Modular function for $SL_2(\mathbb{Z})$ is a field

Question

How do I prove

1) Modular functions for $SL_2(\mathbb{Z})$ is a field K with addition and multiplication defined pointwise.

2) $K= \mathbb{C}(j)$, where $j(z)= \frac{(240 E_4)^3}{\bigtriangleup}$?

Answer 1

• Let $\Gamma = SL_2(\mathbb{Z})$. The set $A_0$ of meromorphic functions on $\mathcal{H} = \{ z \in \mathbb{C},\Im(z) >0\}$ invariant under $f(z) \mapsto f(\gamma z), \gamma \in \Gamma$ and meromorphic at $i\infty$ $(*)$ is clearly a field

  $\qquad$ ($(*)$ $f(\frac{1}{z-1})$ is meromorphic at $z=0$)

• Let $\mathcal{H}^* = \mathcal{H} \cup \{i \infty\}\cup \mathbb{Q}$ then $X =\mathcal{H}^*/\Gamma= \{ \Gamma z, z \in \mathcal{H}^*\}$ is a compact Riemann surface and $A_0 = \mathbb{C}(X)$.

• By the argument principle, if $f \in \mathbb{C}(X)$ then $\# poles(f) = \# zeros(f)$ (counted with multiplicity)

• By the elliptic curve definition of the $j$-invariant or by the product definition of $\Delta$, $j(z)$ and hence $j(z)-c$ has a unique pole of order $1$ on $X$ (at $\Gamma i\infty$), thus so does $\frac{1}{j(z)-c}$ (so that $j$ is bijective $X \to \mathbb{C} \cup \{\infty\}$)

• Take some $f(z) \in \mathbb{C}(X)$. If it has a zero of order $k$ a $i\infty$ replace it by $ f(z)j(z)^k$. Then let $\Gamma a_1,\ldots,\Gamma a_L$ its remaining zeros counted with multiplicity. Thus $$g(z) = f(z)\prod_{l=1}^L \frac{1}{j(z)-j(a_l)} \in \mathbb{C}(X)$$ has no zeros, which means it has no poles too, hence by the maximum modulus principle it is constant and $f \in \mathbb{C}(j) \implies \mathbb{C}(X) = \mathbb{C}(j)$