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Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ 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:

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 for $E_2^*$:

Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference.

Complete Solution for $E_4$ and $E_6$:

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 $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1728$$ with the absolute invariant $j(\tau)=\frac{1728E_4^3}{E_4^3-E_6^2}=\frac{E_4^3}{\eta^{24}}$ which is an algebraic integer (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).

Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ 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:

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 for $E_2^*$:

Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference.

Complete Solution for $E_4$ and $E_6$:

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 $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1728$$ with the absolute invariant $j(\tau)=\frac{1728E_4^3}{E_4^3-E_6^2}=\frac{E_4^3}{\eta^{24}}$ which is an algebraic integer (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).

EDIT: Complete Solution for $E_2^*$:

The answer of Michael Griffin (see below) can now be found in more details in the appendix of this arXiv-preprint.

Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ 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:

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 for $E_2^*$:

Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference.

Complete Solution for $E_4$ and $E_6$:

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 $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1$$ 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 (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).

Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ 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:

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 for $E_2^*$:

Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference.

Complete Solution for $E_4$ and $E_6$:

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 $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1728$$ with the absolute invariant $j(\tau)=\frac{1728E_4^3}{E_4^3-E_6^2}=\frac{E_4^3}{\eta^{24}}$ which is an algebraic integer (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).

Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ 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:

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 $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1$$ 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 (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).

Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ 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:

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 for $E_2^*$:

Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference.

Complete Solution for $E_4$ and $E_6$:

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 $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1$$ 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 (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).