Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 1

For points $P,Q$ on a smooth plane cubic, draw the line through them, using the tangent at $P$ when $P=Q$, and let its third intersection with the cubic be $R$. The chord-and-tangent group law defines $P+Q$ as the reflection of $R$ in the $x$-axis; the point at infinity $O$ is the identity and $-(x,y)=(x,-y)$.

If $P,Q\in E(\mathbb{Q})$, their chord or tangent has rational coefficients. Substitution into the cubic gives a polynomial with rational coefficients for which two intersection roots are rational, so the third is rational as well. The identity and inverse of every rational point are rational, and closure under addition follows. Assuming the elliptic-curve law is a group law, $E(\mathbb{Q})$ is therefore a subgroup of $E(\overline{\mathbb{Q}})$.

For $E:y^2=x^3-9x+9$, the addition formulas usewith $m=(3x_1^2-9)/(2y_1)$ for doubling. They giveand

Let $(x,y)\ne O$ be rational. If the 3-adic valuation satisfies $v_3(y)\ge 2$, first $v_3(x)$ cannot be negative: otherwise $x^3$ is the unique term of least valuation in $x^3-9x+9$, giving $2v_3(y)=3v_3(x)<0$. If $v_3(x)=0$, the right side has valuation zero. If $v_3(x)\ge 1$, its three terms have valuations at least $3$, at least $3$, and exactly $2$, so $2v_3(y)=2$. In every case $v_3(y)\le 1$.

The Lutz–Nagell theorem says that ifhas nonzero discriminant and $P=(x,y)\in E(\mathbb{Q})$ is a torsion point, then $x,y\in \mathbb{Z}$ and either $y=0$ or

For integrality, fix a prime $p$. If a rational point has nonintegral coordinates, its primitive projective coordinates reduce to $O$, so it belongs to the kernel of reduction. The parameter $t=-x/y$ identifies this kernel with the formal group of an elliptic curve over $p{\mathbb{Z}}_p$. The formal logarithm, with the standard separate first-step argument at $p=2$, shows that this group has no nonzero rational torsion. A rational torsion point therefore has nonnegative $p$-adic valuations in both coordinates for every prime $p$, hence integral coordinates.

Suppose now that $y\ne 0$. The point $2P$ is again a nonzero torsion point and hence integral. Its $x$-coordinate is $m^2-2x$, whereThus $m^2$ is an integer. A rational number whose square is integral is integral, so $2y\mid 3x^2+a$ and in particular $y^2\mid (3x^2+a{)}^2$. The curve equation and the identitythen prove $y^2\mid 4a^3+27b^2$. This is the divisibility proof in the Nagell–Lutz theorem.

HereBy the Lutz–Nagell theorem, a nonzero rational torsion point has integral coordinates and either $y=0$ or $y^2\mid 3^6$. Part b gives $v_3(y)\le 1$, so $y\in \{0,\pm 1,\pm 3\}$. The cubic $x^3-9x+9$ has no integral zero. Substitution of $y=\pm 1,\pm 3$ gives exactlyThis proves the required inclusion.

Since $2Q=-Q$, the point $Q$ has order three. The point $P$ is not torsion because $2P=(9/4,3/8)$ is nonintegral, contradicting Nagell–Lutz. If $P+Q$ or $P-Q$ were torsion, adding the torsion point $\mp Q$ would make $P$ torsion; their negatives are excluded in the same way. Hence