# Ideal of cusp forms for $\Gamma_0(4)$ is principal

Let $\Gamma_0(4)$ be a congruence subgroup of $SL(2,\mathbb{Z})$ defined as

$$\Gamma_0(4)=\{M=\begin{pmatrix} a &b\\ c& d \end{pmatrix}\in SL(2,\mathbb{Z})\mid c=0\bmod 4\}.$$
Dedekind eta-function is defined as
$$\eta(z)=q^{1/24}\prod_{n\geq1}(1-q^n).$$

How to prove that the ideal of cusp forms for $\Gamma_0(4)$ is principal and generated by

$$f(z)=\eta(2z)^{12}?$$

#### Solutions Collecting From Web of "Ideal of cusp forms for $\Gamma_0(4)$ is principal"

The form $f(z)$ generates the $\Gamma_0(4)$ cuspforms (as an ideal over the
graded ring of all modular forms for $\Gamma_0(4)$) for much the same
reason that $\Delta(z) = \eta(z)^{24}$ generates the cuspforms for
the full modular group $\Gamma(1) = {\rm SL}_2({\bf Z})$: this $f$ is
a cuspform with no zeros in the finite plane, and it vanishes at each cusp
only to the multiplicity needed to make it a cuspform. Hence if $\varphi$
is any other cuspform, say of weight $w$, then $\varphi_1 := \varphi/f$
is a modular form of weight $w-6$, and we have written $\varphi$ as
$f \varphi_1$ for some modular form $\varphi_1$ as desired.

The fact that there are no zeros in the finite plane is clear from
the product formula for $\eta(z)$. The vanishing order at the cusps
is easier to see if we change variables to $w := 2z$, because
then $f(z) = \eta(w)^{12}$ is a cuspform for the action of $\Gamma(2)$
on $w$, and the three cusps of $\Gamma(2)$ all have the same width
(namely $2$) because they’re transitively permuted by $\Gamma(1)$,
so $\eta(w)^{12}$ vanishes to order $1$ at each cusp.

Similar examples for other groups are $\eta(3z)^8$ for $\Gamma_0(9)$
(equivalently $\eta(z)^8$ for $\Gamma(3)$), and $(\eta(z)\eta(pz))^w$
for $\Gamma_0(p)$ with $p=2,3,5,11$ (i.e. the primes with $p+1\mid12$) and
$(p+1)w=24$. The last of these also works for $p=7$ and $p=23$
if we allow modular forms with quadratic character.