Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / B / 5

Put $U(T)=F_X(0,T)$. The formal group identity $F(X,0)=X$ gives $U(0)=1$, so $U$ is a unit in a ring in $R[[T]]$, even if $R$ has zero divisors.

A differential $h(T)dT$ is invariant under formal translation ifas an identity of formal power series. Setting $X=0$ gives $h(T)U(T)=h(0)$. Consequently every invariant differential of a formal group law, if it exists, must have the form $h(0)U(T{)}^{-1}dT$. In particular there can be only one with constant coefficient one.

Existence follows directly from associativity. Differentiatewith respect to $Z$ at zero, obtainingEvery displayed $U$ is a unit in a ring, soThis is precisely invariance of $U(T{)}^{-1}dT$. ThusNormalization means its constant coefficient is one. This proves both existence and uniqueness over an arbitrary commutative ring, without division by any integer.

Let ${\omega }_G=h_G(T)dT$. The pullback through the formal group homomorphism isThe identity $f(F(X,T))=G(f(X),f(T))$ and the chain rule show that this pullback is invariant under $F$: translation by $T$ before $f$ becomes translation by $f(T)$ afterward. By part (i) it is a scalar multiple of ${\omega }_F$. Its coefficient at zero is $h_G(0)f^{\prime }(0)=f^{\prime }(0)$, so

Apply this to the multiplication-by-$p$ endomorphism $q(T)=[p](T)$. Since the linear coefficient of $F$ is $X+Y$, repeated addition gives $q^{\prime }(0)=p$. If ${\omega }_F=h(T)dT$ with $h(0)=1$, the pullback formula givesbecause $h(q(T))$ is a unit in a ring. Write $q(T)={\sum }_{m\ge 1}{a}_m{T}^m$. Then $m{a}_m\in pR$ for every $m$. If $p\nmid m$, choose integers $u,v$ with $um+vp=1$. It follows thatChoose $b_m\in R$ with $a_m=p{b}_m$ for those $m$ not divisible by $p$, and defineCoefficient comparison now provesNo cancellation of $p$ was used, so this prime-multiplication coefficient splitting for formal groups remains valid when $p$ is a zero divisor in $R$.

The discriminant of $y^2=x^3-9$ is $-16\cdot 27\cdot 81=-2^4{3}^7$, so five and seven are primes of good reduction of an elliptic curve. Count solutions over their residue fields.

Modulo five the values of $x^3-9$ for $x=0,1,2,3,4$ are $1,2,4,3,0$. They yield $2,0,2,0,1$ affine points, respectively, and the point at infinity gives $\#E({\mathbb{F}}_5)=6$.

Modulo seven the values for $x=0,\dots ,6$ are $5,6,6,4,6,4,4$. Only the three values four are nonzero squares, giving six affine points and $\#E({\mathbb{F}}_7)=7$.

For a rational elliptic curve at an odd prime of good reduction of an elliptic curve, the kernel of reduction on ${\mathbb{Q}}_p$ has no torsion element: it is the formal group on $p{\mathbb{Z}}_p$, whose formal logarithm embeds it in the additive group $p{\mathbb{Z}}_p$. Thus the entire torsion subgroup injects into each of the two finite reduction groups. Its order divides both six and seven, hence is one:Equivalently one may use prime-to-residue-characteristic injection at both primes; it gives the same coprime bound. This is a two-prime bound on rational elliptic torsion.

Since neither $n$ nor $-n$ is a square, both cubics have exactly one rational root, zero. Thus each torsion point of an elliptic curve group has exactly one nonidentity 2-torsion, namely $T=(0,0)$.

Neither curve has a rational point of order four. Indeed on $E_m:y^2=x^3+mx$ the duplication formula givesIf $2P=T$, then $P$ is neither the identity nor a 2-torsion, so the denominator is nonzero and $x(P{)}^2=m$. Thus $m$ must be a rational square; an integer that is a rational square is an integer square. Applying this with $m=n$ or $m=-n$ contradicts the hypotheses.

If $3\nmid n$, both curves have good reduction of an elliptic curve at three and exactly four points over ${\mathbb{F}}_3$. To see this, $x^3=x$ there, so the equation is $y^2=(1+m)x$. For $m=1$ the right side runs through all residues, and for $m=-1$ it is always zero; either way there are three affine points. Injection of torsion point of an elliptic curve at this odd good prime bounds each torsion element order by four. An order-four torsion subgroup with just one nonzero 2-torsion would contain an order-four point, which was excluded. Therefore in this case both torsion subgroups equal $\{o,T\}$.

The remaining case is $3\mid n$. The literal hypothesis $15\nmid n$ then ensures $5\nmid n$, so both curves have good reduction of an elliptic curve at five. Direct square counting givesFor example the square residues are $0,1,4$; substituting $x=0,\dots ,4$ gives this table, including the point at infinity. The pair $m=n,-n$ has residues $(1,4)$ or $(2,3)$ in some order. In either pair at least one reduction group has order two or four. The same injection and absence of order-four points then force that curve's torsion point of an elliptic curve to have order two. ThusThis torsion alternative for opposite quartic-twist coefficients uses only $15\nmid n$; it does not strengthen the source hypothesis to coprimality with fifteen.