Exercise 5.7.8 (Inflection points on the elliptic curve $x^3 + 2y^3 - 3$)

Proof. Since $f(1,1)=0$, we have $(1,1)\in {𝒞}_f(\mathbb{Q})$, so ${𝒞}_f(\mathbb{Q})\ne \varnothing$.

Le homogeneous equation of ${𝒞}_f(ℝ)$ is

The curve has no singularity, thus ${𝒞}_f(ℝ)$ is an elliptic curve.

As in Problem 6, we obtain the inflexion points with the Hessian. Le point $(a:b:c)\in {𝒞}_f(ℝ)$ is an inflection point if and only if

This gives the equation $\mathit{abc}=0$, so $(a:b:c)$ is an inflection point of ${𝒞}_f(ℝ)$ if and only if

• If $c=0$, then ${(a∕b)}^3=-2$, where $a,b$ are real numbers, thus $a=-\sqrt[3]{2}b$. The inflection point at infinity is $(-\sqrt[3]{2}:1:0)$.
• if $b=0$, then $a=\sqrt[3]{3}c$, and $(a:b:c)=(\sqrt[3]{3}:0:1)$. The corresponding affine point is $(x_1,y_1)=(\sqrt[3]{3},0)$.
• If $a=0$, then $b=\sqrt[3]{3∕2}c$, and $(a:b:c)=(0:\sqrt[3]{3∕2}:1)$. The corresponding affine point is $(x_2,y_2)=(0,\sqrt[3]{3∕2})$.

So ${𝒞}_f(ℝ)$ has three inflection points (including one at infinity) : they are the intersection points of the curve with the axes (see figure). Since $\sqrt[3]{2},\sqrt[3]{3},\sqrt[3]{3∕2}$ are irrational numbers, ${𝒞}_f(ℝ)$ has no inflection point with rational coordinates. □