Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 22 / 3

An elliptic curve has good reduction of an elliptic curve at $p$ if it admits a Weierstrass equation of an elliptic curve over ${\mathbb{Z}}_p$ whose reduced projective cubic is nonsingular. Equivalently its minimal discriminant is a $p$-adic unit.

For the given integral short equation,It therefore directly supplies good reduction outside $2,3,5$. These three primes cannot be rescued by changing the model. An admissible change of Weierstrass model changes the discriminant by a twelfth power, so its valuation changes by a multiple of twelve. The displayed valuations $4,7,4$ are not congruent to zero modulo twelve. No integral model can have unit discriminant at any of those primes. The good primes are exactly $p\notin \{2,3,5\}$.

At $P_1$ the tangent slope is zero. The elliptic-curve addition formula givesThe chord from $P_1$ to $P_2$ has slope $(-3-15)/(-6)=3$, so its sum has $x$-coordinate $3^2-0-(-6)=15$ and $y$-coordinate $3(0-15)-15=-60$. Thus $P_1+P_2=(15,-60)$.

Reduction at the good prime seven gives $\tilde{E}:y^2=x^3+1$. For $x=0,1,2,3,4,5,6$, the right sides are respectively $1,2,2,0,2,0,0$. There are respectively $2,2,2,1,2,1,1$ choices of $y$. Including the point at infinity gives $\#\tilde{E}({\mathbb{F}}_7)=12$. Its nonzero torsion points of an elliptic curve of order two are $(3,0),(5,0),(6,0)$, so it cannot be cyclic. Its two-primary component is $(\mathbb{Z}/2\mathbb{Z}{)}^2$ and its three-primary component is cyclic of order three; henceIn particular its exponent is six.

The reduction of an elliptic curve is a group homomorphism defined on every ${\mathbb{Q}}_7$-point by projectivity. Thus $6P$ reduces to the identity for every rational $P$. A finite point with integral coordinates reduces to an affine point, whose projective last coordinate is one, so it cannot reduce to the identity at infinity. Therefore every nonzero finite point $6P$ has nonintegral coordinates. If $6P=O$, it has no affine coordinates at all. This is the reduction exponent obstruction to integral multiples.

The Lutz–Nagell theorem says that for a nonsingular short equation $y^2=x^3+ax+b$ with $a,b\in \mathbb{Z}$, every nonidentity rational torsion point of an elliptic curve $(x,y)$ has $x,y\in \mathbb{Z}$, and either $y=0$ orThis is a necessary condition for torsion, not a converse.

To prove integrality, fix a prime $q$. If $v_q(x)<0$, the $x^3$ term dominates the right side of the equation. Hence $2v_q(y)=3v_q(x)$, so $v_q(x)=-2s$, $v_q(y)=-3s$ for some $s\ge 1$. The parameter $t=-x/y$ has valuation $s$ and identifies the point with the formal neighbourhood of the identity, namely the formal group of an elliptic curve evaluated on $q{\mathbb{Z}}_q$.

For odd $q$, this group is a torsion-free group by the logarithm argument of Question 2. For $q=2$, use the special short-model structure. Negation sends $t$ exactly to $-t$, so the formal multiplication series $[2](t)$ is odd and has integral coefficients:If $s=v_2(t)\ge 1$, the first term has valuation $s+1$ and all the others have valuation at least $3s>s+1$. Thus $v_2([2](t))=s+1$, and no iteration of doubling kills a nonzero point. Multiplication by an odd integer has unit linear coefficient and also cannot kill it. This proves the torsion-free formal subgroup for a short Weierstrass equation, including at two. Consequently a rational torsion point of an elliptic curve cannot have $v_q(x)<0$ at any prime. Its $x$ is integral, and the equation then makes its rational $y$ integral too.

For the divisibility conclusion suppose $y\ne 0$. Then $2P\ne O$ and is also a rational torsion point of an elliptic curve, so $x(2P)$ is integral. Its tangent slope $\lambda =f^{\prime }(x)/(2y)$ satisfies $x(2P)={\lambda }^2-2x$. Thus ${\lambda }^2\in \mathbb{Z}$; a rational number with integral square is itself integral. In particular $y^2\mid f^{\prime }(x{)}^2$. Reduce the supplied polynomial identity modulo $y^2=f(x)$: both terms on the left are divisible by $y^2$, so the right side $4a^3+27b^2$ is too. This is the divisibility proof in the Nagell–Lutz theorem.

For the specific curve, $4a^3+27b^2=3^7{5}^4$. The computed equality $2P_1=-P_1$ makes $P_1$ a point of exact order three. The coordinate $y(P_3)=17$ has square not dividing $3^7{5}^4$, so $P_3$ has infinite order. To decide $P_2$, use the already computed point $P_1+P_2=(15,-60)$. Its $y$-coordinate has even square, which cannot divide the odd number $3^7{5}^4$, so that sum has infinite order. Since $P_1$ is torsion, $P_2$ must have infinite order as well. $P_1$ has order three; $P_2$ and $P_3$ both have infinite order.