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 $\mathbb Q$?



Algebraicity of isogenies as maps of lattices
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 $\alpha \in \bar{\mathbb{Q}}$.


