Exercise 5.6.10 (Finite subgroup of an elliptic curve)

Proof.

(a)

Put $p(x)=4x^3+x^2-2x+1$ and $f(x,y)=y^2-p(x)$.

We show that the cubic curve ${𝒞}_f(ℝ)$ is nonsingular. Since for all $M=(x,y)\in {𝒞}_f(ℝ)$

$$\frac{\mathit{\partial f}}{\mathit{\partial x}}(x,y)=-p^{\prime }(x)=-12x^2-2x+2,\frac{\mathit{\partial f}}{\mathit{\partial y}}(x,y)=2y,$$

$M$ in singular iff and only if $p^{\prime }(x)=0$ and $y=0$, that is if $p(x)=p^{\prime }(x)=0$. But $\gcd (4x^3+x^2-2x+1,-12x^2-2x+2)=1$, thus $p(x)=p^{\prime }(x)=0$ has no solution.

(To prove that ${𝒞}_f(ℝ)$ is an elliptic curve, we must also verify that the point at infinity of the curve is non singular.)

(b)

${𝒞}_f(ℝ)$ contains the rational points $A=(1,2),A^{\prime }=(1,-2),B=(0,1),B^{\prime }=(0,-1)$. There is also $C=(-1,0)$, the unique point of intersection of ${𝒞}_f(ℝ)$ with the $x$-axis.

The homogeneous equation of ${𝒞}_f(ℝ)$ is $y^{2}t=4x^3+x^{2}t+2x{t}^2+t^3$. We obtain the points at infinity with $t=0$, thus $x=0$. The unique point at infinity of the curve is

$$O=(0:1:0).$$

We write $P=\mathit{MN}$ the third point of intersection of two points $M,N$ of the curve, and $M+N$ the symmetric point of $P$ relative to the $x$-axis. In section 5.7, the authors prove that the set of rational points of the elliptic curve is a group for this law $\text{+}$, with the point at infinity $O$ as neutral element. The opposite of $P$ is the symmetric of $P$ relative to the $x$-axis, so that $A^{\prime }=-A$, $B^{\prime }=-B$ (and $-C=C$).

Starting from $A=(1,2)$, we compute $2A=A+A$, $3A=A+A+A$, and so on.

• The tangent $T$ at point $A=(1,2)$ has for equation

  $$T:0=\frac{\mathit{\partial f}}{\mathit{\partial x}}(1,2)(x-1)+\frac{\mathit{\partial f}}{\mathit{\partial y}}(1,2)(y-2)=-12(x-1)+4(y-2),$$

  thus the equation of $T$ is $y=3x-1$. Since $-B=(0,-1)\in T\cap {𝒞}_f(\mathbb{Q})$, $\mathit{AA}=-B$ and $A+A=B$.

  $$2A=B=(0,1).$$

• The chord $\mathit{AB}$ has for equation $y=x+1$, thus $C=A+B$, so

  $$3A=C=(-1,0)$$

• Since the tangent line at $C$ is vertical, $\mathit{CC}=O$, that is $C+C=O$, so that

  $$6A=O=(0:1:0).$$

The point $A$ has order 6, and $B^{\prime }=-B=-2A=4A$, $A^{\prime }=-A=5A$. Therefore, the group generated by $A$ is

$$⟨A⟩=\{O,A,2A,3A,4A,5A\}\simeq \mathbb{Z}∕6\mathbb{Z},$$

which contains the initial points $(0,\pm 1),(1,\pm 2)$. So the chord-and-tangent method starting of these points is achieved.

$⟨A⟩=\{O,A,2A,3A,4A,5A\}$ is a subgroup of $({𝒞}_f(\mathbb{Q}),+)$, isomorphic to $\mathbb{Z}∕6\mathbb{Z}$.