a question on CM elliptic curves Let $E$ be an elliptic curve over $\overline{\mathbb{Q}}$ defined by an equation 
$y^2=4x^3-g_2x-g_3$
and let $\omega=\int_\gamma \frac{dx}{y}$ be the integral of the regular differential form $\frac{dx}{y}$ over a cycle $\gamma \in H_1(E(\mathbb{C}, \mathbb{Q})$ not homologous to zero. Assume now that $E$ has complex multiplication by an imaginary quadratic field $K$. 
Question: Why is it true that $\Delta(\omega \cdot \mathfrak{a})$ is algebraic for any fractional ideal $\mathfrak{a}$ of $\mathcal{O}_K$? 
Here $\Delta$ is the usual modular form, defined on lattices by
$\Delta(\Lambda)=g_2(\Lambda)^3-27g_3(\Lambda)^2$
and $g_2(\Lambda)=60\sum_{\lambda \in \Lambda^\ast} \lambda^{-4}$, $g_3(\Lambda)=140\sum_{\lambda \in \Lambda^\ast} \lambda^{-6}$. 
 A: The 1-sentence reason is: scalar extension from an algebraically closed field of characteristic zero to a bigger one (such as $\overline{\mathbf{Q}} \hookrightarrow \mathbf{C}$) has no effect on the isogeny class (objects or maps!) of a given elliptic curve, ultimately because all torsion is found over the smaller such field and the quotient operation can be performed down there too.
Let's now make the link to that idea.
For any elliptic curve $E$ over a field $k$, equipped with a non-vanishing global 1-form $\eta$, one can always find a Weierstrass model
$$y^2 w + a_1 xyw = x^3 + a_2 x^2 w + a_4 x w^2 + a_6 w^3$$
in $\mathbf{P}^2_k$ for $E$ such that $\eta$ corresponds to ${\rm{d}}x/(2y + a_1x)$.  Then one checks that the usual polynomial $\Delta(a_1,\dots,a_6)$ in the $a_i$'s is nonzero and independent of the choice of such Weierstrass model compatible with $\omega$.  We call this value $\Delta^{\rm{alg}}(E,\eta) \in 
k^{\times}$. 
For example, consider $k = \mathbf{C}$ and $(E,\eta)$ over $k$ corresponding to the analytic construction $(\mathbf{C}/L, {\rm{d}}z)$ for a lattice $L$ in $\mathbf{C}$.  We have the Weierstrass model $y^2 w = 4x^3 + g_2(L) x w^2 + g_6(L)$ for which ${\rm{d}}x/y$ corresponds to ${\rm{d}}z$, and $\Delta(a_1,\dots,a_6) = g_2(L)^3 - 27 g_3(L)^2$. Thus, after bookkeeping with the factor of 2 (since the Weierstrass model uses a non-monic cubic in $x$) one concludes that
$$\Delta^{\rm{alg}}(\mathbf{C}/L, {\rm{d}}z) = \Delta(L).$$
In your situation, the (algebraization of the) pair $(\mathbf{C}/L, {\rm{d}}z)$ is equipped with a descent to $\overline{\mathbf{Q}}$.  
But what is $L$ in that case?  It is precisely the image of ${\rm{H}}_1(E(\mathbf{C}),\mathbf{Z})$ under an injection into $\mathbf{C}$ defined by $\gamma \mapsto \int_{\gamma} \eta$.  Now under the CM hypothesis (with CM field $K$ embedded into $\mathbf{C}$ via its $\mathbf{C}$-linear action on the 1-dimensional tangent space at the identity), if $O$ is the CM order then the algebraic $O$-action on $E$ has the effect of multiplication against $O$ inside $\mathbf{C}$ at the level of integrals of 1-forms (this is why it is called "complex multiplication", after all). 
Thus, for any nonzero $\gamma$ in ${\rm{H}}_1(E(\mathbf{C}), \mathbf{Z})$ the set $O \cdot \int_{\gamma}\eta$ is of finite index inside $L$.  Hence, if $q \in \mathbf{Q}^{\times}$ and we define $\omega := q \int_{\gamma}\eta$ (the factor of $q$ accounts for you wish to work with rational homology rather than integral homology) then for any fractional ideal $\mathfrak{a}$ of $K$ we see that $\omega \mathfrak{a}$ as a lattice inside $\mathbf{C}$ is commensurable with $L$.  
The problem is therefore reduced to the following assertion that has absolutely nothing to do with the theory of complex multiplication: if $L$ and $L'$ are commensurable lattices inside $\mathbf{C}$ and the (algebraization of the) pair $(\mathbf{C}/L, {\rm{d}}z)$ can be defined over an algebraically closed subfield $k \subset \mathbf{C}$ (forcing $\Delta(L) \in k$) then we claim the same for $(\mathbf{C}/L', {\rm{d}}z)$ (so $\Delta(L') \in k$).   The main content will be that passing from one algebraically closed field of characteristic 0 to a bigger one does not create "new" isogenies.
Let $L'' = L \cap L'$, so by comparing $L$ to $L''$ and then $L''$ to $L'$ we may assume that one of $L$ or $L'$ is contained in the other. 
First assume $L' \subset L$, so we have an isogeny $f:\mathbf{C}/L' \rightarrow \mathbf{C}/L$ given by the canonical quotient, so $f^{\ast}({\rm{d}}z) = {\rm{d}}z$ (abusing notation slightly, as the two ${\rm{d}}z$'s do not mean the same thing, being 1-forms  on distinct elliptic curves).  If $(E, \eta)$ is a pair over $k$ that descends the algebraization of $(\mathbf{C}/L, {\rm{d}}z)$ then any isogeny to or from $E_{\mathbf{C}}$ (equivalently in the analytic theory: to or from $E_{\mathbf{C}}^{\rm{an}} = \mathbf{C}/L$) descends to one to or from $E$ over $k$ since $k$ is algebraically closed.  Hence, we get an elliptic curve $E'$ over $k$ and an isogeny $f_0:E' \rightarrow E$ over $k$ that descends (the algebraization of) $f$, so $(E', f_0^{\ast}(\eta))$ is a pair over $k$ that descends $(\mathbf{C}/L', {\rm{d}}z)$. 
Next, assume $L \subset L'$.  Then there is an integer $n > 0$ such that $L' \subset (1/n)L$, so if we can handle the lattice $(1/n)L$ then we can apply the preceding case to get the result for $L'$ from the case of $(1/n)L$.  In other words, we just need to check that (the algebraization of) $(\mathbf{C}/(1/n)L, {\rm{d}}z)$ descends to $k$.  But this analytic pair is isomorphic to $(\mathbf{C}/L, n {\rm{d}}z)$ via multiplication by $n$ on $\mathbf{C}$, so it remains to note that this latter pair descends to $(E, n \eta)$.
