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Continuing in this way, we obtain
g
(
τ
)=
f
(
τ
)
− b
n
j
n
−···b
0
∈ q
Z
[[
q
]]
for integers
b
n
,...,b
0
. The function
g
(
τ
) is analytic in
H
and vanishes at
i∞
.
Also,
g
(
τ
) is invariant under the action of
SL
2
(
Z
). Proposition 9.16 says that
if
g
is not identically zero then a sum of the orders of
g
at various points is 0.
But these orders are all nonnegative since
g
is analytic. Moreover, the order
of
g
at
i
is positive. Therefore the sum of the orders must be positive, hence
cannot be zero. The only possibility is that
g
is identically zero. This means
that
∞
b
n
j
n
g
(
τ
)=
f
(
τ
)
−
−···
b
0
=0
,
so
f
(
τ
)
∈
Z
[
j
].
Combining Lemma 10.12 and Proposition 10.14, we obtain the first part of
the following.
THEOREM 10.15
Let
N
be a positive integer.
1. T here isapolynom ialw ithinteger coe cients
Φ
N
(
X, Y
)
∈
Z
[
X, Y
]
su ch that the coe cient of the highest pow er of
X
is1andsuchthat
F
N
(
X, τ
)=Φ
N
(
X, j
(
τ
))
.
2. If
N
is not a perfect square, then
H
N
(
X
)=Φ
N
(
X, X
)
∈
Z
[
X
]
isnonconstant and the coe cient of itshighest pow er of
X
is
±
1
.
PROOF
We have already proved the first part. For the second part, we
know that
H
N
(
j
)=Φ
N
(
j, j
)=
F
N
(
j, τ
)=
S
(
j − j ◦ S
)
∈
S
N
is a polynomial in
j
with integer coecients. We need to look at the coecient
of the highest power of
j
.Let
S
=
ab
∈ S
N
. Ifweexpandthefactor
0
d
S
as a Laurent series in
e
2
πiτ/N
, the first term for
j
is
j
−
j
◦
e
−
2
πiτ
=(
e
−
2
πiτ/N
)
N
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