An Automorphic Classification of Real Cubic Curves

TL;DR

This paper classifies real cubic curves by analyzing the action of ring automorphisms on degree-three polynomials, identifying orbit representatives to understand their equivalence classes.

Contribution

It provides an automorphic classification of real cubic curves, detailing the orbit structure under polynomial ring automorphisms.

Findings

Computed the orbits of degree-three polynomials under automorphisms

Identified a representative for each orbit of real cubic curves

Enhanced understanding of the symmetry and classification of real cubic curves

Abstract

The action of ring automorphisms of the polynomial ring in two variables over the real numbers on real plane curves is considered. The orbits containing degree-three polynomials are computed, with one representative per orbit being selected.

Tables

Table 1. Table 1. For polynomials x 3 + x ​ y 2 + E ​ x 2 + F ​ x ​ y + G ​ y 2 + λ ​ ( f ) superscript 𝑥 3 𝑥 superscript 𝑦 2 𝐸 superscript 𝑥 2 𝐹 𝑥 𝑦 𝐺 superscript 𝑦 2 𝜆 𝑓 x^{3}+xy^{2}+Ex^{2}+Fxy+Gy^{2}+\lambda(f)

$E\ne 3G$	$⟨\left(E-3G\right)x-G,\left(E-3G\right)y-\frac{F}{2}⟩$
$E=3G$	$⟨x-G,y-\frac{F}{2}⟩$

Table 2. Table 2. For polynomials c ​ ( x 3 + x ​ y 2 + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 3 𝑥 superscript 𝑦 2 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{3}+xy^{2}+Hx+Iy+J)

$I\ne 0$	$⟨\sigma \left(J\right){\left|I\right|}^{1/2}x,\sigma \left(J\right)\sigma \left(I\right){\left|I\right|}^{1/2}y⟩$
$I=0$, $H\ne 0$	$⟨\sigma \left(J\right){\left|H\right|}^{1/2}x,\sigma \left(J\right){\left|H\right|}^{1/2}y⟩$
$I=H=0$, $J\ne 0$	$⟨J^{1/3}x,J^{1/3}y⟩$
$I=H=J=0$	$⟨x,y⟩$

Table 3. Table 3. For polynomials x 3 − x ​ y 2 + E ​ x 2 + F ​ x ​ y + G ​ y 2 + λ ​ ( f ) superscript 𝑥 3 𝑥 superscript 𝑦 2 𝐸 superscript 𝑥 2 𝐹 𝑥 𝑦 𝐺 superscript 𝑦 2 𝜆 𝑓 x^{3}-xy^{2}+Ex^{2}+Fxy+Gy^{2}+\lambda(f)

$E\ne -3G$	$⟨\left(-\frac{E}{3}-G\right)x-\frac{E}{3},\left(-\frac{E}{3}-G\right)y+\frac{F}{2}⟩$
$E=-3G$	$⟨x-\frac{E}{3},y+\frac{F}{2}⟩$

Table 4. Table 4. For polynomials c ​ ( x 3 − x ​ y 2 + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 3 𝑥 superscript 𝑦 2 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{3}-xy^{2}+Hx+Iy+J)

$I\ne 0$, $\left|H\right|\le \left|I\right|$	$⟨\sigma \left(J\right){\left|I\right|}^{1/2}x,\sigma \left(J\right)\sigma \left(-I\right){\left|I\right|}^{1/2}y⟩$
$I\ne 0$, $\left|H\right|>\left|I\right|$	$⟨\alpha |I{|}^{1/2}\sigma \left(J\right)(x+\sigma \left(H\right)y),$
	$-\alpha |I{|}^{1/2}\sigma \left(J\right)\sigma \left(IH\right)(3x-\sigma \left(H\right)y)⟩$
$I=0$, $H\ne 0$	$⟨\beta \sigma \left(J\right)\left(x+\sigma \left(H\right)y\right),\beta \sigma \left(J\right)\left(3x-\sigma \left(H\right)y\right)⟩$
$I=H=0$, $J\ne 0$	$⟨J^{1/3}x,J^{1/3}y⟩$
$I=H=J=0$	$⟨x,y⟩$

Table 5. Table 5. For polynomials c ​ ( x 3 − x ​ y 2 − y 2 + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 3 𝑥 superscript 𝑦 2 superscript 𝑦 2 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{3}-xy^{2}-y^{2}+Hx+Iy+J)

	$⟨-\frac{1}{2}\left(x+y\right)-\frac{3}{4},\sigma \left(I\right)\left(-\frac{1}{2}\left(3x-y\right)-\frac{3}{4}\right)⟩$
$H+\frac{3}{4}\ge \left|I\right|$	$⟨-\frac{1}{2}\left(x-y\right)-\frac{3}{4},\sigma \left(I\right)\left(-\frac{1}{2}\left(3x+y\right)-\frac{3}{4}\right)⟩$
$H+\frac{3}{4}\le -\left|I\right|$	$⟨x,\sigma \left(I\right)y⟩$

Table 6. Table 6. For polynomials x 2 ​ y + E ​ x 2 + F ​ x ​ y + G ​ y 2 + λ ​ ( f ) superscript 𝑥 2 𝑦 𝐸 superscript 𝑥 2 𝐹 𝑥 𝑦 𝐺 superscript 𝑦 2 𝜆 𝑓 x^{2}y+Ex^{2}+Fxy+Gy^{2}+\lambda(f)

$G\ne 0$	$⟨x-\frac{F}{2},\frac{1}{G}y-E⟩$
$G=0$	$⟨x-\frac{F}{2},y-E⟩$

Table 7. Table 7. For polynomials c ​ ( x 2 ​ y + y 2 + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 2 𝑦 superscript 𝑦 2 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{2}y+y^{2}+Hx+Iy+J)

$H\ne 0$	$⟨-H^{1/3}x,H^{2/3}y⟩$
$H=0$, $I\ne 0$	$⟨{\left|I\right|}^{1/2}x,\left|I\right|y⟩$
$H=I=0$, $J\ne 0$	$⟨{\left|J\right|}^{1/4}x,{\left|J\right|}^{1/2}y⟩$
$H=I=J=0$	$⟨x,y⟩$

Table 8. Table 8. For polynomials c ​ ( x 2 ​ y + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 2 𝑦 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{2}y+Hx+Iy+J)

$HI\ne 0$	$⟨\sigma \left(-HJ\right){\left|I\right|}^{1/2}x,\sigma \left(-HJ\right)\frac{-H}{{\left|I\right|}^{1/2}}y⟩$
$HJ\ne 0$, $I=0$	$⟨-\frac{J}{H}x,\frac{H^2}{J}y⟩$
$H\ne 0$, $I=J=0$	$⟨x,-Hy⟩$
$H=0$, $IJ\ne 0$	$⟨{\left|I\right|}^{1/2}x,\frac{J}{\left|I\right|}y⟩$
$H=J=0$, $I\ne 0$	$⟨{\left|I\right|}^{1/2}x,y⟩$
$H=I=0$, $J\ne 0$	$⟨x,-Jy⟩$
$H=I=J=0$	$⟨x,y⟩$

Table 9. Table 9. For polynomials x 3 + E ​ x 2 + F ​ x ​ y + G ​ y 2 + λ ​ ( f ) superscript 𝑥 3 𝐸 superscript 𝑥 2 𝐹 𝑥 𝑦 𝐺 superscript 𝑦 2 𝜆 𝑓 x^{3}+Ex^{2}+Fxy+Gy^{2}+\lambda(f)

$G\ne 0$	$⟨\sigma \left(-G\right)x+\frac{F^2-4EG}{12G},\frac{F}{2\left|G\right|}x+\frac{1}{\sqrt{\left|G\right|}}y⟩$
$G=0$, $F\ne 0$	$⟨x-\frac{E}{3},-\frac{1}{F}y⟩$
$G=F=0$	$⟨x-\frac{E}{3},y⟩$

Table 10. Table 10. For polynomials c ​ ( x 3 − y 2 + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 3 superscript 𝑦 2 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{3}-y^{2}+Hx+Iy+J)

$\left|H\right|\ne 0$	$⟨{\left|H\right|}^{1/2}x,{\left|H\right|}^{3/4}y+\frac{I}{2}⟩$
$H=0$, $\frac{I^2}{4}+J\ne 0$	$⟨{\left|\frac{I^2}{4}+J\right|}^{1/3}x,{\left|\frac{I^2}{4}+J\right|}^{1/2}y+\frac{I}{2}⟩$
$H=\frac{I^2}{4}+J=0$	$⟨x,y+\frac{I}{2}⟩$

Table 11. Table 11. For polynomials c ​ ( x 3 − x ​ y + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 3 𝑥 𝑦 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{3}-xy+Hx+Iy+J)

$\gamma \ne 0$	$⟨{\gamma }^{1/3}x+I,\mathrm{\hspace{0.17em}3}{\gamma }^{1/3}Ix+{\gamma }^{2/3}y+3I^2+H⟩$
$\gamma =0$	$⟨x+I,\mathrm{\hspace{0.17em}3}Ix+y+3I^2+H⟩$

Table 12. Table 12. For polynomials c ​ ( x 3 + H ​ x + I ​ y + J ) 𝑐 superscript 𝑥 3 𝐻 𝑥 𝐼 𝑦 𝐽 c(x^{3}+Hx+Iy+J)

$I\ne 0$	$⟨x,-\frac{H}{I}x-\frac{1}{I}y-\frac{J}{I}⟩$
$I=0$, $H\ne 0$	$⟨\sigma \left(J\right){\left|H\right|}^{1/2}x,y⟩$
$I=H=0$, $J\ne 0$	$⟨J^{1/3}x,y⟩$
$I=H=J=0$	$⟨x,y⟩$

Table 13. Table 13. Affine/Automorphic Classification of Cubic Curves

$x^3+x{y}^2+x^2+Hx+Iy+J;H,J\in ℝ,I\ge 0$

$x^3+x{y}^2+y+Hx+J;H\in ℝ,J\ge 0$

$x^3+x{y}^2\pm x+J;J\ge 0$

$x^3+x{y}^2+1$

$x^3+x{y}^2$

$x^3-x{y}^2-y^2+Hx+Iy+J;J\in ℝ,I\ge 0,H+I\le -\frac{3}{4}$

$x^3-x{y}^2-y+Hx+J;J\ge 0,H\in [-1,1]$

$x^3-x{y}^2+1$

$x^3-x{y}^2$

$x^{2}y+y^2-x+Iy+J;I,J\in ℝ$

$x^{2}y+y^2\pm y+J;J\in ℝ$

$x^{2}y+y^2\pm 1$

$x^{2}y+y^2$

$x^{2}y-x\pm y+J;J\ge 0$

$x^{2}y\pm y+1$

$x^{2}y\pm y$

$x^{2}y-x+1$

$x^{2}y-x$

$x^{2}y-1$

$x^{2}y$

$x^3-y^2\pm x+J;J\in ℝ$

$x^3-y^2\pm 1$

$x^3-y^2$

$x^3-y$

$x^3-xy+1$

$x^3-xy$

$x^3\pm x+J;J\ge 0$

$x^3+1$

$x^3$

Table 14. Table 14. For polynomials in Proposition 2.14

$f$	$\text{cusp}\left(f\right)$	$\text{isol}\left(f\right)$	$\text{node}\left(f\right)$	$\text{red}\left(f\right)$
$x^3-y^2+x+J$	$\varnothing$	$\varnothing$	$\varnothing$	$\varnothing$
$x^3-y^2-x+J$	$\varnothing$	$J+\frac{2\sqrt{3}}{9}$	$J-\frac{2\sqrt{3}}{9}$	$\varnothing$
$x^3-y^2\pm 1$	$\pm 1$	$\varnothing$	$\varnothing$	$\varnothing$
$x^3-y^2$	$0$	$\varnothing$	$\varnothing$	$\varnothing$
$x^3-y$	$\varnothing$	$\varnothing$	$\varnothing$	$\varnothing$
$x^3-xy+1$	$\varnothing$	$\varnothing$	$\varnothing$	$1$
$x^3-xy$	$\varnothing$	$\varnothing$	$\varnothing$	$0$

Table 15. Table 15. Cubic polynomials with AutDeg ​ ( f ) < 3 AutDeg 𝑓 3 \text{AutDeg}(f)<3

$f$	$\phi$	$\phi \left(f\right)$
$x^3-y$	$⟨y,y^3-x⟩$	$x$
$x^3-xy$	$⟨y,y^2-x⟩$	$xy$
$x^3-xy+1$	$⟨y,y^2-x⟩$	$xy+1$

Equations

$$x^3+x{y}^2+E{x}^2+Fxy+G{y}^2+\lambda (f)$$

$$c\left(x^3+x{y}^2+x^2+Hx+Iy+J\right),$$

$$c\left(x^3+x{y}^2+Hx+Iy+J\right).$$

$$x^3-x{y}^2+E{x}^2+Fxy+G{y}^2+\lambda (f)$$

$$c\left(x^3-x{y}^2+Hx+Iy+J\right),$$

$$c\left(x^3-x{y}^2-y^2+Hx+Iy+J\right).$$

$$\left({\theta }_1\circ \theta \right)(f)=|I{|}^{3/2}\left(x^3-x{y}^2+\frac{H}{|I|}x-y+\frac{J}{|I{|}^{3/2}}\right).$$

$$\Big{\langle}\>\alpha\left(\,x+\sigma(H)\>\!y\,\right)\>,\>-\alpha\>\!\sigma(-H)\left(\,3x-\sigma(H)\>\!y\,\right)\>\Big{\rangle}\,.$$

$$-8{\alpha }^3\left(x^3-x{y}^2+\left(\sigma (-H)\frac{|H|-3|I|}{|H|+|I|}\right)x-y+\frac{J}{-8{\alpha }^3|I{|}^{3/2}}\right).$$

$$\frac{|H|-3|I|}{|H|+|I|}>\frac{-2|I|}{2|I|}=-1.$$

$$\big{\langle}\,{-\beta\left(\,x+\sigma(H)\>\!y\,\right)}\>,\>{-\beta\left(\,3x-\sigma(H)\>\!y\,\right)}\,\big{\rangle}\,.$$

$$8{\beta }^3\left(x^3-x{y}^2+\frac{H}{-8{\beta }^2}x+\frac{|H|}{-8{\beta }^2}y+\frac{J}{8{\beta }^3}\right).$$

$$⟨-\frac{1}{2}(x+y)-\frac{3}{4},\sigma (I)\left(-\frac{1}{2}(3x-y)-\frac{3}{4}\right)⟩.$$

$$H^{\prime }=-\frac{9}{8}-\frac{H}{2}-\frac{3|I|}{2};I^{\prime }=-\frac{3}{8}-\frac{H}{2}+\frac{|I|}{2}.$$

$$I^{\prime }\ge -\frac{3}{8}-\left(-\frac{3}{8}+\frac{|I|}{2}\right)+\frac{|I|}{2}=0.$$

$$H^{\prime }+I^{\prime }=-\frac{3}{2}-(H+|I|)\le -\frac{3}{4}.$$

$$⟨-\frac{1}{2}(x-y)-\frac{3}{4},\sigma (I)\left(-\frac{1}{2}(3x+y)-\frac{3}{4}\right)⟩.$$

$$H^{\prime }=-\frac{9}{8}-\frac{H}{2}-\frac{3|I|}{2};I^{\prime }=\frac{3}{8}+\frac{H}{2}-\frac{|I|}{2}.$$

$$I^{\prime }\ge \frac{3}{8}+\left(-\frac{3}{8}\right)=0.$$

$$H^{\prime }+I^{\prime }=-\frac{3}{4}-2|I|\le -\frac{3}{4}.$$

$$x^{2}y+E{x}^2+Fxy+G{y}^2+\lambda (f)$$

$$c(x^{2}y+y^2+Hx+Iy+J),$$

$$\displaystyle c(x^{2}y+Hx+Iy+J)\,.\$$

$$x^3+E{x}^2+Fxy+G{y}^2+\lambda (f)$$

$$\left(a^3\right)x^3+\left(3a^{2}r+E{a}^2+Fac+G{c}^2\right)x^2+\left(Fad+2Gcd\right)xy+\left(G{d}^2\right)y^2.$$

$$\frac{F^2-4EG}{4G}+E-\frac{F^2}{2G}+\frac{F^2}{4G}=0,\left(F\sigma (-G)+F\sigma (G)\right)d=0,$$

$$c\left(x^3-y^2+Hx+Iy+J\right),$$

$$c\left(x^3-xy+Hx+Iy+J\right),$$

$$c\left(x^3+Hx+Iy+J\right).$$

$$-6R\mp 6S-9$$

$$12R\mp 4S+6$$

$$2R\pm 2S-1$$

$$\left[\begin{array}[]{c}R\\ S\\ A\end{array}\right]\,=\,\left[\begin{array}[]{ccc}-6&\mp 6&0\\ 12&\mp 4&0\\ 2&\pm 2&-8\end{array}\right]^{-1}\left[\begin{array}[]{c}9\\ {-6}\\ 1\end{array}\right]\,=\,\left[\begin{array}[]{c}{-3/4}\\ {\mp 3/4}\\ {-1/2}\end{array}\right]\,.$$

$$H^{\prime }+I^{\prime }=-\frac{3}{2}-H\mp I\le -\frac{3}{4}.$$

Full text

Special thanks to S.B. Mulay for his patient guidance.

Cubic curve classification has a deep history, dating back to Newton [8], Plücker [9], and Cayley [5]. Many have addressed this problem by studying the action of ring automorphisms of $\mathbb{R}[x,y]$ that are linear in $x$ and $y$ on degree-three polynomials of $\mathbb{R}[x,y]\,$ [4, 6, 7, 10]. One such example was published in this journal [10].

This article serves two primary purposes. First, we wish to complete the classification that was started in [10]. Second, we wish to extend this result to a classification that considers the action of all ring automorphisms of $\mathbb{R}[x,y]\,$.

Throughout, we will refer to the ring automorphisms of real-valued polynomials in two variables by $\text{Aut}\,\mathbb{R}[x,y]\,$. When explicitly defining a particular $\varphi\in\text{Aut}\,\mathbb{R}[x,y]\,$, we will frequently express $\varphi$ in the form \big{\langle}\,p\,,\,q\,\big{\rangle}\,, where $p=\varphi(x)$ and $q=\varphi(y)\,$. The group of automorphisms for which $p$ and $q$ are both linear (ie. the affine group) we will refer to by $\Gamma_{2}(\mathbb{R})\,$. We will also use $\sigma$ to refer to the signum function that maps zero to one.

1. Affine Classification

We will now begin our affine classification by defining the relevant equivalence relation on $\mathbb{R}[x,y]\,$.

Definition 1.1**.**

If $f,g$ are polynomials in $\mathbb{R}[x,y]\,$, then we say $f\sim g$ if there exists some $\theta\in\Gamma_{2}(\mathbb{R})$ and some $c\in\mathbb{R}^{\times}$ such that $\theta(f)\,=\,cg\,$.

With respect to our equivalence relation $\sim\,$, we seek a list of polynomials comprised of exactly one representative from each equivalence class that contains a degree-three polynomial. As in [10], we will assume that the homogeneous degree-three component of our polynomials be in one of four canonical forms: $x^{3}+xy^{2}\,$, $x^{3}-xy^{2}\,$, $x^{2}y\,$, $x^{3}$.

Proposition 1.2**.**

If $f$ is a polynomial in $\mathbb{R}[x,y]$ of the form

[TABLE]

where $E,F,G$ are real numbers and $\lambda(f)$ is linear, then $f\sim g$ for some $g$ listed in Table 13.

Proof 1.3**.**

Let $\theta$ be as defined in Table 1 and consider $\theta(f)\,$. It follows that $\theta(f)$ is in one of the following forms, where $c,H,I,J$ are in $\mathbb{R}\,$.

[TABLE]

Should $\theta(f)$ be of form (1), let $\theta^{\prime}$ be \big{\langle}\,x\,,\,\sigma(I)y\,\big{\rangle}\,. Should $\theta(f)$ be of form (2), let $\theta^{\prime}$ be as defined in Table 2. Consider $\left(\theta^{\prime}\circ\theta\right)(f)$ and the result follows.

Proposition 1.4**.**

If $f$ is a polynomial in $\mathbb{R}[x,y]$ of the form

[TABLE]

where $E,F,G$ are real numbers and $\lambda(f)$ is linear, then $f\sim g$ for some $g$ listed in Table 13.

Proof 1.5**.**

Let $\theta$ be as defined in Table 3 and consider $\theta(f)\,$. It follows that $\theta(f)$ is in one of the following forms, where $c,H,I,J$ are in $\mathbb{R}\,$.

[TABLE]

We will consider four example cases, two where $\theta(f)$ is of form (3) and two where $\theta(f)$ is of form (4) .

Assume that $\theta(f)$ is of form (3) and $I\neq 0\,,\,|H|>|I|\,$. (We will neglect $c\,$, which can be factored out.) Define $\theta_{1}$ to be \big{\langle}\,|I|^{1/2}x\,,\,\sigma(-I)|I|^{1/2}y\,\big{\rangle}\,, and note that

[TABLE]

Subsequently let $\alpha:=-\sqrt{\frac{|H|+|I|}{8\>\!|I|}}\,$, and define $\theta_{2}$ to be

[TABLE]

It follows that $\left(\theta_{2}\circ\theta_{1}\circ\theta\right)(f)$ equals

[TABLE]

It remains to consider the absolute value of $\frac{|H|-3|I|}{|H|+|I|}\,$. If $|H|-3|I|$ is positive, then $|H|-3|I|$ must have a smaller absolute value than $|H|+|I|\,$. Should $|H|-3|I|$ be negative, our assumption $|H|>|I|$ implies that

[TABLE]

Regardless of case, we have that $x$-coefficient of $\frac{\left(\theta_{2}\circ\theta_{1}\circ\theta\right)(f)}{-8\alpha^{3}}$ must be less than $1$ in absolute value. Letting $\theta_{3}$ be defined as \big{\langle}\,\sigma(J)x\,,\,\sigma(J)y\,\big{\rangle}\,, note that the appropriate $\theta^{\prime}$ in Table 4 is exactly $\theta_{3}\circ\theta_{2}\circ\theta_{1}\,$, and the desired result follows from considering $\left(\theta^{\prime}\circ\theta\right)(f)\,$.

Assume that $\theta(f)$ is of form (3) and $I=0\,,\,H\neq 0\,$. (We will neglect $c\,$, which can be factored out.) Let $\beta$ be $-\sqrt{\frac{|H|}{8}}\,$, and define $\theta_{1}$ to be

[TABLE]

It follows that $\left(\theta_{1}\circ\theta\right)(f)$ is equal to

[TABLE]

Note that the $x$- and $y$-coefficient of $\frac{\left(\theta_{1}\circ\theta\right)(f)}{8\beta^{3}}$ must be $\pm 1$ and ${-1}$, respectively. Letting $\theta_{2}$ be defined as \big{\langle}\,\sigma(-J)x\,,\,\sigma(-J)y\,\big{\rangle}\,, note that the appropriate $\theta^{\prime}$ in Table 4 is exactly $\theta_{2}\circ\theta_{1}\,$, and the desired result follows from considering $\left(\theta^{\prime}\circ\theta\right)(f)\,$.

Assume that $\theta(f)$ is of form $(4)$ and $\left|H+\frac{3}{4}\right|<|I|\,$. (We will neglect $c\,$, which can be factored out.) Define $\theta_{1}$ to be

[TABLE]

It follows that $(\theta_{1}\circ\theta)(f)$ is equal to $x^{3}-xy^{2}-y^{2}+H^{\prime}x+I^{\prime}y+J^{\prime}\,$, where

[TABLE]

First, observe that our assumed relation on $H$ and $I$ implies that $H\leq{-\frac{3}{4}}+|I|\,$, which subsequently yields

[TABLE]

Second, observe that our assumed relation on $H$ and $I$ implies that $H+|I|\geq{-\frac{3}{4}}\,$, which subsequently yields

[TABLE]

As such, $(\theta_{1}\circ\theta)(f)$ is of the form of a polynomial from Table 13 .

Assume that $\theta(f)$ is of the form of $(4)$ and $H+\frac{3}{4}\geq|I|\,$. (We will neglect $c\,$, which can be factored out.) Define $\theta_{1}$ to be

[TABLE]

It follows that $(\theta_{1}\circ\theta)(f)$ is equal to $x^{3}-xy^{2}-y^{2}+H^{\prime}x+I^{\prime}y+J^{\prime}\,$, where

[TABLE]

First, observe that our assumed relation on $H$ and $I$ implies that $H-|I|\geq{-\frac{3}{4}}\,$, which subsequently yields

[TABLE]

Second, observe that

[TABLE]

As such, $(\theta_{1}\circ\theta)(f)$ is of the form of a polynomial from Table 13 .

In the remaining cases where $\theta(f)$ is of form (3), let $\theta^{\prime}$ be as defined in Table 4. In the remaining cases where $\theta(f)$ is of form (4), let $\theta^{\prime}$ be as defined in Table 5. Consider $\left(\theta^{\prime}\circ\theta\right)(f)$ and the result follows.

Proposition 1.6**.**

If $f$ is a polynomial in $\mathbb{R}[x,y]$ of the form

[TABLE]

where $E,F,G$ are real numbers and $\lambda(f)$ is linear, then $f\sim g$ for some $g$ listed in Table 13.

Proof 1.7**.**

Let $\theta$ be as defined in Table 6 and consider $\theta(f)\,$. It follows that $\theta(f)$ is in one of the following forms, where $c,H,I,J$ are in $\mathbb{R}\,$.

[TABLE]

Should $\theta(f)$ be of form (5), let $\theta^{\prime}$ be defined as in Table 7. Should $\theta(f)$ be of form (6), let $\theta^{\prime}$ be defined as in Table 8. Consider $\left(\theta^{\prime}\circ\theta\right)(f)$ and the result follows.

Proposition 1.8**.**

If $f$ is a polynomial in $\mathbb{R}[x,y]$ of the form

[TABLE]

where $E,F,G$ are real numbers and $\lambda(f)$ is linear, then $f\sim g$ for some $g$ listed in Table 13.

Proof 1.9**.**

Let $\theta$ be defined as in Table 9 and consider $\theta(f)\,$.

We will consider an example case where $G\neq 0\,$. Noting Table 9, we see that $\theta$ is of the form \big{\langle}\,ax+r\,,\,cx+dy\,\big{\rangle} where $a=\sigma(-G)\,$, $r=\frac{F^{2}-4EG}{12G}\,$, $c=\frac{F}{2\>\!|G|}\,$, and $d=\frac{1}{\sqrt{|G|}}\,$. Subsequently note the terms of $\theta(f)$ of degree greater than one are equal to

[TABLE]

Substituting appropriately for $a,b,c,d$ yields that the $x^{2}$- and $xy$-coefficients in $\theta(f)$ are

[TABLE]

respectively. Moreover, the $x^{3}$- and $y^{2}$-coefficients in $\theta(f)$ are $\sigma(-G)$ and $\sigma(G)\,$, respectively. Hence, $\theta(f)$ is of the form of (7) below.

From inspection of the remaining cases, it follows that $\theta(f)$ is in one of the following forms, where $c,H,I,J$ are in $\mathbb{R}\,$.

[TABLE]

Should $\theta(f)$ be of form (7), let $\theta^{\prime}$ be defined as in Table 10. Should $\theta(f)$ be of form (8), let $\gamma:=I^{3}+IH+J$ and $\theta^{\prime}$ be defined as in Table 11. Should $\theta(f)$ be of form (9), let $\theta^{\prime}$ be defined as in Table 12. Consider $\left(\theta^{\prime}\circ\theta\right)(f)$ and the result follows.

To conclude that the list of polynomials in Table 13 contains only one representative from each equivalence class of degree-three polynomials in $\mathbb{R}[x,y]\,$, it remains to show that the polynomials listed are pairwise inequivalent with respect to $\sim\,$. Observing that the canonical forms $x^{3}+xy^{2}\,$, $x^{3}-xy^{2}\,$, $x^{2}y\,$, and $x^{3}$ are pairwise inequivalent with respect to $\sim\,$, we can proceed by inspecting each canonical form individually.

Proposition 1.10**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{3}+xy^{2}\,$. If $f\sim g\,$, then $f=g\,$.

Proof 1.11**.**

Let $\theta\in\Gamma_{2}(\mathbb{R})$ be such that $\theta(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and express $\theta$ as \big{\langle}\,Ax+By+R\,,\,Cx+Dy+S\,\big{\rangle}\, where $A,B,R,C,D,S$ are in $\mathbb{R}\,$. Given that $x^{3}+xy^{2}$ factors as $x(x^{2}+y^{2})\,$, it follows that $B$ and $C$ must both be zero. Since the $xy$- and $y^{2}$-coefficients of $\theta(f)$ must be zero, it follows that $R=S=0\,$. As such, $\theta$ is of the form \big{\langle}\,Ax\,,\,Dy\,\big{\rangle}\,. The desired result follows from inspection.

Proposition 1.12**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{3}-xy^{2}\,$. If $f\sim g\,$, then $f=g\,$.

Proof 1.13**.**

Let $\theta\in\Gamma_{2}(\mathbb{R})$ be such that $\theta(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and write $\theta$ in the form \big{\langle}\,Ax+By+R\,,\,Cx+Dy+S\,\big{\rangle}\, where $A,B,R,C,D,S$ are in $\mathbb{R}\,$. Since this canonical form factors as $x(x+y)(x-y)\,$, it follows that $\theta(x)$ is of the form $Ax+R\,$, $A(x+y)+R\,$, or $A(x-y)+R\,$. It is a straightforward computation to determine that $\theta(y)$ must (respectively) be of the form $\pm Ay+S\,$, $\pm A(3x-y)+S\,$, or $\pm A(3x+y)+S$. Should $\theta$ be of the form \big{\langle}\,Ax+R\,,\,\pm Ay+S\,\big{\rangle}\,, observe that $R$ and $S$ must be zero since the $x^{2}$- and $xy$-coefficients of $\theta(f)$ must be zero. As such, the desired result follows from inspection. Observing that $\theta(f)=cg$ implies $f(1,y)$ and $g(1,y)$ have equal $y^{2}$-coefficients will assist with the remaining cases. We will consider each individually.

Assume that $f(1,y)$ has a $y^{2}$-coefficient of ${-1}\,$. Should $\theta$ be of the form \big{\langle}\,A(x+y)+R\,,\,\pm A(3x-y)+S\,\big{\rangle}\,, observe that the $xy$- and $x^{2}$-coefficients of $\theta(f)$ being zero implies that $S=R=0\,$. This implies our desired result for $x^{3}-xy^{2}+1$ and $x^{3}-xy^{2}\,$. For the remaining polynomials to consider, since the $x^{3}$- and $y$-coefficients of $\theta(f)$ must be opposites, it follows that $-8A^{3}=-HA\mp A\,$. Since the $x^{3}$-coefficient of $\theta(f)$ must be at least as large as the $x$-coefficient of $\theta(f)$ in absolute value, we also have that $|\pm 3A+HA|\leq|-8A^{3}|\,$. Combining these observations yields $|H\pm 3|\leq|H\pm 1|\,$. Given that $H$ must be contained on the interval $[{-1},1]\,$, it is forced that $H=\pm 1$ and $A=\pm\frac{1}{2}\,$. As such, $\theta(f)\,=\,\pm f$ and we achieve our desired result. The case where $\theta(x)$ is of the form $A(x-y)+R$ is similar.

Assume that $f(1,y)$ has a $y^{2}$-coefficient of ${-2}\,$. Should $\theta$ be of the form \big{\langle}\,A(x+y)+R\,,\,\pm A(3x-y)+S\,\big{\rangle}\,, inspection of the $x^{2}$-, $xy$-, and $y^{2}$-coefficients yields this system of equations. (Please note that within this proof, any use of the notation $\pm$ or $\mp$ will be used such that the top symbol will correspond to the case $\theta(y)=A(3x-y)+S$ and the bottom symbol will correspond to the case $\theta(y)=-A(3x-y)+S\,$.)

[TABLE]

Rearranging these equations and solving yields

[TABLE]

For convenience, let us refer to the $x$- and $y$-coefficients of $\theta(f)$ as $H^{\prime}$ and $I^{\prime}$ respectively. With the values forced upon $R,S,A$ in the expression of $\theta\,$, it now follows that: $H^{\prime}$ must be ${-\frac{9}{8}}-\frac{H}{2}\mp\frac{3I}{2}\,$; $I^{\prime}$ must be ${-\frac{3}{8}}-\frac{H}{2}\pm\frac{I}{2}\,$; and $c$ is $1$ (ie. $\theta(f)=g\,$). Given the restriction on the $x$- and $y$-coefficients of $g$ from Table 13 , it further follows that

[TABLE]

In the case that $\theta(y)$ is equal to $-\frac{1}{2}(3x-y)-\frac{3}{4}\,$, we get the result $-H-I\leq\frac{3}{4}\,$. Given the restriction on $H+I$ from Table 13 , it must be that $H+I$ is equal to $\frac{3}{4}\,$. As such, inspection on the expressions for $I^{\prime}$ and $H^{\prime}$ above yields

[TABLE]

These equalities imply that $(H,I)$ must equal $(H^{\prime},I^{\prime})\,$. It follows that $\theta(f)$ must also equal $f\,$, and the desired result is achieved in this case. In the case that $\theta(y)$ is equal to $\frac{1}{2}(3x-y)+\frac{3}{4}\,$, we get the result $-H+I\leq\frac{3}{4}\,$. Subtracting $2I$ and multiplying by $-1$ yields $H+I\geq{-\frac{3}{4}}+2I\,$. Since $I$ is nonnegative, the restriction on $H+I$ from Table 13 implies that $I$ must be zero and subsequently $H$ must equal $-\frac{3}{4}\,$. Substituting accordingly into the expressions for $H^{\prime}$ and $I^{\prime}$ above, we have that $(H^{\prime},I^{\prime})$ must equal $(H,I)\,$. It follows that $\theta(f)$ must also $f\,$, and the desired result is achieved in this case as well. Should $\theta(x)$ be of the form $A(x-y)\,$, the result is similar.

Proposition 1.14**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{2}y\,$. If $f\sim g\,$, then $f=g\,$.

Proof 1.15**.**

The proof is similar to the that of Proposition 1.10 but with inspection of $xy$- and $x^{2}$-coefficients to yield that $R=S=0\,$.

Proposition 1.16**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{3}\,$. If $f\sim g\,$, then $f=g\,$.

Proof 1.17**.**

Let $\theta\in\Gamma_{2}(\mathbb{R})$ be such that $\theta(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and write $\theta$ in the form \big{\langle}\,Ax+By+R\,,\,Cx+Dy+S\,\big{\rangle}\, where $A,B,R,C,D,S$ are in $\mathbb{R}\,$. Given the canonical form $x^{3}\,$, it follows that $B$ is zero and $D$ is nonzero. As such, the polynomials $f(1,y)$ and $g(1,y)$ must have equal degree. We will consider three cases accordingly.

Should the degree of $f(1,y)$ be less than one, observe that the $x^{2}$-coefficient of $\theta(f)$ being zero implies that $R=0\,$. Hence, $\theta$ is of the form \big{\langle}\,Ax\,,\,Cx+Dy+S\,\big{\rangle} and the desired result follows from inspection. Should the degree of $f(1,y)$ equal one, observe that $f$ is either $x^{3}-xy\,$, $x^{3}-xy+1\,$, or $x^{3}-y\,$. Noting that $B=0$ forces the $xy$-coefficient of $\theta(x^{3}-y)$ to be zero, it follows that $f$ equaling $x^{3}-y$ implies that $f=g\,$. Further, since $x^{3}-xy$ is reducible and $x^{3}-xy+1$ is irreducible, our desired result follows in this case from the fact that $\theta$ is a ring automorphism of $\mathbb{R}[x,y]\,$. Should the degree of $f(1,y)$ be greater than one, observe the the $xy$-coefficient of $\theta(f)$ being zero implies that $C=0\,$. Further inspection of the $x^{2}$- and $y$-coefficients of $\theta(f)$ implies that $R=S=0\,$. As such, $\theta$ must be the of the form \big{\langle}\,Ax\,,\,Dy\,\big{\rangle} and the result follows by inspection.

We have now established an affine classification of cubic curves, specifically one that finishes the work started in [10].

Theorem 1.18**.**

Every degree-three polynomial in $\mathbb{R}[x,y]$ is equivalent with respect to $\sim$ to some polynomial in Table 13. Moreover, the polynomials in Table 13 are pairwise inequivalent with respect to $\sim\,$.

Proof 1.19**.**

The result follows from the propositions of this section.

2. Automorphic Classification

We will use our recently established affine classification to develop an automorphic classification. This approach follows naturally given that the equivalence relation we define below is coarser than $\sim\,$.

Definition 2.1**.**

If $f,g$ are polynomials in $\mathbb{R}[x,y]\,$, then we say $f\approx g$ if there exists some $\varphi\in\text{Aut}\,\mathbb{R}[x,y]$ and some $c\in\mathbb{R}^{\times}$ such that $\varphi(f)\,=\,cg\,$. More generally, if $K$ is a field and $f,g$ are polynomials in $K[x,y]\,$, then we say $f\approx_{K}g$ if there exists some $\phi\in\text{Aut}\,K[x,y]$ that fixes $K$ and some $c\in K^{\times}$ such that $\phi(f)\,=\,cg\,$.

To allude to the additional complexity associated with the equivalence relation $\approx$ in comparison to $\sim\,$, let us consider an example involving $f=x^{3}-y^{2}$ and $g=-x^{3}+y^{2}$. If we suppose that there exists a $\theta$ in $\Gamma_{2}(\mathbb{R})$ such that $\theta(f)=g\,$, we can utilize the fact that $\theta$ fixes the degree of any monomial term to imply contradiction on the forced relation of $\theta(-y^{2})$ being to equal $y^{2}\,$ (as there is no real number whose square is $-1\,$). Instead, if we suppose that there exists a $\varphi$ in $\text{Aut}\,\mathbb{R}[x,y]$ such that $\varphi(f)=g\,$, we can make no such claim about the image of $-y^{2}\,$. In principle, it is feasible that for some positive integer $k\,$: $\varphi(x)$ be of degree $2k\,$; $\varphi(y)$ be of degree $3k\,$; and the expressions for $\varphi(x^{3})$ and $\varphi(-y^{2})$ add to one another to cancel all terms of degree greater than three to yield $-x^{3}+y^{2}\,$. With circumstances as such, additional considerations regarding $f$ and $g$ will be required to arrive at a conclusion (of contradiction) for this argument. In fact, within Proposition 2.14 we will work through details of a proof regarding this precise situation.

Our general strategy within this section will be to consider the equivalence classes with respect to $\approx$ that contain at least one polynomial from Table 13. We will develop a list of those polynomials that contain exactly one representative from each equivalence class. Observing that the canonical forms $x^{3}+xy^{2}\,$, $x^{3}-xy^{2}\,$, $x^{2}y\,$, and $x^{3}\,$ have distinct factorization structures in $\mathbb{R}[x,y]$ and are hence pairwise inequivalent with respect to $\approx\,$, we can proceed by investigating each form individually.

In the two subsequent propositions, the following definition and lemma will prove useful.

Definition 2.2**.**

If $f$ is a polynomial in $\mathbb{R}[x,y]\,$, let

[TABLE]

Lemma 2.3**.**

$\emph{AutDeg}(x^{2}\pm y^{2})\,=\,2\,.$**

Proof 2.4**.**

Observe that $x^{2}-y^{2}=0$ defines crossing lines in $\mathbb{R}^{2}$ and $x^{2}+y^{2}=0$ defines an isolated point in $\mathbb{R}^{2}\,$. Also observe that neither of these curves could be mapped to a line, an empty variety, or all of $\mathbb{R}^{2}$ by a map in $\text{Aut}\,\mathbb{R}[x,y]$ (ie. a bi-polynomial homeomorphism of $\mathbb{R}^{2}\,$). Hence, $\text{AutDeg}(x^{2}\pm y^{2})$ could not be less than two.

Proposition 2.5**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{3}+xy^{2}\,$. If $f\approx g\,$, then $f=g\,$.

Proof 2.6**.**

Let $\varphi\in\text{Aut}\,\mathbb{R}[x,y]$ be such that $\varphi(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and express $\varphi$ in the form \big{\langle}\,p\,,\,q\,\big{\rangle}\,. We will consider three cases. Suppose $\deg(p)$ is greater than $\deg(q)\,$, and observe $\deg(g)$ would be determined by $\deg(p^{3})\,$. This leads to contradiction since $\deg(p)$ must be at least two in this case. Supposing $\deg(p)$ is less than $\deg(q)$ leads to a similar contradiction when inspecting the value of $\deg(pq^{2})\,$. It is left to consider when $\deg(p)$ and $\deg(q)$ are equal. Suppose that $p,q$ are not linear. From inspection of the polynomials in Table 13, it follows that $p(p^{2}+q^{2})$ must be of degree three or less. This demands $\deg(p)=2$ and $\deg(p^{2}+q^{2})=1\,$. But, this is a contradiction since $\text{AutDeg}(x^{2}+y^{2})$ equals two via Lemma 2.3. Hence, $p,q$ must both be linear, and the result follows from Theorem 1.18.

Proposition 2.7**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{3}-xy^{2}\,$. If $f\approx g\,$, then $f=g\,$.

Proof 2.8**.**

Let $\varphi\in\text{Aut}\,\mathbb{R}[x,y]$ be such that $\varphi(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and express $\varphi$ in the form \big{\langle}\,p\,,\,q\,\big{\rangle}\,. Should $f(1,y)$ have a $y^{2}$-coefficient of ${-1}\,$, the result follows similarly to Proposition 2.5 using the fact that $\text{AutDeg}(x^{2}-y^{2})$ equals two via Lemma 2.3. Hence, it remains to consider when $f(1,y)$ has a $y^{2}$-coefficient of ${-2}\,$. We will proceed in three cases. Suppose $\deg(p)$ is greater than $\deg(q)\,$, and observe that $\deg(g)$ would be determined by $\deg(p^{3})\,$. This leads to contradictions since $\deg(p)$ must be at least two in this case. Supposing $\deg(p)$ is less than $\deg(q)$ leads to a similar result upon inspecting the degree of $\deg(pq^{2})\,$. It remains to consider the case when $\deg(p)$ and $\deg(q)$ are equal. Suppose $p,q$ are not linear and note that $\varphi(f)$ is of the form $p(p^{2}-q^{2}+H)+(-q^{2}+Iq+J)\,$. Since $\text{AutDeg}(x^{2}-y^{2})$ is not less than two, the restriction on $\deg(g)$ forces the relation

[TABLE]

Since we are in the case where $\deg(p)$ equals $\deg(q)\,$, it follows that $\deg(p^{2}-q^{2})$ is equal to $\deg(p)\,$. In particular, this implies that $\deg(p+q)$ and $\deg(p-q)$ could not both be equal to $\deg(p)\,$. But, observe that should either $\deg(p+q)$ or $\deg(p-q)$ be less than $\deg(p)\,$, then the conjugate factor must have degree of exactly $\deg(p)\,$. This implies that either $p+q$ or $p-q$ must be a constant, which violates the algebraic independence of $p$ and $q\,$. This contradiction implies that $p,q$ must be linear and the result follows from Theorem 1.18.

Proposition 2.9**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{2}y\,$. If $f\approx g\,$, then $f=g\,$.

Proof 2.10**.**

Let $\varphi\in\text{Aut}\,\mathbb{R}[x,y]$ be such that $\varphi(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and express $\varphi$ in the form \big{\langle}\,p\,,\,q\,\big{\rangle}\,. We will consider three cases. First, supposing $\deg(q)$ is less than $\deg(p)$ implies that $\deg(g)$ must be more than three. This is a contradiction. Second, assume $\deg(q)$ equals $\deg(p)\,$. The fact that $\deg(g)$ equals three forces $p,q$ to be linear, and hence $\varphi$ is in $\Gamma_{2}(\mathbb{R})$ and our result follows from Theorem 1.18. Finally, assume $\deg(q)$ is greater than $\deg(p)\,$, and note from Table 13 that $\varphi(f)$ can be written as $q\,f_{1}+f_{2}\,$, where

[TABLE]

Since $p,q$ are algebraically independent, the value of $\deg(f_{1})$ in $\mathbb{R}[x,y]$ must be at least one. It follows that $\deg(q)=2\,$, $\deg(f_{1})=1\,$, and $\deg(p)=1\,$. Supposing that $f_{1}$ is of the form $p^{2}$ or $p^{2}\pm 1$ leads to contradiction, as $\deg(f_{1})$ could not be one. Hence, $f_{1}$ must be of the form $p^{2}+q+I$ for some $I\in\mathbb{R}\,$. Observing that the curves of $\mathbb{R}[x,y]$ associated with $x^{2}+y+I=0$ and $x=0$ have exactly one intersection point, it follows that the lines of $\mathbb{R}[x,y]$ associated with $f_{1}=0$ and $p=0$ are crossing lines. Hence, we can apply some $\theta\in\Gamma_{2}(\mathbb{R})$ such that $\theta(f_{1})=-y$ and $\theta(p)=x\,$. This implies that $\theta(f_{2})=f_{2}$ and yields

[TABLE]

Hence, $\theta(q)$ is equal to $-y-x^{2}-I$ and it follows that

[TABLE]

As such, $\varphi(f)$ and $f$ are equivalent via $\sim\,$, and Theorem 1.18 implies that $g$ and $f$ must be equal.

In the subsequent proposition, the following definition and lemma will prove useful. This definition is adapted from [1].

Definition 2.11**.**

If $f$ is a polynomial in $\mathbb{R}[x,y]\,$, $K$ is a field, and $g$ is a polynomial in $K[x,y]\,$, let

[TABLE]

Lemma 2.12**.**

Let $f,g$ be polynomials in $K[x,y]\,$. If $f\approx_{K}g\,$, then

[TABLE]

A similar statement can be made about $\emph{sing}(f)\,$. Should $K$ equal $\mathbb{R}\,$, additional similar statements can be made about $\emph{cusp}(f)\,$, $\emph{isol}(f)\,$, and $\emph{node}(f)\,$.

Proof 2.13**.**

Let $\phi\in\text{Aut}\,K[x,y]$ be such that $\phi$ fixes $K$ and $\phi(f)=cg\,$. Observe that $\phi(f-cr)\,=\,c(g-r)\,$.

The information in Table 14 is straightforward to compute. The contents will be useful when proving the proposition below.

Proposition 2.14**.**

Assume $f,g$ are polynomials listed in Table 13 with canonical form $x^{3}\,$. If $f\approx g\,$, then $f=g\,$.

Proof 2.15**.**

Let $\varphi\in\text{Aut}\,\mathbb{R}[x,y]$ be such that $\varphi(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$, and express $\varphi$ in the form \big{\langle}\,p\,,\,q\,\big{\rangle}\,. If $f$ is a polynomial only in $x\,$, then $\varphi(f)$ is determined completely by $p\,$. Supposing $\deg(p)$ is greater than one leads to contradiction since $\varphi(f)$ would then be of degree at least six. Hence, $\deg(p)$ must be one, and the result from Theorem 1.18 applies. It remains to consider the polynomials that are the subject of Table 14. By Lemma 2.12, our desired result follows should $f$ be either $x^{3}-xy\,$, $x^{3}-xy+1\,$, or $x^{3}-y^{2}\,$. Lemma 2.12 also implies that $f,g$ must both be contained in one of the following sets

[TABLE]

Assume $f,g$ are from set (10), and suppose $f$ is not equal to $g\,$. By Lemma 2.12 and the automorphism property of $\varphi\,$, we have that $c$ equals $-1$ and subsequently that $\varphi(x^{3}-y^{2})$ equals $-x^{3}+y^{2}\,$. Since $\varphi$ must map the singular points of $f$ to singular points of $g\,$, it follows that $\varphi$ must fix the origin. Hence, the constant coefficient of $p,q$ must be zero. Expressing $q$ in terms of homogeneous components $q_{i}$ each of degree $i\,$, it follows that ${-q_{1}}^{2}$ equals $y^{2}\,$. This leads to contradiction since ${-1}$ does not have a square root in $\mathbb{R}\,$.

Assume $f,g$ are from set (11), and refer to $f$ by $x^{3}-y^{2}-x+J_{1}$ and to $g$ by $x^{3}-y^{2}-x+J_{2}\,$. By Lemma 2.12, we have the system

[TABLE]

Subtracting (13) from (14) to eliminate $J_{1},J_{2}$ yields that $c$ equals $1$ and our desired result follows.

Assume $f,g$ are from set (12). Viewing these polynomials in $\mathbb{C}[x,y]\,$, it is straightforward to show that $\text{sing}(x^{3}-y)\,=\,\emptyset$ and that

[TABLE]

It follows from Lemma 2.12 that our desired result follows for $x^{3}-y\,$. Henceforth, refer to $f$ by $x^{3}-y^{2}+x+J_{1}$ and to $g$ by $x^{3}-y^{2}+x+J_{2}\,$. Note that $\varphi$ can be uniquely be extended to $\overline{\varphi}\in\text{Aut}\,\mathbb{C}[x,y]$ by defining $\overline{\varphi}(i)=i\,$, and it follows that $f\approx_{\mathbb{C}}g\,$. Hence, Lemma 2.12 applies and two possible systems of equations result. One such system is similar to (13) and (14), resulting in the solution $c=1$ and our desired result follows. The other such system is

[TABLE]

Suppose this system holds. Subtracting (15) from (16) to eliminate $J_{1},J_{2}\,$ yields that $c$ equals $-1$ (ie. $\overline{\varphi}(f)=-g\,$) and $J_{2}$ equals $-J_{1}\,$. Let $\theta$ be in $\text{Aut}\,\mathbb{C}[x,y]$ such that it fixes $\mathbb{C}$ and is of the form \big{\langle}x+\frac{i\sqrt{3}}{3}\,,\,y\big{\rangle}\,. It follows that

[TABLE]

Combining this with $\overline{\varphi}(f)=-g$ yields

[TABLE]

Express $\theta\circ\overline{\varphi}\circ\theta$ in the form \big{\langle}\,P\,,\,Q\,\big{\rangle}\,, and let $P_{i},Q_{i}$ be the homogeneous component of $P,Q$ of degree $i\,$. Observe that all of the coefficients of $Q$ must be real. Also observe that $\theta\circ\overline{\varphi}\circ\theta$ must fix the origin as it sends the singular point of $x^{3}-(i\sqrt{3})x^{2}-y^{2}$ to the singular point of $-x^{3}+(i\sqrt{3})x^{2}-y^{2}$. Hence, $P_{0},Q_{0}$ must both be zero. As such, isolating the degree-two terms of (17) results in

[TABLE]

Express $P_{1}$ as $(a_{1}+a_{2}i)x+(b_{1}+b_{2}i)y$ and $Q_{1}$ as $cx+dy\,$, with $a_{1},a_{2},b_{1},b_{2},c,d\in\mathbb{R}\,$. Inspecting the $x^{2}$-, $xy$-, and $y^{2}$-coefficients of (18) yields the system

[TABLE]

From (21), it follows that ${b_{1}}^{2}$ must equal ${b_{2}}^{2}$ (and both cannot be zero). From (20), it follows that $a_{1}b_{2}$ must equal $a_{2}b_{1}$ (and hence ${a_{1}}^{2}$ and ${a_{2}}^{2}$ are equal). This leads to contradiction, as (19) demands that ${a_{1}}^{2}$ and ${a_{2}}^{2}$ must differ by $1\,$. As a result, the system of (15) and (16) is logically impossible, and we have achieved our desired result.

We now have established an automorphic classification of cubic curves, which incidentally coincides with our affine classification. That said, it is worth noting that some polynomials $f$ in our affine classification were such that $\text{AutDeg}(f)$ was less than three. Such polynomials are listed in Table 15 along with a representative of minimal degree.

Theorem 2.16**.**

Every degree-three polynomial in $\mathbb{R}[x,y]$ is equivalent with respect to $\approx$ to some polynomial in Table 13. Moreover, the polynomials in Table 13 are pairwise inequivalent with respect to $\approx\,$.

Proof 2.17**.**

The result follows from the propositions in this section.

As a concluding result, we will note that the polynomials in Proposition 2.14 of the form $x^{3}-y^{2}+\lambda(f)$ with $\lambda(f)$ being linear could be alternatively handled (more concisely) using the Epimorphism Theorem of Abhyankar and Moh [2]. The subsequent proposition demonstrates this approach.

Proposition 2.18**.**

Let $f,g$ be polynomials from Table 13 of the form $x^{3}-y^{2}+\lambda(f)$ and $x^{3}-y^{2}+\lambda(g)$ where $\lambda(f),\lambda(g)$ are linear. If $f\approx g\,$, then $f=g\,$.

Proof 2.19**.**

Suppose $\varphi$ is in $\text{Aut}\,\mathbb{R}[x,y]\setminus\Gamma_{2}(\mathbb{R})$ such that $\varphi(f)=cg$ for some $c\in\mathbb{R}^{\times}\,$. Express $\varphi$ in the form \big{\langle}\,p\,,\,q\,\big{\rangle} and observe that $\deg(p),\deg(q)$ must equal $2k,3k$ (repectively) for some positive integer $k\,$. Apply $\theta_{a}$ of the form \big{\langle}\,x+ay\,,\,y\,\big{\rangle} for some $a\in\mathbb{R}$ so that $\theta_{a}(p)$ will have a nonzero $y^{2k}$-coefficient and $\theta_{a}(q)$ will have a nonzero $y^{3k}$-coefficient. Such an $a$ is possible since the $y^{2k}$- and $y^{3k}$-coefficients of $p$ and $q$ (respectively) will be polynomials in $\mathbb{R}[a]$ and hence have only finitely many roots in $\mathbb{R}\,$. View $\overline{\theta\circ\varphi}$ as the natural extension of $\theta\circ\varphi$ in $\text{Aut}\,\mathbb{C}[x,y]$ that fixes $\mathbb{C}\,$, and express $\overline{\theta\circ\varphi}$ as \big{\langle}\,P\,,\,Q\,\big{\rangle}\,. It follows that the degree of $P(0,y),Q(0,y)$ equals $2k,3k$ (respectively). This leads to contradiction of Abhyankar-Moh [2] as neither $2k$ nor $3k$ will divide the other.

This work is adapted, in part, from the author’s Ph.D. thesis [3].