Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 27 / 3

In the one-dimensional commutative convention, a formal group law over $R$ is a formal power series $F(X,Y)=X+Y+\text{terms of degree at least two}$ satisfying $F(X,0)=X$, $F(0,Y)=Y$, $F(X,Y)=F(Y,X)$ and $F(F(X,Y),Z)=F(X,F(Y,Z))$. The formal inverse is a series $I(T)=-T+O(T^2)$ with $F(T,I(T))=0$. For $R={\mathbb{Z}}_p$ all these series converge on $p{\mathbb{Z}}_p$, which becomes a group with operation $F$.

Repeated formal addition gives the multiplication isomorphism of a formal group law $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the coefficient $n$ is a unit in the p-adic integers. We can construct an inverse $H(T)=n^{-1}T+{\sum }_{j\ge 2}{h}_j{T}^j$ by coefficient recursion. At degree $j$, the equation $[n{]}_F(H(T))=T$ has the form $n{h}_j+\text{known integral terms}=0$; dividing by the unit $n$ keeps $h_j$ integral. The inverse is two-sided by uniqueness of formal composition inverses. Since integral coefficients multiplied by successive powers of $t\in p{\mathbb{Z}}_p$ tend to zero, both series converge on that set and their formal identities hold there. This proves prime-to-residue-characteristic multiplication on a formal group:The argument also applies at $p=2$ for odd $n$; it does not require a logarithm or torsion-freeness.

For an integral Minimal Weierstrass equation, let $E_0({\mathbb{Q}}_p)$ be the subgroup of points whose reduction is a smooth point of the reduced cubic. Its reduction is a homomorphism to the smooth-locus group of a singular Weierstrass cubic, or to the whole reduced elliptic curve in the good-reduction case. Define the formal kernel of a minimal Weierstrass equation byIt consists of the identity and the points reducing to the identity at infinity. For a nonidentity affine point this is equivalent to $v_p(x)=-2r$ and $v_p(y)=-3r$ for some positive integer $r$.

The local parameter $t=-x/y$, with $t(O)=0$, gives the group isomorphismwhere $\hat{E}$ is the formal group of an elliptic curve. To see the local coordinates, put $w=-1/y$ in the Weierstrass equation of an elliptic curve. The resulting relation is $w=t^3+a_{1}tw+a_2{t}^{2}w+a_3{w}^2+a_{4}t{w}^2+a_6{w}^3$, and its unit linear coefficient gives a unique integral series $w(t)=t^3+O(t^4)$. Thus $x=t/w$ and $y=-1/w$ have valuations $-2v_p(t)$ and $-3v_p(t)$. The chord-and-tangent group law expressed in this parameter is precisely the integral formal group law $\hat{E}$. This description includes bad reduction; only the smooth identity neighborhood is involved.

Direct substitution gives $y^2+y=x^3-x+6$ for both points. The displayed Weierstrass equation of an elliptic curve is minimal at every prime by the permitted assumption.

For $P$, neither $364$ nor $11737$ is divisible by three or five. ThereforeThe formal kernel of a minimal Weierstrass equation description gives $P\in E_1({\mathbb{Q}}_3)\cap E_1({\mathbb{Q}}_5)$. Similarly, neither $2059$ nor $98783$ is divisible by three or seven, so $Q\in E_1({\mathbb{Q}}_3)\cap E_1({\mathbb{Q}}_7)$, with the same respective coordinate valuations $-2,-3$.

If a point $T\in E_1({\mathbb{Q}}_p)$ had finite order of a group element $N=p^{a}n$ with $p\nmid n$, the point $p^{a}T$ would be killed by $n$. By prime-to-residue-characteristic multiplication on a formal group, multiplication by $n$ is injective on this kernel, so $p^{a}T=O$. Its order is therefore a power of $p$. Applying this at two distinct primes is the two-prime formal-kernel test for nontorsion: the order of $P$ would be both a power of three and of five, and the order of $Q$ both a power of three and of seven. Each would have to be the identity, although both are affine points. Both $P$ and $Q$ have infinite order.

We use the canonical height pairing normalization $⟨P,Q⟩=(\hat{h}(P+Q)-\hat{h}(P)-\hat{h}(Q))/2$, so that $⟨P,P⟩=\hat{h}(P)$. The supplied values giveThe convention without the factor $1/2$ instead gives the doubled pairing $B(P,Q)\approx 7.50$; all independence conclusions are the same.

In our convention the Gram matrix is approximatelyThis positive margin is much larger than the rounding uncertainty. Even allowing each of the three stated heights an error of $0.01$, we have $a\ge 6.60$, $b\ge 7.86$ and $|c|\le 3.765$, giving $ab-c^2\ge 6.60\cdot 7.86-3.765^2=37.700775>0$. Thus the actual canonical height pairing matrix is positive-definite, not merely its rounded approximation.

If $rP+sQ=O$ for integers $r,s$, bilinearity and the height parallelogram identity give $0=\hat{h}(rP+sQ)=a{r}^2+2crs+b{s}^2$. Positive definiteness forces $r=s=0$. This is the rank certificate from rounded canonical heights, and provesHere the last brackets denote the generated subgroup, not the scalar height pairing. The argument proves independence; it does not assert that these points generate the entire Mordell-Weil group.