Let $A,B$ be two $\Bbb Q$-algebras. Assume that $A \otimes_{\Bbb Q} \Bbb C \cong B \otimes_{\Bbb Q} \Bbb C$ as $\Bbb C$-algebras. Does it follows that $A \otimes_{\Bbb Q} \overline{\Bbb Q} \cong B \otimes_{\Bbb Q} \overline{\Bbb Q}$ as $\overline{\Bbb Q}$-algebras ?
Here $\overline{\Bbb Q}$ is an algebraic closure of $\Bbb Q$. (Clearly the converse holds, since $- \otimes_{\overline{\Bbb Q}} \Bbb C$ is a functor, so it preserves isomorphisms).
My feeling is that this is true if $A,B$ are finitely generated as $\Bbb Q$-algebras. We could even replace $\Bbb Q$ by any (perfect) field $k$ and considering two algebraically closed fields $K \subset K'$ containing $k$. We would have $A = k[X_1,...,X_r]/I,B=k[X_1,...,X_s]/J$ with a $K'$-algebras isomorphism $\phi : K'[X_1,...,X_r]/I^e \to K'[X_1,...,X_s]/J^e$ ($(\cdot)^e$ denotes the extension of ideals). Then I would like that $\phi$ is actually defined over a finite extension $L \supset k$, so in particular $L \subset K$ and $\phi$ is defined over $K$ and provides an isomorphism $A \otimes_k K \cong B \otimes_k K$ of $K$-algebras.
Looking at this question, I think that this is true if we replace algebras by modules.
This is wrong in larger generality. For instance, let $A = \Bbb Q[x,y] / (y^2-x^2) \cong \Bbb Q \times \Bbb Q$ and $B=\Bbb Q[x,y] / (y^2-2x^2)$. Then $A \otimes_{\Bbb Q} \Bbb Q(\sqrt 2) \cong B \otimes_{\Bbb Q} \Bbb Q(\sqrt 2) \cong \Bbb Q(\sqrt 2) \times \Bbb Q(\sqrt 2)$ as $\Bbb Q(\sqrt 2)$-algebras, but $A \not\cong B$ as $\Bbb Q$-algebras (not even as rings, since $B$ is a domain but not $A$). So we need to make base extensions by algebraically closed fields.