Is it true that $(A[[x]]\otimes_A\mathbb{C})\cap \overline{\mathbb{Q}}[[x]] = A[[x]]\otimes_A\overline{\mathbb{Q}}$? Let $A$ be a subring of $\overline{\mathbb{Q}}$. Is it true that:
$$(A[[x]]\otimes_A\mathbb{C})\cap \overline{\mathbb{Q}}[[x]] = A[[x]]\otimes_A\overline{\mathbb{Q}}$$
where the intersection takes place inside the ambient ring $\mathbb{C}[[x]]$?
(Certainly the left side contains the right side)
 A: Yes, and this is just linear algebra.  An element $\sum a_nx^n\in\overline{\mathbb{Q}}[[x]]$ is in $A[[x]]\otimes_AF$ (for $F=\overline{\mathbb{Q}}$ or $F=\mathbb{C}$) iff there exist elements $b_1,\dots,b_m\in F$ and elements $c_{in}\in A$ (for $1\leq i\leq m,n\in\mathbb{N}$) such that $a_n=\sum_i c_{in}b_i$ for all $n$. 
Note that for any given choice of elements $c_{in}\in A$, the existence of $b_i\in F$ that work just amounts to being able to solve a certain (infinite) system of linear equations over $F$ in finitely many variables.  You can solve such a system by Gaussian elimination, eventually reaching either a contradictory equation or a system where arbitrary values may be chosen for some of the variables and then the rest are uniquely determined.  Enlarging the ambient field doesn't affect how Gaussian elimination works (you're just doing certain computations with the field operations and the coefficients of the system of equations), so the system has a solution for $F=\overline{\mathbb{Q}}$ iff it has a solution for $F=\mathbb{C}$.
So, for any specific choice of $c_{in}\in A$, appropriate $b_i$ exist in $\overline{\mathbb{Q}}$ iff they exist in $\mathbb{C}$.  Thus, there exist $c_{in}\in A$ such that there exist appropriate $b_i\in\overline{\mathbb{Q}}$ iff there exist $c_{in}\in$ such that there exist appropriate $b_i\in\mathbb{C}$.  That is, $\sum a_nx^n\in A[[x]]\otimes_A\overline{\mathbb{Q}}$ iff $\sum a_nx^n\in A[[x]]\otimes_A\mathbb{C}$.
