$\left(\frac{\Delta(6z)\Delta(z)}{\Delta(3z)\Delta(2z)}\right)^{1/2}$ generates the field of modular functions on $\Gamma_1(6)$

Question

I have already shown that the function, defined on the upper half plane $\{z\in\mathbb{C}:\Im(z)>0\}$, $$t(z)=\left(\frac{\Delta(6z)\Delta(z)}{\Delta(3z)\Delta(2z)}\right)^{1/2}$$ is a modular function with respect to $$\Gamma_1(6)=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}):c\equiv0,a\equiv d\equiv 0\mod 6\right\}.$$

Now I have found the statement that $t$ generates the field of modular functions on $\Gamma_1(6)$.

There are now three questions from those I don't know the answer:

1. What does it mean if a function generates the field of modular functions?
2. Why are the modular functions a field?
3. Why does $t$ generates the field of modular functions on $\Gamma_1(6)$?
4. Why do we need the $\cdot^{1/2}$?

I would be really happy if you know the answer(s) of (at least) one question! Thank you very much.

Answer 1

1.) Generating set of a field means the following: A subset $S$ of a field $K$ generates $K$ if the only subfield of $K$ containing $S$ is $K$ itself.

2.) We can associate to $\Gamma \le SL_2(\mathbb{Z})$ a compact Riemann surface $\Gamma/\mathbb{H}^*$, so that the modular functions constitute the field of functions of this Riemann surface, and hence form a field of transcendence degree one over $\mathbb{C}$.

3.) See for example here; Generators of function fields of the modular curves $ X_1(5)$ and $ X_1(6)$, by Kim and Koo, 2010. The generator in the second case is $j_{1.6}$ (which is your $t(z)$).