Let $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ be the Eisenstein Series of weight $2$
and let $E_2^*(\tau) = E_2(\tau) - \frac{3}{\pi\cdot Im(\tau)}$ be the corresponding *almost holomorphic modular form*. Then let
$$\eta(\tau)=e^{\pi i\tau /12}\cdot\prod_{n=1}^\infty(1-q^n)$$
be the Dedekind $\eta$-Function.

**Question: How can I prove the following statement from [here][1]:**

> Let $\tau$ be any Complex Multiplication point. By basic theorems of complex multiplication, if you choose a suitable period
$\omega(\tau)$, $E_4(\tau)/\omega(\tau)^4$, $E_6(\tau)/\omega(\tau)^6$,
and $\sqrt{D}E_2^*(\tau)/\omega(\tau)^2$ (with $E_2^*(\tau)=E_2(\tau)-3/(\pi \cdot Im(\tau))$ and $D$ the discriminant of $\tau$) will be algebraic numbers of known degree, and if you choose
$\omega(\tau)=\eta(\tau)^2$, they will even be algebraic integers.


**Partial Solution:**

The statement that $\frac{E_4}{\eta^8}$ is an algebraic integer follows from
$$\left(\frac{E_4}{\eta^8}\right)^3 = j(\tau)$$
and the statement that $\frac{E_6}{\eta^{12}}$ is an algebraic integer follows from
$$\left(\frac{E_6}{\eta^{12}}\right)^2 = 1-j(\tau)$$
with the absolute invariant $j(\tau)=\frac{E_4^3}{E_4^3-E_6^2}=\frac{E_4^3}{\eta^{24}}$ which is an algebraic integer.


  [1]: https://mathoverflow.net/questions/300640/how-to-compute-coefficients-in-chudnovskys-formula/300656#300656