Let $E$ be an elliptic curve over $\overline{\mathbb{Q}}$ defined by a Weierstrass equation $$ y^2=4x^3+g_2x+g_3. $$ Then $H^1_{dR}(E/\overline{\mathbb{Q}})$ is spanned by the classes of the differential forms $\omega_1=\frac{dx}{y}$ and $\omega_2=x\frac{dx}{y}$ and "the" period matrix of $E$ consists of the four numbers $$ \int_{\gamma_i} \omega_j $$ where $\gamma_1, \gamma_2$ are generators of $H_1(E(\mathbb{C}), \mathbb{Q})$.

**Theorem**. The degree of transcendence of the field spanned by these numbers is at least $2$.

I am trying to understand why this implies that $\int_\gamma \omega_1$ and $\int_\gamma \omega_2$ are algebraically independent whenever $E$ has complex multiplication. This will follow if one shows that the ratio between periods attached to the same differential form is algebraic.

For $\omega_1$ I can see what's going on: by uniformization, $$ E(\mathbb{C}) \simeq \mathbb{C} / \mathbb{Z} \oplus \mathbb{Z} \tau $$ for some $\tau$ in the upper half plane. Complex multiplication forces $\tau$ to be imaginary quadratic. The differential form $\omega_1$ corresponds to $dz$ on the complex torus and the periods become $\int_0^1 dz=1$ and $\int_0^\tau dz=\tau$, so the quotient is $\tau$ which is algebraic.

What about the other differential form? I know that $\omega_2=x \frac{dx}{y}$ corresponds to $\mathcal{P}(z)dz$ where $\mathcal{P}$ is the Weierstrass function attached to the lattice. Then the periods $\int_{\gamma_i} \omega_2$ are minus the periods of $\zeta$, the primitive of $-\mathcal{P}$. Why their quotient is algebraic in this situation?

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