Exercise 5.6.9 (Chord-and-tangent method for the curve $x^3 + y^3 = 1$)

Question

The cubic curve $x^3 + y^3 = 1$ contains the two rational points $(0,1)$ and $(1,0)$. Explain why the chord-and-tangent method does not yield any further points on this curve.

Richard Ganaye · Accepted Answer

\begin{proof} 
These points are simple points on the curve. The chord $L$ passing through $A(0,1)$ and $B(1,0)$ has for equation $y = -x+1$. Therefore the intersection of $L$ with the curve are given by
$$x^3 + (-x+1) ^3 = 1,\qquad y = -x+1.$$
Then 
\begin{align*}
x^3 + (-x+1) ^3 = 1 &\iff x^3 - x^3 + 3x^2 - 3x + 1 = 1\
&\iff x(x-1) = 0
\end{align*}
Then $x = 0$ and $y = -x+1 = 1$, or $x =1$ and $y = -x+1 = 0$. The intersection points are $A$ and $B$, so the the chord-and-tangent method does not yield any further points on this curve (however the curve has other rational points, such as $(1/3, 2/3)$).

Moreover the tangent line to the curve at $A(0, 1)$ is $T : y = 1$ which gives $x = 0$. There are no other point of intersection with the curve. By symmetry, there is the same with $B(0,1)$.
\end{proof}
Note: We expect three points of intersection, counting multiplicity,  between the line $L$ and the third degree curve. The explanation is that the points at infinity on the curve are given by 
$$x^3 + y^3 = t^3,\qquad t = 0,$$
so that $C = (1:-1:0)$ is a point at infinity on the curve. This is also a point of the completion $\overline{L}$ of the chord, with equation $y = -x + t$. In the projective plane, the chord and the curve have three points of intersection $A,B$ and $C$. Same remark for the two tangent lines.

Exercise 5.6.9 (Chord-and-tangent method for the curve $x^3 + y^3 = 1$)

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Exercise 5.6.9 (Chord-and-tangent method for the curve $x^3 + y^3 = 1$)

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