Algebraicity of isogenies as maps of lattices

Let $E_i\colon y^2=4x^3+A_ix+B_i$, for $i=1,2$ be two elliptic curves where $A_i,B_i \in \mathbb C$ are algebraic over $\mathbb Q$. For $i=1,2$ let $\Lambda_i\subseteq \mathbb C$ be the unique lattice that parametrizes $E_i$, namely such that $\displaystyle g_2(\Lambda_i)=60\sum_{0\neq \omega\in \Lambda_i}\frac{1}{\omega^4}=A_i$ and $\displaystyle g_3(\Lambda_i)=140\sum_{0\neq \omega\in \Lambda_i}\frac{1}{\omega^6}=B_i$.

The question goes as follows: suppose that $\psi\colon E_1\to E_2$ is an isogeny. Then I can write down rational maps for $\psi$ with coefficients in $\overline{\mathbb Q}$. Now $\psi$ corresponds to a unique complex number $\alpha$ such that $\alpha\Lambda_1\subseteq \Lambda_2$. Does $\alpha$ need to be algebraic over $\mathbb Q$?

I think it is. The point is that you can identify $\Lambda _i$ with $H_1(E_i,\mathbb{Z})$, embedded into $\Bbb{C}$ by $\gamma \mapsto \int_\gamma \eta _i$, with $\eta _i=\frac{dx}{y}$ (there may be some normalization factor, but it should be in $\mathbb{Q}$ so that doesn't matter). Now multiplication by $\alpha$ on $\mathbb{C}$ induces $\psi _*:H_1(E_1,\mathbb{Z})\rightarrow H_1(E_2,\mathbb{Z})$; since $\int_{\psi _*\gamma }\eta _2=\int_\gamma \psi ^*\eta _2$, this implies $\psi ^*\eta _2=\alpha \eta _1$. But both forms are in the $\bar{\mathbb{Q}}$-vector space $H^0(E_1,\Omega ^1_{E_1})$, so $\alpha \in \bar{\mathbb{Q}}$.