Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 125 / 3

Here a formal group means a one-dimensional commutative formal group law. Over a ring $R$, it is a formal power series $F(X,Y)\in R[[X,Y]]$ satisfyingIn particular, $F(X,Y)=X+Y+$ terms of total degree at least $2$. A formal inverse $\iota (T)=-T+\cdots$ with $F(T,\iota (T))=0$ is obtained recursively. An isomorphism of formal group laws from $F$ to $G$ is a series $h(T)\in TR[[T]]$ with invertible linear coefficient andIts compositional inverse exists over $R$ by coefficient recursion.

For $R={\mathcal{O}}_K$, all these series converge on $\pi {\mathcal{O}}_K$; the operations preserve that set and make it the group $F(\pi {\mathcal{O}}_K)$. Use the uniformizer $\pi$ and normalize the valuation by $v_K(\pi )=1$, and write $e=v_K(p)$.

We explicitly construct its formal logarithm. DefineThen putThis is a formal series over $K$ with linear coefficient $1$. Differentiating associativity in its last variable at $Z=0$ givesIt follows that the $Y$-derivative of ${\log }_F(F(X,Y))$ is $1/f(Y)$, the same as that of ${\log }_F(Y)$. Evaluating at $Y=0$ therefore provesThe differential $dT/f(T)$ is the invariant differential of a formal group law.

Choose an integer $m>e/(p-1)$. In degree $n\ge 2$, the coefficient of ${\log }_F(T)$ has valuation at least $-e{v}_p(n)$. Thus the scaled serieshasHere $v_p(n)\le (n-1)/(p-1)$, and the coefficient valuations tend to infinity. Consequently $H(U)=U+\pi R(U)$ with $R$ a convergent integral power series on ${\mathcal{O}}_K$. Such a series is $1$-Lipschitz. For any $a\in {\mathcal{O}}_K$, the equation $H(U)=a$ is equivalent to $U=a-\pi R(U)$, a strict contraction, to which the contraction mapping theorem applies on the complete ring ${\mathcal{O}}_K$. It has exactly one solution.

The deep logarithm subgroup of a formal group is obtained as follows. The formal group exponential is the formal compositional inverse ${\exp }_F(T)$ of ${\log }_F(T)$, obtained recursively; equivalently it solves ${\exp }_F^{\prime }(T)=f({\exp }_F(T))$, with ${\exp }_F(0)=0$. The preceding contraction, applied to the scaled series, constructs its convergent inverse on ${\pi }^m{\mathcal{O}}_K$ and agrees with that formal inverse. This supplies the required convergence sketch for ${\exp }_F$ as well as a bijectivity proof for the logarithm.

The integral formal law and inverse preserve ${\pi }^m{\mathcal{O}}_K$, so this is a subgroup. The logarithm identity and bijectivity giveReduction of the formal law modulo ${\pi }^m$ has kernel $F({\pi }^m{\mathcal{O}}_K)$. Its underlying image set is $\pi {\mathcal{O}}_K/{\pi }^m{\mathcal{O}}_K$, which has size $q^{m-1}$, where $q=\#({\mathcal{O}}_K/\pi )$. Thus this subgroup has finite index, explicitly $q^{m-1}$.

In fact, $2,3,5,7$ must all be primes of bad reduction.

At a prime of good reduction of an elliptic curve, the kernel $E_1({\mathbb{Q}}_p)$ of reduction is the formal group of an elliptic curve evaluated on $p{\mathbb{Z}}_p$, using its parameter at the identity. This kernel has no prime-to-$p$ torsion: for an integer $\ell$ prime to $p$, the multiplication series is $[\ell {]}_F(T)=\ell T+$ higher-degree integral terms, so for nonzero $T\in p{\mathbb{Z}}_p$ its first term has strictly smaller valuation than the others and $[\ell {]}_F(T)\ne 0$.

For each odd good prime, the entire given torsion subgroup, of order $16$, would therefore inject into $E({\mathbb{F}}_p)$. But the Hasse theorem for elliptic curves givesNone can contain a subgroup of order $16$. Hence $3,5,7$ are bad.

For the prime-two torsion bound for good reduction, use the deeper formal subgroup corresponding to $4{\mathbb{Z}}_2$. Part (i), with $m=2>1/(2-1)$, shows that $E_2({\mathbb{Q}}_2)$ is torsion-free and has index $2$ in $E_1({\mathbb{Q}}_2)$. Therefore any finite torsion subgroup of $E_1({\mathbb{Q}}_2)$ injects into $E_1/E_2$ and has order at most $2$.

If $E$ had good reduction at $2$, its rational torsion subgroup of order $16$ would consequently have reduction image of order at least $8$. The Hasse theorem for elliptic curves instead gives $\#E({\mathbb{F}}_2)\le 5$. This contradiction proves bad reduction at $2$ as well, and thereforeThe special formal-group argument at $2$ is necessary; prime-to-$p$ injectivity alone says nothing about the given $2$-primary torsion there.