Exercise 5.7.6 (Inflection points on the projective curve $X^3 + Y^3 + Z^3 = dXYZ$)

Proof.

1.

(a) Let $f(X,Y,Z)=X^3+Y^3+Z^3-\mathit{dXY}Z$. Then $$\frac{\mathit{\partial f}}{\mathit{\partial X}}(X,Y,Z)=3X^2-\mathit{dY}Z,\frac{\mathit{\partial f}}{\mathit{\partial X}}(X,Y,Z)=3Y^2-\mathit{dXZ},\frac{\mathit{\partial f}}{\mathit{\partial Z}}(X,Y,Z)=3Z^2-\mathit{dXY}.$$

Assume that ${𝒞}_f(ℂ)$ has a singular point $(X:Y:Z)$. Then $f(X,Y,Z)=0$ and

$$\begin{array}{lll}\hfill \{\begin{array}{cc}3X^2-\mathit{dY}Z\hfill & =0,\hfill \\ 3Y^2-\mathit{dXZ}\hfill & =0,\hfill \\ 3Z^2-\mathit{dXY}\hfill & =0.\hfill \end{array}& & \hfill \text{(1)}\end{array}$$

This implies

$$\begin{array}{lll}\hfill \mathit{dXY}Z=3X^3=3Y^3=3Z^3.& & \hfill \text{(2)}\end{array}$$

If $X=0$, then $(X,Y,Z)=(0,0,0)$, but $(0:0:0)$ is not a projective point, so $f$ has no singular point. This contradicts the hypothesis, so $X\ne 0$. Therefore $Y={\omega }^{i}X,Z={\omega }^{j}X$, where $\omega =e^{2\mathit{i\pi }∕3}$ and $i,j\in \{0,1,2\}$. By equation (2), $\mathit{dXY}Z=d{\omega }^{i+j}{X}^3=3X^3$ where $X\ne 0$, so $d{\omega }^{i+j}=3$, and

$$d^3=27.$$

Conversely, suppose that $d^3=27$. Then $d=3{\omega }^k$, where $k\in \{0,1,2\}$. The triple $({\omega }^{-k},1,1)$ is solution of (1), thus $({\omega }^{-k}:1:1)$ is a singular point of ${𝒞}_f(ℂ)$.

So ${𝒞}_f(ℂ)$ is nonsingular if and only if $d^3\ne 27$.

(b)

Suppose that $d=3$. Then $$f(X,Y,Z)=X^3+Y^3+Z^3-3\mathit{XY}Z=(X^2+Y^2+Z^2-\mathit{XY}-\mathit{XZ}-YZ)(X+Y+Z).$$

This shows that ${𝒞}_f(ℂ)$ is the union of the conic $X^2+Y^2+Z^2-\mathit{XY}-\mathit{XZ}-YZ=0$ and the line $X+Y+Z=0$.

(c)

Here $d^3\ne 27$. By part (a), ${𝒞}_f(ℂ)$ is an elliptic curve.

Consider the Hessian $H(a,b,c)$ of $f$ at a point $(a,b,c)$:

$$\begin{array}{lll}\hfill H(a,b,c)& =\left|\begin{array}{ccc}\hfill \frac{{\partial }^{2}f}{\partial X^2}(a,b,c)\hfill & \hfill \frac{{\partial }^{2}f}{\mathit{\partial X\partial Y}}(a,b,c)\hfill & \hfill \frac{{\partial }^{2}f}{\mathit{\partial X\partial Z}}(a,b,c)\hfill \\ \hfill \hfill \\ \hfill \frac{{\partial }^{2}f}{\mathit{\partial Y}\mathit{\partial X}}(a,b,c)\hfill & \hfill \frac{{\partial }^{2}f}{\partial Y^2}(a,b,c)\hfill & \hfill \frac{{\partial }^{2}f}{\mathit{\partial Y}\mathit{\partial Z}}(a,b,c)\hfill \\ \hfill \hfill \\ \hfill \frac{{\partial }^{2}f}{\mathit{\partial Z\partial X}}(a,b,c)\hfill & \hfill \frac{{\partial }^{2}f}{\mathit{\partial Z\partial Y}}(a,b,c)\hfill & \hfill \frac{{\partial }^{2}f}{\partial Z^2}(a,b,c)\hfill \end{array}\right|& \hfill \text{(3)}\end{array}$$

Since all points of ${𝒞}_f(ℂ)$ are non-singular, $(a,b,c)$ is an inflection point if and only if $H(a,b,c)=0$ (see for instance Walker “Algebraic curves“, Theorem 6.3). By (1),

$$\begin{array}{llll}\hfill H(a,b,c)& =\left|\begin{array}{ccc}\hfill 6a\hfill & \hfill -\mathit{cd}\hfill & \hfill -\mathit{bd}\hfill \\ \hfill -\mathit{cd}\hfill & \hfill 6b\hfill & \hfill -\mathit{ad}\hfill \\ \hfill -\mathit{bd}\hfill & \hfill -\mathit{ad}\hfill & \hfill 6c\hfill \end{array}\right|& \hfill & \\ \hfill & =-2\mathit{abc}{d}^3-6a^3{d}^2-6b^3{d}^2-6c^3{d}^2+216\mathit{abc}.& \hfill & \end{array}$$

So we must solve the system of equations, with $a,b,c$ as unknowns,

This implies $\left(\frac{108-d^3}{3d^2}-d\right)\mathit{abc}=0$ Since $d^3\ne 27$, $\frac{108-d^3}{3d^2}-d\ne 0$, thus $\mathit{abc}=0$. So our system is equivalent to

Then $a=0$, or $b=0$, or $c=0$. If $c=0$, then $a\ne 0$, otherwise $a=b=c=0$, so ${(b∕a)}^3=-1$, and $b∕a\in \{-1,-\omega ,-{\omega }^2\}$. We obtain similar results if $b=0$ or $c=0$.

The solutions are $\lambda (1,-{\omega }^k,0),\lambda (0,1,-{\omega }^k),\lambda (-{\omega }^k,0,1)$, where $\lambda$ is an arbitrary complex number and $k\in \{0,1,2\}$.

This proves that the points $(1:-1:0),(0:1:-1),(-1:0:1)$ are inflection points, and that the curve ${𝒞}_f(ℝ)$ has no other real inflection points. (but ${𝒞}_f(ℂ)$ has nine inflection points: the statement doesn’t give explicitly the field). □