4 Torsion Points

The duplication formulae above allow us to consider the $(1 + i)^n$ -torsion points of $R$ , i.e. the points such that acting by $(1 + i)^n$ sends the point back to the origin. There is a slight subtlety here, due do the fact that we are working with $\operatorname{sl}$ , which is the composition $\mathbb {C}/\Lambda \to R \overset {\pi }{\to } \mathbb {P}^1$ . If $\operatorname{sl}(z) = 0$ , then the corresponding point in $R$ need not be the origin. It could be the other point $(0, -1)$ . Fortunately, we already know exactly which point gets mapped to $(0, -1)$ , namely $\Omega$ , which is the unique $(1 + i)$ -torsion point of $\mathbb {C}/\Lambda$ .

We might also worry about the two points at infinity, which is not adequately captured by our formula. Observing that

\operatorname{sl}((1 + i)z) = \frac{(1 + i)\operatorname{sl}(z)}{\sqrt{1 - \operatorname{sl}(z)^4}},

and the fact that $\operatorname{sl}(\frac{\Omega }{2}) = 1$ , we know that the two points at infinity are $\frac{1 \pm i}{2} \Omega$ , which are the remaining $2$ -torsion points. These let us conclude

Proposition

$\operatorname{sl}((1 + i)^n z) = \infty$ iff $z$ is a $(1 + i)^{n + 2}$ -torsion point but not an $(1 + i)^{n + 1}$ -torsion point.

This lets us compute the higher torsion points:

$n$	$(1 + i)^n$ -torsion points
0	$(0, 1)$
1	$(0, -1)$
2	$\infty , \infty$
3	$(\pm 1, 0), (\pm i, 0)$
4	$(\zeta _8^k, \pm \sqrt{2})$ $k = 1, 3, 5, 7$
5	$(i^k \sqrt[4]{3 \pm _a 2 \sqrt{2}}, \pm _b \sqrt{2 \mp _a 2\sqrt{2}})$ $k = 0, 1, 2, 3$

Finding higher-order torsion points will require advanced WolframAlpha techniques.

Recall that we can find ray class fields of $\mathbb {Q}(i)$ by adjoining the torsion points of the curve with CM by $\mathbb {Q}(i)$ . We need to first identify a Weber function, which is given by quotienting out $R$ by the automorphism group. The automorphism group multiplies by powers of $i$ , and so we find that $h(x, y) = y$ is a perfectly good Weber function on $R$ . So we find that

The ray class field modulo $(1 + i)^3$ is still just $\mathbb {Q}(i)$ .
The ray class field modulo $(1 + i)^4 = (4)$ is $\mathbb {Q}(i, \sqrt{2})$ .
The ray class field modulo $(1 + i)^5$ is $\mathbb {Q}(i, \sqrt{2 + 2 \sqrt{2}})$ .

We can compute the ray class groups explicitly, using the definition of the ray class group, and see that they match. Observe that since $\mathbb {Q}(i)$ has class number $1$ , we can decompose

\mathbb {A}_{\mathbb {Q}(i)}^\times = \frac{\mathbb {Q}(i)^\times }{\mu _4} \times \mathbb {C}^\times \times \prod _{v \nmid \infty } \mathcal{O}_v^\times ,

and so

C_{\mathbb {Q}(i)} = \frac{\mathbb {C}^\times \times \prod _{v \nmid \infty } \mathcal{O}_v^\times }{\mu _4}.

The ray class group modulo $\mathfrak {m} = (1 + i)^k$ is then given by

\mathrm{Cl}_\mathfrak {m} = \frac{\mathcal{O}_{(1 + i)}^\times }{\mu _4 \times (1 + (1 + i)^k \mathcal{O}_{(1 + i)})} = \left(\frac{\mathbb {Z}[i]}{(1 + i)^k}\right)^\times \Big/ \mu _4.

The invertible elements in $\mathbb {Z}[i]/(1 + i)^k$ are exactly those $(1 + i)$ does not divide, i.e. those with odd norm. Thus, we see that

\mathrm{Cl}_{(1 + i)^3} = 0,\quad \mathrm{Cl}_{(1 + i)^4} = C_2,\quad \mathrm{Cl}_{(1 + i)^5} = C_2 \times C_2,

in agreement with above (exercise: find the other subfields on $\mathbb {Q}(i, \sqrt{2 + 2\sqrt{2}})$ ).