Proposition 4.6
Let

be the point in

defined by the degree 0 divisor

on

, and
let

be the order of the image of

in

.
Then the denominator of

divides

.
Proof.
Let

be the image of

in

,
and set

. Since

is a newform,
the Hecke operators

, for

, act as 0 or

on

(see, e.g., [
DI95, §6]).
If

, then a standard calculation (see, e.g., [
Cre97, §2.8])
shows that

. There is an injection

sending

to

, where

is
the order-

cyclic subgroup of

generated by

.
With notation as in the proof of Theorem
4.5, we have
The final inclusion follows from two observations.
First, the index
![$ [\Phi(H)^+:I\Phi(e)]$](img388.png)
is an integer because

is exactly
the ideal of elements of

that send

into

,
and

.
Second, there is a surjective map
sending

to

, so
![$ [\mathbf{T}{}\Phi(e):IP(e)]$](img396.png)
divides

.