Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 18 / 1

Let $E=\mathbb{C}/\Lambda$, and let ${\Lambda }^{\prime }$ be the inverse image of the cyclic subgroup $G$ under the quotient map. Then $\Lambda \subset {\Lambda }^{\prime }$ is an inclusion of Euclidean lattices with ${\Lambda }^{\prime }/\Lambda \cong \mathbb{Z}/N\mathbb{Z}$. The Smith normal form supplies a basis $u,v$ of ${\Lambda }^{\prime }$ for whichChoose the orientation of this basis so that $\mathrm{Im}(v/(Nu))>0$. Dividing the coordinate on $\mathbb{C}$ by $Nu$ gives an isomorphism of complex toriHere $u$ maps to $1/N$, and its class generates ${\Lambda }^{\prime }/\Lambda$. Thus the isomorphism carries $G$ to the required cyclic subgroup. Every pair is represented: $\Phi$ is surjective. This argument avoids assuming that an arbitrary integer lift of a primitive vector modulo $N$ is itself primitive.

Suppose ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$, with $ad-bc=1$, and put $\alpha =(c{\tau }_2+d{)}^{-1}$. The identitiesshow that multiplication by $\alpha$ takes one period lattice isomorphically to the other: its basis matrix has determinant $ad-bc=1$. If $N\mid c$, the image of $1/N$ is congruent to $a/N$ modulo the target period lattice. Moreover $ad\equiv 1(\mathrm{mod}N)$, so $a$ is a unit modulo $N$. The image therefore generates the same cyclic subgroup.

Conversely, a holomorphic group isomorphism of complex tori lifts to $z\mapsto \alpha z$, with $\alpha {\Lambda }_{{\tau }_2}={\Lambda }_{{\tau }_1}$. Indeed its lift fixing zero is additive: its failure to be additive is a continuous lattice-valued function, hence zero. An additive holomorphic map is complex linear. WriteThe corresponding integral basis matrix has determinant $1$: complex multiplication preserves the real orientation, and both bases $(1,{\tau }_j)$ are positively oriented. Solving gives ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$. Preservation of the cyclic subgroup means $\alpha /N\equiv u/N(\mathrm{mod}{\Lambda }_{{\tau }_1})$ for some unit $u$ modulo $N$. Comparing the two real basis coefficients gives $N\mid c$. Thus the matrix lies in the Gamma 0 congruence subgroup, and the classification isThis is the noncompact modular curve classification; no cusp points have been added.