j+1}^{j^{\prime}-1}{\bf v}_{k}+t^{\prime}{\bf
v}_{j^{\prime}}$. Note that ${\bf v}_{i},{\bf v}_{j},{\bf v}_{j+1},\dots,{\bf
v}_{j^{\prime}}$ lie in a half plane and are in the clockwise order. So we
must have $j^{\prime}=j+1$ and $s-s^{\prime}=1-t=t^{\prime}=0$. In this case,
$r=A_{i-1,j}+s{\bf v}_{i}\in\overline{A_{i-1,j}A_{ij}}$.

If $j^{\prime}=j$ and $i^{\prime}>i$, then by a similar argument we have
$i^{\prime}=i+1$ and $r\in\overline{A_{i,j-1}A_{ij}}$.

If $i^{\prime}>i$ and $j^{\prime}>j$, then $s{\bf v}_{i}+t{\bf
v}_{j}=\sum_{k=i}^{i^{\prime}-1}{\bf v}_{k}+\sum_{k=j}^{j^{\prime}-1}{\bf
v}_{k}+s^{\prime}{\bf v}_{i+1}+t^{\prime}{\bf v}_{j+1}$, $(1-s){\bf
v}_{i}+(1-t){\bf v}_{j}+\sum_{k=i+1}^{i^{\prime}-1}{\bf
v}_{k}+\sum_{k=j+1}^{j^{\prime}-1}{\bf v}_{k}+s^{\prime}{\bf
v}_{i+1}+t^{\prime}{\bf v}_{j+1}=0$. Note that all ${\bf v}_{k}$ lie in a half
plane, we must have $i^{\prime}=i+1,j^{\prime}=j+1$,
$1-s=1-t=s^{\prime}=t^{\prime}=0$. In this case, $r=A_{ij}$.

If $i^{\prime}>i$ and $j^{\prime}<j$, then $\sum_{k=j^{\prime}}^{j-1}{\bf
v}_{k}+s{\bf v}_{i}+t{\bf v}_{j}=\sum_{k=i}^{i^{\prime}-1}{\bf
v}_{k}+s^{\prime}{\bf v}_{i^{\prime}}+t^{\prime}{\bf v}_{j^{\prime}}$,
$(1-t^{\prime}){\bf v}_{j^{\prime}}+\sum_{k=j^{\prime}+1}^{j-1}{\bf
v}_{k}+t{\bf v}_{j}=(1-s){\bf v}_{i}+\sum_{k=i+1}^{i^{\prime}-1}{\bf
v}_{k}+s^{\prime}{\bf v}_{i^{\prime}}$. Note that ${\bf v}_{i},\dots,{\bf
v}_{i^{\prime}},{\bf v}_{j^{\prime}},\dots,{\bf v}_{j}$ lie in a half plane
and are in the clockwise order, we must have $i^{\prime}=i+1$,
$j^{\prime}=j-1$, $1-s^{\prime}=s=1-t=t^{\prime}=0$. In this case,
$r=A_{i,j-1}$.

This proves all cases for (b). ∎

###  4.2. Construction of broken lines

Next we construct broken lines $T_{D}$’s and $T^{\prime}_{D}$’s.

(a) For every point $D$ on the line segment $\overline{C^{\prime}C}$, we draw
a broken line $T_{D}$ that goes to the left until it reaches the boundary of
$N(F)$, as follows:

Step 1. First goes in the direction $\overrightarrow{A_{\alpha}A_{\alpha-1}}$
until it reaches a point $D_{\alpha-1}$ on the west boundary of
$\mathcal{P}_{r,\alpha}$ for some $r$;

Step 2. Then goes in the direction $\overrightarrow{A_{\alpha-1}A_{\alpha-2}}$
until it reaches a point $D_{\alpha-2}$ on the west boundary of
$\mathcal{P}_{r,\alpha-1}$;

Step 3. then goes in the direction $\overrightarrow{A_{\alpha-2}A_{\alpha-3}}$
until it reaches a point $D_{\alpha-3}$ on the west boundary of
$\mathcal{P}_{r,\alpha-2}$;

$\cdots$

Step $(\alpha-r)$. Finally, goes in the direction
$\overrightarrow{A_{r+1}A_{r}}$ until it reaches a point $D_{r}$ on the west
boundary of $\mathcal{P}_{r,r+1}$.

If $\overline{DD_{\alpha-1}}$ does not contain the north edge of
$P_{r,\alpha}$ for any $r\geq 1$, then define
$T_{D}=DD_{\alpha-1}D_{\alpha-2}\cdots D_{r}$. Note that it now reaches the
boundary of $N(F)$.

If $\overline{DD_{\alpha-1}}$ contains the north edge of $P_{r,\alpha}$ for
some $r\geq 1$, we define $T_{D}$ to be the broken line consisting of
$\overline{DD_{\alpha-1}}$ and the north edges of $P_{r,\alpha-1}$,
$P_{r,\alpha-2}$,…, $P_{r,r+1}$, and ending at $A_{r}$.

(b) Still for every point $D$ on the line segment $\overline{C^{\prime}C}$,
similarly we define a broken line $T^{\prime}_{D}$ that goes down whose linear
pieces are parallel to $\overrightarrow{B_{i}B_{i-1}}$ for $i\leq\beta$. Note
that for $D=C^{\prime}$,
$T_{C^{\prime}}=\frac{1}{2}A_{\alpha}A_{\alpha-1}\cdots
A_{1}=A_{0\alpha}A_{0,\alpha-1}\cdots A_{01}$, and
$T^{\prime}_{C^{\prime}}=\frac{1}{2}B_{\beta}B_{\beta-1}\cdots
B_{1}=B_{0\beta}B_{0,\beta-1}\cdots B_{01}$.

(c) For every $D\in\overline{OC^{\prime}}\setminus\\{O\\}$, define
$T_{D}=D_{\alpha}D_{\alpha-1}\cdots
D_{1}=\frac{||OD||}{||OC||}A_{\alpha}A_{\alpha-1}\cdots A_{1}$ (rescaling the
boundary broken line from $A_{\alpha}$ to $A_{1}$ so that it passes the point
$D=D_{\alpha}$).

(d) For every $D\in\overline{OC^{\prime}}\setminus\\{O\\}$, define
$T^{\prime}_{D}=D_{\beta}D_{\beta-1}\cdots
D_{1}=\frac{||OD||}{||OC||}B_{\beta}B_{\beta-1}\cdots B_{1}$ (rescaling the
boundary broken line from $B_{\beta}$ to $B_{1}$ so that it passes the point
$D=D_{\beta}$).

Note that every point on $N(F)\setminus\overline{OC}$ lies in a unique $T_{D}$
or $T^{\prime}_{D}$.

###  4.3. Proof of Theorem 1.3

Let $P$ be a polynomial given by Lemma 2.6, and let $R=F-P^{2}$. We have the
following generalization of Proposition 3.1.

######  Proposition 4.2.

Suppose that $(F,G)$ has generic boundaries with $a=2$. In the above setting,
$R$ must be a constant; that is, $F=P^{2}+u_{0}$ for some constant $u_{0}\in
K$.

###### Proof.

We consider the finite sequence of broken lines $T_{1},T_{2},\dots,T_{s}$ such
that
$\bullet$ each $T_{i}$ is either $T_{D}$ or $T^{\prime}_{D}$ for some
$D\in\overline{OC}\setminus\\{O\\}$;
$\bullet$ $\text{supp}(F)\setminus\\{O\\}\subset\cup_{i=1}^{s}T_{i}$;
$\bullet$ each $T_{i}$ contains a lattice point;
$\bullet$ and the distance from $O$ to $T_{i}\cap\overline{OC}$ is no less
than the distance from $O$ to $T_{i+1}\cap\overline{OC}$.

To make the order unique, we assume that if
$T_{i}\cap\overline{OC}=T_{i+1}\cap\overline{OC}=D$, then $T_{i}=T_{D}$ and
$T_{i+1}=T^{\prime}_{D}$. In particular,
$T_{1}=\overline{A_{\alpha}A_{\alpha-1}}$ (recall that $\alpha>1$) and
$T_{2}=\overline{B_{\beta}B_{\beta-1}}$ if $\beta>1$.

We will inductively prove that $\text{supp}(R)\cap T_{i}=\emptyset$.

(a) Let $w=(u,v)$ be the normal direction of the edge
$\overline{A_{\alpha}A_{\alpha-1}}$. Let $d=w\text{-deg}(F_{+})$ and
$m=w\text{-deg}(P_{+})$, so $d=2m$. The assumption $[F,G]\in K$ implies that
$F_{d}=P_{m}^{2}$. This is well known (for instance, see [2, 4, 29]), and also
follows from Theorem 1.1 for $\mu=0$. Thus
$\text{supp}(R)\cap\overline{A_{\alpha}A_{\alpha-1}}=\text{supp}(R)\cap
T_{1}=\emptyset$. Similarly,
$\text{supp}(R)\cap\overline{B_{\beta}B_{\beta-1}}=\text{supp}(R)\cap
T_{2}=\emptyset$ if $\beta>1$.

(b) Assume the inductive hypothesis that
$\text{supp}(R)\cap(T_{1}\cup\cdots\cup T_{i-1})=\emptyset$. We consider the
broken line $T_{i}$. There are two cases.

(b1) Suppose $T_{i}=T_{D}$ or $T^{\prime}_{D}$ for $D=(d_{1},d_{2})$ in the
segment $C^{\prime}C$. Without loss of generality assume
$T_{i}=T_{D}=DD_{\alpha-1}D_{\alpha-2}\cdots D_{r}$.

$E_{1}$$E_{2}$$D$$D_{r}$$D_{r+1}$$D_{\alpha-1}$$D_{\alpha}$$O$$x$$y$$E$$D=D_{\alpha}$$D_{\alpha-1}$$D_{2}$$D_{1}$$O$$x$$y$
Figure 6. Case (b1) on the left; Case (b2) on the right.

By Proposition 2.4, we see that $P_{m}^{-1}R_{ud_{1}+vd_{2}}\in K[x^{\pm
1},y^{\pm 1}]$. Denote by $D_{\alpha}$ the intersection of the line segment
$\overline{DD_{\alpha-1}}$ with the boundary of $\mathcal{N}^{\prime\prime}$.
Let $E_{1}$ be the intersection point of the (unbounded) half line
$\overrightarrow{D_{\alpha}D}$ and the boundary of
$\mathcal{N^{\prime\prime}}$. Let $E_{2}$ be the intersection of
$\overrightarrow{D_{\alpha}D}$ with the boundary of $N(F)$. See the picture on
the left in Figure 6.

Since $\text{supp}(R)\cap\mathcal{N}^{\prime\prime}=\emptyset$, we have
$\text{supp}(R)\cap\overline{D_{\alpha}E_{1}}=\emptyset$. Since the lattice
points on $\overline{E_{1}E_{2}}$ must be in some $T_{\ell}$ for some
$\ell<i$, we also have $\text{supp}(R)\cap\overline{E_{1}E_{2}}=\emptyset$.
Hence
$\text{supp}(R_{ud_{1}+vd_{2}})\in\overline{D_{\alpha}D_{\alpha-1}}\setminus\\{D_{\alpha}\\}$,
so $\text{len}(R_{ud_{1}+vd_{2}})<\text{len}(P_{m})$. Then
$\text{supp}(R)\cap\overline{D_{\alpha}D_{\alpha-1}}=\emptyset$ due to
Corollary 2.5. Applying a similar argument, we see that
$\text{supp}(R)\cap\overline{D_{\alpha-1}D_{\alpha-2}}=\cdots=\text{supp}(R)\cap\overline{D_{r+1}D_{r}}=\emptyset$.
Therefore $\text{supp}(R)\cap T_{i}=\emptyset$.

(b2) Suppose $T_{i}=T_{D}$ or $T^{\prime}_{D}$ for
$D=(d_{1},d_{2})\in\overline{OC^{\prime}}\setminus\\{O,C^{\prime}\\}$ .
Without loss of generality assume $T_{i}=T_{D}=D_{\alpha}D_{\alpha-1}\cdots
D_{1}$, where $D_{\alpha}=D$. Let $E$ be the intersection of
$\overrightarrow{D_{\alpha-1}D}$ with the boundary of $N(F)$. See the picture
on the right in Figure 6. Since the lattice points on $\overline{DE}$ must be
in some $T_{\ell}$ for some $\ell<i$, we also have
$\text{supp}(R)\cap\overline{DE}=\emptyset$. Hence
$\text{supp}(R_{ud_{1}+vd_{2}})\in\overline{D_{\alpha}D_{\alpha-1}}$, so
$\text{len}(R_{ud_{1}+vd_{2}})<\text{len}(P_{m})$. From here, we get the same
conclusion as in (b1). ∎

###### Proof of Theorem 1.3.

The same as the proof of Corollary 3.2. ∎

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